Am I a crank?
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From: Chip Eastham on 10 Sep 2006 15:53 Han.deBruijn(a)DTO.TUDelft.NL wrote: > John Schutkeker wrote: > > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote in news:9ed01$44fd392c > > $82a1e228$31509(a)news1.tudelft.nl: > > > > > John Schutkeker wrote: > > > > > >> You're nothing but fatalists. Quitters, even. > > > > > > Perelman is a quitter as well. The next step is to think about why. > > > > In what way is Perleman a quitter? > > Don't you keep up with the news? Did you intended to answer a question with a question? regards, chip
From: Aatu Koskensilta on 10 Sep 2006 16:23 Tony Orlow wrote: > Aatu Koskensilta wrote: >> No. In whatever way, we develop the idea of a series of numbers >> obtained from some starting point by repeatedly applying[1] the >> successor operation (the "add one" operation). >> [1] For some reason the phrase "finitely many times" is added here, as >> if "applying the successor function infinitely many times" made any >> sense. > > I don't think you mean to disagree with standard theory here, so I must > be reading you wrong. In standard theory, aren't there infinitely many > naturals in succession? Certainly. My rather trivial point was just that "applying the successor function infinitely many times" doesn't mean anything in particular, and thus it is somewhat odd to note that the naturals are obtained from 0 (or 1, it doesn't really matter) by a finite number of applications of the successor function. -- Aatu Koskensilta (aatu.koskensilta(a)xortec.fi) "Wovon man nicht sprechen kann, daruber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Virgil on 10 Sep 2006 18:55 In article <65l8g21crkflahh4isra2laon7hiui95pj(a)4ax.com>, Lester Zick <dontbother(a)nowhere.net> wrote: > On 9 Sep 2006 19:22:35 -0700, "David R Tribble" <david(a)tribble.com> > wrote: > > >Lester Zick wrote: > >>> Because you self declaredly proclaim assumptions of truth in lieu of > >>> demonstrations. > >> > > > >Virgil wrote: > >>> Zick frequently does this, but why does he deceive himself that > >>> mathematicians emulate his idiocies? > >> > > > >Lester Zick wrote: > >> Because they don't and probably can't demonstrate their trivial > >> assumptions of truth. > > > >Here's the first Peano axiom: > > 1. 0 is a natural number. > > > >Is it (trivially) true or false? > > Assumptions are always trivial. So the only question is whether the > assumption is true or false. Zero is not a natural number. What constitutes a natural number, or any other sort of number, is a matter of common agreement, not a matter of historic discovery. Of more people choose to agree that zero is a natural number that otherwise, then it is one. It is certainly an ordinal number. If it were > it would have been discovered long before it was. Natural numbers > begin with 1. Zero just represents the difference between any number > and itself. > > ~v~~
From: Proginoskes on 10 Sep 2006 21:39 icahit(a)gmail.com wrote: > Again the same argument applies that is there must be at least two > triangles with a common edge which contains the vertex colored with the > fourth color [in a planar graph which is not 3-colorable]. This is known to be false. There are plenty of counterexamples to this; http://math.asu.edu/~checkman/steinberg5cycle.jpg for one. (V. A. Aksionov, L. S. Melnikov, "Some Counterexamples Associated with the Three-Color Problem", JCT-B 28, 1-9 (1980), is an article that Cahit should read.) In fact, this graph also shows that there is something wrong with Cahit's claims about Spiral Coloring. He claims: (1) If a planar graph is k-colorable, then Spiral Coloring will k-color it. (2) He has a proof of Steinberg's Conjecture, where he concludes that a counterexample has a cycle of length 4 or 5 among some vertex v and its neighbors. Note (1) by itself would show that P=NP, but that's not the real point, which is: these two claims cannot both be true. Again, consider the graph (G) at http://math.asu.edu/~checkman/steinberg5cycle.jpg and let x be the vertex adjacent to all of u, v, and w. Start the Spiral Coloring algorithm with this vertex, and continue it in any way. G-v is 3-colorable, so by (1), the SC algorithm has a 3-coloring of it. Clearly a fourth color is needed for x (whether the algorithm works or not). Now by (2), this means that x is in a 4- or a 5-cycle that consists of x and some of its neighbors. But this is not true; there is only a 3-cycle. Hence, one of (1) or (2) must be wrong. --- Christopher Heckman > But of course if we assume that G, planar graph without > cycles of length four and five is chromatic critical then this > situation occured at the last vertex of the spiral chain. > > Proginoskes wrote: > > icahit(a)gmail.com wrote: > > > 1. Detailed proof on the spiral chain coloring will be given soon > > > (nothing will be changed in the structure of the proof) which is an > > > addition of the use (S,NS)-Kempe chain switching in case of undecided > > > case on the current spiral segment, where S is a "safe color" and NS is > > > a non-safe color. > > > > > > 2. On my proof of the Steinberg's conjecture, I have just added the > > > following justification: In the final step of the spiral chain coloring > > > if the planar graph without cycles of length 4 and 5 is four colorable > > > then the color four has to be assigned to the last node of the spiral > > > chain. But then edges incident to the last node has to be in a cycle of > > > length 4 or 5 which is an contradiction. > > > > What if there are other vertices colored with the fourth color? > > > > --- Christopher Heckman > > > > > Well after my presentation of my 4CT proof at the ICM someone asked me > > > then I will submit to a journal? I made the following joke, if someone > > > has the non-computer proof of the four color theorem then historically > > > (due to Kempe's proof) he has the right to wait at least 10 years. But > > > I am almost sure that no one will be able to submit an counter-example > > > or a serious critics. > > > > > > Cahit > > > > > > > > > Proginoskes wrote: > > > > Jesse F. Hughes wrote: > > > > > John Schutkeker <jschutkeker(a)sbcglobal.net.nospam> writes: > > > > > > > > > > > Isn't Arxiv peer-reviewed? > > > > > > > > > > No. > > > > > > > > There are a couple of Four Color Theorem papers which have obvious > > > > errors in them. (I found them in < 5 minutes.) One is by Friess, and > > > > the other is by an Asian mathematician. I haven't yet decided whether > > > > Cahit's result works, mainly because of the lack of a proof that is > > > > detailed enough. > > > > > > > > You might want to check Cahit's "proof" of Steinberg's Conjecture (any > > > > planar graph without 4- or 5-cycles is 3-colorable). There's a sentence > > > > in the proof which says (paraphrased), "We only need to consider the > > > > one following graph." He offers this statement without any kind of > > > > justification, and the rest of the paper is unnecessary fluff! > > > > > > > > --- Christopher Heckman
From: icahit on 11 Sep 2006 01:36
I don't have that paper. But the example has been already studied with spiral chains at(http://www.flickr.com/photos/49058045(a)N00/201540965/) and shows that the node colored "four" is on the cycle of length five at the last node of the spiral chain. This shows as you indicated the need of cycle of length five in the statement of the conjecture. Furthermore it shows the selfishness of the use of the fourth color in the spiral coloring. If you introduce a cycle of length four in the graph then I say that fourth color would be assigned to the one of the node of that cycle. I will write in more detail soon. Cahit Proginoskes wrote: > icahit(a)gmail.com wrote: > > Again the same argument applies that is there must be at least two > > triangles with a common edge which contains the vertex colored with the > > fourth color [in a planar graph which is not 3-colorable]. > > This is known to be false. There are plenty of counterexamples to this; > http://math.asu.edu/~checkman/steinberg5cycle.jpg for one. (V. A. > Aksionov, L. S. Melnikov, "Some Counterexamples Associated with the > Three-Color Problem", JCT-B 28, 1-9 (1980), is an article that Cahit > should read.) > > In fact, this graph also shows that there is something wrong with > Cahit's claims about Spiral Coloring. He claims: > > (1) If a planar graph is k-colorable, then Spiral Coloring will k-color > it. > (2) He has a proof of Steinberg's Conjecture, where he concludes that a > counterexample has a cycle of length 4 or 5 among some vertex v and its > neighbors. > > Note (1) by itself would show that P=NP, but that's not the real point, > which is: these two claims cannot both be true. Again, consider the > graph (G) at http://math.asu.edu/~checkman/steinberg5cycle.jpg and let > x be the vertex adjacent to all of u, v, and w. Start the Spiral > Coloring algorithm with this vertex, and continue it in any way. > > G-v is 3-colorable, so by (1), the SC algorithm has a 3-coloring of it. > Clearly a fourth color is needed for x (whether the algorithm works or > not). Now by (2), this means that x is in a 4- or a 5-cycle that > consists of x and some of its neighbors. But this is not true; there is > only a 3-cycle. Hence, one of (1) or (2) must be wrong. > > --- Christopher Heckman > > > But of course if we assume that G, planar graph without > > cycles of length four and five is chromatic critical then this > > situation occured at the last vertex of the spiral chain. > > > > Proginoskes wrote: > > > icahit(a)gmail.com wrote: > > > > 1. Detailed proof on the spiral chain coloring will be given soon > > > > (nothing will be changed in the structure of the proof) which is an > > > > addition of the use (S,NS)-Kempe chain switching in case of undecided > > > > case on the current spiral segment, where S is a "safe color" and NS is > > > > a non-safe color. > > > > > > > > 2. On my proof of the Steinberg's conjecture, I have just added the > > > > following justification: In the final step of the spiral chain coloring > > > > if the planar graph without cycles of length 4 and 5 is four colorable > > > > then the color four has to be assigned to the last node of the spiral > > > > chain. But then edges incident to the last node has to be in a cycle of > > > > length 4 or 5 which is an contradiction. > > > > > > What if there are other vertices colored with the fourth color? > > > > > > --- Christopher Heckman > > > > > > > Well after my presentation of my 4CT proof at the ICM someone asked me > > > > then I will submit to a journal? I made the following joke, if someone > > > > has the non-computer proof of the four color theorem then historically > > > > (due to Kempe's proof) he has the right to wait at least 10 years. But > > > > I am almost sure that no one will be able to submit an counter-example > > > > or a serious critics. > > > > > > > > Cahit > > > > > > > > > > > > Proginoskes wrote: > > > > > Jesse F. Hughes wrote: > > > > > > John Schutkeker <jschutkeker(a)sbcglobal.net.nospam> writes: > > > > > > > > > > > > > Isn't Arxiv peer-reviewed? > > > > > > > > > > > > No. > > > > > > > > > > There are a couple of Four Color Theorem papers which have obvious > > > > > errors in them. (I found them in < 5 minutes.) One is by Friess, and > > > > > the other is by an Asian mathematician. I haven't yet decided whether > > > > > Cahit's result works, mainly because of the lack of a proof that is > > > > > detailed enough. > > > > > > > > > > You might want to check Cahit's "proof" of Steinberg's Conjecture (any > > > > > planar graph without 4- or 5-cycles is 3-colorable). There's a sentence > > > > > in the proof which says (paraphrased), "We only need to consider the > > > > > one following graph." He offers this statement without any kind of > > > > > justification, and the rest of the paper is unnecessary fluff! > > > > > > > > > > --- Christopher Heckman |