Second attempt. Apologies for the 1st attempt, didn't check the definitions properly.
Looking forward to useful criticism.
1–6 of 6 posts
Second attempt. Apologies for the 1st attempt, didn't check the definitions properly.
Looking forward to useful criticism.
L_R is not decidable. Deciding if x##y is in L requires simulating y until it produces |x| digits of output. x being anything and y = “while(true) {}” is a counterexample. Another way to look at it is that for any fixed x, "y produces x as its first |x| digits of output" is a nontrivial semantic property of y, so Rice's theorem shows that the property is undecidable.
L_R is decidable by a master algorithm running every y of x#y, e.g. 3.14# runs y_pi(|3.14|), so y runs up to 2 decimals.
The master algorithm runs y(|x|) and then compares the output to x.
It halts correctly for every x#y unless you can provide a counter-example?
post the paper on https://arxiv.org/ or speak with your local university mathematics department. These would be far more qualified to assess a proof than HN.
I only ask because people who do mathematics in isolation are often quite naive and people reacting to such a fantastical proof on social media like here, are not equipped to evaluate it on its merits, (I’m certainly not, and unless you’re an active researching mathematician you’re probably not either)
Please provide a formalization of your argument in any widely-used theorem prover (I'd recommend Lean) configured with options widely accepted to be sound.
BTW, the argument seems trivially bullshit because you say that L is in NP, and then claim that L is not P because "there exists no algorithm that...", but if L is in NP there is of course an exponential-time algorithm for L.