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01/15/2011 12:22:50 AM · #4726 |
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01/15/2011 01:40:05 AM · #4727 |
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01/15/2011 01:52:30 AM · #4728 |
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01/15/2011 01:58:28 AM · #4729 |
| I'm sure the spray they "disinfect" the bowling shoes with helps a lot on soggy, sweaty shoes someone played bowling in while drinking beer and eating chips. And then they hand them to you. |
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01/15/2011 04:51:53 AM · #4730 |
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01/15/2011 09:07:55 AM · #4731 |
| Morning...have my coffee...the world is good |
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01/15/2011 08:18:58 PM · #4732 |
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01/15/2011 08:45:47 PM · #4733 |
| do you have a hoogie? need a Kleenex? it's ok...just step a side and let a lady win... ;) |
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01/15/2011 10:47:50 PM · #4734 |
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01/15/2011 11:01:27 PM · #4735 |
| I've done some reading on both the four-color and five-color map problems this week. The Five Color Theorem is, as it turns out, quite an elegant and efficient result, and it comes rather intuitively. I believe I'm prepared to recite by rote. |
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01/16/2011 12:43:58 AM · #4736 |
| I've got a theorum for you, bvy, but I can't show cuz there's ladies present. :P |
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01/16/2011 09:31:27 AM · #4737 |
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01/16/2011 12:26:36 PM · #4738 |
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01/16/2011 12:32:20 PM · #4739 |
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01/16/2011 12:51:28 PM · #4740 |
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01/16/2011 01:25:34 PM · #4741 |
| Lost me....I so don't get the initials...slow and old here |
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01/16/2011 01:33:32 PM · #4742 |
Originally posted by Art Roflmao:
I've got a theorum for you, bvy, but I can't show cuz there's ladies present. :P |
The aforementioned Four Color Theorem is several hundered pages long. How l...
Oh, never mind. |
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01/16/2011 03:11:43 PM · #4743 |
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01/16/2011 04:55:50 PM · #4744 |
| I'm preparing to present my own version of the Five Color Theorem later this evening. Mine doesn't augment or refine the existing proof in any way, which has been around for over a century. Rather, I've stated it in my own words, and it's been a good exercise. Anyone who's been following my postings of the last several weeks will hopefully find that it follows naturally from the previous discourse. Also, I think it's fairly readable. |
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01/16/2011 05:28:53 PM · #4745 |
| You give us too much credit, bvy. |
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01/16/2011 08:13:19 PM · #4746 |
| sneaking in quietly....yep...I win.. |
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01/16/2011 08:15:44 PM · #4747 |
| What the hell was that loud crashing sound??? It woke me up! |
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01/16/2011 08:27:22 PM · #4748 |
| shhhhhh....go to sleep...go to sleep....while I win the game...go to sleeeeppppp |
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01/16/2011 08:38:56 PM · #4749 |
| FYI - paid up for another year. That means you have to keep posting for at LEAST another 365 days to win. I am really tempted to end this post with "Bring it on bitches!" but that might be construed as a bit cheeky so I will end with "I await your response". |
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01/16/2011 10:08:51 PM · #4750 |
I dedicate this to Art -- who wins.
First note that the following statements about a planar graph G are equivalent:
- G is five-colorable.
- G has chromatic number less than or equal to five.
- G can be colored using five or fewer colors.
For consistency, I will use the syntax of the first statement.
The Five Color Theorem
Every planar graph is five-colorable.
Proof: The proof is by induction on k, the order of the planar graph. The graph of order one is obviously five-colorable. We let G be a planar graph of order k, and assume that all planar graphs of order k - 1 are five-colorable. We've already shown that 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 five-colorable. If deg(v) <= 4, then there is some color available in G - v that we may assign to v obtaining a five-coloring of G. We assume, then, that deg(v) = 5 and that each of the five colors of G - v are used to color the five vertices adjacent to v. We assign them the colors 1, 2, 3, 4 and 5 in a cyclical fashion around v, and call them v1, v2, v3, v4 and v5 respectively.
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 G1 does not contain v3, then we alternate the colors of G1. v1 now has color 3, and the color 1 can be assigned to v, giving us a five-coloring of G.
Suppose, then, that G1 does contain v3. Then there exists a v1-v3 path in G - v. Next we consider the subgraph of G - v containing all vertices colored 2 and 4 and the edges connecting them; the connected component of this subgraph which contains v2, we will call G2. Clearly G2 cannot contain v4, for if it did, there would have to be a v2-v4 path in G - v which did not intersect the v1-v3 path. We can therefore alternate the colors of G2. v2 now has color 4, and the color 2 can be assigned to v, giving us a five-coloring of G.
Hence, G is five-colorable.
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