
A Solution of the P versus NP Problem? - fahrbach
https://arxiv.org/abs/1708.03486
======
ajarmst
This is definitely one of those "I'm going to wait for the peer review"
claims, but it is pretty exciting.

Pros: The author is not a dilettante, and is actively researching in the area
([http://theory.cs.uni-
bonn.de/blum/Forschung/forsch.var](http://theory.cs.uni-
bonn.de/blum/Forschung/forsch.var))

Cons: It's not my area, but I was expecting something a little more novel for
a solution to P?NP. This almost seems too simple (it might almost fit in a
margin...).

Could be pro or con: Single author. It's becoming rare for important new work
to not have multiple authors, especially from professional academics. However,
Andrew Wiles...

~~~
btilly
The paper looks like it is 38 pages and builds on other work. That doesn't
seem too simple to me.

~~~
ethan_g
I tend to agree with GP that this is suspiciously simple. I've only skimmed
it, but the novel part of the proof appears to only be Thm 5-6 which is less
than 10 pages, and it's not especially dense writing. So this would be a
relatively simple proof. Moreover, the technique used appears to be rather
incremental over known techniques, so it's surprising it would be strong
enough to prove PvNP which is so far away from the frontier of known
techniques.

~~~
daveFNbuck
The known technique was a proof of an exponential lower bound on something
extremely similar to an NP-complete problem. That moved the frontier of known
techniques a lot closer to P vs NP.

------
DominikPeters
A list of 116 previous "solutions" to the P versus NP problem:
[http://www.win.tue.nl/~gwoegi/P-versus-
NP.htm](http://www.win.tue.nl/~gwoegi/P-versus-NP.htm)

~~~
valine
After skimming its interesting that the majority of the proofs in your list
claim P equals NP. I would have guessed it would be more common for proofs to
claim the opposite because P != NP makes sense intuitively. That being said it
would certainly be more exciting if P did equal NP.

~~~
mbrubeck
A simple but wrong proof of "P = NP" is easier to write in some ways, since
you "just" need to provide a single algorithm for one NP-hard problem, and
show that it runs in polynomial time. It looks like many or most of the proof
attempts in that list take this form.

A plausible proof of "P != NP" won't be quite as simple to express, since it
needs to prove that _all_ such algorithms do _not_ run in polynomial time.

~~~
sova
Well yes! Two things: You have a successful algorithm that runs in P time that
solves an NP Hard problem and 2) you can map other NP Hard problems to your
problem. Without the second factor, it is only a demonstration of a "range" in
the computational realm in question, where p = np or whatever the declaration.
Being able to show that your pizza slice is actually an ocean of pizza, and
also show that any other shape of pizza slice can be appropriately transformed
into the shape you have means p = np. a "solution." Or perhaps better put, it
is a _funnel_ through which complex computational patterns can either be
reduced or simplified or elegantly correlated, approaching absolutely perfect
parallelization of operations. This is just one way to look at it, but
essentially intractability is an interesting term to consider.

Please forgive me if my liberal use of the language is an offense

~~~
dbaupp
Your 2 is the easy part: "NP-hard" is exactly the set of problems to which any
NP problems can be reduced in polynomial time, and there are many known
existing examples, both within NP (aka NP-complete) and outside it.

~~~
sova
So part of the solution set satisfies P=NP and some satisfy P!=NP?

~~~
dbaupp
No, some NP-hard problems are outside NP (i.e. harder than NP), while some are
inside. The wikipedia article has a good explanation and a good diagram:
[https://en.wikipedia.org/wiki/NP-hardness](https://en.wikipedia.org/wiki/NP-
hardness)

~~~
sova
Thanks many for your kind explanation. It makes more sense now. I forgot that
important detail that mapping is pretty easy as most involvements reduce to
and from 3-SAT

------
Tomte
Scott Aaronson on "suppose someone sends you a complicated solution to a
famous decades-old math problem, like P vs. NP. How can you decide, in ten
minutes or less, whether the solution is worth reading?":
[http://www.scottaaronson.com/blog/?p=304](http://www.scottaaronson.com/blog/?p=304)

~~~
mcguire
" _Yet here’s the bloodied patient, and here we are in the emergency room._ "
" _The lemma encrusted shoulders of giants._ "

Aaronson on writes very well.

~~~
monochromatic
A little too flowery. He's also needlessly political.

~~~
Analemma_
Are you one of those people who equates any mention whatsoever of Trump as
"being needlessly political"?

~~~
monochromatic
Not at all. But bringing him up in an unrelated post on a technical blog just
screams "virtue signaling" to me.

~~~
kybernetikos
When you feel strongly about something and write about it, you aren't virtue
signalling, you're expressing your view. If you have any reason to take the
rather uncharitable stance that his views on Trump are inauthentic, then post
them, otherwise it's better just to assume that people who say things you
disagree with _actually think those things_ rather than are just saying them
to look good.

Accusing others of "virtue signaling" is a much clearer example of "virtue
signaling" in my opinion, since it's a particular population that tends to do
it, and is used primarily to put down those they disagree with and mark them
as belonging to a different group rather than to actually make any kind of
meaningful point.

~~~
monochromatic
I'm curious what I said that sounded to you like "his stated views on Trump
aren't authentic."

~~~
vecter
[https://en.wikipedia.org/wiki/Virtue_signalling](https://en.wikipedia.org/wiki/Virtue_signalling)

    
    
        Virtue signalling is the conspicuous expression of moral
        values done primarily with the intent of enhancing
        standing within a social group.
    

I don't think Aaronson states his views on Trump with the hopes of increasing
his social status. He does it because it's just what he feels and wants to
express himself.

~~~
monochromatic
And which part of that definition has anything to do with the views being
inauthentic?

I feel like I'm taking crazy pills here.

~~~
kybernetikos
When you express a view that you don't hold in order to receive social
approval, or when you express a view that you hold lightly with more intensity
and frequency than you would except for your desire to receive social approval
you are being inauthentic.

The whole value of the phrase 'virtue signalling' is that it accuses those you
disagree with of inauthenticity. If I'm saying a thing I genuinely believe
because I genuinely believe it, I'm not virtue signalling. It's only if the
reason for saying the thing is to gain social approval that it's 'virtue
signalling'. Any assumption that your opponent is doing something for this
reason is uncharitable, and kills rational debate. It's essentially an ad
hominem attack.

It's also ignoring the fairly detailed entry that Scott posted explaining in
his own words why he started talking about Trump in his blog:
[http://www.scottaaronson.com/blog/?p=2777](http://www.scottaaronson.com/blog/?p=2777)
when previously he'd avoided the subject.

Maybe it is virtue signalling, or maybe he really did feel as he claimed, that
he had a moral responsibility to speak out against Trump. If you assume that
it is only 'virtue signalling', you are cutting yourself off from engaging
with his points.

------
ethan_g
Interesting that this is the second P!=NP proof from a University of Bonn
researcher. Other one, by Mathias Hauptmann, is here:
[https://arxiv.org/abs/1602.04781](https://arxiv.org/abs/1602.04781)

I never did hear the status of Hauptmann's proof (I'm not connected to
academia so only know what I've read on the internet), but given it's been
over a year without word, presumably there's something flawed.

I might not get too excited over this proof, either, until another member of
the TCS community can vouch for it. There have been many serious-looking
attempts at PvNP that turn out to have fundamental flaws.

~~~
alderz
How does the paper of Hauptmann prove P != NP?

Sigma_2^p != NP as far as I know and after a brief skimming the paper does not
mention P != NP.

Edit: the paper does indeed mention P != NP in the form of P != Sigma_2^p => P
!= Sigma_1^p = NP. Please disregard my comment.

~~~
modalduality
P = NP implies the polynomial hierarchy collapses, thus P = Sigma2, so by
contradiction P != NP.

------
jjgreen
I would normally sigh and move on seeing such a claim, but this guy is an
established senior researcher at the University of Bonn. A career-ending
disaster or instant and eternal fame, that's some serious cahunas.

~~~
jordigh
> serious cahunas

I think you mean "cojones" (which btw is a very rude word in Spanish). A
kahuna is a kind of Hawai'ian shaman.

~~~
jjgreen
I stand corrected, thank you. I would normally have said "serious bollocks",
but this forum is mostly left-ponders who would probably not have caught my
drift.

~~~
liberte82
North American here, we hear enough Brits speak to get the expression :)

~~~
Myrmornis
"serious bollocks" means "very questionable statement" in the British English
I'm familiar with. Perhaps GP's idiom is from elsewhere. If the reference is
to bravery/daring one would use "balls" as in US English, I believe.

~~~
IronSean
"Bollocks" in Britain and "Balls" in North America both refer to the Testes,
as does "cojones". So all three statements are actually the exact same slang
in regional dialects.

Bollocks, like many slang words, have multiple meanings in different contexts.
It can be used as you stated as well, but "What you said is bollocks" and "You
have bollocks for saying it" are very different statements.

------
yifanlu
Just a half minute skim shows the author claims it passes the natural proof
barrier but makes no claim about it being non-relativizing or non-
algebraizing.

~~~
lvh
For those following along at home: this is important because we have a proof
that relativizing wouldn't work. In this context, relativizing means relative
to an oracle. For example, P^A means "just like a Turing machine would
regularly recognize languages except this one has a magical oracle that can
recognize A in one step". A can be arbitrarily complex, including NP-complete
ones like 3SAT.

This is important because the Baker-Gill-Solovay theorem already demonstrates
that there exist oracles A != B relative to which P^A = NP^A, but P^B != NP^B.
This shows that the problem has contradictory relativizations, and hence can't
be proven that way. This matters because it's a litmus test against quack
proofs.

I don't think this is a problem here; the proof doesn't appear to be doing
that.

~~~
schoen
I first learned about this because there's a decades-old joke in NetHack that
cites this result when you ask for a major consultation from the Oracle when
you can't afford to pay for it. I think the joke is that in this case, the
Oracle can't tell you anything that's useful to you. :-)

------
atemerev
I like the straightforward title. I know that it is politically correct to
christen your paper solving e.g. the Poincare conjecture like e.g. "Ricci flow
with surgery on three-manifolds", but all rules are there to be broken once.

I wish the author best of luck.

~~~
danbruc
At least he couldn't resist making the important point a corollary.

~~~
posterboy
what else, a mere remark? a theorem?

------
mathgenius
John Baez has a bit of a discussion here:

[https://johncarlosbaez.wordpress.com/2017/08/15/norbert-
blum...](https://johncarlosbaez.wordpress.com/2017/08/15/norbert-blum-on-p-
versus-np/)

------
rickbradley
I see a few typos in the wording of the paper (e.g., "spezify", "touchs",
etc.). While this doesn't mean much, I would expect if the paper had gotten a
fine-toothed comb review that these sorts of typos would have been caught.
Moving back to the "not holding my breath" stance unless there start to be
indications from experts in the field that the claims are holding up.

~~~
Houshalter
I can forgive imperfect English. He's not a native English speaker and his
reviewers probably are not be either.

------
RSchaeffer
Can someone ELI5 what this problem is, how likely the proof is to hold up to
scrutiny, and whether P != NP follows?

~~~
teton_ferb
Here is a largely correct ELI5: P != NP asks the question "Are problems that
are easy to _verify_ (NP) also Easy to _solve_? (P)".

Note that the _reverse is obviously true_ : problems that are easy to _solve_
are also easy to _verify_.

Here is an example: Take the problem "Find minimum of 5,6,7,8". You solve the
problem and tell me that the answer is 5. I can verify your answer by solving
the problem myself, getting the the answer 5 and comparing it with your
answer. So we can conclude "Problems that are easy to solve are easy to
verify" In other words, P ⊆ NP.

_Now is the reverse true? Are problems that are easy to verify also easy to
solve?_

Let me give you an example. Let us assume that the question is "Is
1053188576519689 prime?". You come back and tell me, "No it is not prime, it
is divisible by 32,452,867".

1) It is easy to verify your solution. I can divide 1053188576519689 by
32,452,867 and verify that it is indeed divisible. 2) It is hard to solve the
problem, I have to try out numbers from 2,3,...,sqrt(1053188576519689), which
is quite painful. (Or maybe there is as yet undiscovered better algorithm). So
it appears that problems that are easy to verify may not be easy to solve. Or
it appears that NP ⊆ P is not true. In other words, it appears P != NP
(because if P ⊆ NP and NP ⊆ P, P == NP).

NP problems have wide ranging applications in things like cryptography for
example. Let us assume I have a hashing technique. It is easy to hash a
document, but hard to reconstruct the document from the hash. Then this
technique can be used in auctions where you do not trust the auctioneer. You
publicly submit the hash of your bid before the deadline. You do not submit
your bid itself, because you are afraid that the person handing out the
contracts will reveal the number to his brother-in-law who will bid $1 more
than you and win the contract. After the deadline is passed, you send your
actual bid to the Auctioneer.

Now 1) Everyone can verify that the documents have not been altered (the
hashes are posted publicly, each document can be hashed and compared with its
publicly posted hash). So it is easy to _verify_ that the documents have not
been tampered with after the deadline.

2) Nobody can construct the document from the hash. So it is not easy to
_solve_ for the bid document given the hash. So everyone can post the hash
publicly with confidence before the deadline.

If P != NP we can have this type of auctions. If P == NP then there is no
difference between posting the hash publicly and posting the document
publicly.

~~~
cvoss
(You may have omitted this detail on purpose, but I think it's worth pointing
out) The example of trial division as a primality test is a good illustration
of an algorithm that takes a lot of work to run but whose output is easy to
verify. However, the problem of primality testing is actually in P. [1] You
have to use a fancier algorithm than trial division in order to get polynomial
time.

[1]
[https://en.wikipedia.org/wiki/AKS_primality_test](https://en.wikipedia.org/wiki/AKS_primality_test)

~~~
credit_guy
another detail that the GP most likely left out on purpose is that the fact
that easily checking divisibility means that primality is co-NP. Primality
being NP is actualy non-trivial (even before AKS). see wikipedia for more
details:
[https://en.m.wikipedia.org/wiki/Primality_certificate](https://en.m.wikipedia.org/wiki/Primality_certificate)

------
jjgreen
[https://cstheory.stackexchange.com/questions/38803/](https://cstheory.stackexchange.com/questions/38803/)

------
danbruc
After skimming the paper, the methods used look quite basic. There seems to be
no really deep new insight as far as I can tell. Let's see what experts will
have to say.

------
ckugblenu
An insightful interview to give perspective. [http://www.se-
radio.net/2017/07/se-radio-episode-298-moshe-v...](http://www.se-
radio.net/2017/07/se-radio-episode-298-moshe-vardi-on-p-versus-np/)

------
nabla9
If established researcher in the field makes a breakthrough of this magnitude
I would expect rumors to start to circulating first.

Showing the draft to few colleagues to see if they can spot mistakes before
'shaking the world' is probably a good idea.

~~~
sean2
Hmmm, maybe he analyzed the game theory:

If he publishes without consulting peers:

    
    
      If his proof is correct, he gets unending fame, millions in prize money.
    
      If his proof is laughably flawed, he'll promptly be forgotten as one of the 100s who have been wrong before him.
    

If he consults his peers:

    
    
      If his proof is correct, he may end up sharing credit, maybe they'll even publish his work quietly under their own name while he's still waiting for feedback. Maybe that is what we are reading now.
    
      If his proof is laughably flawed, then they'll give him his feedback and only his peers, instead of the whole internet will laugh at him for a day, before it's all forgotten.
    

Seems like with any tiny chance of a correct proof, the dominant strategy is,
by far, to publish without consulting your peers.

~~~
killercup
Entirely possible: The paper's author has actually given lectures on game
theory.

------
choxi
What are the implications of solving the P versus NP problem? What practical
effects would that have? Not trying to belittle the problem, just curious as
an outsider.

~~~
zolthrowaway
The most interesting thing is if P=NP. If that's the case, that means that
there is an algorithm that can solve any NP problem in polynomial time. This
means that things like crypto would be able to be cracked in polynomial time
which presents a huge problem for security.

We basically operate under the assumption that P!=NP currently. Validation
that this is true doesn't really change much. I can't speak to how this may
help a academic researcher in CompSci, but it probably doesn't change much for
most programmers.

~~~
vanjoe
I think it's even less useful, even if P = NP it is possible that no one finds
an algorithm. Creating a (useful) algorithm is independent of proving the
theorem. Also interesting is that someone could create an algorithm that
solves NP complete in polynomial time without proving P=NP. They would be
unable to prove the algorithm correct though.

~~~
Sharlin
If P=NP, then finding an algorithm is, by certain definition, exactly as easy
as proving an algorithm correct! This is, after all, the gist of what P vs NP
means. And this is why it's such a huge deal - the problem is almost self-
referential in nature.

~~~
anarazel
> This is, after all, the gist of what P vs NP means. And this is why it's
> such a huge deal - the problem is almost self-referential in nature.

Could you back that up with some citations? This doesn't ring true. But my
pure CS has withered a bit...

------
ajarmst
I remain optimistic, but some potential holes are starting to show. Scott
Aaronson:
([http://www.scottaaronson.com/blog/?p=3389](http://www.scottaaronson.com/blog/?p=3389))

"To everyone who keeps asking me about the “new” P≠NP proof: I’d again bet
$200,000 that the proof won’t stand, except that the last time I tried that,
it didn’t achieve its purpose, which was to get people to stop asking me about
it. So: the paper claims to give an exponential circuit lower bound for
Andreev’s function, but that function as defined in Section 7 of the paper
seems clearly to have a polynomial-size circuit, based on polynomial
interpolation (thanks to Luca Trevisan for this observation). So I don’t see
how this can possibly stand. Please just stop asking, and if the thing hasn’t
been fully refuted by the end of the week, you can come back and tell me I was
a closed-minded fool."

------
TekMol
Say you wanted to let a computer search for a proof of P!=NP - which axioms
would you start with and which rules to transform the axioms into additional
valid statements?

~~~
brianberns
I wonder if the algorithm to search such a space of axioms and transforms for
a solution would itself be P or NP. :)

~~~
yorwba
Almost certainly so super-exponential that NP appears easy in comparison.

------
fazkan
As someone with an undergrad level knowledge of the problem can someone please
outline the importance of this research. Is this research done in some bounded
domain. From what I have read in the comments, there seems to be other papers
which have also proved P!=NP. So what does this research add...Just curious,
feel free to ignore this...

~~~
CarolineW
_> From what I have read in the comments, there seems to be other papers which
have also proved P!=NP._

None of those papers have been accepted as correct, the problem remains open.

There's a gazillion articles describing the context and importance, and I
suggest you do a web search for something like "P vs NP and why it's
important".

Here's one:
[http://www.scottaaronson.com/blog/?p=459](http://www.scottaaronson.com/blog/?p=459)

HN discussion:
[https://news.ycombinator.com/item?id=1605415](https://news.ycombinator.com/item?id=1605415)

~~~
fazkan
Thanks for the reply. Thats a great article. I have read a lot of articles
like that, in fact was planning to write one myself. Just wanted to know more
about the paper submitted i.e. what it proposes and the methodology adapted. I
tried reading it but couldnt parse the data with all the jargons.

~~~
CarolineW
That's not the question you asked. If you want to know more about the context
of this specific paper then you need to (a) wait for the people in the area to
start writing explanations, or (b) ask more specific questions about specific
terms.

A good place to start to get a sense of what's going on with this specific
paper is this thread, but more particularly, the articles linked from it. One
that I think is a great starting point is this one:

[https://johncarlosbaez.wordpress.com/2017/08/15/norbert-
blum...](https://johncarlosbaez.wordpress.com/2017/08/15/norbert-blum-on-p-
versus-np/)

But you should perhaps start by asking the question you actually want the
answer to - it sounds like you didn't. So take a moment, do some reading, then
come back with particular questions. They may already be answered, but before
you did the outside reading you didn't realise. That's what happens to me.

------
andy_ppp
So at least according to this P != NP. I’ve always been skeptical of this just
because if you find an efficient optimal algorithm the problem moves between
the classes.

~~~
spiznnx
>if you find an efficient optimal algorithm the problem moves between the
classes

This hasn't ever happened.

You can prove an algorithm is in NP-hard by reducing another NP-hard problem
to it. You can prove an algorithm is in P by showing the algorithm.

If you find and efficient algorithm to a problem in NP-hard, you show P == NP.
No one has ever done this.

P != NP is considered by many to be most likely true.

~~~
andy_ppp
> You can prove an algorithm is in P by showing the algorithm.

Thanks for clarifying - I always though that's what the above meant but I
didn't realise that the reduction side was also important.

------
kaffeemitsahne
If true, would this be an actual 100% solution, or just an indicator that P !=
NP? This is all way over my head but the word "approximator" makes me wonder.

~~~
NikolaeVarius
>> This implies P not equal NP.

~~~
bmh_ca
> This implies P not equal NP

I did some graduate level research on P =? NP, specifically in the SAT space
<[https://en.wikipedia.org/wiki/Satisfiability>](https://en.wikipedia.org/wiki/Satisfiability>).

In particular, I helped design MARMOSET (Marmoset Automated Reasoner Mostly
Only Solves Easy Theorems), a competitive SAT problem solver.
<[http://www.cs.unb.ca/research-
groups/argroup/marmoset/>](http://www.cs.unb.ca/research-
groups/argroup/marmoset/>) . (It's a cool name... I didn't come up with it :))

The conclusion I drew was:

1\. P != NP because you can convert in polynomial time every SAT problem down
to Horn clauses, which are P to solve, plus non-Horn clauses that cannot be
converted i.e. have intractable intrinsic NP complexity whose reduction to the
polynomial space requires "clairvoyance" of the quantum computation variety.

2\. Nobody's really interested in a proof that P != NP.

That said, I only spent a couple years at it, and my memory may be faulty and
I might change my mind if I revisited the issue. Part of me has always felt
that the Horn clause reduction is a first step to isolating problems for a
next step, but again — it's been a long time.

~~~
hnaccy
>Nobody's really interested in a proof that P != NP.

I doubt that.

~~~
hervature
I believe what they are saying is that the entire field just assumes P != NP
and so if the proof came to be P != NP, then most people would just have their
inklings confirmed and that's about it. It wouldn't really cause a shift in
research in any of the CS departments. However, if P = NP, you can bet on a
renewed interest in finding polynomial time algorithms.

~~~
x3n0ph3n3
That's like saying no one is interested in a solution to the Riemann Zeta
Hypothesis. It's thought to be true, and mathematics have been derived from
its assumed truth, but there is still major interest in proving it to be true.

~~~
hervature
The difference is that if Riemann Zeta is incorrect, many theorems that are
based on the hypothesis being correct will also be invalidated. Whereas
assuming P != NP is limited to a relatively small area of complexity theory. I
think a much more equivalent conjecture in complexity theory would be the
unique games conjecture. Because the unique games conjecture already provides
certain problems to be completely characterized (see:
[https://en.wikipedia.org/wiki/Unique_games_conjecture#Releva...](https://en.wikipedia.org/wiki/Unique_games_conjecture#Relevance)).
Thus, proving this to be true kind of closes the chapter on many problems.
Whereas proving P != NP true still leaves many gaps open.

------
efm
tl;dr Not crackpot, not proven, interesting math. There's a discussion on
Google+ started by John Baez, which includes a comment by Gowers:
[https://plus.google.com/+johncbaez999/posts/hGshTfdimpZ](https://plus.google.com/+johncbaez999/posts/hGshTfdimpZ)

There's some interesting ideas in the paper, the researcher has done good work
in this area before.

------
waynecochran
If's its legit proof, they author should be able to explain why SAT2 is in P
and SAT3 is NP-complete. That's my BS test.

~~~
21
Why? Proving something doesn't mean we understand all implications of said
proof or how to apply it to a particular situation.

Think about non-constructive proofs.

~~~
waynecochran
It's an important special case. You need some sort of filter for all the crazy
submissions of potential proofs. I am not the only one who uses this problem
as the filter.

------
letitgo12345
Some informed discussion --
[https://www.facebook.com/shachar.lovett/posts/10155282027901...](https://www.facebook.com/shachar.lovett/posts/10155282027901321)

~~~
elnygren
There's really nothing there - 11 likes and 1 comment ?

Also, this very same Facebook link was linked to at
[https://cstheory.stackexchange.com/questions/38803/is-
norber...](https://cstheory.stackexchange.com/questions/38803/is-norbert-
blums-2017-proof-that-p-ne-np-correct) by
[https://cstheory.stackexchange.com/users/24/opt](https://cstheory.stackexchange.com/users/24/opt)

Neither this HN user, that FB user nor that stackexchange user seems like
anyone important, respected or relevant in the field of CS of this article
(correct me if I'm wrong).

Smells like a bad attempt at karma whoring to me.

~~~
letitgo12345
That 1 comment raises very serious doubts about that paper. The paper is
basically assuming that a polynomial time solvable problem takes super-
polynomial time. If it's using that fact in an integral manner then that kills
the proof. And even if not, just the fact that the paper is making such a
basic mistake raises questions as to how solid the rest of the paper is. Also
the guy who made the comment (Luca Trevisan) is a very famous researcher in
theory/crypto...

------
enether
I wonder if crypto values will skyrocket if this is proved

~~~
binarysaurus
Interesting, I would assume it would tank since there is a way of undermining
the system.

~~~
JCzynski
No, that would be if P=NP. This is either a proof that P!=NP, or not a valid
proof of anything.

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mathperson
1\. no google scholar account 2\. three publications on the arxiv.

I am willing to toss this out of hand...but look at yitang zhang or perelman.
who knows? best of luck

~~~
eatplayrove
although not prolific, the guy has published in top cs theory conferences and
journals: [http://dblp.uni-trier.de/pers/hd/b/Blum:Norbert](http://dblp.uni-
trier.de/pers/hd/b/Blum:Norbert)

dblp is the proper place to check computer scientists' record, not
arxiv/google scholar.

~~~
mathperson
I am not really sure I agree. At least in my part of CS in the united states
arxiv/google scholar are pretty omnipresent. Thank you for pointing me towards
a new resource.

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ImadK
Until the singularity P will always be different than NP.

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waldrews
Haven't we seen too many proofs that demonstrate N!=1 but forget to also show
that P!=0?

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zeptomu
If this is true (and the proof has no flaw), it would be huge and pleasant
(many expect it, although there are some who think it's a problem that might
be unprovable in our current logic framework). So let's see how it works out.

However the author missed that there is a special case for N == 1, where
actually P == NP (sorry, I could not resist).

~~~
naturalgradient
Don't forget the case where P == 0

~~~
dom0
You're both wrong.

P is the union of all TIME complexity classes defined by n^k for every natural
k.

NP is the union of all NTIME complexity classes defined by n^k for every
natural k.

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yters
If p != np, then comp sci will lose much of its appeal. There is an underlying
hope behind the field that p = np, otherwise most problems of interest are
intractable, and programmers are no longer masters of the universe.

This is probably why there is not a proof yet, since the truth is undesirable.

~~~
azdavis
I'm not sure what you mean by "appeal." But, it seems to me that if P = NP,
and if we can find a constructive proof of this fact, i.e. someone presents a
P-time algorithm A_L deciding an NP-complete language L, the field of CS in a
sense would get much _less_ interesting, because although we would have
answered arguably the most important question ever posed, there would be much
less of a need to research and develop efficient algorithms or approximation
strategies. We'd probably just focus on trying to improve the runtime of A_L
and the reductions from other NP languages.

~~~
yters
Yes every field is of interest to its specialists. But the popular appeal is
that comp sci promises the ultimate explanation of a materialistic reality. At
least this was its appeal when I chose it for my major, and it drives the
religious transhumanism and the AI hype and arguably the funding of IT. But if
the public learns seemingly trivial problems are inherently intractable or
impossible for computers, that glamorous spectacle will shatter.

