Author | Thread |
|
01/16/2011 10:30:56 PM · #4751 |
your on Craig...365 Days...eish...yippy...need to send Langdon a better bribe to shut this down... |
|
|
01/17/2011 12:34:42 AM · #4752 |
What you say bvy? A 5-color G-string in the Theorem? Nah...not for me. |
|
|
01/17/2011 05:05:59 AM · #4753 |
I found only one flaw in bvy's theorum. |
|
|
01/17/2011 08:55:40 AM · #4754 |
what...it's "his" theorum....is that the flaw? |
|
|
01/17/2011 10:31:50 AM · #4755 |
I too am curious to hear about the flaw that Arthur discovered.
The part of the theorem that discusses the G2 subgraph not containing v4 because it is surrounded by the v1-v3 path is perhaps a little unclear. Would a visual help?
|
|
|
01/17/2011 05:59:17 PM · #4756 |
The only flaw is the fact that you posted it in the thread where MATH IS BANNED! Other than that, it is sound. |
|
|
01/17/2011 06:35:00 PM · #4757 |
Banned....banned...banned....I win |
|
|
01/17/2011 06:36:27 PM · #4758 |
Okay, then. I'll post a visual. Give me a short while to prepare something. |
|
|
01/17/2011 06:40:30 PM · #4759 |
Not to boast, but Discrete Mathematics was one of my fields of research (CS background) - Having investigated graph theory stemming from the seminal works of Kempe/Heawood, I can spot one fundamental flaw in bvy's reasoning. |
|
|
01/17/2011 07:24:32 PM · #4760 |
I also studied discrete mathematics. In fact, it was so discrete it's like I wasn't even in class. I majored in thread winning though. |
|
|
01/17/2011 07:30:22 PM · #4761 |
Originally posted by JH: Not to boast, but Discrete Mathematics was one of my fields of research (CS background) - Having investigated graph theory stemming from the seminal works of Kempe/Heawood, I can spot one fundamental flaw in bvy's reasoning. |
Well, don't hold back... Let's hear it! |
|
|
01/17/2011 08:43:23 PM · #4762 |
Sorry, I was away double-checking my workings. Yup, there's definitely an issue with one assumption you're making. |
|
|
01/17/2011 09:16:11 PM · #4763 |
You're a tease. Out with it! |
|
|
01/17/2011 09:17:21 PM · #4764 |
Is it the dedication? Of course Art doesn't win. He never wins. |
|
|
01/18/2011 02:59:22 AM · #4765 |
Originally posted by bvy: First note that the following statements about a planar graph G are equivalent:
- G is five-colorable.
- G has chromatic number less than or equal to five.
- G can be colored using five or fewer colors. |
These are all correct, but you failed to mention that the 5 color G also has radioactive fire breath.
edit to add: But you did mention that I win and that is also correct.
Message edited by author 2011-01-18 03:00:39. |
|
|
01/18/2011 09:39:48 AM · #4766 |
|
|
01/18/2011 02:46:08 PM · #4767 |
I have now ignored all threads but this one. I am also ignoring all of you. I'm so ignorant. I love it! |
|
|
01/18/2011 03:38:19 PM · #4768 |
All comments posted in this thread copyright me ©2010.
As such no matter what you post I win because I will own the contents on your post.
Should you have any issues with this...tough. |
|
|
01/18/2011 04:08:00 PM · #4769 |
Originally posted by bvy: You're a tease. Out with it! |
G - v sub-graph... hmmm.... |
|
|
01/18/2011 04:19:03 PM · #4770 |
Originally posted by JH: Originally posted by bvy: You're a tease. Out with it! |
G - v sub-graph... hmmm.... |
I was going to mention the same issue, but I went with the psychedelic Godzilla instead. |
|
|
01/18/2011 08:34:17 PM · #4771 |
He's beautiful, Art. Really.
JH, I'm not sure what you're getting at. G - v is a subgraph, although I didn't call it that. By G - v, I mean the original planar graph G, take away v and all edges incident with v. |
|
|
01/18/2011 09:00:32 PM · #4772 |
Originally posted by bvy: He's beautiful, Art. Really.
JH, I'm not sure what you're getting at. G - v is a subgraph, although I didn't call it that. By G - v, I mean the original planar graph G, take away v and all edges incident with v. |
Oh yes, I understand that part, of course. However... this assertion: "Clearly G2 cannot contain v4, for if it did, there would have to be a v2-v4 path in G - v which did not intersect the v1-v3 path." .... |
|
|
01/18/2011 09:27:43 PM · #4773 |
I just entered my 100th Challenge...yea....and I win this blooming thread...done... |
|
|
01/18/2011 09:57:59 PM · #4774 |
Originally posted by JH: Originally posted by bvy: He's beautiful, Art. Really.
JH, I'm not sure what you're getting at. G - v is a subgraph, although I didn't call it that. By G - v, I mean the original planar graph G, take away v and all edges incident with v. |
Oh yes, I understand that part, of course. However... this assertion: "Clearly G2 cannot contain v4, for if it did, there would have to be a v2-v4 path in G - v which did not intersect the v1-v3 path." .... |
Okay, work through this with me. If I can make it clearer for these nice people, I'd like to.
Perhaps a poorly rendered diagram will help.
I define G2 as the connected component subgraph of G - v having, as shown, all blue and green vertices conneceted to v2. The v1-v3 path is shown in red and yellow; it circumscribes G2. G2 cannot contain v4 -- to make the connection, there would have to be a v2-v4 path in G2, and the only way to get to v4 from v2 is to cross the v1-v3 path. (All casually stated.) |
|
|
01/18/2011 10:23:40 PM · #4775 |
Meh, I'm only here for the pictures!! ;p
I graduated years ago, thus math is now boring and somewhat over my head, lol! |
|
Home -
Challenges -
Community -
League -
Photos -
Cameras -
Lenses -
Learn -
Help -
Terms of Use -
Privacy -
Top ^
DPChallenge, and website content and design, Copyright © 2001-2025 Challenging Technologies, LLC.
All digital photo copyrights belong to the photographers and may not be used without permission.
Current Server Time: 07/18/2025 06:23:55 PM EDT.