Andree_Programming_the_IBM_650_Magnetic_Drum_Computer_and_Data Processing_Machine_1958 Andree Programming The IBM 650 Magnetic Drum Computer And Data Processing Machine 1958

Andree_Programming_the_IBM_650_Magnetic_Drum_Computer_and_Data-Processing_Machine_1958 Andree_Programming_the_IBM_650_Magnetic_Drum_Computer_and_Data-Processing_Machine_1958

User Manual: Andree_Programming_the_IBM_650_Magnetic_Drum_Computer_and_Data-Processing_Machine_1958

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RICHARD

ANDREE

ASSOCIATE PROFESSOR OF MATHEMATIC S
UNIVERSITY OF OKLAH O MA

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RICHARD

ANDREE

ASSOCIATE PROFESSOR OF MATHEMATICS
UNIVERSITY OF OKLAHOMA

U U U 11

HOLT,

RINEHART

AND

WINSTON,

INC.

383 MADISON AVENUE, NEW YORK 17, N. Y.

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Copyright @ 1958 by Richard V. Andree
Library of Congress Catalogue Card Number: 58: 12522
Printed in the United States of America
20416-0218

6l 70 11

73 14 '5 7S 77 7S 79 80

This book is a first introduction to the programming of the IBM 650 computer. Most of the
techniques involved are applicable to any medium .. speed or high-speed computer. There are
important advantages in programming for a specific machine, available to the reader, rather
than with a mythical machine on which it is impossible to check out programs once they are
written. The student learns much while debugging his program, and enjoys the thrill of finally seeing it work. These notes can be used whether or not a 650 is available, and they
have been used both ways with marked success. Students use flow charts and write actual
programs from the very beginning. After half a dozen class sessions, they write simple programs with surprising ease.
These notes are designed to take the neophyte and advance him to the stage where he can
use the several available IBM programmer's manuals. Those manuals, are, after all, designed
as reference manuals, not as textbooks. These notes provide a text, with suitable problems,
and are not intended as a reference manual. The first four chapters are self.. contained. Chapters 5 to 9 begin to lean slowly more and more upon the published IBM manuals, which the
student is encouraged to consult.
The main purpose of these notes then, is to provide a teachable text on introductory programming, usable either for self.. study or for classroom work. They have been used in intensive one-week courses, in three .. and four ..week courses, and in semester-long courses.
They have been used by graduate engineers and by high-school students.
After the first four chapters are completed, the student can be turned loose on his own,
although the author much prefers to have the student finish Chapter 5 on the use of SOAP
rather than forcing him to write in machine language using five-ten optimization.
Chapter 6, which emphasizes overall principles stressed elsewhere in the book, concludes
with a fairly extensive set of real.. life problems, the programming of which may develop considerable sophistication on the part of the reader. Chapter 7 discusses the advantages, and
also the disadvantages, of interpretive systems and uses the Belllnterpretive System (FLOPS)
as an example. Chapter 8 is devoted to compilers, with particular mention of IT and For
transit. Since excellent manuals are now (1958) available on the Perlis comp.iler, IT, and on
For transit, no attempt is made to discuss these two all-important techniques in complete detail. Instead, the reader is given enough insight into For transit to overcome his first fear and
to convince him that learning For transit is worthwhile; he is then encouraged to program,
turning to the For transit manual as additional help is needed. Chapter 9 is devoted to several odds and ends with which the student should become familiar if he is seriously interested
in programming.
II;

Iv. PREFACE

A convenient teaching device is to make a series of "bug cards" for a .program with
which the students are familiar (say Problem 9, Set 1-15). These typical errors can be
punched on one-per-card load cards. It is convenient to punch the words "Bug Card" and the
problem number in the remarks columns for future use. If a single bug card is placed at the
end of the program deck, the false instruction will be loaded over the top of the correct instruction. The bug card can then be removed without disturbing the main body of the program.
The author will sincerely welcome discussion from users of this book concerning errors,
or portions that can be improved in the next edition.
It is my sincere hope that you will find this brief text both helpful and enjoyable.

ACKNOWLEDGMENTS
This list of acknowledgments should include the names of hundreds of students and
dozens of colleagues who used these notes in various preliminary forms. That seems impractical. It would, however, be ungracious not to mention individually those people who have
contributed most.
Mr. Frank E. McFarlin, formerly of Oklahoma State University and now with IBM Computer
Research and Design Group, Endicott, has helped with the planning of these notes from the
beginning and has read the entire manuscript, offering numerous excellent suggestions.
Dr. John Hamblen formerly Director of the 650 Computer Laboratory at Oklahoma State
University (currently at the University of Kentucky) and Dr. William Viavant, Director of Scientific Computations at the University of Oklahoma have each given encouragement and assistance far beyond the call of duty.
Dr. Melvin Shader, Manager, IBM University and Research Institute Program has graciously read the entire manuscript. Many improvements have resulted from his thought-provoking suggestions.
Dr. H. A. Meyer, University of Florida and Dr. G. E. Forsythe, U. C. L. A., each made
verbal suggestions for which the author is sincerely grateful.
The award for efficiency and elegance in real-life programming goes to the author's wife,
who never goes into a loop when her instructions are modified in unanticipated ways. Without
her understanding cooperation this book would have been impossible.
To each of the above, and to other colleagues and students, the author expresses sincere
thanks.
Norman, Oklahoma
September, 1958

R. V. A.

CHAPTER 1 - SIMPLE PROGRAMMING ..................................................................................................
1- 1 Introduction . 1- 2 What a Computer Is and Is Not . 1- 3 The IBM 650 .
1- 4 The IBM Card . 1- 5 Word . 1- 6 Drum. 1-7 Accumulator· 1- 8 Distributor·
1- 9 Instructions . 1 - 10 A Problem . 1 -11 OPeration Codes . 1-12 Logical
Test - NZU . 1- 13 Multiplication. 1-14 New Operation Codes· 1-15 Flow
Charts.
I

CHAPTER 2 2222-

THE 650 CONSOLE .......................................................................................................... 24

1 Read and Punch . 2 - 2 Power Control Switch. 2 - 3 Display Lights .
4 Flow of Instructions • 2 - 5 Internal Checking of the 650 . 2 - 6 Load Cards .
7 Board Wiring for Load Cards . 2 ~ 8 Operating and Checking Lights .
9 Address Stop . 2- 10 Program Start . 2 - 11 Cookbook Directions.

CHAPTER 3 - MORE ADVANCED PROGRAMMING .............................................................................. 38
3 - 1 Improving Speed • 3 - 2 Stepping Instructions • 3 - 3 Looping and Stepping •
3 - 4 The Count Box • 3 - 5 Another Terminating Technique . 3 - 6 Branching on a
Code Number (Optional) • 3 -7 Scaling. 3 - 8 Scaling in Multiplication.
3 - 9 Division • 3 - 10 (A + B) . C
D'

T. 3 - 11 Optimum Programming.

3 -12 Index Registers (Optional).

CHAPTER 4 - EASY PROGRAMMING VIA SUBROUTINES ................................................................ 60
4 -1 Accuracy of Computed Results· 4 - 2 Subroutines· 4 - 3 Partial List of
Available Programs and Subroutines· 4-4 A Problem· 4-5 A Word to Those Who
Will Have Others Write Their Programs • 4 - 6 Floating-point Arithmetic.
CHAPTER 5 -

SOAP ...... .............................................................................................................................. 72

5 - 1 SOAP (Symbolic Optimum Assembly Program)

CHAPTER 6 -

A RE-EXAMINATION OF PRINCIPLES ........................................................................ 76

6 - 1 Sophistication in Programming • 6 - 2 Flow Charting . 6 - 3 Loops •
6 - 4 Free Data . 6 - 5 Economy in Loading the Program • 6 - 6 Program Errors Debugging. 6 -7 Special Traces and Debugging Routine· 6 - 8 Experience.
v

CHAPTER 7 7777-

CHAPTER 8 -

INTERPRETIVE SYSTEMS ..................................................................................~...........

1 Interpretive Systems . 7 - 2 Bell System (FLOPS) • 7 - 3 Read and Punch •
4 Additional Bell Instructions • 7 - 5 Loop Operations • 7- 6 Example 1 .
7 Example 2 • 7 - 8 Example 3 . 7 - 9 Summary of Bell Operations •
10 Remarks on Interpretive Systems.

COMPILERS ......................................................................................................................

8 - 1 Easy Programming • 8 - 2 IT . 8 - 3 For Transit· 8 - 4 Programming in For
Transit.

CHAPTER 9 - ODDS AND ENDS .............................................................................................................. 104
9 - 1 Common Courtesy • 9 - 2 Sources of Information • 9 - 3 Organization of a 650
Laboratory • 9 - 4 The 650 Library . 9 - 5· Suggested Group Demonstrations .
9 - 6 Subroutines and Programs for Your Library· 9 -7 Well.

INDEX ................................................................................................;............................................................. 108
SAMPLE PROGRAMMING SHEETS .............................................................................................................. 110

(CHAPTER

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1-1. INTRODUCTION.

There is probably no easier way to lose the friendship of a mathematician than to hand
him a pencil and paper along with the remark, "You are a mathematician, you keep score."
Mathematics is the art of avoiding computation, and it is partly because digital computers
help avoid large amounts of computation that they interest mathematicians.
Instead of making fairly general remarks about computers, these notes will examine one
particular computer, namely, the IBM 650 Magnetic Drum Computer~ in considerable detail. To
a large extent, the techniques illustrated are applicable to other computers.
After one week of classroom course work, or approximately three hours of study from this
book, you should be able to write simple programs for the IBM 6501 At the end of a month's
study, you should be able to do fairly complicated programs on your own. It will, of course,
be desirable to have the guidance of a more experienced programmer for some time, if you are
to learn most efficiently. One important thing to learn is to think in a manner compatible
with the machine's optimum abilities. In evaluating X 4 + 3 X 3 - 5 X + 2 for example, it is
much quicker for the machine to compute

x {X[X (X + 3)] -

5} + 2

than for it to compute

Many persons will never do a great deal of programming themselves~ but must understand the
machine well enough to direct intelligently those who will be doing the actual programming
for their problems. Chapters 1 to 6 will suffice for such purposes,but experience shows that
even readers who do not plan to do their own programming will probably read the entire book
once they get started. The 650 is like a virus which gets into the blood stream-it is hard to
shake loose once you are inoculated. (F AIR WARNING!) These notes may be used either for
a short intensive course, or for a semester-long course. They are also excellent for selfstudy. It is much easier to study computers if you have a computer available; but it is quite
possible to have an elementary course in programming in the complete absence of an actual
computer. For those who do have a 650 available, a series of programs will be given, along

2 • SIMPLE PROGRAMMING

with certain "bug cards" which will make the type of mistakes in the program which you, as
a beginning programmer,are most apt to make. Your problem will be to f.ind, with the aid of
your instructor, just where the mistakes are. The debugging process takes a good deal ofmachine time in many programs, particularly in the type of one-time-through programs which occur often in research applications.

1-2. WHAT A COMPUTER IS, AND IS NOT.

A computer is not a giant brain, in spite of what some of the Sunday supplements and
science fiction writers would have you believe. It is a remarkably fast and phenomenally accurate moron. It will do exactly what you tell it to do-no more, no less. There are about 60
different instructions you can give an IBM 650. It can add, subtract, multiply,and divide. It
can determine whether or not a given 10-digit number is zero, and it can determine whether or
not a 10-digit number is negative. It can also determine whether or not any given digit in a
10-digit number is an 8 or a 9.
Actually, it is somewhat misleading to say that it can add, subtract, multiply, and divide, since it does its multiplication by repeated addition and it does its division by repeated
subtraction, just as one does on a desk calculator. Actually, it does not even subtract. When
you tell the machine to subtract, it first takes the complement of the number and then adds the
complement; so, in the last analysis, a computer adds!
Let us see what it is that distinguishes a computer from a desk calculator. In the first
place, in the desk calculator, an auxiliary piece of paper is usually used to store initial data
or numbers and intermediate results, ,and either paper or the operator's memory is used to keep
track of the sequence of steps to be performed. On a computer, both data and the sequence of
instructions are stored in the storage or memory of the machine, although not necessarily in
the same type of storage. For example, data might be stored on a magnetic drum and the instructions punched on paper tape. In a "stored-program" machine, both data and instructions
are stored in the same memory, and a given m.emory location may be used to store either data
or instructions. The 650 is a stored-program machine and,as will be illustrated later, such a
machine can be programmed to generate its own instructions. This is possibly the most important identifying characteristic of a stored-program computer. The main difference between
a computer and a desk calculator is that once the computer is started it will follow a list of
instructions without further guidance, since it is merely necessary to change data. Another
advantage of most of the large and medium-scale computers is their speed of calculation. For
example, the IBM 650 can add two 10-digit numbers in 125~OOO of a minute, and can multiply
two 10-digit factors, obtaining a 20-digit product, in 60~O of a minute. A more meaningful
example is to note that the 650 can extract the square root of a 10-digit number, accurate to 9
or 10 significant digits, in about 0.152 seconds. In so doing, the 650 does about 60 program
steps ranging from storing a 10-digit number to dividing a 20-digit number by a 10-digit
number.

1-3. THE IBM 650.
This book contains an introduction to the programming of the IBM 650 Magnetic Drum
Computer (data processing machine). No previous knowledge of computing machines, nor of
IBM equipment, is assumed.
Figure 1.1 shows the three units which comprise the basic 650. The center is the 650
Con8ole, which houses the computing units. On the right is the 533 input-output unit, through

Input Cards

•
POWER UNIT

650 CONSOLE
Board Location

533 INPUT-OUTPUT UNIT

FIGURE 1-1
MAGNETIC DRUM DATA-PROCESSING MACHINE
TYPE 650

4 • SIMPLE PROGRAMMING

which punched cards are read into the 650 and from which results are punched out. The unit
in the background is the power unit. The 533 input-output unit contains a wired control panel,
by means of which all or part of the data on incoming cards may be fed into the 650 in various
combinations. The form of the output cards may also be altered by control-panel wiring.
Control-panel wiring is more varied in data-processing applications than in scientific computation, where two or three "standard boards" are often used. Elementary comments on board
wiring will be given as needed.

1-4. THE IBM CARD.

Each IBM card may contain 80 columns of information. Each column may be punched in
any of 12 posi tions:
+ (sometimes called 12, or High punch),

- (sometimes called 11 or X),
0,1, 2, 3, 4,5,6,7, 8, 9.
A combination of two punches in a given column is used to represent alphabetic data. A
glance at the sample card gives the alphabetic code key, so there is no reason to describe it
verbally. The 650 computes with numerical data. Normally, alphabetic information may be
transferred from input cards to output cards. Special usage is described under the SOAP
programming system (Chapter 5).

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FIGURE 1-2

1-5. WORD.

The basic numerical unit with which the 650 computer works is called a "WORD." It
consists of 10 decimal digits and a sign. Everything which goes into, or comes out of, the
650 will be in the form of "10-digit and a sign" words. An IBM card can contain 8 words (80
digits). The sign of each word is usually punched in the same column as the units digit of
the word (a double-punched column). Other arrangements are also possible.

SIMPLE PROGRAMMING. 5

1-6. DRUM.

The storage (formerly called "memory") unit of the 650 is a cobalt-nickel-plated cylinder
about 4: inches in diameter and 16 inches long called a Magnetic Drum or just Drum. The drum
revolves at about 12,500 rpm. Information is stored on the drum in the form of magnetized
spots. The drum is divided into 2000 word locations, each word containing 10 digits and a
sign. It is convenient to think of the surface of the drum as divided into 2000 rectangles,
each containing 10 digits and a sign. Each rectangle is then a "drum location." Figure 1.3
shows a layout of a portion of the drum. Each location is given a "street address" or "drum
location address" which is a 4:-digit number running from 0000 to 1999. Words may be thought
of as stored in 4:0 bands of 50 words each, running around the drum. Since the drum rotates at

FIGURE 1·3
MAGNETIC DRUM LAYOUT

6 • SIMPLE PROGRAMMING

12,500 rpm, the largest possible access time-Le., time to read out one word of information
from a specific location-would be .0048 second, with the average time being much less if a
sensible programming procedure is followed. A word stored on the drum remains there until
another word is stored on top of it, at which time the old word is first automatically canceled
by the new word. Words remain on the drum even if the power is cut off.

1-7. ACCUMULATOR.

The arithmetical unit is a 20-digit accumulator (like a 20-digit desk calculator). For convenience, it is thought of as divided into three parts, the upper half of the accumulator (10
digits), the lower half of the accumulator (10 digits), and a sign. In some operations it is
UPPER HALF

LOWER HALF

SIGN

I I II I I II I I III II I IIII I I
±

FIGURE 1·'ACCUMULATOR

desirable to use only the upper half of the accumulator; in some, to use only the lower half;
and, in some, to use both parts. However, the machine always behaves as if the entire 20digit accumulator were used, with zeros usually being placed on the unused section. A result
that is in the accumulator will remain" there until it is erased by a "reset" operation, or altered by an arithmetical operation, or until the power is shut off. (Capacitor storage which is
used in the accumulator does not "hold"-Le., remain indefinitely in the absence of regenerative power-as" does the magnetic storage of the drum.)
1-8. DISTRIBUTOR.

The distributor has a capacity of one word (10 digits and a sign). Any word transferred
between the drum and the accumulator passes through the distributor and remains there until
it is replaoed by another word. A word located in the distributor is available more quickly
IBM CARD
INPUT-OUTPUT

10 DIGITS AND SIGN
ACCUMULATOR 20 DIGITS AND SIGN
(ARITHMETIC UNIT)

FIGURE 1·5

SIMPLE PROGRAMMING. 7

than one on the drum, and should be used when appropriate. A word will remain in the distributor until another word is transferred from the drum to the accumulator, or from the accumulator to the drum, or until the power is shut off.
The schematic diagram in Figure 1.5 will help fix this important "routing pattern" in
your mind.

1-9. INSTRUCTIONS.
The IBM 650 is a stored-program machine. The instructions are stored on the drum in the
same form as data. Since the only thing which the drum can store is a set of 10-digit signed
numbers, the instructions to the machine are coded as 10-digit numbers. The 10 digits of an
instruction are sectioned into three groups as shown:
10

876

432

DIGIT POSITION

I I
NEXT

DATA

INSTRUCTION

CODE

ADDRESS

SIGN

GROUP
NAME

In general:
OPeration: The two digits on the left (positions 10,9) tell the machine what operation to perform and where to perform it. (Example: 65 means "Reset the entire 20-digit accumulator
to zero, and then add the contents of that storage location specified by the data address
to the lower half of the accumulator.")
Data Address: The four digits in positions 8-5 (Data Address) usually give the storage location needed in performing the OPeration code of the instruction.
Instruction Address: The four right-hand digits (positions 4-1) usually give the storage location of the next instruction,-i.e., tell the machine where to go for its next instruction
after the 0 Peration has been performed.
Sign: The sign of the instruction (±) is quite useful in connection with subroutines, but will
not be discussed until needed. A sign must appear on each instruction in storage, or the
machine will stop, but it is of no importance here what sign appears. For the present we
shall make them all +. The sign has no effect on the operation code.

1-10. A PROBLEM.
Let us assume that a deck of data cards, data deck 1, contains 600 IBM cards. The
first card word (Cols. 1-10) consists of a 10-digit signed number, of which the first three
digits are always zero; hence, the first card word contains a 7-digit number, Xi. The second
card word (Cols. 11-20) contains a 10-digit word of which the six high-order digits are all
zeros; hence, the second word contains a 4-digit number, Yi , with its sign. The other six
words on the card will not be discussed currently, since only the first two words are used in
this example.
Although the 650, itself, receives its instructions through numerical codes, it is convenient to have a symbolic letter code (mnemonic) which suggests the operations implied.
There are two of these code systems in common use. One is the "old code" found in the old
650 Manual, and the second is the SOAP (Symbolic Optimum Assembly Program) code. Since

8 • SIMPLE PROGRAMMING

WORDI

OOOXXXXXXX

0
0

WO~O::z

0
0

WORD 3

0
0

WORD4

0
0

WORD 5

WORD6

0
0

WORD 7

0
0

WORD

0
0

OOOOOOyyyy

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the reader will almost certainly wish eventually to learn to program using SOAP, we shall
adopt these symbols throughout. The main difference between the two codes is that each
SOAP operation code contains exactly three letters, while the old code has from two to five
letters.
EXAMPLE 1.

Compute the sum Zi = Xi + Yi for each of the 600 pairs of Xi' Yi which are given in deck 1,
words one and two. How would you explain to someone what you wish him to do, using a desk
calculator, if you were giving him this assignment? Roughly, you would probably tell him to
read the card and determine Xi and Yi • Next, he should compute Xi + Yi == Zi and mark Zi on a
blank card, then go to the next card and repeat the instructions.

READ A CARD
TO DETERMINE Xi AND Y

+Yi

COMPUTE Zi = XI

WRITE Z. ON A BLANK CARD

l
1

GO BACK TO STEP

This might be adequate if the person knew how to add on a desk calculator.
A more detailed set of .instructions might be: Read a card for Xi and Yi ; clear (reset) the
desk calculator (accumulator) to all zeros; add X i into the accumulator; add Yi into the accumulator; read out the answer Zi from the accumulator and remember the answer long enough
to write it in the desired columns of a blank card. Repeat these instructions with the next
card.
This detailed sequence of instructions is approximately what one does on a 650 computer.
The flow chart giving the general overall idea of the program is similar.

SIMPLE PROGRAMMING. 9

READ A CARD
TO DETERMINE Xi AND Y i

,l.

COMPUTE Xi

+ Yi

= Zj

1
PUNCH Zj ON A BLANK CARD

J~-----I

When writing a program, in addition to writing down instructions (10-digit words), it is also
necessary to keep track of the drum location in which the instruction is stored. A form similar
to that which the student has on his IBM 650 Planning Chart (Form 22-6151-2) is used for
this. We shall arbitrarily choose to locate our first instruction in drum location 0500 and each
successive instruction in a location 10 larger than its predecessor, e.g., 0500, 0510, 0520,
0530 ....

INSTRUCTION
-r-rDRUM LOCA TION
NEXT
OF
INSTRUCTION
INSTRUCTION
OPERATION DATA ADDRESS
ADDRESS
COMMENTS
0500

RCD 70

0510

1851

This first instruction ReD (Read CarD, or, sometimes, RD1-ReaD input one) 70 1851 0510is
to be located in drum location 0500. It tells the machine to read a card (8 words, 80 digits)
and to store the
contents of these 8 words in drum
locations
1851, 1852, ..• , 1858; then to go
to drum location 0510 for its next instruction. Note that both X.1 and Y.1 are now on the drum.
Xi is in location 1851, while Y i is in 1852. Drum locations 185~-1858 also contain data, but
they will not be used in this problem. Drum locations 1859, 1860 have zeros stored in them
through the board wiring used on the 533-input-output unit. (Caution: Unless an experienced
person has wired the board, these zeros may not have a sign with the word. Do not try to use
them unless you are certain they also have a sign. The program will stop if a number without
a sign is used.)
The next instruction is to Reset (clear) the entire accumulator and Add Xi (now in location 1851) into the Lower half of the accumulator (Note: The upper half of the accumulator
might have been used. The choice was arbitrary.)

INSTRUCTION
-r-

DRUM LOCATION

-r-

NEXT
INSTRUCTION
ADDRESS

OF
INSTRUCTION

OPERATION

0500

RCD 70

1851

0510

RAL 65

1a51

0520

DATA ADDRESS

COMMENTS

After performing these two operations, the machine now has Xi in its lower accumulator (and
distributor) and, according to the instruction address, is seeking its next instruction in drum
location 0520.

70 • SIMPLE PROGRAMMING

The program continues:
INSTRUCTION
DRUM LOCATION
OF
INSTRUCTION

-r-

OPERATION

-r-

DATA ADDRESS

NEXT
INSTRUCTION
ADDRESS

0500

RCD 70

1851

0510

OS10

RAL 55

18S1

0520

ALO

1852

0530

COMMENTS

The last instruction (in 0520) tells the machine to Ad~ (without reset) the contents of drum
location 1852 (namely, Y) to whatever is in the LOwer accumulator (it contains Xi)' thus forming Zi = Xi + Yi there, anA then to go to drum location 0530 for its next instruction.
We now wish to punch out Zi (the contents of the lower accumulator). To do so, it is
necessary to store Zi on the drum, since it is impossible to punch directly from the accumulator. Furthermore, there are only certain drum locations from which data may be punched.
(See shaded portion of Figure 1.3) Quite arbitrarily we select 1530 (one such location) for
this purpose. The program now reads:
INSTRUCTION
DRUM LOCATION
OF
INSTRUCTION

OPERATION

DATA ADDRESS

NEXT
INSTRUCTION
ADDRESS

0500

RCD 70

1851

0510

RAL

1851

0520

ALO

1852

0530

STL

1530

0540

which tells the machine to store the contents of the lower half of the accumulator in drum
location 1530 and to go to drum location 0540 for its next instruction. (Note: The number Zi
is now in three locations in the 650: in drum location 1530; in the lower accumulator; and,
since it just passed from accumulator to drum, it also appears in the distributor.) We now tell
the 650 to punch out the contents of drum location 1530 and to go back to location 0500 for
its next instruction, which is (ReD) 70 1851 0510-i.e., read another card and store it in drum
locations 1851 to 1858, then go to location 0510 for the next instruction.
INSTRUCTION
DRUM LOCATION
OF
INSTRUCTION

NEXT
OPERATION

DATA ADDRESS

INSTRUCTION
ADDRESS

0500

RCD 70

1851

0510

RAL

1851

0520

ALO

1852

0530

STL

1530

0540

PCH

1530

oseo

If this 5-step program is loaded onto the drum by some means (actually this, too, is done
through punched cards) and the set of 600 XiYi data cards placed· in the read hopper of the
533-read-punch unit, the machine will read the Xi and Yi from the first card when the program
is started, then reset (clear) the lower accumulator and add in Xi; add Yi to form Zi = Xi + Yi ;
store Zi in location 1530; punch Zi on a blank card in the output hopper of the 533 read-punch

SIMPLE PROGRAMMING. 77

unit; read the Xi and Yi from next card, etc., etc. When all 600 cards have been processed,
the 650 will receive the final instruction 70 1851 0510, but there will be no card to read, so
the machine will stop. The total time to read, compute, and punch these 600 cards will be
about 6 minutes. However, this is not to be interpreted that it takes the machine 6 minutes to
compute and store these 600 sums. Most of the time the computer. is merely sitting idle, waiting for the punch unit which has a maximum output capacity of 100 cards per minute. Actually,
the 650 could have computed and punched not only Xi + Yi , but also several other functions,
such as Vi = X~ - 4X iYi - Xi and Wi = X~Y~ - 3X i + Yi - 17, in addition to Zi = Xi + Yi , in the
same 6-minute period. It makes no difference, in time, whether only one word, Zi' or all eight
words of the output card are punched! -It would, of course, require a longer program to compute all three, Zi' Vi' and Wi' but the 650 would be able to compute and punch all of them in
the same 6 minutes it would take to compute and punch the Zi'S alone. This may give you a
clue as to why you must learn a little about how to program the 650, even if you only plan to
use it by having someone else write your programs. Often you can secure much more output
information for the same "price" if you plan for it in advance.
If the above program is used, the results will be a series of 600 cards, each of which will
contain a Zi in the fourth word (Col. 31-40) of the card, and zeros elsewhere. The output
card contains the contents of the eight drum locations 1527-1534 in words 1 to 8 of the card,
respectively, if the "standard" 8- to 10-digit word f>33 control board is used. The kth output
card will contain the sum of the two numbers on the kth input card.
If someone should happen to drop the answer pack of cards, there would be no way to get
them back into order. This is embarrassing and it is also costly. To avoid this difficulty,
you, the student, are asked (Problem 3, Set 1-11) to reprogram this problem in such a manner
that not only ZI but also ~ and Yi are punched on each output card. Store the Xi in drum location 1527 and the Y1 in location 1528. Your punch instruction can still be 71 1530 0500, although it will be located in a different drum location. Using one of the "standard control
boards," which we assume our 533 read-punch unit contains, any data address between 1500
and 1549 used with OP code 71 will cause all eight drum locations 1527 to 1534 to be punched
out in words 1 to 8, respectively, of the output card.

1-11. OPERATION CODES.

The operation codes needed in the following problem set are described here.
70 RCD (Read CarD). You may use RD1 in place of RCD: This operation code causes the
machine to read the 80 columns of an IBM card into the 10 read words (xxOl to xxl0, or
xx51 to xx60; see Fig. 1.3) in the band selected by the Data Address. The xx means any
valid number or address-for example, 0001 to 0010 or 1951 to 1960. With the" standard
5 33 board," we are assuming that card-word 1 (Cols. 1-10) becomes the first word in the
chosen read section, card-word 2 (Cols. 11-20) becomes the second word in the chosen
read section ... card-word 8 (Cols. 71-80) becomes the eighth word (i.e., xx08 or xx58) in
the chosen read section and 0000000000 is automatically placed in words 9 and 10. Anything previously stored in these read locations is lost when the contents of a new card
are read in. It takes 300 milliseconds (.3 second) to read a card into the machine, but of
these 300 milliseconds, 270 milliseconds are available for computing as a result of an
irigenious read buffer storage discussed in Chapter 2.
71 PCH (PunCH) You may use WRl (W Rite output one) in place of PCB: This causes 80 of
the 100 digits In 10 words of the punching section (xx27 to xx36, or xx77 to xx86; see Fig.

12 • SIMPLE PROGRAMMING

1.3) to be punched into an IBM card. Board wiring selects the digits and order in which
they are punched. Our' 'standard board" takes the first eight words of the punching
section of the drum (xx27 to xx34, or xx77 to xx84) into the eight words of the card, ignoring the last two words of the punching section.
24 STD (STore Distributor): The word (10 digits and sign) in the distributor is stored into the
drum location specified by the Data Address. The word also remains in the distributor.
15 ALO (Add to L Ower): The 10-digit-and-sign word in the location specified by the Data
Address is added to the lower 10 digits of the 20-digit accumulator. The word is found
in the distributor, and also in its former location, after the operation is complete.
16 SLO (Subtract from LOwer): The word in the location specified by the Data Address is subtracted (algebraic subtraction) from the lower 10 digits of the 20-digit accumulator. The
word is, of course, in the distributor and still in its former (drum) location after the
operation is complete.
65 RAL (Reset and Add into Lower):IReset (clear) the entire 20-digit accumulator to zero, and
add the word specified in the Data Address to the lower 10-digits of the accumulator.
The word is on the drum, in the distributor, and in the lower half of the accumulator after
the 65 operation is completed.
20 STL (STore Lower onto drum): The lower half of the accumulator (10 digits and sign) is
stored into the drum location specified by the Data Address, replacing whatever was already in the drum location. The contents of the lower half of the accumulator are undisturbed, and the stored word is in the distributor after the operation is complete.
69 LDD (LoaD Distributor): The word in the location specified. in the Data Address is brought
into the distributor. It also remains, unchanged, on the drum.
These eight operation codes will enable you to program a number of problems. In doing
so, you will not only gain familiarity with 650 technique, but certain properties of the operations themselves will be brought out in the problems. The problems in this manual are designed to help you learn how the computer works; and if you are to achieve maximum learning
in the shortest possible time, each problem should be attempted and the results discussed
with fellow students before the next class session.

Problem Set 1-11

1. The numbers given with each problem show the contents of the accumulator, distributor, and of a specific drum location, before the instruction is given. Write the contents of
each after the command has been executed.
Accumulator

OP
Code

Data
Address

Upper
(9003)

Lower
(0002)

Distributor
(9001)

Loc

(a)

LDD S9

1978

+ 0000098765

0000605679

"'70000000099

1978

+ 1234500000

(b)

STD 24

0030

+ 0 456558903

0000000000

-0000089997

0030

+0987999765

(c)

RAL 65

1944

- 9999999999

4444444444

+ 9999999955

1944

+ 9ee9999889

(d)

ALO 15

1776

-1111111111

889899999R

+098765432

1776

+ 55555$5555

(e)

SLO

1492

+ 1492177622

0000000000

- 9999988875

1492

+ 3333333333

(f)

SLO

1S98

+ 0000010000

0000088988

+0000000009

1588

+ 9000000009

Drum Storage
Contents

SIMPLE PROGRAMMING. 13
Accumulator

Data

Code

Address

Upper
(8003)

Lower
(8002)

Distributor
(8001)

Loc

Drum Storage
Contents

(g)

AL.O

0024

+ 0000000080

0000000777

+ 0006000777

0024

-0000033333

(h)

RAL.

0945

+9999999999

9999977786

+ 0908070706

0945

- 0001213140

(i)

STL.

1973

-0008800001

5555555555

- 3456723456

1973

+ 2345645376

(j)

STL.

1000

+ 0000000000-

0000000043

- 0987223434

1000

+ 3333343334

2. (a) Take a 650 Planning Chart (Form 22-6151-2) and write the 5-step program given in
Example 1. As you write each step, also note the contents of the upper accumulator, lower
accumulator~ and distributor for the first card which contains Xl = 0005555555 +,
y 1 = 0000003333 +. Assume that after the first operation~ 70 1851 0510, the upper half of
the accumulator contains the digits of your telephone number, the lower half contains the
digits of your girl's telephone number with a (+) sign, and that the distributor contains
1492177657 -, all of which are left over from a previous problem. The execution of a Read
instruction does not alter the contents of the distributor or accumulator.
(b) Show the contents of the upper accumulator, lower accumulator, and distributor after
each step of the program on the second time through, assuming that the second card read contains X 2 = 0005769121 +, Y 2 = 0000009342 -.
\.
3. Reprogram the problem of Example 1, Section 1-10, so that X i and Y i as well as Z i
will be punched in the output card, as suggested in Section 1-10.
4. Show the contents of the upper accumulator, lower accumulator, and distributor after
each step of the program you wrote in Problem 3. Do not use specific numbers, but fill in
known as zeros as zeros, and write OOO+--Xi~' OOOOOO+--Y~, OO+--Xi + Yi~'
etc., to show the general nature of contents as well as the range of digits which it will
invol vee For example:
INSTRUCTION
OP
RAL

UPPER ACCUMULATOR

1851

0510

0000000000

LOWER ACCUMULATOR
Ooo~Xi-~

DISTRIBUTOR

OOO+--Xj----..

5. Write a program which will use the same input Xi' Yi cards as in Example 1, but which
will compute T i = 2 X i - Y i and which will punch X i in word 1, Y i in word 2, and T i in word 6
of the output card.

STORE X,Y,
{ T IN PUNCH

6. Fill in the Accumulator and Distributor columns of your planning chart for the program
of Problem 5, using the type of notation described in Problem 4.
7. If the Xi'S were 2-place decimals, OOOxxxxx.xx, and Yi's were also 2-place decimals,
OOOOOOyy.yy, what, if any,difference would it make in your program for: (a) Problem 3
(b) Problem 5?

74 • SIMPLE PROGRAMMING

8. If the Xi'S were 2-place decimals, as in Problem 7, but the Yi's were whole numbers
(integers), what directions would you give your key-punch operator so that she could punch
the Y i'S in such a form that you could use the program of Problem 3 to obtain the correct
Zi=Xi+Y I ?
9. Alter the situation in Problem 3 by requiring that additional steps be written into the
program so that drum location 1534 will contain the sum of the Z /s from the first card to the
present card, in each case. What will appear on the output cards? (Hint: Remember 1534 is
part of the punchout band.)
k

10. In addition to placing

Z i in drum location 1534 on the kth card, also alter your

ia 1

program to compute LXi in location 1532 and
i=1

Y1 in location 1533. What will the 7th

1:01

output card contain? What will the 600 th output card contain? What will each accumulator
and the distributor contain when the machine stops after the 600th card? What will each contain after the power is shut off?
600

11. Is it possible that any of the sums

600

Zp LXi'

i= 1

600

might contain more than 10

1= 1

digits? Why must this possibility be considered?
12. In Section 1-5 we decided that, for the present, the sign of each instruction should
be +. However, no such restriction was placed on the data. Suppose some of the 600 cards
in Problems 3 and 5 contain negative valu~s for either X i or Y i ' or both. What, if any, changes
need be made in the programs?
k

13. Reprogram Problem 5 so that

is also punched on the k th card. Discuss the

i= 1
k

possibility that

might be larger than 10 digits, for some k in the range 1 ~ k ~ 600.

i= 1

14. Devise a program which will enable you to number 600 cards consecutively within the
numbers placed in the 8th word of the card. This could be combined with another program, if
desired.
15. Devise a program which will reproduce the first seyen card words (Col. 1-70) of each
card in Data Deck 1 on output cards, and will place a card number k in the 8th word· of the kth
card. (Hint: Use LDD-69 and STD-24 to move the words and use the program of Problem 14 to
number the cards.)

1-12. LOGICAL TEST-NZU.

In working Problem 3 of the last set, when did you place the Xi and Y1 into storage locations 1527 and 1528? It would be possible to do this after ZI was computed, but both computing time a.nd storage locations can be saved by storing a needed word when it is already in
the distributor. Although there is no time saving on this particular program (why not?), it is a
good habit to program economically, even when this is not essential. One possible program
would be:

SIMPLE PROGRAMMING. 15
INSTRUCTION
DRUM LOCATION

NEXT
INSTRUCTION

OF
INSTRUCTION

OPERATION

DATA ADDRESS

ADDRESS

0500

RCD 70

1851

0510

RAL

1851

0520

STD

1527

0530

ALO

1852

0540

STD

1528

0501*

STORE X FOR
PUNCH.
Zi = Xi + Y i
IN LOWER.
STORES Y j FOR

0501

STL

1530

0511 *

PUNCH.
STORES Zj FOR

0511

PCH

1530

0500

COMMENTS
{READS 8 WORDS
INTO 1851-58.
Xi IN LOWER.

PUNCH.
PUNCH AND
REPEAT.

*By using 0501 rather than 0550 as the address of our next instruction, the program is kept in the same
band on the drum.

Under the conditions of the given problem we were assured that the sum would not exceed
a 10-digit number. However, once a program is written, it may be used by others who fail to
observe this restriction; perhaps we, ourselves, know that we shall wish later to use the same
program to add a series of 10-digit numbers which might have an ll-digit sum. If an ll-digit
sum is developed in the lower accumulator, the 11th digit will be found in the low-order position of the upper half of the accumulator. (The reader should recall that there is, really, only
one accumulator, and that it is a 20-digit accumulator. It is merely convenient to speak of its
upper half and its lower half as separate entities.) An output card from the above program
looks like this:
OOO+-Xi~

000000
~Yi-+

~ZEROS""'"

LAST 10
DIGITS OF
X. + Y.
I

COL. NO.
CARD WORD
CORRESPONDING
DRUM LOCATION

WORD 1
1527

WORD 2

WORD 3

WORD 4

ZEROS

WORD 5 WORD 6 WORD 7 WORD 8

1530

1528

The zeros in word 3 and in words 5 to 8 are the result of having zeroed the drum before loading the program. We shall alter the program so that it will test the contents of the upper accumulator, and if it is not all zeros-i.e., if an overflow into the upper accumulator has occurred-will punch the contents of the upper accumulator into word 3 of the output card and
also punch a special warning by punching a 9 into Column 80 if, and only if, an ll-digit sum
was developed.
Operation code 44 NZU (Branch on Nonzero in Upper half of accumulator) is an entirely
new type of operation, namely, a logical test. This code performs no ari thmetical operation.
Instead, it tests the upper accumulator to see whether or not it is zero. If the upper accumulator is nonzero, the machine goes to the drum location given in the data address for its next
instruction-i.e., it branches on nonzero in the upper accumulator. If the upper accumulator
is zero, the machine goes to the drum location of the instruction address for its next.instruction as usual. Thus,
NZU 44

0621
IF NONZERO

0521
IF ZERO

16 • SIMPLE PROGRAMMING

will send the machine to drum location 0621 for its next instruction if the upper is nonzero,
and to 0521 if the upper is zero. We may use this in our program as follows:
INSTRUCTION
DRUM LOCATION
OF
INSTRUCTION

OPERATION

DATA ADDRESS

0500
0510
0520
0530
0540
0501
0511

RCD 70
RAL 65
STD 24
AL015
STD 24
STL 20

0521
0621

PCH 71
STU 21

1851
1851
1527
1852
1528
1530
0621
(IF NONZERO
UPPER)
1530
1529

NZU 44

NEXT
INSTRUCTION
ADDRESS
0510
0520
0530
0540
0501
0511
0521 (NEW INSTRUCTION)
(IF ZERO
UPPER)
0500
0631 (NEW INSTRUCTION)

Let us now suppose that drum location 1998 contains a special code word, say
0000000009, which we wish to have punched into word 8 of each card containing an ll-digit
sum. This may be accomplished as follows:
0631
0641

LDD 69
STD 24

1998
1534

0941
0521

If the sum in the accumulator contains 10 or fewer significant digits, the program will proceed
as before. If it contains 11 digits, we shall go from drum location 0511 to 0621. The instruction in 0621 places the digits from the upper accumulator into punch storage location 1529,
which contained all zeros in the program. The instructions in 0631 and 0641 place the special code word which identifies an ll-digit-sum card in punch storage location 1534, before
continuing with the old program in location 0521, as before. The instruction in 0521 (e.g.,
711530 0500) causes all eight words in 1527-1534 to be punched. Notice that many of the
instructions need not be written in any particular sequence. The important thing is that the
drum locations and data addresses be correct.
The reader should fill in this program on a 650 planning chart, filling in the accumulator
and distributor entries both when a 10-digit and when an ll-digit sum is developed. This is
a vital part of this course, the purpose of which is to teach you to think in a pattern compatible with the machine's operation-a quality which is essential in the successful use of
computers. This suggested program oontains a serious blunder. You can find it, if you will
fill in a planning chart. If an ll-digit sum is developed, all subsequent outputs will show 11digit sums and will contain a 9 punch in Col. 80. Why? See if you can discover the source of
this difficulty before continuing. Consider what the difficulty is-namely, once an 11-digit
sum is developed, an ll-digit sum and a 9 in Col.· 80 appear even when only a 10-digit sum is
developed subsequently-and discover why the 650 behaves so. Remember that the 650 does
exactly what you tell it to do-no more, and no less. Try your hand at "debugging" this program before you continue!
The source of the difficulty is that, when the overflow digit is placed into drum word 1529
and an identifying 0000000009 in drum word 1534, they remain there until they are replaced by
something else; and since neither location is disturbed when a 10-digit sum is developed, the
contents remain and are punched even though unwanted. If we want to avoid this difficulty,
we should store zeros (possibly from drum location 1859) into each of these locations (before
we punch) when the 44 test finds the upper all zeros. This is called "housekeeping."

SIMPLE PROGRAMMING. 77

EXAMPLE 1:

A flow chart of our proposed program looks like this (It is customary although not essential in
a flow chart, to indicate a branch by a long oval and two circles:

READ A CARD

COMPUTE

+ Y.I

= Z.I

Zj (LOWER 10 DIGITS)
IN PUNCH BAND

TEST:
IS UPPER HALF
OF ACCUMULATOR ZERO?

STORE UPPER IN 1529

STORE ZEROS INTO

AND STORE CODE

WORDS 1529

WORD IN 1534.

AND 1534.

The astute reader will be able to suggest other possible programs which will accomplish the
desired ends.
LOCATION
OF
INSTRUCTION

INSTRUCTION
NEXT
OPERATION

DATA
ADDRESS

INSTRUCTION
ADDRESS

COMMENT

0500

RCD

1851

0510

READS CARD INTO

0510

RAL

1851

0520

X. I NTO LOWER

0520

STD

1527

0530

STORE Xi IN
PUNCH LOCATION

0530

ALO

1852

0540

X. + Y. = Z.
I
I
I
IN LOWER

0540

STD

0501

STL

24
20

1528

0501
0511

1530

0511

NZU

0621
(IF NONZERO)

STORE Y. IN PUNCH
I
STORE Z. IN PUNCH
I
BRANCH

1851-58

0521
(IF ZERO)
0631

0621

STU

1529

0631
0641

LDD
STD

69
24

1998
1534

0641
0502

IF NONZERO UPPER
BRANCH
CODE WORD FROM 1998
CODE WORD INTO
PUNCH

0521

LDD

1859

0531

IF ZERO UPPER BRANCH

0531
0541

STD

1529

0541

STORE ZEROS IN 1529

STD

1534

0502

STORE ZEROS IN 1934

0502

PCH

1527

0500

PUNCH AND REPEAT

78 • SIMPLE PROGRAMMING

Another way to guard against developing an 11-digit sum is available if you perform the
operations in the upper half of the accumulator. When an 11-digit number is developed in the
upper half of the accumulator, ,an overflow occurs. When it does, the overflow light on the
console comes on, indicating that an overflow has taken place. If the overflow switch (also
on the console) is set to STOP, the program will stop. There is also a "branch on overflow"
operation, 47 BOV, which may be used to reroute the program if an overflow occurs, provided
the overflow switch is set in the SENSE position. As usual on branch operations, if the
specified condition occurs (here, an overflow), the next instruction executed will be in the
location given by the Data Address; otherwise the instruction address is used as usual.

1-13. MUL TIPLICA TION.

The operation mUltiply (19 MPY) is somewhat different in nature from those already discussed. The accumulator is reset and one of the 10-digit factors is placed in the upper ac-

cumulator before the instruction

MPY

is given. The

location of the other 10-digit factor is specified in the Data Address of the instruction
NEXT
INSTRUCTION
ADDRESS
I

• The 20-digit product stretches across both

the upper and lower halves of the accumulator. If it is known that the product will not contain more than 10 significant digits-say, the product of the two 4-digit factors, (OOOOOOxxxx).
(OOOOOOyyyy)-then the upper accumulator will contain all zeros, and only the lower accumulator is pertinent.
We have already discussed 2000 of the 2004 possible addressable locations of the IBM
650. (What are they?) The other four are the console switches (8000), the distributor (8001),
the lower accumulator (8002), and the upper accumulator (8003), each of which is a valid and
convenient data address or instruction address for most operations. The 800x series addresses are the most quickly accessible on the 650, and should be used where possible. The
650 will not accept an 800x series address as the data address of a store operation (20,21,22,
23,24); in fact, the machine will stop if an 800x series address is used with a store instruction. Storage of a word in an 800x address may be achieved by using OP codes 60,65,69.
EXAMPLE 1:

In a correlation analysis, 600 pairs of 5-digit integers (Xi' Y i) are available on a set of 600
IBM cards, with the X's in word 3 and the Y's in word 4. Either or both values may be
negative.
- 69999 ;£ Xi

OOOOOxxxxx ;£ 69999

69999 ;£ Y i

OOOOOyyyyy ;£ 69999.

We wish to compute the necessary results, and punch cards containing the following:

~WOR~
x.

Y.I

x~I

XIYi

WORD 2

WORD 3

WORD 4

x~I

- X.Y.
I

WORD 5

Y~I

y2_ X.Y.

WORD 6

WORD 7

y2_ x2

WORD 8

SIMPLE PROGRAMMING. 79

A flow chart follows:
READ A CARD TO
OBTAIN X. AND Y.
I

J.
COMPUTE Xi Y ,
i
STORI NG X. AND Y.
I

AND X.Y.
I

I N PUNCH-OUT BAND

!COMPUTE X2 AND X2 -

XY
WE DROP THE

AND STORE IN PUNCH-OUT BAND

SUBSCRIPTS HERE,

SINCE NO
COMPUTE y2 AND y2 -

XY AND
CONFUSION

STORE IN PUNCH-OUT BAND
WILL ARISE.

~
COMPUTE y2 - X2 AND STORE
IN PUNCH-OUT BAND.

1
PUNCH

The top of your 650 planning chart should be filled in to indicate that word 3 of input contains
a 5-digit Xi to be stored into drum location 0303, while word 4 contains a 5-digit Yi to be
stored into location 0304. The output words will be:
WORD

MEMORY
ADDRESS

1077

1078

1080

OUTPUT

1081
x2 -

1082

1083

1084

Y2_XY

y2_X2

A possible program follows: The command RAU 60 0303 1010 resets the entire 20-digit accumulator and adds the contents of drum location 0303 to the upper half of the accumulator before going to drum location 1010 for its next instruction.
INSTRUCTION
LOCATION
OF

INSTRUCTION

OPERATION

DATA
ADDRESS

1100
1105
1010
1015
1020
1025
1030
1035

70
60
STD 24
MPY 19
STD 24
STL 20
RAU 60
MPY 19

0301
0303
1077
0304
1078
1080
0303
8001

RCD

RAU

INSTRUCTION
ADDRESS

1105
1010
1015
1020
1025
1030
1035
1040

COMMENTS

STORE X
COMPUTE XY
STORE Y
STORE XY
COMPUTE x

(Continued on next page)

20 • SIMPLE PROGRAMMING
INSTRUCTION
LOCATION
OF
INSTRUCTION

DATA
ADDRESS

OPERATION

1040

STL

104S
1001
1006

SLO
STL
RAU

1011
1016

MPY
STL

1021

SLO

1026

STL

1031
1036
1041

SLO
STL
PCH

20
16
20
60
19
20
16
20
16
20

1079

1045

1080
1081
0304
8001

1001
1006
1011
1016

1082

1021

1080
1083

1026

1081

1036
1041

1084
1077

NEXT
INSTRUCTION
ADDRESS

1031

COMMENTS
STORE X2

COMPUT~

STORE X

X2 - XY
- XY

COMPUTE y2
STORE y2
COMPUTE y2 - XY
STORE y2 - XY
COMPUTE y2 _ X2

1100

The reader is expected to fill in the accumulator and distributor columns at each step of the
program, as he copies it on to a 650 planning chart. Attention is called to the fact that, if an
ll-digit product or a sum were developed, the program would ignore the most significant digit
and no one would be the wiser. Unless more is known about the nature of the data~ it is quite
possible that any of X 2 - XY~ y2 - XY or y2 - X2 might be ll-digit numbers. (Why?) In the
next problem set the reader is asked to reprogram the problem to take care of this possibility,
either by including 44 NZU operations which will stop the 650 by sending it to an invalid address, if an ll-digi t d ifferenceis developed~ or by using the upper half of the accumulator and
overflow stop. (Note: The difference between two 10-digit numbers can be an ll-digit number,
if the terms are opposite in sign.)
j

1-14. NEW OPERATION CODES•.
10 AUP (Add to UPper): The "lO-digit and sign" word in the location specified by the Data
Address is added to the upper half of the 20-digit accumulator. If an overflow occurs, the
overflow circuit will be activated. If the overflow switch is set to STOP, the program
will stop when an overflow occurs.
11 SUP (Subtract from UPper): Similar to 10, but subtracts.
60 RAU (Reset accumulator and Add to UPper half): Resets the entire 20-digit accumulator to
plus zero and adds the contents of the Data Address to the upper 10 digits of the
accumulator.
61 RSU (Reset and Subtract from Upper): Resets the entire 20-digi t accumulator and subtracts
the contents of the location given in the Data Address from upper half.
66 RSL (Reset and Subtract from Lower): Similar to 61,but subtracts from the lower half.
21 STU (STore Upper): Causes the upper half of the accumulator, with sign of the accumulator,
to be stored in the drum location specified by the Data Address. The contents of the
accumulator are unaffected by the operation. The word also appears in the distributor,
of course.
44 NZU (Branch on NonZero in Upper): If the upper half of the accumulator contains all zeros,
the next instruction to be performed will be found in the Instruction Address. If the upper
half is nonzero, the address of the next instruction performed will be found in the Data
Address. This operation does not affect the arithmetical units nor the storage units of
the 650, but provides a logical test by means of which different branches of a program
may be selected.
45 NZ E (Branch on NonZero Entire accumulator): Similar to 44, but the entire 20-digit accumulator is examined.
46 BMI (Branch on Minus): If the accumulator has a negative sign, the next instruction performed will be that in the location speci fled by the Data Address. Otherwise the Instruction Addres s will be used, as usual.

SIMPLE PROGRAMMING. 21

47 BOV (Branch on OVerflow): When the overflow switch on the console is set to SENSE, the
program will branch to the location specified in the Data Address if an overflow occurs,
but use the Instruction Address as usual if no overflow is present. If the console switch
is set to STOP the program will stop on overflow. (Note': If the console switch is in the
sense position and an overflow occurs, but the BOV is not used, the overflow light will
come on, but the 650 will continue to run ignoring the fact that an overflow occurred. Always leave the overflow switch in the STOP position unless you are using the BOV
OPeration code.)
00 NO P (No OPeration): The 650 merely goes to the Instruction Address for its next instruction. This code is useful for switching the path of the program if an instruction must be
deleted or added. Some valid Data Address (usually 0000) must be included in the Data
Address location or the machine will stop.
01 Hl T (Hal T or Stop): If the console programmed stop switch is set to STOP, the 650 will
stop at this operation. If the switch is set to RUN, the 01 code is treated as a NO OPertion (00) code. This is useful in debugging programs and will be discussed in greater detail later.
j

1-15. FLOW CHARTS.

The flow chart or block diagram is the heart of all true programming-be it for the 650 or
for another computer. No program of any complexity should even be discussed, let alone programmed, without a careful flow chart. A flow chart should show the exact sequence of operations including each branch and its alternatives. It should also indicate how and when loops
are tested. A flow chart does not show individual steps used in computing. The object of a
flow chart is to present the overall method of attack, without details. There is little or no
mention of specific 650 operation in a flow chart. A competent coder can take a flow chart
and translate it into a program for an IBM 650 or for a DATATRON or for one of the IBM 700
series, or almost any other computer. Flow charting is often used to explain processes which
do not involve any computer.
The bit of whimsey on the following page which has been making the rounds of computer
laboratories for several years may illustrate the point. It purports to be a program for getting to
8 o'clock class.
It would be hard to overemphasize the importance of flow charts. Indeed, the time may
come when you will make only the flow chart, leaving the details of programming to someone
(or something) else. "For transit" is an example of a "something."
j

Problem Set 1-15

1. Copy the program given at the end of Section 1-12 on a planning chart, filling in the
accumulator and distributor columns.
2. Reprogram Example 1, Section 1-12, using the upper accumulator so that, if an overflow
occurs, the program will stop.
3. Explain what the 650 will do upon receipt of each of the following instructions (beware
of booby traps). At the beginning of each of the operations, the following situation prevails:
8003
3333333333

(a) RAU 60

(b) SLO
(c) STU

16
21

a001
8002
a001

8002
2222222222 +

XXXX
XXXX
XXXX

8001
1111111111-

(d) LOO 69

(e) PCH 71
(f) SLO 16

a003
a002
a003

XXXX
XXXX
xxxx

Show the contents of each accumulator and the distributor at the· end of the operation.

22 • SIMPLE PROGRAMMING

TURN
OFF
AL.ARM

GROAN

SHAKE
WIFE

CRAWL.
OUT

KISS
WIFE

RESOL.VE TO
SET AL.ARM
EARLIER

BATHROOM
CHORES

WAL.K OUT
TO CAR
BREAKFAST
GO BACK
FOR THEM

DRIVE TO
GIVE HER

CL.ASS

$10.00

4. Explain the effect of each of the following operations on the entire accumulator which,
at the beginning of the operation, contains 20 nonzero digits and a plus sign.
(a) sUP

11 8003 XXXX

(d) RAL 65 8002

(b) RAU

60 8003

XXXX

(e) SUP

16 8003

XXXX

(f) RAU

60 8002 XXXX

XXXX

8002 XXXX

5. The following programs are intended to form a product of two 3-digit factors located in
drum locations 0133 and 0134. However, when the cards are punched and the first word of the
output card -examined, the desired product is not there. Can you "debug" the program?

SIMPLE PROGRAMMING. 23
(0)

LOCATION
OF

(b)

INSTRUCTION
0500
0505
0510
0515

OP
RAU
MPY
STU
PCH

INSTRUCTION

0328
0338
0337
0345

RAU 60
MPY 19

0734
0733

NZU
STU

0300
0727

0337
0345
0301

0345

PCH

0715

0300

60
19

0733
0734

0505
0510

21
71

0727

0515
XXXX

0730

LOCATION
OF

44
21
71

IA
0338

6. With the Xi' Yi from the set of 600 cards having X = OOOOOOxxxx in word one, -and Y
OOOOOOyyyy in word two, make up a flow chart and program which will accomplish the
following:
Form the necessary computations to compute and punch output cards of the following
form for each input card giving X and Y.

~O:D 1

Y
WORD 2

2X - Y

2x2 - XY

WORD 3

WORD 4

- X2y

WORD 5

t--

ZEROS

WORDS 6, 7, 8

7. Modify your program of Problem 6 before reading the next card, so that it will also
punch a digit 9 in Col. 80 if 2X a - X 2y + X is positive.
8. Make up a segment of a program which will accomplish the following:
Obtain A-B in the lower accumulator where A was stored in 0576 and B in 1576. If
A-B < 0, go to an instruction located in 1426; if A-B = 0, go to an instruction in 1626; and if
A-B > 0, go to an instruction in 1926. (Hint: Use both 45 NZE and 46 EM! operation codes.)
9. Write a flow-chart and a program which will use the 600 cards of Problem 6, and give
the following on the output card:
(a) Punch X in word 1 and Y in word 2.
(b) If X 2 < XY, punch the difference X 2 - XY in word 4.
(c) If X 2 = XY, punch an 8 in Column 80.
(d) If X 2 > XY, punch a 9 in Column 80 and punch the difference X 2 - XY in word 5. Be
sure to include the necessary housekeeping to erase unwanted previous results from the punch
band on the drum if nothing is stored there later.
10. Devise a program which will enable you to count the number of drum revolutions the
650 makes in a i-minute period, assuming that you can start and stop the program at will.
(Which you can do by pushing buttons on the console.)

(CHAPTER

21
11[H]~ ®~@
~@[M~@[L~

2-1. READ AND PUNCH.

Before continuing with our discussion of the IBM 650, it is essential that you learn more
of how it works. This will greatly improve your programming ability as well as assisting you
immeasurably in discovering flaws in your program.
You already know that the code 70 ReD causes the contents of an input card to be read
into the 650, and that it reads the entire card (80 digits) into either xxOl - xxl0 or xx·51 xx60 locations. We have assumed that ,the 80 columns of the card were read into the first 8
words of the read strip on the drum, and that zeros were fed into the other locations. Although
this may be done, it is not necessary, since control-panel wiring in the 533 read-punch unit
can be used to place any digit desired in any given position in the 100 (10 words) available
drum digits and any sign desired in any of the 10 available sign positions of these words.
This data may be taken from the input cards, or may consist of constants supplied by the 533
unit. The control panel may even be so wired that the location of a High Punch (+ or 12
punch) iIi certain columns will change the entire arrangement in which data are fed into the
drum. The wiring of these boards is a separate study, and need not concern us at present,
except that even nonprogramming users of the 650 must realize that the input is quite
versatile! Similar remarks apply to the output. This is especially important in business data
processing such as inventory, accounting, payroll applications, etc. Students interested in
these fields should study board wiring.
The 650 has an input rate of 200 cards per minute, but this does not mean that the machine is unable to compute during the 60/200 = 0.3 seconds = 300 milliseconds, during which
a card is read. An ingenious arrangement called a buffer storage makes 270 of these 300
milliseconds available for computation! If properly programmed, the 650 could perform about
300 10-digit additions and subtractions in the 257 milliseconds available.
The punch buffer acts in a similar fashion but twice as much time is available (110 drum
revolutions).
2-2. POWER CONTROL-SWITCH.

If a 650 is available, go and look at it. If not, turn to the front cover of this book where
the console control panel is shown. A word of warning to beginners. On the lower right-hand
24

THE 650 CONSOLE • 2S

edge is a recessed "Master Power" switch. This is for e:JJtreme emergency use only! If it is
turned to OFF, all power to the 650, including that to the cooling blowers, is cut off and
severe damage to the machine may result. Once this switch is turned to OFF, a special locking device is engaged which makes it impossible to turn the 650 on again. A customer engineer must be called in order to release the lock and, since damage to the 650 often results,
the machine may be out of order for a week or more. Stay away from the master power switch
unless a real emergency arises. The "power-off" and "power-on" buttons at the top of the
console are used to turn the power off and on. If the' 'power-off" button is pushed, the
"power-on" button should not be pushed until the drum has stopped rotating (about 5. minutes);
otherwise a belt may be snapped, again causing delay until the customer service engineer can
come to replace it.

2-3. DISPLAY LIGHTS.
Numerical data is displayed in a simple biquinary (2-5) code, requiring one light of a
pair of lights, and one light of a set of five lights to be lit to represent flo number.

•
•
~
•
•

•

•
•

?If

•

The pair of lights at the top tells which column of digits to read, and the set of five lights
between the two columns of digits tells which digit in the column to read. When the 650 is
computing, it is usually impossible to read these lights; they change so fast that the impression remains that almost all of the lights are continuously lit.
The top bank of lights contains 10 digits and a sign (see cover). By suitable positioning
of the display switch, the contents of either accumulator, the distributor, the program register,
or any drum location may be displayed there.
The lower bank of numerical display lights consists of a 2-digit operation register and a
4-digit address register. The remaining lights tell what the machine is doing and, if stopped
by an internal validity check, where an error was sensed. These lights are the best friend you
have when trouble arises (which it does-and almost invariably it is your fault, not the
machine's.)

2-4. FLOW OF INSTRUCTIONS.
Let us wander through the machine, as it executes a few steps of the program of Example
1, Section 1-13.

1100
1105
1010

OP
Reo 70
RAU 60
STD 24

DA
0301
0303
1077

1105
1010
1015

26 • THE 650 CONSOLE

The 650 is told (by means of a card, or the storage entry switches) to seek an instruction in
location 1100. Here is what happens:
OP LIGHTS

ADDRESS LIGHTS

BLANK

1100

The 650 now goes to location 1100, finding the instruction 70 03011105 which is brought into
the program register (this is inside of the 650, but can be displayed on the upper display
lights by proper setting of the display switch.) (Note: the sign is not shown when reading the
program register.)
After the instruction has been stored in the program register, the firs t 6 digits are then
moved into the operation and address registers. At this time the lower lights show the first
part of the instruction.
OP

ADDRESS

0:301

The 650 now sends the 10 words, which are already waiting in the buffer storage, over to
words 0301-0310. It then moves the last 4 digits (address of next instruction) of the word
stored in the program register into the address register, blanks out the operation register, and
displays this in the lower display lights
OP

ADDRESS

BLANK

1105

The 650 goes to location 1105, finding instruction 60 0303 1010 which is then brought into the
(internal) program register. While it is doing this, and for the next couple of hundred steps, it
is also independently reading another card onto the read buffer storage to be ready for the next
70 code read instruction when it comes. This is the slow part that takes about 27/100 of a
second, but the 650 can compute merrily, if it has computing to do, while this is being done.
After the instruction was brought into the program register, the first 6 digits were automatically brought into the operation and address registers, so that the lower display lights now
show:
OP

ADDRESS

0:30:3

and the 650 executes this instruction (reset the entire 20-digit accumulator and add the 10digit-plus-sign word stored in drum location 0303 to the upper half of the accumulator). The
last 4 digits from the program register are displayed next:
OP

ADDRESS

BLANK

1010

which sends the 650 scurrying to drum location 1010 for its next instruction, 24 1077 1015. It
places this in the internal program register and the lower lights
OP

ADDRESS

1077

THE 650 CONSOLE • 27

Next, this instruction is executed (store the 10-digit-plus-sign word now in the distributor
in.to drum location 1077). The location of the next instruction is then displayed, and the
machine goes there to find it, etc.
op

ADDRESS

BLANK

1015

At the same time the 533 read-punch unit is slowly (very slowly, considering the speed of the
operations going on in the arithmetic unit) reading another card into the read buffer storage.
The arithmetic unit may do several hundred arithmetic operations on 10-digit numbers in the
27/100 second needed for the 533 read-punch unit to get the 80 digits of information sorted
into the 100 digits of drum space (plus signs) in the read-buffer storage.
The progress of a program through the 650 can be followed step by step on the 650
console by using the half-cycle switch.
It is interesting to note the three quite distinct forms which a number has in its course
through the 650:
(a) a punched hole in an IBM card, which is "read" by the length of time a current is
interrupted while the card separates a brush from a charged roller as it passes between them.
(b) magnetized spots on the drum (permanent until remagnetized in a new pattern).
(c) capacitor storage (transient).

2-5. INTERNAL CHECKING OF THE 650.

One of the strong points of the IBM 650 Magnetic Drum Computer is the excellent system
of internal checking which has been engineered into its design. Every digit that leaves the
distributor or the accumulator (which means any time a word moves in the machine) is automatically checked to see if it has exactly one binary and exactly one quinary bit. * If not, the
machine will stop and the error-sense light will come on. The program register also has a
separate validity check built into it. In addition to this, the machine will also stop if any of
the following conditions occur:
(a) An invalid address is given on an instruction (i.e., other than 0000-1999, 8000, 8001,
8002, 8003, the latter four not being valid on store, read, or punch instructions).
(b) An invalid operation code is used (say, 29 1726 1344).
(c) An ll-digit number is developed in the upper accumulator (overflow), with the overflow switch set on STOP.
(d) Division by zero, or by an improperly positioned divisor, is attempted (causing what
is called overflow).
The reader should consult Figure 2-1 to help assimilate these facts.

Problem Set 2-5

1. (A class problem). Assign one member of the class to take the part of each of the
following portions of the 650 at the blackboard.
(a) OP and Address register.
(b) Distributor.
(c) Upper accumulator.
*Internally the 650 is not biquinary, but the distinction is not important here and now.

28 • THE 650 CONSOLE

INTERNAL FlOW SCHEMATIC
OUTPUT

INPUT

."".--___ ------

------

."".---

."".""'"

GENERAL STORAGE

FIGURE 2-1. INTERNAL FLOW SCHEMATIC

(d) Lower accumulator.
(e) Drum (show only location being used at the time).
(f) Read buffer and punch buffer.
Carry out the steps of Example 1, Section 1-13, showing exactly what happens at each step.
2. (A class problem). (a) Use, the technique of Problem 1 to determine what will happen
on the following program. Assume data input cards having X = OOOOXXXXXX in word 1 and
Y = 0000000 YYY in word 2 are used.
LOCATION
OF
INSTRUCTION

0190
0199
0207
0208
0209
0210
0211

70
60
MPY 19
STD 24
STU 21
STL 20
PCH 71
RCD
RAU

1801
1802
1801
1831
1828
1827
1830

0199
0207
0208
0209
0210
0207
0190

sic.

(b) Assume that the input words contained 8 significant digits each. What effect would
this have on the machine's behavior? Assume that, if an ll-digit number is developed in the
upper accumulator, the 650 will stop (overflow), but if developed in the lower, it will merely
use the upper to store the extra digits. (c) Answer the same questions assuming that 0210
contains 20 1827 0211, and 0211 contains 71 1830 0207.
3. What happens if the following program is used, assuming that the cards of Problem 2
are used? Describe the output cards, if any. Code RAM 67 is Reset the entire accumulator and Add into the lower half the Magnitude (absolute value) of the 10-digit word
given in the location specified by the Data Address.

THE 650 CONSOLE. 29
LOCATION
OF
INSTRUCTION
0200
0209
0217
0235
0267
0284
0201
0218
0220
0225

OP
RCD
RAM
STL
RAM
RAU
STD
MPY
STU
STL
PCH

70
67
20
67
60
24
19
21
20
71

1851
1852
1878
1851
8002
1877
1878
1879
1880
1852

0209
0217
0235
0267
0284
0201
0218
0220
0225
0200

(LOWER)
(LOWER)

4. What will the following program do, assuming that location 1099 contains the constant
0000000007, that 1098 contains 0000000001, and that location 0200 is the first instruction of
a valid program which will multiply the number, X, in location 1027 by the number 0000000002,
store the product back into 1027, ·and then go to 0399 for its next instruction. Describe output
cards, if any.
LOCATION
OF
INSTRUCTION

1901
0501
RCD 70
0502
1901
0503
0502
LDD 69
1027
0504
0503
STD 24
0504
STD 24
1028
0200
0200 {PROGRAM DESCRIBED ABOVE, }
MANY STEPS
0399
0399
RAL 65
1099
0400
0401
0400
1098
SLO 16
0402
0401
0200
NZE 45
1027
0501
0402
PCH 71

SUBROUTINE

5. Explain how the program above would need to be modified, both in the portion given
and in the subroutine, if the output card were to contain both X and X(2) 71 assuming that X is
of the form OOOOOOXXXX.
6. Would the program of Problem 4 punch X and X(24)7 if X were a 4-digit number and if
the drum location containing the 0000000002 contained 0000000024 instead?
7. (a) Describe the effect of the three following programs. Assume drum location 1999
contains the number 0000000001, and that 1998 contains the number 1080018002
PROGRAM A

PROGRAM B

LOCATION

0120
0200

RAU 60
AUP 10

1999
1999

0200
0200

1319
1320

RAU 60
AUP 10

1999
8001

1320
1320

PROGRAM C
LOCATION
1520
1521

OP
RAL 65
AUP 10

1998
1999

1521
8002

(b) Which program will have a higher number in the upper aooumulator at the end of 1
minute? Why? Explain what each does.
(N ote: The remainder of Chapter 2 may be read quickly now, and referred to in more detail as
the need arises.)

30 • THE 650 CONSOLE

2.6. LOAD CARDS.
A wide variety of card forms, both for input and output, is available as a result of board'
wiring. No matter what board wiring is used, a card having a "high punch" (12 punch) in
Column 1 will be known as a load card. It would be quite possible to use a high punch in any
other column to designate a load card, but we shall use Column 1. (Note: A high punch is, in
appearance, the same as a "12" or plus punch. Since it is used to control the way the card
is read in, rather than as the sign of a word, ,it is given the name of "high punch.") This load
(high) punch in Column 1 does two things:
1. It instructs the 533 to ignore all board wiring on the control panel and to read the contents of the card in as 80-80-i.e., just as we have been doing so far, with zeros entered into
words 9 and 10.
2. In a read instruction, ReO 70 DA lA, if the card read is a load card, the next instruction will be obtained from the location specified by the Data Address, -rather than by the Instruction Address (i.e., a Branch on load card is built into the machine.)
The primary use of load cards is to load instructions and constants onto the drum. However, load cards may be used at any time, supplying a branching operation. All 10 columns of
a word which ,will be used in the 650 must be punched, ·and each word must have a sign (+, ,-)
punch over the units positions of every word used. If a word is not used in the program, this
restriction does not apply.
Standard "prepunched" cards of the form shown below are often used for loading in structions onto the 650 drum.
Cards are prepunched with 69 1954 1953 in word 1, 24 - - -'- 8000 in word 3, and high punches
(+ signs) placed in Columns 1, 10, 20, 30, 40,
Card word 1 (Cols. 1-10)
Card word 2 (Cols. 11-20)
Card word 3 (Cols. 21-30)

Card word 4 (Cols. 31-40)
Card words 5 to 8 (Cols. 41-80

69 1954
1953
(Prepunched)
Blank now. Will contain problem reference data if des ired.
24 - - - - 8000 The blank DA will contain the drum address into which the
instruction will be stored.
Will contain instruction to be punched.
Blank now. Used to punch descriptive
comments if desired.

Set storage entry switches (8000) on the console to 70 1951 9000. Let us examine what
happens.
LOCATION
OF
INSTRUCTION

1951

9000

8000
1951

RCD

LDD

1954

1953

STD

xxxx

8000

1954

THIS IS THE 10-DIGIT INSTRUCTlON WHICH IS TO BE STORED.

(WHERE xxxx REPRESENTS THE
DRUM LOCATION IN WHICH THE
INSTRUCTION IN 1954 IS TO
BE STORED.)

What does this program do?
8000 (console switches) 70 1951 9000 says, "Read a card into drum locations 1951 to
1958 and, since it is a load card,go to 1951 for the next instruction."
1951 is the first word on the card (Cols. 1-10) which says, "Load distributor with the
contents of 1954 (the instruction to be placed on the drum) and go to 1953 for the next instruction. ' ,

THE 650 CONSOLE • 37

1953 says, "Store distributor into xxxx (the desired drum location) and go back to the
console 8000 for your next instruction.
Read another card into drum location 1951-1958 and •••• "
Eventually we may come to ::t nonload card (no high punch in Col. 1). In this case the
650 goes to drum location 9000 for its next instruction and the card is also read in through
board wiring. Since 9000 is an invalid address, the program stops on a data-address storageselection error and does not proceed until sent to an address containing a valid instruction.
This is easily done using a transfer card, described later. The transfer card, a load card,
contains 00 0000 NNNN in word 1 where NNNN is the location of the first instruction in the
program. How does it work?
This is a one-instruction-per-card load routine. Each card carries its own complete loading program. Two hundred instructions, constants, etc., may be fed into the 650 in 1 minute.
This method is almost universally used to debug programs, -since corrections can be made by
merely inserting another card at the end of the deck; there is no need to remove the card which
is in error, since the correction card fed in at the end of the deck will store the correct instruction in the drum location, erasing whatever was there before. After all errors are corrected, it is possible to shorten the input time on a long program by using load routines which
store more than one instruction per card. There are 4:'per-card, 5-per-card, 6-per-card, 7-percard and even 8-per-card load routines, each of which has certain advantages. Now, we shall
use only the l-per-card load routirie.

2-7. BOARD WIRING FOR LOAD CARDS.

A person familiar with general board wiring for IBM machines will find the 533 board
familiar in most details. Others may find it quite awesome. We merely observe here that in
any installation it will always be possible to have a board wired to meet your own personal
needs by someone who understands this phase of the work, and that "standard boards" will
be available in permanent form. At present we describe only the wiring needed for load cards.
This consists of one wire running from the hub marked 1 in the "first-reading" section to
either of the two hubs marked "load," as shown on the diagram below. If you are
interested in board wiring, consult the IBM 650 Programmers Manual for further details.
10

FIRST READING

25
41

0
ENTRY B

0-0

0
45

0
65

2-8. OPERATING A:ND CHECKING LIGHTS.

The best friends you will have in debugging a program on the 650 will be the operating
lights and the checking lights on the console.
The six operating lights tell what the 650 is doing.
1. The Data Address and, 2. Instructions Address Lights, indicate which half-cycle of
the program is ready to be executed (see Section 2-3).
3. The H Program" Light indicates a programmed stop (01), manual stop (button on console), or address stop ( console).
4. The Accumulator Light is on whenever the accumulator is in use.

32 • THE 650 CONSOLE

5. The Input-Output Light indicates that the machine is ready to use one of these facilities. If the 650 stops with the in put-output lights on, look for the following reasons:
(a) You forgot to press the 533 read or punch feed button, hence no cards are available when the program calls for them.
(b) The read or punch hopper has run out of cards.
(c) Someone pressed the read or the punch stop key.
(d) A 'feed failure (often the result of a bent or torn card) has occurred. Be careful of
this, or a card jam may result.
6. The Overflow Light will come on if overflow is detected. An overflow condition can
be caused by one of the following:
(a) An excessive accumulation.
(b) Trying to develop a quotient of more than 10 digits.
(c) Trying to exceed the number of shifts called for in a shift and count operation.
The eight checking lights show up the following causes of program stoppage:
1. The Program Register Light indicates the detection of a validity error at the output of
the program register.
2. The Storage Selection Light indicates errors of the following type:
(a) A store operation with a data address of the 8000 series.
(b) An invalid D or I address (other than 0000 through 1999 and 8000 through 8003).
(c) An attempted manual entry from storage entry switches to 8001, 8002, or 8003.
3. The Distributor Light will come on if a validity error is detected at the output of the
distributor.
4. The Accumulator Light comes on if a validity error is detected at the output of the
accumulator.
5. The Clocking Light lights if an error is detected in the clocking (timing) circuitry.
This is a machine error. If it occurs, try to run your program again. If trouble persists, give
the 650 a rest and tell your service man about it.
6. The Error-Sense Light is operative when the error-sense switch is in the SENSE position and will come on when one of the following errors has been detected:
(a) Validity error
(1) Program register
(2) Accumulator
(3) Distributor
(b) Clocking (timing) error
This light will remain on until reset by the operator. It indicates machine trouble. The
same comments as given in 5 apply •. Overheating due to inadequate venting is a common
trouble maker.
7. and 8. The Control Unit and Auxiliary' Lights (not on all units) indicate difficulties in
these sources.
One of the most common troubles is to "get into a loop or repeated sequence of instructions and not get out of it." This may occur if an error is made in an instruction address,
sending the program back to a previous instruction. For example, if the next to the last instruction of Example 1, Section 1-12, ,had been
DRUM LOCATION
0530

1530

0510

(THE IA SHOULD HAVE BEEN 0540)

the program would loop back to 0510 and repeat .itself over and over again without ever reaching the 0540 step of "punch and go to read card. ,.,
If, through an error, the 650 is given an IA of a drum location not used in this program, it
will "take off" on whatever it finds there. If someone else has a program in that location,

THE 650 CONSOLE. 33

you may find yourself working his problem with your data-an unsatisfactory arrangement. To
prevent this, it is usual to load the entire drum with all zeros before starting to load in a
program. Several good "drum zero routines" are available. The author prefers to "stop
code" the drum rather than zero it. This places 01 NNNN 0000 in drum location NNNN.
If the 650 is sent to a drum location containing 00 0000 0000, it behaves as follows:
Since 00 says "no operation, go to Instruction Address," it does. If drum location 0000 also
contains 00 0000 0000, the machine will very rapidly do
"no op on 0000 and go to 0000; no op on 0000 and go to 0000; no op on 0000 and go to
0000; etc."
The lights will give the impression that the machine has stopped, since it will be repeating
"no op and go to 0000" at the rate of 12,500 times per minute. (Query: Since it takes less
than .003 second to perform a "no op," why are there not more than 12,500 repetitions per
minute?) The reader should decide what happens if 01 NNNN 0000 is in drum location NNNN.
HLT 01 is the operation code which halts the program if the program switch is on STOP.
(Note: Drum zero routines usually clear the drum to zero with a minus sign. There is a very
good reason for this. In most mathematical operations resulting in a zero, the computer produces a zero with a plus sign. Hence, a plus zero in a given storage location indicates that
it is the result of computation, not of the drum-clearing routine.)

2·9. ADDRESS STOP.

In debugging programs it is often useful to check and see how far a program runs before
getting into trouble. The Address Stop permits this. Set the control switch on the console to
Address Stop. Set the desired address at which you wish to stop into the four address selection switches. Press the program start button. Every time the program uses the stop address
as either a Data Address or an Instruction Address, the computer will stop. When Program
Start is depressed, the computer will continue with the program.

2-10. PROGRAM START.

The address selection switches are also useful in other ways. If you wish to be"gin computing on a program with a certain instruction (possibly, but not necessarily, the first instruction in the program), the following steps are useful.
1. Set the control switch to manual.
2. Set drum location [ i ] of the first instruction to be executed into the address
switches.
3. Depress the Program Reset (or Computer Reset) key.*
4. Depress the Transfer key (the address register will now show the address of the drum
location to which you desire to go).
5. Set the control switch to Run or to Address Stop depending upon your wishes.
6. Depress the Program Start key.
Another method of starting the program involves the set of 11 storage entry switches:
1. Set 00 0000 [ i ] in the storage entry switches where [ i ] is the drum location of
the first instruction to be executed.
2. Depress the Program Reset or the Computer Reset* key.
3. Set the control switches in either Run or Address Stop positions.
4:. Depress the Program Start Key.
*The Program Reset will reset the program register only. The Computer Reset will reset both the
program register and the accumulator.

34 • THE 650 CONSOLE

2.11. COOKBOOK DIRECTIONS.
For the convenience of the beginner running his first 650 program, we present the following "cookbook" directions for running the 650. This approach may be justified during the
first two weeks of 650 use, but, after that, understanding of the 650 and its operations should
replace such methods. If you are to learn to use the 650 intelligently, a general understanding
of its construction is essential.

650 Console Operations

To Load and Start Program (Using transfer card).

O. Punch in on the time clock, if one is used.
1. Be sure that the correct control panel is in 533.
2. Obtain drum zero cards from your instructor or the library file. Place the load cards
behind (i.e., so they are read later ~y the 533) the zero drum card{s).
3. Place the transfer card behind the load cards. (Transfer card: Cols. 1-10, 00 0000
xxxx, where xxxx is the location of the first instruction overpunches in Columns 1 and 10.)
4. Place the data cards behind the transfer card. (Data cards are cards which will be
read in by program.)
5. Place the above deck in the read hopper, 12 edge first, face down.
6. Place blank cards in the punch hopper. (Be sure that the punch feed has been cleaned
out-i.e., hold the punch start button down 3 or 4 cycles while the cards are removed from the
hoppe~ ifin doubL)
7. Set the Storage Entry switches to 70 1951 9000 •
. s. Set the Control switch on RUN.
9. Set the Display switch to ACCUMULATOR, DISTRIBUTOR, OR PROGRAM REGISTER.
10. Set the Half cycle switch on RUN.
11. The Programmed Stop, Overflow, and Error switches can be set according to desire.
12. Push the Computer Reset. (This destroys contents of accumulators and distributor.)
13. Push the Program Start.
14. Push the Read Start button on 533.
15. Push the Punch Start button on 533.
To read the contents of a given drum location (say 0546):
1. Push the Program Stop button.
Set the Address Selection switches to 0546.
Set the Control switch on MANUAL.
Set the Display switch to READ-OUT STORAGE.
Push the Program Reset button. (This destroys the contents of distributor.)
Push the Trans fer button.
Push the Program Start button.

2.
3.
4.
5.
6.
7.
S.

Read the contents of 0546 in the DISPLAY lights.

Problem Set 2-11

1. The following simple program may be punched on a single card; the overpunches (plus
signs) are placed in Columns 1, 10, 30, 40, 50, 60, 70, so. A minus overpunch is used in
Column 20. Console switches are set to 70 0004 xxxx.

THE 650 CONSOLE. 35
LOCATION OF
INSTRUCTION

8000
0004

RCD 70
RSU 61

9900
0007

0007

LDD 69

0004
0008
0006

0005

STD 24

'0000
1999

+
+
+
+

8003

STL 20

[xxxx]

0003

AUP 10
00

0000
0001
0001

0000
8003
0000

0002

0000

0006

HLT 01
STL 20

0000
1999

8000
0003

8002
0003
0001

0008

0005
8003

COMMENT
CONSOLE SWITCH ES
CARD WORD 4
CARD WORD 7
CARD WORD 5
{NOT ON DRUM. IN UPPER ACCUMULATOR. ADDRESS
[xxxx] WILL CHANGE IN PROGRAM. IT STARTS AS

+
+

1999.
NOT ON DRUM, BUT IN LOWER ACCUMULATOR
CARD WORD 3
CONSTANT, CARD WORD 1
CONSTANT, CARD WORD 2

+
+

CARD WORD 6
CARD WORD 8

Since the card is a load card, the 650 goes to drum location 0004 for its first instruction after
having read the card into words 0001 to 0008. This routine will clear all drum locations to
00 0000 0000 - except locations 0000 and 0001; these will contain 01 0000 8000 and
00 0001 0000, respectively. Explain how it works. Make out a l?l'~nning chart and fill in the
accumulator and distributor columns. This is not the best drum-clearing routine available, by
any means, but it is simple and satisfactory. If you have a 650 available, time it to see how
long it takes. ,Can you think of a possible advantage of all zeros and a minus sign in almost
every location on the drum before beginning to load a program?
2. A load card contains a high punch (+) in Columns 1, 10, 20, 30, 40, 50, 60, 70, 80, and
the following information:
Card word 1 60 0002 8003
Card word 2 20 LLLL 0003 (fill in LLLL with lower clearing limit)
Card word 3 10 0004 0005
Card word 4 00 0001 0000
Card word 5 11 0006 0007
Card word 6 20 UUUU 0003 (fill in UUUU with upper clearing limit)
Card word 7 44 0008 8000
Card word 8 10 8001 8003
Set the console switches to 70 0001 9000. The program will then clear the drum to zero
between the limits (drum locations) given in the Data Address of words 2 and 6.
(a) How does it work?
(b) Will the drum locations specified in the limits also be cleared to zero?
(c) Will the cleared locations contain zero and a plus sign or zero and a minus sign?
(d) What happens if LLLL 0001 and UUUU = 1492?
3. Make up a drum-clearing routine similar to that in Problem 2 but having the following
characteristics: (a) It uses console switch settings of 70 1951 9000 and (b) clears between
limits, but (c) the, upper limit must be less than 1900.
4. What will happen if someone uses the program you have written in Problem 3, but
places 1999 as the upper limit?
5. The reader is already familiar with the example of Section 1-12. Your instructor will
furnish you with a set of load cards supposedly for this program, but which actually contain
an error. Different errors may be given to different students. Debug the program.
6. The following program is supposed to re'ad a card into the 650, compute the product of
the second and the seventh card words and punch them and their product into an output card.
Debug this routine before the cards are punched, punch the cards, and test the revised program
on the 650.
::%

36 • THE 650 CONSOLE
LOCATION OF
INSTRUCTION

0901
0902
0903
0904
0916
0927
0928
0934

RCD 70
AUP10
STD 24
MPY19
STD 24
STL 20
STU 21
PCH 71

1535
1502
1377
1507
1378
1381
1380
1377

0902
0903
0904
0916
0927
0928
0934
0901

In Problems 7-15 certain common troubles encountered in running the 650 are described.
See if you can diagnose the possible sources of trouble, and explain its cause and cure.
7. (a) The .program deck is placed in the 533 read hopper and the switches on the console
are properly set. David pushes the Computer Reset and the Program Start buttons but the
cards are not read into the 650. Since the input,.output light is on, he reads the OP register
which shows 70 1951. What should he do next?
(b) David now pushes the read button on the 533 and two or three cards are fed into the
533, but then action stops again. This time the following lights are lit: Instruction Address,
Program, Storage Selection. Can you diagnose David's trouble? (If not, 'see Step 1 in the
cookbook instructions)
8. David fixes the trouble of Problem 7 and begins again. All goes serenely for a while
and about 30 or 40 cards read in and load onto the desired positions on the drum when suddenly action stops again. The OP register is blank and the address register shows 9000. The
following lights are lit: Instruction, Program, Storage Selection. Since the trouble is indicated
in the Program register, David displays it, finding 70 1951 9000. Can you suggest at least
three possible sources of David's trouble?
9. David's instructor removes the cards which remain in the 533 read hopper and holds
down the start button to clean out the cards inside the 533 read. He then counts back four
cards from the last one which comes through and shows it to David, who notices that it does
not have a high (load) punch in Column 1. David punches a high punch in Column 1 and returns it to the pack and begins again. Exactly the same thing happened as in Problem 8 (but
from a different cause). David repeats his instructor's action and finds the fourth card from
the end is the same card he had trouble with before. He's puzzled at first, but then realizes
what the trouble is. He returns the card to the pack, glances quickly at a few cards on each
side of it, and starts again. This time the entire program loads into the 650 and the transfer
card starts the program which runs perfectly. What was David's trouble?
10. Alice punches and verifies a program from an old listing. The program and the cards
are both correct. She then places three blank cards on the end of the program (which requires
no data cards) and loads it onto the 650 drum in the usual fashion. Everything goes fine down
to the last card, but then the action stops. The address register shows 9000 and the instruction, program, and storage selection lights are on. Alice reads the cookbook instructions and
finds her trouble. Can you?
+
11. Alice punches an 8-word card containing 00 0000 1346 in word 1, with words 2 to 8
blank to send the 650 to drum location 1346 where her program starts. She reloads the program but her trouble of Problem 10 is repeated. What did she fail to do to her load card,and
why did it cause trouble?
12. Lois finds an error in her program. An instruction reads 60 0813 1645 but should
read 60 0813 0645. She "duplicates" the first 36 columns of the card using the key punch
"duplicate" button and punches 0645 in columns 37-40. When she reloads her program,action stops with 69 1954 in the operation lights and the Data, Program, and Distributor lights
lit. What did Lois overlook on the card she repunched?

THE 650 CONSOLE • 37

13. Lois corrects her error of Problem 12 by punching one additional punch, and then
loads her program. Since she has a valid transfer card between the program cards and the data
cards, the 650 transfers into her program, reads a data card, 'and computes a while. Suddenly
the data address and the input-output lights come on and the operation register shows
11 0112. What did Lois forget to do?
14. Betty punches the necessary button to let Lois' program continue. It does so, ,reading, computing, and punching merrily along for about two or three minutes when suddenly the
same lights appear as in Problem 13. What happened this time?
15. Betty was so pleased with the result of her button punching that she punched a few
more 533 buttons. The "end of file" light was lit when Lois returned and the action had
stopped. Lois quite properly depressed the stop button on the 533, turning off the end of file
light, before depressing the read and punch buttons. Her program ran fine and, since she had
placed three blank cards at the end of the data cards, all the data cards fed into the 650 and
all the results were punched. However, ,when she listed (407 or 402) her output cards, the
answers corresponding to the last one or two input cards were missing. Where are they?

(CHAPTER

31
[M)@OO~ IA@W IAOO~~@

[f)OO@~OOIA[M)[M)~OO~

3-1. IMPROVING SPEED.

In Problem 3 Set 1-11, Z I Xl + Yl' it took six minutes to compute Z I and punch Xl' Y I
and Z I for 600 values of i. The time needed to obtain the 600 Z I may 'be cut in two by punching two sets of results (X 2i - 1 Y2i - 1 Z2i_t :X 21 Y 2i Z2i for i = 1,2, •.• ,300) into each output
card. No further increase in speed is possible unless the input cards are also altered. It may
be both feasible and economical to alter the input cards. If a less expensive machine, like a
summary punch, were used to put 4 pairs of Xi' YI onto each input card, the program could be
speeded up even more .. If, for example, a program were written which would read in and compute Z i for 8 pairs Xi Y i from two of the new cards, and punch all 8 answers plus some identifying mark (quite possible, since the 8 answers will need only 64 of the 80 available columns),
it would be possible to compute all 600 values in 45 seconds-a savings of $7 (at $80 per
hour for 650 time).
Actually, the amounts involved in the above problem are small, and probably not worth
the effort of reprogramming, but the ideas are essential. The given problem, Z 1 = Xi + Y l' is
not even a reasonable 650 problem, but would be relegated to a smaller, less expensive machine in most production schedules. However, it is an excellent instructional problem, since
it offers much leeway for discussion without the more involved considerations of advanced
programming.
\
:II:

3-2. STEPPING INSTRUCTIONS.

In previous problems the data have been read from cards as needed, and punched into
cards as soon as they were computed. In actual programming, part of the data may be on cards
and part on the drum from a previous calculation. A very simple example, which serves to
introduce the next topic, would be to compute SI = Xi + Yi - C where the single constant C is
already on the drum (in location 1001) while Xi' Y 1 are read into the locations 1953 and 1954
from words 3 and 4 of cards. The Si are to be punched into output cards as computed. A possible program will be the following:
38

MORE ADVANCED PROGRAMMING. 39
LOCATION OF
OP

RCD 70
RAU 60
AUP 10
SUP 11
STU 21
PCH 71

1951
1953
1954
1001
1982
1977

0510
0520
0530
0540
0501
0500

INSTRUCTION

0500
0510
0520
0530
0540
0501

COMPUTE AND
STORE
SI = XI

+ VI -

COMMENT
READ X AND VI
X
I
I
XI +Y I
SI = Xi + V I - C
STORE S.'·
I

PUNCH S.
I

This is not a particularly efficient program, but it serves as an introduc~ion to our main problem. In the above program the same C was subtracted from each Xi + Y i' Consider now the
more realistic problem in which 600 values of C i are stored in drum locations 1001 to 1600,
and it is desired to compute Pi = Xi + Y i - C i obtaining C i from the drum and Xi and Yi from
words 3 and 4: of cards, as before, and to store the Pi in locations 1001 to 1600 (Le., store
P k in drum location which had contained C k ). Your first thought may be:
LOCATION OF
INSTRUCTION
(READ CARD 1

0500

RCD 70

1951

0510

READ X, AND

0510

RAU

1953

0520

AUP

1954

0530

SUP

t001

0540

P, = X, + Y, - C,

0540

STU

1001

0501

STORE P,

0501

RDS

1951

0511

RAU

1953

0521

AUP

1954

0531

!
COMPUTE

P, = Xl

+ V,

- C,

STORE

P, IN 1001

COMPUTE

P2 = X2

+V2-

COMMENT

0531

SUP

1002

0541

STU

1002

0502

RDS

1951

ETC.

+ V1

SAME FOR P

STORE

IN 1002

RAU 60

COMPUTE

P 3 = X3

+ V3-

STU

STORE

IN 1003

!
ETC. FOR 1BOO BLOCKS

ETC. FOR 3000 STEPS

(Continued on next page)

40 • MORE ADVANCED PROGRAMMING

continued

..
RCD 70

COMPUTE
S600 = X 600

+ Y 600 -

RAU 60
AUP 10
SUP 11

C 600

J.
STU

STORE

1600

9999

S600 IN 1600

INVALID I

~ADDRESS
TO STOP

HALT

PROGRAM.

However, this will be a long, long program-containing about 3000 instructions-which is not
feasible with only 2000 drum locations available, especially since 600 are needed for storing
the C I and PI values. Even if it were po&sible to store this long program, it would be very
tedious to write, since much of the work is repetitious. Repetitious work is the very thing the
650 was built to do! Let us use our ingenuity to make the 650 do the repetitious part of its
own programming. Essentially, the steps of this program are very similar except that the value
of i is changed in each successive set ~f five instructions.
EXAMPLE h

It seems reasonable, then, to replace the 1800 blocks (3000 instructions) of the above diagram
with a program of the following type where the program "loops" around after modifying the
instructions to repeat the same operations with new data.·

SET i = 1 TO BEGIN WITH

J COMP.UTE PI

0:::

XI +Y I - C I

l
STORE PI IN DRUM LOCATION 1000

!
REPLACE i BY i

1 IN THOSE INSTRUCTIONS

L...-----1

INVOLVING C. AND P.
I

The first three blocks of the loop are similar to the first three blocks in the other flow
diagram, and lead to a similar program, namely, the five instructions:
LOCATION OF INSTRUCTION

COMMENT

0500
0510
0520
0530
0540

RCD 70
RAU 60
AUP 10
SUP 11
STU 21

1951
1953
1954
[ 1001]
[ 1001]

0510
0520
0530
0540
0545

Xi
Xi + Y i
Pi = Xi + Y i - C i
STORE Pi

READ X. AND Y.
I

Now we use our ingenuity. The addresses enclosed in [
] should be increased by one each
time the program reuses the set of instructions. Since an instruction is, after all, a 10-digit
nUll'ber, we may operate on it arithmetically, just as with data. One plan is to load the 10-

MORE ADVANCED PROGRAMMING. 47

digit number (instruction) into the accumulator, and then add 00 0001 0000 to it and store the
. result back into the proper drum location. We do this by continuing the program (assuming that
00 0001 0000 is in drum location 0906).
LOCATION
OF
INSTRUCTION

0545

RAU

0530

0501

0501
0511

AUP 10
STU 21
RAU 60
AUP 10
STU 21

0906
0530

0521

0521
0531
0541

0540
0906
0540

0511

053]
0541
0500

COMMENT
INSTR. FROM 0530
INTo ACCUM.
INCREASE DA BY ONE
STORE NEW INSTR. INTO 0530
SIMILARLY
STEP INSTR.
IN 0540

With the given program, the computation proceeds as fast as the cards can be read. The
new program requires a dozen program steps instead of 3000, and the 650 now has been set up
to do the tedious portions of the programming. The machine is not smart; you, who gave it its
instructions, are smart. The machine merely follows accurately the program you have set to
develop its further instructions. The thoughtful reader may have begun to gain some small
glimmering of the power that lies ahead. The use of symbolic assembly programs (Chapter 5),
interpretive systems (Chapter 7), and compilers (Chapter 8) absorbs much programming drudgery. First, however, the fundamental principles must be learned, as they are common to all
systems.

Problem Set 3-2

1. Write out the program of Section 3-2 which involves the loop. Fill in the distributor and accumulator entries. Decide whether or not the program will run before you put it
on the 650. Debug as necessary. Be sure you "initialize" it by setting i = 1 to start with.
2. What happens in the problem just discussed if an overflow occurs? Can an overflow occur if the data are as specified? What data specifications will insure you against overflow?
3. (a) Reprogram Problem 3 of Set 1-11 with the input bunched 3 sets per input card plus
an identifying number, and the output bunched 6 answers per card plus both identifying
numbers.
(b) As an extra-credit problem, devise a method whereby 4 sets of data can be put on each
input card and 8 answers on each output card, plus some identifying number on each so the
problems and answers may be collated later.
4. Explain the action of the 650 upon receiving the instruction 00 0000 0000 (a) from location 0924, (b) from location 0000.
S. What will the following program do, if anything?
LOCATION OF
INSTRUCTION
1951
1952
1953
1954

1952
8001
1954
0000

1953
8003
8003
0001

15
15
00

Punch these on one (load) card in words 1, 2, 3, 4 and set console to 70 1951 9999 and start
program.

42 • MORE ADVANCED PROGRAMMING

3-3. LOOPING AND

sre PPING.

The word "loop" is ,used to describe a portion of a program which is repeated within the
program and hence the flow chart contains a closed loop. It is quite usual to modify the instructions each time a loop is repeated. Even intermediate programming introduces loops
within loops within loops. Indeed, it is the ability to modify looped instructions which provides the stored-program machine with much greater versatility than is possessed by an externally programmed machine. Care should be taken to program an inner loop as efficiently
as possible, even though it may seem to involve only a few steps in a much longer program.
In actual computation, more than half of the computing time is often spent in the small loop!
As a matter of fact, a 2000 instruction program which completely filled the drum could be executed in less than 2 seconds if no loop were involved. It is not surprising that a great many
different techniques have been developed for looping. The one described in Section 3-2 adds
the constant 00 0001 0000 to certain instructions to modify the data address before repeating
the loop.
Another important technique which may save time is to "step" the first instruction
(called key instruction), and then to develop the other modified instructions by adding constants to the key instruction.
In the program discussed in Section 3-2, only two instructions are to be changed:
0530
0540

11[1001]
21 [1001]

0540~

0545~

DIFFERENCE

= ~ = 10

0000 0005

If ~ = 10 0000 0005 is added to the instruction in 0530, the instruction in 0540 will result,
no matter what common number is used to replace [1001]. It is, therefore, feasible to step the
instruction in drum location 0530, as before, and then replace the last three steps (stepping
the instruction in 0540) by the two steps given below, since the accumulator already contains
the result of stepping the instruction from 0530. Assume 0926 contains ~ = 1000000005.
0521
0531

AUP 10
STU 21

0926
0540

0531
0500

In this particular problem no saving in time is achieved (Why not?-consider read speed.),
but in other programs the savings may be very worthwhile. The technique is also important in
instances when stepping is not involved.
Readers who are seriously interested in becoming expert programmers should read the
IBM 650 operation manual on operation 22 SDA and consider the possibility of using this
operation in the above program.

Problem Set 3-3

1. Write a program which will compute Sl - Xl + Y1 - C i where the Xi and Y i are on cards
and the Cion the drum, as in the example of Section 3-2. When 6 values of S i have been computed, punch the 6 values plus a card number onto output cards. Let the card number be the i
of the last S i on that card. Will all of the answers be punched out with your program? Would
they be, if there were 500 rather than 600 values of Sl?
2. Write a program which will find the sum of 1000 10-digit numbers which are stored in
drum location 0501-1500 and which will punch this total on a single card. Be sure that your
program punches the entire total, which may easily contain a dozen digits.
3. (a) Design a program which will form the sum of one 1, two 2's, three 3's, four 4's,
five 5'8, ,••• , one hundred 100's, ••• , k k's for any preassigned number k - OOOOOOOkkk < 10 3

MORE ADVANCED PROGRAMMING. 43

where k is placed in location 1954. In other words, form Sk - 1 + (2 + 2) + (3 + 3 + 3) +
(4 + 4 + 4 + 4) + • ~ ~ + (k + k + k + k + • ~ ~ "+ k).
(b) Does the following program yield the same final result? If so, which takes longer to
compute? How does each work? Assume that OP 14 divides the 20-digit number in the' accumulator by the 10-digit number whose address is specified by the DA, leaving the quotient
in the lower half of the accumulator.
0400
0410
0420
0430
0440
0401
0411
0421
0431
0402

RAU 60
AUP 10
STU 21
AUP 10
MPY 19
RAU 60
MPY 19
DIV 14
STL
PCH

1954
1915
0415
1954
8001
8002
0415
1926

20
71

1977
1977

0410
0420
0430
0440
0401
0411
0421
0431
0402
9999

k + 1 [NOTE CONSTANT LISTED AT BOTTOM]
2k + 1
{2k + Ok
(2k + 1)k(k + 1)
(2k+1)k(k+1)/6

CONSTANTS
1915
1926
0415

00 0000

0001

00 0000 0006
TEMPORARY STORAGE (k + 1)

(Hint: Consult a mathematical handbook for a convenient formula.)
4. What happens in the following program? Will either eventually overflow? Why? Assume location 1955 contains 00 0000 0010.
(a)

0400

(b)

0410
0420
0500

RAU 60
MPY 19
RAU 60
RAL 65
ALO 15

0510

1955
8003
8002
1955
8001

0410
0420
0410
0510
0510

5. Describe the effect of the following program: Explain how many times the contents of
0502 is added and what, jf anything, is punched. How many output cards are there for each input card? Drum location 0502 contains a 7-digit positive-number Y OOOyyyyyyy+.
:II:

1400
1410
1420
1430
1440
1401
1471
1421
1431

Reo 70
LDD 69
STD 24
RAU 60
SUP 11
ALO 15
NZU 44
STL 20
PCH 71

0501
0501
0827
0671
0672
0502
1440
0828
0827

1410
1420
1430
1440

SEE CONSTANTS AT

1401
1471
1421
1431
1400

END OF PROGRAM.

CONSTANTS
0671
0672

00 0000 0005
00 0000 0001

3-4. THE COUNT BOX.

Let us now assume that all the data are on the drum, and that the results are to be placed
on the drum-Le., that this is but one small fragment of a larger program.
Xi IN 1101 TO 1190
Y I IN 1201 TO 1290

44 • MORE ADVANCED PROGRAMMING

where Xl are 5-digit integers 'OOOOOxxxxx and the Y1 are 4-digit integers OOOOOOyyyy. Compute

Tl-XlYi-Y~
and store T k in location [1700 + k] .. It is quicker for the 650 to compute (X - Y) Y than to
compute XY - Y 2, ,and we shall use the quicker form. A possible computation block for T 1
would be:
0700
0710

RAU 60
SUP 11

0720

MPY

0730

STL

19
20

[ 1101]
[1201]

0710
0720

8001

0730

[1701]

0740

Xi
Xi -Vi
(Xi-Yj)Y j
STORE LOWER 10-DIGITS
OF PRODUCT

Since the 20-digit product contains at least 10 leading zeros, it is unnecessary to examine the
upper half of the accumulator. It is, .of course, necessary to modify (step) the instructions before T 2 is computed.
In previous examples, the computation stopped when it was finished, 'either because an
invalid next instruction address was given, or because the 533 unit received a 70 RCD instruction when no further cards were in the read hopper. Since this is merely a program fragment, ,neither course is desirable. Instead we shall make a count box("B box") or index
register, which will register k when T k is computed and will test after each computation to
see if k = 90. If k does not equal 90, the instruction will be modified and the loop continued.
If k does equal 90, the next instruction in the major program will be taken. For reasons to appear later, we shall make a count box of the form 00 OOkk 0000 in drum location 1915 by adding l's of the form 00 0001 0000 from drum location 1925:
0510
0520

RAU 60
AUP 10

1915
1925

0520
0530

COUNT BOX INTO ACCUMULATOR

0530

STU

1915

0600

STORE NEW COUNT INTO 1915

ADD 1

A possible flow diagram follows.
FROM PROGRAM

INCREASE COUNT BOX BY 1

COMPUTE TI = (X j - YI) Yj AND STORE
TI IN DRUM LOCATION [1700

DOES j

= 901

+ I]
~----------------------~ NO

It is quite feasible to make the desired modification of instructions by adding
00 0001 0000 to those instructions containing a [ ] in the data address position, but we
prefer to demonstrate another method which has proved desirable in repeated sUbloops. The
three instructions to be modified are:

MORE ADVANCED PROGRAMMING. 45
0700
0710
0730

60
11
20

1100 + I
1200 + I
1700

0710}
0720
0740

!1 = - 48
!1 = + 09

9899 9990
0500 0020

Store the number 60 1100 0710 on the drum, say in location 1905. If the count box
00 OOii 0000 is added to 60 1100 0710, the first desired instruction 60 [1100 + i] 0710
results. In order to do this easily, ,we made the count box (in location 1915) by adding l's in
the fifth digit position-i.e., 1 = 00 0001 0000. The desired i will always be in position
such that 60 1100 0710 plus 00 OOii 0000 is the instruction needed in drum location 0700.
The instruction desired in 0710 may be obtained by adding !1 = - 48 9899 9990 (from drum
location 1935) to 60 [1100 + i] 0710 which is in the accumulator. Finally, the instruction
needed in 0730 is obtained by adding !1 = + 09 0500 0020 (from drum location 1906) to
11 [1201 + i] 0720, now in the accumulator, using the technique described in Section 3-3.
0600
0610

RAU

60
10

1905
1915

0610
0620

MASTER INSTRUCTION TO ACCUMULATOR
ADD COUNT BOX

STU 21
AUP 10

0700
1935

0630

0640

STORE NEW 0700 INSTRUCTION
ADD

0640
0601

STU
AUP

0710

0601

STORE NEW 0710 INSTRUCTION

1906

0611

ADD

0611

STU

0730

0700

STORE NEW 0730 INSTRUCTION

0620

AUP

This method has the advantage of being self-setting if the fragment is reused later in the {frogram for different data. We conclude by making the test, "Does i = 90?" In this we assume
that drum location 1924 contains the constant 00 0090 0000, ,and that drum location 1946
contains all zeros, 00 0000 0000.
0810
0820
0830
0840
0800

RAU
SUP
NZU
LDD
STD

1915
1924

0820

11
44

0510

0840

1946

1915

0800
NEXT INSTRUCTION
IN MAIN PROGRAM

0830

COUNT Bdx INTO ACCUMULATOR
SUBTRACT 90
BRANCH ON NON-ZERO UPPER
RESET COUNT BOX TO ZERO
RESET COUNT BOX TO ZERO

The reader is asked to assemble this program on a 650 planning sheet, ,improving it where
possible. For example, a step is saved by using 10 1905 0620 in place of the instruction
now given in 0600, ,since the count box is already in the accumulator. Doing such assembly
work is an important part of learning to program. Be certain you understand each step.

Problem Set 3-4

1. Assemble the program just discussed and debug it. It may be necessary to modify
some of the instruction addresses to connect the program pieces.
2. If drum location 0740 contained the instruction 00 0000 0810, what effect would it
have on the program suggested in the text of this section?
3. Write a program of your own using a loop instruction.

3 ... 5. ANOTHER TERMINATING TECHNIQUE.

As a subloop of a larger program we have 45 sets of values of each of 4 variables
(Wi' Xl' Yp Zi) stored on the drum in the locations shown:

46 • MORE ADVANCED PROGRAMMING

Wi
Xi
Yi
Zj

stored
stored
stored
stored

in
in
in
in

1101
1201
1301
1401

--4
--4
--4

1145
1245
1345
1445

(i.e.,
(i.e.,
(i. e.,
(i.e.,

Wk in 1100
X k in 1200
Y k in 1300
Zk in 1400

+ k)
+ k)
+ k)
+ k)

Each Wi is a 3-digit integer (whole number) stored in the form 00 0000 OWWW •. Each of the
numbers Xi' .y I',Z i is an integer having not more than 6 significant digits. ' Each is stored in
the word form 00 OOdd dddd.
Our problem is to form
Q i == (2X. + ¥i)W i

and to punch 45 output cards of the following form:

COLUMN NOS. L0-____- L________L -______

______

____

In programming this problem, it will either be necessary to write a separate program to
compute each Qi(since the WI' Xl' YI' Zl are in differe~t locations for various values of i),
or else to loop, modifying the instructions before repeating them. The latter method will be
used.

JSTORE Xj' Y j
liN PUNCH BAND

STORE W.
{ IN

PUNC~

BAND

COMPUTE
Qj

= (2X j + Yj)W,- Zj

STORE Z. AND Q.
I

{

IN PUNCH BAND

MODIFY
INSTRUCTIONS

OF Q j BEEN PUNCHED?
HALT

We could make a "count box" which would register k when Qk is computed, .and test after
each card is punched to see if k =45. However, since the instructions must be modified, it
is feasible to check whether or not the modified instruction'is in its final form. In this case
one of the final ins tructions will be
SLO 16 1445 0240 (Subtract Z 415)
Place the instruction 16 [1400 + k] 0240 ink> the accumulator and subtract 16 1445 0240.
If the result is zero, 45 values of Qi have been punched; if not, repeat the computation loop.

MORE ADVANCED PROGRAMMING. 47

Before considering the program, let us examine the top lines on a 650 planning sheet.
Let us elect to punch from drum locations 1827-1834. This decision is quite arbitrary at
this stage, ,except that we must use permissihle punching locations and we must not use
1101 --4> 1145, 1201 - 4 1245, 1301 - 4 1345, 1401 - 4 1445. (Why not?)
Take a fresh 650 planning chart. Note that across the top is -a line 'reading "Storage
Exit Words. U We shall use only 8 of these words with our control-panel wiring, so cross out
words 9 and 10. Fill in the form as follows
1

STORAGE-EXIT WORD
MEMORY ADDRESS
OUTPUT CARD

1827

1828

1829

1830

Z.I

5
1831

1832

1833

1834

Recall that the distributor (8001), lower accumulator (8002), and upper accumulator (8003)
may be used as valid addresses for data and instructions, just as are drum locations. (A
Store into operation such as 24 STD may not be used to store into an 800n address. This is
accomplished by another instruction. For example, 60 8001 IA (RAU) will, in effect, ·store
the contents of the distributor into the upper accumulator.)
Enclose in square brackets those portions of instructions which will be modified later.
(It is convenient to use red, if color is available.)
LOCATION
OF
INSTRUCTION
0200
0205
0210
0215
0220
0225
0230

0235
0240
0245
0201

OP
RAU
STD
AUP
AUP
STD
MPY
STD
SLO
STD
STL
PCH

60
24
10
10
24
19
24
16
24
20
71

[ 1201]
1828
8001
[ 130,]
1829
[ 1101]
1827
[ 1401]
1830
1832
1827

0205
0210
0215
0220
0225
0230
0235

0240
0245
0201
0216

8003
UPPER

8002
LOWER

8001
DIST.

0000 X
0000 X
0002X
000 2X + Y
000 2X + Y
0000000000
0000000000
0000000000
0000000000
0000000000

0000000000
0000000000
0000000000
0000000000
0000000000
(2X + Y)W
(2X + Y)W
(2X+ Y)W- Z
(2X + Y)W - Z
(2X + Y)w- Z

0000 X
0000 X
0000 X
0000 Y
0000 Y
00000000 W
00000000 W
0000 Z
0000 Z
0000 Q

The four computation blocks are complete. We have computed Q i = (2Xi + Yi)W i - Zp
and have stored Wi' Xi' Yp and Qi in the designated punch locations and punched the output
card. The reader should study this portion of the program until he understands it, before
continuing.
, Let us assume that drum location 1926 contains the constant 16 1445 0240. We test to
see if the modified instruction in drum location 0235-namely, 16 [1400 + i] 0240-is equal
to 16 1445 0240.
0216
0221
0231

RAU 60
SUP 11
NZU 44

0235
1926
0241

0221
0231
NNNN
NEXT INSTRUCTION
IN MAJOR PROGRAM

This sends the program to 0241 for its next instruction if i < 45, and to the next instruction in
the major program if 45 values of Qi have been punched.
Our next task is to step the four data addresses containing [
]. This is left as an exercise for the student. Suitable methods were explained in Sections 3-2 and 3-3.

48 • MORE ADVANCED PROGRAMMING

Problem Set 3-5

1. Complete the program of this section using the stepping method of Section 3-2. Debug
the program. Check to see that no storage location is used for two different instructions-a
common error.
2. Complete the program of this section using the modified instruction method of Section
3-3. Debug the program.
3. Assume that storage location 1777 contains a particular value of x, while 1778 contains eX, location 1779 contains e- x for the specified value of x. Prepare a program which
will punch the following data from punch locations 1777-1781.
1777

LOCATION

CONTENTS

1778
x
e

1779
-x
e

1780

1781

SINH X

COSH X

4. Bring a simple problem of your own to class and block diagram it.
5. Spend 5 minutes explaining your problem and block diagram to another member of the
class, and then listen to his explanation of your problem to see if he understood it. Do not
interrupt him!

3-6. BRANCHING ON A CODE NUMBER (Optional.)

It is often desirable to be able to direct a program into one of several channels depending
upon a code number in the input card. This could be done by machine sorting the cards before
putting them on the 650, or by board wiring, but it is often more convenient to use the 650 itself to make the switch, by having the key number appear as 00 0000 OOkk on the drum. (If
you are short of card space, board wiring can place the key number into word 9 or 10 of the
drum read band using only two card columns.) If the key number is added to a master instruction before, it is performed, the next instruction will depend upon the key number,and the program may be branched into any of several paths. Other variations will occur to you as you
program. This book is not a list of tricks, but rather a study of methods. Develop tric.ks of
your own. A good programmer first ,decides what he would like to have the machine do, and
then invents tricks to do it. New tricks are constantly being invented and the better you
understand the 650, the more ingenious you can be.
A progr.am fragment will make the method clear. We 'assume that drum location 1937 contains 00 0000 1300, and that the code number 00 0000 OOOk with k = 1, 2, 3, 4, 5, 6, 7, or 8
is in drum location 1947.
LU

FROM LAST INSTRUCTION
L..l...LLJ
0432

LOCATION
OF
INSTRUCTION

0432
0442

OP
RAL

ALO

COMMENT

1937
1947

0442
8002

MASTER I NSTR.
ADD CODE NO.

MORE ADVANCED PROG.RAMMING .49

The next instruction received will be in the accumulator 8002-namely, 00 0000 130k-and
the program will branch to the desired location.
Many variations are possible.
• Problem Set 3-6

1. Devise a step in a program which will branch in drum location 0464 to any of 5 positions 120k, with k = 1, 2, 3, 4, 5 depending upon which pay code 1, 2, 3, 4, 5 is stored in
drum location 1954.
2. (a) In devising a program to play tic-tac-toe, it is desirable to have several "branches
into one of nine positions. I t Devise such a branch.

3-7. SCALING.

Thus far, all our problems have dealt with integers. The 650 deals only with integersbut really, so does all computation. We now study modifications of the program needed when
elements are decimal fractions rather than integers. A method of describing the location of
the decimal point :without great verbiage is convenient. One common notation is to speak of
4: 73.2908600 as a "3-7 number," meaning that there are three significant digits before the
decimal point and seven after it; xxx.xxxxxxx is another common notation for the same concept. The latter is particularly useful if some of the digits are known to be zeros-for example, xxx.xxxxxOO. This does not imply that the digits represented by x's are all identical,
but only that the decimal point follows the third digit and that the ninth and tenth digits are
always zeros. The number 004.3200000 is of the form xxx.xxxxxOO. So is 989.9743400.
If a 2-8 number 11.22222223 and 3-7 number 444.5555555 are to be added on the
650, it is necessary to align the decimal points, just as one would do if h'e were using pencil
and paper.
WRONG

CORRECT

o 1 1.2 2 2 2 2 2

2+-(L~5T)

4 4 4.5 5 5 5 5 5

4 5 5.7 7 7 7 7

Since the 650 has no decimal point, it is quite willing to add either way. It is you, the programmer, who must be alert. Even problems in which the data are all in integers will often
involve decimal output, since a division is necessary-in computing an average, for example.
We must find a way of aligning decimal points for add and subtract operations. In other
words, a shift operation is needed. The 650 is provided with four different types of shift
operations. In the problem under discussion, all that is needed is to shift the 2-8 number
11.22222223 one digit to the right, giving 011.2222222(3). Shifting is done in the accumulator
only. Since only 10 digits can be handled in each half of the accumulator, a "shift to the
right one" instruction will place the low-order digit of the upper accumulator into the highorder position of the lower accumulator, while the low-order digit of the word in the lower accumulator is lost completely. Zeros are automatically supplied into the "blanks" which occur
on a shift operation, but the reader should remember that he is, as always, ,dealing with one
20-digit accumulator, not two separate 10-digit accumulators.
VVVVVVVVVV

LLLLLLLLLL

± before Shift
Right one

OVVVVVVVVV

VLLLLLLLLL(L)

± after SRT 0001

50 • MORE ADVANCED PROGRAMMING

In a shift operation code, the units position of the .Data Address indicates the number of
plaoe8 to be shifted (a new use of the DA). Any valid data address 0000.... 1999, 8000, 8001,
8002, 8003 may be used, since only the unit8 digit is analyzed by the machine.
30 OOON xxxx SRT (Shift Right) will Shift the 20 digit accumulator to the
Right N digits, for 0 ~ N ~ 9.
If N = 0, no shift results.
31 OOON xxxx SRD (Shift and Round) will Shift the 20-digit accumulator Right
N digits for 1 ~ N ~ 10, and will "Round" the new
units ,digit by adding 5 to the last digit lost. If
N = 0, a 10-digit shift and round results.
35 OOON xxxx SLT (Shift Left) will Shift Left N digits for 0 ~ N ~ 9.
IfN = 0, no shift results. Numbers shifted off the left
end are lost, but the overflow circuit is not activated.
36 OOON xxxx SCT (Shift and Count). This useful operation is discussed in more advanced work. Consult the IBM programmers manual if interested.
The shift operations are essential in "scaling" numbers for arithmetic operations, but they
also are useful in other' capacities. For example; if X is a 5-digit number located in the lower
5 positions of drum location 1851 (i.e. l OOOOOxxxxx) and Y is a 5-digit number OOOOOyyyyy
similarly located in location 1852, then, if both X and Yare positive, the program,
RAL 65
SL T 35
ALO 15
STL 20

1851
0005
1852
1527

will place both X and Y in punch location 1527 as xxxxxyyyyy. This technique permits more
output on one card even if the "standard board" is used.
The ~sual axiom in scaling is "expect the worst and plan for it." Thus X + Y = Z
becomes

o oxx.xxxxxx
oooO.OOOOVY
ozzz.zzzzzz
Even though the sum may have two (or more) initial zeros, only one can be guaranteed. (Why?)
Scaling is another spot where the person who knows most about the problem c,an do the best
job. If, for example, he knew that X ~ 0099.999900, it would then follow that Z could be
guaranteed to have two leading zeros instead of one. (Why?)
The more that is known about the nature of the data, the more efficient the resulting
program is likely to be.

P robl em Set 3-7

1. Assume that the following is on the drum:
Xi "., Oxx.xxOOOOO in [0900 + i] for i
Y i = .yyyOOOOOOO in [1000 + i] for i

1, 2, ••• , 45.

= 1, 2, ••• , 45.

'(a) Design a program which will punch Xi' Y i' Xi + Y i' and Xi - Y i from drum locations
1977, 1978, 1980, 1982', respectively.

MORE ADVANCED PROGRAMMING. 57

(b) Design a program which will also punch Xi + (i " Y i) from drum location 1979. Consider the problem of possible overflow.
2. Drum locations 0201-0300 contain one hundred 9-digit values of X i' The corr~sponding
Y 1 and Zi are stored in drum locations 0301-0400 and 0401-0500 respectively, each being a
10-digit number. Make a block diagram which will store (X. Y . Z), rounded off to 10 digits, in
locations 0701-0800. (Hint: Multiply X iby Y i' Since a 19-digit product is developed, it is
necessary to reduce this to a 10-digit product before multiplying by Z i' [Why?]) This may be
done as follows where drum location 1999 contains the constant 5000000000.
SL T 35
ALO 15
RAU
MPY 19

0001
1999
8003

It will be desirable to modify (step) instructions in this problem. Could SLT 35 0002 be
used?
3. The accumulator contains the 20-digit number
11111223334445566666+
before each of the following commands is received. Explain what it will contain after the
command is executed.
(a) 30
(b) 30
(c) 30
(d) 31
(0) 31
(f) 31
(9) 31

0002

1956

(h) 35
( i) 35
( i) 35
(k) 31
( I) 31
(m) 31
(n) 30

8002 1956
0010
0002

8002
0010
0000

1951S
1956
1956
1951S
1956

0002 1956
0010 1956
0742 1956
0007 1956
0008 1951S
0009 1956
1492 1956

3-8. SCALING IN MULTIPLICATION.

The scaling problem in multiplication is not one of aligning decimal points, but rather of

keeping track of decimal points. For example:
DEC. PT.
MPY

3 3 3.3 3 3 3 3,3 3
1 1 1 1 1 1.1 1 1 1

370 3 7 0

3 7.0

DEC. PT.

{3-7}
{6-4}

2 9 G 2 9 6 2 9 IS 3
{9-11}

MPY
0

3 3.3 3 3 3 3 3 3 3
1.1 1 1 1 1 1 1 1 1

{2-S}

J'-9}

3 7.0 370 3 7 0 2 9 6 2 9 6 2 9 IS 3
{3-17}

In the first example a 3-7 number is multiplied by a 6-4 number, giving a 9-11 number as
product. In the second example a 2-8 number is multiplied by a 1-9 number, giving a 3-17
number as product.
In general, if an h-k number is multiplied by a l-m nllmber, the result will be an
(h + l)-(k + m) number, if the correct number of leading (nonsignificant) zeros is supplied.
EXAMPLE 11

In computing an hourly payroll, it is necessary to multiply the number of hours worked by the
hourly pay rate, and then round to the nearest cent. Assume
DRUM LOCATION

CONTENTS

1201
1214

HOURS WORKED
PAY RATE/HR

TYPE

FORM

SAMPLE

OOOOOOOhhh
OOOOOOrrrr

000000024.7
0000001.795

52 • MORE ADVANCED PROGRAMMING

Obtain the amount earned to the nearest cent, as an 8~2 number in drum location 1977. Upon
multiplying the numbers as they stand, a 16-4 number results. (Why?) This will be rounded
to the nearest cent by using the SRD 31 0002 - - - - operation, giving an 18-2 number, the first 15
digits of which will be zeros. (Check this last statement.) Hence we shall store the lower,
which will be in the form ooOOOeeeee.
LOCATION
OF
INSTRUC1l ON
0500
0510
0520
0530

OP
RAU
MPY
SRD
STL

.SO

19
31
20

1201
1214
0002
1977

0$10
0520
0530

xxxx

Let us examine the multiplication 444.99 to see how the location of the zeros needed to make
the word affects the location of the product.
o 0 0 0 0 0 0 444
MPY

0 0 0 0 0 000 9 9

00000000000000043956
lower

upper

MPY

4 440 0 0 0 000
000000009 9

00000000439560000000
upper

lower

4 4 4 0 0 0 0 0 0 0
00000 9 9 000

00000439560000000000
upper

lower

Some people use the notation 5/7-3/3 to represent the number OOOOOxx.OOO, where the digits
before and after the / indicate numbers of leading and terminal zeros. The author finds it
simpler to do the analysis without using the notation.
Two basic facts are the 650 computes the product of two 10-digit numbers, obtaining a
20-digit product, and the decimal point can be located using the h-k notation as described
above. A little practice on a desk calculator will help you to understand this.

3-9. DIVISION.
In the 650 computers, division is accomplished by repeated subtraction, just as it is on a
desk calculator. Scaling precautions are necessary, as on a desk calculator, to prevent developing a quotient having more digits than the 10-digits which the accumulator can handle.
For the benefit of readers unfamiliar with the desk calculator, we state the rule. The absolute
value of the divisor must be greater than the absolute value of that portion of the dividend
which i8 in the upper half of the accumulator.
In division, a 20-digit dividend (the entire accumulator) is divided by a 10-digit divisor
(on drum) producing a 10-digit quotient in the lower half of the accumulator. If operation
DIV 14 (Divide) is used, the remainder will be found in the upper half of the accumulator.
If operation DVR 64 (Divide and Reset Upper) is used, no remainder will be retained. In
Q
10-digit quotient

QIN

10-digit divisor/20-digit dividend
many problems the dividend will be a 10-digit or smaller number, and if this is put into the
lower half of the accumulator with all zeros in the upper (RAL 65), no overflow trouble will
occur unless a division by zero is attempted. (Attempted division by zero always stops the
program, no matter how the overflow switch is set.) The use of RAL 65 may at first appear to

MORE ADVANCED PROGRAMMING. S3

provide a very simple "outU on division scaling difficulties. Like most "easy methods" it requires a "price. I t In this case the price may be the accuracy of the result. One thing that
often helps when a number N of fewer than 10 digits is being divided by D is to put the spare
zeros in the low-order positions on N but on the high order positions on D. For example, if
N -= 37.259, place it in the lower accumulator as the 2-8 number 37.25900000 rather than as a
7-3 number 0000037.259. The student who fails to grasp the reason for this will probably be
well advised to find a desk calculator (there should be at least one in every 650 lab) and
carry out a few divisions. Perhaps your instructor will demonstrate in class. Briefly, here is
what happens in machine division using 3-digit in place of lO-digit 'lCords to conserve space,
N 3.4
Q:a:- _ - .
D 1.2

o 2.8

o 3.4:

3.4: 0

lower

2.8 3

o 1.2

overflow

r------.-----

0 0 3
upper
overflow

.4 0 0
lower

0 12

0 3 4: 0 0 0
upper lower

~
o00
o 3.4:
lower

upper
o 0 2.

1.2 0 0 0 0
upper

3.4: 0
lower

lower

1.2 0 0 0 3.
upper
overflow

o 2.8

2.8 3

4: 0 0
lower

000
lower

1.2 0 3.4: 0 0 0 0
upper lower
The question of which answer is the most desirable leads to the most vexing question in computer work,-namely, error analysis.
If the original numbers are each accurate to the nearest .1 (2 significant digits), then the
"true value" may lie anywhere in the range:
N = 3.4: ± .05,

D = 1.2

± .05

3.35
N
3.45
- - ::; "true value --" ::; - 1.25 D - 1.15
N
"true value --" ::; 3.00
2 • 68 ::;
D -

The mere fact that the 650 gives the answer as '2.833333333 does not alter the fact that the
true value T may lie anywhere in the range 2.68 ~ T ~ 3.00, and that it is quite likely that the
10-digit quotient is incorrect in its second digit, as well as in all following digits. This is
not a problem of the 650, but is a problem in the mathematical theory of error analysis which
pervades all arithmetic involving approximate numbers, no matter how the arithmetic is done.
We shall have more to say about this later, but prefer to discuss the matter of scaling first.
In general, if a 16-4 number is divided by a 9-1 number, the result is a 7-3 number.
The decimal point in a product was located by using addition with the h-k notation. In
a similar fashion, the decimal point in a quotient may be located by using subtraction and the
h-k notation.
16-4
9-1
7-3

S4 • MORE ADVANCED PROGRAMMING

If the problem of scaling seems a bit difficult at this point, take heart, since there are three
ways in which much of the difficulty of scaling can be sidestepped:
1. By careful positioning of the data on the input cards, or by adequate use of board
wiring.
2. By use of floating-point arithmetic (see Chapter 4).
3. By use of the "shift and count" operation 3 6 SCT.
The latter is probably the most satisfactory, although the present trend is to favor floatingpoint arithmetic because of its simplicity in programming. In business work and data handling, on the other hand, scaling is usually simple enough so that direct scaling is used.

3-10.

(A + B). C
D

= T.

EXAMPLE 1:

As an example of a problem involving multiplication and division, we program
(_A_+_B_)_._C = T
D
assuming that the input cards are the 600 cards of the data deck and A = word 1 (OOOxxxxxxx);
B = word 3 (OOOOOxxxxx); C = word 2 (OOOOOOOxxx); D = word 4 (OOOOOxxxxx) where all are
10-0 numbers (integers). Later we shall consider the same problem with A, B,. C, D decimal
fractions.
r--~{ READ

A CARD INTO 1351-13581

I
(A
COMPUTE T

+ B)·C
D

STORE

= ---

A, B,C, D,T
{
IN PUNCH BAND

' - - - - - - i PU NC H 1

Since A is a 7-digit number and B is a 5-digit number, (A + B) may. contain 8 digits. When
one 8-digit number is mUltiplied by a 3-digit number, an ii-digit (A + B). C may develop. This
is of no concern, since we are not storing the sum and it is unlikely that the 5-digit number D
is smaller in absolute value than the eleventh digit-i.e., that part of the dividend in the
upper. * We shall not worry about division overflow here. If an overflow does occur, the overflow circuit will be activated and the program will stop.
The reader should fill in a 650 planning chart, giving particular attention to the location
of nonzero digits in the accumulator columns for the following program. Fill in the storage
entry and storage exit words as well.

*The reader may feel that a 5-digit number is never smaller in absolute value than a 1.digit number,
but that will be because he has overlooked the fact that the 5-digit number may have four leading
zeros, 00001, or even all zeros, 00000.

MORE ADVANCED PROGRAMMING. 55
LOCATION
OF
OP

INSTRUCTION

RCD

0101
0102

70
60

1351
1351

COMMENT

1377

0102
0106
0132

AUP
STD

1353

0108

24
MPY 19
STD 24

1378
13S2

0133
0109

1379

0134

DVR

1354

0134
0110

0110

STD

1380

0135

STORE D FOR PUNCH

0135
0139

STL
PCH

20
71

1382
1377

0139
0101

STORE RESULT T FOR PUNCH
PUNCH

RAU
STD

0106
0132
0108
0133
0109

READ
A
STORE A FOR PUNCH
A+B
STORE B FOR PUNCH
A PLUS B TIMES C
. STORE C FOR PUNCH
DIVIDE BY D

Problem Set 3-10

1. Fill in a 650 planning chart for the above program, showing the accumulator and distributor columns.
(A + B) C

2. Flow chart and program: T

where

A = OOOxxxx.xxx
B = OOOOxxxxx.x
C
D

=
=

OOOOOOOx.xx
OOOOOxxx. xx > 100

Be sure to pay careful attention to the scaling and indicate where the decimal point will be in
the answer.
3. Reprogram Problem 2 with D > 1 in place of D > 100. What difference does this make
on the likelihood of overflow? Can you plan for this possibility?
4. The following represents the contents of the accumulator, distributor, and an appropriate drum location before the instruction occurred. What will it contain after the instruction
is carried out?
INSTRUCTION
DA
OP CODE

ACCUMULATOR
UPPER(8003)
LOWER(8002)

DISTRIBUTOR
(8001)

DRUM STORAGE
LOC.
CONTENTS

(a)
(b)
(c)
(d)
(e)

MPY

1942

+ 9922222221

0000000000

+ 222222222 1

1942

- 0009220000

DIV

0135

+ 0625000000

0000000000

+ 0000000094

0135

+ 1250000000

DIV

0135

+ 6250000000

0000000000

DIV
14
DVR 64

0135
0135

+ 0000625000

0000000087
0000000087

+ 0000000087
+ 0000000087
+ 0000000087

0135
0135

+ 1250000000
- 1250000000 .

0135

+ 1250000000

(f)

DVR

0000000087

0782

0000000125

(g)
(h)
(i)

SRT
SLT

30
35

0742
0004

SLT

- 0000625000

0000086542

+ 0000700000

0697

+ 0000086542

0003

+ 0000625000
+ 0000700000
+ 0000700000

0000086542

+ 0000700000

0697

+ 0000086542

0002

+ 1234500000

0000006000

+ 0000006000

0697

+ 0000006000

5. What sets of instructions will take the accumulator in Problem 4(b) into each of the
forms given in 4(c), 4(d), 4(e), 4(f)?
6. Rework Problem 4 parts (g), (h), (i), where the given command is replaced by SRD 31.
7. The following data are on a card:
WORD 1

IDENTI FYING MAN It

WORD 2

HOURS WORKED

WORD 3

HOURLY RATE

OOOOOOmmmm
OOOOOOhhhh
OOOOOOrrrr

10--0
8--2
7--3

56 • MORE ADVANCED PROGRAMMING

Block diagram and write a program which will compute the earnings and punch it, along with
the input data, on an output card. Earnings should be rounded to the nearest cent.
8. In an inventory problem, the identifying part number P is stored in drum location 1851;
the total cost T, ,as an 8-2 number < 10 15 , is stored in drum location 1855; and the total
quantity Q, as a 10-0 number < 10 6 , is stored in drum location 1856. All extra zeros are
leading zeros. Compute the unit price U = T/Q rounded to the nearest mill, and punch out a
card containing

(wo:o

WORD 2

WORD 3

WORD 4

WORD 5

WORD 6

WORD 7

WORD 8

BLANK

Be very careful of your scaling. When you have finished the problem, follow your directions
exactly using a desk caclulator.
9. Design a program which will compute a numerical grade based on 100 as perfect from
the following data: 17 short-quiz grades Qi where 0 ~ Qi ~ 10, four l-hour examinations Ei
where 0 ~ Ei ~ 100, and one final examination F with 0 ~ F ~ 100, if the final examination is
to be weighted twice as much as an hour examination, and the average of the 17 short quizzes
is to be weighted as much as an hour examination.
10. Design a program which will compute a student's grade-point average to date using
the following available data: the number of grade points earned and the number of credit hours
taken during the last semester; the total number of grade points earned and the total number
of credit hours taken during all semesters prior to the last semester.
11. Bring to class a problem of your own interest and at least block diagram a possible
program for it.
12. Discuss the results of Problem 11 with a classmate.

3-11. OPTIMUM PROGRAMMING.

You should pay serious attention to the relative merits of various drum locations in positioning instructions and constants. If instructions are placed in consecutive drum locationssay locations 0201, 0202, 0203, etc.-then by the time instruction in location 0201 is
executed, the drum has already passed drum location 202 and must wait a full revolution (or
perhaps even two revolutions, depending upon the data address used) before it can pick up
the next instruction. Ideally, if the instruction is located in drum location 201, the data address of that instruction should be just far enough around the drum so that its drum location
will come under the read head just as needed, and the address of the next instruction should
be just enough further around so that it will be under the read heads at exactly the time it is
called for by the program register. Since the execution time for different instructions varies,
it is necessary to keep track of them with care in order to secure an optimum program. This
will be discussed later, but a good rule to follow at this stage is the 5-10 rule, which states
that whenever possible the data address should b~ 5 greater (modulo 50) than the address in
which the instruction is stored, and the next instruction address should be 5 greater than the
data address just used. In the case of a shift operation it is usual to make the next instruction address 10 greater than the location in which the instruction is stored. This does not
give an optimum program, but it will decrease the running time of most programs. Often programs set up using this rule will take only 1/5 or even 1/10 of the time required by sequential
programming, unless, of course, the program requires a great deal of punching or reading.
A glance at Figure 1-3 or at a drum layout chart discloses that drum locations 0201,
0251, 0301, and in fact (0001 + 50k) for any integer k - 0, 1, 2, ••• , 39, are located the same
distance around the drum and hence are accessible at the same instant. In fact, for any drum

MORE ADVANCED PROGRAMMING. 57

location N, the 40 drum locations contained in the set (N ± 50k) are all accessible at the
same time.

Problem Set 3-11
1. Each of the following programs adds 0000000001's to the upper half of the accumulator
from the drum. Put each on the drum and let it run for 1 minute. Examine the totals in the
upper accumulator at the end of each run.
(a)

0200
0201

RAU 60
AUP 10

(b)

0201
0201

0202
0202

0500
0507

RAU 60
AUP 10

0503
0510

0507
0515

0515

AUP

0518

0523

0523
0531

AUP
AUP

10
10

0526
0534

0531
0539

0539
0547

AUP
AUP

0542
0501

0547

CONSTANTS
0202

0000000001

0507

CONSTANTS 0000000001 IN EACH
OF LOCATIONS 0503,0510,0518:
0526, 0534, 0542, AND 0501.

3-12. INDEX REGISTERS (OPTIONAL).
An index register (sometimes called B-box or Z-box) is a small accumulator (4 digits and
a sign). The 650 has available (on special order) three such index registers, known as index
registers A, B, and C. They may be used as ordinary accumulators, but are more frequently
used as count boxes, or in modifying addresses. Special circuitry permits the use of the index
register for direct modification of addresses. If, for example, the instructions
SUP

(1001)

0540

SUP

(3001)

0540

were replaced by

then the actual command executed by the 650 will be
SUP

(1001

+ CONTENTS

OF INDEX REGISTER A)

0540;

that is, if either an instruction address or a data address is increased by 2000, then the 650
will perform as if, instead of adding 2000 to the address, the contents of index register A had
been added to it.
IN DEX REGISTER A
0012+

COMMAND EXECUTED

COMMAND RECEIVED
SUP

3001

0540

1013

0540

Only one command is required to change (step) the contents up on the index register.
For example:
DA

OP
(ADD TO INDEX REGISTER A)

AXA

0002

58 • MORE ADVANCED PROGRAMMING

will add the number 0002 to the contents of index register A. The command
(SU BTRACT FROM INDEX REGISTER A)

SXA

0001

will cause the number (0001) given in the data address to be subtracted from index register A.
Note that the numbers added or subtracted are not the numbers in drum locations 0002 and
0001, but the actual number given in the data address (providing this number lies between
0000 and 1999 inclusive).
The command
DA

(RESET AND ADD TO INDEX REGISTER A)

•
RAA

1437

resets index register A and adds the constant 1437 to it.
Similarily (RESET AND SUBTRACT FROM INDEX REGISTER A) RSA 81 resets index register A to
zero and subtracts the four-digit number « 1999) from it.
There exist Branch operations for index registers just as for any accumulator.
DA

(GO HERE IF IRA IS
NONZERO)
(GO HERE I F IRA IS -)

(GO HERE IF IRA IS ZERO)

OP
(BRANCH ON NON ZERO IN INDEX
REGISTER A)
(BRANCH ON MINUS IN INDEX
REGISTER A)

NZA

BMA 41

(GO HERE IF IRA IS NOT
MINUS)

There are similar operation codes for index registers Band C. Consult back cover for
numeric codes.
Let 400 numbers be stored in drum locations' 1201 to 1600. A simple program forms the
sum of these 400 numbers and then passes to the next step in the main program as index
register A indicates that 400. passes have been made. The addition is done by starting with
drum location 1600 followed by' 1599 .. 1201 + 0398, etc., until the number in drum location
1201 .. 1201 + 0000 is finally added in when index register A shows 0000, where the next step
.in the main program is completed. The 650 will complete the program (four hundred 10-digit
additions plus test) in less than one-half second.
PROGRAM
0100
01qo
0120
0130
0140

TEST IS IRA = 01

NEXT STEP
IN MAIN
PROGRAM

RAA 80
RAL 65
SXA S1
ALO 15
NZA 40

0399
1600
0001
(3201)
0120

SET I RA TO 400
RESET AND ADD INTO UPPER
SUBTRACT 0001 FROM IRA
0130
ADD (STEPPED)
0140
(NEXT STEP IN MAIN PROGRAM)

0110
0120

MORE ADVANCED PROGRAMMING. 59

The index register is of such great utility that it should certainly be used if available.
Consult the IBM pUblication on index registers for additional information.

Problem Set 3-12

1. As a speed demonstration, the following program is prepared. Explain what it does
and how many operations are performed on the 650 between the punching of each card. (The
cards will not contain any desirable output data, they merely serve as a timing device.) Make
a "guesstimate" of how long will elapse between punching the second and third output cards.
Place the program on the 650 and run it to see how long it really does take between cards.
0201
0208
0202
0203
0205
0207

RAA
RAU
SXA
ALO
NZA

80
60
51
15
40
PCH 71

1999
0202
0001
0205
8003
0000

0208
0203
8002
8003
0207
0201

2. (a) See if you can arrange to load and run the entire program given above on one input
card.
(b) Can you rewrite the program so as to use the standard 70 1951 9000 console setting?
3. Rewrite some of the programs of this chapter using index registers.
4. Use index register~ to obtain the sum of the first 1000 terms of the (divergent) series

L
00

i=l+"2+3"+"4+ ....

i ... 1

5. Use index registers to write a program to form the "sum of products"

600

Xi Yi where the one output card, containing the sum Z (20 digits), is punched into

words 1 and 2. Assume the X i are in drum locations 0601-1200 and the Y i in 0001-0600.
6. Devise and code a routine which will read two integers A and B (with A < B) from
words one and two of a card, compute the greatest common divisor of A and B, G =: (A, B), and
which will punch A, B, and G on an output card. Hint: Consult any text on Modern Abstract
Algebra or on Number Theory under "Euclid's Algorithm" for a method of computing the greatest common divisor of two integers-for example, see p. 26 of Selections from Modern Abstraat
Algebra (Holt, 1958) •.
7. (a) Write a program which will take a 0-10 number X =: .xxxxxxxxxx which is in the upper half of the accumulator and form X 12 9 (accurate to not more than 10 decimal places, probably less), placing X 129 in the upper accumulator when it is computed, and. going to drum location 1327 for the next command. In drum location 1327 place the command 01 0000 1327
for the present.
(b) Note how many fewer steps 'the computer will take to compute X129
«««X2)2)2)2)2)2)2 • X rather than as X 129 = X • X • X· •••• X.

(CHAPTER

41
~ t!\~ Yf ~ 00 @((ii 00 t!\ [M) [M) ~ [ro ((ii
W~t!\ ~(wrn3oo@(wI1~ [ro~~

4-1. ACCURACY OF COMPUTED RESULTS.

One of the most serious problems in any computation is that of determining the accuracy
of the computed result. This problem is not confined to the 650-the same problem exists on
a desk calculator or in hand computation. Since almost all scientific and/or engineering
computations involve approximate numbers, the difficult subject of error analysis looms large
on the scientific horizon. When few people did calculations, error analysis was usually
ignored in college courses, except for a few crude (and usually incorrect) remarks concerning
significant digits in multiplication and division, and decimal places in addition and subtraction. Now that large-scale computers are available to all, error analysis becomes a must.
The subject is strange (hence difficult) for two reasons:
1. No even reasonably simple rules on the a.ccuracy of computed results have yet been
devised.
2. Error analysis shakes our faith in the fundamentals of arithmetic, since certain basic
algebraic rules such as (a+ b)c = ac + bc are not valid when approximate computation is used.
If the two members (a + b) c and ac + bc are each computed, different results may be obtained!
You have heard the old' 'rule" that in addition and subtraction the result is accurate to
one fewer decimal place than the least accurate data, and that in multiplication and division
the result has one less significant digit than the least accurate piece of data. It has been
known for a long time that the'se rules are not valid, but we point out two simple examples
here to make this fact more vivid.
EXAMPLE 1:
1200

Find

where at

:z:

x.xx, accurate to the second decimal place-i.e., (true a i ) =- x.xx

i_I

.005. If most of 1200 pieces of data happened to be .005 too small, then the correct answer
1200

1200

would be about

i_I

ai +

i. 1

1200

.005 =

a i + 6.00. In other words, not only would the

i-I

result not be accurate to the nearest 1/10, as the rule states, but even the units digit, and
60

EASY PROGRAMMING VIA SUBROUTINES. 67

possibly even the 10's digits, would be incorrect. The error here is quite apt to be greater
than many of the individual pieces of data. You may think it unlikely that the majority of the
errors will be of the same type, but can you be sure? Before you trust a result, you need
some estimate of how accurate it is likely to be.
EXAMPLE 2:

An even more striking example is found in computing T ::: (A +B). C. If A, B, C are each accurate to 10 digits, how much accuracy will T have?
Let A ::: + 12345.67899
with a possible error of ± .000005
B ::: - 12345.67898
with a possible error of ± .000005
C·::: + 77777777.33
accurate
Then (A + B) . C ::: (00000.00001) . (77777777.33),
::: 777.7111733
computed answer.
ERROR ANALYSIS
POSSIBLE "TRUE VALUES" OF
A
SMALLEST "TRUE Ttl
LARGEST "TRUE TOO

+ 12345.678985

-12345.678985

12345.678995

- 12345.678975

(A+B)
ZERO
00000.000020

T = (A+B)·C
ZERO
1555.5555466

In other words, although computation gives the answer 771.1177133, the correct answer may
lie anywhere between 0 and more than 1555. Do you begin to see why some sort of error
analysis is vital?
It is improbable that an answer of 717.8 will give satisfactory results if the correct
answer is 0.52 or 1471.8. A more striking, but less typical, example may be obtained by
using A = 9999999999, B ::: ..:. 9999999998, C ::: 9999999999, each with possible error of ± 0.5.
Some indication of the possible variation in the computed result is essential, but not easy to
obtain. All this has been known for years, but it is only recently, since large-scale computation has become widespread, that it is making itself felt among nonprofessionals.
It would be imprudent to attempt to give a course in numerical analysis as part of these
notes, but it does seem essential that the vital need for error analysis at least be mentioned.
One colleague suggested including an outline of courses which would be appropriate.
This sounds like an excellent idea until one tries to do so. The only conclusion is that those
courses which are app~opriate depend largely upon the use you plan to make of the 650. A
person who planned to use the 650 only for inventory control, for example, really needs no
college mathematics. A student of psychology, on the other hand, needs quite a bit, as does
any other scientist or an engineer. The study of modern abstract algebra has come to the fore
so much in all science and engineering during the last ten years that it probably needs no
further mention except to say that modern abstract algebra is the heart not only of approximate
computation, but also of the electrical circuitry involved in the computer itself (Boolean
algebra)*. Computation which eventually resolves itself into a finite iteration of an infinite
series requires a knowledge of Taylor's series uith remainder and similar more advanced
estimates of truncation error.
"N umerical analysis," like" statistics," means different things to different people. It
is to be hoped that numerical analysis will not become the varied hodgepodge of courses which
statistics presents today, where a single university may offer "statistics" courses in half a
*More detailed applications of Boolean Algebra to circuit theory will be foupd in the author's Selections
from Modern Abstract Algebra (Holt, 1958).
.

62 • EASY PROGRAMMING VIA SUBROUTINES

dozen different departments. The variation which already exists between universities is
startling. One midwestern university offers numerical analysis with no prerequisite other than
calculus; another midwestern university has as its minimum prerequisite "two courses in real
variable, two courses in modern algebra, and two courses in point-set topology." Some
universities also require advanced probability and statistics as prerequisites for numerical
analysis. It is difficult to say what courses constitute a reasonable program for a research
scientist or e~gineer who wi~hes to use a computer. Certainly courses in advanced calculus,
modern abstract algebra, and numerical methods are a portion of the bare minimum.
The reader now has a sufficient command of 650 programming to attempt programs of his
own devising. Not all the possible 650 operations have been studied, but this need not deter
you. The remaining operations, including 84 TLU (Table Look Up) and 90-99 BD(k) (Branch
on 9 in Distr,ibutor) as well as work on 36 SCT (Shift and Count), are included the IBM
Programmer's Manual. Many beginning programs can be programmed without their use, ,just as
you can program without understanding board wiring. If you are to do serious programming,
you will eventually want these tools, as weHas more knowledge of floating-point numbers,
and index registers.

4-2. SUBROUTINES.

Perhaps the secret of becoming a capable (if not expert) programmer in, a short time is to
write only the simple parts of programs. This can be done either by letting the 650 more or
less write its own program (see Chapters 5, 7, 8) or by using subroutines which someone else
has already written. It is the latter alternative which will 'be discussed next. Before considering a partial list of some of the hundreds of subroutines now available, let us examine
the directions for using two specific subroutines. The subroutines themselves consist of a
pack of mM cards which are available in your computer library.
1. Square Root-0070 (esentiallySR2 of Technical Newsletter No.9). Computes VA for A
in the range 0 ~ A ... xxxxxxxxxx ~ .9999999999. Uses location 0070 and 22'locations in
range 0158-0197.
Linkage: Put A into upper accumulator.
Load distributor with next instruction to be performed after .JA has been computed, and transfer control to 0070-'Le., LD 69 (address next instruction) 0070. 'VA will be placed i~ upper
accumulator after it is computed, and the program will continue with the instruction whose
location was specified in the data address of 69 (next) 0070.

lIE

.xxxxxxxxxx

Both A and .fA are 0-10 numbers.
2. eX E-to-the-X-0073 (essentially Exponential Subroutine of Technical Newsletter
'No.9).
Computes eX for X in the range -1 < X "" .xxxxxxxxxx < 1. Uses locations 0073, and ,270 to
295.
Linkage: Put X in upper accumulator.
'Load distributor with next instruction to be performed after eX has been computed.
Transfer control to 0073-i.e., LD 69 (next) 0073.
The pack of mM cards in the subroutine takes care of computing eX.
eX will be placed in upper accumulator after it is computed.
eX _ x.xxxxxxxx

X is a 0-10 number and eX is a 1-9 number.

EASY PROGRAMMING VIA SUBROUTINES. 63

EXAMPLE 1:

To compute Y = eVA for a set of 200values of A, where the A's are positive 6,;,digit decimals
of the form .xxxxxxOOOO, which are found in word 5 of a set of data cards, we may proceed,
using these two subroutines, as follows. We shall read into 0401-0410 and punch from 04270434.
LOCATION OF
INSTRUCTION

COMMENTS

0411
0412
0413

RCD 70
RAU 60
STD 24

0412
0413
0414

0414
0415
0416

LDD 69
LDD 69
STU 21

0417

PCH 71

0401
0405
0427
0415
0416
0430
0427

READ IN A
PUT A INTO UPPER
STORE A IN PUNCH BAND
TO SQUARE ROOT SUBROUTINE
TO E TO THE X SUBROUTINE
STORE RESULT. EVA
PUNCH

0070
0073
0417
0411

By loading this 7-step program plus the two subroutine decks into the 650, a program which
computes y = eVA results.
.
.EXAMPLE 2:

For given values of K when 0 < K < 10 obtained from word 1 of input data cards, solve
X sin X = K for some one positive solution X s ' and print out K and Xs on output cards.
In essence, we ask, "Where does the curve Y = X sin X cross the line Y = K?"
We shall find only some one positive (first) intersection, but others can be found by
slight program modification. The technique used is one which would be crude for hand
computation, but is well adapted to machine computation. Set f(X) "'" X sin X - K. Note that
f(O) "'" - K is negative. We shall compute f(O), f(l), f(2), ••• until we find a value of X, say
X = 8, such that f(8) is positive. When this happens, we have passed the desired Xs value,
so we back up one step, to X = 7, and compute f(7.1), f(7.2), ••• 'until f(X) again becomes
positive. We then back up one step (this time a .1 step, not a 1 step) and proceed taking
steps only .01 long, etc., until Xs is obtained accurate to 9 or 10 significant digits~ whereupon K, X s are punched on an output card and the process repeated with a new value of K.
We assume that a subroutine for computing sin X for 0 ~X ~ 5 is available which takes
x.xxxxxxxxx from the upper as a 1-9 number and computes sin X when the command 69 (next)
0071 is given, where (next) is the address of the drum location containing the next
instruction to be performed after sin X is computed and placed in the upper. Sin X will be a
1-9 number in the upper when the subroutine is completed.
A program could be wri tten as:
LOCATION OF
INSTRUCTION

0500
0551
0557
0513
0519

RCD
LDD
STD
LDD
STD

70
69
24
69
24

0501
0554
0560
0516
0522

0525

LDD 69
STD 24

IA
0551
0557
0513
0519
0525
0604
0580

RAU 60

0501
0527
0522

0577
0630

LDD 69
MPY 19

0630
0522

0542
0555

SUP 11
BMI 46

0501

0577
0071
0542
0555

0558
0627

RAU 60
AUP 10

0558
0522
0560

0559
0627
0515

0604
0580

t::. IS STORED IN 0560
X IS STORED IN 0522

(Continued on next page)

64 • EASY PROGRAMMING VIA SUBROUTINES
LOCATION OF
INSTRUCTION

0515
0559

STU
RAU
SUP
STU
RAU
MPY
STU
NZU
LDD
STD
PCH

0677

0565
0575
0615
0538
0563
0568
0625
0581
0554
0516
0518

{

CONSTANTS

DA
21
60
11
21
60
19
21
44
69
24
71

ONE 10

ZERO 00
ONE-TENTH 10

0522
0522
0560
0522
0560
0518
0560
0558
0522
0528
0527
0000
0000
0000

IA
0577
0677
0565
0575
0615
0538
0563
0568
0625
0581
0500
0000
0000
0000

A flow chart would be:
RCD

INITIALIZE

~ "'" 1

STORE K
IN PUNCH

COMPUTE
X SIN X - K

REPLACE X
BY X-~

REPLACE X
REPLACE ~

PUT X IN
PUNCH

BY X

PUNCH

EASY PROGRAMMING VIA SUBROUTINES. 65

Note that it has been necessary to pay some attention to the scaling problem: X is a 1-9
number, hence to form X-~ we need ~ as a 1-9 number. We wish ~/10 to be a 1-19 number
so it truncates to a 1-9 number. Hence we take 1/10 as a 0-10 number so that the product
(1/10) (~) is a 1-19 number. It happens that 1/10 as a 0-10 number has the same form as
the number 1 has as a 1-9 number, but both numbers have been kept to make it easier to
understand. In actual practice only one need be used. Since X and' sin X are both 1-9 numbers, ,X sin X is a 2-18 number, and if we are to subtract K without shifting, ,we must put K
in as a 2-8 number.
Since there are many expert 650 programmers working out subroutines, more of these
routines are being added all the time. A partial list of available subroutines and full programs
is included below to give an indication of how broad the collection is. With the help of these
subroutines and a knowledge of basic programs, you can write programs to take care of
complicated problems. The only real difficulty remaining in many problems is that of proper
scaling. This is often avoided by the use of floating-point arithmetic.

4-3. PARTIAL LIST OF AVAILABLE PROGRAMS AND SUBROUTINES.
Matrix:'vector Multiplication
Matrix Addition
Matrix Inversion
Matrix Inversion by Gaussian Elimination
Large-scale Matrix Multiplication
Double Precision Matrix Inversion
Complex and Real Eigenvalues
Small-scale Matrix Multiplication
Multiple Regression Analysis
Simple Correlation Coefficients
Correlation Coefficient Routine
Analysis of Variance Program
Auto-correlation Program
Polynomial of Best Fit by 'Least Squares
Method
Multiple Correlation for 50 Variables
-Unbiased Standard Error of the
Regression Coefficients
Weighted Least Square Polynomial
Approximation
Real and Complex Roots of Algebraic
Equations
Roots of a Function of a Real Variable
Optical Ray Tracing
Neutron Diffusion
Critical Reactor Assembly
One-s pace-dimensional Multigroup
Lost, A Cross-section Averaging Program
Survey Traverse
Cut and Fill Program
Survey Traverse with Balancing
Calculation of Piping System Expansion
Stresses
Square Root

Cube Root
Exponential
Exp A
Sinh A, Cosh A, Exp A
Sinh X and Cosh X
Sine or Cosine
Arc Sin A
Log loA, LneA
Natural Logarithm
Polar to Cartesian Coordinates
Square Root IX I
Arctan X and Ln I X I
Sum and Sum of Squares
Circular and Hyperbolic Functions
Regular Bessel Functions
Irregular Bessel Functions
Simultaneous First Order DHferential
Equations
Multiple Numerical Integration
Fourier Synthesis
LaPlace Transformation
Systems of First Order Ordinary Differential Equations
Solution of Simultaneous Linear Equations
Complex Arithmetic Matrix Inversion
Well Bore Deviation Record
Linear Programming
Transportation Problem
Numerical Integration (Runge-Kutta or Milne)
Chi Square
t-test
Numerous special traces and debugging
programs.

66 • EASY PROGRAMMING VIA SUBROUTINES

The subroutine library of your own installation will list the subroutines availableLand
will undoubtedly furnish you with "entry data summaries" -similar to those given for
=
square root, and eX = E to the X, found in Section 4-2.
A few words of warning concerning subroutines are pertinent.
1. It is usual to have the sign of each instruction in a subroutine negative, since this
permits the use of a punch-trace routine which will trace positive instructions at a speed of
100 instructions per minute (since a card is punched for each instruction, giving the instruction, the contents of both the accumulators and of the distributor, and other data), but which
will trace negative instructions (known to be correct) at what is 8:1most normal machine speed
without punching cards.
2. In general it is possible to relocate subroutines if this should be necessary.
3. It is always necessary to know what locations the subroutine uses, even though you
may never know how it works, since the same drum location must not be used for two instructions.
4. It is important, in using a subroutine or a program that someone else has set up, that
you ascertain the range of permissible input values and the form of the input and output
values. For example, if the input X on the eX subroutine were a form of a 4-6 number
OOOO.xxxxxx instead of a 0-10 number .xxxxxxxxxx, it would have been necessary to shift
left 4 before going into the subroutine to secure the desired answer, even though it w~s known
that the"first four digits were all zeros (i.e., - 1 < X < 1).

4-4. A PROBLEM.
The following is an actual problem which occurs in the planetary gear systems currently
used in automatic transmissions on automobiles. The search for smooth automatic shifting
using planetary gears rather than the inefficient double turbine system is still (1958) in
progress. The problem of Example 1 below is solved repeatedly for various values of wand a.
It is quicker to compute the answer anew each time it occurs, using infinite series, than to
look it up in a table, even if such a table were availahle.
E~ample

The curves y = sin wt and y = e- at are plotted on the same set of y, t-axes. Determine the
smallest positive value of t at which they intersect.

Before continuing, the reader should decide how he would solve this problem without using
the computer, and should actually obtain one solution accurate to 5 decimal places for a
specific pair of value of a and w, say -a = -.625 and w = .475. It would, of course, be .ideal
if the pair of equations could be solved in general, obtaining some neat formula involving a
.
and w for the intersection t, much as the quadratic formula x

- B ± ../B 2
=

- 4AC
2A
is a general

EASY PROGRAMMING VIA SUBROUTINES. 67

solution of Ax 2 + Bx + C = O. Persons who are skilled in transcendental analysis may wish
to attempt this, but it is neither easy nor necessary.
The problem reduces to finding the smallest positive root of the equation
f(t) = e- at
for fixed values of a, w. At t = 0, f(O) = ef(t)

sin wt = 0

at -

sin wt = 1 - 0 > 0, while at t

sin wt < O.

Somewhere between 0 and--is a value of t which makes f(t)
2w
continuous in the region 0

=-,
2w

0 1 since f(t) is

-2 • (Would this be true if f(t) were not continuous?) How

can you be sure that f(t) is continuous on the given region? What does "continuous over a
region" mean? Name a function which is not continuous for the region 0 ~ t ~ 1.
One possible technique is to compute f(ti + ~t) with ~t = 1 and ti = ti _ 1 + ~t starting
wi th to ... 0 i.e., f(O), f(l), f(2), ••• -until a value tl + ~ t is found such that f(ti + ~ t) < o.
Assume that this happens at ti + ~t = 3. We then back up one step to ti = 2 and mUltiply the
old ~t by 1/10 and continue obtaining, say, f(2.1), f(2.2), ••• , f(2.8) before a change in sign
in f is found at ti + ~t = 2.8. We again back up one ~t step to ti = 2.7, multiply ~ t by 1/10 1
and continue. This yields f(2.72), ••• ,etc. Since the machine computation of f(t) is rapid,
this method will be quite satisfactory.
.
If 0 < w < .1, -it may be desirable to take the original ~t as 10 or even 100 rather than 1.
(Why?)
A possible flow chart for this problem appears on the next page.

4... 5. A WORD TO THOSE WHO WILL HAVE OTHERS WRITE THEIR PROGRAMS.
17

If w > .1, then the desired solution t < - - < 16 for all values of - a < O•. If 0 < w "< .1, :

.
2(.1)
.
.
it may be desirable to take the original ~t as 10 or even 100 rather than 1.. Thus, the more the
person who writes the program knows about the nature of the data, the easier it is ·for him to
write a good program. Persons who depend upon others to write their programs often could
save considerable money and time by giving the programmer more details concerning the data.
In-this case it makes a difference in programming whether you state w > 0, w > .001, or w > .1.
In other problems the difference may be even more important. Certain inventory problems are
impossible as 650 problems (without tape or Ramac) unless the data (issues and receipts) are
sorted into two sets and the new receipts read into the 650 before the issues are read. Advance planning will always make this possible, ·and it is usually possible anyway, but the
programmer must know this. The usual axiom of the professional programmer is "expect the
worst and plan for it. U If he plans for conditions which are much worse then actually will occur, it is costly not only in programming time but also in machine time.
You must also be careful in specifying what results you want. On a recent research problem in psychology, the running time on the problem was cut from 1 hour 23 minutes to just 7
minutes by cutting out the punching of certain unneeded intermediate results-a saving of
over $100 each time the problem was run.
Moral: If you plan to have someone else do your programming for you, give him as complete details as possible-even the seemingly trivial details may be vital. Also, learn enough

68 • EASY PROGRAMMING VIA SUBROUTINES

USE PU NCH WORD 3
{ TO STORE T

COMPUTE
F(T)

= eAt -

SIN WT

REPLACE T

BY T -~T

REPLACE T

BY T

+ ~T

about programming so that you know what output data could be obtained at no extra cost, and
what output data would entail a considerable saving if omitted.
Give your programmer some leeway. Tellhim (a) results you must have; (b) results you
would like to have, if available at not more than $25 extra per run; and (c) results that would
be handy if they could be obtained at no extra running time cost. (This is often possible.)

Problem Set 4-5

1. Solve e-· 625t - sin .475t = 0 by hand methods, obtaining the answer correct to 5 decimal places. Use tables to speed your work, and keep track of your time, including time spent
seeking numbers in the table. The 650 can perform the entire computation in less than 20
seconds •.
2. Write a program which will read x and a from a card (the cards are not yet punched,so
you may specify the form in which each is to be punched), compute·
x + a - .j2X2 + .001

and punch x, a, y in an output card and continue. Assume that 0 < x < .6 and 1 < a < 3 and
that the square root subroutine 0070 described in Section 4-2 is available.

EASY PROGRAMMING VIA SUBROUTINES. 69

3. Write a program which will read a value of x (you decide what the permissible range
and form of x shall be), from a card, and punch x, eX, sinh x, cosh x on an output card. Assume the eX subroutine of Section 4-2 is available.
4. Modify the program of Problem 3 so that T == (1 - cosh 2 + sinh 2 x) will also be punched
in word 8 of the output card. Of what value is the test function T? (Hint: If exact values of
cosh x and sinh x were obtained, in place of decimal approximations, what constant value
would T have?)
5. Your calculus tables will remind you that
d Arc cos x

-1

J1- x 2

d Arc cos x
in an outdx
.75. Discuss your program with a classmate before running

Devise a program which will read x from an input card and punch x and
put card. Assume that 0
each on the 650.

dx
=
;:::;:1T===;:: if a > b. Devise a program
a + b cos x Ja 2 - b 2
which will compute the given integral for suitable values of a and b. Then write up your program indicating the limits and form of a and b which may be used with your program.
6. It is possible to show that ('"

J).J ... vx:-

7. Design a program which will compute
hy taking
Be sure
that you loop the correct number of times.
8. Obtain a list of the subroutines currently available in your 650 library and devise a
program of your own interest, using some of these subroutines.
9. Examine one specific subroutine, a short one, and see how it meets the problem of
getting hack into the main program after the subroutine is completed.
10. Use the trapezoidal rule, with t = 0.01, ,to approximate the area under the curve
y= -

y'2;

e- t 2/ 2 between t = 0 and t = X for values of X, read from word 7 of a set of input

cards. The input X is of the form x.xxOOOOOOO and 0 ~ X ~ 4. This is, of course, the normal
probability integral and is of great importance in scientific and engineering work as well as
in the social sciences.

4-6. FLOATiNG.POINT ARITHMETIC.
Your most troublesome problem at present is probably scaling. The difficulties of scaling
may have certain other (presumably lesser) difficulties substituted for them by using floatingpoint arithmetic. Recall the "scientific notation" for writing numbers as O.nnnn x 10 k • (See
p. 142 ofPuooamentals of Oollege Mathematics by Brixey and Andree, Holt, 1954.)
DECIMAL
NOTATION

SCIENTIFIC
NOTATION

- 375.67899 = - .37567899 X 10 3
.0000123456 =
.123456
X 10-4

37567899 (+ 3)
123456
(- 4)

There is, ,actually, no need to write the 10, since the last expression on each line is clearly
derived directly from the scientific notation and may be so interpreted. Since both the number
and the exponents have signs, we must either store two signs (inconvenient) or devise a new

70 • EASY PROGRAMMING VIA SUBROUTINES

method of expression. The new method of expression used is simply to add 50 to the exponent,
giving
DECIMAL

SCIENTIFIC

- 375.67899 = - 37567899 X 1 0 3
.0000123456 = .123456
X 10-4

FLOA TI NG POI NT

~
~

(-)37567899~
(+)1234560042

Some systems place the (exponent plus 50) in the two right·hand digits (low order) while
others use the two left-hand (high order) positions. Currently most of the IBM 650 floatingpoint routines favor the low-order position, and this notation will be used here even though the
other has certain advantages.
Floating-point notation allows one to write any number N whose absolute value lies in
the range 10- 50 ~ IN I < 10 50 with an accuracy of eight significant digits. This range has
been found quite adequate for most problems, ,providing the number zero can also be represented. Zero is represented in floating point by 00000000 OO-i.e., the "exponent plus 50"
is always taken as 00, .along with the significant digit zeros.
SAMPLES:

DECIMAL

SCIENTIFIC

+ 3333333.3 =+.33333333 X 10 7 ~
- 444.44444 = - .44444444 X 1 0 3 ~
+.0000777777 = +.7777777 X 10- 4 ~

FLOATING POI NT

+3333333357
- 4444444453
+7777777746

This system appears to be the true answer to the scaling problem, particularly since both
subroutines and hardware have been devised which perform the floating-point operations.
However, there are two serious drawbacks-namely, accuracy and execution times. We discuss each briefly.
It may, at first, seem that an accuracy of eight significant digits is adequate for all normal problems. True accuracy of eight significant digits probably would be. However, consider the following type of problem:
(22222.222-22222.221) • (5555555.5).
If the subtraction is performed by hand, one obtains

(.001) (5555555.5) = 5555.5555.
Since (.001) is accurate to only one significant digit, it is immediately apparent that the eight
5 's certainly do not represent eight significant digits. In fact, since
(.001 ± .0005) . (5555555.5) ::: (5555.5555 ± 2777.77775),
the correct answer may lie anywhere between 2778 and 8333 rather than near 5555. The same
difficulty arises in fixed-point arithmetic (either hand or machine), but the programmer is more
apt to be aware of the difficulty, since he has considered the scaling of numbers rather carefully. It has long been recognized that the old rules concerning the reliability of computed
results being "a little less accurate than the least accurate piece of data" were unfounded,
but use of the desk calculator with all intermediate results visible permitted the substitution
of the (also invalid) rule u a little less accurate than the least accurate number that appears
in the computation. U High-speed stored-progr~m computers have made it necessary to reexamine the entire subject of accuracy. Much new theory has been developed within the last
five years, and no one who plans to work with a large-scale computer can afford to be without a modern, -up-to-date course in numerical analysi s!
An appalling example of the drastic effect of "slight round-off error" is obtained by expanding the polynomial

tea:) ::: (x - l)(x - 2)(x - 3) (x - 4) ... (x - 19) (x - 20)

EASY PROGRAMMING VIA SUBROUTINES. 77

whose roots are the real integers 1, 2, 3, ... , 20. Since the coefficients are large, floatingpoint notation is, used. When the resulting polynomial is then solveq, roots are indeed approximately 1, 2, 3, and 20, but many of the remaining roots prove to be comple[JJ numbers,
a + bi, with b not at all near zero! In other words, the round-off error has drastically changed
the basic nature of the polynomial.
The second possible disadvantage of the floating-point operations is that if floatingpoint arithmetic is performed by subroutines, it is much slower than fixed-point arithmetic.
If your 650 is equipped with an automatic floating-decimal arithmetic device (an auxilliary
feature in the 653 unit), mUltiplication and division execution times will be about the same
as for fixed-point operations, ,while floating-point addition and subtraction take about twice
as long as the corresponding fixed-point operations. However, if your 650 is not equipped
with automatic floating-point operations, then it is necessary to use floating-point subroutines,
in which case the arithmetic operations take between 10 and 20 times as long to perform as
fixed-point operations. Interestingly enough, the general trend in computers seems to be toward the use of floating-point arithmetic in scientific computation, in spite of the time and accuracy disadvantages, since it is so much easier to use. Some of the modern subroutines automatically check each operation to see if a loss of more than two significant digits has occurred. If it has, the program will stop on a program stop (OP code 01) until the operator
again pushes the program start key. This is a highly desirable feature and it is hoped that
it will eventually be available on the automatic floating-point operations.
All floating-point operations are done in the upper half of the accumulator. The lower
and the distributor may be disturbed in the process, but the final result will be in the upper.
The operation codes are
FAD 32

FLOATING ADD

FSB 33
FDV 34

FLOATING SUBTRACT
FLOATING D'IVIDE

FAM 37

FLOATING ADD MAGNITUDE
FLOATING SUBTRACT MAGNITUDE

FSM 38
FMP 39
UFA 02

FLOATING MULTIPLY
UNNORMALIZED FLOATING ADD. (THIS MAKES IT POSSIBLE TO ATTACH THE SAME
EXPONENT TO A GROUP OF NUMBERS FOR FIXED-POINT OUTPUT, BY SUPRESSING
THE NORMALIZATION WHICH OCCURS AFTER ADDING.)

Subroutines are available which use the floating point numbers. An example is the "Elementary Transcendental Function Package-Floating Point" which accepts floating point
values of X from the upper accumulator and computes any of the following values, placing the
result in the upper: tan -1 X, cos X, sin X, eX,
In X, log X. The commands to be given are

rx,

RAU

LDD

when X is the address of the floating point number X, NEXT is the address of the next instruction to be performed after the subroutine is comp~eted and ENTRY is the number specified below, which determines what function of X is computed:
FUNCTION

ENTRY

tan -IX

0050

cos X

0076

sin X
X

0176
0250

In X

0331
0439

logX

0457

This packet uses drum locations 0000 to 0459.

(CHAPTER 5

5-1. SOAP (SYMBOLIC OPTIMUM ASSEMBLY PROGRAM).

As was pointed out·iIi Chapter 3, a program written serially, in consecutive drum locations, lis about as poorly constructed as possible, ,since the drum must make at least one
complete revolution between each program step. The 5-10* rule speeds up a program considerably, but still is not the best (optimum) program. To write an appreciably better program
requires knowledge of exactly how long each operation takes. Tables giving this data are
available in the IBM manual, but experience shows that the work is so tedious that most
programmers are reluctant to spend the time necessary, and consequently fall back on some
pet variant of the 5-10 rule. However, tedious arithmetic and memory work are exactly what
the 650 is designed to do! Several programs exist which take a given unoptimized program
and use the 650 to produce an optimized program. One of the best known of these programs
is IBM's Symbolic Optimum Assembly Program, S.O.A.P. Not only does SOAP optimize a
program, but it will do so from a program in which most of the addresses have been omitted,
provided the steps are in order. Furthermore, SOAP utilizes the alphabetic device of the 650
so that the alphabetic mnemonic operation codes (RAL) may be used in place of the corresponding numeric (65) codes. Alphabetic names may also be used to designate storage locations if desired. Even with limited understanding of the 650 programming, you will find SOAP
programming much easier than machine-language programming.
Special SOAP programming sheets and special SOAP cards are available. Three versions
(I, II, and III) of SOAP are available, but the theory is the same in each case. SOAP II is
used here. After loading the SOAP deck (your lab has one), it is possible to "process"
(i.e., SOAP) as many programs as desired. The input (SOAP) cards consist of the unoptimized
program written in acceptable SOAP symbols. The output cards will contain the optimized
program expressed in machine language. Each output card also contains the program instruction in SOAP language from which it was obtained, for ease in debugging. The output cards
contain the standard 650 loading code and the machine-language instructions in words 1 to 4,
and are treated like the program (load) cards we have been using. Reload them, using the
"standard board," and debug as usual. A special SOAP board is used in the 533 while the
SOAP program is being optimized and converted into machine language. Let us now re*The 5-10 rule advises using a data address 5 greater than the location of the instruction and a next
instruction address 5 greater than the data address whenever possible.

SOAP.73

examine the program of Example 22, Section 4-2, in which we computed Xn such that
X sin X z:: K for 0 < K < 10. The flow chart follows:

INITIALIZE

6=1

X=O

STORE K
IN PUNCH

COMPUTE
XSIN X - K

REPLACE X

x-6

REPLACE X

DOES

x+6

6=0

PUT X
PUNCH

IN PUNCH

Let us write the program in SOAP, first reserving drum location 0000 to 0499 and 1000 to
1999 for other purposes-i.e., the program can be placed in locations 0500 to 0999, but not
elsewhere. Tb.is is done by placing Block Reserve
BL.R
BLR

0000
1000

0499
1999

on the first two SOAP cards. We shall next reserve Regional Read (0501-0510) and Punch
(0527-0534:) areas thus:
REG
REG

ROS01
POS27

0510
0534

Even though we only use one read location (0501) we reserve the entire block, since otherwise the SOAP program might place an instruction in one of these locations (say 0506) only
to have it erased when a data card was later read in the program. We may now address drum

74 • SOAP
location

0501

either as

0501

ROOOl if desired.

or as

venient when programming.

sample

SOAP

Experience shows this is very con-

program follows:

T
A
LOC
I

I
I

B/LR

1000

999

0000

t 0510

R/EG PI 0527

, 0534

RICo

0001

LIDo

OINE

sir 0

o'ELTA

1.1 00

stro

11ERO
xl

StrD

RAU

I
I
I
I

~EGI
I

I
I

POSI
I

I
I
I

I
I

END

I
I

T,ENTH

LOOP

M/py

RI 0001

pbsl

ElMI

tJEGI
I

~AU

DIELTA

~UP

S~U

Lbop
I

~AU
SUP
I
s;ru

I
I

DELTA

D,ELTA

MPY
I
S~U

~ENTH

NIZU

NIEGI

L'DO
I
SpO

Xl
I
PI 0002

RAU

D,ELTA

I
END'
I

I
I

FfCH P,I 0001
I 10

' 10

START

0000

0000
1

I 0000

1 0000

TO SINE X

, 0071

slJp

169

SET

pi 0001

ONE
I
ZERO

~OO R 1 0001

READ CARD

LOOP

COMMENTS

I- 0499

I
I

650

REG R: 0501

I
START

The

T
A

10000
I

0000

will translate and optimize this program giving the program as:

OP
0500
0551
0557
0513
0519
0525
0604
0580

RCD
LDD
STD
LDD
STD
LOD
STD
RAU

70
69
24
69
24
69
24
60

0501
0554
0560
0516
0522
0501
0527
0522

0551
0557
0513
0519
0525
0604
0580
0577

SOAP.75
OP
0577
0630
0542
0555
0558
0627
0515
0559
06n

0565
0575
0615
0538
0563
0568
0625
0581

CONSTANTS
.

f554
0516
0518

LDD
MPY
SUP
8MI
RAU
AUP
STU
RAU
SUP
STU
RAU
MPY
STU
NZU
LDD
STD
PCH
1
0
1

69
19
11
46
60

10
21
60
11
21
60
19
21
44
69
24
71
10
00
10

0630
0522
0501
0558
0522
0560
0522
0522
0560
0522
0560
0518
0560
0558
0522
0528
0527
0000
0000
0000

0071
0542
0555
0559
0627
0515
0577
0677
0565
0575
0615
0538
0563
0568
0625
0581
0500
0000
0000
0000

This program has better optimization than most programmers would achieve in two hours
time, but it is not the best that can be achieved by expert hand optimization. For most purposes SOAP is an excellent method of writing good programs with a minimum of effort, without using For Transit (see Chapter 8). There is more to be said about using SOAP, but the
general rules are simple.
If you wish to use a machine-language numeric operation or address (drum location),
leave the left-hand column (separated by a dotted line on the program sheet) blank. If you
wish to use a symbolic (usually alphabetic) operation or address, be sure that the left column
is not blank. Symbolic addresses may contain from ope to five symbols. Typical examples
are:
0
5
3
G

D
D
D
J
F

NE
RD
ROSS
ELTA
X
T
OSIE
INIS

S
L
L
M
N
X
Y
E
X

TART
OOOP
oOPZ
ARY
ET

ND
IX

Whenever a data address or an instruction address desired is the location of the ne~t instruction, it may be left blank, provided the desired LOCATION column is also blank. Thus the
address END used here is unnecessary, but harmless. The address NEG 1 and LOOP, however, are essential, since they are used later in the program as addresses to which the 650 is
sent in a loop.
Any desired comments may be placed in the comments column. If more than 10 characters
are needed for remarks, they should be placed on the next line of the program sheet with a 1
indicated in the type column (Col. 41). Several type 1 (i.e., 30 character comment) cards may
be used in succession to identify different portions of a program, if desired. It is often desirable to begin with a type 1 (comment) card identifying the problem and the program. A
"blank" type 1 card can be used to make the 402 (or 407) listing skip a space, thus providing
a more readable output. (For example, put a blank type 1 card between your last block reserve
and the beginning of your program.)
What you need now is practice-GO TO IT.
Obtain a copy of the SOAP n manual and refer to it as needed.

(CHAPTER

~~~~b{IA~~rrolA 1J~@rro
@~ [f)~~ rro~~[f)[l~~

6-1. SOPHISTICATION IN PROGRAMMING.

Experience provides the opportunity to become more sophisticated in the matter of programming. Genuine sophistication cannot be learned from books, but it may help to mention certain
indications of sophistication which are usually observed in artful programming.
It must be remembered that elegant programming, like most creative work both in the
sciences and outside of them, is more of an art than a science in the usual meaning of the
words. As such, there must develop a "feeling for" programming rather than a set of rules.
One way to develop this feeling rapidly is to work out your own solution to a specific problem
and compare your solution with that of others. A set of situations designed to help improve
your programming technique is presented at the end of this chapter.
First, a few comments gleaned from experience may serve as guideposts.

6-2. FLOW CHARTING.

Never program without a flow chart!
Never debug without a flow chart.
Never discuss a program w.ith which you are unfamiliar unless you have a flow chart
before you.
Never file a program you have written without also filing its flow chart. If the program is
worth saving, so is the flow chart.
In discussing problems with non programmers , use a flow chart rather than the program
itself.
A flow chart should show the exact sequence of operations, including each branch and its
alternatives. It should also indicate how and when loops are tested. It does not show
individual steps used in computing; the overall method of attack, without details, is the
object of flow charting. A typical portion of a flow chart is shown on the opposite page.
The actual program for the block compute f(x) ... eX sin 3x + 10 would involve two subroutines
plus a dozen program steps.
There are some convenient conventions which you may (or may not) wish to adopt in
writing flow charts. Boxes of specific shapes are often used for particular types of
76

A RE·EXAMINATION OF PRINCIPLES. 17

COMPUTE

f(x) _ eX sin 3x + 10

opera'tions-for example, a box with a truncated corner
used for read and punch operations. A rectangle

(a~

on an IBM card) (PU~CH

COMPUTE

I is often

is used for arithmetic

f (x) _ eX sin 3x + 10

operations. Branch tests may be indicated by an elongated oval with circles indicating the
possible decisions

IS f{x) NEGATIVE

of the instruction or the data

sent to the console

€>

• A fat oval is often used to indicate a modification

• A diamond is used to indicate that the program is

or to a deliberately invalid address to halt it. If a com·

pleted flow chart shows any portion which is not a closed loop or does not end in a

type block, be very suspicious and check it before continuing.
Block diagramming (flow charts) of complex problems is often done in three stages.
Stage 1: The overall flow chart. representing the general flow of the program. We have
not, as yet, worked any problems of sufficient complexity to merit this" flow chart of the flow
chart" technique.
Stage 2: The semidetailed flow chart is used to explode each of the stage 1 blocks into
more detail.
Stage 3: The detailed flow chart explodes each block of stage 2 into a series of logical
operations. It should be noted that even at stage 3 there will be little or no mention of the
specific 650 operations involved; indeed it is at this stage that the decision is often made
whether to place the problem on a 650, relegate it to a lesser computer such as the 407 or
602A, or perhaps program it for one of the larger 700 series machines. A programmer can take
the detailed (stage 3) block diagram and translate it into a program for whichever of these
machines is suitable. So far, the flow charts we have used have all been of the stage 3
variety, since the problems have been easy to visualize.
The booklet "Planning for IBM 65,0 Ca.rd Systems" (form F28-4016) may be helpful at this
sta.ge. The following diagrams of a "payroll problem" were taken from this source.

78 • A RE-EXAMINATION OF PRINCIPLES

STAG E 1: OVERALL FLOW CHART

BLOCK 00

r-----------I
I
I.

BLOCK 10
DEVELOP AND
STORE
~-----4---;
GROSS PAY

BLOCK 20
DEVELOP AND
STORE
~----~---4
GROSS PAY

BLOCK 30
DEVELOP AND
STORE
~-----4---4

BLOCK SO

GROSS PAY

L ___________________ J
BLOCK 40
DEV. AND STORE
DEDUC., NET
PAY, YR.-TODATE, CRSFT.

BLOCK 50

r-------I~-___.

PUNCH

PERFORM
PAYROLL
DISTRIBUTION

Stage 1 is omitted in uncomplicated problems.

BLOCK 70

A RE-EXAMINATION OF PRINCIPLES. 79

STAGE 2: SEMI-DETAILED

FROM PREVIOUS ROUTINES
BLOCKS: 10, 20, OR 30
BLOCK 41
r-----'''----..

DEVELOP AND
STORE
WITHHOLDING
TAX

BLOCK 42
DEVELOP AND
STORE
OASI

BLOCK 43
DEVELOP AND
STORE
NET PAY

BLOCK 44

DEV. AND STORE
NEW YR.-TODATE BAL.

BLOCK 45

TO: ERROR 2
BLOCK 70

CROSSFOOT
RESULTS

TO: DISTRIBUTION ROUTINE
BLOCK 50

There will, of course, be similar stage 2 charts for the other blocks of the stage 1 flow chart.
In many of our problems both stage 1 and stage 2 have been omitted.

80 • A RE.EXAMINATION OF PRINCIPLES
STAGE 3: DETAILED

FROM BLOCKS 10, 20, OR ~

BLOCK 41.1

-EXEMPTION
AMOUNT
XTAX
CLASS

BLOCK 41.2

+GROSS

IS THERE TAXABLE
INCOME?
BLOCK 41.3

YES

BLOCK 41.4
TAXABLE
BALANCE X
TAXABLE %
1/2 ADJ.

BLOC K 41.6

BLOCK 41.5
STORE
WITHHOLDING
TAX FOR
PUNCHING

LOAD AND
STORE
ZEROS FOR
PUN9HING

TO OASI CALCULATION
BLOCK 42

There will, of course, be similar stage 3 charts for each block of each stage 2 chart.

A RE-EXAMINATION OF PRINCIPLES. 87

Note: Practice suggests that one of the most useful devices in a detailed flow chart is
the use of a "remarks box." The author uses a triangle/ ....~ and a dotted line
,.
(indicating that it is not an actual part of the flow chart) for this purpose. These "remarks"
greatly simplify the reading of an unfamiliar flow chart (one you have not looked at for a month
or more, or one written by another person). Any time you use an "ingenious trick," a counter
which was initialized when the program was loaded, or a program which needs a bit of explanation, the reader will be well advised to insert it on the flow chart in a "remarks box."
Few things help more than a dozen well-chosen remarks on a flow chart you are examining.
A good flow chart has many uses. It helps identify the exact problem, which, surprisingly
enough, is often the most difficult part of writing a program. It permits later modification of
the problem with a minimum of effort. Since the flow chart breaks the problem up into logical
segments, it often is used to assign program sections to several programmers, thus reducing
the total time needed to complete a complicated program. A flow chart is almost essential in
discussing a program, and in the final write-up. One can easily justify the statement,
"Programming and flow charting are synonymous-the remainder is mere coding."

6-3. LOOPS.
If your program has no loops, it probably should be relegated to a less expensive machine
than the 650.
While looping is the technique which provides the 650 with so much power, it is also a
serious source of programming errors~ Always recheck the loop-testing steps of your program
before using up valuable 650 time on a debugging run. A good technique is to pretend that
each loop was to be executed only once, and analyze what your program would do (i.e., pretend the looping constants were different from those actually used in the programs). The most
-common errors in programming a loop result in going through the loop either k -: 1 or k + 1
times instead of k times. Other common errors result in never getting out of the loop at all.
It is usually worthwhile to initialize a loop (place the correct number in the loop counter)
just before it is used, rather than to depend upon its being there. It is often desirable to
reuse (steal) a loop, or even an entire program section as part of a long program, and there is
no telling what may be in a count box if it has been used previously. It is sometimes possible
to write a few extra instructions in initializing a loop and remove them from the loop itself
(i.e., from the part the program which repeats). Since three quarters of the machine time on a
500-step program may be spent in repetitions of a lO-step loop, a real savings may be effected
by dropping two steps from the loop.
Another machine-time-saving technique on a program which involves loops within loops
(as most advanced programs do) is to "unwind" the inner loop-i.e., to write out the innermost loop and carefully hand optimize it rather than modify the instructions. This results,
of course, in a program which uses more drum locations, but may save considerable overall
machine time, since often half the time on an advanced program is spent in the "inner loops."
In a loop which is repeated frequently, ~t may even make a difference whether instructions
are "stepped" (i.e., modified) before or after passing through the loop. Which is more
desirable depends upon the program involved. Sometimes it may be worthwhile,to take six or
seven extra steps in the nonlooping part of the program in order to keep the looping constant
(count box) the same as either the number of iterations performed, or as the number still to be
performed. This is particularly true if the program is to be used by many people or over a long
time, thus increasing the likelihood that the "tricks" involved will be forgotten. On the other

82 • A RE.EXAMINATION OF PRINCIPLES

hand, it is more often desirable to program as efficiently as possible and indicate the tricks
in a "remarks

box'~ ~

on the flow chart.

,"
6-4. FREE DATA
A sophisticated program makes much more output available for the same price (i.e., machine
time). Any time you punch out a card with fewer than 80 columns of information, you should
look around to see either if it is worthwhile combining several answers on a single output
card (see section 3-1), or if other output data are available which might be useful. If such
data are available store them in the punch band-they are free. In a program with too much
reading or punching for the amount of computation, the console lights seem static for a moment
at regular intervals while the 533 is operative. The 650 is waiting for the 533. This is often
unnecessary (see Chapters 2 and 3) and wasteful, although sometimes it is unavoidable in a
given portion of a program-for example, when the program is being loaded.

6-5. ECONOMY IN LOADING THE PROGRAM.
After a program is debugged and is running well, if it is to be used often it is worthwhile
to place it on 7-per-card (or at least on 4- or 5-per-card) load cards in place of single-word
load cards. Thus a given program will take only one seventh as long to load, in addition to
taking less card storage room. There are routines which your laboratory has (or can get) which
will automatically punch out a given program from the drum in any of these forms. Ask your
instructor about it when you are ready. Never try to debug from a multiple load card.

6-6. PROGRAM ERRORS-DEBUGGING.
Inexperienced people are apt to consider it a sign of incompetence when a program fails
to work the first time it.is tried (or even the tenth time, if the program is complex). This is
unjustified. It is a rare program which works correctly the first time! The sign of incompetence, then, is not that an undebugged program contains errors. If the programmer uses up a
lot of machine time debugging the program on the 650, that is a sign of lack of experience. An
experienced operator will almost always be off the machine in 10 or 15 minutes on a debugging
run, since, if he doesn't find the error quickly, a trace routine will be used and the results
taken to the quiet of his office for unhurried deliberation without tying up valuable 650 time.
Some debugging is done on the 650, of course, but most is better don"e in the privacy of your
own study.
Although the following suggestions are self-evident, experience suggests that even good
programmers are prone to these slips. lt may be well for a beginner to check theru over, just
as a pilot runs his checklist before taking off or landing a plane.
1. Do not place instructions (nor allow SOAP to do so) in drum locations 1951-1960 nor
in any 10-word read band which your program uses. (Why not?)
2. Always list your "program (on a 402 or 407), and check the program carefully before attemping to put it on the 65 O.
3. Have your key-punch operator verify the program cards before you accept them. (If a
card has been verified, it will have a small semicircular nick on the right-hand edge.) A
competent key-punch operator will often question your program. By all means encourage this,
even if you happen to be correct this time. She may save you $20 worth of machine time in
debugging your next program.

A RE-EXAMINATION OF PRINCIPLES. 83

4. Find out the drum locations used by the trace routines in your installation and do not
put program steps or data in these locations. If you use SOAP., block reserve (BLR) these
locations.
5. If you use subroutines, be sur~ they do not use the same drum locations as each other
or as your program or data. It is amazing how messy debugging becomes when the program
starts to take a square root and ends up using instructions from a sine subroutine.
6. Always" zero" or "stop code" the drum before placing a program on it for debugging.
Current policy seems to favor putting 0000000000- in each drum location. The author prefers
putting 01 0000 nnnn in drum location nnnn. Then if the program progresses to an unused
address, it will stop, and also show the unused drum location to which it was sent. If you do
zero the drum, then be sure drum location 0000 does not contain an instruction. (Why?)
7. Check the scaling required by every subroutine used, and the permissible range of
data.
8. Be sure that you have looped the desired number of times. Assuming that you had only
wanted to loop once, what would your program have done?
9. If you are doing engineering or scientific computing rather than data processing, then
learn as much as you can about numerical analysis! It is not easy, but it is vital! A worth- .
while course in numerical analysis will have at least a complete course in calculus (including.
Taylor's series with remainder and improper integrals) as a prerequisite. If the Monte Carlo
method is to be considered, and it should be, then probability theory is also a prerequisite.
10. Learn some modern abstract algebra-you'l1 need it. Modern algebra provides the
vocabulary and many of the concepts needed in applied mathematics today.
11. Alwa~s have a flow chart and a program listing on the console shelf when you sit
down to debug a program. Use trace routines to get off the 650 as quickly as possible (see
Section 6-7).
12. Explain your program (using a flow chart) to someone who knows less about it than
you do. Spouses, girl friends, and beaus make ideal victims.
13. Always debug a complex program using data for which the answer is known (run it out
on a desk calculator using one- or two-digit numbers if necessary).
14. If you suspect you are "in a loop," use the address stop feature of the 650 to locate
the loop.

6-7. SPECIAL TRACES AND DEBUGGING ROUTINE.
There are many debugging and tracing routines, and more are being develpped. Your
laboratory will have several. They are all easy to use. General descriptions are included.
here, while specific details will be found in the program write-up.
In general a "trace routine" punches a card for each step of the program performed.
This card contains, 'usually, an identifying number, the location of the instruction, the
instruction, the contents of the upper accumulator, the contents of the lower accumulator, the
contents of the distributor, and, if appropriate, the contents of the three index registers.
(Note: These may be the contents either before or after the instruction is performed, depending upon the trace used. See the program write-up in your lab for complete details about the
trace being used.)
The simplest (and most wasteful) trace is one which starts at the first instruction of your
program and traces each step. Since a card is punched for each step of the program, the 650
can proceed at only 100 program steps per minute-a slow walk compared to its usual more
than 60,000 steps per minute. For this reason it becomes imperative to speed up the tracing
routine. One of the simplest methods is not to trace subroutines, which are assumed to be
correct. This is one reason many subroutines in your 650 library have "negative instruc-

84 • A RE·EXAMINA TION OF PRINCIPLES

tions." By properly setting the sign knob on the console switches, the trace can be made to
trace only positive instructions. The next most obvious requirement is to have a trace which
will run at high speed until it gets to a certain address, then begins the slower punching until
a second address is reached, etc. Further refinements include· traces that trace a loop only
once or twice, and then let the machine run at high speed until the looping is finished, whereupon it begins tracing again.
A "logical trace" which punches cards out only for branch instructions is available.
The snapshot trace and flow trace will punch cards only for certain key drum locations which
the programmer designates as "bus stops." These locations are punched on cards and loaded
with the trace routine. There are many variations available, and others will spring up.
Usually it is good policy to begin with a "bus-stop" type trace and then do a "+ only trace"
on that portion of the program where trouble seems to occur.
Your laboratory write-up will give specific directions for using each trace, but sample
instructions are included here so that you may be familiar with what to expect.
Trace Routine: Uses 100 drum locations located in any two successive bands.
A maximum of 21 "bus stops" are permitted, and these must not be instructions uhich
are modified in the program.
Contents of the distributor and both accumulators prior to the execution of the "bus-stop"
instruction are punched along with the card number, instruction location, and instruction
specified by the bus stop.
Send the program to the drum location of the first address FFFF in the trace routine,
either by setting the console to 00 0000 FFFF or by using a transfer card. Push the start
button.

6-8. EXPERIENCE.

Frankly, what you need now is experience! Hop to it. The remainder of this book is
devoted to a brief discussion of various more advanced techniques, but what you need now is
not more reading, but about two week's experience on the 650. Then you can read the rest of
this book at your leisure, and also the related IBM manuals. There is much to learn, but your
most urgent need is experience. The following set of problems will get you started.

Problem Set 6-8

In each case compare your results with those of other students in your class.
1. The following questions are ones which you can answer by actual experiment with the
650.
(a) If the upper accumulator contains a positive number, what will the sign of the accumulator be after the instruction 11 8003 9000 is executed?
(b) Same as (a) but with a negative number in the upper accumulator.
(c) If your 650 is equipped with index registers, determine what sign the register has
when k-k is registered, for k > 0, and for k < 0 (i.~., HAA OOOk, SXA OOOk).
(d) Row can you load a constant into the drum using the storage entry switches? (Find
two ways-one using Read In Storage, and one not using this feature.)
(e) What happens if you try to read a SOAP program into the 650 with a standard loan
board in the 533? Why?
(f) Does the overflow light come on if an overflow occurs with the overflow switch in (1)
sense position? (2) stop position?

A RE·EXAMINATION OF PRINCIPLES. 85

(g) If an instruction has a positive sign, what sign will show on the display of the
program register?
(h) Same as (g) for negative sign.
(i) Same as (h) for no sign. (Booby trap.)
2. (a) Let drum locations 1101 to 1110 contain ten positive integers (whole numbers).
Design a program which will place the difference between the largest and the smallest of
these ten integers in drum location 1100. This technique is very valuable in problems involving handling and scaling of statistical data.
(b) Rework part (a) with the assumption that the ten integers may contain both positive
and negative numbers.
3. (a) Devise a program which will take the ten integers of Problem 2 and rearrange them
in descending order of size in drum locations 1127 to 1136.
(b) Write a program which will arrange 700 integers from drum locations 0001 to 0700 in
descending order of size in drum locations 1001 to 1700. Be careful that your program does
not use any data position or 1951-1960.
4. Write a program which will read a 10-digit integer K from a card and factor it into prime
factors. Punch out cards containing Pi and n i where Pi is a prime number which divides K
and n i is the highest power of Pi dividing K. Do not attempt to store primes on the drum.
Instead:

FACTOR OUT
HIGHEST POWER
OF 2 AND PUNCH
IF SUITABLE.
STORE QUOTIENT IN K-BOX

FACTOR OUT
HIGHEST POWER
OF P AND PUNCH
IF SUITABLE.
STORE QUOTIENT INK-BOX

TEST (ARE WE DONE?)
DOES QUOTIENT = 11

5. (a) Write a program which will compute the coefficients of the straight line of best, fit
through a set of 10 points.
(b) Do the same for an unspecified number of points where the last point is identified by
an 8-punch in Column 80 of the input card. Each input card is to contain x in word 1 and y in
word 2 as fixed-point numbers.

86 • A RE·EXAMINATION OF PRINCIPLES

(c) Read the IBM 650 manual on the use of the "Branch on 8 Distributor" codes 90-99
and use this in programming (b).
6. (a) Define K! = 1 • 2 ·3 ·4 ••• K for an integer K > o. Devise a program which will read
a card containing two integers M and N with M < N, and will punch out cards containing K!
for M ~ K ~ N. Discuss the limits on the size of N.
(b) Reprogram the problem of part (a) so that at least four answers are punched on each
output card. Did you use 10 or 20 digits for K!? Discuss the limits of the size of K in either
case.
(c) If your 650 has automatic floating-point arithmetic and/or index registers, use them in
reworking (a) and (b). Again discuss limitations on the size of N.
7. Write a program which will read eight fixed-point constants x, ao' at' a 2 , aa' a 4 , arS'
a 6 (you decide on permissible range of size for x and for the a's), and will compute and punch
f( X ) == a o + a1x + a 2 x 2 + aa x a + a 4 x 4 + at\x t\ + a 6 x 6 •
8. (a) Write a program which will read a card giving x min, x max, ~, A, and B, and will
x
punch out a card giving this data follo~ed by cards giving x and f(x) == A sin x + B e- for
x == x min, x min + ~, x min + 2~, ••• , x max. Discuss the restrictions on the input data
necessary in your program.
(b) Reprogram (a) so that at least four sets of answers are punched on each output card.
(c) Reprogram for floating point.
9. (a) A method for computing cube roots is the following: Let the zeroth approximation
be 10. Then

Y +1

~~ ~
[

+ 2y

Write a flow chart for a subroutine which will compute the cube root of a floating-point number
and exit to the instruction originally in the distributor. Assume that A is in floating point
and is in the upper. The cube root should be in the upper when the routine exits. The
answer should have at least 6-digit accuracy.
(b) Read the two articles on variations of Newton's method which appeared in the American
Mathematical Monthly Vol. 51, pp. 36-8, 1944, and Vol. 52, pp. 212-4, 1945, from the viewpoint
of digital computer programming.
10. Assume that eleven floating-point coefficients for a polynomial are stored sequentially
in SOAP Region 0 (C0001 through C0011). The following SOAP program is supposed to
evaluate the polynomial for the value of X (in floating point) in the upper accumulator. There
are two errors in the program and'one unnecessary line. The program exits to the i'nstruction
originally in the distributor (like a normal subroutine). The method for evaluating the polynomial is called "nesting. ,,' For example, to evaluate a polynomial, lao + a 1x + a 2 x 2 + aa x a + a 4 x4,
with 5 coefficients, one writes:
y

a o + x[a 1 + x {a 2 + x(aa + xa 4 )}]

In this way one repeats a sequence of two steps-"multiply by x, add a new coefficient,
repeat. "
T

LOC
POLYX

DATA

INSTR

ADD

STD

LEAVE

STU

RAU

C0011

RAA

ELEVN

RETRN

A RE.EXAMINATION OF PRINCIPLES. 87
T

I..OC

RETRN

EI..EVN

DATA

INSTR

ADD

ADD T

FMP

AUP

COOOO

SXA

0001

NZA

RETRN

I..EAVE

0000

0011

Debug the program.
11. Write a flow chart and a SOAP program to evaluate the following series
2
3
4
i+1
1 2 . 3 + 22 . 4 + ~ + ••• + (i 2) (i + 2) + • • •

to at least 3 significant figures. How many terms will this take? You can estimate the error
you make by cutting off at the nth term as follows: The sum of the terms left out is

L+

(i + 1)
i2(i+2)

i _ n

If n is large, a good approximation to E is made by replacing the sum by an integral:
1

E:::::

2 dx

n+l X

Since you want E < .001, you can determine n. Note that it is much larger than your first
guess!
12. Ten numbers are stored in locations 0100-0109. Write a flow chart and SOAP program
which will punch these numbers, one to a card, in descending order.
13. Write a flow chart to solve a set to three simultaneous linear equations. Assume that
the twelve coefficients are stored sequentially on the drum, and that the three constants b i
are read from a card.
Let the equations be:

All Xl - Al 2X 2 - A 1aX a = b 1
A 21 X 1 - A 22 X 2 - A2a X a

A a1 X 1 -A a2 X 2 - AaaXa

(Suggestion: Decide on a method and try a sample calculation on paper, noting each step you
must take. Then make a rough flow chart and from it make the final one.)
14. Two polynomials representing truncated power series are given by the following
equations:
m

A(x) =

B(x) =

i _ 0

i -

The product of these two series is
C(x) = A(x)
m

If C(x)

j -

B(x)

88 • A RE-EXAMINATION OF PRINCIPLES

then
j

a.b
..
1 .l-l

i .. 0

For example: Co = aob o ,

c l = a Ob l + a l b o ,

c 2 = a ob 2 + a l b l + a 2 b o
Assume that the coefficients of A(x) and of B(x) are stored in consecutive storage locations.
There a.re (n + 1) a's and (m + 1) b's. Note that, in the summation for c j ' any a i with subscript
greater than n is considered zero, as is any b i with subscript greater than m. Write a flow
chart to evaluate C(x) for x = 0, 1, ••• 10, punching a card for each value.
Result card format: Word 1 = x
Word 2 = C(x)
15. Write a SOAP n program to find the difference between the largest and smallest of the
1011 numbers stored in locations 0783-1793. Write comments to indicate what each step or
small group of steps does. Be sure to include your flow chart.
16. Discuss a significant application of digital computers to your major field. Outline
the problem, the nature and sources of data, the method of solution, and write a general flow
chart. Indicate the storage and speed required to make the application practical. The application should be something which is not being done at present, to your knowledge.
17. Devise an "alternator" which will send the 650 alternately to drum locations 0602
and 0702 on successive passes through the loop •. Try to invent several ways to accomplish
this.
18. Read "Computation of Common Logarithms by Repeated Squaring," by O. E.. Brown,
American Mathematical Monthly, Vol. 65, pp. 118-19 (February, 1958), and write a program.
Note that the article contains a suitable flow chart.
19. (a) Read "The Solution of Equations by Continued Fractions," by J. S. Frame,
American Mathematical Monthly, Vol. 60, pp. 293-305 (May, 1953).
(b) Read some of the references given at the end of the above article.
20. Consult recent issues of the Journal, of the Association for Computinq Machinery or
Communication of the Association for Computing Machinery and report on one article.
21. Devise two different methods of obtaining "20 zeros and a minus sign" in the accumulator. Try your proposals on an actual 650 to be sure that they will work.
22. Devise a short program which will permit you to specify a number N, 0 < N ~ 1999,
using the four Address Selection switches on the console, and will place this number into
index register A.
23. (For mathematicians only.) Use the available matrix inversion· routlnes in your computer library to form A- 1 for the following (ill-conditioned) matrix

A [~
=

7 5

Now form (A- 1 l and compare it with A, or form both A .A- 1 and A- 1 .A and compare
each with the identity matrix. Life is now always as simple as it appears.

24. Same as problem 23 ,using A =.
bert matrix of order six.

1/2

1/3

1/4

1/5

1/6

1/2

1/3

1/4

1/5

1/6

1/7

1/3

1/4

1/5

1/6

1/7

1/8

1/4

1/5

1/6

1/7

1/8

1/9

1/5

1/6

1/7

1/8

1/9

1/10

1/6

1/7

1/8

1/9

1/10

1/11

, the so-called Hil-

(CHAPTER

7.. 1. INTERPRETIVE SYSTEMS.
If a number of subroutines are used frequently, it becomes natural to leave them on the
650 drum more or less permanently. The next step is to load certain subroutines, along with
programs to get into and out of these subroutines, onto the drum, and soon an "interpretive
system" comes into existence. It is not our purpose to discuss all interpretive systems, but
rather to discuss one in detail. After this, the reader should have little difficulty adjusting to
other interpretive systems. One of the nicest interpretive systems available to date is the
"653 Rocket Package" programmed by V. N. Huff, D. N. Turner, and O. N. Reese for the
Lewis Flight Propulsion Laboratory, Cleveland. The "653 Rocket Package" makes use of the
automatic floating-point and index-register features of the 650, as well as using immediateaccess (core) storage. It is, therefore, not available for use on all 650's.
A year ago the author favored interpretive systems. Now, with the availability of For
transit, interpretive systems no longer seem as attractive. For transit offers even more ease
of programming than does an interpretive system. Furthermore, -while an interpretive system
is slow each time it is used, For transit uses extra machine time only the first time it is run
and proceeds at full machine speed on subsequent runs. Interpretive systems will probably
continue in favor for "one-time problems," but the author prefers the use of For transit or IT
even for these. Chapter 8 gives further information on For transit, and may be read next if
desired.

7-2. BELL SYSTEM (FLOPS).
The Bell Interpretive system, and its many variants, converts the 650 into a different
machine which is slower than the 650, but much eas.ier to program. The Bell Routine is
loaded into drum locations 1000-1999. Drum locations (0) 001 to (0) 999 are available to the
programer for use in storing instructions and data. Location (0) 000 has a special use as a
"previous result box" in which the result of the last operation is also available. If this
"previous'rosult" is needed on the next operation, time is saved in getting it from 000 rather
than from another storage location. Note that since there are only 1000 storage locations,
three digits 000-999 suffice as addresses. A Bell instruction consists of a signed 10-digit
number (word) which is broken up as follows:
89

90 • INTERPRETIVE SYSTEMS
ADDRESS

ADDRESS

9 10

For example, the instruction
ADDRESS

ADDRESS

057

621

483

says "Multiply the number in A drum location, 057, by the number in B drum location, 621,
and store the result in C drum location, 483." All three numbers are in floating-point form
and hence have eight significant digits and a 2-digit (mod 50) exponent. The old Bell system
uses floating-point numbers of the form x.xxxxxxx.l0 yy-ti 0 = xxxxxxxxyy rather than having
the decimal point before the first X as does machine automatic floating point.
To make the discussion clearer, the convention of using a bar over a drum location to
indicate the 10-digit-sign contents of that drum location will be adopted. Thus:
B
B

=
=

3-digit drum location
the floating-point number located in drum location B.

Some of the more important Bell operations are:

OP
ADD

SUB
MPY
DIV
NG MPY

2
3
4·
5

__li_

--~--~-

_!3__

Go to O2

-~--~-

_A._

_~_~r_ Q2 B

Add, in flo~ting point, the number, A,
stored in drum location A to the number, B,
stored in drum location B, and store the
result, C, in drum location C. This is
abbreviated: A + B = C.

_9_-

A-B=C

_£-

- A. B = C

_9__
_C__

X.B = C
XIB = C
negative multi ply

If the l-digit operation code 0 is used, this indicates that only two addresses are needed,
so a 3-digit operation code can be given in address A (now called O2 ) with Band C as addresses.
B

0 1
0

300

EXP E

301

LOGE

302

LOGe

SINR

303

SIN § -

C, B

COSR

304

COS § -

ART R

305

ARC TAN

ABS

350

lsi

SQRT

ys-c
=0
B= C

IN RADIANS

§ IN RADIANS

B - C, C

IN RADIANS,

lei

<~
2

The system contains similar codes for exponents and logarithms to the base 10, and for
functions in decimal degrees rather than radians. These will be listed later.

INTERPRETIVE SYSTEMS. 91

We illustrate a portion of a program which computes:
f(x)

x, in radians, is assumed to be in drum
location 500 and the constant, 1, in location 600 (1 = 1000000050)

sin x 3
1 + e- x

f(x) is to be placed in location 502, with 501 and 400 used to store e- x and .../1 + e-X 3
respectively. Note that the "previous results box," 000, is used as much as possible.
LOCATION OF

INSTRUCTION

N0 2

COMMENTS

401

MPY

500

000

402

NG MPL

000

500

000

-x

403

EXP E

301

060

501

a- x3 STORED IN 501

404

ADD

000

600

000

405

SQRT

300

000

400

+ a- x3
+ e- x3

STORE TEMPORARILY
IN 400

406

SIN R

303

500

000

SIN X

407

DIV

000

400

502

f (x) INTO 502

The extensive use of the "previous results box," 000, which has a quicker access time than
other storage, is worthy of note. A proper "read" and "punch" instruction is all that is
needed to complete the above program.
Another noteworthy thing about the program given above is the absence of a "next
instruction address. tt An integral part of the functioning of most interpretive routines is that
once the program is started, the next instruction executed is the one in the next higher drum
location. (This is also true of the larger JBM-700 ser:ies computers.) If it is desirable to go
from drum location k to a drum location other than k + 1, it is necessary to specifically command the machine to do so. The command

OP
TR

000
429
transfers control to C (i.e., the next
instruction executed will be C which
is stored in C). The B address is ignored, but
must be filled in with something, or the 650
will stop on a validity check.
7-3. READ AND PUNCH.

The card form used on Bell is not the same as that usually used for machine-language
programming. Six (or sometimes 5) words are read or punched at a time, in addition to certain
identifying numbers. The arrangement is as follows:
Columns
1-6
7-9

Contents
Identification
Drum location into which first word
on card is to be stored
(Continued on next page)

92 • INTERPRETIVE SYSTEMS

Columns
10
11

12-21
22
23-32
33
34-43
44
45-54
55
56-65
66
67-76
77-80

Contents
Number of words on card
Sign of word 1
Word 1
Sign of word 2
Word 2
Sign of word 3
Word 3
Sign of word 4
Word 4
Sign of word 5
Word 5
Sign of word 6
Word 6
More identification

Both input (Read) and output (Punch) cards have the same form. Cards may be read into or
punched out from any location, not merely read and punch locations. The operation codes are:

READ

400

PCH

410

A block of consecutive storage locations beginning at B and ending
at C is read into the drum from one or several cards. Numerous
checks are included here. For example, the sum of the word
counts of all cards (Col. 10) must be C - B + 1, and the cards must
be in consecutive order.
A block of consecutive words beginning at B and ending at C (inclusive) is punched. Six words (some programs use 5 words) and
the corresponding word count are punched into each card,except
possibly the last. Words remain in drum storage after punching.

7-4. ADDITIONAL BE LL INSTRUCTIONS.
Alpha.
Set A

Numer.
= 500

Set B

Set C

O2 == 005

Add A

Add B

= 060

Add C

...,

006

MOVE

:0::

= 050

600

Function
Set the A-address. The A-address of the instruction (B) specified by
B is set equal to C.
Set the B-address. The B-address of the instruction (B) specified by
B is set equal to C.
Set the C-address. The C-address of the instruction (B) specified by
B is set equal to C.
Add to the A-address. The A-address of the instruction (B) specified
by B is increased by C.
Add to the B-address. The B-address of the instruction (B) specified
by B is increased by C.
Add to the C-address. The C-address of the instruction (B) specified
by B is increased by C.
Move. If A 1'= 000, the block of A consecutive words beginning at B is
moved into the set of A consecutive locations beginning at C. The
words in the original locations are not destroyed, 'except where the
two regions overlap. The number in location 000 ("previous
result") is not affected when C F 000. Both C > Band C < Bare
permissible. An error stop occurs if C + A - 1 > 1000. If A 000,
the word (B) specified by the B-address is moved into location C
and into 000. It also remains in location B.
:oK

INTERPRETIVE SYSTEMS. 93

Alpha.

Numer.

Function

MOVE with A = 000 differs from MOVE with A = 001 only in that the execution time with A
000 is shorter and the previous result location is affected.
COND STOP O2

= 2~0

Stop conditionally and transfer. The machine stops if the Programmed
Stop switch on the console is in the Stop position. The number
1120 is displayed on the address lights and B on the display lights.
When the Program Start key is depressed, control is transferred to
C. If the Programmed Stop switch is in the Run position, control is
transferred to C without stopping.
(Caution: If the Programmed Stop switch is on Run, the stops for loss
of accuracy in sine, co-sine and logarithm, and the stop in the Cond
Stop operation will not occur.)
This operation may be used for stopping at check points in the early running stages of a
problem, with the option of avoiding the stops during later runs.
7-5. LOOP OPERATIONS.
A highly repetitive character is required of any problem to be economically handled on an
automatic computer. In certain instances, such as Newton's iteration procedure for the solution of equations, a repetitive process or "loop" is conveniently programmed, merely using
conditional transfer operations. In many cases, however, some of the instructions to be
repeated must be slightly modified in a systematic way before each new repetition. For
N

example, in the evaluation of a linear expression
i

aix i with the a i and the Xi stored in
1

blocks of consecutive locations, the addresses of a i and x i must be increased by 1 each time
a new term is to be computed. To facilitate programming of this kind, the system provides
two methods of so-called address modification. The simpler-but less general-of these
methods employs a special 10-digit index register called the loop box, which is stored in a
location normally inaccessible to the programmer. If an instruction carries a minus .sign~ the
current contents of the loop box will he added to the instruction (in fixed-decimal arithmetic
and without regard to the sign) before it is executed. If, for example, the instruction
-1 531 600 901 is given and the loop box contains + 0 009 000 009, the instruction actually
executed by the machine would read 1 540 600 910. The original instruction remains unchanged in its storage location. At the end of a calculation, an O2 instruction called LOOP
enables the programmer to increase the contents of the loop box by 1 in one or several address
positions and to transfer control back to the beginning of the calculation. Hence, the calculation may be carried out repeatedly, each time with different addresses used in the execution
of instructions with minus signs. A test provision included in the LOOP order stops the repetition after a specified number of executions and resets the loop box'to zero for future use.
An example will be given after the following list of LOOP operations.

Alpha.
LOOP A

Numer.

Function
Loop on A. The contents of the loop box are increased by
o 001 000 000. In other words, the A-segment of the loop box
is increased by 1. After the increase, the A-segment of the
loop box is compared to the B-address of the LOOP instruction. If the A-segment is less than B, control is transferred to
C. If the A-segment is equal to B (or greater, which will
(Continued on next page)

94 • INTERPRETIVE SYSTEMS

Alpha.

Numer.

Function
never be the case in normal use), the loop box is reset to
zero and control proceeds to the next instruction.
Loop on B. Analogous ''to LOOP A, with the B-segment of the
loop box now being increased and compared to the B-address
of the LOOP instruction.
Loop on C. Analogous to LOOP A, with the C-segment of the
loop box being increased and compared to B.
Loop on A and B. Analogous to LOOP A. The A- and B-segmentsof the loop box are increased by 1 and the A-segment is
compared to B.
Loop on A and C. Analogous.
Loop on Band C. Analogous. The B-segment is used for the
comparison.
Loop on A, Band C. Analogous. The A-segment is used for the
comparison.

LOOP B

LOOP C
LOOP AB

LOOP AC
LOOP BC
LOOP ABC

To illustrate the use of a LOOP order, consider the evaluation of the linear expression
20

L(x)

aix p where the at and the

are stored in memory. In choosing storage locations

i .. 1

for numbers, it is wise to plan in advance how they are to be used in the program. In this
case, since the a t and the x i are to be reached using the LOOP operation, it is advantageous
to store them in blocks of consecutive locations, say the a i in 800 + i and the Xi in 900 + i,
(i = 1, 2, ••• , 20). Suppose L(x) is to be stored in 700. For simplicity, assume that register
700 contains zero at the beginning of the calculation and that the loop box has been reset.
The entire program for this calculation might be written as follows:
INSTR. NO

ALPHA.

101

MPY

102
103
104

ADD

SING.

A OR O
2
801

~01

000

700

C
000
700

020

101

0
110
LOOP AB
+
NEXT INSTRUCTION IN THE PROBLEM.

Note that the B-address of the LOOP order simply indicates the number of times the arithmetic
calculation is to be performed, including the first time when the addresses are actually unmodified (modified by adding zero). The practice of starting the instruction numbering at, for
examples 101, rather than 001, facilitates later additions to the beginning of a program.
The loop box is automatically reset at the beginning of a new problem and whenever a
transfer out of a loop is effected by a loop order (as stated in the definitions above). Hence,
the resetting of the loop box need not concern the programmer under normal conditions. If the
need for resetting the loop box should arise, however, this is easily done by giving, for example, the order LOOP A with the B-address 000. According to the definition of LOOP A,
this will cause control to proceed to the next instruction with a resetting of the loop box.

7-6. EXAMPLE 1.
As an introductory example, the summation in the section on LOOP operations will be
programmed again using address change methods. This would be an inefficient choice in an
actual problem, but it will illustrate the difference, as well as the analogy, between the
methods. It is again assumed that register 700 contains zero at the start, but the steps
analogous to the resetting of the loop' box will be included.

INTERPRETIVE SYSTEMS. 95
INST.

ALPHA.

SIGN

101
102
103
104
105
106
107
108

SET A
SET B

+
+
+
+
+
+
+

0
0
3
1
0
0
6

MPY
ADD
ADD A
ADD B

A OR O

103
103

500
050

[

]

[

000
600
060
103

]

700
103
103
821

TR A
NEXT INSTRUCTION IN THE PROBLEM.

801
901
000
700
001
001
103

The brackets in the A- and B-addresses of instruction 103 are used to indicate that these
addresses are variable and will be supplied by the program before the instruction is executed,
hence what is written there when the program is loaded into the machine is irrelevant. At the
end of the program when instruction 108 is reached,storage location 103 will contain
+ 3 821 921 000. It is assumed that the summation just programmed is part of a larger problem
in which it is used repeatedly. This is the reason for the SET A and SET B instructions. If
801 and 901 were simply loaded into their respective positions in instruction 103 initially, the
summation would be performed correctly the first time it is used, but the next time the summation is called for, instruction 103 would read + 3 821 921 000 and erroneous calculations would
result. The SET instructions could, of course, have been inserted after the completion of the
summation, restoring instruction 103 to its proper value for the next application. This procedure is not recommended, however,because, in case of an interruption (e.g., error stop)
during the loop, it makes it more difficult to restart the problem from the beginning without reloading the program.

7-7. EXAMPLE 2.
A more realistic example of the use of address-change methods would be a calculation involving more than one summation index or parameter. In such a case, one of the fast and convenient LOOP orders would normally be used in the "inner loop" (i.e., the loop occurring
most frequently), with address-change operations controlling the "outer loop" or loops. SUp10

pose, for example, that it is desired to calculate Sj =

aj i Xi for j = 1, 2, ••• 5, where the

i:= 1

a ji are stored in 800 + 10 j + i (i.e., all is in 811, a 12 in 812~ etc.; a 21 in 821, a 22 in 822,
and so forth), the Xi in 900 + i, and the Sj are to be stored in 700 + j. It will be assumed that
register 500 contains zero. For completeness, the setting of all variable addresses to their
initial values for repeated use of the summation program will be included.
INSTR.

101

ALPHA.
SET A

SIGN

A OR O
2

500

104

COMMENTS

811 }

SET VARIABLE
ADDRESSES TO
THEIR INITIAL

102
103

SET C

104
105
106
107

MPY
ADD

108
109

MOVE

+
+

0
9

005
000

107
500

701.
400

VALUES

OOO}
400

"INNER LOOP"-

104

ON I
STORE THE

LOOP AB
MOVE

+
+
+

0
9

000

901
400
010
400

ADD A
ADD C

+
+

0
0

600
006

104
107

3
000
110

[

]
010 }
001

SET REGISTER
400 TO ZERO
I.E., SUMMATION

RESULT
INCREASE ADDRESSES
FOR NEXT REPETITION
IN THE OUTER LOOP
(J-LOOP)

(Continued on next page)

96 • INTERPRETIVE SYSTEMS
INSTR.

ALPHA.

110

111

PCH

SIGN

A OR O2

107

706

103

410

701

705

COMMENTS
TEST FOR END
OF JeLOOP

1-8. EXAMPLE 3.
In conclusion, a problem will be programmed in order to illustrate the use of many of the
operations and methods described above.
First,suppose it is desired to evaluate the "error function"
1. ¢ (x)

= -

IX e- t2 dt

YTi

for a set of values x = a, a + ~, a + 2~, ••• , a + 10~, uS 1ng the RAND approximation
2. ¢* (x) = 1-(a 1 n + a 2 n 2 + a g n 8 + a 4 n 4 + a5n5) ¢'(x),
3. n = 1/(1 + px) (p is a numerical constant),
and
4. ¢'(x)

Vrr

_x2

and to punch out the results as well as to store them for later use. The evaluation of the
polynominal in n will be faster if (2) is written in the form
5. ¢ (x) = 1- (n (a 1 + n(a 2 + n(aa + n(a4 + na5 ))))) ¢'·(x).
To make it possible to use the loop order in evaluating ¢* (x) this way, the coefficients a i
will be stored in consecutive locations in decreasing order. The LOOP program will be
given a form applicable to an arbitrary polynomial by including a "dummy" coefficient
a o = O. Storage locations will be chosen as follows:
LOCATION

CONTENTS

101-119

INSTRUCTIONS

(CARDS

200
201
202
203
204

r~}

CONSTANTS

221
222
223

CARD

LOC.

(CARD 5)

TEMPORARY

¢' (x)

NOT ON

STORAGE

CARD

("ERASABLE")

301
302
303
304
305
306

a. }

401-410

cp*(x}

ALPHA.

1-4)

0
4
0
8
O
2
0 1
0 0 =0

COEFFICIENTS
IN ¢* (x)
(CARD 6)

RESUL TS

SIGN

A OR

O
2

COMMENTS

101

SET C

005

114

40.1

SET ADDRESS OF FIRST cp*(x)

102

MOVE

000

202

221

FI RST x IS x = a

103

NG MPY

000

-x

104

EXP E

301

000

105

MPY

000

200

222

cp' (x) = .fiT.
2 e -x

-x

2
2

INTERPRETIVE SYSTEMS. 97
CARD

LOC.

{O6

107

108
109
110

ALPHA.

+
+
+
+
+

MPY
ADD
DIV
MOVE
MPY

LOOP B

112
114

NG MPY
ADD

115

ADD
ADD C

117
118

TR C
PCH

119

COND STOP

A OR O2

COMMENTS

204

221

000

201
000
301

000
223
000

000
201
000

000

223

000

302

000

ADD NEXT COEFF.

+
+
+
+

010

110

000
000
221

005
222
201

LOOP IN POLYNOMIAL EVAL.
POL YN. cp'(x)
cp'(x) 1 + PREVo RE,.S.

+
+
+
+

0
8
0
0

200

ADD

112

{'6

SIGN

000

]

[

= 1 I( 1 + px)

as INTO 000 FOR LOOP
PREVo RES. n

+ 6. = NEXT

203

221

006

114

001

NEXT cp*(x) ADDRESS

114
410

412
401
221

103
410

TEST FOR END
PUNCH TWO CARDS
END; STOP, DISPLAY LAST x,

500

GO TO 500 ON PROGRAM
START

SOO

NEXT INSTRUCTION IN THE PROBLEM.

An important remark should be made: If there is no shortage of storage locations and if
the programmer does not mind writing a somewhat larger number of instructions, the running
time for many problems can be decreased and the logic simplified by "unwinding" the innermost loop,-i.e., by writing out the mathematical instructions in the loop in a straight sequence instead of using the LOOP operation. In the present problem, a sequence containing
five MPY and four ADD instructions could replace the instructions 109-112 and also eliminate the use of the dummy coefficient a o ' The execution time for the polynomial loop would
be reduced by nearly one third, and the LOOP operation could be used to replace the addresschange operations in the outer loop. The polynomial evaluation accounts for about half of the
total running time of this problem! In many large problems, the innermost loop consumes an
even larger' fraction of the running time, making it important to program the innermost loop
efficiently even at the expense of apparent inefficiencies elsewhere.
7-9. SUMMARY OF BELL OPERATIONS.
BELL OPERA TIONS
O

0 , OPERATIONS

OPERA TIONS

000
200

ALPHA.
UNC STOP
COND STOP

NUM.

ALPHA.

ALPHA.
GO TO O
2
ADD

SUB

201

TR SGN

302

MPY

TR EXP

303

TR
TR SUBR

304

SIN R
COS R

DIV
NG MPY

202
203
204

305

ART R

TR A

205

TR PUT

350

ASS

7
8

TR S
TRC

100

LOOP A

351
352

EXP10
LOG 10

MOVE

010
001

LOOP B
LOOP C

353
354

SIN D'
COS D'

110

LOOP AB

355

ART D'

101

LOOP AC

NUM.
0

300
301

SQRT
EXP E
LOG E

011

LOOP BC

400

READ

111

LOOP ABC

401

CONS

410

PCH

500
050

SET A
450

START TR

SET B

(Continued on next page)

98 • INTERPRETIVE SYSTEMS
02 OPERA TlONS

0, OPERATIONS
NUM.

ALPHA.

NUM.

SET
ADD
ADD
ADD
SUB

005
600
060
006
700
070

C
A
B
C
A

NUM.

ALPHA.

451
452

STOP TR
ST TR ERAS

454

NOOP

SUB B

007

SUB C

A programming sheet of the following type is convenient for use with the Bell interpretive
system.
PROGRAMMER:
BELL SYSTEM
PROJECT NO.: _ _ _ _ _ _ DATE: _ _ __
650 PROGRAM FORM
DEPT

LOC

NO.
WD

MNE

°2/A

NO·

REMARKS

~~
~~
~
~
~

---

~~
~~
~~
~~
~
~/

7-10. REMARKS ON INTERPRETIVE SYSTEMS.

It should be remembered that there are over a dozen variations of the Bell system alone,
not to mention SIR, MITILAC, SIS, and similar independent routines. Each has advantages,
and each is slow on .~he "machine. The use of For transit seems likely to replace interpretive
routines for scientific and engineering work in many established laboratories, ,although there
will certainly be inertia to the change.

8-1. EASY PROGRAMMING.
In the discussion of looping (Chapter 3) you discovered that it was quite feasible and desirable to let the 650 take over some of the drudgery of writing a long program by instructing
it to modify its own instructions on each loop. Chapter 5 was a discussion of the Symbolic
Optimum Assembly Program SOAP, in which the 650 was programmed to do the tedious work
of optimizing a program and also the job of translating from an alphabetic to a machinelanguage code. In using SOAP it is still necessary to write one instruction in the SOAP program for each instruction in the final program. In Chapter 7 interpretive systems which sacrificed speed for ease in programming were discussed. Essentially, ·an interpretive system uses
ingenious internal programming to convert the 650 into a machine considerably different in
nature-a three-addresses machine, in the case of the Bell system. We now turn our attention
to two types of compilers whose purpose is to further simplify programming by making programs
more nearly parallel to the normal mathematical language used in stating problems.

8-2. IT.
There are a number of compilers in existence. Perhaps the best known of these is the InTranslator, IT, composed by A. 'J. 'Perlis, J. W. Smith, and H. R. VanZoeren at the Computation Center of Carnegie 'Institute of Technology. It is based on an earlier compiler developed. for the Datatron at Purdue. Having this machine-independent language, IT, for the
two most popular medium-speed computers should help solve the curre.nt problem of babble of
machine languages" In use,a programmer writes his program in the IT language, observing the
necessary conventions and restrictions. An IT program deck is then read into the 650 and the
program, written in IT language~ is placed in the READ hopper. When the Start button is
punched, the IT statemerttS which the programmer has written are taken into the 650 and a resulting SOAP program is punched out. Each IT command may yield several SOAP commands.
After the SOAP program is obtained as the output of the 533, this is run through in a normal
fashion and SOAPed to obtain an optimized machine-language program which is the final program usod. Two runs were necessary on the original IT compiler, but a "one pass compiler"
was released at the time this book was being printed (late 1958). Since ,excellent manu
for
the IT compilor are now available, it seems unnecessary to describe IT here in more de
If
your laboratory usos IT, .they will have a write-up describing the version used.
ter~al

700 • COMPILERS

8-3. FOR TRANSIT.

Fortran is the language used on the IBM 700 series computers. A simplified version of
Fortran, called For transit, is available (in five forms) for use on the 650. The basic advantages of For transit is twofold:
1. It is even easier to program in For transit than in IT.
2. A program written in For transit has the advantage that it will also run on a 700 level
machine, if such a machine becomes available to the writer.
For transit is essentially a compiler's compiler. For transit translates the programmer's
statement into the language of the Perlis compiler, IT. IT then translates the statement into
a series of SOAP statements. The SOAP program then translates the SOAP statements into
machine-language statements and, at the same time, optimizes the program. Since each For
transit program requires three passes on the 650 to produce a machine-language program, the
reader may feel this is a lengthy process. For a program in which the programmer merely
writes fifteen or twenty steps tying together several subroutines, ,this may be the case, and
SOAP language is probably the best tool to use in such an instance. However, in actual
practice, it turns out that the For transit language is an excellent one to use. Furthermore,
the amount of machine time spent in translation is actually quite negligible because once the
For transit cards are punched, they are placed in a For transit box somewhere in the lab.
Sometime when the 650 is not being used otherwise, a laboratory employee translates all the
programs from For transit to IT with one loading of the For.transit deck. Then the output, the
entire group of programs, will be translated into SOAP with one loading of the IT deck. Again
one loading of the SOAP deck is sufficient to produce machine-language programs from all of
the original For transit programs which were in the box. This is usually done at a time when
the 650 would otherwise be idle, and the programmer has no connection with it. He puts his
cards in the to-be-translated box and later picks up his program in machine language along
with the origin~l For transit cards from a translated box, without having bothered about them
in the interim. Most laboratories also list the programs, if requested .. The programmer then
schedules 650 time in the normal way. The biggest advantage of the For transit program is
the ease with which programs may be written and the reduction in the amount of machine time
needed for debugging. It may well be that, in a few years, the casual programmer will learn
only For transit and never have any great understanding of the machine-language program. It
seems unlikely, ,however, that this will extend to serious programmers who wish to take full
advantage of the machine, since this comes only by a thorough understanding. If the 650 in
your installation does not have the optional features necessary for use of any of the available
varieties of For transit, you should inquire about other programs from your representative. In
the meantime, secure a copy of the internal translator, IT, which is available in the 650
library program under file number 2.1.001, and use IT, or the newer one-pass compiler.

8-4. PROGRAMMING IN FOR TRANSIT.

If you will watch your parentheses and commas, it is fairly easy to learn to write 650 pro-

grams in For transit language. Three important conventions must be understood:
1. The symbol * denotes multiplication. That is, A * B means "A times B. U
2. The symbol ** denotes raising to a power. That is, A ** 5 means A 15.
3. The equals sign, when used in For transit, has nothing to do with an equation. It
means "compute the value of the right-hand member and store that value in the storage location which is designated by the left-hand member."
With these conventions in mind, the reader can see that the For transit statemont
ROOT

= (-

B + SQRTF (8

** 2 -

4.0

* A '" C»/(2.0 * A)

COMPILERS. 707

will eventually, after triple translation, give a machine-language program which will compute
the value of the expression
[- B + v'B2 - 4ACJ/2A
and will store this in the drum location called ROOT. This is, of course, not a complete program since there is no way to get the result from the machine after it has been computed. A
more reasonable problem would be:
EXAMPLE 1:

Read a card from which A, B, and C are obtained, then compute the larger root of
AX2 + Bx + C = 0, punch A, B, C, and the root on an output card,and repeat. The following
flow chart illustrates the problem in mind.

(

RCD
L

COMPUTE
ROOT=

(

- e +..Je'l. - 4AC
2A

1
PUNCH
A, e, C, ROOT

The For transit program given below will accomplish this.
1
2
3
4
5

READ 1, A, B, C
ROOT = (- B + SQRTF (B ** 2 - 4.0
PUNCH1,A,B,C,ROOT
GO TO 1
STOP

* A * C))/(2.0 * A)

The first statement causes the computer to read in a data card on which values of A, B, and C
have been punched in floating-point form. The second statement computes the root. The third
statement punches A, B, C, and the root. Notice that there is a comma after the phrase
PUNCH 1. This is essential. The fourth statement causes the program to return to the statement numbered 1, and hence completes the loop. Each statement must have a statement number, but if it is not referred to in the program the number can be zero unless the preceding
statement is a "DO" statement (not described here-see the For transit programmer's
manual). The last statement, STOP, is required at the end of each For transit program, even
though it is never used in the program. It is essential in the translation. The last line of
every For transit program will read STOP. Placing a 1 after the word punch and before the
comma is unnecessary as far as the basic 650 goes, but is essential if compatibility with the
700 series is desired. If more than one result is to be punched, they can be listed in the same
punch command and separated by commas. Fixed-point variables are indicated by expressions
invol ving from one to five letters and/or numbers, the first of which is I, J, K, L, M, or N.
Floating-point variables are indicated by expressions involving from one to five letters and/or
numbers, the first of which is a letter, but not I, J, K, L, M, or N. Either fixed- or floatingpoint variablos may be subscripted, but it is well to a void this complication in your first few
program8.

102 • COMPILERS

The For transit command

IF (A

* T) m, z,

is a branch command. It is translated into a series of program steps which computes and tests
the sign of the expression in parentheses, A . T in this case. If (A . T) is less than zero, the
next command executed will be the command with statement number m. If (A. T)equals zero,
the next command executed will be the command with statement number z. If (A. T) is greater
than zero, the next command will be the command with statement number p, where m, z, ,and p
are any integers which are used as statement numbers. The parentheses may contain any desired expression.
If you have a good flow chart, the For transit program is fairly easy to write in most
cases. Let us consider a refinement (since variable accuracy is permitted) of the example of
Section 4-4.
EXAMPLE 2:

Read a card giving, in floating point, A, W, and P where .0000000001 S P ~ .1 represents the
number of decimal places desired in the answer. Compute one solution of e- At - sin wt - 0,
obtaining a result to as many decimal places as indicated by P.
A possible flow chart is:

COMPUTE
F :::: e- At - sin Wt

PCH A. W. p. T

COMPILERS. 703

The corresponding For transit program could be
1
0
0
2
0
3
0
4
0
0
5
6
7
8

READ,A,W,P
T =< 0.0
DELTA == 1.0
F = EXP F (- A*T) - SIN F (W*T)
IF (F) 4, 4, 3
T == T + DELTA
GO TO 2
T = T - DELTA
DELTA :z 0.1* DELTA
IF (DELTA - P) 5, 3, 3
CONTINUE
PUNCH,A, W,P,T
GO TO 1
STOP

The 650 will translate this into a suitable machine-language program for use. The resulting machine-language program contains over 180 program steps exclusive of the subroutines. The reader is now ready to turn to the For Transit Programmer's Reference Manual
(32-7842) for additional instruction. Try a few programs. They are fun!
As you become more sophisticated you may wish to write a program to evaluate an integral
using Simpson's Rule (see any calculus text). The following program purports to evaluate
1
dx
- - 2 by Simpson's Rule using h = .05.

1
o

l+x

Will it work? If not, ,why not?
1
0
0
0
2
0
3
0
0
4
0
0
5
0
0
6
7

K=l
X- 0
SUM. 1.5
H "'" .05
X.X+H
IF (X - 1.0) 3, 6, 6
Y :: 1.0/(1 + X ** 2)
GO TO (4, 5) I{
PAUSE
SUM = SUM + 4 * Y

K-2

GO TO 2
SUM '"" SUM + 2

K.l
GO TO 2
Z .. SUM * H/3
PUNCH,H, Z
STOP

(CHAPTER

rn~[W~

9-1. COMMON COURTESY.

Devoting space to remarks concerning the etiquette of a computer laboratory may, at first,
seem unusual, but certainly such comments are not out of place in a book on programming. As
in all of life, courtesy is an oil which makes the machinery run more smoothly. In any laboratory there are bound to be certain rules which must be obeyed. The number of these rules can
be decreased sharply by observing a little common courtesy. For example, it is quite usual in
a 650 laboratory to have one key punch located reasonably close to the console of the 650. It
is common courtesy and indeed almost an unwritten rule that no one uses this key punch without the permission of the person who is signed up for the 650. A moment's thought will make
the reason fairly obvious. Time on the 650 is worth between $70-$100 per hour. An unmanned
key punch is worth perhaps a $1.25 per hour. It doesn't make economic sense to have the 650
idle while its user is forced to wait for you to remove your cards from the key punch, change
the drum roll, insert his own cards, and punch a correction for a program which he is debugging. If the user of the 650 is on a production run rather tha.n a debugging session, he will
almost always give permission to use the key punch, but it is certainly boorish to use the key
punch adjacent to the 650 without permission. Another item of common courtesy is not to
leave your cards lying around on the 533 or the 650 or anywhere "else in the laboratory after
you have finished your turn. Many laboratories enforce an automatic rule that before starting
a session with the 650 any loose cards found lying around are discarded in the wastebasket
or the "used-card box" for salvage sale.
Certainly no one would be so unthinking as to change a wire on a public board for the 533
or some other machine. Nevertheless, the author has seen it happen. Usually the person who
makes the change has every intention of replacing the wire in its original hole, but sometimes
the press of other tasks makes him forget to replace it, or he forgets which is the correct
hole. It is perhaps more common sense than courtesy which dictates that no beverages be
placed on the console shelf or on any of the auxiliary equipment. An upset cup can cause untold damage.
There is another unwritten convention among users of IBM equipment which is both commendable and remarkable. That is the practice of sharing programs. If you are visiting a
laboratory which has a program you desire, they will almost always reproduce a program deck
704

ODDS AND ENDS. 705

and give you a write-up. Experience suggests that the time to acquire such a program is when
you are on the scene, rather than writing later for it, although most installations and particularly IBM World Headquarters are willing to send out programs through the mail. Requesting
programs you do not actually need, or which your laboratory already has, will not make you
popular. If you go to a nearby installation with the intention of acquiring several programs
from them, naturally you take blank cards with you.
These items, as in the case of any matter of true courtesy, are each so obvious that it
should not be necessary to mention them. Nevertheless, experience suggests that a word is
in order.

9-2. SOURCES OF INFORMATION.

The field of high-speed computing is undergoing so many changes that it becomes rather
difficult to secure up-to-date information in book form. Your laboratory will certainly have the
Journal of the Association for Computing Machinery, a quarterly publication, the Oommunications of the Association for Computing Machinery, a monthly publication, and the American
Mathematical Monthly, which is published ten times a year. If you are seriously interested in
computing, you should join the Association for Computing Machinery ($6.00 a year) and the
Mathematical Association of America ($5.00 a year). You will then receive the three abovementioned journals without additional cost. The journal Mathematical Tables and 0 ther Aids
to Computation may also be consulted by interested persons. Most companies that make computing machines have fairly regular pUblications available. For example, the International
Business Machines Corporation puts out a Quarterly Journal of Research and Development,
the Technical Newsletters, and The 650 Data PlOcessing Bulletins, all of which you may
read at the laboratory.
A readily available source of information on Boolean algebra, number systems (including
binary numbers), matrices, systems of linear equations, groups, etc., is the author's Selections from Modern Abstract Algebra (Holt, 1958)-a text designed primarily for undergraduate
students.

9-3. ORGA",UZATION OF A 650 LABORATORY.

The organization of a 650 laboratory, or any other computing center, will naturally depend
upon the type of installation in which it is used. Two basic types of organization are widely
used. These have come to be known as the "Open-shop" and the "Closed-shop" systems.
In the open-shop system, everyone who uses the computer is expected to write his own program and do his own debugging and running of the programs. Usually the laboratory has keypunch operators and a machine-room supervisor available, in addition to laboratory assistants
who can give you aid. In a closed-shop system, the programs are written by employees of the
laboratory and the 650 is run only by laboratory employees. Obviously, many installations
function somewhere between the two extremes. Most university computing centers operate on
the open-shop system, since the variety of problems is almost too great to enable the laboratory to have competent personnel in the various fields. Irrespective of what system is used,
programs should always be u;ritten by a person who i8 quite familiar u;ith the problem, since
additional information is often available without cost, if the programmer knows the problem.
Furthermore, by cutting out the punching of intermediate results and by complete knowledge of
the data (Scaling), as has been emphasized previously, a considerable saving in machine time
may somotirnos be realized.

706 • ODDS AND ENDS

The physical set-up of the laboratory should include programming rooms and conference
rooms which are quiet and not affected by the noise of a busy laboratory. Usually the machine-room supervisor will take care of scheduling time on the 650 as well as on the auxiliary
equipment. One method which has been found quite successful is to give precedence during
the morning hours to the debugging of programs, and not permit anyone person to reserve more
than 30 minutes during the morning. Afternoon priority is given to production runs, and the
maximum advanced reservation for anyone person is one hour. Longer production runs are
usually scheduled during the evening or Saturday afternoon and Sunday. Laboratory schedules
for instrl,lctional purposes must, of course, be fitted into local requirements.

9-4. THE 650 LIBRARY.

A 650 installation will actually have two libraries. One consists primarily of books,
pamphlets, and journals. The second is a library of punched cards containing subroutines and
programs for the 650. It is usually the latter which is meant when one speaks of a "computer
library." A small library of books can become quite jumbled and still be useful because the
prospective reader can browse if he wishes. However, a library consisting of sets of punched
cards is of very little use unless it is properly catalogued. A helpful practice is to have a
loose-leaf notebook containing both abstracts and complete write-ups of each program and subroutine. Programs and subroutines which are used frequently should have duplicated descriptions for use in the programming room or the quiet of an individual's study. The IBM abstracts
of programs provide a useful guide, along with their more complete write-ups. The write-ups
furnished by IBM are master copies, suitable for reproduction by standard methods.

9-5. SUGGESTED GROUP DEMONSTRATIONS.

Every laboratory will develop its own policies concerning group demonstrations. This
brief paragraph is merely a plea that:
1. You do give such group demonstrations for interested organizations, school groups, and
industrial groups
2. You make such demonstrations informative and realistic rather than merely "cute and
clever. t.7
It is a fairly simple matter to program the 650 so that it will answer questions when
asked. This merely requires a branching operation on some specific columns in the input
data. For example, ,the 650 can be programmed so that whenever the first letter in card column
1 is a W the answer will be "yes"; if the letter in card column 1 is H, the answer will be
.
"no"; if the letter in card column 1 is D, .the answer will be "not at present but probably in
the near future"; if the letter in card column 1 is M, the answer will be "Of course the best
team will win"; etc. Then by carefully phrasing the first word of the questions when they are
typed onto input cards, ,the desired answer can be obtained from the output. This can be modified and improved considerably by listing a dozen different affirmative answers and having the
650 select them in sequence, ·thus giving different appearing, but equivalent, outputs for the
answer "yes." This is "cute and clever," but it is not informative. I sincerely hope that
your laboratory will not be guilty of using this type of demonstration for neophytes. It is a lot
of fun for experienced programmers, ,but it gives a very wrong impression to the general public
who, perhaps not surprisingly in view of the current science-fiction Sunday-supplement type of
literature, show considerable awe and fear of the power possessed by a computer. Instead of
reinforcing this fear, let us rather explain that the 650, or any other computer, is in essence a
supersonic moron, not a giant brain. It has truly'remarkable power. Its accuracy and phenom-

ODDS AND ENDS • 707

enal rapidity do merit awe, but there is no reason to endow it with mental properties it does
not possess. Let the programmer have the credit he deserves.
There are available, through the IBM library, several rather interesting programs of a nonmathematical nature, such as the playing of tic-tac-toe or of Nim. It is also simple to devise
a drum-count routine which will count the number of revolutions which the 650 drum makes in
one minute by adding l's to the accumulator. This short program can be placed on a single
load card for convenience in demonstrating (the observer always receives a thrill out of being
permitted to start and stop the 650 by punching buttons). A similar one-card program can be
designed (see Problem 7, Set 2-5) which will place its instruction in the upper and do additions in the lower, -taking the added 1 from the distributor. This program is of course independent of the drum, and will form a sum in excess of 100,000 in one minute. The solution of
a set of six simultaneous linear equations in six unknowns, or the evaluation of a polynomial
of high degree for a value of x selected by the observer also provides an easily understood
demonstration. It is desirable to have a short statistical problem, say the computation of the
means and standard deviations and possibly also the correlations for a set of 100 observations on four variables. These latter three provide understandable "workhorse" problems
which are meaningful to many observers who are potential users of the 650.

9·6. SUBROUTINES AND PROGRAMS FOR YOUR LIBRARY.

The International Business Machines Corporation distributes cards and write ups for programs and subroutines. If your installation develops a particularly useful or interesting program, you are invited to submit a copy of the program and write up to IBM World Headquarters
in New York. It will then be made available to anyone desiring it. If you desire a certain
subroutine ~r program, your laboratory can write to World Headquarters giving the file number.
A complete write up and a deck of program cards will be mailed to you by IBM. This is merely
one of many services which IBM makes available to you without additional charge. All requests from a given laboratory or institution should be channeled through one person rather
than having three or four people write for the same program. Once your laboratory has a copy
of the program, it is simple to run off duplicates on your reproducing punch.

9-7. WELL.

Why are you looking here? Get busy, ,write programs. You are mature enough to be on
your own now. Consult the various IBM programmers' manuals when you want information.
Experience is what you need now! Have fun!

The numbers refer to pages

Accumulator, 6, 18
Accuracy, 60, 70
Address
data, 7
instruction, 7
invalid, 27
stop, 33, 83
symbolic, 75
Algebra
abstract, 61, .83
Boolean, 61, 105
modern, 61, 83
Alphabetic code key, 4
Amerioan MathematioaZ Monthly,
105
Analysis
multiple regression, 65
numerical, 61, 70, 83
Approximate numbers, 60
Association for computing machinery, 105
"B-box," 44, 57
Bell system, 89
Bessel functions, 65
Biquinary, 25
Block diagramming, 8, 22, 76, 77
Boards, 4, 24, 31
"standard," 31
wiring for load cards, 31
Boolean algebra, 61, 105
Box
loop, 93
"previous results," 91
"remarks," 81
Branching, 15, 21, 30, 48, 58
on code number, 48
on index register, 58
on load card, 30
on overflow (47 BOV), 21
"Bus stops," 84
Button
"power-off," 25
"power-on," 25
Canned programs, 107
Capacitor storage, 6, 27
Card, IBM, 4
7-per-card load cards, 31, 82
Chart
flow, 8, 21, 22, 76, 77
planning, 9
Checking of the 650, internal, 27
Checklist, 82
Chi square, 65
"Closed-shop," 1 05
Code key, alphabetic, 4
Codes, operation, 11, 12, 20,
back cover
see Operation codes for specific page references
Common courtesy, 104
Communioations of the A.C.M.,
105

708

Computer laboratory, etiquette
of, 104
Console, front cover, 2, IB, 24,
36
operation, 36
switches (8000), 18
Continued fractions, 88
Continuous function, 67
Control
panel wiring, 4
switch, power, 24
Cookbook directions
to load and start program, 34
to read contents of given drum
location, 34
Count box, 43, 46, 57
Data, free, 82
Data Prot3.essing Bulletins, The
650, 105
DATATRON,21
Debugging, 81, 82, 83
Diagram, block, 8, 21, 22, 76,
77
Directions, cookbook
drum location, 34
to load and start program, 34
to read contents of given drum
location, 34
Display lights, 25
Distributor, 6, 10, 18
Division, 52
by- zero, 27
DiVIsor, improperly positioned,
27
Drum, 5
clear, 33, 35
location address, 5
to read contents of given location, 34
zero routines, 33, 35
Equations, simultaneous linear,
87
Error
analysis, 53, 60, 70, 82, 87
estimation of, 87
"function," 96
program, 82
"round-off," 70
sense-light, 27
Etiquette of a c9mputer laboratory, 104
First-reading, 31
Fixed-point variables, 101
Floating-poi nt
arithmetic, 69
operations, 71
variables, 101
Flops, 89

Flow
chart, 8, 22, 76, 77
of instructions, 25, 28
FORTRAN, 100
"For transit," 21, 89, 100
Fractions, continued, 88
Free data, 82
Function, error, 96
Group demonstrations, 106
hOok numbers, 49, 51
High punch, 30
Housekeeping, 16
Huff, V. N., 89
IBM card, 4
Improperly positioned divisor,
27
Improving speed, 38
Index register, 44, 57
Information, sources of, 105
input-output unit, 533, 2
Instruction, sign of, 7, 83
Instructions, 7
flow of, 25, 28
stepping, 38, 81
Integral, normal probability, 69
Integration, numerical, 65
Internal checking of the 650, 27
Interpretive systems, 89
IT, 99

JournaZ of the Assooiation for
Computing Maohinery, 105
Laboratory, 650, organization
of, 105
Library, the 650, 106
Light
accumulator, 31, 32
auxiliary, 32
checking, 32
clocking, 32
control unit, 32
data address, 31
display, 25
distributor, 32
error sense, 27, 32
input-output, 32
instructions address, 31
operating and checking, 31
overflow, 32
program, 31, 32
storage selection, 32
Linkage, 62
List of available programs
and subroutines, 65
Load cards, 3
board wiring for, 31
Load routine
one-instruction-per-card, 31
4, 5, 6, 7, and 8-pcr-card, 31,
82
Location address, drum, 5
Locations, addressable, 18

INDEX. 709

Logical
test, 15
"trace," 84
Loop, 32, 40, 42, 46, 81, 93
box, 93
unwinding innermost, 81, 97
Looping and stepping, 42
Lower accumulator (8002), 18
Magnetic drum layout, 5
Mathematical Association of
America, 105
Mathematical Tables and Other
Aids to Computation, 105
Matrix, 65
Memory (or storage), 2
MITILAC,98
Modern algebra, 61, 83
Modifying addresses, 40, 46,
57, 81
Monte Carlo, 83
Multiple regression analysis, 65
Multiplication, 18
"Negative instructions," 83
Newton's
iteration, 93
method, 86
Nim, 107
Normal probability integral, 69
Numbers
approximate, 60
h-k, 49, 51
Numerical analysis, 61, 70, 83
"Open-shop, I t 105
Operating and checking lights,
31
Operation, shift, 49
OPeration, 7
Operation code, invalid, 27
Operation codes, list (see back
cover for complete)
15 ALO (Add to LOwer), 12,
59
10 AUP (Add to UPper), 20
50 AXA (Add to indeX register
A),57
90-99 BD (k) (Branch on 9 in
Distributor), 62
41 BMA (Branch on Minus in
index register A), 59
46 BMI (Branch on MInus), 20
47 BOY (Branch on OVerflow),
18, 21
32 FAD (Floating ADd), 71
37 F AM (Floating Add Magnitude), 71
34 FDV (Floating DiVide), 71
39 FMP (Floating MultiPly),
71
33 FSB (Floating SuBtract),
71
38 FSM (Floating Subtract
Magnitude), 71
01 HLT (HaLT) or stop, 21,
33
69 LDD (LoaD Distributor), 12
00 NOP (No OPeration), 21
40 NZA (br!Ulch on Non Zero
in index register A), 58, 59

45 NZE (branch on Non Zero
Entire accumulator), 20
44 NZU (branch on Non Zero
in Upper half of accumulator), 14, 15, 20
71 PCH (PunCH), 11, 24, 59
80 RAA (Reset and Add to index register A), 58, 59
65 RAL (Reset and Add into
Lower), 12
60 RAU (Reset accumulator
and Add to Upper half), 20,
59
70 RCD (Read CarD), 11, 24
81 RSA (Reset and Subtract
from index register A), 58
66 RSL (Reset and Subtract
from Lower), 20
61 RSU (Reset and Subtract
from Upper), 20
36 SCT (Shift and CounT), 50,
62
16 SLO (Subtract from LOwer),
12
35 SLT (Shift LefT), 50
31 SRD (Shift and RounD), 50
30 SRT (Shift RighT), 50
24 STD (STore Distributor), 12
20 STL (STore Lower onto
drum), 12
21 STU (STore Upper), 20
11 SUP (Subtract from UPper),
20
51 SXA (Subtract from indeX
register A), 58, 59
84 TLU (Table Look Up), 62
02 UFA (Unnormalized Floating Add), 71
Operation codes, new, 20
Optimum programming, 56, 72
Organization of a 650 laboratory,
105
Overflow, 18
switch, 27
PerIis, A. J., 99
Permissible values, range of, 66
planning chart, 650, 47
power control switch, 24
"power-off" button, 25
"power-on" button, 25
Power series, truncated, 87
Power unit, 2
"Previous results" box, 91
Program
errors, 82
to load and start, 34
optimum, 56, 72
start, 33
Programming
linear, 65
sophistication in, 76
Programs, sharing of, 104
Programs and subroutines, list
of available, 65
Punch, 11, 24, 59
"high," 30
summary, 38
Punched hole, 27

Quarterly Journal of Research
and Development, 105

Read, 24
Reese, O. N., 89
Register,
address, 25
operation, 25
Relocatable subroutines, 66
"Remarks box, It 81
"653 rocket package," 89
"Round-off error," 70
Routine,
drum zero, 33
trace, 82, 83
5-10 rule, 56, 72
Scaling, 49, 51, 59, 65,
in multiplication, 51
SENSE, 18
Sharing programs, 104
Shift operation, 49, 50
Significant digits, 70
Simpson's Rule, 103
Simultaneous linear equations,
87
SIR,98
SIS, 98
Smith, J. W., 99
SOAP (Symbolic Optimum Assembly Program), 7, 72ff
card, type 1, 75
Sources of information, 105
Standard board, 4
Stepping instructions, 38, 81
"Stop code," 33
Storage or memory, 2
buffer, 24
capacitor, 6, 27
unit, 2, 5
Subroutines, 62, 65, 83, 107
list of available, 65
Summary punch, 38
Switch
, 'master power," 25
overflow, 27
power control, 24
Switches, console (8000), 18
Symbolic addresses, 75
Taylor, M. A., 83
Taylor's series, 61
Technical Newsletters, 105
Terminating technique, 43, 45
Test, logical, 14
Tic-tac-toe, 107
Time, saving of, 81, 97
Trace routine, 66, 82, 83, 84
Trapezoidal rule, 69
Truncated power series, 65, 87
Turner, D. N., 89
Type 1 SOAP card, 75
"Unwinding innermost loop, I t
81,97
Van Zoeren, H. R., 99
Variables,
fixed-point, 101
floating-point, 101
Wiring, control pane 1, 4
"WORD," 4
Z-box, 57

IBM

Problem Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

INTERNATIONAL BUSINE

650

TRADE MARK

Storage Entry Words

PLAN

Memory Address

load Card
Input Card

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3-Sheet reading
4-Additional description

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8003
Upper Accumulator

Operation
Abbrev.

STANDARD REFERENCE CODES
a-Normal instruction
1 -Instruction developed by 650
and Temporary Storage

• •

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.CHINES CORPORATION

Form No. 22-6151-2
Printed in U.S.A.

Problem No. _ _ _ _ Page _ _ _ of _ __

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PROGRAM SHEET

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PROBlEM _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ PROGRAMMER _ _ _ _ _ _ _ _ DATE _ _ _ _ PAGE _ _ OF_

Form No. 22-6181-3
Printed in U. S. A.

650 PROGRAM SHEET

IBM
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