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06/13/2011 04:18:23 AM · #6676 |
Now I'll have nightmares!
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06/13/2011 06:36:36 AM · #6677 |
Just to be sure...
Good night ladies. :) |
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06/13/2011 09:33:49 AM · #6678 |
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06/13/2011 02:18:56 PM · #6679 |
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06/13/2011 02:37:35 PM · #6680 |
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06/13/2011 02:54:06 PM · #6681 |
| ...it's Quagmire... Quagmire! |
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06/13/2011 02:56:21 PM · #6682 |
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06/13/2011 02:58:04 PM · #6683 |
Did someone say Eeeeek? |
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06/13/2011 02:58:17 PM · #6684 |
Originally posted by Art Roflmao:
quite what? |
Yeah, that's what I get for posting that time of the morning when half my brain is shutdown. LOL. Quiet...not quite. :) |
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06/13/2011 03:03:59 PM · #6685 |
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06/13/2011 03:05:20 PM · #6686 |
Simple mistake... Similar to how I ALWAYS... ALWAYS... mess up sense and since...
I have been told a million times which means which but for the life of me I can't get it right... EVER... Oh well...
I may eventually get it... Who knows... |
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06/13/2011 03:08:48 PM · #6687 |
| Typos, Spelling and Grammatical errors will not be toleratted. |
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06/13/2011 03:19:59 PM · #6688 |
| reelie? hay Art... y dontchya lyke speling errers? |
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06/13/2011 03:29:24 PM · #6689 |
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06/13/2011 03:51:24 PM · #6690 |
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06/13/2011 06:18:28 PM · #6691 |
| ironically, after 4 dyeing sessions in the past week that is a more accurate representation of my current hair color |
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06/13/2011 10:40:05 PM · #6692 |
| Can't pick a hair color or are you running from police? |
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06/13/2011 11:10:05 PM · #6693 |
| no, i can't accept that you can't really go from dark brown to blond in one step. or 4. i'm up to light orange creamsicle currently |
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06/13/2011 11:35:45 PM · #6694 |
| So, you can't even be a redhead, lol! Too bad for you...good for us! :P |
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06/14/2011 01:52:56 AM · #6695 |
| Give it up, Joe. You'll never have as much fun as Kat. |
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06/14/2011 07:26:40 AM · #6696 |
Let G be a group and S a set of generators of G. We construct the Cayley color digraph of G, Cay(S:G), as follows
- Associate a vertex of Cay(S:G) with each element of G.
- Assign a distinct color to each element of S.
- For each g in G and s in S, draw an s-colored arc (directed edge) from g to gs in Cay(S:G).
Cay(S:G) is connected: Observe that we may find the value of any product of n elements of S and their inverses, x1*x2*â€Â¦*xn, by traversing Cay(S:G) as follows: Start at the vertex associated with the identity element e. Follow the x1 arc. From this vertex, follow the x2 arc. Any time an inverse element is encountered, follow the arc in reverse. Continue following each arc in succession in the manner described. The vertex at which you finally arrive corresponds to the value of the desired product. Since S is a generating set of G, any vertex is reachable from e (more formally, a path exists between the two) and Cay(S:G) is connected.
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06/14/2011 09:05:13 AM · #6697 |
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06/14/2011 09:09:29 AM · #6698 |
| Find the value of this! * firmly raising middle finger * :P |
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06/14/2011 10:46:19 AM · #6699 |
| No finger gestures, lol. So, you can count to one...biggie!!!! :P |
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06/14/2011 02:37:20 PM · #6700 |
Thankfully, I have this feature...
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