Open Problem Garden - Comments

beals conjecture (re: DIS-PROOF OF BEALS CONJECTURE)

lalitha — Wed, 24 Apr 2013 13:45:03 +0200

why dont we take A(pwr)x=2n+1 and B(pwr)y=2n.he said to add 2 co-primes dnt even&odd numbers are co-prime.iam representing odd number as 2n+1 and even number as 2n

mistake (re: DIS-PROOF OF BEALS CONJECTURE)

eddybob123 — Wed, 24 Apr 2013 01:18:35 +0200

For those interested, the statement of the problem is wrong and the step "let us take A(pwr)x=2n+1 B(pwr)y=2n" is incorrect (there may exist no such n). The following Aaronson signs apply: 1, 2, 3 (would imply no triples rather than no coprime triples), 6, 7, and 10.

This conjecture has recently (re: Does every subcubic triangle-free graph have fractional chromatic number at most 14/5?)

fhavet — Sat, 02 Mar 2013 07:49:47 +0100

This conjecture has recently been proved by Dvorak, Sereni and Volec.

Frederic Havet

Lovasz Theta (re: Hedetniemi's Conjecture)

Anonymous — Wed, 27 Feb 2013 21:58:11 +0100

I believe that Robert Samal has conjectured a version of this for the Lovasz function, i.e. that

where . I can't find it in the Garden, but it is in this presentation by Samal: http://iuuk.mff.cuni.cz/research/cmi/cmi-I-Samal.pdf

N != n (re: Discrete Logarithm Problem)

Anonymous — Mon, 25 Feb 2013 00:43:58 +0100

The above paper solves the discrete logarithm in time O(N^3) not O(n^3), two very different things. N being the size of the modulus, n being log_2 of N (the binary length). There are many algorithms that will solve the discrete log problem much faster than this method, brute force search runs at a worst case of O(N), or in other words O(2^n). The provided algorithm in the above paper runs in (2^(3*n)).

history and application (re: Transversal achievement game on a square grid)

mshj — Fri, 22 Feb 2013 21:39:08 +0100

i'm not sure but i think to solve this problem, i was wondering if any body gives me some information about the history and application of this problem

First question (almost) solved (re: Good Edge Labelings)

Anonymous — Sat, 16 Feb 2013 12:06:12 +0100

Mehrabian, Mitsche, and Pralat showed that any -vertex graph with a good edge-labelling has at most edges, and that for each there is an -vertex graphs with a good edge-labelling having edges.

http://arxiv.org/abs/1211.2641

Are there a simple solution? (re: Transversal achievement game on a square grid)

Anonymous — Wed, 13 Feb 2013 10:11:54 +0100

I suspect, there are no simple answer and it can be solved only by heavy calculations, that is essentally there is no solution to this problem.

All primes are (re: Special Primes)

Anonymous — Thu, 07 Feb 2013 15:19:33 +0100

All primes are p=(q-1)/(order of 2 mod q)

Finite Three-Chromatic (0,2)-graphs (re: Three-chromatic (0,2)-graphs)

Anonymous — Wed, 06 Feb 2013 17:48:03 +0100

An infinite three-chromatic (0, 2)-graph is easy to construct. See Payan.

This supposed proof is incorrect (re: The Riemann Hypothesis)

eric — Sat, 26 Jan 2013 23:37:05 +0100

Dear Zeraoulia Elhadj,

The proof detailed in http://arxiv.org/pdf/1210.1517v10.pdf is wrong since eq. 9 (alpha=1/2) cannot be deduced at all from eq. (8), hence invalidating the whole proof.

To make a more general comment, I don't think this section should be used to propose attempts of a proof, but only a fully validated proof if there's one some day (and I hope so). There are indeed numerous invalid proofs every year for this mathematical problem, and regular forums are much more adapted to discuss on these attempts. This site is much more a collection of open mathematical problems with their current status and would be rapidly obfuscated by long standing discussions about various attempts of proof on each problem... Of course I'm not the webmaster of the site and he may confirm or contradict this personal opinion

Regards, Eric Chopin

Euler-Masqueroni contant (re: Euler-Mascheroni constant)

Anonymous — Mon, 21 Jan 2013 18:35:04 +0100

If n = 2^k then log(n) is irrational. But Sum{1/k} is ever rational . Then Sum{1/k} - Log(n) is ever irrational.. Ludovicus

No (re: Graham's conjecture on tree reconstruction)

leshabirukov — Wed, 16 Jan 2013 10:34:25 +0100

Consider L^i(T) is a graph with "even triangle"(triangle with even degrees of vertices) subgraph. Edges of even triangle produce new even triangle in L^(i+1)(T). And if there is an odd degree vertex adjacent to parent triangle, there would be another one adjacent to child. So, irregular subgraph remains.

If G is a star graph of (re: Graham's conjecture on tree reconstruction)

Anonymous — Tue, 08 Jan 2013 22:36:31 +0100

If G is a star graph of order 5, then L(G) = K_5.

A positive answer to the Riemann hypothesis: A new result predic (re: The Riemann Hypothesis)

zeraoulia — Thu, 27 Dec 2012 15:06:25 +0100

Dear Prof,

I am Prof. Zeraoulia Elhadj from the university of Tébessa, Algeria. Please see this link http://vixra.org/pdf/1210.0176v7.pdf http://arxiv.org/pdf/1210.1517v10.pdf

for a Solution of the Riemann Hypothesis: A positive answer to the Riemann hypothesis: A new result predicting the location of zeros. I think that this is a fine solution. Please let me know about your opinion on it. I think that your opinion is the final decison to accept or reject this solution. Any furhter comments are welcome. With kind regards. Elhadj