DPChallenge: A Digital Photography Contest You are not logged in. (log in or register
 

DPChallenge Forums >> General Discussion >> Last wins
Pages:   ... [185] [186] [187] [188] [189] [190] [191] [192] [193] ... [1227]
Showing posts 4701 - 4725 of 30665, (reverse)
AuthorThread
01/13/2011 02:52:53 PM · #4701
*crumple crumple rip shred chomp chomp*
01/13/2011 06:08:11 PM · #4702
so...I guess I'm still winning...
01/13/2011 07:25:03 PM · #4703
guess again.
01/13/2011 07:26:42 PM · #4704
We all know I'll win, nice try Art.
01/13/2011 07:27:11 PM · #4705
what are you two talking about!!! I WIN...no need to respond
01/13/2011 07:30:20 PM · #4706
If I ever waited for a "need" to respond, I would disappear from this site completely. :P
01/13/2011 07:56:22 PM · #4707
You do have a point Ken...
01/13/2011 07:59:20 PM · #4708
Must....resist....

The needs of the many outweigh the needs of the few. Or the one.
01/14/2011 01:56:46 AM · #4709
resistance is futile. bow to your superior "Last wins" competitor.
01/14/2011 02:30:01 AM · #4710
this is not some silly square dance...there will be no bowing!
01/14/2011 02:43:02 AM · #4711
Damn. Then I better change my shoes.
01/14/2011 02:49:54 AM · #4712
How about bowling? I'm sure Ken wouldn't mind being a pin.

01/14/2011 02:57:28 AM · #4713
Alllllrighty. Now I have to go change my shoes once again.
01/14/2011 08:22:21 AM · #4714
Good morning. A planar graph has a vertex of degree less than or equal to five.

Proof: The result is obvious for planar graphs with six or fewer vertices since the maximum degree of any vertex is five.

Let G(p, q) be a planar graph with p >= 7, and suppose G has no vertices of degree less than or equal to five. Then every vertex of G has degree six or greater. The sum of the degrees of the vertices is 2q which must be greater than or equal to 6p, since each of p vertices has degree six or greater. So 2q >= 6p or q >= 3p. However, a corollary of Euler’s formula (p – q + r = 2 where r is the number of regions of a planar graph) states that q <= 3p – 6. Combining these results, we get 3p <= q <= 3p – 6, or 3p <= 3p – 6, which is a contradiction. Hence, G must have some vertex of degree less than or equal to five.

We will use this result later.
01/14/2011 09:04:52 AM · #4715
oh, good grief Brian...it's just way to early for this....I will have to say "I win" again
01/14/2011 01:00:57 PM · #4716
I REALLY LOVE bowling!
01/14/2011 01:35:50 PM · #4717
Well I do too...maybe we should all meet there...I win
01/14/2011 05:12:46 PM · #4718
Win again, I do.
01/14/2011 06:35:52 PM · #4719
yes...I do win again...thanks Art...
01/14/2011 06:45:21 PM · #4720
(>.<)
01/14/2011 07:02:57 PM · #4721
and again....
01/14/2011 07:07:40 PM · #4722
Tennis, anyone?
01/14/2011 07:08:23 PM · #4723
stomps foot!!!! I WIN...NO NEED TO RESPOND....HUMF
01/14/2011 07:10:46 PM · #4724
01/14/2011 07:35:39 PM · #4725
damn it...I win...now shoooo
Pages:   ... [185] [186] [187] [188] [189] [190] [191] [192] [193] ... [1227]
Current Server Time: 07/18/2025 04:00:25 PM

Please log in or register to post to the forums.


Home - Challenges - Community - League - Photos - Cameras - Lenses - Learn - Help - Terms of Use - Privacy - Top ^
DPChallenge, and website content and design, Copyright © 2001-2025 Challenging Technologies, LLC.
All digital photo copyrights belong to the photographers and may not be used without permission.
Current Server Time: 07/18/2025 04:00:25 PM EDT.