Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 274 — #282
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Chapter 8
Infinite Sets
Let’s begin by showing how we could use a recognizer for No-halt- to define a
recognizer for No-halt. That is, we will “reduce” the weird problem of recognizing
No-halt to the more understandable problem of recognizing No-halt-. Since there
is no recognizer for No-halt, it follows that there can’t be one for No-halt- either.
Here’s how this reduction would work: suppose we want to recognize when a
given string s is in No-halt. Revise s to be the declaration of a slightly modified
procedure P s 0 which behaves as follows:
P s 0 applied to argument t 2 ASCII , ignores t, and simulates P s applied
to s.
So, if P s applied to s halts, then P s 0 halts on every string it is applied to, and if P s
applied to s does not halt, then P s 0 does not halt on any string it is applied to. That
is,
s 2 No-halt IMPLIES lang.P s 0/ D ;
IMPLIES … lang.P s 0/
IMPLIES s 0 2 No-halt-;
s … No-halt IMPLIES lang.P s 0/ D ASCII
IMPLIES 2 lang.P s 0/
IMPLIES s 0 … No-halt-:
In short,
s 2 No-halt IFF s 0 2 No-halt-:
So to recognize when s 2 No-halt all you need to do is recognize when s 0 2
No-halt-. As already noted above (but we know that remark got by several students,
so we’re repeating the explanation), this means that if No-halt- was recognizable,
then No-halt would be as well. Since we know that No-halt is unrecognizable, then
No-halt- must also be unrecognizable, as claimed.
(a) Conclude that Finite-halt is unrecognizable.
Hint: Same s 0 .
Next, let’s see how a reduction of No-halt to Always-halt would work. Suppose we
want to recognize when a given string s is in No-halt. Revise s to be the declaration
of a slightly modified procedure P s 00 which behaves as follows:
When P s 00 is applied to argument t 2 ASCII , it simulates P s applied
to s for up to jtj “steps” (executions of individual machine instructions).
If P s applied to s has not halted in jtj steps, then the application
of P s 00 to t halts. If P s applied to s has halted within jtj steps, then the
application of P s 00 to t runs forever.