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P=NP solved in the affirmative?

+ 2 like - 0 dislike
932 views

The question whether $P=NP$ is one of the famous unsolved problems, one of the Clay Millennium problems (with a prize each worth 1 Million Dollars). Apart from its fundamental relevance for theory and practice of computer science, its positive solution may have very important consequences for the tractability of many computations in physics. It may also shed a new light on the future of quantum computing.

After many years of very hard work, I finally found the correct approach to simplifying the problem. I was able to reduce the problem to the significantly simpler question of whether $P=0$ or $N=1$.

My most successful attempt to prove $P=0$ (which would have settled the conjecture in the affirmative) was by noting that for any $X$, we have $P(X−X)=PX−PX=0.$ Thus after division by $X−X$, we find that $P=0.$ Unfortunately, this argument proved to have a small gap, since for the argument to work, the divisor must be nonzero. Thus I would have to find an $X$ such that $X−X$ is nonzero. Unfortunately again, I could prove that this is never the case. Thus I couldn't close this gap in my argument.

If any of you is able to close the gap, we could share the prize....

asked Apr 1, 2016 in Chat by Arnold Neumaier (12,570 points) [ revision history ]
recategorized Apr 2, 2016 by Arnold Neumaier

Slightly related MO thread: http://mathoverflow.net/q/235008/30967

@dilaton: You could cross-link there, directly as a comment to the question, which also mentions P=NP. My near proof is much simpler, and the little gap shoud be not too hard to close....

I would have thought April's fool will go unnoticed here. ;-)

moved to chat since its now April 2.

2 Answers

+ 3 like - 0 dislike

I think that I have found a way to close your gap. I will let you take a look at it before I send you my bank information for my half of the prize money. My idea is to use the geometric sum and combine it with a particular value of the Riemann zeta function.

Indeed we have

$$\frac{1}{1+x} = 1-x+x^2-x^3+x^4- \dots  $$

In particular (for $x = 1$),

$$\frac{1}{2} = 1-1+1-1+1 - \dots = (1-1)\left[1+1+1+1+1+\dots\right] = (1-1) \zeta(0) \, .$$

Since

$$\zeta(s) = 1 + \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+ \frac{1}{5^s}+ \dots$$

Then, using the fact that $\zeta(0) = -1/2$ we can write

$$\frac{X-X}{X} = (1-1) = \frac{1}{2\zeta(0)} = -1\, .$$

This is clearly incompatible with $X-X = 0$.

answered Apr 1, 2016 by Steven Mathey (350 points) [ no revision ]

Well, we'll need to wait for two years until the proof has been published and accepted as correct. Sometimes this is the most difficult part!

+ 1 like - 0 dislike

An even simpler way to fix the gap (compared to the fix by Steven Mathey) was communicated to me by Maurice de Gosson:

Assign X:=X+1. Then we may conclude that $X-X=1$, and division by $X-X$ is no longer a problem.

answered Apr 2, 2016 by Arnold Neumaier (12,570 points) [ no revision ]

Such an assignment is called renormalization ;-)

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