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Apr 23, 2013 at 18:03 comment added Joey Eremondi I've edited my answer to reflect the flaws Raphael found.
Apr 23, 2013 at 18:02 history edited Joey Eremondi CC BY-SA 3.0
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Apr 23, 2013 at 18:01 comment added Joey Eremondi I guess I should clarify. The idea is that you can't brute-force the halting problem the way you can, say, 3-SAT. Or how, when you run something on a PDA, you can try all possible paths even though it's non-deterministic. But for a TM, you can't because it could run infinitely long, so the set of "things to try" i.e. the possible paths through the program, is not finite.
Apr 23, 2013 at 17:46 comment added Raphael @Steven Yes, but that's an even harder proof. (The set you are referring to is typically denoted with R, the set of recursive(ly decidable) languages.)
Apr 23, 2013 at 17:45 comment added Raphael Also: "Anything with a finite "set of possibilities" to try will be decidable" -- no. The halting problem has two possible answers but is undecidable.
Apr 23, 2013 at 17:45 comment added Steven Ah, that sparks an idea. We know that $NP \subseteq \{L|\ L\ Decidable\}$, so that would imply that if we can show a problem to be in NP (or P for that matter) then it simply follows from that. That could help narrow it down.
Apr 23, 2013 at 17:44 comment added Raphael "If the solution space is countably infinite, then we know that the problem is recognizable." -- Not necessarily. First, the solution space may not be effectively enumerable (Example: "Is there a TM that computes a total function and so that [non-trivial predicate]?"). Second, the decision whether the considered object is a solution may be undecidable itself (Example: "Find a TM that does not halt on 77.").
Apr 23, 2013 at 17:34 history answered Joey Eremondi CC BY-SA 3.0