Computability and Logic

52 ABACUS COMPUTABILITY
Figure 5-7. Addition in standard format.
machine A in standard form, then for each register n that we might specify as holding
the result of the computation there are infinitely many functions A r n that we have
specified as computed by the abacus: one function for each possible number r of arguments.
Thus if A is determined by the simplest chart for addition, as in Example 5.2,
with n = 1 and m = 2, we have
A 2 1 (x, y) = x + y
for all natural numbers x and y, but we also have the identity function A 1 1 (x) = x
of one argument, and for three or more arguments we have Ar 1(x 1,...,x r ) = x 1 + x 2 .
Indeed, for r = 0 arguments we may think of A as computing a ‘function’ of a sort,
namely, the number A 0 1
= 0 of strokes in box 1 when the computation halts, having
been started with all boxes (‘except the first r’) empty. Of course, the case is entirely
parallel for Turing machines, each of which computes a function of r arguments in
standard format for each r = 0, 1, 2, ..., the value for 0 being what we called the
productivity of the machine in the preceding chapter.
Having settled on standard formats for the two kinds of computation, we can turn
to the problem of designing a method for converting the flow chart of an abacus A n ,
with n designated as the box in which the values will appear, into the chart of a
Turing machine that computes the same functions: for each r, the Turing machine
will compute the same function A r n of r arguments that the abacus computes. Our
method will specify a Turing-machine flow chart that is to replace each node of type
n+ with its exiting arrow (as on the left in Figure 5-1, but without the entering arrow)
in the abacus flow chart; a Turing-machine flow chart that is to replace each node
of type n− with its two exiting arrows (as on the right in Figure 5-1, again without
the entering arrow); and a mop-up Turing-machine flow chart that, at the end, makes
the machine erase all but the nth block of strokes on the tape and halt, scanning the
leftmost of the remaining strokes.
It is important to be clear about the relationship between boxes of the abacus and
corresponding parts of the Turing machine’s tape. For example, in computing A 4 n (0,
2, 1, 0), the initial tape and box configurations would be as shown in Figure 5-8.
Boxes containing one or two or ...stones are represented by blocks of two or three
or ...strokes on the tape. Single blanks separate portions of the tape corresponding to
successive boxes. Empty boxes are always represented by single squares, which may
be blank (as with R 5 , R 6 , R 7 , ...in the figure) or contain a single 1 (as with R 1 and