Computability and Logic

54 ABACUS COMPUTABILITY
If it is scanning a blank, there can be no more strokes to the right, and it therefore
returns to standard position. But if it is scanning a stroke at that point, it has more
work to do before returning to standard position, for there are more blocks of strokes
to be dealt with, to the right on the tape. These must be shifted one square rightwards,
by erasing the first 1 in each block and filling the blank to the block’s right with a
stroke—continuing this routine until it finds a blank to the right of the last blank it
has replaced by a stroke. At that point there can be no further strokes to the right, and
the machine returns to standard position.
Note that node 1a is needed in case the number r of arguments is 0: in case the
‘function’ that the abacus computes is a number A 0 n . Note, too, that the first s − 1
pairs of nodes (with their efferent arrows) are identical, while the last pair is different
only in that the arrow from node sb to the right is labelled B:1 instead of B:R. What
the general s+ flow chart looks like in the case s = 1 is shown in Figure 5-10.
Figure 5-10. The special case s = 1.
The second step in our method of converting abacus flow charts into equivalent
Turing machine flow charts can now be specified: replace each s− node (with the
two arrows leading from it) by a copy of an s− flow chart having the general pattern
shown in Figure 5-11.
Readers may wish to try to fill in the details of the design for themselves, as an
exercise. (Our design will be given later.) When the first and second steps of the
method have been carried out, the abacus flow chart will have been converted into
something that is not quite the flow chart of a Turing machine that computes the same
function that the abacus does. The chart will (probably) fall short in two respects,
one major and one minor. The minor respect is that if the abacus ever halts, there
must be one or more ‘loose’ arrows in the chart: arrows that terminate in no node.
This is simply because that is how halting is represented in abacus flow charts: by an
arrow leading nowhere. But in Turing-machine flow charts, halting is represented in
a different way, by a node with no arrows leading from it. The major respect is that
in computing A r n (x 1,...,x r ) the Turing machine would halt scanning the leftmost 1
on the tape, but the value of the function would be represented by the nth block of
strokes on the tape.Evenifn = 1, we cannot depend on there being no strokes on the
tape after the first block, so our method requires one more step.
The third step: after completing the first two steps, redraw all loose arrows so they
terminate in the input node of a mop-up chart, which makes the machine (which will
be scanning the leftmost 1 on the tape at the beginning of this routine) erase all but
the first block of strokes if n = 1, and halt scanning the leftmost of the remaining
strokes. But if n ≠ 1, it erases everything on the tape except for both the leftmost