IBM 610 Instruction Manual

m
IBM 610
Auto-Point Computer

©1957 by
International Business Machines Corporation
590 Madison Avenue, New York 22, N. Y.
Printed in U. S. A.
Form 23-6335-0

CONTENTS
INTRODUCTION
Speciajl Features
Auto-Point —
Program Preparation
Decimal-Octal
Multiple Operation _
Checking
Address System
Register
Word
Selected Register
Entry of Data
BASIC ARITHMETIC OPERATIONS
Resuming Automatic Operation 25
Physical Arrangement of Tapes 25
Tape Loading 26
Tape Code 28
Control Panel 28
Transferring Control to the Control Panel 30
Sequence Operation 3
Function Hubs 3
1
Program Skip Hubs 31
Type Suppression Hubs 32
Sequence Hubs 32
Selectors 3 3
Distributors 36
Response Class Control Hubs 37
Miscellaneous Hubs 38
Addition
Subtraction
Changing the Sign
Clearing
9
1
1
Transfer of Data within the Machine 10
Setting aRegister to Zero 10
Clear Right Half 10
Shifting 1
One Place Shifts 11
Multiple-Place Shifts 11
Rounding 1
2
Special Registers 1
3
Multiplication 1
4
Division 1
4
Simultaneous Divide and Multiply 1
5
Square Root 1
5
Programming Example 15
OUTPUT DEVICES
CONTROL DEVICES
Manual Keyboard
Program Tape
Instruction Classes
Program Tape Response Class
Program Tape Example
Data Tape
Reading-In from Data Tape -
Deleting Characters on Tape
Incorrect Key Depression
19
20
20
21
22
23
23
24
24
Automatic Typewriter
Punched Tapes
Data Tape Correction
Tape Word-End and Group-End Symbols _
Tape Codes Related to Teletype Operation
Tape Usage Suggestions
Cathode Ray Tube
OTHER MODES OF OPERATION
Octal-Decimal
Fixed-Point
43
45
47
47
48
48
49
51
51
Interrupting Automatic Operation 24
SUMMARY OF OPERATION
Manual Keyboard 54
Keys 54
Switches 57
Checking Lights 58
Display Tube 59
Computer Light Panel 59
Checking Features 60
PROGRAMMING THE 610
Timing 61
Planning the Program 62
Preliminary Planning 62
Most-Frequent Types of Calculations 63

IBM 610 Computer

IBM 610 AUTO-POINT COMPUTER
THIS MANUAL is intended as areference for pro-
grammers of the IBM 610. All information about a
given topic is assembled in one place with the refer-
ence purpose in mind. However, the organization is
such that abeginner can gain complete knowledge
of the machine through careful study of this manual.
The IBM 610 is amobile, general-purpose computer
specifically developed for awide range of industrial
and engineering organizations, from the small devel-
opment laboratory where it can function as afast
and automatic primary calculator, to the giant re-
search facility where it will be auseful helper to large-
scale computers. It has many of the arithmetic and
logical facilities usually associated with modern large-
scale automatic computers; yet it is as easy to use as
adesk calculator.
The machine consists of three physical units: the
keyboard, the typewriter, and the computer itself.
The computer unit contains paper-tape punching
and reading units and apluggable control panel, as
well as all of the arithmetic and control circuitry.
Numbers up to fifteen digits in length may be
entered into the 610 either manually from the key-
board or automatically under computer control from
punched paper tape. Data are internally stored in
eighty-four registers on the magnetic drum. Each
digit is identified by arecognizable pattern of mag-
netic spots. Any information recorded on the drum
will remain there permanently, or until it is erased
by recording other information in the same location.
The machine may be turned off completely without
any danger of loss of data.
Control of the machine may be exercised in three
ways:
1. manually from the keyboard,
2. automatically from apunched paper tape pro-
gram,
}. automatically by control panel wiring.
Transfer between control devices is automatic.
Final or intermediate answers may be obtained on
the typewriter, on punched paper tape, or both simul-
taneously.
SPECIAL FEATURES
Some outstanding and unusual features have been in-
corporated into the IBM 610.
Auto-Point
Atruly unique feature of the machine is the ability
to automatically handle adecimal point. When oper-
ating in the auto-point mode, data entered are auto-
matically positioned within storage locations so that
the point is centered using one-half of the register for
integers and the remainder for fractions. Every com-
putation causes an accurate automatic re-alignment
of the point. In comparison to widely used floating-
point systems, this feature needs no lengthy pro-
gramming routines or separate storage locations to
remember where the point is. The machine carries an
actual decimal-point notation, and within machine
capacities, the programmer is completely freed from
scaling input numbers and remembering decimal posi-
tion during computation.
Program Preparation
While aproblem is being solved manually, the com-
puter is able to create aprogram tape at the same time.
Asubsequent problem of the same type will require
only the input of new data; control of the machine
can then be transferred to the program tape for com-
pletion of the calculations and the printing of an-
swers.
Decimal-Octal
It is possible to perform either decimal or octal
arithmetic. This means that test programs for large-
scale binary computers can be developed in decimal
form and then executed in octal arithmetic for direct
comparison to the binary machine results.
While operating in the octal mode, the machine
will perform all the functions as in the decimal mode
including auto-point octal.

IBM 610
Multiple Operation
The machine provides multiple function instruc-
tions for programming simplicity. The Square Root
instruction accomplishes in one operation what is nor-
mally amany instruction subroutine. Divide-Multiply
combines into one operation the division of two fac-
tors and multiplication of the quotient by athird.
Checking
The IBM 610 has been designed with reliability as
aprimary consideration. Conservative circuit design
and components chosen for their high reliability are
employed throughout. This computer also uses built-
ia self-checking features, which provide further as-
surance of accurate results.
ister is 9999 99999 99999.9 in auto-point. (Note that
these two statements are not equivalent. The number
10000 00000 00000. is afifteen-digit number, but
it may not be entered into the machine because it is
greater than the maximum allowable magnitude.)
The register capacity of thirty-one digits allows
the machine to position every number with its point
in the middle of the register for operation in the
auto-point mode. Astorage register may be visualized
as follows:
15 Positions 16 Positions
.t
sign high-order half of
storage register
low-order half of
storage register
ADDRESS SYSTEM
The machine is concerned with two basic types of
communication —problem data and control infor-
mation. Problem data are entered into the machine
and stored as magnetic spots on the surface of the
drum. In order for these data to be used they must be
stored and retrieved from the same physical location.
Asystem of numerical addresses plus checking and
timing circuitry allows the operator to locate data
for input, output and computations. The idea is anal-
ogous to an automatic parking lot in which cars are
mechanically stored in numbered slots and retrieved
again from the same slot.
Register
Within the machine, data are stored on the drum
in 84 addressable locations called registers. Each of
the registers has an address: These addresses are 00,
01 79, plus special registers A, MP, MC, DIV
(to be explained later). This permits retrieval and
storage of information during calculation. Each of
these locations has acapacity of up to 31 digits plus
algebraic sign and radix point.
Numbers up to fifteen digits in length, plus point
and sign, may be entered into any storage register.
The largest magnitude that can be entered into areg-
The first position to the left contains the sign of
the number. If this position contains a0, the sign of
the number is positive; if this position contains a9,
the sign of the number is negative.
The storage register may be thought of as divided
into two parts as indicated by the space in the middle
of the register. When the machine is operating
in auto-point, the integer portion of anumber is
in the left half or high-order half of the register. The
fractional portion of the number is contained in the
right half or the low order of the register. The decimal
point is stored in combination with the first digit in
the lower half of the register. The number stored in
this manner is said to be in right-hand standard posi-
tion. During operation in the auto-point mode, it is
permissible to move anumber out of the right-hand
standard position. However, before any arithmetic
operations can be correctly performed in the auto-
point mode, the number must be returned to the
right-hand standard position.
Anumber is said to be in left-hand standard posi-
tion if the first significant (non-zero) digit has been
moved to the first position to the right of the sign
position, or if the decimal point has been moved to
the first position to the right of the sign position.

INTRODUCTION
Figure 1. Keyboarx)
Word
Tke basic unit of problem data is called aword,
which is 15digits in length plus algebraic sign and
decimal point. The words may enter the machine via
paper tape or the keyboard, and once stored on the
drum they can be addressed for computations.
The 31 -digit register size allows automatic double-
precision arithmetic. For most purposes, word size
need not concern the operator.
Selected Register
There are normally two factors necessary for the
execution of any arithmetic instruction. The 610 gives
special attention to the address of the first or prime
factor by designating it as the selected register. AH
610 registers are capable of performing arithmetic
and logical operations, and therefore are recipients of
results. The result for an operation is placed in the
selected register. This emphasis on the prime factor
gives aprogramming advantage, because aregister re-
mains selected until aprogramming change is speci-
fied. Obviously, afollowing instruction could be
written supplying only an operation and asecond
factor, the machine using the selected register con-
tents as the first factor. The conventions used to ex-
press this and the few exceptions will be explained
in alater section.
When aproblem is processed in aone-time calcu-
lation, instructions are given to the machine by de-
pressing the keys on the keyboard. Problem solution
is therefore accomplished in amanner similar to a
desk calculator. (See Figure 1.)
This manual first explains machine performance
in terms of manual control and subsequently pro-
gresses to automatic control.
Because of the advantage of operation in auto-
point mode, all problems will be in the auto-point
mode unless otherwise stated.
ENTRY OF DATA
Let us consider the entry of data, remembering that
depression of akey activates the machine. We first
select one of the registers by keying in its address on
the keyboard (00-79). For this operation ent is
depressed, followed by the required digits, decimal
point, and finally the algebraic sign. The auto-point

IBM 610
feature inserts zeros in unused digit positions when
the number is automatically positioned, requiring the
programmer to enter only significant digits.
s(Enter)
The depression of the ent key on the keyboard
clears the selected register to zeros and enters the
digits next specified into the selected register. The
actual data may contain from one to fifteen digits,
plus point and sign. Depression of the +or —key
gives the numerical data its appropriate sign, causes
the auto-point feature to position the data in the
register, and completes entry.
If aprogram is being prepared to process several
sets of data under automatic control, provision must
be made for the entry of each set of data. To accom-
plish this, machine calculation is stopped long enough
to enter the data manually or to allow instructions
for data to be read from the tape.
Example: The instruction 07 ent 62.731+ causes the
register 07 to be cleared and the number 62.731-)- to be
placed in the register. Register 07 remains selected. The
process takes place as follows:
1. The key is depressed.
2. The 7key is depressed.
3. The ENT key is depressed, resulting in the register 07
being cleared to all zeros.
0000000000000000 .0000000000000000
4. The 6key is depressed giving
0600000000000000 .0000000000000000
5. The 2key is depressed giving
0620000000000000 .0000000000000000
6. The .key is depressed giving
062.0000000000000 0000000000000000
7. The 7key is depressed giving
062.7000000000000 0000000000000000
8. The 3key is depressed giving
062.7300000000000 0000000000000000
9. The 1key is depressed giving
062.7310000000000 0000000000000000
10. The -(- key is depressed to give the number asign
causing ato be placed in the sign position or highest-
order position of the high-order half of the register and
the number is moved to the center of the register giving
0000000000000062 .731000000000000
If anumber -|- .0076 is entered into the machine, ., 0, 0,
7, 6, and +keys are depressed, and the number appears
in the register as:
0000000000000000 .0076000000000000
In the example used with the ENT instruction, the
number entered in this manner would be used in aone-
time calculation from the keyboard, or as aproblem con-
stant. If aprogram is being prepared to process several sets
of data under automatic control, provision must be made
for the entry of each set of data. In order to accomplish
this, the machine calculation is stopped long enough to en-
ter the data manually from the keyboard or instructions
are inserted calling for data from the tape.
At the moment, attention will be directed to one-time,
keyboard calculations. Later discussion will be concerned
with multiple sets of data.

BASIC ARITHMETIC OPERATIONS
THIS SECTION deals with the basic arithmetic op-
erations in 610 language along with an explanation
of their functions.
Note: The IBM 610 automatically obeys all of the
usual sign rules of elementary algebra.
ADDITION
SUBTRACTION
©(Plus)
The first occasion for the depression of this key is
giving asign to apositive number being entered into
the 610. Depression of this key terminates the entry
of the number and places ain the left-most position
of the register (a machine symbol for a+sign), and
automatically positions the number in the register
for calculation.
During the calculation part of aproblem, the de-
pression of this key tells the machine to add the con-
tents of the next addressed register to the register
already selected, and to replace the contents of the
already selected register by the sum. The contents
of the addressed register remain unchanged.
Example 1:
Register 01 plus Register 02 (01 selected, 02 addressed)
Register
Before Ol-j-OOOOOOOOOOOOOOl .7300000000000000
02+000000000000256 .0033000000000000
After 01-1-000000000000257 .7333000000000000
02-1-000000000000256 .0033000000000000
Following the addition register 01 is selected.
Example 2:
05-1-17
Register
Before O5-f00OOOOOOO0OOO02 .7600000000000000
17-000000000000003 .0700000000000000
After 05-000000000000000 .310000a;p00000000
17 -000000000000003 .0700000000000000
©(Minus)
Following the addition register 05 is selected.
The first occasion for the depression of this key is
in giving asign to anegative number being entered.
Depression of this key terminates the entry of the
number and places a9in the left-most position of
the register (the machine symbol for a-sign), and
automatically positions the number in the register
for calculation.
During the calculation part of aproblem, the de-
pression of this key tells the machine to subtract the
contents of the next addressed register from the se-
lected register, and to replace the contents of the
selected register by the difference. The contents of the
addressed register remain unchanged.
In general, the operator need not concern himself
with the actual representation of the number within
the machine. The operator will find it best to think
of the number as being stored with aminus sign and
proceed as he would if he were writing the problem
on awork sheet for hand calculation. "When anega-
tive number is read out of the machine to the type-
writer, it appears as atrue number with aminus sign.
Example 1:
Register 06 Minus Register 73
Register
Before 06+000000000000005 .7600000000000000
73-1-000000000000003 .1500000000000000
After 06+000000000000002 .6100000000000000
73-f-000000000000003 .1500000000000000
Following the subtraction register 06 is selected.
Example 2:
07-14
Register
Before 07+000000000000013 .6500000000000000
14-f-000000000000017 .7500000000000000
After 07-000000000000004 .1000000000000000
14+000000000000017 .7500000000000000
07 Selected. '

10 BM 610
CHANGING THE SIGN Example: 01 CPY OJ
(cNn (Convert)
It is sometimes necessary, during acalculation, to
change the sign of anumber from plus to minus or
vice-versa. On most computers this necessitates find-
ing aregister containing all zeros and subtracting the
number from the zero register.
However, in the IBM 610 it is necessary only to
give a single command—convert.
The depression of this key tells the machine to re-
place the contents of the selected register by the nega-
tive of the original contents. The register remains
selected.
Example 1:
06 CNV
Register
Before 06+000000000000013 .6J00000000000000
After 06-000000000000013 .6500000000000000
06 Remains Selected
17 CNV
.0700000000000000
.0700000000000000
Example 2:
Before 17 -000000000000003
After 17-J-OOOO00O0O000003
17 Remains Selected
CLEARING
Transfer of Data within the Machine
COPY) (Copy)
In order to move anumber from astorage register
to another location in storage we use copy.
The depression of this key tells the machine to erase
the contents of the selected register, to replace it by
aill zeros with aplus sign, and then add the contents
of fekc next addressed register to the zeros of the se-
lected regi«ter. The contents of the addressed register
Fcmain unchanged. The selected register remains se-
lected.
One important use of this instruction is for the
storage of intermediate results during calculation.
Register
Before 01+000000000000001 .7300000000000000
05 +000000000000002 .6700000000000000
After 01+000000000000002 .6700000000000000
05-f-000000000000002 .6700000000000000
01 Selected
Setting aRegister to Zero
(Clear)
In order to clear astorage location previous to an
accumulation we use clear.
The depression of this key tells the machine to erase
the contents of the selected register and replace the
original contents with all zeros and aplus sign.
Example: 22 CLR
Register
Before 22+000021563185733 .4210000000000000
After 22+000000000000000 .0000000000000000
22 Selected
Clear Right Half
^CLR^
JH. (Clear Right Half)
The depression of this key causes the last sixteen
positions of the selected register to be replaced by
zeros. The machine takes no cognizance of decimal
position or the contents of the right half of the regis-
ter. The selected register remains selected. The decimal
point is not erased.
Example: 29 CLR RH
Register
Before 29+000000000001765 .4180021004000000
After 29+000000000001765 .0000000000000000
29 Selected
SHIFTING
If the operator wishes to shift the contents of areg-
ister to the left or to the right, he has four operations
available to help him accomplish this. As aresult of
shifting, the sign position never changes, but numbers
shifted beyond register limits are lost.

BASIC ARITHMETIC OPERATIONS 11
One-Place Shifts
(Shift Right One Place)
The depression of this key tells the machine to shift
the contents of the selected register to the right one
place. The point moves with the digits, and the regis-
ter shifted remains selected. This operation may be
given several times in succession. For instance, SR SR SR
will cause the contents of the selected register to be
moved to the right three places. In using this instruc-
tion on auto-point numbers, care must be taken that
the number is returned to the right-hand standard
(point in the middle) before other operations are car-
ried out on it; otherwise, errors will result.
Example:
24 SR
Register
Before 24+000621053162181 .2560000000000000
After 24+000062105316218 1.256000000000000
24 Selected
When apositive number is shifted to the right in
aregister, the vacated high-order positions of the
upper half of the register are automatically filled with
zeros.
When anegative number is shifted to the right in
aregister, the vacated high-order positions of the up-
per half of the register are automatically filled with
nines.
SL HShift Left One Place)
The depression of this key tells the machine to shift
the contents of the selected register to the left one
place. The point moves with the digits, and the regis-
ter remains selected. This operation may be given
several times in succession. Care must be taken that
auto-point numbers shifted away from the right-hand
standard position with this instruction are returned
before the number is used again in an arithmetic op-
eration.
Example:
25 SL
Register
Before 25 +000000067142187 .2180000000000000
After 25+00000067142187.2 1800000000000000
25Selected
When anumber is shifted to the left in aregister,
the vacated low-order positions of the lower half of
the register are automatically filled with zeros.
Multiple-Place Shifts
There are two additional shifting operations. The
operations they cause will be according to whether
the 610 is in the auto-point or the fixed point mode:
SL15) (Shift Left 15, Auto-Point)
The depression of this key tells the machine to shift
the contents of the selected register to the left until
one of the following conditions is satisfied:
1. The most significant (non-zero) digit is next
to the sign position.
2. The decimal point is next to the sign position.
3. Amaximum shift of 15 places has been made.
Ashift of exactly fifteen places is not implied by
this instruction when the 610 is in the auto-point
mode.
Example 1:
auto-point
26 SL 15
Register
Before 26+000010620151843 .2146000000000000
After 26+10620151843.2146 0000000000000000
26 Selected
Example 2:
auto-point
30 SL 15
Register
Before 30+000000000000000 .0076511821750000
After 30+.007651182175000 0000000000000000
30 Selected
Reminder: In the auto-point -mode, if anumber
is positioned in aregister in such away that no zeros
appear to the left of the high-order significant digit,
OR if anumber is positioned with the decimal point
in the high-order position of the register, the number
is said to be in the left-hand standard position.
The operation of the machine as a result of the
depression of the SL 15 and SR 15 keys is different
in the case of the fixed-point mode.

12 IBM 610
SL15) (Shift Left 15, Fixed Point)
The depression of this key tells the machine to shift
the contents of the selected register to the left fifteen
places.
Example 1:
fixed-point26 SL 15
Register
Before 26+000000000000000 07.62051831210000
After 26+07.6205183121000 0000000000000000
26 Selected
Example 2:
fixed-point32 SL 15
Register
Before 32+0017700051.02015 7600000000000000
After 32+760000000000000 0000000000000000
32 Selected
Note: If during calculation of aproblem, trans-
fer from fixed-point to auto-point mode is made, and
no decimal point appears in aregister to be shifted,
the machine shifts exactly 15places.
SRI 5J(Shift Right 15, Auto-Point)
The depression of this key tells the machine to
shift the contents of the selected register to the right
until the decimal point is in the left-most digit posi-
tion of the lower half of the register (right-hand
standard position). If the decimal point is already
in that position or to the right of that position,
the operation has no effect on the contents of the
register. This instruction does not imply ashift of
exactly fifteen places in the auto-point mode.
Example 1:
auto-point28 SR 15
Register
Before 28 +000010.775251847 2146000000000000
After 28 +000000000000010 .7752518472146000
28 Selected
Example 2:
auto-point29 SR 15
Register
Before 29+000050607088123 .7717000000000000
After 29+000050607088123 .7717000000000000
29 Selected
Reminder: In the auto-point -mode, when anum-
ber is positioned so that the decimal point is in the
high-order position of the lower half of the register,
the number is said to be in right-hand standard posi-
tion.
SRI 5) (Shift Right 15, Fixed-Point)
The depression of this key tells the machine to shift
the contents of the selected register to the right fif-
teen places.
Example:
fixed-point
30 SR 15
Register
Before 30+000010620151772 4186000000000000
After 30+000000000000000 0000106201517724
30 Selected
Rounding
Because of the large capacity of the 610 registers
(31 digits) and the auto-point mode of operation, it
will seldom be necessary to round anumber. Should
it become desirable, however, it can be done as fol-
lows:
1. Enter aconstant with the 5in the correct posi-
tion (one position to the right of that desired correct
in the results) into avacant register.
2. Add the register containing the 5to the register
to be rounded.
3. Shift the number to the left so that the portion
of the number to be eliminated is in the lower half
of the register and the portion of the number to be
retained is in the upper half of the register.
4. Clear the right half of the register.
5. Shift Right (SR 15) to return the number to
the right-hand standard position for further compu-
tations.
Example:
Suppose register 28contains
+000000076512136 .7718154000000000
and we wish to round to 3decimal places.
We proceed as follows:
1. Select an unused register, e. g., 37, and give the fol-
lowing instruction:
37 ENT .0005 +
The operator's programming of the problem will tell the
machine whether the number he wishes to round is positive
or negative. If the number is positive, the constant will be
+.0005; if negative, —.0005. The rounding 5may be en-
tered from the program tape, data tape, or manually from
the keyboard.

BASIC ARITHMETIC OPERATIONS 13
2. The next program step is:
28 +37
The result in register 28 is:
+000000076512136 .7723154000000000
3. We now give the next steps:
SL SL SL
And the result in register 28 is:
+000076512136.772 3154000000000000
4. The next instruction step is:
CLR RH
+000076512136.772 0000000000000000
5. If we wish to use this number in further processing,
we give the instruction: SR 15
and the result in register 28 is:
+000000076512136 .7720000000000000
It is entirely possible that aprogram may develop and
call for retention of alarger number. If the number in
register 31 to be rounded is:
+017631542187721 .7727121000000000
If we wish to round and retain four decimal places, we
would add .00005+. The result of adding in register 31
would be:
+017631542187721 .7727621000000000
However, we cannot perform the usual shifts to the left
because the instructions
SL SL SL SL
would give in register 31
:
+31542187721.7727 6210000000000000
and we have lost the high-order digits of our answer.
Instead of the left shifts and clr rh, we must find an
alternate procedure. The alternate procedure calls for a
series of right shifts. In this case the instructions are:
SR SR SR SR SR SR SR SR SR SR SR SR
Register 31will now contain:
+000000000000017 631542187721.7727
To replace the decimal point in the regular auto-point
position, we must give the instructions:
SL 15
with the result
+17631542187721.7 7270000000000000
Then SR 15
with the following final result:
+017631542187721 .7727000000000000
The 80 registers (00 .... 79) used with the opera-
tions explained above required no additional register
during the calculation itself, because the operations
were actually performed within the selected register.
When X, ^, ^X or Vare performed, extra
intermediate storage is required; and to supply this
need, the machine has 4special registers A(Answer),
MP (Multiplier) ,DIV (Divisor) and MC (Multipli-
cand) .
The special registers are 31 digits in length plus
algebraic sign and decimal point.
Though normally used for these functions, these
registers are also addressable bringing to atotal 84
addressable registers in this machine.
These registers are addressed as follows:
MP
DIV 1
A2
MC 3
Note that the decimal point replaces the units digit
in the address.
When addressing these four special registers, it
must be remembered that their contents are destroyed
during the process of X, -^, -^X and V.There-
fore, these registers may not be used for storing data
when these arithmetic operations are to be performed.
However, for all other operations, such as +, —
,
COPY, etc., the contents of these registers are not
destroyed, and will therefore have the properties of
the other 80 registers.
fA1(Answer)
SPECIAL REGISTERS
Register selection can be changed only as the result
of intentionally selecting anew register at the com-
pletion of an instruction, or after the operations of
multiplication, division, or square root when the at-
tention of the machine is focused on the special A
(Answer) register containing the answer or result
of the calculation just performed.
The depression of this key allows the operator to
address the A(Answer) register.
Upon completion of the operation of X, -i-, -^X,
and Vthe result appears in the Aregister and
the Aregister is selected. For this reason the Aregister
is addressed more often than the other special regis-
ters.
Hence the special Akey is provided to eliminate
the need for the use of the (2.) address.

14 IBM 610
MULTIPLICATION DIVISION
©(Multiply)
The depression of this key tells the machine to mul-
tiply the contents of the selected register by the con-
tents of the next addressed register. The product
appears in the Aregister. In the process, the contents
of the selected register are placed in the MP register,
and the contents of the next addressed register are
placed in the MC register. After the movement of
the contents of both registers is completed, the mul-
tiplication takes place. The contents of both the origi-
nal registers remain unchanged. At the end of the
operation, the Aregister is selected.
Example 1:
Before
Register 08 XRegister 18
Register
08-1-000000000000007 .0000000000000000
18-1-000000000000012 .0000000000000000
Axxxxxxxxxxxxxxxx .xxxxxxxxxxxxxxxx
After 08 +000000000000007 .0000000000000000
18-1-000000000000012 .0000000000000000
A-1-000000000000084 .0000000000000000
ARegister Selected
Example 2:
Before
Register 8XRegister A
Register
08-f-000000000000007 .0000000000000000
a4-0OOOO0000OOOOO3 .0700000000000000
After 08 +000000000000007 .0000000000000000
A-f-000000000000021 .4900000000000000
ARegister Selected
In the auto-point mode, the machine will per-
form amultiplication, provided the product is
99999999999999.9 or smaller. Exceeding this limit
causes the machine to stop, and the keyboard over-
flow light comes on.
C^J (Divide)
The depression of this key tells the machine to divide
the contents of the selected register by the contents
of the next addressed register. The quotient will ap-
pear in the Aregister. In the process the contents of
the selected register are placed in the MP register,
and the contents of the next addressed register are
placed in the DIV register. After both registers are
filled, division takes place. The contents of both origi-
nal registers remain unchanged. The Aregister stands
selected.
Example 1:
Register 01 ^Register 08
Register
Before 01+000000000000105 .0000000000000000
08+000000000000007 .0000000000000000
Axxxxxxxxxxxxxxxx .xxxxxxxxxxxxxxxx
After 01 +000000000000105 .0000000000000000
08-1-000000000000007 .0000000000000000
A-j-000000000000015 .0000000000000000
ARegister Selected
Example 2:
19 -f- 18
Register
Before 19-000000000000132 .0000000000000000
18 +000000000000012 .0000000000000000
Axxxxxxxxxxxxxxxx .xxxxxxxxxxxxxxxx
After 19-000000000000132 .0000000000000000
18 +000000000000012 .0000000000000000
A-000000000000011 .0000000000000000
ARegister Selected
In the auto-point mode, the machine will perform
adivision, provided the quotient is 99999999999999.
9
or smaller. Exceeding this limit causes the machine
to stop, and the keyboard overflow light comes on.
When an operator is performing adivision of a
small number by arelatively large number, the result
will be an even smaller number. It is entirely normal
that the quotient may be so small that it will be be-
yond the range of the machine. In the event this
happens, the contents of the Aregister will be all
zeros.

BASIC ARITHMETIC OPERATIONS 15
The operator should be careful not to interpret this
as a failure by the machine to perform the operation.
Complete knowledge of the problem and the range
of accuracy desired must constantly be kept in mind.
Note: In auto-point, if division by zero is at-
tempted, the machine will stop and the keyboard
overflow light will come on.
In the auto-point mode, the machine will per-
form adivide-multiply, provided the result is
9999999999999.9 or smaller. Exceeding this limit
causes the machine to stop, and the keyboard over-
flow light comes on.
SIMULTANEOUS DIVIDE AND MULTIPLY SQUARE ROOT
-T-X) (Divide Multiply)
The depression of this key tells the machine to divide
the contents of the selected register by the contents
of the next addressed register, and to multiply the
quotient by the second addressed register. The result
will appear in the Aregister. All of the numbers are
moved to special registers before the operation be-
gins; therefore, the contents of the three registers
addressed remain unchanged. In the process of di-
vision, the dividend (any of the eighty registers)
register has its contents transferred to the MP reg-
ister. The register containing the divisor has its con-
tents transferred to the DIV register. The contents
of the two addressed registers are thus undisturbed
as division proceeds. The Aregister is cleared just
prior to the placement of the quotient in the Areg-
ister. The Aregister stands selected at the end of
the operation.
Example 1:
10 -=-X 08 18
Register
Before lO+OOOOOOOOOOOOlOJ
08+000000000000007
18-1-000000000000012
Axxxxxxxxxxxxxxxx
After 10+OOOOOOOOOOOOlOy
084-000000000000007
184-000000000000012
A-i-000000000000180
Example 2:
19~X A17
Register
Before 19-000000000000132
A+000000000000012
17-1-000000000000003
After 19-000000000000132
A-000000000000033
17-1-000000000000003
.0000000000000000
.0000000000000000
.0000000000000000
•xxxxxxxx xxxxxxxx
.0000000000000000
.0000000000000000
.0000000000000000
.0000000000000000
.0000000000000000
.0000000000000000
.0700000000000000
.0000000000000000
.7700000000000000
.0700000000000000
(Vj (Squa re Root)
The depression of this key tells the machine to take
the square root of the contents of the selected regis-
ter. The result will appear in the Aregister. The num-
ber in the selected register remains unchanged. The
machine will place the point correctly in the answer in
Awhen the machine is operating in the auto-point
mode, and the Aregister will be selected.
Example 1
:
20 V
Register
Before 20-f000000000000025 .0000000000000000
Axxxxxxxxxxxxxxxx .xxxxxxxxxxxxxxxx
After 20-1-000000000000025 .0000000000000000
A-j-000000000000005 .0000000000000000
Example 2:
21 V
Register
Before 21-f000000000000003 .0000000000000000
Axxxxxxxxxxxxxxxx .xxxxxxxxxxxxxxxx
After 21+000000000000003 .0000000000000000
A-l-OOOOOOOOOOOOOOl .7320508075688700
If the operator directs the machine to take the
square root of anegative number, the machine will
treat the number as if it were apositive number and
will give apositive answer.
Programming Example
As an example of the mechanics of using the 610
instructions described up to now, suppose we con-
sider asimple problem in which we want to find the
values of xthat satisfy the equation 2x^ —llx +12
=0. We recall that this equation may be written

IBM 610
Instr.
No. CI. First
Reg. Oper. Second
Reg.
Reg. Sel.
After REMARKS
01 £/vr 01 2.* -]
oz ENT 02- II.- ... Th-fiiM—t^ -f oaJ
03 EHT 03 /2.+ {—'-•
zo ENT 20 V. i-Ccofis-fAh-t) J
Figure 2. Program Steps for Entering Data
in the general form ax^ +bx +c=0. Practically
all algebra textbooks contain the development of a
formula to solve this equation:
-b± Vb^-4ac
2a
This is the formula we shall use to solve this simple
example. To begin, we must enter the data into the
machine. Suppose we plan to place the coefficient of
x^ (a =2) in the magnetic-drum storage location 01.
This would be equivalent to writing an a=2at the
top of awork sheet when we are doing ahand solu-
tion. In order to accomplish this, we depress the '0'
and '1' keys on the keyboard to direct the attention
of the 610 to register 01. We next depress the ent
key, clearing the 01 register and preparing it to re-
ceive data. "We are now ready to enter the desired
value into register 01. "We now depress the 2, .and +
keys, and the value of the coefficient of x^ is entered
into the storage register. The machine was actually
ready to receive up to fifteen digits, decimal point.
and sign, but any lesser number of digits may be in-
serted. The depression of the sign key terminates the
entry (Figure 2).
In summary, our key depressions were:
01 ENT 2. +
In the same manner we enter b=-11in storage
register 02:
02 ENT 11. -
And, finally, we enter c=12 into storage register 03:
03 ENT 12. ^-
We now have all of our problem data in the ma-
chine, and we are ready to begin our calculations.
Let us develop the discriminant (b^ -4ac) first. We
will develop 4ac, then b^ and take the difference. As
we begin to develop 4ac, we discover that we should
like to have aconstant 4to use as amultiplier. We
select aregister, say 20, and enter afour into the
machine:
20 ENT 4. -f
Now we develop the product 4a (Figure 3)
:
20 X01
ZO X0/ ADevelop ptoJucf ¥a. it, A)-eiii^e>^
X03 ADs<felop p^oJ«ct y<to />, At-eqisU*-
II COPY AII SUre V«C' in h&qiiii> II
02. y02. ADevelop ii^ "I Ata<fis'ft.y ^Atithn\eti(.
_II ADevelop h^-i^at, in Ai-eqisiet- Oevttofmtt
r- ADevelop Sb*--¥(u, i» At-eq'ff-ef
02. CA/V 0% Chinije. hh-t ih i-tijiiiti- 02.
01 COPY 02. 09 SUrt. -i> In »-e<j'»t«.)" 0? J
Figure 3. Arithmetic Development

BASIC ARITHMETIC OPERATIONS 17
The product is in the Aregister, and the Aregister
is selected. We multiply by cin register 03, recalling
that the product is in A, with Aselected:
X03
We can combine these two steps on one line
20 X01 X03
and have the answer (4ac) in A, with Aselected.
We wish to store 4ac for later use in astorage reg-
ister, say 11
:
11 COPY A
We next wish to develop b". We recall that bstands
in storage register 02. Hence:
02 X02
gives b^ in the Aregister, with Aselected.
Because we want b^ —4ac, we subtract 4ac (in 11)
from register A: -11
The difference stands in the Aregister, with Ase-
lected.
We now wish to take the square root of the differ-
ence. We notice that if we take the square root of
the difference by simply depressing the Vkey,
we would destroy the difference (b^ —4ac) and re-
place it by Vb^ —4ac. Care must be exercised not to
destroy any intermediate results that -may be needed
later. In this case, we do not need to retain the dif-
ference, b^ —4ac; so we depress the "v/ key and get
Vb* —4ac in the Aregister, with the Aregister se-
lected.
Because we have completed usage of the constant
(-f-b) for this problem, it would be convenient to
change it to —b. This is accomplished by:
02 CNV
At this point, we notice that both values of xare
developed in almost the same manner, the exception
being the sign of the radical. Let us now store —bin
one additional register for use in developing the nu-
merators of the two fractions. We do this by selecting
aregister (09) and putting —bin it:
09 COPY 02
09 now stands selected, and \/b^ —4ac is in A:
The key depressions: +A
gives -b +\/b^ -4ac in 09.
To develop -b -Vb^ -4ac in 02, we instruct the
machine:
02 -A
We are now ready to divide the two developed nu-
merators by 2a in order to obtain the two values of x.
However, we note that we have only the value of an
ain location 01. In order to obtain 2a, we might in-
sert aconstant 2into an arbitrary storage register by
an ENT instruction and multiply. However, it is
easier to merely say:
01 -f 01
This would give 2ain 01 ;the value of the awould be
destroyed in the process.
Now we are ready to divide numerator by denomi-
nator to obtain our values of x. For the first value:
02 -f- 01
The quotient stands in A, with Aselected. Let us store
the first value of xmomentarily:
15 COPY A
Now we are ready to obtain the second value of x:
09 H- 01
The quotient stands in A, with Aselected (Figure 4).
+Aoq Develop -b+U^'i'ae. /n n^is1k> 09 ^
02. —A02. Develop -b-U^-¥cu. intt<iidt*OZ
01 +01 01 Deifelop 2* fh heqisfe*- 01
oa. ^01 AUevelof ^b-Yb'-Zae, />, AypquT^Y
2.IK
/r COPY AIS- Sfohe, -L- yi>'--4aju /h yeaisfty IS" »v»eiit*
ia.
of -;- Ol ADevelop -h +U^-¥eu^ /* At-fiif«>
2*1 J
Figure 4. Answer Development

18 IBM 610
The entire program is shown in Figure 5.
IBM 610 PROGRAMMING SHEET
Storage Usage
00 °'^ 32^ 03 /C 04 05 06 07 08 09
10 11 12 13 14 15 16 17 18 19
20 ^21 22 23 24 25 26 27 28 29
s30, ^.2J ^
First
Reg.
^2_^J33 _-J^..i5—>._ _36 ^_ 37 38^ 39
1
Instr.^
No. CI. Oper. "Second
Reg.
RegTSel.
After REMARKS
01 £t^T 2.+
01 INT //.-
03 €NT /Z.+
20 tUT ¥.+
20 X1At/fi. /N A
X03 AVac/ /a a
// copy AII y^u/ /vTo //
OZ XOZ A/« /y»^A
—// Ajg'-Vae/ /A* 4
r— Afji'^-^OA^ /MA
OZ CA/V 02 .ji /Al OZ.
09 copy <?2 07 -j& /MTa 09
+A0? -/,*t&''-ya^ /x^y
01 _AOZ -^_>^*. V*o //V'^-i!
01 +01 ol 2«- /A <?/
02 -i- 1A-.A->B».VA^ /^^
/6 copy A/S <' /AfTO /5
^""^
-N
„^ 07 -r ii_l A-j^yjti^^^ ^^ A
•»———'"^ 1«^' ^^^——^^ ^^~^
Figure 5. Programming for Solution of Quadratic Equation

CONTROL DEVICES
CONTROL of the computing system is exercised by
the following four devices:
Manual keyboard
Program tape
Data tape
Control panel
They may serve as input devices for numerical
information and may also transmit instructions caus-
ing the computer to execute any of the functions of
which it is capable. (The data tape reader is normally
used for the entry of numerical data, but on occa-
sion it may be used in functional control.) See block
diagram in Figure 6.
No more than one control unit may be in operation
at one time; however, the unit in control can shift
automatically from itself to any other control unit as
the needs of the problem in process dictate.
MANUAL KEYBOARD
The keyboard is always in control of the machine
until acommand is issued via the keyboard to cause
automatic control by one of the other units. There-
after, automatic transfer from one control to another
may be made.
Many of the functions of the manual keyboard
have already been explained. The remainder of key-
board functions will be covered later in the manual
in context with their application.
PAPER TAPE
PAPER TAPE
'/programX
-*—(TAPE I
yREADER J
CONTROL
PANEL
T^
INSTRUCTION FLOW
DATA FLOW
COMPUTER
Figure 6. Internal Flow Schematic
19

20 BM610
PROGRAM TAPE
There are many occasions when it is desirable to
process several sets of data for the same problem in
some automatic manner. One method of doing this
is by using the Program Tape Reader.
While the operator is working aproblem for the
first time, he may set switches on the keyboard that
cause the machine to punch codes into apaper tape
equivalent to the keys being depressed. Later he may,
by changing switch settings and depressing appro-
priate keys, turn over control of the machine to this
program tape. The only activity required of the
operator when the program tape is being used will
be the entering of new sets of data into the machine
during the processing of each problem.
In order to get the program into the tape during
the processing of the first set of data, the operator
simply sets the punch switch on the keyboard to the
ON position. Afull discussion of the switches and
lights on the keyboard will be found in alater section.
The operator now proceeds in amanner similar to
the one-time calculation. He must always be cog-
nizant of the source of the data. The only difference
will be that the program tape punch is active. At the
end of the processing of the first calculation, the
program is entirely punched into the tape and the
tape is accumulated (up to fifty feet) in the machine.
Now the operator is ready to allow the program
tape to take over control of the machine. He first
turns off the punch switch and turns on the dup
switch. The dup switch causes the code symbols read
by the program tape reader to be duplicated back into
program tape that is passing by the program tape
punch. The program tape thus continuously recreates
itself sequentially so that it may be used again and
again. The program tape is placed in control of the
machine by depressing the ptr key.
®(Program Tape Reader)
The depression of this key tells the machine to turn
control over to the instructions punched into the pro-
gram tape, which is standing ready to be read at the
Program Tape Reader. The first character encoun-
tered by the reader will be the first step in the pro-
gram to be executed.
TkB J(Keyboard)
When the 610 is under the control of the program
tape unit, it may be necessary to give an instruction
to return control to the keyboard.
This instruction causes control of the machine to
be transferred from the unit in control to the key-
board. One use of this instruction is to return control
from the program tape reader to the keyboard for the
entry of data. An instruction sequence 05 ent kb
will cause the machine to transfer control to the key-
board for data entry. After the number has been en-
tered, depressing ptr will return control to the pro-
gram tape reader.
Instruction Classes
In many problems, it may be necessary to choose
one of several possible methods of solution after some
point in the calculation, depending on conditions
generated up to that point.
For example, in the calculation of the roots of a
quadratic equation (previously used example), it is
possible that the discriminant (b^ -4ac) may be
negative. If the discriminant is negative, the roots of
the quadratic equation will be complex numbers and
it will be necessary to separate the real and imaginary
parts for printing. The possibility of having complex
roots was not considered in our previous example.
In solving for the roots of aseries of such equa-
tions, it is usually inconvenient to separate by visual
inspection the equations having real roots, from the
equations having imaginary roots, and then solve
them with separate programs. It would be extremely
desirable to have aprogram that would solve for
either condition, dependent upon the sign of the dis-
criminant. Because all symbols on the program tape
must be read in sequence, aproblem is posed having
both the series of steps for finding the roots of an
equation with apositive discriminant and the series
of steps for finding the roots of an equation with a
negative discriminant on the same tape. It would be
very convenient to have the machine make alogical
decision and use only one series of steps while ignoring
the other. The IBM 610 is able to make such alogical
decision.