Proof of the hypothesis Edmonds's, not polynomial of NPC-problems and classification of the problems with polynomial certificates

Proof of the hypothesis Edmonds's, not polynomial of NPC-problems and classification of the problems with polynomial certificates Abstract We show that the affirmation P ⊆ N P (in computer science) erroneously and we prove the justice of the hypotesis J.Edmonds's P = N P. We show further that all the N P-complete problems is not polynomial and we give the classification of the problems with the polynomial certificates. concept of complexity class P. Definition 1 [1,2]. A language L belong to P if there is an algorithm A that decide L in polynomial time (≤ O(n k)) for a constant k. Class of problems P is called polynomial. According [3] J.Edmonds has entered also the complexity class NP. This is the class of the problems (langages) that can be verified by a polynomial-time algorithm. Definition 2 [3]. A langage L belongs to NP if there exists a two-input polynomial-time algorithm A and such polynomial p(x) with whole coefficients that L = {x ∈ {0, 1} * : there exists a certificate y with | y |≤ p(| x |) and A(x, y) = 1}. In this case we say that the algorithm A verifies the language L in polynomial time. length of the certificate not polynomial from length x, then L / ∈ NP. J. Edmonds has conjectured also that P = NP. In [4] we builded one class of the polynomial problems with not polynomial certificates. According of the considerations see above immediately it follows that P = NP. In [3,5,6] authors consider only two version P ⊆ NP In 1971 S.A.Cook has put the question: " whether can the verification of correctness of the decision of a problem be more long than the decision itself independently of algorithm of verification? " This problem have a relation to cryptography. In other formulation this problem look so: whether can to build a cipher such that his decipher algorithmically more complicated than find of cipher?