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Showing posts 17251 - 17275 of 30665, (reverse)
AuthorThread
 10/18/2012 10:11:50 PM · #17251
Why not fully drunk...fully drunk is so much better
 10/18/2012 10:14:14 PM · #17252
infact, I plan on getting "fully drunk" at my boss's house tomorrow. He is having an end of summer swimming party...Thank god he has a hot tub and a heated pool
 10/18/2012 10:36:30 PM · #17253
You must be in a warmer climate than me.

And I would probably be fully drunk, but I have to work tomorrow. Blech.
 10/18/2012 10:40:23 PM · #17254
I am in Louisiana...but it has been in the 40/50' range at night and high of like 80
 10/18/2012 10:42:14 PM · #17255
I'm in PA. Hate it. HATE. I cannot stand winter. I have oil heat. It's insane.
 10/18/2012 10:48:22 PM · #17256
It gets COLD here.....I have had friends from New York come down and they nearly froze to death...The humidity makes it soo much cooler
 10/18/2012 11:16:43 PM · #17257
Originally posted by cowboy221977:
infact, I plan on getting "fully drunk" at my boss's house tomorrow.

I'm sure I am not alone in thinking this is a very BAD idea.
 10/18/2012 11:59:11 PM · #17258
be careful Adam
 10/19/2012 01:10:44 AM · #17259
Tis been a while since i was winning..
 10/19/2012 01:39:52 AM · #17260
Deja Vu IAmEliKatz You are not winning again for a while. :)
 10/19/2012 01:40:46 AM · #17261
Originally posted by Art Roflmao:
Originally posted by cowboy221977:
infact, I plan on getting "fully drunk" at my boss's house tomorrow.

I'm sure I am not alone in thinking this is a very BAD idea.


OOohh yeah. That's a bad one... good luck with that!
 10/19/2012 01:46:37 AM · #17262
Originally posted by TrollMan:
Originally posted by Art Roflmao:
Originally posted by cowboy221977:
infact, I plan on getting "fully drunk" at my boss's house tomorrow.

I'm sure I am not alone in thinking this is a very BAD idea.


OOohh yeah. That's a bad one... good luck with that!


See, now you have three people saying it's a BAD idea.

Message edited by author 2012-10-19 01:50:25.
 10/19/2012 01:46:53 AM · #17263
And I'm winning
 10/19/2012 01:49:26 AM · #17264
Originally posted by beatabg:
See, now you have tree people saying it's a BAD idea.


Just to be perfectly clear: I am NOT a tree hugger! :P
 10/19/2012 01:50:50 AM · #17265
Originally posted by TrollMan:
Originally posted by beatabg:
See, now you have tree people saying it's a BAD idea.


Just to be perfectly clear: I am NOT a tree hugger! :P


Corrected! Sorry I'm tired :) At least I made myself laugh at myself (thanks to your comment :)

Message edited by author 2012-10-19 01:56:54.
 10/19/2012 01:57:17 AM · #17266
Tree's a crowd. I win.
 10/19/2012 02:08:29 AM · #17267
Turd time's the charm - I win
 10/19/2012 02:10:36 AM · #17268
I'm looking up some stuff for school, but I'll be here... winning.
 10/19/2012 02:46:23 AM · #17269
never.

Spork wins.

that is all.
 10/19/2012 02:56:57 AM · #17270
Originally posted by beatabg:
I'm looking up some stuff for school, but I'll be here... winning.

Here is the info you are looking for:
Glucose
aniphalactic shock
Placebo
Backgammon
Knit sweater
Art wins
 10/19/2012 03:39:01 AM · #17271
Originally posted by Art Roflmao:
Originally posted by beatabg:
I'm looking up some stuff for school, but I'll be here... winning.

Here is the info you are looking for:
Glucose
aniphalactic shock
Placebo
Backgammon
Knit sweater
Art wins


anAphYlactic shock was covered long time ago :)
I'm actually looking for some articles on Photovoice used in patients with severe mental illness. The ones that look good aren't free :(
Anyway I win.

Message edited by author 2012-10-19 03:39:25.
 10/19/2012 05:27:40 AM · #17272
Sleeeeep. You evade me yet again, you son of a bitch.
 10/19/2012 05:49:24 AM · #17273
Originally posted by beatabg:
I'm actually looking for some articles on Photovoice used in patients with severe mental illness.

Can't help you there without giving away my secret. And I win. Me too. So do I. All of us do. Me, that is.
 10/19/2012 08:13:00 AM · #17274
Let G be a k-critical graph with k >= 2. Then G is (k-1)-edge connected.

Proof: The only 2-critical and 3-critical graphs are K2 (the complete graph on two vertices) and the odd cycles, respectively. Clearly K2 is 1-edge connected, and the odd cycles (all cycles, actually) are 2-edge connected. So we assume k >= 4.

Let G be a k-critical graph, and suppose it is not (k-1)-edge connected. Then we can partition the vertices of G into sets V1 and V2 such that the induced subgraphs G1 = G[V1] and G2 = G[V2] have at most (k-2) edges connecting them. Since G is k-critical, G1 and G2 are both necessarily (k-1)-critical. Let E be the edge set connecting vertices of G1 and G2. There should then be some edge in E that connects vertices of the same color -- for if there is not, then G could be properly colored with (k-1) colors and, as such, would not be k-critical. But this is the crux of our proof; by showing that it's possible to permute the colors of the vertices of G1 in such a way that every edge in E connects vertices of different colors, we arrive at the contradiction needed to complete our proof.

We start by partitioning the vertices of G1 by their assigned colors. There are at most (k-2) partitions that have neighbors in V2, and we denote these U1, U2, ..., U_t (t <= k-2). Let k_i be the number of edges in E incident with a vertex in U_i. Then 1 <= k_i <= (k-2), and the sum of all k_i is (k-2). Now consider the vertices of U1 that have neighbors in V2. If every such vertex has a different-colored neighbor, then we leave U1 alone. If, however, some vertex of U1 has a neighbor in V2 of the same color, then it is possible to permute the colors of the various U_i and assign U1 a color such that every vertex of U1 has a different colored neighbor in U2. This is possible because the vertices of U1 have at most (k-2) neighbors in V2, yet there are (k-1) possible colors to assign to U1.

Next we take U2. Again, if every vertex therein has a different-colored neighbor in G2, then we move on. If, though, some vertex of U2 has a like-colored neighbor in G2, then we permute the colors of the various U_i (i>=2) and assign U2 a color that erases any conflicts. This is possible because there are (k-1) - (k_2 + 1) colors to avoid, and (k-1) - (k_2 + 1) >= (k-1) - (k_2 + k_1) >= (k-1) - (k-2) = 1.

Continuing in this manner, we can find a (k-1)-coloring for G, which is the desired contradiction.

10/19/2012 09:14:04 AM · #17275
AM+BER = WIN+NER

Message edited by author 2012-10-19 09:14:45.
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