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# NP - Non deterministic polynomial time

By : Hiu Fat Tse
Date : November 20 2020, 11:01 PM
wish helps you In fact, coin flip is about randomness and some people mistakenly describe non-determinism with randomness. Let’s say you have the following problem:
There are n doors, and behind one of them there is prize that you want to find. Now, let’s analyze different approaches:
code :

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## What are the consequences of saying a non-deterministic Turing Machine can solve NP in polynomial time?

By : Kayron Lopes
Date : March 29 2020, 07:55 AM
I wish this helpful for you A non-deterministic Turing machine is a tricky concept to grasp. Try some other viewpoints:

## NP complete - solvable in non-deterministic polynomial time

By : miziguchi_san
Date : March 29 2020, 07:55 AM
I wish this help you Finding the solution to an NP-complete problem can be done in polynomial time on a non-deterministic Turing machine. Given a candidate solution of an NP-complete problem, it can be verified whether it is indeed a solution or not, i.e. checked, in polynomial time on a deterministic Turing machine.
So the difference is in finding a solution and checking a solution. The former usually requires some kind of search for NP-complete problems while the latter is just verifying the assignments to your variables.

## Running time of a program on deterministic and non-deterministic Turing machine

By : fitrat
Date : March 29 2020, 07:55 AM
around this issue The deterministic Turing Machine simply emulates the non-deterministic Turing machine. Each time the NDTM takes a fork, the DTM pushes one branch and takes the other. When it has followed one possible chain for p(S) steps without reaching an accepting state, it backtracks to a previous branch point.
This assumes the NTDM only does two-way branches. If it can take up to k branches, rewrite it as a machine that only does two-way branches, increasing its running time to O(log_2(k) p(S)), which makes it still technically O(p(S)). There's a little bit of sloppiness here. O(2^{x p(S)} is larger than O(2^{p(S)}) if x > 1, so although we can ignore constant factors multiplying the full expression, we can't ignore them in the exponent.

## non-deterministic polynomial solutions over deterministic polynomial solution

By : krittermall
Date : March 29 2020, 07:55 AM
this will help Every deterministic polynomial solution can be translated to a non-deterministic polynomial one [since P is a subset of NP]
We do not know if the oposite is true or not [we do not know if P=NP or P!=NP], so if P!=NP, there are problems [all NP-Complete problems] , which we have non deterministic polynomial solutions, but not polynomial solutions.

## Is this solvable in polynomial (or pseudo-polynomial) time?

By : mat smalec
Date : March 29 2020, 07:55 AM
With these it helps This is at least as complex as the Knapsack problem - consider a case where all balls are red and there is only one red box.
In the case when balls that have the same combination of colors must have the same weights and values consider a case when you have red/blue, red/green, etc. balls and only one red box.