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Old 05-29-2006, 04:30 PM   #16
tomolesonjr
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Registered: Mar 2006
Location: Iowa
Distribution: Suse 10.0
Posts: 70

Original Poster
Rep: Reputation: 15

The question is non determinsitic polynomial time the same as polynomial time?[NP?P]

There are machines, algorithms that resided in np, but only have known exponential algorithms. if you agree I found a NP machine/algorithm in 2^n. Then NP!=P.

I can elaborate on NP if needed.
 
Old 05-29-2006, 04:35 PM   #17
slantoflight
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Registered: Aug 2005
Distribution: Smoothwall
Posts: 283
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Rep: Reputation: 35
Talking
Quote:
Originally Posted by sundialsvcs
Ma <y 3be m0r(e) words?

Unless someone was accidentally delivered a few truckloads of non-character symbols ... this might win the 'most incomprehensible thread' award.

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Old 05-29-2006, 04:53 PM   #18
tomolesonjr
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who ever solves it gets probably a nobel! an 1 million $$
 
Old 05-29-2006, 05:47 PM   #19
AdaHacker
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Quote:
Originally Posted by tomolesonjr
who ever solves it gets probably a nobel! an 1 million $$
Seeing how there is no Nobel prize in mathematics or computer science, I wouldn't count on that. However, it most likely would win you a Turing award. On the down side, the ACM "only" gives out $100K for that.
 
Old 05-29-2006, 06:56 PM   #20
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my error
Quote:
Originally Posted by AdaHacker
Seeing how there is no Nobel prize in mathematics or computer science, I wouldn't count on that. However, it most likely would win you a Turing award. On the down side, the ACM "only" gives out $100K for that.
Thank you for the correction.

The proof I have constructed is rather short, relies on boolean logic and an algorithm I call a length algorithm. L(M)=" on input w nd select x,|x|=|w|. run x on M".

I'm pretty sure on everything but the last statement.
(L(MN) in 2^n=>M in 2^n and n in 2^n) or (L(MN) in 2^n => M in 2^n and N in NP)

go with me that this is false. that means both sides of the or have to be false for the same m,n right? does that make m,n in np, both of them? if so it was assumed that L(MN) in 2^n. but by construction L(MN) in np.

I guess did I prove it or just write what I wished?
 
Old 05-29-2006, 07:39 PM   #21
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(L(MN) in 2^n=>M in 2^n and n in 2^n) or (L(MN) in 2^n => M in 2^n and N in NP)

a=>b is only false if b is false. so on the left A and B is false if one of those is false. so the possibilities for the left to be true are a mix of one in 2^n the other in np, or both in np. on the right.. the and... the possibilities are 2^n and 2^n, but that would make the right side true. and it's an or. only one of the two sides has to be true for the entire statement to be true. so, ok. the right is already a mix of 2^n and np, so if those are the values then the right side is true. that leaves the values of n in np and m in np that are possibilites for making the statement false. and it is known the statement is false.

it's a long statement and room for error. but that is my thought on that statement.
 
Old 06-02-2006, 10:03 AM   #22
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Smile boolean logic
(L(MN) in 2^n=>M in 2^n and n in 2^n) or (L(MN) in 2^n => M in 2^n and N in NP) or (L(MN) in 2^n => M in np and N in 2^n)

is the same as:

for every x, y:
(f(x,y) in AC => x in AC and y in AC) or (f(x,y) in AC => x in AC and y in A)or(f(x,y) in AC => x in A and y in AC)

the complement of that statement is:
there exists x,y such that

((f(x,y) in AC xor (x in AC and y in AC)) and f(x,y) in AC) and
((f(x,y) in AC xor (x in A and y in AC)) and f(x,y) in AC) and
((f(x,y) in AC xor (x in AC and y in A)) and f(x,y) in AC)

if this is true then (L(MN) in 2^n=>M in 2^n and n in 2^n) or (L(MN) in 2^n => M in 2^n and N in NP) or (L(MN) in 2^n => M in np and N in 2^n) is false. then there is M,N in NP such that L(MN) in NP and 2^n
 
Old 06-02-2006, 11:06 AM   #23
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yeah baby?

the assumption is L(MN) in 2^n is true.... when it is false and implication will always be true... so think of the conjucate f(x,y) in AC, AC for A complement. I have to think that over... But if you assume f(x,y) in AC then that complement above is always true????

Tom
 
Old 06-02-2006, 05:56 PM   #24
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Back to:
there exists x,y such that

((f(x,y) in AC xor (x in AC and y in AC)) and f(x,y) in AC) and
((f(x,y) in AC xor (x in A and y in AC)) and f(x,y) in AC) and
((f(x,y) in AC xor (x in AC and y in A)) and f(x,y) in AC)

This is true if f(x,y) in AC if one of the three conditions is met. it's a there exists. L=f.

there exists: x in 2^n and y in 2^n => L(x,y) in 2^n is true. take the conjugate of this statement! it is false.

so (L(MN) in 2^n=>M in 2^n and n in 2^n) or (L(MN) in 2^n => M in 2^n and N in NP) or (L(MN) in 2^n => M in np and N in 2^n) is false.

there is a m in np, n in np such that L(MN) in NP and in 2^n. which were assumed to separate by the assumption np=p. np<>p
 
Old 06-05-2006, 03:14 PM   #25
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article is pending for journal of the acm. 8 pages. 4 theorems.

I know I messed up on the 4th proof, but they should see it is fixable. it's pretty obvious.

Tom
 
Old 06-06-2006, 01:48 AM   #26
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I solved it, Its the Simpsons..........
 
Old 06-06-2006, 09:13 AM   #27
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If you messed up the fourth proof, why didn't you fix it before you wrote an article for the acm?

Threads like this make me whish I had paid attention in my Calc, Linear Algebra, and Comp Sci classes.
 
Old 06-07-2006, 02:00 PM   #28
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I didn't catch it, I had no reviewers.

ok, if you have A=>B. if A is false the statement is false. so I added a stronger condition (A=>B) and A. well in the fourth proof A was c and d. well I wrote ((c and d)=>B) and c, it should of been ((c and d)=>B) and (c and d).

The article is still waiting for admission to acm. 2 days later.

Tom
 
Old 06-07-2006, 07:15 PM   #29
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I believe they are waiting for my consent to turn over the copyright. and they probably get a lot of articles. I got a pdf back with my manuscript and lines numbered, saying peer review.

if they don't accept it, I'll discuss it here.

Tom
 
Old 06-08-2006, 06:33 AM   #30
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Quote:
I know I messed up on the 4th proof, but they should see it is fixable. it's pretty obvious.
Quote:
got a pdf back with my manuscript and lines numbered, saying peer review.
Logic dictates you withdraw your paper and resubmit with corrections.
If logic does not, ethics does (in my books anyway)

PS: you do not need to post & post, you can use the edit button when necessary
 
  


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