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This paper will illustrate two versions of an algorithm for finding prime number up to N, which give the first version complexity
where c1, c2 are constants, and N is the input dimension, and gives a better result for the second version. The method is based on an equation that expresses the behavior of not prime numbers. With this equation it is possible to construct a fast iteration to verify if the not prime number is generated by a prime and with which parameters. The second method differs because it does not pass other times over a number that has been previously evaluated as not prime. This is possible for a recurrence of not prime number that is (mod 3) = 0. The complexity in this case is better than the first. The comparison is made most with Mathematics than by computer calculation as the number N should be very big to appreciate the difference between the two versions. Anyway the second version results better. The algorithms have been
In this paper we will give an algorithm that in the worst case solve the question about the primality of a number in but that gives better result if the number is not prime (constant operation). Firstly, we will introduce an equation on which are based not prime numbers. With this equation it is possible to deduce the prime number that generates a not prime number and to establish an equation in which if exists a certain integer the number is not prime and therefore vice versa to deduce if it is prime.
This works aims to give an answer to the problem P = NP? The result is
positive with the criteria that solve the Traveling Salesman Problem in polynomial cost of the input size and a proof is given. This problem gets a solution because a polyhedron, with a cut
flower looking, is introduced instead of graph (e.g. tree).