Computability and Logic

4.2. THE PRODUCTIVITY FUNCTION 43
Figure 4-1. Increasing productivity by 1.
the productivity of the most productive (n + 1)-state machines is at least greater by
1 than the productivity of the most productive n-state machine.
Example 4.6. We can take i = 11. To see this, plug together an n-state machine for
writing a block of n strokes (Example 3.1) with a 12-state machine for doubling the
length of a row of strokes (Example 3.2). Here ‘plugging together’ means superimposing
the starting node of one machine on the halting node of the other: identifying
the two nodes. [Number the states of the first machine 1 through n, and those of the
second machine (n − 1) + 1 through (n − 1) + 12, which is to say n through n + 11.
This is the same process we described in terms of lists of instructions rather than flow
charts in our proof of Theorem 4.2.] The result is shown in Figure 4-2.
Figure 4-2. Doubling productivity.
The result is an (n + 11)-state machine with productivity 2n. Since there may
well be (n + 11)-state machines with even greater productivity, we are not entitled
to conclude that the most productive (n + 11)-state machine has productivity exactly
2n, but we are entitled to conclude that the most productive (n + 11)-state machine
has productivity at least 2n.
So much for the pieces. Now let us put them together into a proof that the function
p is not Turing computable. The proof will be by reductio ad absurdum: we deduce an
absurd conclusion from the supposition that there is a Turing machine computing p.
The first thing we note is that if there is such a machine, call it BB, and the number
of its states is j, then we have
(1)
p(n + 2 j) ≥ p(p(n))
for any n. For given a j-state machine BB computing p, we can plug together an
n-state machine writing a row of n strokes with two replicas of BB as in Figure 4-3.
Figure 4-3. Boosting productivity using the hypothetical machine BB.
The result is an (n + 2 j)-state machine of productivity p(p(n)). Now from
Example 4.5 above it follows that if a < b, then p(a) < p(b). Turning this around,