SIMON’S COMPUTING AND REASONING

Now so far the computing registers in Simon are a mystery. We have said that C1, C2, and C3 take in numbers 00, 01, 10, 11, that C4 takes in an operation 00, 01, 10, 11, and that C5 holds the result. What process does Simon use so that he has the correct result in register C5?

Let us take the simplest computing operation first and see what sort of a circuit using relays will give us the result. The simplest computing operation is negation. In negation, a number 00, 01, 10, 11 goes into the C1 register, and the operation 01 meaning negation goes into the C4 register, and the correct result must be in the C5 register. So, first, we note the fact that the C4-2 relay must not be energized, since it contains 0, and that the C4-1 relay must be energized, since it contains 1.

Now the table for negation, with c = -a, is:

ac
0 0
13
22
31

Negation in machine language will be:

ac
00 00
0111
1010
1101

Now if a is in the C1 register and if c is in the C5 register, then negation will be:

C1C5
00 00
0111
1010
1101

But each of these registers C1, C5 will be made up of two relays, the left-hand or 2 relay and the right-hand or 1 relay. So, in terms of these relays, negation will be:

C1-2 C1-1 C5-2 C5-1
0000
0111
1010
1101

Now, on examining the table, we see that the C5-1 relay is energized if and only if the C1-1 relay is energized. So, in order to energize the C5-1 relay, all we have to do is transfer the information from C1-1 to C5-1. This we can do by the circuit shown in [Fig. 6]. (In this and later diagrams, we have taken one more step in streamlining the drawing of relay contacts: the contacts are drawn, but the coils that energize them are represented only by their names.)

Fig. 6. Negation—
right-hand digit.

Fig. 7. Negation—
left-hand digit.

Taking another look at the table, we see also that the C5-2 relay must be energized if and only if:

C1-2
HOLDS:		 AND C1-1
HOLDS:
0 1
1 0

A circuit that will do this is the one shown in [Fig. 7]. In [Fig. 8] is a circuit that will do all the desired things together: give the right information to the C5 relay coils if and only if the C4 relays hold 01.

Fig. 8. Negation circuit.

Let us check this circuit. First, if there is any operation other than 01 stored in the C4 relays, then no current will be able to get through the C4 contacts shown and into the C5 relay coils, and the result is blank. Second, if we have the operation 01 stored in the C4 relays, then the C4-2 contacts will not be energized—a condition which passes current—and the C4-1 contacts will be energized—another condition which passes current—and:

If the number in C1 is:then C1-1:and C1-2:and the C5 relays energized are:
0does not closedoes not closeneither
1closesdoes not closeC5-2, C5-1
2does not closeclosesC5-2 only
3closesclosesC5-1 only

Thus we have shown that this circuit is correct.

We see that this circuit uses more than one set of contacts for several relays (C1-2, C4-1, C4-2); relays are regularly made with 4, 6, or 12 sets of contacts arranged side by side, all controlled by the same pickup coil. These are called 4-, 6-, or 12-pole relays.

Fig. 9. Addition circuit.

Fig. 10. Greater-than circuit.

Circuits for addition, greater than, and selection can also be determined rather easily (see Figs. [9], [10], [11]). (Note: By means of the algebra of logic, referred to in [Chapter 9] and [Supplement 2], the conditions for many relay circuits, as well as the circuit itself, may be expressed algebraically, and the two expressions may be checked by a mathematical process.) For example, let us check that the addition circuit in [Fig. 9] will enable us to add 1 and 2 and obtain 3. We take a colored pencil and draw closed the contacts for C1-1 (since C1 holds 01) and for C2-2 (since C2 holds 10). Then, when we trace through the circuit, remembering that addition is stored as 00 in the C4 relays, we find that both the C5 relays are energized. Hence C5 holds 11, which is 3. Thus Simon can add 1 and 2 and make 3!

Fig. 11. Selection Circuit.