@arbdistress5592
i really like his convex optimization course in edx
@1matzeplayer1
Such a good speaker!
@ehfo0
I hope I enjoy CVX as much as Prof Boyd one day!
@degso0
I need to look up more about this professor.
@frankvega5473
P versus NP is considered one of the great open problems of science. This consists in knowing the answer of the following question: Is P equal to NP? This incognita was first mentioned in a letter written by John Nash to the National Security Agency in 1955. Since that date, all efforts to find a proof for this huge problem have failed. I show a solution to that problem as follows: Given a number x and a set S of n positive integers, MINIMUM is the problem of deciding whether x is the minimum of S. We can easily obtain an upper bound of n comparisons: find the minimum in the set and check whether the result is equal to x. Is this the best we can do? Yes, since we can obtain a lower bound of (n – 1) comparisons for the problem of determining the minimum and another obligatory comparison for checking whether that minimum is equal to x. A representation of a set S with n positive integers is a Boolean circuit C, such that C accepts the binary representation of a bit integer i if and only if i is in S. Given a positive integer x and a Boolean circuit C, we define SUCCINCT-MINIMUM as the problem of deciding whether x is the minimum bit integer which accepts C as input. For certain kind of SUCCINCT-MINIMUM instances, the input (x, C) is exponentially more succinct than the cardinality of the set S that represents C. Since we prove that SUCCINCT-MINIMUM is at least as hard as MINIMUM in order to the cardinality of S, then we could not decide every instance of SUCCINCT-MINIMUM in polynomial time. If some instance (x, C) is not in SUCCINCT-MINIMUM, then it would exist a positive integer y such that y < x and C accepts the bit integer y. Since we can evaluate whether C accepts the bit integer y in polynomial time and we have that y is polynomially bounded by x, then we can confirm SUCCINCT-MINIMUM is in coNP. If any single coNP problem cannot be solved in polynomial time, then P is not equal to coNP. Certainly, P = NP implies P = coNP because P is closed under complement, and therefore, we can conclude P is not equal to NP. You could read the details in the link below… https://hal.archives-ouvertes.fr/hal-01509423/document
@mhenimerzouki1285
What are the properties that qualify a problem as np-hard? If you have a problem X what does X need to verify in order to say that it's an np-hard problem ?
@hanwujee4630
Thank for this lecture
@testchannelpleaseignore2452
P!=NP is also a literal insurance policy. B/c if you can prove that P=NP then you can do funky shit like break all encryption
@nossonweissman
Beautiful!
@wenirahayu491
how did you do it can you share with me , thank you
@nhunghong-sr8um
The video sound is pretty good, beyond my imagination
@thorH.
If someone solved P=NP, wouldn’t our whole encryption technology be kinda useless?
@Muuip
I think finding a synergetic way to solve exponential growth problems has more purpose to life then solving (P=NP or P!=NP?)
@thereGoMapo
p=np when n=1
@oretou7290
probably a Russian :D
@candymintz
P = 3.16 not 3.14. NP = RT.155
@WeichenLi
Russians scientists are also hardcore
@christinamariehicks1078
no children ..nieces n nephews..
@Teagle
This is the answer set it to 0 so u get pn=0 think dumb
@yashtibrewal4259