| Author | Thread |
|
|
02/05/2016 10:35:36 AM · #27426 |
Originally posted by bvy:
Originally posted by kawesttex:
hi |
Hello. |
Whatttttssssssuppppppppp!!!!!! |
|
|
|
02/05/2016 03:10:27 PM · #27427 |
|
|
|
02/05/2016 09:36:57 PM · #27428 |
|
|
|
02/05/2016 09:38:12 PM · #27429 |
| Next up, I'd like to show that the center of a tree -- that is, the subgraph induced by those vertices of least eccentricity -- is always either K1 or K2. |
|
|
|
02/05/2016 10:13:57 PM · #27430 |
| trees and K9s are an ancient verity. |
|
|
|
02/06/2016 12:33:19 AM · #27431 |
K1 is a battle tank
of the South Koreans;
K2 is a mountain peak
that's been around for eons.
What's this nonsense, guys? Lay off
the mathing and the graphing
before the rest of us go weak-
kneed from spastic laughing!
Message edited by author 2016-02-06 00:35:57. |
|
|
|
02/06/2016 10:13:42 AM · #27432 |
|
|
|
02/06/2016 12:18:51 PM · #27433 |
| Doggerel, dude; even better! |
|
|
|
02/08/2016 03:44:57 PM · #27434 |
Originally posted by bvy:
Next up, I'd like to show that the center of a tree -- that is, the subgraph induced by those vertices of least eccentricity -- is always either K1 or K2. |
As promised, I'd like to submit the following for your consideration.
First some terminology. The eccentricity of a vertex v in some graph G, notated e(v), is the maximum distance between v and any other vertex of G. The radius of G, notated rad(G), is the least of all the vertex eccentricities in G. The center of a graph G is the set of vertices whose eccentricity equals rad(G).
Consider the longest path P = u1, u2, ..., un in T. For any vertex ui on P, e(ui) = i if i > n/2 and e(ui) = n - i otherwise. In other words, the eccentricity of ui is the distance to the farthest leaf node (u1 or un) on P. This must be true, for if there is a vertex farther away than the farthest leaf node, then P is not the longest path in T. So e(u) >= floor(n/2) for every u on P, and there is at least one vertex v on P with e(v) = floor(n/2).
For every other vertex v in T (not on P), necessarily e(v) > floor(n/2). To see this, observe that v has a unique path to exactly one vertex, ui, on P. This distance is at least one, and the distance from ui to its farthest leaf node on P is at least floor(n/2).
Therefore rad(T) = floor(n/2), and the center of T lies on P. If the length of P is odd, then exactly one vertex on P has rad(T) = floor(n/2), and T is centered on this one vertex (K1). If the length of P is even, then two adjacent vertices on P have rad(T) = floor(n/2), and T is bicentered on these two vertices (K2).
|
|
|
|
02/08/2016 04:05:19 PM · #27435 |
|
|
|
02/08/2016 04:24:56 PM · #27436 |
 Kawesttex has a sebentary brain... |
|
|
|
02/08/2016 04:31:23 PM · #27437 |
|
|
|
02/08/2016 06:57:01 PM · #27438 |
| But it's a bridge over troubled waters, not a tree! |
|
|
|
02/08/2016 07:13:21 PM · #27439 |
|
|
|
02/08/2016 08:22:59 PM · #27440 |
|
|
|
02/08/2016 09:42:15 PM · #27441 |
|
|
|
02/08/2016 09:42:17 PM · #27442 |
.
Message edited by author 2016-02-08 21:42:33. |
|
|
|
02/08/2016 09:47:48 PM · #27443 |
| folks we not trouble in River City |
|
|
|
02/10/2016 06:20:18 PM · #27444 |
| I'm in trouble everywhere I go... Opposition more corner time for me... |
|
|
|
02/16/2016 12:25:35 AM · #27445 |
|
|
|
02/16/2016 08:38:30 AM · #27446 |
|
|
|
02/17/2016 01:38:33 PM · #27447 |
|
|
|
02/17/2016 09:21:09 PM · #27448 |
|
|
|
02/18/2016 09:33:22 AM · #27449 |
| Way to beat the dead horse. |
|
|
|
02/18/2016 09:33:30 AM · #27450 |
Double Post
Message edited by author 2016-02-18 09:33:49. |
|
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: 08/28/2025 08:56:17 AM EDT.
