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Raphael
  • Member for 12 years, 9 months
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Why hasn't there been an encryption algorithm that is based on the known NP-Hard problems?
@Acccumulation That's switching the computation model (including randomization), IMHO in a not helpful way. Yes, brute force always works but is prohibitive. The problem with easy instances is that given the public information, deterministic solvers can find the key quickly. That would make the best case easier than guessing, which we should try to avoid.
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How do I show that a DFA accepts only one word?
@User584322 Why don't you try it out? And which step do you suspect to loop?
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Why do Tarjan's and Kosaraju's algorithms for finding strongly connected components have the same running-time complexity?
@theprogrammer Good point. I probably intended it to be vertex count, but input size would also work for the argument.
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What is the difference between expected cost and average cost of an algorithm?
@Stef True. The principles carry over: you can talk about average case (weighted sum over all computations) and expected case (potentially different probabilities per computation).
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What is the correct halt status for an input to a simulating termination analyzer that calls its own termination analyzer?
"All that I have to do to refute these conventional proofs" -- is that your goal? Because it's utterly unclear from your posts what you're trying to achieve there. "how this "impossible" counter-example is correctly decided" -- which you simply haven't done; all you've done is identified a class of functions that can't be $H$. (or tried to, at least).
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What is the correct halt status for an input to a simulating termination analyzer that calls its own termination analyzer?
Ah, we're getting closer. "The undecidability of the halting problem is conventionally proved with a single counter-example." -- nope! The conventional proof by contradiction constructs infinitely many counter-examples, one for each candidate for $H$ (using the terminology from Linz). Well, zero counter-examples, if you're exact, since no $H$ actually fulfills the criterion, but by assumption there would be infinitely many such $H$.
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Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?
"When H recognizes what would be infinitely nested simulation it aborts its simulation of this input and halts" -- No. You don't know anything about H.
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Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?
By the way: even if you managed to come up with some machine that can solve its own halting problem, that does not solve the halting problem.
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Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?
There is no material here to spend more time on. I'm sorry you wasted all that time.
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Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?
No: cs.stackexchange.com/questions/141080/… Hard as it is, at some point you'll have to reconsider the idea that you acually haven't understood the proof.
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Does the Linz Ĥ applied to ⟨Ĥ⟩ correctly transition to its final reject state?
@polcott You don't get to make assumptions on how H works.
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