
How not to prove that P is not equal to NP  - dominotw
http://gowers.wordpress.com/2013/10/03/how-not-to-prove-that-p-is-not-equal-to-np
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StavrosK
Ugh, I don't see what all the fuss is about. Here, let me do you all a favor:

P = NP

P/P = NP/P

N = 1

Which is absurd, because N is a letter. Therefore, P != NP.

You're welcome.

~~~
11001
Alternative proof:

Suppose, P = NP

P² = P _(NP)

(subtracting (NP)²)

P² - (NP)² = P_(NP) - (NP)²

(dividing both sides by P-(NP)):

P + (NP) =(NP)

since P = NP (initial assumption):

2*P = P

(dividing by P):

2 = 1

which is a contradiction, therefore P \not= NP

~~~
hausen
Gaaah! My eyes! You assumed that P = NP, therefore you cannot divide by P - NP
= P - P = 0.

I think that mistake shows that even finding a mock proof for the fact that P
!= NP is hard.

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Achshar
Can any one explain the article? It looks too technical for me and I am not
good with advanced maths. I have an idea of what p != NP is but not _very_
knowledgeable about it.

~~~
j2kun
In the quest to prove P != NP a great number of techniques have been tried,
and many of these techniques have attempted to use circuit complexity. The
circuit complexity of a function with n inputs is, roughly, the size of the
smallest circuit that computes it. This is a function depending on n, and so
we can talk about polynomial size circuits, etc. The important fact is that if
we can find a specific problem (a problem in NP) which doesn't have a
polynomial size circuit, then P is not NP.

Unfortunately many of the elementary techniques that try to show this don't
work, and for a long time nobody knew why. Then in the 90's Alexander Razborov
and Steven Rudich proved a general theorem that these elementary techniques
could never prove P != NP.

What Gowers explains in his blog post (and the follow-up) is the Razborov-
Rudich theorem, called natural proofs. Roughly, it says that all of the
attempted techniques use a "natural" property P of boolean functions. A
natural property P asserts one specific function needs large circuits, while
all small circuits cannot satisfy the property (and P is efficiently
computable, in a sense).

The insight is that all of these properties also apply to TOO MANY other
functions. Indeed, if P is natural for polynomial size circuits, and it
applies to too many functions, then it can be used to violate another well
known conjecture in cryptography, a conjecture on the existence of a certain
kind of secure pseudorandom number generator. This is the heart of R-R's
theorem, and it's quite a beautiful proof.

Nowadays there are three known "barriers" to proving P is not NP, and any
proposed method for solving the problem first needs to argue that it bypasses
all of these barriers. The first is called relativization (having to do with
oracles), the second naturalization (this natural proofs barrier), and the
third algebrization (which I'm not familiar with).

Tim Gower's is actually using this post as a platform to propose a new method
for separating P from NP, and so he's studying these barriers to argue why his
method bypasses them. For more on his new proposal, see:
[http://gowers.wordpress.com/2013/10/24/what-i-did-in-my-
summ...](http://gowers.wordpress.com/2013/10/24/what-i-did-in-my-summer-
holidays/)

~~~
tzs
> Indeed, if P is natural for polynomial size circuits, and it applies to too
> many functions, then it can be used to violate another well known conjecture
> in cryptography, a conjecture on the existence of a certain kind of secure
> pseudorandom number generator

Has this completely stopped the search for a natural proof that P!=NP, or are
there (real--not crackpot) researchers who now simply think "Cool...now I've
got a shot at both proving that P!=NP and disproving a widely believed
conjecture that is vital to modern cryptography! That gets me tenure and a
Field's medal for sure!"

~~~
j2kun
The open conjecture (officially, the existence of subexponentially strong
pseudorandom function families), is a weaker variant of some stronger
conjectures (the existence of polynomially-strong pseudorandom number
generators) which actually imply P != NP. So in some sense the problems are
very close to equivalent. But yes I think you're right, that nobody would take
such an approach seriously.

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jackmaney
Wow, a double negative! Does this mean that the article shows how to prove
that P _is_ equal to NP? :P

~~~
foldr
No, because the the two negatives don't cancel out in this case. "I didn't
prove that 3 isn't prime" doesn't mean the same thing as "I proved that 3 is
prime" (because one negation is in the matrix clause and the other is in the
embedded clause).

~~~
jackmaney
Yes, hence the emoticon.

~~~
Helianthus
>P is equal to NP? :P

I thought it was a poorly constructed ternary operator? :D

