Author | Thread |
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01/28/2014 03:24:50 PM · #23526 |
Spork sent you in his place...there or the re-education camp
Spork wins. |
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01/28/2014 03:52:35 PM · #23527 |
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01/28/2014 04:00:41 PM · #23528 |
You, as usual, are wrong.
Spork wins. |
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01/28/2014 04:50:53 PM · #23529 |
Two wrongs make me right. I win. |
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01/28/2014 05:58:24 PM · #23530 |
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01/28/2014 06:02:54 PM · #23531 |
I was thinking that instead of a DP challenge for digital photography that this has actually become a who does the best photo shopping contest. Did I win? Lol |
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01/28/2014 06:38:21 PM · #23532 |
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01/28/2014 09:32:15 PM · #23533 |
Yes, your reign as winner was so brief it was less than the time it takes for light to travel 1mm through ordinary space.
Spork wins for eternity |
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01/28/2014 10:59:06 PM · #23534 |
Eternity? Nice name :P
I win. |
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01/28/2014 11:36:40 PM · #23535 |
eternity is a girly man with a steely gaze, clad in stylish underwear and some strong perfume |
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01/29/2014 12:50:37 AM · #23536 |
I wouldn't really like to meet him but I enjoyed your description. You should write a book! I kind of want to keep reading.
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01/29/2014 01:21:00 AM · #23537 |
I. Am. Eternity. (but without the stylish underwear) |
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01/29/2014 01:52:25 AM · #23538 |
Originally posted by Art Roflmao: I. Am. Eternity. (but without the stylish underwear) |
Eternity should be going to bed soon or tomorrow will be today and even strong perfume won't help you. :O)
Ray |
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01/29/2014 04:07:41 AM · #23539 |
Take a whiff. I am in it to win it. |
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01/29/2014 05:14:55 AM · #23540 |
... and I will see your Axe and raise you a Pepe ... hehehe.
Ray |
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01/29/2014 07:31:27 AM · #23541 |
I win, that's what's up. :P |
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01/29/2014 08:39:42 AM · #23542 |
My what big ears you have...
I win
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01/29/2014 08:43:05 AM · #23543 |
Wabbit season.
Spork wins. |
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01/29/2014 01:47:43 PM · #23544 |
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01/29/2014 03:25:30 PM · #23545 |
Da wabbit fwoze.
Spork wins. |
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01/29/2014 03:34:30 PM · #23546 |
bbbbbbrrrrrr It is cold down here but we missed the snow...It all went south of me
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01/29/2014 04:46:23 PM · #23547 |
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01/29/2014 06:59:16 PM · #23548 |
I dont think so .... |
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01/31/2014 03:08:28 PM · #23549 |
Originally posted by bvy: Originally posted by bvy: Hi. Sorry for my absence. I've been thinking a lot about the next problem, and I finally cracked it last night during a fitful sleep. Given k >= 0, let P_k denote the family of all k-degenerate graphs, and for some graph G, let X_(P_k)(G) represent the minimum number of partitions of the vertices of V(G) necessary for each partition to induce a k-degenerate subgraph of G. A graph G is said to be l-critical with respect to X_(P_k) if X_(P_k)(G) = l but X_(P_k)(G - v) = l - 1 for every v in G. I intend to show that d(G) >= (k + 1)(l - 1). |
Wow, no one caught that. The last statement is unqualified. It should read:
I intend to show that if G is l-critical with respect to X_(P_k) then d(G) >= (k + 1)(l - 1). |
Perhaps there's some confusion about k-degenerate graphs and l-criticality. Consider, for example, K7, the complete graph on seven vertices. Any two partitions of the vertices of K7 will necessarily include one that contains four or more vertices. Hence the induced subgraph will necessarily contain K4 which is 3-degenerate. Therefore X_(P_2)) = 3 as shown.
Further, K7 is 3-critical with respect to X_(P_2)); the removal of any vertex of K7, results in K6 which is 2-critical with respect to X_(P_2)) -- i.e. we could remove (without loss of generality) the single vertex partition and still need at least two, not three, partitions to induce 2-degenerate subgraphs. Observe, though, that K8 is not 3-critical with respect to X_(P_2)). Three partitions of K8 are also needed to induce 2-degenerate subgraphs, but the removal of any one vertex brings us back to the case of K7 for which X_(P_2) = 3. We can generalize that for p >= 3, that Kp (the complete graph on p vertices) is [ceil(k/3)]-critical with respect to X_(P_2)) whenever p % 3 = 1. We can further generalize that for r >= 0, Kp is [ceil(k/r)]-critical with respect to X_(P_(r - 1))) whenever p % r = 1.
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01/31/2014 03:33:23 PM · #23550 |
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