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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.
:) |
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04/22/2010 05:29:19 PM · #52 |
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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. |
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04/22/2010 07:00:10 PM · #54 |
Originally posted by Runzamukk:
~:: ( cough ) ::~ |
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04/22/2010 08:00:57 PM · #55 |
| Really? This thread is still going? |
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04/22/2010 08:16:02 PM · #56 |
Originally posted by Covert_Oddity:
Really? This thread is still going? |
Yes, and you just lost.. :) |
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04/22/2010 08:16:32 PM · #57 |
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04/22/2010 08:20:26 PM · #58 |
Originally posted by Five_Seat:
As did you. |
Not just yet my fiendish friend.. |
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04/22/2010 08:43:59 PM · #59 |
| You know, we're not that different, you and I... |
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04/22/2010 08:46:33 PM · #60 |
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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? |
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04/22/2010 09:41:18 PM · #62 |
| I need to go blow my nose. |
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04/22/2010 10:14:16 PM · #63 |
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04/22/2010 10:20:12 PM · #64 |
| just a friendly reminder that I *could* win this easily. :P |
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04/22/2010 11:08:36 PM · #65 |
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04/23/2010 04:19:39 AM · #66 |
This thread is officially locked. |
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04/23/2010 05:13:37 AM · #67 |
| Then I'll inofficially open it up again. |
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04/23/2010 07:02:59 AM · #68 |
This thread has now been officially laid to rest...
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04/24/2010 09:08:38 AM · #69 |
| Nice try...the spirit has risen again! |
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04/24/2010 12:20:40 PM · #70 |
 <---- This thread |
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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. |
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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?? |
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04/24/2010 04:11:01 PM · #73 |
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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.
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04/24/2010 08:00:11 PM · #75 |
Originally posted by PixelKing:
P.S. I win. |
Yes, yes you do. |
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