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Showing posts 4851 - 4875 of 30665, (reverse)
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01/22/2011 02:27:43 AM · #4851
You said Janine was snorting??
I don't think so, unless she is responding to your personal proclamation that you have reclaimed this thread.
Back to the basement with you!
01/22/2011 02:49:32 AM · #4852
Ooooh...are we beeing kept in the basement by the woman? Exciting...
01/22/2011 02:55:13 AM · #4853
Wait until you see what's for dinner...if you're good, we'll give you two bowls!
01/22/2011 03:29:46 AM · #4854
It's cold down here. Can I at least have blanket?
01/22/2011 03:40:13 AM · #4855
After I eat, can you then give me bowl cut? Pretty please?


01/22/2011 04:20:37 AM · #4856
Oh, such a handsome chap you'd be then!
**ignores Ken**
01/22/2011 04:23:36 AM · #4857
Think again, Kat.



and I don't know why you're crying, TM, YOU asked for it.
01/22/2011 04:27:44 AM · #4858
That's just a creepy haircut on anyone!
As I said, back to your basement! (Anne is a good friend of mine, muahaha!) :P
01/22/2011 05:19:13 AM · #4859
LOL - but hey. I think I look good! I will definitely get me one of those cuts.
01/22/2011 05:43:46 AM · #4860
OK I am in (not sure wg=hat but who cares)
01/22/2011 04:31:36 PM · #4861
Margaret wants a bowl cut!!!!
Funzapoppin!!
01/22/2011 06:01:46 PM · #4862
As promised, I am presenting Kempe's version of the four color theorem. The proof is false, but it is instructive all the same. First a prerequisite:

In my proof of the Five Color Theorem which appeared earlier, I used this verbiage:

Consider the subgraph of G - v which contains all vertices colored 1 and 3 and all edges connecting them; the connected component of this subgraph which contains v1, we will call G1.

If we instead call our colors a and b and refer to some vertex v, then we have defined an (a, b)-Kempe chain of G containing v. For brevity, I will use this terminology going forward in my restatement of Kempe's theorem. Consistent with this, I will assign letters to the colors instead of numbers as before.

Kempe's False Proof of the Four Color Theorem

Proof: Again, the proof is by induction on k, the order of the planar graph. The graph of order one is obviously four-colorable. We let G be a planar graph of order k, and assume that all planar graphs of order k - 1 are four-colorable. G must have some vertex v such that deg(v) <= 5, and we consider the graph G - v. By our induction hypothesis, G - v is four-colorable. If deg(v) < 4, then there is some color available in G - v that we may assign to v obtaining a four-coloring of G.

Suppose deg(v) = 4, and the neighbors of v in cyclical order are v1, v2, v3 and v4, colored a, b, c and d respectively. If the (a, c)-Kempe chain containing v1 does not contain v3, then we can alternate the colors of the chain without changing the color of a neighbor of v3. Hence, v1 is now colored c, and v can be colored a. If the chain does contain v3, then it isolates v2 from v4, and the (b, d)-Kempe chain containing v2 cannot contain v4. This allows us to alternate the colors of v2's (b, d)-Kempe chain. v2 is now colored d, and v can be colored b. In either case, we have achieved a four-coloring of G.

Suppose then that deg(v) = 5. If the neighbors of v are colored in three or fewer colors, then there is some color available in G - v that we may assign to v obtaining a four-coloring of G. Suppose then that the neighbors of v in cyclical order are v1, v2, v3, v4 and v5, and are colored a, b, c, b and d respectively. As before, if v1's (a, c)-Kempe chain or v5's (c, d)-Kempe chain does not contain v3, then the colors in that chain can be alternated to make a color (either a or d) available for v. So we assume that both chains contain v3 -- that is, there is an (a, c)-Kempe chain which contains both v1 and v3 and surrounds v2, and there is a (c, d)-Kempe chain which contains both v3 and v5 and surrounds v4. Since v2, colored b, is isolated from v5, colored d, there is a (b, d)-Kempe chain containing v2 which does not contain v5. Similarly, v4, colored b, is isolated from v1, colored a, and there is a (b, a)-Kempe chain containing v4 which does not contain v1. Since both chains are isolated, we can alternate the colors of both; v2 is now colored d, and v4 is now colored a. v, then, can be colored b, and we have achieved a four-coloring of G.

This proof was published toward the end of the nineteenth century and unquestioned for about ten years. It has a flaw (otherwise it wouldn't be a "false" proof). Do you see what it is?
01/23/2011 04:25:28 AM · #4863
theorems are for suckers.
01/23/2011 07:04:31 AM · #4864
The attempted proof was written by a man who was imprisoned in the cellar by a woman who invited her friends. Cut him some slack.
01/23/2011 10:37:34 AM · #4865
come on guys...let a gal get her coffee...Morning...

ETA: Damn it....who hid the coffee...???? ART!!!!!!!!!!!!!!!!!!!

Message edited by author 2011-01-23 10:38:43.
01/23/2011 06:45:14 PM · #4866
@ raish....did you ever see the movie The Lady Killers w/Tom Hanks?
Hmmm
01/23/2011 07:13:24 PM · #4867
No, no. The original with Alec Guinness and Peter Sellers is the one to see.
01/23/2011 08:26:42 PM · #4868
I could use a pint of Guinness...
01/23/2011 08:42:43 PM · #4869
Waiter!

Not much feedback on Kempe's theorem. Maybe Kempe was a sucker, as Art suggested. Would a diagram help?
01/23/2011 09:36:33 PM · #4870
I'll have to check out the original movie...I love Peter Sellers movies! (And I love Guinness, lol!)
01/24/2011 01:07:50 AM · #4871
Originally posted by bvy:

Would a diagram help?

It would help decorate more dartboards and urinals.
01/24/2011 10:12:12 AM · #4872
Morning...had my coffee...now I'm off...
01/24/2011 01:53:54 PM · #4873
Originally posted by Ja-9:

Morning...had my coffee...now I'm off...

I thought you've always been a little off. :P
01/24/2011 02:34:40 PM · #4874
Originally posted by Art Roflmao:

Originally posted by Ja-9:

Morning...had my coffee...now I'm off...

I thought you've always been a little off. :P


ha ha ha...ha ha...

ETA:...been talking to my family again have ya...

Message edited by author 2011-01-24 14:35:00.
01/24/2011 02:36:55 PM · #4875
I like turtles!!
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