Author  Thread 

04/22/2010 04:58:15 PM · #51 
Originally posted by Five_Seat: I am a weiner. 
agreed...and I still need that cowbell.
:) 


04/22/2010 05:29:19 PM · #52 


04/22/2010 05:53:18 PM · #53 
Originally posted by Jdroullard: Pi
3.1416 ...
Message edited by GeneralE  DPC stats only go to 4 decimal places. 
Freaking classic. 


04/22/2010 07:00:10 PM · #54 
Originally posted by Runzamukk: ~:: ( cough ) ::~ 



04/22/2010 08:00:57 PM · #55 
Really? This thread is still going? 


04/22/2010 08:16:02 PM · #56 
Originally posted by Covert_Oddity: Really? This thread is still going? 
Yes, and you just lost.. :) 


04/22/2010 08:16:32 PM · #57 


04/22/2010 08:20:26 PM · #58 
Originally posted by Five_Seat: As did you. 
Not just yet my fiendish friend.. 


04/22/2010 08:43:59 PM · #59 
You know, we're not that different, you and I... 


04/22/2010 08:46:33 PM · #60 


04/22/2010 08:56:12 PM · #61 
Yes, Art is the end of the beginning. But it begs the question: Is it the beginning of the end? 


04/22/2010 09:41:18 PM · #62 
I need to go blow my nose. 


04/22/2010 10:14:16 PM · #63 


04/22/2010 10:20:12 PM · #64 
just a friendly reminder that I *could* win this easily. :P 


04/22/2010 11:08:36 PM · #65 


04/23/2010 04:19:39 AM · #66 
This thread is officially locked. 


04/23/2010 05:13:37 AM · #67 
Then I'll inofficially open it up again. 


04/23/2010 07:02:59 AM · #68 
This thread has now been officially laid to rest...



04/24/2010 09:08:38 AM · #69 
Nice try...the spirit has risen again! 


04/24/2010 12:20:40 PM · #70 
< This thread 


04/24/2010 03:28:53 PM · #71 
I will now demonstarte that a tournament T (a complete digraph) is transitive if an only if it contains no cycles. Let T be transitive. This means that for any three vertics of a tournament, a, b and c, with a > (directed to) b, and b > c, that a > c. Now suppose T has a cycle, C: v1 > v2 > v3 > ... vn > v1. Since v1 > v2, and v2 > v3, then v1 > v3 and the cycle can be reduced to v1 > v3 > ... vn > v1. Continuing in this manner, C is reduced to v1 > v(n1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n1) > vn suggests that v1 > vn is a directed edge of T. Yet the cycle suggests that vn > v1. Hence T can contain no cycles.
I will leave it as an exercise to the reader to prove the converse. If no takers, I'll follow up later. 


04/24/2010 03:30:43 PM · #72 
Originally posted by bvy: I will now demonstarte that a tournament T (a complete digraph) is transitive if an only if it contains no cycles. Let T be transitive. This means that for any three vertics of a tournament, a, b and c, with a > (directed to) b, and b > c, that a > c. Now suppose T has a cycle, C: v1 > v2 > v3 > ... vn > v1. Since v1 > v2, and v2 > v3, then v1 > v3 and the cycle can be reduced to v1 > v3 > ... vn > v1. Continuing in this manner, C is reduced to v1 > v(n1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n1) > vn suggests that v1 > vn is a directed edge of T. Yet the cycle suggests that vn > v1. Hence T can contain no cycles.
I will leave it as an exercise to the reader to prove the converse. If no takers, I'll follow up later. 
My shed must be T, cos it doesn't contain a cycle?? 


04/24/2010 04:11:01 PM · #73 


04/24/2010 05:25:27 PM · #74 
And I used to have such a high opinion of this site..............
Your all freaks......LOL
P.S. I win.



04/24/2010 08:00:11 PM · #75 
Originally posted by PixelKing: P.S. I win. 
Yes, yes you do. 
