Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  P=NP solved in the affirmative?

+ 3 like - 0 dislike
2168 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 (15,787 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

+ 4 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!

+ 2 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 (15,787 points) [ no revision ]

Such an assignment is called renormalization ;-)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...