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Mathematician Claims Proof of Riemann Hypothesis
TheSync points to this press release about a Purdue University mathematician, Louis de Branges de Bourcia, who claims to have "proven the Riemann hypothesis, considered to be the greatest unsolved problem in mathematics. It states that all non-trivial zeros of the zeta function lie on the line 1/2 + it as t ranges over the real numbers. You can read his proof here. The Clay Mathematics Institute offers a $1 million prize to the first prover."
Karma-whoring free.
Ha! They've already found an error in the proof! All that he posted was his apology! :-)
Yes, I was actually confused at first. For the non-math geeks like myself, who are feeling stupid, look at definition 2a of apology.The paper is called, Apology for the proof of the Riemann hypothesis (in pdf format). To find the apology you have to read through to page 4 where he talks briefly about the problems that the solution of a celebrated problem creates for others who weren't expecting it. Basically the title is, "a form of Mathematical smack talk" (to quote a co-worker).
Most of the paper appears to be history, and the results leading up to his proof. Only a few pages at the end make up the actual new proof, so the novel material is far shorter than 23 pages.
I wouldn't be surprised if there is a fairly final verdict on his proof very quickly. This is not like Wiles' proof of Fermat that was very long and nobody had the background to understand. This proof looks reasonably short and straightforward.
Cheers,
Ben Tilly
Apology - 2: a formal written defense of something you believe in strongly
This should at most have earned a "Funny", or is there something I'm missing here?
Note to mods: Mod parent funny, not interesting! This is a play off a quote from the beginning credits sequence in Monty Python and the Holy Grail. As for the pdf link, it's the first link in the purdue page referenced in the article. RTFA, people!
Shameless self promotion
The Riemann Hypothesis, among other things, implies that the Prime Number Theorem is off in the distribution of primes by no more than O(sqrt(n)*log(n)). However even without the full result, we already had very good error bounds for the approximation of the prime number theorem for "small" numbers, including numbers far larger than any which come up in cryptography.
So, in short, no, no help for cracking crypto based on primes...though the article does mention possible crypto applications down the line. I'm not sure what, exactly, those would be.
The art of cryptography can be summed up as: Easy to encode, but hard to decode.
Prime numbers are easy to multiply together. Little CPU needed.
But it's hard to do the reverse: Factor a big number into two separate prime numbers. Lots of CPU needed.
It's based on that principle.
Sorry to burst the bubble, but some usenetting shows:
The same guy claimed to have solved the same problem at least 4 years ago.
The guy has a reputation for sometimes getting it wrong.
(Probably because he has published flawed proofs of other well-known problems.)
He could be right, but I wouldn't get my hopes up.
The magic of PKI occurs through the use of extremely long prime numbers, called keys. Two keys are involved - a private key, which only you have access to, and a public key, which can be accessed by anyone. The two keys work together, so a message scrambled with the private key can only be unscrambled with the public key and vice versa. The more digits in these keys, the more secure the process. --Public-key encryption for dummies
Not the best explanation, I prefer this
--
The problem is simple enough to understand, assuming you know some math basics. As most of you know, any function f(X) where f(Xo)=0 is said to have a zero at Xo. For functions of complex numbers f(z) where z=x+iy and x,y are real numbers, you obviously have the function taking on different values for every x and y, so the zeros can be anywhere on the x-y plane. For the zeta function, "trivial zeros" occur at the negative even integers (z=-2+i0,-4+i0,...) and also at points on the line x=1/2 (i.e 1/2 +iy for certain y).The Riemann Hypothesis says that all zeros that aren't negative even integers lie on this line.
Most of you have who have taken basic calculus courses have probably seen a simplified definition of the zeta function for real intergers greater than 1. when z=n, a natural number, the zeta function reduces to the infinite series Zeta(n)= SUM (k=1-->inf) 1/k^n
Nope, probably not.
Most mathematicians felt that the Riemann Hypothesis was true so that this view has been taken into consideration into mathematics for a long time. Perhaps if he developed some new methods for playing with numbers in the proof, but it doesn't seem like it to me.
There's a ton of math papers that begin with "Assume the extended riemann hypothesis...".
At least that's my guess.
One of the fallout corollaries from a proof of the Riemann hypothesis is that there exists a simple algorithm for factorization (read: p-time).
No. GRH implies that isprime() is in P (by bounding the cost of a strong pseudoprime test); but we already knew that, thanks to AKS.
Tarsnap: Online backups for the truly paranoid
It's actually a little more complex than that.
Riemann was investigating the distribution of prime numbers. Along the way he devised (discovered?) the Zeta Function, which describes with considerable accuracy the distribution of prime numbers. While working with the Zeta Function, he discovered an interesting property: It appeared that all the non-trivial zeroes of the function had a real part of one-half. However, since this property of the function was not related to the prime-distribution work he was doing, he did not bother to prove this apparent property, which came to be known as the "Riemann Hypothesis" (presumably, once it is proven it will be known as the Riemann Theorem, or some such).
Thus, the Riemann Hypothesis is in fact tangential to (and possibly unrelated to) the distribution of prime numbers. Riemann's notes on the Zeta Function, regarding his work on prime distribution, are pretty explicit about this.
Any sufficiently well-organized community is indistinguishable from Government.
does it's proof have any impact on crypto?
No. Almost all mathematicians have assumed for years that GRH is true anyway; proving it would mean that all those footnotes ([1] Under the assumption of the Riemann Hypothesis) could be removed, but that's the only practical effect it would have.
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Dude, what the fuck are you talking about?
Mathematicians tackle difficult problems all of the time, regardless of the (lack of) money involved.
I don't know why you say that interest in "theoretical" mathematical proof is waning. It certainly isn't where I come from. (And what is ultra-math??!)
This usage of "apology" is fashionable in math circles; a prime example is the title of G. H. Hardy's memoir : A Mathematician's Apology.
The 23 page "apology" is not the actual purported proof, contrary to what the article and press release say. The actual proof is the manuscript "Riemann zeta functions", the third link on de Branges' home page, which weighs in at 124 pages!
So if his "proof" isn't obviously wrong, it'll likely take quite a while for the experts to verify.
I think you are going a bit overboard here. The Riemann hypothesis is the greatest open problem in mathematics right now and solving it would be HUGE :-). However, famous open problems usually do not advance mathematics that much and I suspect that a proof of the Riemann hypothesis would not introduce new techniques which would have wide (or even slightly wide) use in math. Look at some of the Fields metal papers (e.g. restricted Burnside problem - Zelmanov - 1994 metal) and tell me how they changed mathematics.
For influences on math, consider Dirac (crazy British scientist who predicted to existence of the positron) whose ideas led L. Schwartz, L. to write "Théorie des distributions. Tome I,II"; distribution theory has had a huge influence on analysis.
Unlike what the other 2 people say, this is *not* an impossible ("undecidable") problem. Computers *can* formally verify any proof. The problem is that the complexity of such a program would be huge, and it would take a lot of time to make.
*Creating* proofs is the hard part, and was proven undecidable. But verifying ones that exist is not.
Melissa <3
The 23-page "Apology" referred to in the press release is also apparently mentioned in this 1996 Usenet post. So is there a new proof? No one seems to know yet.
A cool overview of why this is such an interesting hypothesis.
If nothing else check out the animation.
mind-boggling
The Riemann Hypothesis is a pain to explain before getting into some complex stuff about complex analysis (calculus involving complex numbers; IE numbers involving i = sqrt(-1)). I'm going to show the Riemann Hypothesis in the simplest terms possible, something that calculus students could understand (though not prove, of course!).
The Riemann hypothesis states that if this function:
sum from n=1 to infinity ((-1)^(n+1) / n^(x+iy))
equals zero, then x = 1/2 . x and y are variables; x+iy is a complex number normally labeled "z".
If you don't want to deal with complex numbers, you can use the equivalent statement about real numbers: If the following function:
(sum from n=1 to infinity ((-1)^(n+1) * cos(y*ln(n)) / n^x))^2 + (sum from n=1 to infinity ((-1)^(n+1) * sin(y * ln(n)) / n^x))^2)
equals zero, then x = 1/2 . The following URL is a picture of the above in normal notation so it's easier to read:
http://www.geocities.com/myriachan/riemann.png
Melissa <3
This should at most have earned a "Funny", or is there something I'm missing here?
Uh, yeah. I'm going to assume that the thing you missed was "Monty Python and the Holy Grail." Or, at least the opening credits.
I signed this
First: complex numbers, explained. You may have heard the question asked, "what is the square root of minus one?" Well, maths has an answer and we call it i. i*i = -1. If the real number line ...-4, -3, -2, -1, 0, 1, 2, 3, 4... is represented as a horizontal line, then the numbers ...-4i, -3i, -2i, -i, 0, i, 2i, 3i, 4i... can be thought of as the *vertical* axis on this diagram. The whole plane taken together is then called the complex plane. This is a two-dimensional set of numbers. Every number can be represented in the form a+bi. For real numbers, b=0.
Right. Now the Riemann Zeta Function is a function/map (like f(x)=x^2 is a function) on the complex plane. For any number a+bi, zeta(a+bi)will be another complex number, c+di.
Now, a zero of a function is (pretty obviously) a point a+bi where f(a+bi)=0. If f(x)=x^2 then the only zero is obviously at 0, where f(0)=0. For the Riemann Zeta Function this is more complicated. It basically has two types of zeros: the "trivial" zeroes, that occur at all negative even integers, that is, -2, -4, -6, -8... and the "nontrivial" zeroes, which are all the OTHER ones.
As far as we know, *all* the nontrivial zeroes occur at 1/2 + bi for some b. No others have been found in a lot of looking... but are they ALL like that? The Riemann Hypothesis suggests that they are... but until today nobody has been able to prove it.
qntm.org
The truth (or falsity) of the Riemann Hypothesis is intimately related to the distribution of the primes. Specifically, if the RH is true, then the primes are distributed about as regularly as possible.[1]
Carl
[1] See, for example, equation (2) of Riemann Hypothesis
Actually, as with most things Euler was the first to study it. The zeta function is also the simplest of a class of functions that Dirichlet studied Dirichlet L-series. There is also a Generalized Riemann Hypothesis that states that no Dirichlet L-series has zero with real part greater than 1/2.
The Riemann Hypothesis is more than tangential to the study of the distribution of primes. There is a function derived from the distribution of the primes that can be expressed in terms of the non-trivial zeros of the zeta function. The Prime Number Theorem is also equivalent to the statement that the zeta function has no zeros with real part 1. The Generalized Riemann Hypothesis implies the weak form of Goldbach's conjecture (i.e. that any odd number greater than 7 can be expressed as the sum of three odd primes).
My only political goal is to see to it that no political party achieves its goals.
This review of Karl Sabbagh's book The Riemann Hypothesis contains some background on De Branges. http://www.maa.org/reviews/sabbaghRH.html He sounds like quite a character, from that and from his "apology"... given recent trends, I wonder if someone might write a novel or play about him?
No, cryptosystems are based off of simple math based on a pair of very large pre-selected (read not random) prime numbers that make up your public and private key.
Cheers, Ben
cheers, ben
Never miss a good chance to shut up -- Will Rogers
Not to spoil your joke or anything, but actually, AFAIK, NP-complete problems are perfectly solvable. The problem is how long it takes to solve them in general (a certain instance of a problem could prove easy). They cannot be solved deterministically in polynomial time (i.e., quickly).
See "Riemann Zeta Functions" lower for the actual paper in question. The bibliography has a date of May 25th, 2004 at the top. This looks like a modification of a previous paper he had posted there for several years (his previous attempt.)
The reasons why most specialists doubt that his approach can ever yield the result are well described in this paper from 1998:
(i.e., despite the name, the "generalized RH" proved by de Branges actually did not include the standard RH as a special case.)This is...
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Mathworld has a link to a paper which claims de Branges' method of proof is invalid. Here is a direct link to the site with the paper.
Berkeley Groks has an interview that aired today with John Derbyshire discussing the Riemann Hypothesis. He states that after talking with many mathematicians in the field, the prospects for a solution any time soon are quite low.
mathworld.wolfram.com
Riemann Hypothesis "Proof" Much Ado About Noithing A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing
The proof (or, better said, the sketch of the proof) actually starts at the end of page 21, very close to the last page. The original work is actually pretty hard to find since it is buried in so many unrelated side notes.
/. until now :-)
Here is the general outline:
1) At the end of page 19 he mentions that "The positivity condition which is introduced implies the Riemann hypothesis if it applies to Dirichlet zeta functions."
2) After some introduction of the quantum gamma functions that lasts two pages, the actual proof starts at the end of page 21 with the phrase "A quantum gamma function is obtained when is nonnegative. A proof of positivity is given from properties of the Laplace transformation."
3) The proof ends in the middle of page 23 with the a verification that W(z) is a quantum gamma function with quantum q = exp(-2*pi), obtained from a spectral theory of the shift operator.
Overall this is just a very brief sketch of the whole proof.
BTW, to add gas on fire, here is an exceprt from mathworld.com, which surprisingly was missed by
http://mathworld.wolfram.com
Riemann Hypothesis "Proof" Much Ado About Noithing (sic)
A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing.
The counterexample to Brangles approach can be reached here: http://arxiv.org/abs/math.NT/9812166
Don't try to use the force. Do or do not, there is no try.
It should be noted that while mathematicians often assume non-trivial things, they always need to keep track of their assumptions. You can't just say that we know XYZ theorem is true, you must always note that what you're stating is only true if Riemann's hypothesis is assumed. Keeping track of what assumptions you are relying on, is actually one of the most important skills a mathematician must have, and it sure would be nice if non-scientists would learn to do it.
It would appear that mathworld.com agrees with you...
----------------
Riemann Hypothesis "Proof" Much Ado About Noithing
A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing.
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NP-Complete problems are by definition problems that can't be solved in polynomial time(at least not by a Turing machine?). However, most problems that are considered NP-Complete are not mathematically proven to be so. Some are though, and the thing with NP-Complete problems is that you can always translate one NP-Complete problem to another NP-Complete problem.
:)
So in practice, NP-complete problems can be solved (you can solve just about anything by just trying every single solution) but for very big instances you will need several times the age of the universe etc.
Several other posters in this thread seem to be mistaken about what the term actually means, but they were being so vague I thought I'd write this up.
The definition of NP-Complete has nothing to do with whether the problem can be solved in polynomial time. NP-Complete problems are the hardest of the class of problems in the set NP. The set NP is defined as all problems whose solutions can be verified in polynomial time. It hasn't been proven whether or not NP-Complete problems can or cannot be solved in polynomial time. Because of the way the proofs have been constructed, if you can solve any of the NP-Complete problems in polynomial time, you can solve all of them in polynomial time. That's why it is considered unlikely that any of the NP-Complete problems has a polynomial time solution, but it hasn't been proven.
The definition of NP-Complete has nothing to do with whether the problem can be solved in polynomial time.
...is one of *two* equivalent definitions of the class NP. The other is:
Not quite correct, because:
The set NP is defined as all problems whose solutions can be verified in polynomial time
"Set of all problems which can be solved in polynomial time by a nondeterministic Turing Machine."
So the question "Does P equal NP?" is also the question "Does an NTM have the same computional power(in time) as a TM, or does it have more?" (It is already known that as far as decidiblity is concerned, TMs and NTM are equivalent.)
Because of the way the proofs have been constructed, if you can solve any of the NP-Complete problems in polynomial time, you can solve all of them in polynomial time.
Some more detail here:
An NP-hard problem is a problem to which any problem in NP can be reduced in polynomial time.
(Essentially, it can be used as a subroutine for any NP problem, with only a polynomial number of calls. Thus a solution to it is a solution to any problem in NP.)
An NP-complete problem is one that is:
1. NP-hard
2. In the set NP.
Thus if a polynomial-time solution exists to an NP-complete problem, then P=NP, because a polynomial number of calls to a function that terminates in polynomial time is O({polynomial}*{polynomial}) = O({polynomial}) .
Please note, however, that not all NP-hard problems are NP-complete.
NOTICE: This notice will appear at the bottom of all my slashdot posts.
NP problems are problems that can be solved by a Nondeterministic Turing machine in Polynomial time. NP-Complete problems are the class of "hardest" problems in NP. All the usual suspects (Traveling Salesman, 3-SAT, SAT, ...) are proven
to be in NPC.
We know that we can solve NPC problems in exponential time (as we can simulate a non-deterministic Turing maschine on deterministic hardware with exponential overhead). What we do not know is if there is any smarter way. That is the P=?=NP question.
Stephan
Sorry, but...
.... Now, this is where I admit that I do not really understand that area of math, and have not been closely following the status of (alliteration alert) Perelman's proposed proof. Still, Perelman is a real mathematician, and even if the proof is (was?) wrong, it has real ideas of value in it.
It is not proved; he is not at the top of his field; this "paper" will be quickly forgotten among professional mathematicians; and I doubt any professional mathematician is going over the proof with any sort of comb.
L. de Branges first achieved fame for proving the Bieberbach conjecture. His proof went through strange and abstract methods. He went on the road to present his proof at various seminars in France, Russia, etc; IIRC a bunch of Russian students got very excited and basically rewrote his proof. Their new proof was much shorter and avoided the use of strange methods. Nowadays, their proof is remembered and his is not, but the proof still bears his name, since after all he was the first to come up with *some* kind of proof, and their proof did more or less come out of his.
So he deserves credit for that, and it was quite an achievement to prove the Bieberbach conjecture. But even then he was using unwieldy proofs with unnecessarily abstract methods.
For many years he has been claiming to have a proof of the Riemann Hypothesis. Professional mathematicians stopped listening a long time ago.
This guy is washed-up.
I whole-heartedly agree that this short article is hilarious, but I would like to add the adjective condescending. What kind of asshole apologizes for solving a problem? Does he think he lives on some higher plane, and therefore must take direct, personal responsibility for every aspect of our lives?
Look at how G. Perelman submitted his ideas on proving the Poincare conjecture just a little while ago. He didn't waste anyone's time by rehashing the already-available history of the problem or its wider context in mathematics. Nor did he apologize for having an idea. Rather, he submitted his ideas for consideration, with the full awareness that there may have been a mistake.
de Branges is so full of crap, it makes me sick.
zach
NP-Complete problems are by definition problems that can't be solved in polynomial time(at least not by a Turing machine?).
.
No, you're wrong. NP problems are, by definition, problems that can be solved in polynomial time by a nondeterministic Turing machine
Essentially this means that a Turing machine could solve the problem in polynomial time, if it had some magic 'oracle' which instructed it on the right computational path to follow for a given input.
Obviously there are problems out there that would require exponential time for even a nondeterministic Turing machine to solve. An example from the Wikipedia link I provided is finding the best move in a chess or Go game.
Such problems are not in NP, and proving P=NP would not suddenly give us algorithms for solving these problems deterministically in polynomial time.
However, most problems that are considered NP-Complete are not mathematically proven to be so.
What? Sorry, if there ain't a proof, it ain't NP-complete. There are a lot of problems that are described as "believed to be NP-complete", but that's different.