the four color theorem...
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From: noemata on 17 May 2010 10:45 On May 17, 11:44 am, riderofgiraffes <mathforum.org...(a)solipsys.co.uk> wrote: > To emphasise to the original poster, this reply from > David Rusin is excellent and deserves close study. > > RiderOfGiraffes write: > > > Your argument is that because you cannot have five > > mutually joined points, five colours cannot be > > necessary. That means that if I have a graph without > > four mutually joined points, four colours won't be > > necessary. > > David Rusin replied: > > > Well, maybe you should say "In the spirit we might > > expect..." instead of "That means..." . > > > It might be helpful to the OP to demonstrate a graph > > that really requires 5 colors but contains no K_5. > > Building on the examples of other posts we might begin > > with a ring of 5 points (the vertices of a pentagon) > > and add two more vertices, each connected to all 5 > > of the original points as well as to each other. That > > way the pentagon requires 3 colors and each of the > > additional points requires an additional color. On > > the other hand any set of 5 mutually adjacent points > > would have to include at least 3 points of the pentagon, > > and no such set of 3 mutually adjacent points exists. > > > This particular graph is not a counterexample to the > > 4CT because it is not planar, but the fact that it is > > not planar is not especially obvious, and moreover > > there is no obvious way to use the planarity of the > > graph to rule out a similar but more complex example > > that IS planar. > > > dave > > I understand the example finally if the last line says 5 instead of 3, because sets of 3 mutually adjacent points exist. From the comments I've had to modify the original idea to be "a graph that needs n colours can be /reduced/ to contain K_n", which turns out to be Hadwiger's conjecture given the proper "reduction" (contracting the edges of its k disjoint connected subgraphs to a K_n as minor). regards, Bjørn
From: Gerry Myerson on 17 May 2010 19:09 In article <d14e44d8-8667-47de-9ad8-c427d811b66f(a)f13g2000vbm.googlegroups.com>, noemata <noemata(a)kunst.no> wrote: > On May 17, 11:44�am, riderofgiraffes <mathforum.org...(a)solipsys.co.uk> > wrote: > > To emphasise to the original poster, this reply from > > David Rusin is excellent and deserves close study. > > > > RiderOfGiraffes write: > > > > > Your argument is that because you cannot have five > > > mutually joined points, five colours cannot be > > > necessary. That means that if I have a graph without > > > four mutually joined points, four colours won't be > > > necessary. > > > > David Rusin replied: > > > > > Well, maybe you should say "In the spirit we might > > > expect..." instead of "That means..." . > > > > > It might be helpful to the OP to demonstrate a graph > > > that really requires 5 colors but contains no K 5. > > > Building on the examples of other posts we might begin > > > with a ring of 5 points (the vertices of a pentagon) > > > and add two more vertices, each connected to all 5 > > > of the original points as well as to each other. That > > > way the pentagon requires 3 colors and each of the > > > additional points requires an additional color. On > > > the other hand any set of 5 mutually adjacent points > > > would have to include at least 3 points of the pentagon, > > > and no such set of 3 mutually adjacent points exists. > > > > > This particular graph is not a counterexample to the > > > 4CT because it is not planar, but the fact that it is > > > not planar is not especially obvious, and moreover > > > there is no obvious way to use the planarity of the > > > graph to rule out a similar but more complex example > > > that IS planar. > > > > > dave > > > > > > I understand the example finally if the last line says 5 instead of 3, > because sets of 3 mutually adjacent points exist. Yes, but no *such* sets exist, where "such" means 3 points of the pentagon. No 3 points of the pentagon that you begin with are mutually adjacent, therefore no set of 5 mutually adjacent points exists. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: spudnik on 18 May 2010 01:16 an argument by infinite descent would take a long time, if you actually have a way of encoding a planar graph as some thing; yeah. - Hide quoted text -
From: spudnik on 18 May 2010 01:22 yes; a "complete graph on five points" is a what-you-call-it, having diagonals that (by definition?) cross each-other, which is not a "simple graph" that is dual to the map, whose edges & vertices also constitute a simple graph, simply connected, with (connected) oceans counting as "no color" (or a color .-) > no set of 5 mutually adjacent points exists. thus & so: what is a reasonable investment in rebinding, iff "the book is of value to me, myself & y'know," if y'know; or what is the best DIY format? I'm sure there are at least two of them in the old catalog-can-you-fax-it-over-to-here? what's the URL? > I asked them years ago when they started gluing their signatures. No > answer ever came. (Amusingly enough, I still have a signature-sewn > catalogue from them.) thus: prove and/or define the most canonical "law of cosines" in trgionometry, you can; you can define "canonical," two. well, I just read the definition of the law, or the supposed outcome of formula -- which is completely standard, as far as I can tell -- in a large dictionary (of English). thus: I haven't proven that the Bible Code was a hoax; only a hueristical argument about any ring of "letters of all of letters" ... not the Object or Bunny Rings, neccesarily. however, the biblical topic is "skip codes." > Where..Easter bunny..him? --Light: A History! http://wlym .com
From: spudnik on 20 May 2010 23:35
hey; maybe tehy'd let you look at your trophy with those 3d glasses! thusNso: dood, my valu of pi is lots simpler to calculate than yours -- seven cans of beer & a string! thusNso: nice cartoon; is there only one beamsplitter in Sagnac? --Pi, the surfer's canonical value, is not constructible with a pair of compasses .. but, could be with a pair and a half of compasses; dyscuss. |