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Showing posts 51 - 75 of 28973, (reverse)
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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
You all lose, I WIN!
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
As did you.
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
This ends now.
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
snot!
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
Yeah yeah, whatever.
04/23/2010 04:19:39 AM · #66
Copyrighted_Image_Reuse_Prohibited_872200.gif
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...

Copyrighted_Image_Reuse_Prohibited_92554.jpg
04/24/2010 09:08:38 AM · #69
Nice try...the spirit has risen again!
04/24/2010 12:20:40 PM · #70
torched_horse.gif <---- 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(n-1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n-1) > 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(n-1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n-1) > 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
T is for Tricycle
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.
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