What are the consequences of saying a nondeterministic 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 nondeterministic Turing machine is a tricky concept to grasp. Try some other viewpoints:

NP complete  solvable in nondeterministic polynomial time
By : miziguchi_san
Date : March 29 2020, 07:55 AM
I wish this help you Finding the solution to an NPcomplete problem can be done in polynomial time on a nondeterministic Turing machine. Given a candidate solution of an NPcomplete 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 NPcomplete problems while the latter is just verifying the assignments to your variables.

Running time of a program on deterministic and nondeterministic Turing machine
By : fitrat
Date : March 29 2020, 07:55 AM
around this issue The deterministic Turing Machine simply emulates the nondeterministic 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 twoway branches. If it can take up to k branches, rewrite it as a machine that only does twoway 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.

nondeterministic 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 nondeterministic 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 NPComplete problems] , which we have non deterministic polynomial solutions, but not polynomial solutions.

Is this solvable in polynomial (or pseudopolynomial) 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.

