Talk:K-edge-connected graph

This should NOT redirect from "k-connected graph". The two are very distinct concepts, and the latter NEEDS an entry...


 * You're right. Fixed. —David Eppstein 16:49, 22 October 2006 (UTC)

This seems to be redundant, given the Connectivity (graph theory) article. Radagast3 (talk) 00:18, 1 May 2008 (UTC)


 * There is plenty of material to expand this and K-vertex-connected graph into two separate long articles. —David Eppstein (talk) 00:20, 1 May 2008 (UTC)
 * Agreed, but I'm too busy to do it. ;) Radagast3 (talk) 00:36, 1 May 2008 (UTC)

Complexity
Gabow's paper states $$O(k m \ln (n^2/m))$$ complexity, not $$O(n^3)$$. Reminiscenza (talk) 11:40, 2 June 2015 (UTC)
 * There's no contradiction between those two bounds. k is necessarily O(n) and m is necessarily O(n2). So if you eliminate those two variables and express everything in terms of n, you get O(n3).—David Eppstein (talk) 15:20, 2 June 2015 (UTC)
 * Yes, but why we need to express everything in terms of n? For graph algorithms, complexity is usually expressed in terms of m and n, and whatever other parameters available - so that the bound is useful for both sparse and dense graphs. Bound $$O(n^3)$$ is at least misleading for sparse graphs, where it is possible that $$k = O(1)$$, $$m = O(n \ln n)$$ and overall complexity would be $$O(n \ln^2 n)$$, much better than $$O(n^3)$$. — Preceding unsigned comment added by Reminiscenza (talk • contribs) 06:21, 3 June 2015 (UTC)

Proper math

 * In graph theory, a graph $$G$$ with edge set $$E(G)$$ is said to be $$k$$-edge-connected if $$G \setminus X$$ is connected for all $$X \subseteq E(G)$$ with $$\left| X \right| < k$$.
 * Is it syntactically correct to say $$G \setminus X$$, as G is a graph (= a set of vertices and a set of edges connecting some vertices) whereas X is a set of edges? --Abdull (talk) 15:33, 27 July 2008 (UTC)

Stronger

 * If a graph $$G$$ is $$k$$-edge-connected then $$k \le \delta(G)$$, where $$\delta(G)$$ is the minimum degree of any vertex $$v \in V(G)$$.
 * Can we make this theorem stronger by saying $$k < \delta(G)$$, as we could delete all but one edges from a vertex and still have all vertices of this connected graph connected? --Abdull (talk) 15:55, 27 July 2008 (UTC)

Consistency
It is usual, even on this WIKI, to write a graph as an ordered pair (V,E) with vertex set V and edge set E. This article does it the other way around. Is there a compelling reason for this? Leen Droogendijk (talk) 11:31, 28 June 2012 (UTC)

Contradiction between the formal definition and the next section
Consider a graph G of a single vertex and no edge. According to the formal definition, G is k-edge-connected for all k. On the other hand, the minimum vertex degree of G is 0, thus the statement of the second section does not hold for this example. --Jalpar75 —Preceding undated comment added 15:59, 9 August 2013 (UTC)