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Showing posts 23551 - 23575 of 30665, (reverse)
AuthorThread
 01/31/2014 09:52:25 PM · #23551
I don't know if Tiny shared with you all how awfully mean I am... but yeah... just look :P a W I N for me :)

01/31/2014 09:55:35 PM · #23552
Spork throws a mean bean ball.

Spork wins.
 01/31/2014 09:58:08 PM · #23553
Oh yeah... well Art had his pet do this to me....... and I am still here.

Guess I win huh?

Ray
 01/31/2014 10:16:17 PM · #23554
Originally posted by RayEthier:


Guess I win huh?

Ray


No Ray.

Spork winsâ€Â¦you do not.
 01/31/2014 11:10:21 PM · #23555

Oh yeah... well take a gander at this

Me wins again right? :O)

Ray
 02/01/2014 12:14:47 AM · #23556
Sorry Ray, but no.

Spork wins.
 02/01/2014 12:51:54 AM · #23557
Art had this happen to me... :D
 02/01/2014 01:01:38 AM · #23558
Strongarming still can't bring you victory. It is mine.

eta: perhaps Beata can use those arms to rip bvy's degenerate graphs to pieces.

Message edited by author 2014-02-01 01:03:08.
 02/01/2014 01:04:13 AM · #23559
V I C T O R Y
I S
M I N E
:)
 02/01/2014 01:06:42 AM · #23560
Ok ok. You win.

...oh wait, apparently my good-natured generosity backfired. Eh, whatcanyado.
 02/01/2014 05:45:56 AM · #23561
Omg it's Madonna!

Originally posted by beatabg:
Art had this happen to me... :D
02/02/2014 02:18:28 AM · #23562
02/03/2014 06:12:21 PM · #23563
Originally posted by Art Roflmao:
eta: perhaps Beata can use those arms to rip bvy's degenerate graphs to pieces.

We refer to that process as "partitioning" not "ripping the graph to pieces." Sheesh.

I'll post the final proof tomorrow.
 02/03/2014 06:17:35 PM · #23564
Man when did Beata get so ripped????????????? She been eatin her spinach I can tell

Message edited by author 2014-02-03 18:17:57.
 02/04/2014 12:17:42 AM · #23565
Originally posted by bvy:
I'll post the final proof tomorrow.


That's some real suspense. I can tell everyone is waiting. I know I am.
 02/04/2014 12:36:01 AM · #23566
Just a friendly reminder...

Originally posted by Cory:
I was first, and I will be last... No other can reign over this domain.
02/04/2014 01:32:47 AM · #23567
thank you
 02/04/2014 01:40:26 AM · #23568
Originally posted by bvy:
Originally posted by Art Roflmao:
eta: perhaps Beata can use those arms to rip bvy's degenerate graphs to pieces.

We refer to that process as "partitioning" not "ripping the graph to pieces." Sheesh.

I'll post the final proof tomorrow.

And I'll very quickly rip it to pieces. Math brings out the Mr. Hyde in me. You've been warned.
 02/04/2014 02:24:03 AM · #23569
ok, now I am hiding you. ha ha.
 02/04/2014 03:13:37 AM · #23570
I once was lost, but now I'm found. Winning.
 02/04/2014 03:18:58 AM · #23571
Not here but I hope to see you winning elsewhere in the future.
For now I will win here,Thank you.
 02/04/2014 03:29:24 AM · #23572
I'm sorry, Tiny, but I really need as many wins as I can get. So if you'll kindly step aside...
 02/04/2014 03:50:51 AM · #23573
Should you not be getting your beauty sleep ?
 02/04/2014 04:20:42 AM · #23574
Sleep is for the lazy and the lazy never win.
 02/04/2014 07:54:49 AM · #23575
Originally posted by beatabg:
Originally posted by bvy:
I'll post the final proof tomorrow.


That's some real suspense. I can tell everyone is waiting. I know I am.


Finally, some enthusiasm. Okay, here we go:

For 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.

If G is l-critical with respect to X_(P_k), then d(G) >= (k + 1)(l - 1).

Proof: Let u be a vertex of least degree in G. Since G is l-critical with respect to X_(P_k), we can partition the vertices of G as follows: Place u in a partition by itself, and partition the remaining vertices of G into (necessarily) l - 1 subsets. Each of these l partitions induces a k-degenerate subgraph, including the single vertex u which is k-degenerate for all k >= 0. Observe that u must be incident with some vertex in each of the l - 1 partitions, for otherwise it could be included in any other partition as a disconnected vertex, and the induced subgraph would still be k-degenerate. Hence, deg(u) >= (l - 1). Expanding on that, u must be incident with at least k + 1 vertices in any partition such that it requires a partition all its own, and deg(u) >= (k + 1)(l - 1). Hence, d(G) >= (k + 1)(l - 1).

This concludes my exploration of the properties of k-degenerate graphs. Thank you all for your support. I next plan to share some fundamental results about edge colorings. At this point, I'm studying Vizing's Theorem and working through an interesting proof that uses induction. (I don't plan to restate it here as it's rather long, though not terribly complicated.) More to come...
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