Wikipedia:Reference desk/Archives/Mathematics/2016 February 6

= February 6 =

question in graph theory
Hi,

Suppose we have a graph $$G$$ where $$|V(G)|\ge k+1$$ ($$k\in\mathbb{N}$$), and for all two non-neighbors vertices it holds that $$d(u)+d(v)\ge2k$$. How can we prove that the average degree of this graph is at least $$k$$?

Tnanks — Preceding unsigned comment added by 217.132.96.145 (talk) 19:51, 6 February 2016 (UTC)


 * Just sum over all vertices, and show that the sum is greater than $$|V|\cdot 2k$$.
 * The idea is to sum over couples of non-neighbors vertices.
 * If all the vertices in the graph are neighbors, we're done, since $$|V|\ge k+1$$, so each node has degree $$\ge k$$.
 * Also, if all the vertices have degree $$\ge k$$, we're done.
 * Otherwise, there're at least 2 non-neighbors vertices, v and u, that one of which has degreek.
 * We know that $$d(u)+d(v)\ge 2k$$. WLOG $$d(v) > d(u)$$. So, $$d(v)\ge k+1$$
 * Now, for every vertex, u, which is not a neighbor of v, it holds that $$d(u)+d(v)\ge2k$$. So, $$d(u)\ge2k-n$$.
 * Now, we remain only with the neighbors of v.
 * If they're all neighbors, then we know that their degree $$\ge n-1 \ge k$$.
 * Otherwise, there are two vertices that are not neighbors - fix one of which and continue this way recursively.
 * Since the statement (that the average of the degrees over the fixed vertex and its non-neighbors vertices is >= k) holds all the time during the recursion, so the statement is correct.
 * Notice that this method of recursion is similar to inducion, that you're probably more familiar with. עברית (talk) 10:39, 8 February 2016 (UTC)
 * I'm not clear on everything here but I'm pretty sure there is a flaw in this argument. The statement that there must be a pair of non-neighbors u and v with d(u)k does not follow; take k=3 and consider the complete bipartite graph K2,4. Also note that you only need to show that the sum of the degrees is at least |V|k, not 2|V|k. --RDBury (talk) 22:43, 8 February 2016 (UTC)
 * Here's what I came up with. First, to simplify notation, let n=|V| = number of vertices in G, and let s be the degree sum = twice the number of edges. So
 * $$s = \sum_u d(u).$$
 * We need to show s≥kn. Write
 * $$sn = \sum_v d(v) \sum_u 1 = \sum_{u,v} d(u)$$
 * so
 * $$2sn = \sum_{u,v} (d(u)+d(v)).$$
 * Split this sum according to u=v, u adjacent to v and u not adjacent to v. I'll use $$\sim$$ and $$\nsim$$ for adjacent and non-adjacent. (Is there a standard notation for this or do graph theorists have to write "adjacent" all the time?)
 * $$2sn = \sum_{u = v} (d(u)+d(v)) + \sum_{u \sim v} (d(u)+d(v)) + \sum_{u \nsim v} (d(u)+d(v)).$$
 * The first sum is simply
 * $$\sum_u 2d(u) = 2s.$$
 * The second sum is
 * $$\sum_{u \sim v} (d(u)+d(v)) = 2\sum_{u \sim v} d(u) = 2\sum_u d(u)^2 \ge \frac{2}{n} \left ( \sum_u d(u) \right )^2 = \frac{2s^2}{n}$$
 * by a well known inequality I can't remember the name of at the moment. (Please fill this in if you know.) There are n2-n-s terms in the third sum so by assumption this is
 * $$\sum_{u \nsim v} (d(u)+d(v)) \ge 2k(n^2-n-s).$$
 * Putting this together gives
 * $$2sn \ge 2s + \frac{2s^2}{n} + 2k(n^2-n-s)$$
 * $$sn^2 \ge sn + s^2 + kn(n^2-n-s)$$
 * $$sn^2 - sn - s^2 - kn(n^2-n-s) \ge 0$$
 * $$(s-kn)(n^2-n-s) \ge 0.$$
 * As pointed out above, there must be at least one non-edge so
 * $$n^2-n-s > 0$$
 * and so
 * $$s-kn \ge 0.$$
 * Note, the inequality used above is strict unless the d(u)'s are all equal, so the average degree is strictly greater than k unless G is k-regular. --RDBury (talk) 00:30, 9 February 2016 (UTC)
 * Re the inequality used above, the Cauchy–Schwarz inequality says
 * $$\left( \sum_{i=1}^n x_i y_i \right)^2 \leq \sum_{j=1}^n {x_j}^2 \sum_{k=1}^n {y_k}^2$$
 * and with yi≡1 this is
 * $$\left( \sum_{i=1}^n x_i \right)^2 \leq n \sum_{i=1}^n {x_i}^2,$$
 * but I thought there was another name for this special case. --RDBury (talk) 00:57, 9 February 2016 (UTC)

Thank you all! — Preceding unsigned comment added by 217.132.96.145 (talk) 16:38, 10 February 2016 (UTC)