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Algorithm Galois Theory and P vs NP

Name: Anonymous 2011-04-18 12:24

Galois theory associates to each polynomial a chain of groups, if each group happens to be cyclic then the polynomial is solvable by radicals.

There is also a "differential" Galois theory which associates chains of groups to differential equations, solvability by radicals is replaced by existence of a solution by exponentials and logarithms.

I have invented a new theory of computation called "Algorithmic" Galois theory. It associates to (a restricted syntax of) program specifications a lattice of groups. I apply standard results from Model theory to show that the lack of certain "shapes" inside the graph of groups is exactly equivalent to the existence of a polynomial time algorithm implementing the program specification.

A rather difficult and long calculation of the graph for the subset sum (NP complete) shows that its graph of groups contains certain symbols obstructing polytime algorithms form existing. This results the P=NP problem.

I have also been able to re-prove the "Primes in P" result as an example of my theory. Although algorithms cannot be "extracted" from my theory, it is useful in guiding algorithm design (I have developed several apparently new for basic problems in CS - one of which, matrix multiplication, beats the current best known algorithm).

Name: Anonymous 2011-04-18 13:07

Paper where?

Name: Anonymous 2011-04-18 14:02

Matrix multiplication has been so widely studied until now that it is impossible for anyone to actually improve the asymptotic time of the best algorithm. Come on.

Name: Anonymous 2011-04-18 15:37

>>3
implying the best possible complexity is already known

Name: Anonymous 2011-04-18 15:49

>>4
Are you fucking kidding me?

Name: Anonymous 2011-04-18 16:29

>>5 trolled NP-hard

Name: Anonymous 2011-04-19 5:56

I'm not a group of mathematicians, I'm single mathematician with experience of 1000 mathematicians, I'm single algorithmist with experience of 1000 algorithmists, I'm single tenured professor with tenure of 1000 tenured professors, so you are right, it was proven by a group of mathematicians, but it was only I with experience of 1000 mathematicians.

Name: Anonymous 2011-04-19 7:06

sage: Everyone's doing it!

Name: Anonymous 2011-04-19 7:08

>>8
Go back whence you came, imageboardish creature!

Name: Anonymous 2011-04-19 7:33

>>9
fuck off and die, faggot ,,creature''

Name: Anonymous 2011-04-20 5:45

obviously a wake

Name: Anonymous 2011-04-20 21:55

>Galois theory associates to each polynomial a chain of groups, if each group happens to be cyclic then the polynomial is solvable by radicals.

Each group is cyclic iff the top (biggest) group is cyclic, and having a cyclic splitting field is a sufficient but not necessary condition for solvability by radicals.

At least put some effort into your trolling next time.

Name: Anonymous 2011-04-21 8:00

>>12
For finite groups, an equivalent definition is that a solvable group is a group with a composition series all of whose factors are cyclic groups of prime order

he thinks he knows Galois Theory

Laughing Sussman.jpg

Name: Anonymous 2011-04-21 16:42

>>13
doesn't understand the difference between cyclic and solvable.

Name: Anonymous 2011-04-22 6:21

>>14
doesn't know what a composition series is

Name: Anonymous 2011-04-22 18:17

>>15
doesn't understand the difference between a composition series with cyclic factor groups and (as in OP) a composition series of cyclic groups.

Name: Anonymous 2011-04-24 18:52

>>16
does

Name: Anonymous 2011-04-24 19:12

>>13-16
think this is the imageboards

Name: Anonymous 2011-04-24 20:10

>>18
thinks he's not a huge fucking flaming faggot

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Name: Anonymous 2011-04-26 9:49


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