It’s not so simple, even finite halting code has to handle the self recursion reversal trick somehow. Aka the finite bounds must exclude the “if H(X,I) loop forever; else halt” code because H would still be broken.

You could try defining H to only work on input machines smaller than itself. However there’s still problems.

If you don’t limit input size (aka the I in H(M,I) does machine M halt on input I), your code would still get tricked if M is just a tiny emulator and the input is the whole self defeating H program. So M and I both must be limited.

Anyway that’s not to say a finite H would work even if the input size is limited. You’d have to prove it’s impossible to self recurse - so you’d need good compression.

That’s all assuming self recursion is the only problem with the halt programs. Maybe it’s not, and it seems difficult to rule out other problems and see how they’d affect finitely bound TMs.

I’m just spitballing here. For a while I was infatuated by the idea of snuffing out self recursion or otherwise somehow making a variant of halting programs possible. But self referential poison seems really, really tough to prove the non existence of.

Anyways, I’m of the attitude that the proof for halting problem undecidability is a dumb trick. I’m simply fine with a halting program that doesn’t work correctly in self referential cases. As above though, that just leaves the whole problem vague aka don’t know about other issues and there’s the problem of actually solving it anyways (aka solving all of mathematics at once).