This note gives yet another example of an undecidability proof via
problem reduction.

Consider the following problem.

Problem B: Given a Turing machine M', determine if there is any input
           for which M' will halt.

We want to show that this problem is undecidable. 

We will prove this via reduction. That is, given an undecidable
problem A, we will reduce A to B. 

We will reduce the halting problem, which we know is undecidable, to 
problem B.

**NOTE: It is important to perform the reduction in the correct
        direction. This is an easy thing to mess up. It would not have
        been correct to reduce problem B to the halting
        problem. Intuitively, such a reduction shows that the halting
        problem is at least as hard as problem B (we can use an
        algorithm for the halting problem to solve problem
        B). However, to show that B is undecidable we nee to show that
        B is at least as hard as the halting problem.

First lets review the input and desired outputs for our two problems.
The input to the halting problem is a pair (M,w) where M is a machine
and w is an input to M. We want the output to be "yes" if M halts when
applied to w, and otherwise be "no". The input to Problem B is a
single Turing machine M' and the output is "yes" if M halts for some
string and is otherwise "no".

To reduce the halting problem to B we must show how to convert any halting 
problem input (M,w) to an input M' for problem B such that an algorithm 
for problem B when applied to M' will answer "yes" iff M halts on w.

It is relatively easy to construct such an M'. Given (M,w) we will 
construct M' such that it erases its input and then writes w on the tape. 
Next M' simulates M running on w. Notice that M' effectively ignores its 
own input, i.e. it does the same thing regardless of its input. 

From this construction we see that M' will halt on some input iff M halts 
on w. (In particular, M' will halt on all inputs if M halts on w, and 
M' will halt on no inputs if M does not halt on w.)

Thus given an algorithm that correctly decides problem B we can correctly 
decide the halting problem.

Since we know that the halting problem is undecidable we know that
problem B must also be undecidable. 

This completes the proof.