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02/15/2011 02:36:38 AM · #5051 |
| Waz! You win. ...well not any more obviously. Welcome back, however brief. Happy VD. |
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02/15/2011 09:31:33 AM · #5052 |
| Morning....was a bit busy yesterday with my Daughter's BD Dinner...quite the production...in our family you get to pick your favorite menu and I make it...so I guess this means that I win...yet again |
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02/15/2011 12:41:53 PM · #5053 |
| Your guess would be wrong. It is me, I win. You get to make me dinner. I'll PM you my 6 course order. |
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02/15/2011 05:09:45 PM · #5054 |
Originally posted by Art Roflmao:
Your guess would be wrong. It is me, I win. You get to make me dinner. I'll PM you my 6 course order. |
now your turning into my daughter.... |
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02/15/2011 05:56:28 PM · #5055 |
| Does that mean I can borrow some money? |
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02/15/2011 06:08:55 PM · #5056 |
| NO....they don't even ask...I win |
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02/16/2011 02:09:09 AM · #5057 |
| Ok, well... what about that meal? |
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02/16/2011 02:35:18 AM · #5058 |
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02/16/2011 06:35:56 AM · #5059 |
| Claudia, Claudia, Claudia! I win. Now I go eat. |
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02/16/2011 12:06:47 PM · #5060 |
| Ken, Ken, Ken! and take a long nap after lunch... |
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02/16/2011 05:44:54 PM · #5061 |
| Yaawwwwwwn. I'm back. I'm full. I'm rested. I win. |
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02/16/2011 06:13:52 PM · #5062 |
what a wonderful world !!
Oh yeah
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02/17/2011 05:23:10 PM · #5063 |
Originally posted by bvy:
Later this week, I'll be demonstrating that all planar graphs of order greater than three, have at least four vertices of degree five or less. |
I'll bet you thought I forgot.
If G is a planar graph with p >= 4 vertices, then G has at least four vertices of degree less than six.
Observe that if we triangulate G to produce a new graph, Gâ, and prove the result for Gâ, then the result also holds for G. (The degree of each vertex of G is either less than or equal to what it is in Gâ since Gâ adds only edges to G.)
Consider the maximum number of vertices of Gâ that have degree six or greater. We maximize this by considering how many vertices of Gâ can have degree exactly six, given that vertices of Gâ have minimum degree three. Because the total degree of all vertices of Gâ is 6p â 12, the difference of 12 is spread out over no fewer than four vertices. Hence, at least four vertices of Gâ has degree less than six, and the same holds for G.
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02/17/2011 06:03:23 PM · #5064 |
You are right.. I know, at least 3 G for triangulate. Your thesis has been approved
Congrats! |
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02/17/2011 06:25:26 PM · #5065 |
| wow...aren't we all getting slow...guess I win again |
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02/17/2011 10:32:04 PM · #5066 |
maybe yesterday, not anymore.
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02/17/2011 10:40:09 PM · #5067 |
| what's wrong with today???? |
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02/18/2011 02:07:30 AM · #5068 |
| Today? Today is fine. In fact today is a good day. |
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02/18/2011 02:33:42 AM · #5069 |
| Can't believe you lot are still yabbering on in this thread. Go over to rant and have a read of the thread about 'infinity' - and then think about your chances of winning. |
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02/18/2011 02:34:34 AM · #5070 |
| my chances of winning are 1 in 1. (for the time being) |
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02/18/2011 02:41:03 AM · #5071 |
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02/18/2011 02:42:26 AM · #5072 |
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02/18/2011 02:43:23 AM · #5073 |
But not on Tuesday.
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02/18/2011 02:48:02 AM · #5074 |
| is it tuesday already? i coulda swore it was still happy hour. |
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02/18/2011 03:25:45 AM · #5075 |
| Ya well - diff'rent time zone over here... little bit ahead of you. Every time I win, you're are far behind no matter ho soon you respond. |
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