3 ms·
Hm, but the definition is not fine if we didn't already know the status of the halting problem?
by calf 2y ago
Hm, but the definition is not fine if we didn't already know the status of the halting problem?
- foldr 2y agoIn that scenario I think the definition itself is still fine, technically. It's just that we wouldn't know what value the function had. It would be a bit like defining f(n) = n + the number of hairs Julius Caesar had when he died. There's nothing inherently wrong with the definition. You just can't do much with it.
- mrkeen 2y ago> It's just that we wouldn't know what value the function had. I think there should be a word for that...
- foldr 2y agoI don't think so. It's a pretty uninteresting property of a function whether or not a particular person happens to know its value for a given input. It does not entail that the function is not computable, if that's what you're suggesting.
- karatinversion 2y agoTo tie this back to TFA, even before we knew that the Halting problem was uncomputable, we could have defined f(n) = { 1 if there is a Turing machine with at most n states that solves the Halting problem; 0 otherwise } and we can easily show that f(n) is computable without proving that the Halting problem is undecideable. Viz., f is either constant 0; or equal to a function of the form g_k(n) = { 1 if n >= k; 0 if n < k }, and both the constant 0 function, and all the g_k functions, are computable; thus f is computable.