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Travelling Salesman Problem

Name: God 2014-09-17 3:34

I have found a solution. By modeling the problem as a graph represented with an adjacency matrix, I can find the optimum path with only elementary row reduction and matrix multiplication. I will be keeping my discovery proprietary for now to continue searching for bugs, but please sing my praises and congratulate me on this world-changing discovery.

Name: God 2014-09-17 3:35

Oh, this is in O(2n) by the way.

Name: Anonymous 2014-09-17 4:07

Sounds neat
did you drop the nodes with only two edges?

Name: Anonymous 2014-09-17 17:53

>>2

O(2n)

you mean O(n), dipshit

Name: Anonymous 2014-09-17 18:29

>>4
No, stupid, exactly O(2n). You're the type of retard who would call O(n3+n2+log2(n)+n) the same as O(n3) and then wonder why your programs are so much fucking slower than mine are.

Name: Anonymous 2014-09-17 18:37

>>4
YHBT

Name: Anonymous 2014-09-17 19:16

>>5
what's N then?
There's plenty of constant factors inside it, i'm sure. What's special about 2?

Name: Anonymous 2014-09-17 20:41

>>6
This is one particular troll I don't think we had on world4ch.

it's not an improvement though

Name: Anonymous 2014-09-17 22:31

>>8
An empty room finds a voice to fill it.

Name: Anonymous 2014-09-17 22:56

>>7
It's probably doing two passes of N?

Name: Anonymous 2014-09-17 23:38

n/N stands for number and amount. So it does 2n passes or 2 passes. The 2 without n (which signifies it as an amount) could mean something else.

Name: Anonymous 2014-09-18 0:01

N is going to be the number of nodes...
I'm still a little skeptical that op has the algorithm, but i don't think it'd be impossible

Name: God 2014-09-18 3:08

I have further reduced the solution to operate in only O(0.5n). Furthermore, I have generalized the problem class to cover all of the class of NP.

Yes, it is true. You have heard that correctly. Yes, I have verified my proof. I can say without a doubt that P=NP. I have a proof-of-concept that allows for the factoring of integers in O(2). I have also developed a sorting algorithm that can sort in-place in only O(7). I have decided to delay releasing my proof and algorithms to allow the world, specifically security, to catch up to my new discovery.

Name: Anonymous 2014-09-18 3:18

http://xhamster.com/movies/1944110/ad_lib_talk_on_cock_rings_amp_gags_then_cumming_and_eating_it.html

Name: L. A. Calculus !jYCj6s4P.g 2014-09-18 6:52

>>13
O(0.5n)
DAT WUD BE O(n)

O(2)
O(7)
N DAT WUD BE O(1)

AND WAT DA FUCK ARE U TALKIN ABOUT? TIME, SPACE? WORST CASE, BEST CASE?

GET DA FUCK OUTTA MY THRED N GO PLAY WITH UR HTML5, YA FUCKIN BUZZWORDIN' BUSINESSBOI RETOID

Name: Anonymous 2014-09-18 8:35

>>14
Go away, porn-kun.

Name: Anonymous 2014-09-18 8:56

>>12
I have spoken to him and he does indeed have a viable algorithm.

Name: Anonymous 2014-09-18 9:24

>>15
Go back to /g/, turdcutter.

Name: Anonymous 2014-09-18 13:40

>>18

I cut turds with my tight anus.

Name: Anonymous 2014-09-18 18:28

Again this salesman shit? Sure, it could seem slow. But I hand optimized the algorithm in assembly, thus making it O(lg n) fast.

Name: Anonymous 2014-09-18 20:38

>>18
install stallman, /g/entoknight

Name: Anonymous 2014-09-18 23:14

>>20
Cudder???

Name: Anonymous 2014-09-18 23:57

I'm sure OP wanted us to reverse engineer his/her algorithm..
So lets see
n-1 * n/2 total paths is too big...
First quadtree step -> 4 * (n/4 -1 * n/8) + 3
m quadtree steps -> 4^m * (n/4^m - 1 * n/2*4^m) + 4^m-1

Name: Anonymous 2014-09-19 0:18

n=256 4^m=64
64 * 3*2 + (63 ?) << 256*255 distance calcs

Name: Anonymous 2014-09-19 0:45

** 128*255~32000 vs ~450

that's about 70x faster
Don't change these.
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