Is /sci able to define if [math]\mathbf{P=NP}?[/math]

>>8815920
Divide NP by P to isolate N
N = undefined

thank me later

>>8815920
P=NP iff N=1

>>8815920
>>8815931
>>8816009
Stale meme.

>>8816064
i agree. i miss all the -1/12 shitposts.

Sure

Entropy doesn't exist, therefor P=NP

>>8816076

Explain why entropy has any holding on whether P is or is not equal to NP.

>>8816095

P=/=NP is a practical problem, no entropy, no problem

>>8816119

If I give you a single fixed instance of an NP-complete problem, there is absolutely zero entropy present. It is a fixed distribution with single-element support.

Explain to me how entropy has any holding on whether or not a deterministic Turing machine is able to correctly decide this instance (while remaining correct for all other instances) in polynomial time.

>>8816120

Sure

Time is relative

No entropy, full relativity, no problem

>>8816009
what if P=0

>>8815920

NP-complete problems are a subset of problems that can be checked in polynomial time for a correct (or incorrect) solution, but presumably cannot be solved in polynomial time.

One such problem is called the subset sum problem. Given a sum S and a set of numbers, find the subset such that all numbers in the subset sum up to S.

It can be checked in polynomial time but there's no efficient way to solve for it in polynomial time as it scales. It supposedly is exponential complexity which makes it 'difficult'.

An important property to note with NP-complete problems is that if a solution is found to one of them, a solution is found to all of them, since they can all be reduced to the same problem. This means any one solution for the subset sum problem, the hamiltonian path, the boolean satisfaction problem, the clique problem, the knapsack problem, the vertex cover problem, and more, means that a solution is found for every one of them.

>>8816126
0 = 1x0

Still correct

practical comp sci does not care whether P=NP or not because no comp is turing equivalent of infinity.

therefore the P=NP will never be part of the question as a whole. you will always be solving subset of it which is already proven to be always "pseud" P=NP within the bounds.

also the correct answer is P=NP and simultaneously P is not equal NP
and at the same time the axioms for truth value of the operation allow contradictory multi-solution space.

>>8815920
The proof of this is trivial and can be proven by any undergraduate in mathematics. The proof uses an ancient technique employed by the greats such as Carl Gauss, Leonard Euler and even Neil deGrasse Tyson.

Let the proposition [math]\rho[/math] denote the statement, "The series, 1 + 2 + 3 + 4 + .... diverges". Now,

[math](1) \alpha [/math] is true, by common results in analysis, simply using the limit comparison test, moreover it is known that [math]1 + 2 + 3 + \dots = \frac{-1}{12} [/math] (proof of this is a familiar exercise to the reader), and therefore we have [math]\lnot \alpha [/math].

Therefore we have, [math]\alpha \land \lnot \alpha [/math]. Hence by the principle of explosion we have, [math] \alpha \land \lnot \alpha \vdash (P = NP) [/math]. Since we have that [math]\alpha \land \lnot \alpha [/math] is a theorem, we have: [math]P = NP [/math] as a theorem, proof is complete.

There I solved your brainlet problem """"computer""""" """"""scientists""""", go make an app and stop bothering real mathematicians with dumb problems like this