k99

I am not sure what your knowledge regarding this subject is, but what we were conversing about was the fact that a universal touring machine operated on itself often attempts to trace it's own computations. Combine this with a preprocessor that copies input, and give the composite machine it's own encoding and you have an infinite regress. So you are repeatedly referring to this trivial notion that the halting machine is supposed to be able to operate on itself in the setup the "proof" provides. This doesn't do or mean anything signifigant. You could make halting algorithms and machines all day long and they just wouldn't be able to operate on themselves. This scenario is not what the halting proof is aiming for. The author wants to be able to say that the halting problem is undecidable the power set of Touring Machines, or just undecidable period. Because the entire purpose of a Halting Machine would be to ask the question "What does Machine M do with input R", tracing computations can be considered to be part of it's definition (although it may not be possible to prove this using the language of mathematics) If a Halting Machine traces computations by definition, Touring's proof is utterly trivial and says nothing about the ability to create halting algorithms in general. The scheduler you mentioned is not an algorithm that can tell if the code is going to halt, because it would use inductive reasoning instead of deductive reasoning. IE it runs it, watches what happens, and then could record the results if it chose to. This is a different subject. The halting problem involves deductively reasoning based on the form of the code to see if it is going to halt or not. It can use things that look like inductive reasoning, only if it can also be represented with deductive reasoning (like mathematical induction). This is why the scheduler is not a counterexample to the halting problem. It solves the "Does the machine halt by the kth transition problem" And then guesses that if it doesn't it probably won't halt at all. There is no proof that an algorithm cannot exist to solve the halting problem - only that the algorithm couldn't function on itself in the setup Touring created for his "proof".

Yes. But then all that has been proven is that the Halting Problem is undecidable for T, but may be decidable for T - {H}. But this proof has been sold as the idea that it is impossible in general to make a machine/algorithm that tells when another machine/algorithm will halt/accept. Well you seem to realize in the setup given by the proof that if it did happen to trace computations, an infinite recursion could occur as it copied and traced over and over again. You simply are claiming that you don't think it is going to trace. This is an assumption since it is not proven... and honestly not a very strong one. As the form of the Halting proof is a proof by contradiction, this assumption could be the cause of the contradiction. Your final statement is a similar claim to what the above poster said, IE we simply assume that the halting machine can operate on itself. This limits the result of the proof to saying that there is no halting machine that can retain it's normal function in this setup. There could still be a halting machine that works fine in other cases. I don't fool around with silly programs like Lisp/Prolog, but the situation there is the same as any version of the proof. A succesful universal algorithm (can operate on others) very well may need to trace part of the computation of the input algorithm to come to a conlcusion. This idea that it is only data that it recieves is a bit silly, because if it is designed to take algorithm encodingns as data then it knows exactly how to turn that data back into an algorithm. Any such (that traced computatitons) universal algorithm in the setup the proof provides would infinintely recurse.

Microsoft only doesn't do stuff like that when they are afraid of funding the government for the year.

How does it increase the quality of conversation? If someone really was out of control, it wouldn't really bother anyone other than the vulgarity. Their anger fits would at most make other people laugh at or feel sorry for them. It makes everyone angry when their opponent has a good point, wins the argument, or unfortunately when they resort to specious arguments and sarcasm when they know they cannot win the debate. Neither of these things are ever directly moderated, and only the specious arguments and sarcasm ever should be. That would be hard and would require some logical based evaluation of everyone's arguments. Instead "heavily moderated" just seems to refer to some sort of cult forum where whenever the moderator's opinions are challenged, the moderator's resort to the above behavior and then interpret any response as being "out of control". If you want an example look at any religous forum where the rules state something as innocent sounding as "Behave in a reasonable manner" or "Treat others with respect" and then watch what happens if you question a core belief. Everyone there would go beserk and make sarcastic comments about your mother, then anything you said would be characterized as violating the rules that 30 seconds ago seemed reasonable. Why don't they just label it as a cult forum and say "No questioning of christian ideals allowed?" Because no one would go there to discuss anything knowing that they just censored anyone who expressed something that the backwards reasoning of the forum members couldn't handle. A forum for discussion is supposed to be just that.

There appears to be a clear problem with these proofs. A halting machine attempting to operate upon itself would cease it's normal function and simply always loop. A function which gives 0 if the input function is defined would cease it's normal function and always loop if operated upon itself. The simple reasoning for this is either of these actions would cause an infinite regress whereas normally an infinite regress of this type would not occur... So these two functions could work fine on everything else, but then if they were operated on themselves they would not work. Every Halting problem undecidability proof asks us to assume that such functions could operate on themselves.