if this is true, np!=p

tomolesonjr · 06-08-2006, 08:00 PM

too late, it's been under review today. 5:30 california time, still under review. good, bad, normal?

tomolesonjr · 06-10-2006, 01:47 PM

still under review as of noon saturday, the weekend of course. I would love to post it, but I did sign the copyright over to acm. the fourth theorem can be ditched if needed. although, like I mentioned earilier it is an easy fix.

tomolesonjr · 06-10-2006, 04:02 PM

I've looked at it too many times. Hence, why I decided to write it up for ACM. I think the next step is either it gets turned down, or it gets passed on to an associate editor. The Journal of the ACM looks for three things:
1. best paper
2. broad audience
3. effective presentation

I think that is the first round, those three things.

tomolesonjr · 06-11-2006, 05:55 AM

I will say this. I know the odds are against me that I solved np?p. But I've been working with L(M) for six years and this proof has the best opportunity at being a solution. I've went over it this proof over and over, I'm either missing my own error, or it's good!

tomolesonjr · 06-11-2006, 04:39 PM

If it is right. I found something in NP that is intractable. that's how you prove np<>p though.

tomolesonjr · 06-12-2006, 06:16 PM

The article is still under review. which I guess is better than being rejected, so it's wait and see

tomolesonjr · 06-14-2006, 01:12 PM

article is still under review.

if jacm rejects it. I'm going to see if it is fixible and then submit it to Journal of Algorithms, with Dr. David Johnson as an editor. for those of you familiar with Computers and Intractability: A Guide to the Theory of NP-Completeness.

tomolesonjr · 06-15-2006, 03:34 AM

I unsubmitted the article to the Journal of the ACM. I made the corrections and added a little more explanation and submitted it to Transactions on Algorithms(ACM).

tomolesonjr · 06-15-2006, 05:45 PM

ok, dsj just destroyed by paper.
2^n = set of intractable problems.

if you can prove this then NP<>P

there exists M, (M in NP => L(M) in 2^n) and M in NP. Because for M in NP, L(M) is in np. to prove this we have to use the conjugate for every M, .... and prove that is false, then the first part is true.

tomolesonjr · 06-15-2006, 06:00 PM

the statement is (A=>B) and A, which is (B or not A) and A

the conjugate is not A or (not B and A)

for every m... M in 2^n or (L(M) in NP and M in 2^n) Take M=SAT? L(SAT)is in P. SAT in NP not 2^n... the complement of NP. therefore this statement is false.

there exists M, (M in NP => L(M) in 2^n) and M in NP. but L(M) for m in np is in np. np<>P.

logic error?

tomolesonjr · 06-15-2006, 07:08 PM

np<>P because 2^n and np were assumed to not intersect. the above says for a M in NP then L(M) in NP and 2^n(the set of intractable algorithms)

tomolesonjr · 06-16-2006, 07:05 AM

ERROR.

well, now I'm working the other way, towards np=p. I think you can show that L(NP) in np and L(2^n) in 2^n. that is the "length spaces" are closed. it can be shown that L(m U n)=L(m) U L(n). complement doesn't work that way.

tomolesonjr · 06-16-2006, 08:52 AM

L(NP) is closed I believe, and so is L(2^n). that is x in NP <=> L(x) in np.

L(x) closed => np=p and l(x) closed

np=p or L(X) not closed and L(x) closed. true if np=p. then l(x) is closed. t=>f. if np<>p. this is false. l(x) is closed because that statement is true. L(x) is closed whether np=p or not.

tomolesonjr · 06-16-2006, 09:28 AM

Let's take:

There exists x,y such that ((x in np and y in np)=>L(x,y) in 2^n) and (x in np and y in np) where L(x,y) is concatenation, the L.

(L(x,y) in 2^n or (x in 2^n or y in 2^n)) and (x in np and y in np)
the conjugate:
for every x,y, (L(x,y) in np and (x in np and y in np)) or (x in 2^n or y in 2^n))

this statement is true so that

and There exists x,y such that ((x in np and y in np)=>L(x,y) in np) and (x in np and y in np) is false.

it seems that L(x) in np <=> x in np.

tomolesonjr · 06-16-2006, 01:04 PM

ok, is there an intractable function?

I just need one. Take that one called M. let sigma="a", a not in sigma of M. M*=M U sigma+. then M* in 2^n(intractable) but L(M*) accepts something of every length and is in P.

there exists M in 2^n=>L(M) in NP.
the contrapositive is then true:
there exists L(M) in 2^n => m in np. but m in np=> L(M) in np. np<>p