NP problem algorithm that can solve more than normal algorithms,will it be P=NP for some cases or limiter?

assume that all algorithms can find optimum in problem like travel selesman problem for 25 cities which mean 25! possibility in polynomial time with using power of supercomputer then if there algorithm that can solve for 50 cities which mean 50! possibility with using power of supercomputer this mean it could to close range of many possibility to small number that could solve by like 25 cities

so what this algorithm mean ,p=np in limit of 50 cities, or just it can get answer under limit but it can not solve higher numbers like 100! although it could throw big number of possibility from 100!

How to prove the optimization version problem (whose decision version is NP-complete) can be solved in poly-time iff P=NP?

I have proved the decision version of my problem be $$\mathcal{NP}$$-complete. And I know that if I can solve the optimization version in poly-time, then I can just to compare the obtained minimum (or maximum) with target value in decision version. Thus, the decision version can be solved in poly-time as well. Since the decision version is $$\mathcal{NP}$$-hard, so is the optimization version, i.e., the optimization version is $$\mathcal{NP}$$-hard.

My question is how to prove the converse direction: if the decision version can be solved in poly-time, can the optimization version be solved in poly-time as well?

I in advance thank you for any suggestions!

Longest simple circuit and P=NP relation

Given the following function: $$\:f\left(G,v\right)\:=\:size\:of\:the\:longest\:simple\:circuit\:in\:a\:directed\:graph\:G\:that\:contains\:v$$

Output: Function returns a natural number or 0, which is the largest simple circuit in directed graph G.

However, I don’t understand the following claim:

if it is possible to compute in a polynomial time the function g(G,v) and it is guaranteed that: $$f(G,v) -5 ≤ g(G,v) ≤ f(G,v) +5$$ , then $$P=NP$$.

I don’t understand why. as far as I know, the longest simple circuit can be proved NP-complete using LongestSimplePath(by adding remaining vertices to a new graph G’ and two directed edges, and then running it) or HamCycle(by constructing an instance of longestsimplecycle(H,|V|) based on hamcycle(H) (can be done in polynomial time) and checking if the longest simple cycle equals to |V|(contains hamiltonian cycle) as the basis for the reductions. So if we know that it is possible to compute in polynomial time g(G,v), I don’t understand how the fact that it is guaranteed that $$f(G,v) -5 ≤ g(G,v) ≤ f(G,v) +5$$ helps deduct that P=NP.

There’s probably some trick to it that I cannot see, And would appreciate your help with it. How is it possible to determine that?

Is This A Solution for does P=NP problem?

Thank you all for your input! It is my belief that if we continue finding things wrong with it, we can eventually find the right! This is an updated proof with conversations I’ve been having all across the web. I do appreciate all constructive input. Let me know what you think. 🙂

The Proof of Everything & Nothing (The Solution to Complexity Theory)

Assume The Set of Everything & Call it {EXPSPACETIME}:

Assume The Empty Set {∅}:

Assume non means negation of:

The Time Part:

Assume deterministic time problems {P} is true. This implies non-deterministic time problems {NP} is the negation of {P}. So, {P}=/={NP}. As a result, there is only {P}.

Assume {NP} is true. This implies {P} is the negation of {NP}. As {P} is already assumed true, {P}={NP}.

This creates a paradox where: {P}={NP} & {P}=/={NP}. However, they can’t equal and not equal at the same time. Therefore, they must only be true in different sets. The set {{P}={NP}} & the set {{P}=/={NP}}.

However, there also needs to be another set where the deterministic time paradox is true because both are included in the everything set. Therefore, there are 3 temporal deterministic sets. {{P}={NP}}, but not {{P}=/={NP}}, one where {{P}=/={NP}}, and one where both are true at once {{{P}={NP}},{{P}=/={NP}}} or {EXPTime} for short. The other possibility can’t be true because both {P}={NP} and {P}=/={NP} are false making absolute nothingness, or the set with nothing it {∅}.

The Space Part:

Assume deterministic Space problems {Space} is true. This implies non-deterministic {Space} problems {NSpace} is the negation of {Space}. So, {Space}=/={NSpace}. As a result, there is only {Space}.

Assume {NSpace} is true. This implies {Space} is the negation of {NSpace}. As {Space} is already assumed true, {Space}={NSpace}.

This creates a paradox where: {Space}={NSpace} & {Space}=/={NSpace}. However they can’t equal and not equal at the same time. Therefore, they must only be true in different sets. The set {{Space}={NSpace}} & the set {{Space}=/={NSpace}} are separate.

However, there also needs to be another set where the deterministic space paradox is true because both are included in the everything set. Therefore, there are 3 spatial deterministic sets. One where {{Space}={NSpace}}, one where {{Space}=/={NSpace}}, and one where both are true at once or {{{Space}={NSpace}},{{Space}=/={NSpace}}} or {EXPSpace} for short. The other possibility can’t be true because both {Space}={NSpace} and {Space}=/={NSpace} are false making absolute nothingness, or the set with nothing it {∅}.

The The SpaceTime Part:

The Everything set {EXPSPACETIME} is the union of the 2 subsets {EXPSpace, EXPTime}

The Everything & Empty Set as the union of Everything Set & Empty Set {{EXPSPACETIME},{∅}} which I would like to call {EXPSPACETIME∅}.

However, the union of the set of everything and the empty set is a set. Which means it is it’s own set of {{EXPSPACETIME∅},∅}, or in other words {EXPSPACETIME∅∅}.

This continues ad infinitum, which means the only thing that exists is the set, which is also everything. Therefore {EXPSPACETIME}={EXPSPACETIME∅,…∅n} where ∅n is any number of ∅.

~Michael Barry “I Got Clap But Im Not A Clapper”

Does P=NP? (The big question) [on hold]

If you know much about theoretical computer science, you have heard the question before.

This site has plenty of questions posted about the implications of one answer or another to this question (some examples here), questions about definitions, dead-end approaches, and strategies for answering this question, as well as several question-posts attempting answers to this question, but I couldn’t find the question itself being asked directly.

It seems a little incomplete for a theoretical computer science question-answering community to not have the million-dollar question (at the floor of its value) posted as an open question.

So, is every language accepted by some nondeterministic algorithm in polynomial time is also accepted by some (deterministic) algorithm in polynomial time (within the definition constraints posted here)?

Alternatively, can you write a polynomial-time algorithm which determines whether a given state of what purports to be a Minesweeper game is logically consistent, or prove that no such algorithm can be written?

Try to avoid known answer attempts that don’t really work.

(“Accepted answer” will be awarded after the Millennium Prize, assuming this post author is still alive and able to do so.)

#P=NP: All satisfying solutions are valid answers by programs for NP

In fact, the implication looks like an equivalence too.

All satisfying solutions to a boolean formula
are valid answers by programs for NP
is equivalent to saying
the class #P equals the class NP.

Any agreement? Looks trivial. However, complexity classes are untrivial, so… negating both sides of the equivalence also produces an equivalence: Not all solutions are valid is equivalent to #P!=NP.

I have a program for counting solutions to boolean formulas, and after some years of experience, I observed that SAT coNP and #P are roughly equivalent in time complexity. I have published previously a related result called #P=#Q, where the number of satisfying assignments equals the number of valid quantifications of any original boolean formula (Satisfiability 2002, Introduction to Qspace). My program runs fast on small to moderate sizes, and when #P is linear, I can decide all qbfs of the original formula.

My theoretical position is:
coNP=NP=#P=#Q=qbf=allQBFs=?Exp
with #P=#Q being the central thesis.

source code for program bob is available, send me email at gmail, pehoushek1. Sincerely, Daniel Pehoushek, 1983 GREs: 2380/2400