From: Subject: Preliminary discussion of the logical design of an electronic computing instrument Date: Thu, 21 Sep 2006 22:54:12 -0400 MIME-Version: 1.0 Content-Type: multipart/related; type="text/html"; boundary="----=_NextPart_000_0000_01C6DDD0.DB3186F0" X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2962 This is a multi-part message in MIME format. ------=_NextPart_000_0000_01C6DDD0.DB3186F0 Content-Type: text/html; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable Content-Location: http://www.cs.unc.edu/~adyilie/comp265/vonNeumann.html Preliminary discussion of the logical design of an = electronic computing instrument

Preliminary discussion of the logical design of an = electronic=20 computing instrument1

Arthur W. Burks / Herman H. Goldstine / = John von=20 Neumann

PART I

1. Principal components of the machine

1.1. Inasmuch as the completed device will be a = general-purpose=20 computing machine it should contain certain main organs relating to = arithmetic,=20 memory-storage, control and connection with the human operator. It is = intended=20 that the machine be fully automatic in character, i.e. independent of = the human=20 operator after the computation starts. A fuller discussion of the = implications=20 of this remark will be given in Sec. 3 below.

1.2. It is evident that the machine must be capable = of storing=20 in some manner not only the digital information needed in a given = computation=20 such as boundary values, tables of functions (such as the equation of = state of a=20 fluid) and also the intermediate results of the computation (which may = be wanted=20 for varying lengths of time), but also the instructions which govern the = actual=20 routine to be performed on the numerical data. In a special-purpose = machine=20 these instructions are an integral part of the device and constitute a = part of=20 its design structure. For an all-purpose machine it must be possible to = instruct=20 the device to carry out any computation that can be formulated in = numerical=20 terms. Hence there must be some organ capable of storing these program = orders.=20 There must, moreover, be a unit which can understand these instructions = and=20 order their execution.

1.3. Conceptually we have discussed above two = different forms=20 of memory: storage of numbers and storage of orders. If, however, the = orders to=20 the machine are reduced to a numerical code and if the machine can in = some=20 fashion distinguish a number from an order, the memory organ can be used = to=20 store both numbers and orders. The coding of orders into numeric form is = discussed in 6.3 below.

1.4. If the memory for orders is merely a storage = organ there=20 must exist an organ which can automatically execute the orders stored in = the=20 memory. We shall call this organ the Control.

1.5. Inasmuch as the device is to be a = computing machine=20 there must be an arithmetic organ in it which can perform certain of the = elementary arithmetic operations. There will be, therefore, a unit = capable of=20 adding, subtracting, multiplying and dividing. It will be seen in 6.6 = below that=20 it can also perform additional operations that occur quite = frequently.

The operations that the machine will view as = elementary are=20 clearly those which are wired into the machine. To illustrate, the = operation of=20 multiplication could be eliminated from the device as an elementary = process if=20 one were willing to view it as a properly ordered series of additions. = Similar=20 remarks apply to division. In general, the inner economy of the = arithmetic unit=20 is determined by a compromise between the desire for speed of = operation-a=20 non-elementary operation will generally take a long time to perform = since it is=20 constituted of a series of orders given by the control-and the desire = for=20 simplicity, or cheapness, of the machine.

1.6. Lastly there must exist devices, the input and = output=20 organ, whereby the human operator and the machine can communicate with = each=20 other. This organ will be seen below in 4.5, where it is discussed, to=20 constitute a secondary form of automatic memory.

2. First remarks on the memory

2.1. It is clear that the size of the memory is a = critical=20 consideration in the design of a satisfactory general-purpose = computing

1From A. H. Taub (ed.), "Collected Works of = John von=20 Neumann," vol. 5, pp. 34-79, The Macmillan Company, New York, 1963. = Taken from=20 report to U. S. Army Ordnance Department, 1946. See also Bibliography = Burks,=20 Goldstine and von Neumann, 1962a, 1962b, 1963; and Goldstine and von = Neumann=20 1963a, 1963b, 1963c, 1963d.

92

machine. We proceed to discuss what quantities the = memory=20 should store for various types of computations.

2.2. In the solution of partial differential = equations the=20 storage requirements are likely to be quite extensive, In general, one = must=20 remember not only the initial and boundary conditions and any arbitrary=20 functions that enter the problem but also an extensive number of = intermediate=20 results.

a For equations of parabolic or hyperbolic type in = two=20 independent variables the integration process is essentially a double = induction.=20 To find the values of the dependent variables at time t + D t one integrates with respect to x from one = boundary to the=20 other by utilizing the data at time t as if they were coefficients which = contribute to defining the problem of this integration.

Not only must the memory have sufficient room to = store these=20 intermediate data but there must be provisions whereby these data can = later be=20 removed, i.e. at the end of the (t + D t) = cycle, and=20 replaced by the corresponding data for the (t + 2D t)=20 cycle. This process of removing data from the memory and of replacing = them with=20 new information must, of course, be done quite automatically under the = direction=20 of the control.

b For total differential equations the memory = requirements=20 are clearly similar to, but smaller than, those discussed in (a) = above.

c Problems that are solved by iterative = procedures such as=20 systems of linear equations or elliptic partial differential equations, = treated=20 by relaxation techniques, may be expected to require quite extensive = memory=20 capacity. The memory requirement for such problems is apparently much = greater=20 than for those problems in (a) above in which one needs only to store=20 information corresponding to the instantaneous value of one variable [t = in (a)=20 above], while now entire solutions (covering all values of all = variables) must=20 be stored. This apparent discrepancy in magnitudes can, however, be = somewhat=20 overcome by the use of techniques which permit the use of much coarser=20 integration meshes in this case, than in the cases under = (a).

2.3. It is reasonable at this time to build a machine = that can=20 conveniently handle problems several orders of magnitude more complex = than are=20 now handled by existing machines, electronic or electro-mechanical. We=20 consequently plan on a fully automatic electronic storage facility of = about=20 4,000 numbers of 40 binary digits each. This corresponds to a precision = of=20 2- 40 ~ 0.9 x 10-=20 12, i.e. of about 12 decimals. We believe that this memory = capacity=20 exceeds the capacities required for most problems that one deals with at = present=20 by a factor of about 10. The precision is also safely higher than what = is=20 required for the great majority of present day problems. In addition, we = propose=20 that we have a subsidiary memory of much larger capacity, which is also = fully=20 automatic, on some medium such as magnetic wire or tape.

3. First remarks on the control and code

3.1. It is easy to see by formal-logical methods that = there=20 exist codes that are in abstracto adequate to control and cause = the=20 execution of any sequence of operations which are individually available = in the=20 machine and which are, in their entirety, conceivable by the problem = planner.=20 The really decisive considerations from the present point of view, in = selecting=20 a code, are more of a practical nature: simplicity of the equipment = demanded by=20 the code, and the clarity of its application to the actually important = problems=20 together with the speed of its handling of those problems. It would take = us much=20 too far afield to discuss these questions at all generally or from first = principles. We will therefore restrict ourselves to analyzing only the = type of=20 code which we now envisage for our machine.

3.2. There must certainly be instructions for = performing the=20 fundamental arithmetic operations. The specifications for these orders = will not=20 be completely given until the arithmetic unit is described in a little = more=20 detail.

3.3. It must be possible to transfer data from the = memory to=20 the arithmetic organ and back again. In transferring information from = the=20 arithmetic organ back into the memory there are two types we must = distinguish:=20 Transfers of numbers as such and transfers of numbers which are parts of = orders.=20 The first case is quite obvious and needs no further explication. The = second=20 case is more subtle and serves to illustrate the generality and = simplicity of=20 the system. Consider, by way of illustration, the problem of = interpolation in=20 the system. Let us suppose that we have formulated the necessary = instructions=20 for performing an interpolation of order n in a sequence of data. = The=20 exact location in the memory of the (n + 1) quantities that = bracket the=20 desired functional value is, of course, a function of the argument. This = argument probably is found as the result of a computation in the = machine. We=20 thus need an order which can substitute a number into a given order-in = the case=20 of interpolation the location of the argument or the group of arguments = that is=20 nearest in our table to the desired value. By means of such an order the = results=20 of a computation can be introduced into the instructions governing that = or a=20 different computation. This makes it possible for a sequence of = instructions to=20 be used with different sets of numbers located in different parts of the = memory.

Section 1 Processors with one address per = instruction

To summarize, transfers into the memory will be of = two=20 sorts:

Total substitutions, whereby the quantity = previously stored=20 is cleared out and replaced by a new number. Partial substitutions = in=20 which that part of an order containing a memory = location-number-we=20 assume the various positions in the memory are enumerated serially = by memory=20 location-numbers-is replaced by a new memory location-number.

3.4. It is clear that one must be able to get numbers = from any=20 part of the memory at any time. The treatment in the case of orders can, = however, b more methodical since one can at least partially arrange the = control=20 instructions in a linear sequence. Consequently the control will be so=20 constructed that it will normally proceed from place n in the = memory to=20 place (n + 1) for its next instruction.

3.5. The utility of an automatic computer lies in the = possibility of using a given sequence of instructions repeatedly, the = number of=20 times it is iterated being either preassigned or dependent upon the = results of=20 the computation. When the iteration is completed a different sequence of = orders=20 is to be followed, so we must, in most cases, give two parallel trains = of orders=20 preceded by an instruction as to which routine is to be followed. This = choice=20 can be made to depend upon the sign of a number (zero being reckoned as = plus for=20 machine purposes). Consequently, we introduce an order (the = conditional=20 transfer order) which will, depending on the sign of a given number, = cause=20 the proper one of two routines to be executed.

Frequently two parallel trains of orders terminate in = a common=20 routine. It is desirable, therefore, to order the control in either case = to=20 proceed to the beginning point of the common routine. This = unconditional=20 transfer can be achieved either by the artificial use of a = conditional=20 transfer or by the introduction of an explicit order for such a = transfer.

3.6. Finally we need orders which will integrate the=20 input-output devices with the machine. These are discussed briefly in = 6.8.

3.7. We proceed now to a more detailed discussion of = the=20 machine. Inasmuch as our experience has shown that the moment one = chooses a=20 given component as the elementary memory unit, one has also more or less = determined upon much of the balance of the machine, we start by a = consideration=20 of the memory organ. In attempting an exposition of a highly integrated = device=20 like a computing machine we do not find it possible, however, to give an = exhaustive discussion of each organ before completing its description. = It is=20 only in the final block diagrams that anything approaching a complete = unit can=20 be achieved.

The time units to be used in what follows will = be:

1 m sec =3D 1 microsecond = =3D 10- 6 seconds

1 msec =3D 1 millisecond =3D 10-=20 3 seconds

4. The memory organ

4.1. Ideally one would desire an indefinitely large = memory=20 capacity such that any particular aggregate of 40 binary digits, or = word=20 (cf. 2.3), would be immediately available-i.e. in a time which is = somewhat=20 or considerably shorter than the operation time of a fast electronic = multiplier.=20 This may be assumed to be practical at the level of about 100 m sec. Hence the availability time for a word in = the memory=20 should be 5 to 50 m sec. It is equally = desirable that=20 words may be replaced with new words at about the same rate. It does not = seem=20 possible physically to achieve such a capacity. We are therefore forced = to=20 recognize the possibility of constructing a hierarchy of memories, each = of which=20 has greater capacity than the preceding but which is less quickly=20 accessible.

The most common forms of storage in electrical = circuits are the=20 flip-flop or trigger circuit, the gas tube, and the electromechanical = relay. To=20 achieve a memory of n words would, of course, require about 40n = such=20 elements, exclusive of the switching elements. We saw earlier (cf. 2.2) = that a=20 fast memory of several thousand words is not at all unreasonable for an=20 all-purpose instrument. Hence, about 105 flip-flops or = analogous=20 elements would be required! This would, of course, be entirely = impractical.

We must therefore seek out some more fundamental = method of=20 storing electrical information than has been suggested above. One = criterion for=20 such a storage medium is that the individual storage organs, which = accommodate=20 only one binary digit each, should not be macroscopic components, but = rather=20 microscopic elements of some suitable organ. They would then, of course, = not be=20 identified and switched to by the usual macroscopic wire connections, = but by=20 some functional procedure in manipulating that organ.

One device which displays this property to a marked = degree is=20 the iconoscope tube. In its conventional form it possesses a linear = resolution=20 of about one part in 500. This would correspond to a (two-dimensional) = memory=20 capacity of 500 x 500 =3D 2.5 x 105. One is = accordingly led to=20 consider the possibility of storing electrical charges on a dielectric = plate=20 inside a cathode-ray tube. Effectively such a tube is nothing more than = a myriad=20 of electrical capacitors which can be connected into the circuit by = means of an=20 electron beam.

Actually the above mentioned high resolution and = concomitant=20 memory capacity are only realistic under the conditions of = television-image=20 storage, which are much less exigent in respect to

the reliability of individual markings than what one = can accept=20 in the storage for a computer. In this latter case resolutions of one = part in=20 20 to 100, i.e. memory capacities of 400 to 10,000, would seem to = be more=20 reasonable in terms of equipment built essentially along familiar = lines.

At the present time the Princeton Laboratories of the = Radio=20 Corporation of America are engaged in the development of a storage tube, = the=20 Selectron, of the type we have mentioned above. This tube is also = planned=20 to have a non-amplitude-sensitive switching system whereby the electron = beam can=20 be directed to a given spot on the plate within a quite small fraction = of a=20 millisecond. Inasmuch as the storage tube is the key component of the = machine=20 envisaged in this report we are extremely fortunate in having secured = the=20 cooperation of the RCA group in this as well as in various other=20 developments.

An alternate form of rapid memory organ is the = acoustic=20 feedback delay line described in various reports on the EDVAC. (This is = an=20 electronic computing machine being developed for the Ordnance = Department, U.S.=20 Army, by the University of Pennsylvania, Moore School of Electrical=20 Engineering.) Inasmuch as that device has been so clearly reported in = those=20 papers we give no further discussion. There are still other physical and = chemical properties of matter in the presence of electrons or photons = that might=20 be considered, but since none is yet beyond the early discussion stage = we shall=20 not make further mention of them.

4.2. We shall accordingly assume throughout the = balance of this=20 report that the Selectron is the modus for storage of words at = electronic=20 speeds. As now planned, this tube will have a capacity of 212 = =3D 4,096=20 =BB 4,000 binary digits. To achieve a total = electronic=20 storage of about 4,000 words we propose to use 40 Selectrons, thereby = achieving=20 a memory of 212 words of 40 binary digits each. (Cf. again 2.3.)

4.3. There are two possible means for storing a = particular word=20 in the Selectron memory-or, in fact, in either a delay line memory or in = a=20 storage tube with amplitude-sensitive deflection. One method is to store = the=20 entire word in a given tube and then to get the word out by picking out = its=20 respective digits in a serial fashion. The other method is to store in=20 corresponding places in each of the 40 tubes one digit of the word. To = get a=20 word from the memory in this scheme requires, then, one switching = mechanism to=20 which all 40 tubes are connected in parallel. Such a switching scheme = seems to=20 us to be simpler than the technique needed in the serial system and is, = of=20 course, 40 times faster. We accordingly adopt the parallel procedure and = thus=20 are led to consider a so-called parallel machine, as contrasted = with the=20 serial principles being considered for the EDVAC. (In the EDVAC the = peculiar=20 characteristics of the acoustic delay line, as well as various other=20 considerations, seem to justify a serial procedure. For more details, = cf. the=20 reports referred to in 4.1.) The essential difference between these two = systems=20 lies in the method of performing an addition; in a parallel machine all=20 corresponding pairs of digits are added simultaneously, whereas in a = serial one=20 these pairs are added serially in time.

4.4. To summarize, we assume that the fast electronic = memory=20 consists of 40 Selectrons which are switched in parallel by a common = switching=20 arrangement. The inputs of the switch are controlled by the = control.

4.5. Inasmuch as a great many highly important = classes of=20 problems require a far greater total memory than 212 words, = we now=20 consider the next stage in our storage hierarchy. Although the solution = of=20 partial differential equations frequently involves the manipulation of = many=20 thousands of words, these data are generally required only in blocks = which are=20 well within the 212 capacity of the electronic memory. Our = second=20 form of storage must therefore be a medium which feeds these blocks of = words to=20 the electronic memory. It should be controlled by the control of the = computer=20 and is thus an integral part of the system, not requiring human=20 intervention.

There are evidently two distinct problems raised = above. One can=20 choose a given medium for storage such as teletype tapes, magnetic wire = or=20 tapes, movie film or similar media. There still remains the problem of = automatic=20 integration of this storage medium with the machine. This integration is = achieved logically by introducing appropriate orders into the code which = can=20 instruct the machine to read or write on the medium, or to move it by a = given=20 amount or to a place with given characteristics. We discuss this = question a=20 little more fully in 6.8.

Let us return now to the question of what properties = the=20 secondary storage medium should have. It clearly should be able to store = information for periods of time long enough so that only a few per cent = of the=20 total computing time is spent in re-registering information that is = "fading=20 off." It is certainly desirable, although not imperative, that = information can=20 be erased and replaced by new data. The medium should be such that it = can be=20 controlled, i.e. moved forward and backward, automatically. This = consideration=20 makes certain media, such as punched cards, undesirable. While cards = can, of=20 course, be printed or read by appropriate orders from some machine, they = are not=20 well adapted to problems in which the output data are fed directly back = into the=20 machine, and are required in a sequence which is non-monotone with = respect to=20 the order of the cards. The medium should be capable of remembering very = large=20 numbers of data at a much smaller price

96 Part 2 =F7 The instruction-set = processor:=20 main-line computers

Section 1 =F7 = Processors with one=20 address per instruction

than electronic devices. It must be fast enough so = that, even=20 when it has to be used frequently in a problem, a large percentage of = the total=20 solution time is not spent in getting data into and out of this medium = and=20 achieving the desired positioning on it. If this condition is not = reasonably=20 well met, the advantages of the high electronic speeds of the machine = will be=20 largely lost.

Both light- or electron-sensitive film and magnetic = wires or=20 tapes, whose motions are controlled by servo-mechanisms integrated with = the=20 control, would seem to fulfil our needs reasonably well. We have = tentatively=20 decided to use magnetic wires since we have achieved reliable = performance with=20 them at pulse rates of the order of 25,000/sec and beyond.

4.6. Lastly our memory hierarchy requires a vast = quantity of=20 dead storage, i.e. storage not integrated with the machine. This storage = requirement may be satisfied by a library of wires that can be = introduced into=20 the machine when desired and at that time become automatically = controlled. Thus=20 our dead storage is really nothing but an extension of our secondary = storage=20 medium. It differs from the latter only in its availability to the = machine.

47. We impose one additional requirement on our = secondary=20 memory. It must be possible for a human to put words on to the wire or = other=20 substance used and to read the words put on by the machine. In this = manner the=20 human can control the machine's functions. It is now clear that the = secondary=20 storage medium is really nothing other than a part of our input-output = system,=20 cf. 6.8.4 for a description of a mechanism for achieving this.

4.8. There is another highly important part of the = input-output=20 which we merely mention at this time, namely, some mechanism for viewing = graphically the results of a given computation. This can, of course, be = achieved=20 by a Selectron-like tube which causes its screen to fluoresce when data = are put=20 on it by an electron beam.

4.9. For definiteness in the subsequent discussions = we assume=20 that associated with the output of each Selectron is a flip-flop. This=20 assemblage of 40 flip-flops we term the Selectron = Register.

5. The arithmetic organ

5.1. In this section we discuss the features we now = consider=20 desirable for the arithmetic part of our machine. We give our tentative=20 conclusions as to which of the arithmetic operations should be built = into the=20 machine and which should be programmed. Finally, a schematic of the = arithmetic=20 unit is described.

5.2. In a discussion of the arithmetical organs = of a=20 computing machine one is naturally led to a consideration of the number = system=20 to be adopted. In spite of the longstanding tradition of building = digital=20 machines in the decimal system, we feel strongly in favor of the binary = system=20 for our device. Our fundamental unit of memory is naturally adapted to = the=20 binary system since we do not attempt to measure gradations of charge at = a=20 particular point in the Selectron but are content to distinguish two = states. The=20 flip-flop again is truly a binary device. On magnetic wires or tapes and = in=20 acoustic delay line memories one is also content to recognize the = presence or=20 absence of a pulse or (if a carrier frequency is used) of a pulse train, = or of=20 the sign of a pulse. (We will not discuss here the ternary possibilities = of a=20 positive-or-negative-or-no-pulse system and their relationship to = questions of=20 reliability and checking, nor the very interesting possibilities of = carrier=20 frequency modulation.) Hence if one contemplates using a decimal system = with=20 either the iconoscope or delay-line memory one is forced into a binary = coding of=20 the decimal system-each decimal digit being represented by at least a = tetrad of=20 binary digits. Thus an accuracy of ten decimal digits requires at least = 40=20 binary digits. In a true binary representation of numbers, however, = about 33=20 digits suffice to achieve a precision of 1010. The use of the = binary=20 system is therefore somewhat more economical of equipment than is the=20 decimal.

The main virtue of the binary system as against the = decimal is,=20 however, the greater simplicity and speed with which the elementary = operations=20 can be performed. To illustrate, consider multiplication by repeated = addition.=20 In binary multiplication the product of a particular digit of the = multiplier by=20 the multiplicand is either the multiplicand or null according as the = multiplier=20 digit is 1 or 0. In the decimal system, however, this product has ten = possible=20 values between null and nine times the multiplicand, inclusive. Of = course, a=20 decimal number has only 1og102 ~ 0.3 times as many digits as = a binary=20 number of the same accuracy, but even so multiplication in the decimal = system is=20 considerably longer than in the binary system. One can accelerate = decimal=20 multiplication by complicating the circuits, but this fact is irrelevant = to the=20 point just made since binary multiplication can likewise be accelerated = by=20 adding to the equipment. Similar remarks may be made about the other=20 operations.

An additional point that deserves emphasis is this: = An=20 important part of the machine is not arithmetical, but logical in = nature. Now=20 logics, being a yes-no system, is fundamentally binary. Therefore a = binary=20 arrangement of the arithmetical organs contributes very significantly = towards=20 producing a more homogeneous machine, which can be better integrated and = is more=20 efficient.

The one disadvantage of the binary system from the = human point=20 of view is the conversion problem. Since, however, it is completely = known how to=20 convert numbers from one base to

another and since this conversion can be effected = solely by the=20 use of the usual arithmetic processes there is no reason why the = computer itself=20 cannot carry out this conversion. It might b argued that this is a time=20 consuming operation. This, however is not the case. (Cf. 9.6 and 9.7 of = Part II.=20 Part II is a report issued under the title Planning and Coding of = Problems=20 for an Electronic Computing Instrument.1) Indeed a=20 general-purpose computer, used as a scientific research tool, is called = upon to=20 do a very great number of multiplications upon a relatively small amount = of=20 input data, and hence the time consumed in the decimal to binary = conversion is=20 only a trivial percentage of the total computing time. A similar remark = is=20 applicable to the output data.

In the preceding discussion we have tacitly assumed = the=20 desirability of introducing and withdrawing data in the decimal system. = We feel,=20 however, that the base 10 may not even be a permanent feature in a = scientific=20 instrument and consequently will probably attempt to train ourselves to = use=20 numbers base 2 or 8 or 16. The reason for the bases 8 or 16 is this: = Since 8 and=20 16 are powers of 2 the conversion to binary is trivial; since both are = about the=20 size of 10, they violate many of our habits less badly than base 2. (Cf. = Part=20 II, 9.4.)

5.3. Several of the digital computers being built = or=20 planned in this country and England are to contain a so-called "floating = decimal=20 point". This is a mechanism for expressing each word as a characteristic = and a=20 mantissa-e.g. 123.45 would be carried in the machine as = (0.12345,03),=20 where the 3 is the exponent of 10 associated with the number. There = appear to be=20 two major purposes in a "floating" decimal point system both of which = arise from=20 the fact that the number of digits in a word is a constant, fixed by = design=20 considerations for each particular machine. The first of these purposes = is to=20 retain in a sum or product as many significant digits as possible and = the second=20 of these is to free the human operator from the burden of estimating and = inserting into a problem "scale factors"- multiplicative constants which = serve=20 to keep numbers within the limits of the machine.

There is, of course, no denying the fact that human = time is=20 consumed in arranging for the introduction of suitable scale factors. We = only=20 argue that the time so consumed is a very small percentage of the total = time we=20 will spend in preparing an interesting problem for our machine. The = first=20 advantage of the floating point is, we feel, somewhat illusory. In order = to have=20 such a floating point one must waste memory capacity which could = otherwise be=20 used for carrying more digits per word. It would therefore seem to us = not at all=20 clear whether the modest advantages of a floating binary point offset = the loss=20 of memory capacity and the increased complexity of the arithmetic and = control=20 circuits.

There are certainly some problems within the scope of = our=20 device which really require more than 2- = 40=20 precision. To handle such problems we wish to plan in terms of words = whose=20 lengths are some fixed integral multiple of 40, and program the machine = in such=20 a manner as to give the corresponding aggregates of 40 digit words the = proper=20 treatment. We must then consider an addition or multiplication as a = complex=20 operation programmed from a number of primitive additions or = multiplications=20 (cf. =F2 =F2 9, = Part II). There=20 would seem to be considerable extra difficulties in the way of such a = procedure=20 in an instrument with a floating binary point.

The reader may remark upon our alternate spells of = radicalism=20 and conservatism in deciding upon various possible features for our = mechanism.=20 We hope, however, that he will agree, on closer inspection, that we are = guided=20 by a consistent and sound principle in judging the merits of any idea. = We wish=20 to incorporate into the machine-in the form of circuits-only such = logical=20 concepts as are either necessary to have a complete system or highly = convenient=20 because of the frequency with which they occur and the influence they = exert in=20 the relevant mathematical situations.

5.4. On the basis of this criterion we definitely = wish to build=20 into the machine circuits which will enable it to form the binary sum of = two 40=20 digit numbers. We make this decision not because addition is a logically = basic=20 notion but rather because it would slow the mechanism as well as the = operator=20 down enormously if each addition were programmed out of the more simple=20 operations of "and", "or", and "not". The same is true for the = subtraction.=20 Similarly we reject the desire to form products by programming them out = of=20 additions, the detailed motivation being very much the same as in the = case of=20 addition and subtraction. The cases for division and square-rooting are = much=20 less clear.

It is well known that the reciprocal of a number a = can=20 be formed to any desired accuracy by iterative schemes. One such scheme = consists=20 of improving an estimate X by forming X' =3D 2X - aX2. = Thus the=20 new error 1 - aX' is (1 - aX)2, which is the = square of=20 the error in the preceding estimate. We notice that in the formation of = X',=20 there are two bona fide multiplications-we do not consider = multiplication by=20 2 as a true product since we will have a facility for shifting right or = left in=20 one or two pulse times. If then we somehow could guess 1/a to a = precision of=20 2-5, 6 multiplications-3 = iterations-would suffice=20 to give a final result good to 2-40. = Accordingly=20 a small table of 24 entries could be used to get the initial = estimate=20 of 1/a. In this way a reciprocal 1/a

1See Bibliography [Goldstine and von Neumann, 1963b, 1963c, = 1963d].=20 References in this chapter are all to this report.

Section 1 Processors with one address per = instruction

could be formed in 6 multiplication times, and hence = a quotient=20 b/a in 7 multiplication times. Accordingly we see that the = question of=20 building a divider s really a function of how fast it can be made to = operate=20 compared to the iterative method sketched above: In order to justify its = existence, a divider must perform a division in a good deal less than 7=20 multiplication times. We have, however, conceived a divider which is = much faster=20 than these 7 multiplication times and therefore feel justified in = building it,=20 especially since the amount of equipment needed above the requirements = of the=20 multiplier is not important.

It is, of course, also possible to handle square = roots by=20 iterative techniques. In fact, if X is our estimate of = a1/2,=20 then X' =3D 1/2(X + a/X) is a better estimate. We se that = this scheme=20 involves one division per iteration. As will be seen below in our more = detailed=20 examination of the arithmetic organ we do not include a square- rooter = in our=20 plans because such a device would involve more equipment than we feel is = desirable in a first model. (Concerning the iterative method of = square-rooting,=20 cf. 8.10 in Part II)

55. The first part of our arithmetic organ requires = little=20 discussion at this point. It should be a parallel storage organ which = can=20 receive a number and add it to the one already in it, = which is=20 also able to clear its contents and which can transmit what it contains. = We will=20 call such an organ an Accumulator. It is quite conventional in = principle=20 in past and present computing machines of the most varied types, e.g. = desk=20 multipliers, standard IBM counters, more modem relay machines, the = ENIAC. There=20 are of, course, numerous ways to build such a binary accumulator. We = distinguish=20 two broad types of such devices: static, and dynamic or pulse-type = accumulators.=20 These will be discussed in 5.11, but it is first necessary to = make a few=20 remarks concerning the arithmetic of binary addition. In a parallel = accumulator,=20 the first step in an addition is to add each digit of the addend to the=20 corresponding digit of the augend. The second step is to perform the = carries,=20 and this must be done in sequence since a carry may produce a carry. In = the=20 worst case, 39 carries will occur. Clearly it is inefficient to allow 39 = times=20 as much time for the second step (performing the carries) as for the = first step=20 (adding the digits). Hence either the carries must be accelerated, or = use must=20 be made of the average number of carries or both.

5.6. We shall show that for a sum of binary = words, each of=20 length n, the length of the largest carry sequence is on the = average not=20 in excess of 2log n. Let pn(v) = designate the=20 probability that a carry sequence is of length v or greater in = the sum of=20 two binary words of length n. Then clearly pn(v) = -=20 pn(v + 1) is the probability that the largest carry = sequence=20 is of length exactly v and the weighted average

Indeed, pn(v) is the probability = that the sum=20 of two n-digit numbers contains a carry sequence of length =B3 v. This probability obtains by adding = the=20 probabilities of two mutually exclusive alternatives: First: = Either the=20 n - 1 first digits of the two numbers by themselves contain a = carry=20 sequence of length =B3 v. This has the = probability=20 pn-1(v). Second: The n = - 1=20 first digits of the two numbers by themselves do not contain a carry = sequence of=20 length =B3 v. In this case any = carry=20 sequence of length =B3 v in the total = numbers (of=20 length n) must end with the last digits of the total sequence. = Hence=20 these must form the combination 1, 1. The next v - 1 digits must=20 propagate the carry, hence each of these must form the combination 1, 0 = or 0, 1.=20 (The combinations 1, 1 and 0, 0 do not propagate a carry.) The = probability of=20 the combination 1, 1 is 1/4, that one of the alternative = combinations 1,=20 0 or 0, 1 is 1/2 . The total probability of this sequence is therefore=20 1/4(1/2)v-1 =3D (1/2)v+1. The remaining = n -=20 v digits must not contain a carry sequence of length =B3 v. This has the probability 1 -=20 pn-v(v). Thus the probability of the second = case is [1=20 - pn-v(v)]/2 v+l. Combining these two cases, the = desired=20 relation

obtains. The observation that pn(v) =3D 0 if = v=20 >n is trivial.

We see with the help of the formulas proved = above=20 that pn(v) - pn-1(v) is always =A3 1/2 v+1, and hence that the = sum


is not in excess of (n - v + l)/2 = v+1 since=20 there are n - v + 1 terms in the sum; since, moreover, = each=20 pn(v) is a probability, it is not greater than = 1. Hence=20 we have

This last expression is clearly linear in n in = the=20 interval 2K =A3 n =A3 2=20 K+1, and it is =3DK for n =3D 2K = and=20 =3DK+ 1 for n =3D 2 K+1 , i.e it is =3D=20 2log n at both ends of this interval. Since the = function=20 2log n is everywhere concave from below, it follows = that our=20 expression is =A3 2log n = throughout this interval. Thus an =A3=20 2log n. This holds for all K, i.e. for all = n,=20 and it is the in equality which we wanted to prove.

For our case n =3D 40 we have an = =A3 log240 ~5.3, i.e. an average = length of=20 about 5 for the longest carry sequence. (The actual value of a40 = is=20 4.62.)

5.7. Having discussed the addition, we can now go on = to the=20 subtraction. It is convenient to discuss at this point our treatment of = negative=20 numbers, and in order to do that right, it is desirable to make some=20 observations about the treatment of numbers in general.

Our numbers are 40 digit aggregates, the left-most = digit being=20 the sign digit, and the other digits genuine binary digits, with = positional=20 values 2-1, 2-2 , . . . , 2-39 (going from left to right). Our accumulator = will,=20 however, treat the sign digit, too, as a binary digit with the = positional value=20 20-at least when it functions as an adder. For numbers = between 0 and=20 1 this is clearly all right: The left-most digit will then be 0, and if = 0 at=20 this place is taken to represent a + sign, then the number is correctly=20 expressed with its sign and 39 binary digits.

Let us now consider one or more unrestricted 40 = binary digit=20 numbers. The accumulator will add them, with the digit-adding and the = carrying=20 mechanisms functioning normally and identically in all 40 positions. = There is=20 one reservation, however: If a carry originates in the left-most = position, then=20 it has nowhere to go from there (there being no further positions to the = left)=20 and is "lost". This means, of course, that the addend and the augend, = both=20 numbers between 0 and 2, produced a sum exceeding 2, and the = accumulator, being=20 unable to express a digit with a positional value 21, which would now be = necessary, omitted 2. That is, the sum was formed correctly, excepting a = possible error 2. If several such additions are performed in succession, = then=20 the ultimate error may be any integer multiple of 2. That is, the = accumulator is=20 an adder which allows errors that are integer multiples of 2-it is an = adder=20 modulo 2.

It should be noted that our convention of placing the = binary=20 point immediately to the right of the left-most digit has nothing to do = with the=20 structure of the adder. In order to make this point clearer we proceed = to=20 discuss the possibilities of positioning the binary point in somewhat = more=20 detail.

We begin by enumerating the 40 digits of our numbers = (words)=20 from left to right. In doing this we use an index h =3D 1, . . . = , 40. Now=20 we might have placed the binary point just as well between digits j = and=20 j + 1, j =3D 0, . . . , 40. Note, that j =3D .0 = corresponds to=20 the position at the extreme left (there is no digit h =3D j = =3D 0);=20 j =3D 40 corresponds to the position at the extreme right (there = is no=20 position h =3D j + 1 =3D 41); and j =3D 1 corresponds to = our above=20 choice. Whatever our choice of /, it does not affect the = correctness of=20 the accumulator's addition. (This is equally true for subtraction, cf. = below,=20 but not for multiplication and division, cf. 5.8.) Indeed, we have = merely=20 multiplied all numbers by 2j-1 (as = against our=20 previous convention), and such a "change of scale" has no effect on = addition=20 (and subtraction). However, now the accumulator is an adder which allows = errors=20 that are integer multiples of 2j it is an adder modulo = 2j.=20 We mention this because it is occasionally convenient to think in terms = of a=20 convention which places the binary point at the right end of the digital = aggregate. Then j =3D 40, our numbers are integers, and the accumulator = is an=20 adder modulo 240. We must emphasize, however, that all of = this, i.e.=20 all attributions of values to j, are purely convention-i.e. it is = solely=20 the mathematician's interpretation of the functioning of the machine and = not a=20 physical feature of the machine. This convention will necessitate = measures that=20 have to be made effective by actual physical features of the = machine-i.e. the=20 convention will become a physical and engineering reality only when we = come to=20 the organs of multiplication.

We will use the convention j =3D 1, i.e. our = numbers lie=20 in 0 and 2 and the accumulator adds modulo 2.

This being so, these numbers between 0 and 2 can be = used to=20 represent all numbers modulo 2. Any real number x agrees modulo 2 with = one and=20 only one number x between 0 and 2-or, to be quite precise: 0 =A3 x < 2. Since our addition functions modulo = 2, we see=20 that the accumulator may be used to represent and to add numbers modulo = 2.

This determines the representation of negative = numbers: If x=20 < 0, then we have to find the unique integer multiple of 2, 2s

Section 1 Processors with one address per = instruction

(s =3D 1, 2, . . . ) such that 0 =A3 ~ < 2 for ~ =3D = x + 2s=20 (i.e. - 2s =A3 x = < 2(1=20 - s)), and represent x by the digitalization of ~.

In this way, however, the sign digit character of the = left-most=20 digit is lost: It can be 0 or 1 for both x =B3 0 and x=20 < 0, hence 0 in the left-most position can no longer be associated = with the +=20 sign of x. This may seem a bad deficiency of the system, but it is easy = to=20 remedy-at least to an extent which suffices for our purposes. This is = done as=20 follows:

We usually work with numbers x between -1 and 1-or, = to be quite=20 precise: -1 =A3 x < 1. Now the ~ with 0 = =A3 ~ < 2, which differs from x by an integer = multiple of 2,=20 behaves as follows: If x =B3 0. then 0 =A3 x < 1. hence ~ =3D x. and so 0 =A3=20 ~ < 1. the left-most-digit of ~ is 0. If x <0, then -1=A3 x< 0, hence ~ =3D x + 2, and so 1 =A3 ~ < 2, the left-most digit of ~ is 1. = Thus the=20 left-most digit (of ~) is now a precise equivalent of the sign (of x): 0 = corresponds to + and 1 to - .

Summing up:

The accumulator may be taken to represent all real = numbers=20 modulo 2, and it adds them modulo 2. If x lies between -1 and 1 = (precisely: - 1=20 =A3 x < 1)-as it will in almost all = of our uses=20 of the machine-then the left-most digit represents the sign: 0 is + and = 1 is -=20 .

Consider now a negative number x with -1 =A3 x < 0. Put x =3D - = y, 0=20 < y =A3 1. Then we digitalize x by = representing=20 it as x + 2 =3D 2 - y =3D 1 + (1 - y). That = is, the=20 left-most (sign) digit of x =3D - y is, as it should be, 1; and the = remaining 39=20 digits are those of the complement of y =3D - x =3D =F7 x=F7 , ie. = those of 1 -=20 y. Thus we have been led to the familiar representation of = negative=20 numbers by complementation.

The connection between the digits of x and those of - = x is now=20 easily formulated, for any x =B3 0. Indeed, - = x is=20 equivalent to

(This digit index i =3D 1, . . . , 39 is related to our previous = digit index=20 h =3D 1 , . . . , 40 by i =3D h - 1. Actually it is = best to=20 treat as if its domain included the additional value i =3D = 0-indeed i =3D 0=20 then corresponds to h =3D 1, i.e. to the sign digit. In any case = expresses=20 the positional value of the digit to which it refers more simply than = h=20 does: This positional value is 2-i = =3D=20 2-(h-1) . Note that if we had = positioned the=20 binary point more generally between j and j + 1, as = discussed further above, this positional value would have been=20 2-(h-j) . We now have, as = pointed=20 out previously, j =3D 1.) Hence its digits obtain by subtracting = every=20 digit of x from 1-by complementing each digit, i.e. by replacing 0 by 1 = and 1 by=20 0-and then adding 1 in the right-most position (and effecting all the = carries=20 that this may cause). (Note how the left-most digit, interpreted as a = sign=20 digit, gets inverted by this procedure as it should be.)

A subtraction x - y is therefore performed by the = accumulator,=20 Ac, as follows: Form x + y', where y' has a digit 0 or 1 where y has a = digit 1=20 or 0, respectively, and then add 1 in the right-most position. The last=20 operation can be performed by injecting a carry into the right-most = stage of=20 Ac-since this stage can never receive a carry from any other source = (there being=20 no further positions to the right).

5.8. In the light of 5.7 multiplication requires = special care,=20 because here the entire modulo 2 procedure breaks down. Indeed, assume = that we=20 want to compute a product xy, and that we had to change one of = the=20 factors, say x, by an integer multiple of 2, say by 2. Then the product = (x + 2)y=20 obtains, and this differs from the desired xy by 2y. 2y, however, = will=20 not in general be an integer multiple of 2, since y is not in general an = integer.

We will therefore begin our discussion of the = multiplication by=20 eliminating all such difficulties, and assume that both factors x, y lie = between=20 0 and 1. Or, to be quite precise: 0 =A3 x = < 1,=20 0 =A3 y < 1.

To effect such a multiplication we first send the = multiplier x=20 into a register AR, the Arithmetic Register, which is essentially = just a=20 set of 40 flip-flops whose characteristics will be discussed below. We = place the=20 multiplicand y in the Selectron Register, SR (cf. 4.9) and use = the=20 accumulator, Ac, to form and store the partial products. We propose to = multiply=20 the entire multiplicand by the successive digits of the multiplier in a = serial=20 fashion. There are, of course, two possible ways this can be done: We = can either=20 start with the digit in the lowest position-position 2-39-or in the highest position-position = 2-1-and proceed successively to the left or right,=20 respectively. There are a few advantages from our point of view in = starting with=20 the right-most digit of the multiplier. We therefore describe that = scheme.

The multiplication takes place in 39 steps, which correspond to the = 39=20 (non-sign) digits of the multiplier x =3D 0, x=20 1,x 2, . . . , x 39 =3D (0. x 1x 2, . . . ,x = 39),=20 enumerated backwards: x 39, . . . = ,x 2,x = 1.=20 Assume that the k - 1 first steps (k =3D 1,..., 39) = have already=20 taken place, involving multiplication of the multiplicand y with the = k -=20 1 last digits of the multiplier: x=20 39, . . . , x 41-k ; and that we are now at the kth step, = involving=20 multiplication with the kth last digit: x = 40-k . Assume furthermore, that Ac now = contains the=20 quantity Pk-1, the result of the k = -=20 1 first steps. [This is the (k - 1)st partial product. = For k=20 =3D 1 clearly P0 =3D 0.1 We now form=20 2pk =3D Pk-1=20 + x 40-ky,=20 i.e.

That is, we do nothing or add y, according to whether = x 40-k =3D 0 or 1. = We can then=20 form pk by halving 2pk.

Note that the addition of (1) produces no carry = beyond the=20 20 position, i.e. the sign digit: 0 =A3=20 ph < 1 is true = for h=20 =3D 0, and if it is true for h =3D k - 1 , then (1) = extends it to=20 h =3D k also, since 0 =A3 = yk < 1. Hence the sum in (1) is =B3 0 and <2, and no carries beyond the = 20=20 position arise.

Hence pk = obtains=20 from 2pk by a simple = right shift,=20 which is combined with filling in the sign digit (that is freed by this = shift)=20 with a 0. This right shift is effected by an electronic shifter that is = part of=20 Ac.

Now

Thus this process produces the product xy, as = desired.=20 Note that this xy is the exact product of x and y.

Since x and y are 39 digit binaries, their exact = product xy=20 is a 78 digit binary (we disregard the sign digit throughout). = However, Ac=20 will only hold 39 of these. These are clearly the left 39 digits of xy. = The=20 right 39 digits of xy are dropped from Ac one by one in the = course of the=20 39 steps, or to be more specific, of the 39 right shifts. We will see = later that=20 these right 39 digits of xy should and will also be conserved = (cf. the=20 end of this section and the end of 5.12, as well as 6.6.3). The = left 39=20 digits, which remain in Ac, should also be rounded off, but we will not = discuss=20 this matter here. (cf. loc. cit. above and 9.9, Part II).

To complete the general picture of our multiplication = technique=20 we must consider how we sense the respective digits of our multiplier. = There are=20 two schemes which come to one's mind in this connection. One is to have = a gate=20 tube associated with each flip-flop of AR in such a fashion that this = gate is=20 open if a digit is 1 and closed if it is null. We would then need a = 39-stage=20 counter to act as a switch which would successively stimulate these gate = tubes=20 to react. A more efficient scheme is to build into AR a shifter circuit = which=20 enables AR to be shifted one stage to the right each time Ac is shifted = and to=20 sense the value of the digit in the right-most flip-flop of AR. The = shifter=20 itself requires one gate tube per stage. We need in addition a counter = to count=20 out the 39 steps of the multiplication, but this can be achieved by a = six stage=20 binary counter. Thus the latter is more economical of tubes and has one=20 additional virtue from our point of view which we discuss in the next=20 paragraph.

The choice of 40 digits to a word (including the = sign) is=20 probably adequate for most computational problems but situations = certainly might=20 arise when we desire higher precision, i.e. words of greater length. A = trivial=20 illustration of this would be the computation of p to=20 more places than are now known (about 700 decimals, i.e. about 2,300 = binaries).=20 More important instances are the solutions of N linear equations = in N=20 variables for large values of N. The extra precision becomes = probably=20 necessary when N exceeds a limit somewhere between 20 and 40. A=20 justification of this estimate has to be based on a detailed theory of = numerical=20 matrix inversion which will be given in a subsequent report. It is = therefore=20 desirable to be able to handle numbers of 39k digits and signs by means = of=20 program instructions. One way to achieve this end is to use k words to = represent=20 a 39k digit number with signs. (In this way 39 digits in each 40 digit = word are=20 used, but all sign digits excepting the first one, are apparently = wasted; cf.=20 however the treatment of double precision numbers in Chapter 9, Part = II.) It is,=20 of course, necessary in this case to instruct the machine to perform the = elementary operations of arithmetic in a manner that conforms with this=20 interpretation of k-word complexes as single numbers. (Cf. 9.8-9.10, = Part II.)=20 In order to be able to treat numbers in this manner, it is desirable to = keep not=20 39 digits in a product, but 78; this is discussed in more detail in = 6.6.3 below.=20 To accomplish this end (conserving 78 product digits) we connect, via = our=20 shifter circuit, the right-most digit of Ac with the left-most non-sign = digit of=20 AR. Thus, when in the process of multiplication a shift is ordered, the = last=20 digit of Ac is transferred into the place in AR made vacant when the = multiplier=20 was shifted.

5.9. To conclude our discussion of the multiplication = of=20 positive numbers, we note this:

As described thus far, the multiplier forms the 78 = digit=20 product, xy, for a 39 digit multipler x and a 39 digit = multiplicand y. We=20 assumed x =B3 0, y =B3 0 and=20 therefore had xy =B3 0 and we will = only depart=20 from these assumptions in 5.10. In addition to these, however, we also = assumed x=20 <1, y <1, i.e. the x, y have their binary points both immediately = right of=20 the sign digit, which implied the same for xy. One might question = the=20 necessity of these additional assumptions.

Prima facie they may seem mere conventions, which = affect=20 only the mathematician's interpretation of the functioning of the = machine, and=20 not a physical feature of the machine. (Cf. the corresponding situation = in=20 addition and subtraction, in 5.7.) Indeed, if x had its binary point = between=20 digits j and j + 1 from the left (cf. the discussion of = 5.7=20 dealing with this j; it also applies to k below), and y between k = and k +=20 1, then our above method of multiplication would still give the correct = result=20 xy, provided that

Section 1 Processors with one address per = instruction

the position of the binary point in xy is = appropriately=20 assigned. Specifically: Let the binary point of xy be between = digits l=20 and l + 1. x has the binary point between digits j and = j=20 + 1, and its sign digit is 0, hence its range is 0 =A3 x < 2j-1. = Similarly y has=20 the range 0 =A3 y < 2k-1, and xy has the range 0 =A3 xy <2l-1. Now the=20 ranges of x and y imply that the range of xy is necessarily 0 = =A3 xy < 2j-1=20 2k-1 =3D 2 j+k-2 . Hence l =3D j + k - = 1. Thus it=20 might seem that our actual positioning of the binary point-immediately = right of=20 the sign digit, i.e. j =3D k =3D 1-is still a mere = convention.

It is therefore important to realize that this is not = so: The=20 choices of j and k actually correspond to very real, = physical,=20 engineering decisions. The reason for this is as follows: It is = desirable to=20 base the running of the machine on a sole, consistent mathematical=20 interpretation. It is therefore desirable that all arithmetical = operations be=20 performed with an identically conceived positioning of the binary point = in Ac.=20 Applying this principle to x and y gives j =3D k. Hence = the position=20 of the binary point for xy is given by j + k - 1 = =3D 2j- 1.=20 If this is to be the same as for x, and y, then 2j - 1 =3D j, = i.e. j=20 =3D 1 ensues-that is, our above positioning of the binary point = immediately=20 right of the sign digit.

There is one possible escape: To place into Ac not = the left 39=20 digits of xy (not counting the sign digit 0), but the digits j to = j=20 + 38 from the left. Indeed, in this way the position of the binary = point of=20 xy will be (2j - 1) - (j - 1) =3D j, the same as = for x and=20 y.

This procedure means that we drop the left j - = 1 and=20 right 40 + j digits of xy and hold the middle 39 in Ac. = Note that=20 positioning of the binary point-means that x < 2j-1, y <2 j-1 = and=20 xy can only be used if xy <2 j-1. Now the assumptions secure only xy=20 <22j-2. Hence xy = must be=20 2j-1 times smaller than it might be. = This is just=20 the thing which would be secured by the vanishing of the left j - = 1=20 digits that we had to drop from Ac, as shown above.

If we wanted to use such a procedure, with those = dropped left=20 j - 1 digits really existing, i.e. with j =B9=20 1, then we would have to make physical arrangements for their = conservation=20 elsewhere. Also the general mathematical planning for the machine would = be=20 definitely complicated, due to the physical fact that Ac now holds a = rather=20 arbitrarily picked middle stretch of 39 digits from among the 78 digits = of=20 xy. Alternatively, we might fail to make such arrangements, but = this=20 would necessitate to see to it in the mathematical planning of each = problem,=20 that all products turn out to be 2j-1 = times=20 smaller than their a priori maxima. Such an observance is not at = all=20 impossible; indeed similar things are unavoidable for the other = operations. [For=20 example, with a factor 2 in addition (of positives) or subtraction (of = opposite=20 sign quantities). Cf. also the remarks in the first part of 5.12, = dealing=20 with keeping "within range".] However, it involves a loss of significant = digits,=20 and the choice j =3D 1 makes it unnecessary in = multiplication.

We will therefore make our choice j =3D 1, = i.e. the=20 positioning of the binary point immediately right of the sign digit, = binding for=20 all that follows.

5.10. We now pass to the case where the = multiplier x and=20 the multiplicand y may have either sign + or -, i.e. any = combination of=20 these signs.

It would not do simply to extend the method of 5.8 to = include=20 the sign digits of x and y also. Indeed, we assume -1 =A3=20 x < 1, -1 =A3 y < 1, and = the=20 multiplication procedure in question is definitely based on the =B3 0 interpretations of x and y. Hence if x = < 0,=20 then it is really using x + 2, and if y < 0, then it is really using = y + 2.=20 Hence for x < 0, y =B3 0 it forms

(x + 2)y =3D xy + 2y

for x =B3 0, y <0 it = forms

x(y + 2) =3D xy + 2x

for x < 0, x < 0, it forms

(x + 2)(y + 2) =3D xy + 2x + 2y + 4

or since things may be taken modulo 2, xy + 2x = + 2y.=20 Hence correction terms -2y, -2x would be needed for x < 0, y = <=20 0, respectively (either or both).

This would be a possible procedure, but there is one=20 difficulty: As xy is formed, the 39 digits of the multiplier x = are=20 gradually lost from AR, to be replaced by the right 39 digits of xy. = (Cf.=20 the discussion at the end of 5.8.) Unless we are willing to build an = additional=20 40 stage register to hold x, therefore, x will not be available at the = end of=20 the multiplication. Hence we cannot use it in the correction 2x of = xy,=20 which becomes necessary for y < 0.

Thus the case x < 0 can be handled along = the above=20 lines, but not the case y <0.

It is nevertheless possible to develop an adequate = procedure,=20 and we now proceed to do this. Throughout this procedure we will = maintain the=20 assumptions -1 =A3 x < 1, -1 =A3 y < 1. We proceed in = several=20 successive steps.

First: Assume that the corrections necessitated = by the=20 possibility of y <0 have been taken care of. We permit therefore y = =B3 0. We will consider the corrections = necessitated by the=20 possibility of x < 0.

Let us disregard the sign digit of x, which is 1, = i.e. replace=20 it by 0. Then x goes over into x' =3D x - 1 and as -1 =A3 x< 0, this x' will actually behave like (x = - 1) +=20 2 =3D x + 1. Hence our multiplication procedure will produce = x'y =3D=20 (x + l)y =3D xy + y,

and therefore a correction - y is needed at the end. = (Note that=20 we did not use the sign digit of x in the conventional way. Had we done = so, then=20 a correction -2y would have been necessary, as seen above.)

We see therefore: Consider x ~0. Perform first all = necessary=20 steps for forming x'y(y ~ 0), without yet reaching the sign digit = of x=20 (i.e. treating x as if it were =B3 0). When = the time=20 arrives at which the digit x 0 of = x has to=20 become effective-i.e. immediately after x = 1=20 became effective, after 39 shifts (cf. the discussion near the end of = 5.8)-at=20 which time Ac contains, say, ~ (this corresponds to the p39 = of=20 5.8). then form

This ~ is xy. (Note the difference between = this last=20 step, forming ~, and the 39 preceding steps in 5.8, forming=20 p1, p2 , . . . , p39.)

Second: Having disposed of the possibility x < = 0, we may=20 now assume x =B3 0. With this assumption we = have to treat=20 all y ~ 0. Since y =B3 0 brings us back = entirely to the=20 familiar case of 5.8, we need to consider the case y <0 = only.

Let y' be the number that obtains by disregarding the = sign=20 digit of y' which is 1, i.e. by replacing it by 0. Again y' acts not = like y - 1,=20 but like (y - 1) + 2 =3D y + 1. Hence the multiplication = procedure of=20 5.8 will produce xy' =3D x(y + 1) =3D xy + = x, and=20 therefore a correction x is needed. (Note that, quite similarly to what = we saw=20 in the first case above, the suppression of the sign digit of y replaced = the=20 previously recognized correction - 2x by the present one - x.) As we = observed=20 earlier, this correction - x cannot be applied at the end to the = completed=20 xy' since at that time x is no longer available. Hence we must = apply the=20 correction - x digitwise, subtracting every digit at the time = when it is=20 last found in AR, and in a way that makes it effective with the proper=20 positional value.

Third: Consider then x =3D 0, x=20 1, x 2 , . . . , x 39 =3D x = 1,=20 x 2 . . .x=20 39). The 39 digits x 1 = . . .=20 x 39 of x are lost in the course = of the 39=20 shifts of the multiplication procedure of 5.8, going from right = to left.=20 Thus the operation No. k + 1 (k =3D 0, 1 38, cf. 5.8) = finds=20 x 39-k in the = right-most stage=20 of AR, uses it, and then loses it through its concluding right shift (of = both Ac=20 and AR). After this step 39 - (k + 1) =3D 38 - k further = steps, i.e.=20 shifts follow, hence before its own concluding shift there are still 39 = - k=20 shifts to come. Hence the positional values are 239-k = times=20 higher than they will be at the end. x = 39-=20 k should appear at the end, in the correcting term - = x, with=20 the sign - and the positional value 2-(39-k) . = Hence we may inject it during the step k + 1 (before its = shift)=20 with the sign - and the positional value 1. That is to say, - x 39-k in the sign digit.

This, however, is inadmissible. Indeed, x 39-k might cause carries = (if x 39-k=3D 1), which would have nowhere = to go from=20 the sign digit (there being no further positions to the left). This = error is at=20 its origin an integer multiple of 2, but the 39 - k subsequent = shifts=20 reduce its positional value 239-k times. Hence it might = contribute to=20 the end result any integer multiple of 2-(38-k) =BE and this is a genuine error.

Let us therefore add 1 - x = 39-k to the sign digit, i.e. 0 or 1 if x=20 39-k is 1 or 0, respectively. We will show = further=20 below, that with this procedure there arise no carries of the = inadmissible kind.=20 Taking this momentarily for granted, let us see what the total effect = is. We are=20 correcting not by -x but by =E5 = i39=3D1=20 2-i - x =3D 1 - 2-39 - x. = Hence a final=20 correction by -1 + 2-39 is needed. Since this is done at the = end=20 (after all shifts), it may be taken modulo 2. That is to say, we must = add 1 +=20 2-39 i.e. 1 in each of the two extreme positions. Adding 1 in = the=20 right-most position has the same effect as in the discussion at the end = of=20 5.7 (dealing with the subtraction). It is equivalent to injecting = a carry=20 into the right-most stage of Ac. Adding 1 in the left-most position, = i.e. to the=20 sign digit, produces a 1, since that digit was necessarily 0. (Indeed, = the last=20 operation ended in a shift, thus freeing the sign digit, cf. = below.)

Fourth: Let us now consider the question of the = carries=20 that may arise in the 39 steps of the process described above. In order = to do=20 this, let us describe the kth step (k =3D 1, . . . , 39), which = is a=20 variant of the kth step described for a positive multiplication in = 5.8,=20 in the same way in which we described the original kth step bc. = cit.=20 That is to say, let us see what the formula (1) of 5.8 has = become. It=20 is clearly 2pk =3D pk-1 + (1 - x=20 40-k) + x 40-ky' i.e. =

That is, we add 1 (y's sign digit) or y' (y without = its sign=20 digit), according to whether x = 40-k =3D 0 or=20 1. Then pk should obtain from = 2pk=20 again by halving.

Now the addition of (2) produces no carries beyond = the=20 20 position, as we asserted earlier, for the same reason as = the=20 addition of (1) in 5.8. We can argue in the same way as there: 0 =A3 ph <1 is = true for h=20 =3D 0, and if it is true for h =3D k - 1, then (1) = extends it to=20 h =3D k also, since 0 =A3=20 y'k =A3 1. = Hence the sum=20 in (2) is =B3 0 and <2, and no carries = beyond the=20 20 position arise.

Fifth: In the three last observations we assumed = y < 0.=20 Let us now restore the full generality of y ~ 0. We can then = describe

Section 1 Processors with one address per = instruction

the equations (1) of 5.8 (valid for y =B3 = 0) and (2)=20 above (valid for y <0) by a single formula,

Thus our verbal formulation of (2) applies here, too: = We add=20 y's sign digit or y without its sign, according to whether x 40-k =3D 0 or 1. All = pk are =B3 0 and = <1, and the=20 addition of (3) never originates a carry beyond the 20 = position.=20 pk obtains from = 2pk by a right shift, filling the sign digit = with a 0.=20 (Cf. however, Part II, Table 2 for another sort of right shift that is = desirable=20 in explicit form, i.e. as an order.)

For y =B3 0, xy is=20 p39, for y <0, xy obtains from p39 by = injecting=20 a carry into the right-most stage of Ac and by placing a 1 into the sign = digit=20 in Ac.

Sixth: This procedure applies for x =B3 0. For x <0 it should also be applied, since = it makes=20 use of x's non-sign digits only, but at the end y must be subtracted = from the=20 result.

This method of binary multiplication will be = illustrated in=20 some examples in 5.15.

5.11. To complete our discussion of the = multiplicative organs=20 of our machine we must return to a consideration of the types of = accumulators=20 mentioned in 5.5. The static accumulator operates as an adder by = simultaneously=20 applying static voltages to its two inputs-one for each of the two = numbers being=20 added. When steady-state operation is reached the total sum is formed = complete=20 with all carries. For such an accumulator the above discussion is = substantially=20 complete, except that it should be remarked that such a circuit requires = at most=20 39 rise times to complete a carry. Actually it is possible that the = duration of=20 these successive rises is proportional to a lower power of 39 than the = first=20 one.

Each stage of a dynamic accumulator consists of a = binary=20 counter for registering the digit and a flip-flop for temporary storage = of the=20 carry. The counter receives a pulse if a 1 is to be added in at that = place; if=20 this causes the counter to go from 1 to 0 a carry has occurred and hence = the=20 carry flip-flop will be set. It then remains to perform the carries. = Each=20 flip-flop has associated with it a gate, the output of which is = connected to the=20 next binary counter to the left. The carry is begun by pulsing all carry = gates.=20 Now a carry may produce a carry, so that the process needs to be = repeated until=20 all carry flip-flops register 0. This can be detected by means of a = circuit=20 involving a sensing tube connected to each carry flip-flop. It was shown = in 5.6=20 that, on the average, five pulse times (flip-flop reaction times) are = required=20 for the complete carry. An alternative scheme is to connect a gate tube = to each=20 binary counter which will detect whether an incoming carry pulse would = produce a=20 carry and will, under this circumstance, pass the incoming carry pulse = directly=20 to the next stage. This circuit would require at most 39 rise times for = the=20 completion of the carry. (Actually less, cf. above.)

At the present time the development of a static = accumulator is=20 being concluded. From preliminary tests it seems that it will add two = numbers in=20 about 5 m sec and will shift right or left in = about 1=20 m sec.

We return now to the multiplication operation. In a = static=20 accumulator we order simultaneously an addition of the multiplicand with = sign=20 deleted or the sign f the multiplicand (cf. 5.10) and a complete carry = and then=20 a shift for each of the 39 steps. In a dynamic accumulator of the second = kind=20 just described we order in succession an addition of the multiplicand = with sign=20 deleted or the sign of the multiplicand, a complete carry, and a shift = for each=20 of the 39 steps. In a dynamic accumulator of the first kind we can avoid = losing=20 the time required for completing the carry (in this case an average of 5 = pulse=20 times, cf. above) at each of the 39 steps. We order an addition by the=20 multiplicand with sign deleted or the sign of the multiplicand, then = order one=20 pulsing of the carry gates, and finally shift the contents of both the = digit=20 counters and the carry flip-flops. This process is repeated 39 times. A = simple=20 arithmetical analysis which may be carried out in a later report, shows = that at=20 each one of these intermediate stages a single carry is adequate, and = that a=20 complete set of carries is needed at the end only. We then carry out the = complement corrections, still without ever ordering a complete set of = carry=20 operations. When all these corrections are completed and after = round-off,=20 described below, we then order the complete carry mentioned above.

5.12. It is desirable at this point in the discussion = to=20 consider rules for rounding-off to n-digits. In order to assess the=20 characteristics of alternative possibilities for such properly, and in=20 particular the role of the concept of "unbiasedness", it is necessary to = visualize the conditions under which rounding-off is needed.

Every number x that appears in the computing machine = is an=20 approximation of another number x', which would have appeared if the = calculation=20 had been performed absolutely rigorously. The approximations to which we = refer=20 here are not those that are caused by the explicitly introduced = approximations=20 of the numerical-mathematical set-up, e.g. the replacement of a = (continuous)=20 differential equation by a (discrete) difference equation. The effect of = such=20 approximations should be evaluated mathematically by the person who = plans the=20 problem for the machine, and should not be a direct concern of the = machine.=20 Indeed, it has to be handled

by a mathematician and cannot be handled by the = machine, since=20 its nature, complexity, and difficulty may be of any kind, depending = upon the=20 problem under consideration. The approximations which concern us here = are these:=20 Even the elementary operations of arithmetic, to which the mathematical=20 approximation-formulation for the machine has to reduce the true = (possibly=20 transcendental) problem, are not rigorously executed by the machine. The = machine=20 deals with numbers of n digits, where n, no matter how large, has = to be a=20 fixed quantity. (We assumed for our machine 40 digits, including the = sign, i.e.=20 n =3D 39.) Now the sum and difference of two n-digit numbers are = again=20 n-digit numbers, but their product and quotient (in general) are not. = (They=20 have, in general, 2n or =A5 -digits, = respectively.)=20 Consequently, multiplication and division must unavoidably be replaced = by the=20 machine by two different operations which must produce n-digits under = all=20 conditions, and which, subject to this limitation, should lie as close = as=20 possible to the results of the true multiplication and division. One = might call=20 them pseudo-multiplication and pseudo-division; however, the accepted=20 nomenclature terms them as multiplication and division with round-off. = (We are=20 now creating the impression that addition and subtraction are entirely = free of=20 such shortcomings. This is only true inasmuch as they do not create new = digits=20 to the right, as multiplication and division do. However, they can = create new=20 digits to the left, i.e. cause the numbers to "grow out of range". This=20 complication, which is, of course, well known, is normally met by the = planner,=20 by mathematical arrangements and estimates to keep the numbers "within = range".=20 Since we propose to have our machine deal with numbers between -1 and 1, = multiplication can never cause them to "grow out of range". Division, of = course,=20 might cause this complication, too. The planner must therefore see to it = that in=20 every division the absolute value of the divisor exceeds that of the=20 dividend.)

Thus the round-off is intended to produce = satisfactory n-digit=20 approximations for the product xy and the quotient x/y of = two=20 n-digit numbers. Two things are wanted of the round-off: (1) The = approximation=20 should be good, i.e. its variance from the "true" xy or x/y = should=20 be as small as practicable; (2) The approximation should be unbiased, = i.e. its=20 mean should be equal to the "true" xy or x/y.

These desiderata must, however, be considered in = conjunction=20 with some further comments. Specifically: (a) x and y themselves are = likely to=20 be the results of similar round-offs, directly or in directly inherent, = i.e. x=20 and y themselves should be viewed as unbiased n-digit approximations of = "true"=20 x' and y' values; (b) by talking of "variances" and "means" we are = introducing=20 statistical concepts. Now the approximations which we are here = considering are=20 not really of a statistical nature, but are due to the peculiarities = (from our=20 point of view, inadequacies) of arithmetic and of digital = representation, and=20 are therefore actually rigorously and uniquely determined. It seems, = however, in=20 the present state of mathematical science, rather hopeless to try to = deal with=20 these matters rigorously. Furthermore, a certain statistical approach, = while not=20 truly justified, has always given adequate practical results. This = consists of=20 treating those digits which one does not wish to use individually in = subsequent=20 calculations as random variables with equiprobable digital values, and = of=20 treating any two such digits as statistically independent (unless this = is=20 patently false).

These things being understood, we can now undertake = to discuss=20 round-off procedures, realizing that we will have to apply them to the=20 multiplication and to the division.

Let x =3D (.x = 1 . . .=20 x n) and y =3D (.h 1 . . . h=20 n) be unbiased approximations of x' and y'. Then the "true" = xy =3D=20 (.x 1 . . . x=20 nx n+1 . . . x=20 2n) and the "true" x/y =3D (.w = 1 .=20 . . w nw = n+1w n+2 . . .) (this goes on ad infinitum!) are = approximations of x'y' and x'/y'. Before we discuss how to round = them=20 off, we must know whether the "true" xy and x/y are = themselves=20 unbiased approximations of x'y' and x'/y'. xy is indeed an = unbiased approximation of x'y', i.e. the mean of xy is the = mean of=20 x( =3D x') times the mean of y( =3D y'), owing to the = independence=20 assumption which we made above. However, if x and y are closely = correlated, e.g.=20 for x =3D y, i.e. for squaring, there is a bias. It is of the = order of the=20 mean square of x - x', i.e. of the variance of x. Since x has n = digits,=20 this variance is about l/22n (If the digits of x', beyond n = are=20 entirely unknown, then our original assumptions give the variance=20 l/l2.22n.) Next, x/y can be written as = x.y-1, and since we have already discussed the = bias of=20 the product, it suffices now to consider the reciprocal y-1. Now if y is an unbiased estimate of y', then = y-1 is not an unbiased estimate of y'-1, i.e. the mean of y's reciprocal is not the = reciprocal of=20 y's mean. The difference is ~y-3 times = the=20 variance of y, i.e. it is of essentially the same order as the bias = found above=20 in the case of squaring.

It follows from all this that it is futile to attempt = to avoid=20 biases of the order of magnitude l/22n or less. (The factor = 1/12=20 above may seem to be changing the order of magnitude in question. = However, it is=20 really the square root of the variance which matters and =D6 (1/12 ~ 0.3 is a moderate factor.) Since we = propose to use=20 n =3D 39, therefore l/278(~3 =B4=20 l0-24) is the critical case. Note that = this=20 possible bias level is 1/239(~2 x 10-12) times our last significant digit. = Hence we will=20 look for round-off rules to n digits for the "true" xy =3D = (.x 1 . . . x = nx n+1 . . . x = 2n)=20 and x/y =3D (.w 1 . . . = w nw n+1w n+2 . . . ). The desideratum (1) which we = formulated=20 previously, that the variance should be small, is still valid. The

Section 1 Processors with one address per = instruction

desideratum (2), however, that the bias should be = zero, need,=20 according to the above, only be enforced up to terms of the order=20 1/22n.

The round-off procedures, which we can use in this = connection,=20 fall into two broad classes. The first class is characterized by its = ignoring=20 all digits beyond the nth, and even the nth digit itself, which it = replaces by a=20 1. The second class is characterized by the procedure of adding one unit = in the=20 (n + 1)st digit, performing the carries which this may induce, = and then=20 keeping only the n first digits.

When applied to a number of the form (.v1 . . .=20 vnv n+1v n+2 . . . ) (ad = infinitum!),=20 the effects of either procedure are easily estimated. In the first = case we=20 may say we are dealing with (.v1, . . ., v n-1) = plus a=20 random number of the form (.0 . . . , 0vnv n+1v=20 n+2 . . .), i.e. random in the interval 0, 1/2n-1 . Comparing with the rounded off=20 (.v1v2 . . . vn-1l), we therefore have a difference random in = the interval=20 - l/2n, l/2n. Hence its mean is 0 and its variance = 1/3=20 . 22n. = In the=20 second case we are dealing with (.v1 . . . vn) = plus a=20 random number of the form (.0 . . . 00vn+1 = vn+2=20 . . . ), i . e . random in the interval 0, 1/2n. The=20 "rounded-off" value will be (.v1 . . . = vn)=20 increased by 0 or by 1/2n, according to whether the = random number=20 in question lies in the interval 0, 1/2n+1, or in the = interval=20 1/2n+1, 1/2n. Hence comparing with the = "rounded-off"=20 value, we have a difference random in the intervals 0, = 1/2n+1,=20 and 0, - l/2n+1, i.e. in the interval = -1/2n+1,=20 1/2n+1. Hence its mean is 0 and its variance=20 (1/12)22n.

If the number to be rounded-off has the form = (.v1 .=20 . . vn vn+1 vn+2 . . . vn+p) = (p=20 finite), then these results are somewhat affected. The order of = magnitude of the=20 variance remains the same; indeed for large p even its relative change = is=20 negligible. The mean difference may deviate from 0 by amounts which are = easily=20 estimated to be of the order 1/2n .1/2p =3D 1/2 n+p.

In division we have the first situation, x/y = =3D (.w 1 . . . w = nw n+1w = n+2 . . . ),=20 i.e. p is infinite. In multiplication we have the second one, xy = =3D=20 (.x 1 . . . x=20 nx n+1 . . . x=20 2n), i.e. p =3D n. Hence for the division both methods = are=20 applicable without modification. In multiplication a bias of the order = of=20 1/22n may be introduced. We have seen that it = is=20 pointless to insist on removing biases of this size. We will therefore = use the=20 unmodified methods in this case, too.

It should be noted that the bias in the case of = multiplication=20 can be removed in various ways. However, for the reasons set forth = above, we=20 shall not complicate the machine by introducing such corrections.

Thus we have two standard "round-off" methods, both = unbiased to=20 the extent to which we need this, and with the variances 1/3 = . 22n, and = (1/12)22n=20 that is, with the dispersions (1/=D6 = 3)(1/2n)=20 =3D 0.58 times the last digit and (1/2 =D6=20 3)(1/2n) =3D 0.29 times the last digit. The first one = requires no carry=20 facilities, the second one requires them.

Inasmuch as we propose to form the product x'y' = in the=20 accumulator, which has carry facilities, there is no reason why we = should not=20 adopt the rounding scheme described above which has the smaller = dispersion, i.e.=20 the one which may induce carries. In the case, however, of division we = wish to=20 avoid schemes leading to carries since we expect to form the quotient in = the=20 arithmetic register, which does not permit of carry operations. The = scheme which=20 we accordingly adopt is the one in which w = n=20 is replaced by 1. This method has the decided advantage that it enables = us to=20 write down the approximate quotient as soon as we know its first (n = - 1)=20 digits. It will be seen in 5.14 and 6.6.4 below that our = procedure for=20 forming the quotient of two numbers will always lead to a result that is = correctly rounded in accordance with the decisions just made. We do not = consider=20 as serious the fact that our rounding scheme in the case of division has = a=20 dispersion twice as large as that in multiplication since division is a = far less=20 frequent operation.

A final remark should be made in connection with the = possible,=20 occasional need of carrying more than n =3D 39 digits. Our = logical control=20 is sufficiently flexible to permit treating k (=3D2, 3, . . . = words as one=20 number, and thus effecting n =3D 39k. In this case the round-off = has to be=20 handled differently, cf. Chapter 9, Part II. The multiplier produces all = 78=20 digits of the basic 39 by 39 digit multiplication: The first 39 in the = Ac, the=20 last 39 in the AR. These must then be manipulated in an appropriate = manner. (For=20 details, cf. 6.6.3 and 9.9-9.10, Part II.) The divider works for 39 = digits only:=20 In forming x/y, it is necessary, even if x and y are available to = 39k=20 digits, to use only 39 digits of each, and a 39 digit result will = appear. It=20 seems most convenient to use this result as the first step of a series = of=20 successive approximations. The successive improvements can then be = obtained by=20 various means. One way consists of using the well known iteration = formula (cf.=20 5.4). For k =3D 2 one such step will be needed, for k = =3D 3, 4,=20 two steps, for k =3D 5, 6, 7, 8 three steps, etc. An alternative = procedure=20 is this: Calculate the remainder, using the approximate, 39 digit, = quotient and=20 the complete, 39k digit, divisor and dividend. Divide this again by the=20 approximate, 39 digit, divisor, thus obtaining essentially the next 39 = digits of=20 the quotient. Repeat this procedure until the full 39k desired digits of = the=20 quotient have been obtained.

5.13. We might mention at this time a = complication which=20 arises when a floating binary point is introduced into the machine. The=20 operation of addition which usually takes at most 1/10 of a

multiplication time becomes much longer in a machine = with=20 floating binary since one must perform shifts and round-offs as well as=20 additions. It would seem reasonable in this case to place the time of an = addition as about 1/3 to 1/2 of a multiplication. At this rate it is = clear that=20 the number of additions in a problem is as important a factor in the = total=20 solution time as are the number of multiplications. (For further details = concerning the floating binary point, cf. 6.6.7.)

5.14. We conclude our discussion of the arithmetic = unit with a=20 description of our method for handling the division operation. To = perform a=20 division we wish to store the dividend in SR, the partial remainder in = Ac and=20 the partial quotient in AR. Before proceeding further let us consider = the=20 so-called restoring and non-restoring methods of division. = In=20 order to be able to make certain comparisons, we will do this for a = general base=20 m =3D 2, 3

Assume for the moment that divisor and dividend are = both=20 positive. The ordinary process of division consists of subtracting from = the=20 partial remainder (at the very beginning of the process this is, of = course, the=20 dividend) the divisor, repeating this until the former becomes smaller = than the=20 latter. For any fixed positional value in the quotient in a = well-conducted=20 division this need be done at most m - 1 times. If, after = precisely k=20 =3D 0, 1, . . . , m - 1 repetitions of this step, the partial = remainder has=20 indeed become less than the divisor, then the digit k is put in = the=20 quotient (at the position under consideration), the partial remainder is = shifted=20 one place to the left, and the whole process is repeated for the next = position,=20 etc. Note that the above comparison of sizes is only needed at k = =3D 0, 1,=20 . . . , m - 2, i.e. before step 1 and after steps 1, . . . , m- = 2. If the=20 value k =3D m-1, i.e. the point after step m - 1, is at = all reached=20 in a well-conducted division, then it may be taken for granted without = any test,=20 that the partial remainder has become smaller than the divisor, and the=20 operations on the position under consideration can therefore be = concluded. (In=20 the binary system, m =3D 2, there is thus only one step, and only = one=20 comparison of sizes, before this step.) In this way this scheme, known = as the=20 restoring scheme, requires a maximum of m - 1 comparisons = and=20 utilizes the digits 0,1,..., m - 1 in each place in the quotient. = The=20 difficulty of this scheme for machine purposes is that usually the only=20 economical method for comparing two numbers as to size is to subtract = one from=20 the other. If the partial remainder rn were less than the = dividend=20 d, one would then have to add d back into = r~=20 - d in order to restore the remainder. Thus at every stage an = unnecessary operation would be performed. A more symmetrical scheme is = obtained=20 by not restoring. In this method (from here on we need not assume = the=20 positivity of divisor and dividend) one compares the signs of=20 rn and d; if they are of the same sign, = the=20 dividend is repeatedly subtracted from the remainder until the signs = become=20 opposite; if they are opposite, the dividend is repeatedly added to the=20 remainder until the signs again become like. In this scheme the digits = that may=20 occur in a given place in the quotient are evidently +1, = +2 , . .=20 . ,+(m - 1) the positive digits corresponding to=20 subtractions and the negative ones to additions of the dividend to the=20 remainder.

Thus we have 2(m - 1) digits instead of the = usual m=20 digits. In the decimal system this would mean 18 digits instead of = 10. This=20 is a redundant notation. The standard form of the quotient must = therefore be=20 restored by subtracting from the aggregate of its positive digits the = aggregate=20 of its negative digits. This requires carry facilities in the place = where the=20 quotient is stored.

We propose to store the quotient in AR, which has no = carry=20 facilities. Hence we could not use this scheme if we were to operate in = the=20 decimal system.

The same objection applies to any base m for = which the=20 digital representation in question is redundant-i.e. when 2(m - = 1) >=20 m. Now 2(m - 1) > m whenever m > 2, = but 2(m=20 - 1) =3D m for m =3D 2. Hence, with the use of a = register which we=20 have so far contemplated, this division scheme is certainly excluded = from the=20 start unless the binary system is used.

Let us now investigate the situation in the binary = system. We=20 inquire if it is possible to obtain a quasi-quotient by using the = non-restoring=20 scheme and by using the digits 1, 0 instead of 1, -1. Or rather we have = to ask=20 this question: Does this quasi-quotient bear a simple relationship to = the true=20 quotient?

Let us momentarily assume this question can be = answered=20 affirmatively and describe the division procedure. We store the divisor=20 initially in Ac, the dividend in SR and wish to form the quotient in AR. = We now=20 either add or subtract the contents of SR into Ac, according to whether = the=20 signs in Ac and SR are opposite or the same, and insert correspondingly = a 0 or 1=20 in the right-hand place of AR. We then shift both Ac and AR one place = left, with=20 electronic shifters that are parts of these two aggregates.

At this point we interrupt the discussion to note = this:=20 multiplication required an ability to shift right in both Ac and AR (cf. = 5.8).=20 We have now found that division similarly requires an ability to shift = left in=20 both Ac and AR. Hence both organs must be able to shift both ways=20 electronically. Since these abilities have to be present for the = implicit needs=20 of multiplication and division, it is just as well to make use of them=20 explicitly in the form of explicit orders. These are the orders 20,21 of = Table=20 1, and of Table 2, Part II. It will, however, turn out to be convenient = to=20 arrange some details in the shifts, when they occur explicitly under the = control=20 of those orders,

Section 1 Processors with one address per = instruction

differently from when they occur implicitly under the = control=20 of a multiplication or a division. (For these things. cf. the discussion = of the=20 shifts near the end of 5.8 and in the third remark below on one hand, = and in the=20 third remark in 7.2, Part II, on the other hand.)

Let us now resume the discussion of the division. The = process=20 described above will have to be repeated as many times as the number of = quotient=20 digits that we consider appropriate to produce in this way. This is = likely to be=20 39 or 40; we will determine the exact number further below.

In this process we formed digits x=20 'i =3D 0 or 1 for the quotient, when the digit should = actually have=20 been x i =3D -1 or 1, with x 'i =3D 2x = 'i -=20 1. Thus we have a difference between the true quotient z (based = on the=20 digits x i) and the quasi-quotient = z'=20 (based on the digits x 'i), = but at the=20 same time a one-to-one connection. It would be easy to establish the = algebraical=20 expression for this connection between z' and z directly, = but it=20 seems better to do this as part of a discussion which clarifies all = other=20 questions connected with the process of division at the same time.

We first make some general remarks:

First: Let x be the dividend and y the divisor. = We assume,=20 of course, -1 =A3 x < 1, -1 =A3=20 y < 1. It will be found that our present process of division = is=20 entirely unaffected by the signs of x and y, hence no further = restrictions on=20 that score are required.

On the other hand, the quotient z =3D x/y = must also=20 fulfil -1 =A3 z < 1. It = seems somewhat=20 simpler although this is by no means necessary, to exclude for the = purposes of=20 this discussion z =3D -1, and to demand =F7=20 z=F7 <1. This means in terms of the = dividend x=20 and the divisor y that we exclude x =3D - y and assume =F7=20 x=F7 <y.

Second: The division takes place in n = steps, which=20 correspond to the n digits x = 'i=20 , . . .,x 'n of the = pseudo-quotient=20 z', n being yet to be determined (presumably 39 or 40). Assume = that the=20 k - 1 first steps (k =3D 1, . . . , n) have already = taken=20 place, having produced the k - 1 first digits: x=20 'i , . . . , x 'k-1; = and that we=20 are now at the kth step, involving production of the kth digit; x 'k. Assume furthermore, that Ac now = contains the=20 quantity rk-1, the result of the k - 1 first steps. = (This is=20 the (k - 1)st partial remainder. For k =3D 1 clearly = r0=20 =3D x.) We then form rk =3D = 2rk-1=20 + y, according to whether the signs of rk-1 and y do or do not agree, i.e.

Let us now see what carries may originate in this procedure. We can = argue as=20 follows: =F7 rh=F7=20 < =F7 y=F7=20 is true for h =3D 0(=F7 r0 = =F7 =3D =F7 = x=F7 <=F7 = y=F7 ), and if it is true for h =3D k = - 1, then (4)=20 extends it to h =3D k also, since rk-1 and =C5 y = have opposite=20 signs. The last point may be elaborated a little further: because of the = opposite signs

=F4 r=F4 =3D2=F4 rk-1=F4=20 - =F4 y=F4 <2=F4 y=F4 - =F4=20 y=F4 =3D =F4 = y=F4

Hence we have always =F4 = rk=F4 <=F4=20 y=F4 , and therefore = afortiori =F4 rk=F4=20 < 1, i.e. -l< rk <=20 l.

Consequently in equation (4) one summand is = necessarily >=20 -2, <2, the other is =B3 1, <1, and the = sum is=20 >-1, <1. Hence we may carry out the operations of (4) modulo 2,=20 disregarding any possibilities of carries beyond the 20 position, = and the=20 resulting rk will be = automatically correct=20 (in the range > - 1. <1).

Third: Note however that the sign of = rk-1, which plays an important role in (4) = above, is=20 only then correctly determinable from the sign digit, if the number from = which=20 it is derived is =B3 - 1 <1. (Cf. = the=20 discussion in 5.7.) This requirement however is met, as we saw above, by = rk-1, but not necessarily = by=20 2rk-1. Hence the sign of rk-1 (i.e. its sign digit) as required by (4), must = be sensed=20 before rk-1 is = doubled.

This being understood, the doubling of rk-1 may be performed as a simple left shift, in = which the=20 left-most digit (the sign digit) is allowed to be lost-this corresponds = to the=20 disregarding of carries beyond the 20 position, which we = recognized=20 above as being permissible in (4). (Cf. however, Part II, Table 2, for = another=20 sort of left shift that is desirable in explicit form, i.e. as an = order.)

Fourth: Consider now the precise implication of = (4) above.=20 x 'i =3D 1 or 0 corresponds to = =C5 =3D - or +, respectively. Hence (4) may be = written

Fifth: If we do not wish to get involved in more=20 complicated round-off procedure which exceed the immediate capacity of = the only=20 available adder Ac, then the above result suggests that we should put = n +=20 1 =3D 40, n =3D 39. The x '1, . . = . , x '39 are then 39 digits of the = quotient,=20 including the sign digit, but not including the right-most digit.

The right-most digit is taken care of by placing a 1 = into the=20 right-most stage of Ac.

At this point an additional argument in favor of the = procedure=20 that we have adopted here becomes apparent. The procedure coincides = (without a=20 need for any further corrections) with the second round-off procedure = that we=20 discussed in 5.12.

There remains the term -1. Since this applies to the = final=20 result, and no right shifts are to follow, carries which might go beyond = the=20 20 position may be disregarded. Hence this amounts simply to = changing=20 the sign digit of the quotient ~: replacing 0 or 1 by 1 or 0, = respectively.

This concludes our discussion of the division scheme. = We wish,=20 however, to re-emphasize two very distinctive features which it=20 possesses:

First: This division scheme applies equally for = any=20 combinations of signs of divisor and dividend. This is a characteristic = of the=20 non-restoring division schemes, but it is not the case for any simple = known=20 multiplication scheme. It will be remembered, in particular, that our=20 multiplication procedure of 5.9 had to contain special correcting = steps=20 for the cases where either or both factors are negative.

Second: This division scheme is practicable in = the binary=20 system only; it has no analog for any other base.

This method of binary division will be illustrated on = some=20 examples in 5.15.

5.15. We give below some illustrative examples of = the=20 operations of binary arithmetic which were discussed in the preceding=20 sections.

Although it presented no difficulties or ambiguities, = it seems=20 best to begin with an example of addition.

Note that this deviates by 1/64, i.e. by one unit of = the=20 right-most position, from the correct result - 3/8. This is a = consequence of our=20 round-off rule, which forces the right-most digit to be 1 under all = conditions.=20 This occasionally produces results with unfamiliar and even annoying = aspects=20 (e.g. when quotients like 0:y or y:y are formed), but it is nevertheless = unobjectionable and self-consistent on the basis of our general=20 principles.

6. The control

6.1. It has already been stated that the computer = will contain=20 an organ. called the control, which can automatically execute the orders = stored=20 in the Selectrons. Actually, for a reason stated in 6.3, the orders for = this=20 computer are less than half as long as a forty binary digit number, and = hence=20 the orders are stored in the Selectron memory in pairs.

Let us consider the routine that the control performs = in=20 directing a computation. The control must know the location in the = Selectron=20 memory of the pair of orders to be executed. It must direct the = Selectrons to=20 transmit this pair of orders to the Selectron register and then to = itself. It=20 must then direct the execution of the operation specified in the first = of the=20 two orders. Among these orders we can immediately describe two major = types: An=20 order of the first type begins by causing the transfer of the number, = which is=20 stored at a specified memory location, from the Selectrons to the = Selectron=20 register. Next, it causes the arithmetical unit to perform some = arithmetical=20 operations on this number (usually in conjunction with another number = which is=20 already in the arithmetical unit), and to retain the resulting number in = the=20 arithmetical unit. The second type order causes the transfer of the = number,=20 which is held in the arithmetical unit, into the Selectron register, and = from=20 there to a specified memory location in the Selectrons. (It may also be = that=20 this latter operation will permit a direct transfer from the = arithmetical unit=20 into the Selectrons.) An additional type of order consists of the = transfer=20 orders of 3.5. Further orders control the inputs and the outputs of the = machine.=20 The process described at the beginning of this paragraph must then be = repeated=20 with the second order of the order pair. This entire routine is repeated = until=20 the end of the problem.

6.2. It is clear from what has just been stated that = the=20 control must have a means of switching to a specified location in the = Selectron=20 memory, for withdrawing both numbers for the computation and pairs of = orders.=20 Since the Selectron memory (as tentatively planned) will hold = 212 =3D=20 4,096 forty-digit words (a word is either a number or a pair of orders), = a=20 twelve-digit binary number suffices to identify a memory location. Hence = a=20 switching mechanism is required which will, on receiving a twelve-digit = binary=20 number, select the corresponding memory location.

The type of circuit we propose to use for this = purpose is known=20 as a decoding or many-one function table. It has been developed in = various forms=20 independently by J. Rajchman [Rajchman, 1943] and P. Crawford [Crawford, = 19??].=20 It consists of n flip-flops which register an n-digit binary = number. It=20 also has a maximum of 2n output wires. The flip-flops = activate a=20 matrix in which the interconnections between input and output wires are = made in=20 such a way that one and only one of 2n output wires is = selected (i.e.=20 has a positive voltage applied to it). These interconnections may be = established=20 by means of resistors or by means of non-linear elements (such as diodes = or=20 rectifiers); all these various methods are under investigation. The = Selectron is=20 so designed that four such function table switches are required, each = with a=20 three digit entry and eight (23) outputs. Four sets of eight = wires=20 each are brought out of the Selectron for switching purposes, and a = particular=20 location is selected by making one wire positive with respect to the = remainder.=20 Since all forty Selectrons are switched in parallel, these four sets of = wires=20 may be connected directly to the four function table outputs.

6.3. Since most computer operations involve at least = one number=20 located in the Selectron memory, it is reasonable to adopt a code in = which=20 twelve binary digits of every order are assigned to the specification of = a=20 Selectron location. In those orders which do not require a number to be = taken=20 out of or into the Selectrons these digit positions will not be = used.

Though it has not been definitely decided how many = operations=20 will be built into the computer (i.e. how many different orders the = control must=20 be able to understand), it will be seen presently that there will = probably be=20 more than 25 but certainly less than 26. For this = reason=20 it is feasible to assign 6 binary digits for the order code. It thus = turns out=20 that each order must contain eighteen binary digits, the first twelve=20 identifying a memory location and the remaining six specifying an = operation. It=20 can now be explained why orders are stored in the memory in pairs. Since = the=20 same memory organ is to be used in this computer for both orders and = numbers, it=20 is efficient to make the length of each about equivalent. But numbers of = eighteen binary digits would not be sufficiently accurate for problems = which=20 this machine will solve. Rather, an accuracy of at least 10-10 or 2-33 is = required.=20 Hence it is preferable to make the numbers long enough to accommodate = two=20 orders.

As we pointed out in 2.3, and used in 4.2 et seq. = and=20 5.7 et seq., our numbers will actually have 40 binary digits = each. This=20 allows 20 binary digits for each order, i.e. the 12 digits that specify = a memory=20 location, and 8 more digits specifying the nature of the

Section 1 Processors with one address per = instruction

operation (instead of the minimum of 6 referred to = above). It=20 is convenient, as will be seen in 6.8.2. and Chapter 9, Part II to group = these=20 binary digits into tetrads, groups of 4 binary digits. Hence a = whole word=20 consists of 10 tetrads, a half word or order of 5 tetrads, and of these = 3=20 specify a memory location and the remaining 2 specify the nature of the=20 operation. Outside the machine each tetrad can be expressed by a base 16 = digit.=20 (The base 16 digits are best designated by symbols of the 10 decimal = digits 0 to=20 9, and 6 additional symbols, e.g. the letters a to f. Cf. Chapter 9, = Part II.)=20 These 16 characters should appear in the typing for and the printing = from the=20 machine. (For further details of these arrangements, cf. loc. cit.=20 above.)

The specification of the nature of the operation that = is=20 involved in an order occurs in binary form, so that another many-one or = decoding=20 function is required to decode the order. This function table will have = six=20 input flip-flops (the two remaining digits of the order are not needed). = Since=20 there will not be 64 different orders, not all 64 outputs need be = provided.=20 However, it is perhaps worthwhile to connect the outputs corresponding = to unused=20 order possibilities to a checking circuit which will give an indication = whenever=20 a code word unintelligible to the control is received in the input=20 flip-flops.

The function table just described energizes a = different output=20 wire for each different code operation. As will be shown later, many of = the=20 steps involved in executing different orders overlap. (For example, = addition,=20 multiplication, division, and going from the Selectrons to the register = all=20 include transferring a number from the Selectrons to the Selectron = register.)=20 For this reason it is perhaps desirable to have an additional set of = control=20 wires, each of which is activated by any particular combination of = different=20 code digits. These may be obtained by taking the output wires of the = many-one=20 function table and using them to operate tubes which will in turn = operate a=20 one-many (or coding) function table. Such a function table consists of a = matrix=20 as before, but in this case only one of the input wires are activated. = This=20 particular table may be referred to as the recoding function table.

The twelve flip-flops operating the four function = tables used=20 in selecting a Selectron position, and the six flip-flops operating the = function=20 table used for decoding the order, are referred to as the Function = Table=20 Register, FR.

6.4. Let us consider next the process of transferring = a pair of=20 orders from the Selectrons to the control. These orders first go into = SR. The=20 order which is to be used next may be transferred directly into FR. The = second=20 order of the pair must be removed from SR (since SR may be used when the = first=20 order is executed), but cannot as yet be placed in FR. Hence a temporary = storage=20 is provided for it. The storage means is called the Control Register, = CR,=20 and consists of 20 (or possibly 18) flip-flops, capable of receiving a = number=20 from SR and transmitting a number to FR.

As already stated (6.1), the control must know the = location of=20 the pair of orders it is to get from the Selectron memory. Normally this = location will be the one following the location of the two orders just = executed.=20 That is, until it receives an order to do otherwise, the control will = take its=20 orders from the Selectrons in sequence. Hence the order location may be=20 remembered in a twelve stage binary counter (one capable of counting=20 212) to which one unit is added whenever a pair of orders is=20 executed. This counter is called the Control Counter, CC.

The details of the process of obtaining a pair of = orders from=20 the Selectron are thus as follows: The contents of CC are copied into = FR, the=20 proper Selectron location is selected, and the contents of the = Selectrons are=20 transferred to SR. FR is then cleared, and the contents of SR are = transferred to=20 it and CR. CC is advanced by one unit so the control will be prepared to = select=20 the next pair of orders from the memory. (There is, however, an = exception from=20 this last rule for the so-called transfer orders, cf. 3.5. This may feed = CC in a=20 different manner, cf. the next paragraph below.) First the order in FR = is=20 executed and then the order in CR is transferred to FR and executed. It = should=20 be noted that all these operations are directed by the control = itself-not only=20 the operations specified in the control words sent to FR, but also the = automatic=20 operations required to get the correct orders there.

Since the method by means of which the control takes = order=20 pairs in sequence from the memory has been described, it only remains to = consider how the control shifts itself from one sequence of control = orders to=20 another in accordance with the operations described in 3.5. The = execution=20 of these operations is relatively simple. An order calling for one of = these=20 operations contains the twelve digit specification of the position to = which the=20 control is to be switched, and these digits will appear in the left-hand = twelve=20 flip-flops of FR. All that is required to shift the control is to = transfer the=20 contents of these flip-flops to CC. When the control goes to the = Selectrons for=20 the next pair of orders it will then go to the location specified by the = number=20 so transferred. In the case of the unconditional transfer, the transfer = is made=20 automatically; in the case of the conditional transfer it is made only = if the=20 sign counter of the Accumulator registers zero.

6.5. In this report we will discuss only the = general method=20 by means of which the control will execute specific orders, leaving the = details=20 until later. It has already been explained (5.5) that when a = circuit is=20 to be designed to accomplish a particular elementary operation (such as=20 addition), a choice must be made between a

static type and a dynamic type circuit. When the = design of the=20 control is considered, this same choice arises. The function of the = control is=20 to direct a sequence of operations which take place in the various = circuits of=20 the computer (including the circuits of the control itself). Consider = what is=20 involved in directing an operation. The control must signal for the = operation to=20 begin, it must supply whatever signals are required to specify that = particular=20 operation, and it must in some way know when the operation has been = completed so=20 that it may start the succeeding operation. Hence the control circuits = must be=20 capable of timing the operations. It should be noted that timing is = required=20 whether the circuit performing the operation is static or dynamic. In = the case=20 of a static type circuit the control must supply static control signals = for a=20 period of time sufficient to allow the output voltages to reach the = steady-state=20 condition. In the case of a dynamic type circuit the control must send = various=20 pulses at proper intervals to this circuit.

If all circuits of a computer are static in = character, the=20 control timing circuits may likewise be static, and no pulses are needed = in the=20 system. However, though some of the circuits of the computer we are = planning=20 will be static, they will probably not all be so, and hence pulses as = well as=20 static signals must be supplied by the control to the rest of the = computer.=20 There are many advantages in deriving these pulses from a central = source, called=20 the clock. The timing may then be done either by means of = counters=20 counting clock pulses or by means of electrical delay lines (an RC = circuit is=20 here regarded as a simple delay line). Since the timing of the entire = computer=20 is governed by a single pulse source, the computer circuits will be said = to=20 operate as a synchronized system.

The clock plays an important role both in detecting = and in=20 localizing the errors made by the computer. One method of checking which = is=20 under consideration is that of having two identical computers which = operate in=20 parallel and automatically compare each other's results. Both machines = would be=20 controlled by the same clock, so they would operate in absolute = synchronism. It=20 is not necessary to compare every flip-flop of one machine with the=20 corresponding flip-flop of the other. Since all numbers and control = words pass=20 through either the Selectron register or the accumulator soon before or = soon=20 after they are used, it suffices to check the flip-flops of the = Selectron=20 register and the flip-flops of the accumulator which hold the number = registered=20 there; in fact, it seems possible to check the accumulator only (cf. the = end of=20 6.6.2). The checking circuit would stop the clock whenever a difference=20 appeared, or stop the machine in a more direct manner if an asynchronous = system=20 is used. Every flip-flop of each computer will be located at a = convenient place.=20 In fact, all neons will be located on one panel, the corresponding neons = of the=20 two machines being placed in parallel rows so that one can tell at a = glance=20 (after the machine has been stopped) where the discrepancies are.

The merits of any checking system must be weighed = against its=20 cost. Building two machines may appear to be expensive, but since most = of the=20 cost of a scientific computer lies in development rather than = production, this=20 consideration is not so important as it might seem. Experience may show = that for=20 most problems the two machines need not be operated in parallel. Indeed, = in most=20 cases purely mathematical, external checks are possible: Smoothness of = the=20 results, behavior of differences of various types, validity of suitable=20 identities, redundant calculations, etc. All of these methods are = usually=20 adequate to disclose the presence or absence of error in toto; = their=20 drawback is only that they may not allow the detailed diagnosing and = locating of=20 errors at all or with ease. When a problem is run for the first time, so = that it=20 requires special care, or when an error is known to be present, and has = to be=20 located-only then will it be necessary as a rule, to use both machines = in=20 parallel. Thus they can be used as separate machines most of the time. = The=20 essential feature of such a method of checking lies in the fact that it = checks=20 the computation at every point (and hence detects transient errors as = well as=20 steady-state ones) and stops the machine when the error occurs so that = the=20 process of localizing the fault is greatly simplified. These advantages = are only=20 partially gained by duplicating the arithmetic part of the computer, or = by=20 following one operation with the complement operation (multiplication by = division, etc.), since this fails to check either the memory or the = control=20 (which is the most complicated, though not the largest, part of the=20 machine).

The method of localizing errors, either with or = without a=20 duplicate machine, needs further discussion. It is planned to design all = the=20 circuits (including those of the control) of the computer so that if the = clock=20 is stopped between pulses the computer will retain all its information = in=20 flip-flops so that the computation may proceed unaltered when the clock = is=20 started again. This principle has already demonstrated its usefulness in = the=20 ENIAC. This makes it possible for the machine to compute with the clock=20 operating at any speed below a certain maximum, as long as the clock = gives out=20 pulses of constant shape regardless of the spacing between pulses. In=20 particular, the spacing between pulses may be made indefinitely large. = The clock=20 will be provided with a mode of operation in which it will emit a single = pulse=20 whenever instructed to do so by the operator. By means of this, the = operator can=20 cause the machine to go through an operation step by step, checking the = results=20 by means of the indicating-lamps connected to the flip-flops. It will be = noted=20 that this design principle does not exclude the use of delay lines to = obtain=20 delays as long as these

Section 1 Processors with one address per = instruction

are only used to time the constituent operations of a = single=20 step, and have no part in determining the machine's operating repetition = rate.=20 Timing coincidences by means of delay lines is excluded since this = requires a=20 constant pulse rate.

6.6. The orders which the control understands may be = divided=20 into two groups: Those that specify operations which are performed = within the=20 computer and those that specify operations involved in getting data into = and out=20 of the computer. At the present time the internal operations are more = completely=20 planned than the input and output operations, and hence they will be = discussed=20 more in detail than the latter (which are treated briefly in 6.8). The = internal=20 operations which have been tentatively adopted are listed in Table 1. It = has=20 already been pointed out that not all of these operations are logically = basic,=20 but that many can be programmed by means of others. In the case of some = of these=20 operations the reasons for building them into the control have already = been=20 given. In this section we will give reasons for building the other = operations=20 into the control and will explain in the case of each operation what the = control=20 must do in order to execute it.

In order to have the precise mathematical meaning of = the=20 symbols which are introduced in what follows clearly in mind, the reader = should=20 consult the table at the end of the report for each new symbol, in = addition to=20 the explanations given in the text.

6.61. Throughout what follows S(x) will denote the = memory=20 location No. x in the Selectron. Accordingly the x which appears in S(x) = is a=20 12-digit binary, in the sense of 6.2. The eight addition operations = [S(x) =AE Ac +, S(x) =AE=20 Ac-, S(x) =AE Ah = +, S(x)=20 =AE Ah -, S(x)=AE Ac +=20 M, S(x)=AE Ac- M, S(x)=AE Ah + M, S(x)=AE Ah -=20 M] involves the following possible four steps:

First: Clear SR and transfer into it the number = at=20 S(x).

Second: Clear Ac if the order contains the symbol = C;=20 do not clear Ac if the order contains the symbol h.

Third: Add the number in SR or its negative (i.e. = in our=20 present system its complement with respect to 21) into Ac. If = the=20 order does not contain the symbol M, use the number in SR or its = negative=20 according to whether the order contains the symbol + or - . If the order = contains the symbol M, use the number in SR or its negative according to = whether=20 the sign of the number in SR and the symbol + or - in the = order do=20 or do not agree.

Fourth: Perform a complete carry. Building the = last four=20 addition operations (those containing the symbol M) into the control is = fairly=20 simple: It calls only for one extra comparison (of the sign in SR and = the + or=20 - in the order, cf. the third step above), and it = requires,=20 therefore, only a few tubes more than required for the first four = addition=20 operations (those not containing the symbol M). These facts would seem = of=20 themselves to justify adding the operations in question: plus and minus = the=20 absolute value. But it should be noted that these operations can be = programmed=20 out of the other operations of Table 1 with correspondingly few orders = (three=20 for absolute value and five for minus absolute value), so that some = further=20 justification for building them in is required. The absolute value order = is=20 frequently in connection with the orders L and R (see = 6.6.7),=20 while the minus absolute value order makes the detection of a zero very = simple=20 by merely detecting the sign of - =F4=20 N=F4 . (If - =F4 = N=F4 =B3 0 then N = =3D 0.)

6.6.2. The operation of S(x) =AE=20 R involves the following two steps:

First: Clear SR, and transfer S(x) to it.

Second: Clear AR and add the number in the = Selectron=20 register into it. The operation of R =AE = Ac=20 merits more detailed discussion, since there are alternative ways of = removing=20 numbers from AR. Such numbers could be taken directly to the Selectrons = as well=20 as into Ac, and they could be transferred to Ac in parallel, in = sequence, or in=20 sequence parallel. It should be recalled that while most of the numbers = that go=20 into AR have come from the Selectrons and thus need not be returned to = them, the=20 result of a division and the right-hand 39 digits of a product appear in = AR.=20 Hence while an operation for withdrawing a number from AR is required, = it is=20 relatively infrequent and therefore need not be particularly fast. We = are=20 therefore considering the possibility of transferring at least partially = in=20 sequence and of using the shifting properties of Ac and of AR for this.=20 Transferring the number to the Selectron via the accumulator is also = desirable=20 if the dual machine method of checking is employed, for it means that = even if=20 numbers are only checked in their transit through the accumulator, = nevertheless=20 every number going into the Selectron is checked before being placed = there.

6.6.3. The operation S(x) =B4 R =AE Ac involves the following six = steps:

First: Clear SR and transfer S(x) (the = multiplicand) into=20 it.

Second: Thirty-nine steps, each of which consist = of the two=20 following parts: (a) Add (or rather shift) the sign digit of SR into the = partial=20 product in Ac, or add all but the sign digit of SR into the partial = product in=20 Ac-depending upon whether the right-most digit in AR is 0 or 1-and = effect the=20 appropriate carries. (b) Shift Ac and AR to the right, fill the sign = digit of Ac=20 with a 0 and the digit of AR immediately right of the sign digit = (positional=20 value 2') with the previously right-most digit of Ac. (There are ways to = save=20 time by merging these two operations when the right-most digit in Ar is = 0, but=20 we will not discuss them here more fully.)

Third: If the sign digit in SR is 1 (i.e. -), = then inject a=20 carry

into the right-most stage of Ac and place a 1 into = the sign=20 digit of Ac.

Fourth: If the original sign digit of AR is 1 = (i.e. -),=20 then subtract the contents of SR from Ac.

Fifth: If a partial carry system was employed in = the main=20 process, then a complete carry is necessary at the end.

Sixth: The appropriate round-off must be = effected. (Cf.=20 Chapter 9, Part II, for details, where it is also explained how the sign = digit=20 of the Arithmetic register is treated as part of the round-off = process.)

It will be noted that since any number held in Ac at = the=20 beginning of the process is gradually shifted into AR, it is impossible = to=20 accumulate sums of products in Ac without storing the various products=20 temporarily in the Selectrons. While this is undoubtedly a disadvantage, = it=20 cannot be eliminated without constructing an extra register, and this = does not=20 at this moment seem worthwhile.

On the other hand, saving the right-hand 39 digits of = the=20 answer is accomplished with very little extra equipment, since it means=20 connecting the 2-39 stage of Ac to the=20 2-1 stage of AR during the shift = operation. The=20 advantage of saving these digits is that it simplifies the handling of = numbers=20 of any number of digits in the computer (cf. the last part of 5.12). Any = number=20 of 39k binary digits (where k is an integer) and sign can be = divided into=20 k parts, each part being placed in a separate Selectron position. = Addition and subtraction of such numbers may be programmed out of a = series of=20 additions or subtractions of the 39-digit parts, the carry-over being = programmed=20 by means of Cc =AE S(x) and Cc' =AE S(x) operations. (If the 20 = stage of Ac=20 registers negative after the addition of two 39-digit parts, a = carry-over has=20 taken place and hence 2-39 must be = added to the=20 sum of the next parts.) A similar procedure may be followed in = multiplication if=20 all 78 digits of the product of the two 39-digit parts are kept, as is = planned.=20 (For the details, cf. Chapter 9, Part II.) Since it would greatly = complicate the=20 computer to make provision for holding and using a 78 digit dividend, it = is=20 planned to program 39k digit division in one of the ways described at = the end of=20 5.12.

6.6.4. The operation of division Ac =B8=20 S(x) =AE R involves the following four = steps:

First: Clear SR and transfer S(x) (the divisor) = into=20 it.

Second: Clear AR.

Third: Thirty-nine steps, each of which consists = of the=20 following three parts: (a) Sense the signs of the contents of Ac (the = partial=20 remainder) and of SR, and sense whether they agree or not. (b) Shift Ac = and AR=20 left. In this process the previous sign digit of Ac is lost. Fill the = right-most=20 digit of Ac (after the shift) with a 0, and the right-most digit of AR = (before=20 the shift) with 0 or 1, depending on whether there was disagreement or = agreement=20 in (a). (c) Add or subtract the contents of SR into Ac, depending on the = same=20 . alternative as above.

Fourth: Fill the right-most digit of AR with a 1, = and=20 change its sign digit.

For the purpose of timing the 39 steps involved in = division a=20 six-stage counter (capable of counting to 26 =3D 64) will be = built into=20 the control. This same counter will also be used for timing the 39 steps = of=20 multiplication, and possibly for controlling Ac when a number is being=20 transferred between it and a tape in either direction (see 6.8.).

6.6.5. The three substitution operations [At =AE S(x), Ap =AE S(x), = and Ap' =AE S(x)] involve transferring all or part of = the number=20 held in Ac into the Selectrons. This will be done by means of gate tubes = connected to the registering flip-flops of Ac. Forty such tubes are = needed for=20 the total substitutions, At =AE S(x). The = partial=20 substitution Ap =AE S(x) and Ap' =AE S(x) requires that the left-hand twelve = digits of the=20 number held in Ac be substituted in the proper places in the left-hand = and=20 right-hand orders, respectively. This may be done by means of extra gate = tubes,=20 or by shifting the number in Ac and using the gate tubes required for At = =AE S(x). (This scheme needs some additional = elaboration, when=20 the order directing and the order suffering the substitution are the two = successive halves of the same word; i.e. when the latter is already in = FR at the=20 time when the former becomes operative in CR, so that the substitution = effected=20 in the Selectrons comes too late to alter the order which has already = reached=20 CR, to become operative at the next step in FR. There are various ways = to take=20 care of this complication, either by some additional equipment or by = appropriate=20 prescriptions in coding. We will not discuss them here in more detail, = since the=20 decisions in this respect are still open.)

The importance of the partial substitution operations = can=20 hardly be overestimated. It has already been pointed out (3.3) that they = allow=20 the computer to perform operations it could not other wise conveniently = perform,=20 such as making use of a function table stored in the Selectron memory.=20 Furthermore, these operations remove a very sizeable burden from the = person=20 coding problems, for they make possible the coding of classes of = problems in=20 contrast to coding each individual problem separately. Because Ap =AE S(x) and Ap' =AE = S(x) are=20 available, any program sequence may be stated in general form (that is, = without=20 Selectron location designations for the numbers being operated on) and = the=20 Selectron locations of the numbers to be operated on substituted = whenever that=20 sequence is used. As an example, consider a general code for nth order=20 integration of m total differential equations for p steps of = independent=20 variable t, formulated in advance. Whenever a prob-

Section 1 Processors with one address per = instruction

lem requiring this rule is coded for the computer, = the general=20 integration sequence can be inserted into the statement of the problem = along=20 with coded instructions for telling the sequence where it will be = located in the=20 memory [so that the proper S(x) designations will be inserted into such = orders=20 as Cu =AE S(x), etc.]. Whenever = this=20 sequence is to be used by the computer it will automatically substitute = the=20 correct values of m, n, p and D t, as = well as=20 the locations of the boundary conditions and the descriptions of the=20 differential equations, into the general sequence. (For the details of = this=20 particular procedure, cf. Chapter 13, Part II.) A library of such = general=20 sequences will be built up, and facilities provided for convenient = insertion of=20 any of these into the coded statement of a problem (cf. 6.8.4). When = such a=20 scheme is used, only the distinctive features of a problem need be = coded.

6.6.6. The manner in which the control shift = operations=20 [Cu=AE S(x), Cu' =AE=20 S(x), Cc =AE S(x), and = Cc' =AE S(x)] are realized has been = discussed in 6.4 and=20 needs no further comment.

6.6.7. One basic question which must be decided = before a=20 computer is built is whether the machine is to have a so-called floating = binary=20 (or decimal) point. While a floating binary point is undoubtedly very = convenient=20 in coding problems, building it into the computer adds greatly to its = complexity=20 and hence a choice in this matter should receive very careful attention. = However, it should first be noted that the alternatives ordinarily = considered=20 (building a machine with a floating binary point vs. doing all = computation with=20 a fixed binary point) are not exhaustive and hence that the arguments = generally=20 advanced for the floating binary point are only of limited validity. = Such=20 arguments overlook the fact that the choice with respect to any = particular=20 operation (except for certain basic ones) is not between building it = into the=20 computer and not using it at all, but rather between building it into = the=20 computer and programming it out of operations built into the computer. = (One=20 short reference to the floating binary point was made in = 5.13.)

Building a floating binary point into the computer = will not=20 only complicate the control but will also increase the length of a = number and=20 hence increase the size of the memory and the arithmetic unit. Every = number is=20 effectively increased in size, even though the floating binary point is = not=20 needed in many instances. Furthermore, there is considerable redundancy = in a=20 floating binary point type of notation, for each number carries with it = a scale=20 factor, while generally speaking a single scale factor will suffice for = a=20 possibly extensive set of numbers. By means of the operations already = described=20 in the report a floating binary point can be programmed. While = additional memory=20 capacity is needed for this, it is probably less than that required by a = built-in floating binary point since a different scale factor does not = need to=20 be remembered for each number.

To program a floating binary point involves detecting = where the=20 first zero occurs in a number in Ac. Since Ac has shifting facilities = this can=20 best be done by means of them. In terms of the operations previously = described=20 this would require taking the given number out of Ac and performing a = suitable=20 arithmetical operation on it: For a (multiple) right shift a = multiplication, for=20 a (multiple) left shift either one division, or as many doublings (i.e.=20 additions) as the shift has stages. However, these operations are = inconvenient=20 and time-consuming, so we propose to introduce two operations (L = and=20 R) in order that this (i.e. the single left and right shift) can = be=20 accomplished directly. These operations make use of facilities already = present=20 in Ac and hence add very little equipment to the computer. It should be = noted=20 that in many instances a single use of L and possibly of R = will=20 suffice in programming a floating binary point. For if the two factors = in a=20 multiplication have no superfluous zeros, the product will have at most = one=20 superfluous zero (if 1/2=A3 X < 1 and = 1/2 =A3 Y < 1, then 1/4 =A3 XY=20 < 1). This is similarly true in division (if 1/4 =A3 X < 1/2 and 1/2 =A3 Y < 1, then 1/4 <X/Y < 1). In = addition and=20 subtraction any numbers growing out of range can be treated similarly. = Numbers=20 which decrease in these cases, i.e. develop a sequence of zeros at the=20 beginning, are really (mathematically) losing precision. Hence it is = perfectly=20 proper to omit formal readjustments in this event. (Indeed, such a true = loss of=20 precision cannot be obviated by any formal procedure, but, if at all, = only by a=20 different mathematical formulation of the problem.)

6.7. Table 1 shows that many of the operations which = the=20 control is to execute have common elements. Thus addition, subtraction,=20 multiplication and division all involve transferring a number from the=20 Selectrons to SR. Hence the control may be simplified by breaking = some of=20 the operations up into more basic ones. A timing circuit will be = provided for=20 each basic operation, and one or more such circuits will be involved in = the=20 execution of an order. The exact choice of basic operations will depend = upon how=20 the arithmetic unit is built.

In addition to the timing circuits needed for = executing the=20 orders of Table 1, two such circuits are needed for the automatic = operations of=20 transferring orders from the Selectron register to CR and FR, and for=20 transferring an order from CR to FR. In normal computer operation these = two=20 circuits are used alternately, so a binary counter is needed to remember = which=20 is to be used next. In the operations Cu' =AE = S(x)=20 and Cc=AE S(x) the first order of a pair is = ignored, so=20 the binary counter must be altered accordingly.

The execution of a sequence of orders involves using = the=20 various

timing circuits in sequence. When a given timing = circuit has=20 completed its operation, it emits a pulse which should go to the timing = circuit=20 to be used next. Since this depends upon the particular operation being=20 executed, these pulses are routed according to the signals received from = the=20 decoding and recoding function tables activated by the six binary digits = specifying an order.

6.8. In this section we will consider what must be = added to the=20 control so that it can direct the mechanisms for getting data into and = out of=20 the computer and also describe the mechanisms themselves. Three = different kinds=20 of input-output mechanisms are planned.

First: Several magnetic wire storage units = operated by=20 servo-mechanisms controlled by the computer.

Second: Some viewing tubes for graphical = portrayal of=20 results.

Third: A typewriter for feeding data directly = into the=20 computer, not to be confused with the equipment used for preparing and = printing=20 from magnetic wires. As presently planned the latter will consist of = modified=20 Teletypewriter equipment, cf. 6.8.2 and 6.8.4.

6.8.1. Since there already exists a way of = transferring numbers=20 between the Selectrons and Ac, therefore Ac may be used for transferring = numbers=20 from and to a wire. The latter transfer will be done serially and will = make use=20 of the shifting facilities of Ac. Using Ac for this purpose eliminates = the=20 possibility of computing and reading from or writing on the wires=20 simultaneously. However, simultaneous operation of the computer and the=20 input-output

Section 1 Processors with one address per = instruction

organ requires additional temporary storage and = introduces a=20 synchronizing problem, and hence it is not being considered for the = first=20 model.

Since, at the beginning of the problem, the computer = is empty,=20 facilities must be built into the control for reading a set of numbers = from a=20 wire when the operator presses a manual switch. As each number is read = from a=20 wire into Ac, the control must transfer it to its proper location in the = Selectrons. The CC may be used to count off these positions in sequence, = since=20 it is capable of transmitting its contents to FR. A detection circuit on = CC will=20 stop the process when the specified number of numbers has been placed n = the=20 memory, and the control will then be shifted to the orders located in = the first=20 position of the Selectron memory.

It has already been stated that the entire memory = facilities of=20 the wires should be available to the computer without human = intervention. This=20 means that the control must be able to select the proper set of numbers = from=20 those going by. Hence additional orders are required for the code. Here, = as=20 before, we are faced with two alternatives. We can make the control = capable of=20 executing an order of the form: Take numbers from positions p to p + = s on=20 wire No. k and place them in Selectron locations v to v = + s.=20 Or we can make the control capable of executing some less = complicated=20 operations which, together with the already given control orders, are = sufficient=20 for programming the transfer operation of the first alternative. Since = the=20 latter scheme is simpler we adopt it tentatively.

The computer must have some way of finding a = particular number=20 on a wire. One method of arranging for this is to have each number carry = with it=20 its own location designation. A method more economical of wire memory = capacity=20 is to use the Selectron memory facilities to remember the position of = each wire.=20 For example, the computer would hold the number t1 = specifying=20 which number on the wire is in position to be read. If the control is = instructed=20 to read the number at position p1 on this wire, = it will=20 compare p1 with t1 and if they = differ, cause=20 the wire to move in the proper direction. As each number on the wire = passes by,=20 one unit is added or subtracted to t1 and the = comparison=20 repeated. When p1 =3D t1 numbers will be = transferred=20 from the wire to the accumulator and then to the proper location in the = memory.=20 Then both t1 and p1 will be increased by 1, = and the=20 transfer from the wire to accumulator to memory repeated. This will be = iterated,=20 until t1 + s and p1 + s are reached, = at=20 which time the control will direct the wire to stop.

Under this system the control must be able to execute = the=20 following orders with regard to each wire: Start the wire forward, start = the=20 wire in reverse, stop the wire, transfer from wire to Ac, and transfer = from Ac=20 to wire. In addition, the wire must signal the control as each digit is = read and=20 when the end of a number has been reached. Conversely, when recording is = done=20 the control must have a means of timing the signals sent from Ac to the = wire,=20 and of counting off the digits. The 26 counter used = for=20 multiplication and division may be used for the latter purpose, but = other timing=20 circuits will be required for the former.

If the method of checking by means of two computers = operating=20 simultaneously is adopted, and each machine is built so that it can = operate=20 independently of the other, then each will have a separate input-output=20 mechanism. The process of making wires for the computer must then be = duplicated,=20 and in this way the work of the person making a wire can be checked. = Since the=20 wire servomechanisms cannot be synchronized by the central clock, a = problem of=20 synchronizing the two computers when the wires are being used arises. It = is=20 probably not practical to synchronize the wire feeds to within a given = digit,=20 but this is unnecessary since the numbers coming into the two organs Ac = need not=20 be checked as the individual digits arrive, but only prior to being = deposited in=20 the Selectron memory.

6.8.2. Since the computer operates in the binary = system, some=20 means of decimal-binary and binary-decimal conversions is highly = desirable.=20 Various alternative ways of handling this problem have been considered. = In=20 general we recognize two broad classes of solutions to this = problem.

First: The conversion problems can be regarded as = simple=20 arithmetic processes and programmed as sub-routines out of the orders = already=20 incorporated in the machine. The details of these programs together with = a more=20 complete discussion are given fully in Chapter 9, Part II, where = it is=20 shown, among other things, that the conversion of a word takes about = 5=20 msec. Thus the conversion time is comparable to the reading or = withdrawing=20 time for a word- about 2 msec-and is trivial as compared to the solution = time=20 for problems to be handled by the computer. It should be noted that the=20 treatment proposed there presupposes only that the decimal data = presented to or=20 received from the computer are in tetrads, each tetrad being the binary = coding=20 of a decimal digit-the information (precision) represented by a decimal = digit=20 being actually equivalent to that represented by 3.3 binary digits. The = coding=20 of decimal digits into tetrads of binary digits and the printing of = decimal=20 digits from such tetrads can be accomplished quite simply and = automatically by=20 slightly modified Teletype equipment, cf. 6.8.4 below.

Second: The conversion problems can be regarded = as unique=20 problems and handled by separate conversion equipment incorporated = either in the=20 computer proper or associated with the

mechanisms for preparing and printing from magnetic = wires. Such=20 converters are really nothing other than special purpose digital = computers. They=20 would seem to be justified only for those computers which are primarily = intended=20 for solving problems in which the computation time is small compared to = the=20 input-output time, to which class our computer does not belong.

6.8.3. It is possible to use various types of cathode = ray=20 tubes, and in particular Selectrons for the viewing tubes, in which case = programming the viewing operation is quite simple. The viewing = Selectrons can be=20 switched by the same function tables that switch the memory Selectrons. = By means=20 of the substitution operation Ap =AE S(x) and = Ap' =AE S(x). six-digit numbers specifying the abscissa = and=20 ordinate of the point (six binary digits represent a precision of one = part in=20 26 =3D 64, i.e. of about 1.5 per cent which seems reasonable = in such a=20 component) can be substituted in this order, which will specify that a=20 particular one of the viewing Selectrons is to be activated.

6.8.4. As was mentioned above, the mechanisms used = for=20 preparing and printing from wire for the first model, at least, will be = modified=20 Teletype equipment. We are quite fortunate in having secured the full=20 cooperation of the Ordnance Development Division of the National Bureau = of=20 Standards in making these modifications and in designing and building = some=20 associated equipment.

By means of this modified Teletype equipment an = operator first=20 prepares a checked paper tape and then directs the equipment to transfer = the=20 information from the paper tape to the magnetic wire. Similarly a = magnetic wire=20 can transfer its contents to a paper tape which can be used to operate a = teletypewriter. (Studies are being undertaken to design equipment that = will=20 eliminate the necessity for using paper tapes.)

As was shown in 6.6.5, the statement of a new problem = on a wire=20 involves data unique to that problem interspersed with data found on = previously=20 prepared paper tapes or magnetic wires. The equipment discussed in the = previous=20 paragraph makes it possible for the operator to combine conveniently = these data=20 on to a single magnetic wire ready for insertion into the computer.

It is frequently very convenient to introduce data = into a=20 computation without producing a new wire. Hence it is planned to build = one=20 simple typewriter as an integral part of the computer. By means of this=20 typewriter the operator can stop the computation, type in a memory = location=20 (which will go to the FR), type in a number (which will go to Ac and = then be=20 placed in the first mentioned location), and start the computation = again.

6.8.5. There is one further order that the control = needs to=20 execute. There should be some means by which the computer can signal to = the=20 operator when a computation has been concluded, or when the computation = has=20 reached a previously determined point. Hence an order is needed which = will tell=20 the computer to stop and to flash a light or ring a bell.

References

BurkA62a, BurkA62b; CrawP??; GoldH63a, b, c, d;=20 RajcJ43.

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