Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 257 — #265
8.2. The Halting Problem 257
be some default string procedure—say one that never halts on anything it is applied
to.
Focusing just on string procedures, the general Halting Problem is to decide,
given strings s and t, whether or not the procedure P s recognizes t. We’ll show
that the general problem can’t be solved by showing that a special case can’t be
solved, namely, whether or not P s recognizes s.
Definition 8.2.1.
No-halt WWD fs j P s applied to s does not haltg D fs … lang.P s /g: (8.3)
We’re going to prove
Theorem 8.2.2. No-halt is not recognizable.
We’ll use an argument just like Cantor’s in the proof of Theorem 8.1.12.
Proof. By definition,
s 2 No-halt IFF s … lang.P s /; (8.4)
for all strings s 2 ASCII .
Now suppose to the contrary that No-halt was recognizable. This means there is
some procedure P s0 that recognizes No-halt, that is,
Combined with (8.4), we get
No-halt D lang.P s0 / :
s 2 lang.P s0 / iff s … lang.P s / (8.5)
for all s 2 ASCII . Now letting s D s 0 in (8.5) yields the immediate contradiction
s 0 2 lang.P s0 / iff s 0 … lang.P s0 / :
This contradiction implies that No-halt cannot be recognized by any string procedure.
So that does it: it’s logically impossible for programs in any particular language
to solve just this special case of the general Halting Problem for programs in that
language. And having proved that it’s impossible to have a procedure that figures
out whether an arbitrary program halts, it’s easy to show that it’s impossible to have
a procedure that is a perfect recognizer for any overall run time property. 3
3 The weasel word “overall” creeps in here to rule out some run time properties that are easy
to recognize because they depend only on part of the run time behavior. For example, the set of
programs that halt after executing at most 100 instructions is recognizable.