Combinatorics (Fall 2010)/Extremal graphs: Difference between revisions
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{{Prooftitle|Second proof. (Cauchy-Schwarz inequality)| | {{Prooftitle|Second proof. (Cauchy-Schwarz inequality)| | ||
For any edge <math>uv\in E</math>, no vertex can be a neighbor of both <math>u</math> and <math>v</math>, or otherwise there will be a triangle. Thus, for any edge <math>uv\in E</math>, <math>d_u+d_v\le n</math>. It follows that | |||
:<math>\sum_{uv\in E}(d_u+d_v)\le n|E|</math>. | |||
Note that <math>d(v)</math> appears exactly <math>d_v</math> times in the sum, so that | |||
:<math>\sum_{uv\in E}(d_u+d_v)=\sum_{v\in V}d_v^2</math>. | |||
Applying Chauchy-Schwarz inequality, | |||
:<math> | |||
n|E|\ge\sum_{v\in V}d_v^2\ge\frac{\left(\sum_{v\in V}d_v\right)^2}{n}=\frac{4|E|^2}{n}, | |||
</math> | |||
where the last equation is due to Euler's equality <math>\sum_{v\in V}d_v=2|E|</math>. The theorem follows. | |||
}} | }} | ||
Revision as of 13:29, 13 October 2010
Extremal Graph Theory
Mantel's theorem
We consider a typical extremal problem for graphs: the largest possible number of edges of triangle-free graphs, i.e. graphs contains no [math]\displaystyle{ K_3 }[/math].
Theorem (Mantel 1907) - Suppose [math]\displaystyle{ G(V,E) }[/math] is graph on [math]\displaystyle{ n }[/math] vertice without triangles. Then [math]\displaystyle{ |E|\le\frac{n^2}{4} }[/math].
First proof. (pigeonhole principle) - [math]\displaystyle{ \square }[/math]
Second proof. (Cauchy-Schwarz inequality) For any edge [math]\displaystyle{ uv\in E }[/math], no vertex can be a neighbor of both [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math], or otherwise there will be a triangle. Thus, for any edge [math]\displaystyle{ uv\in E }[/math], [math]\displaystyle{ d_u+d_v\le n }[/math]. It follows that
- [math]\displaystyle{ \sum_{uv\in E}(d_u+d_v)\le n|E| }[/math].
Note that [math]\displaystyle{ d(v) }[/math] appears exactly [math]\displaystyle{ d_v }[/math] times in the sum, so that
- [math]\displaystyle{ \sum_{uv\in E}(d_u+d_v)=\sum_{v\in V}d_v^2 }[/math].
Applying Chauchy-Schwarz inequality,
- [math]\displaystyle{ n|E|\ge\sum_{v\in V}d_v^2\ge\frac{\left(\sum_{v\in V}d_v\right)^2}{n}=\frac{4|E|^2}{n}, }[/math]
where the last equation is due to Euler's equality [math]\displaystyle{ \sum_{v\in V}d_v=2|E| }[/math]. The theorem follows.
- [math]\displaystyle{ \square }[/math]
Third proof. (inequality of the arithmetic and geometric mean) Assume that [math]\displaystyle{ G(V,E) }[/math] has [math]\displaystyle{ |V|=n }[/math] vertices and is triangle-free.
Let [math]\displaystyle{ A }[/math] be the largest independent set in [math]\displaystyle{ G }[/math] and let [math]\displaystyle{ \alpha=|A| }[/math]. Since [math]\displaystyle{ G }[/math] is triangle-free, for very vertex [math]\displaystyle{ v }[/math], all its neighbors must form an independent set, thus [math]\displaystyle{ d(v)\le \alpha }[/math] for all [math]\displaystyle{ v\in V }[/math].
Take [math]\displaystyle{ B=V\setminus A }[/math] and let [math]\displaystyle{ \beta=|B| }[/math]. Since [math]\displaystyle{ A }[/math] is an independent set, all edges in [math]\displaystyle{ E }[/math] must have at least one endpoint in [math]\displaystyle{ B }[/math]. Counting the edges in [math]\displaystyle{ E }[/math] according to their endpoints in [math]\displaystyle{ B }[/math], we obtain [math]\displaystyle{ |E|\le\sum_{v\in B}d_v }[/math]. By the inequality of the arithmetic and geometric mean,
- [math]\displaystyle{ |E|\le\sum_{v\in B}d_v\le\alpha\beta\le\left(\frac{\alpha+\beta}{2}\right)^2=\frac{n^2}{4} }[/math].
- [math]\displaystyle{ \square }[/math]
Turán's theorem
Theorem (Turán 1941) - Let [math]\displaystyle{ G(V,E) }[/math] be a graph with [math]\displaystyle{ |V|=n }[/math]. If [math]\displaystyle{ G }[/math] has no [math]\displaystyle{ k }[/math]-clique, [math]\displaystyle{ k\ge 2 }[/math], then
- [math]\displaystyle{ |E|\le\left(1-\frac{1}{k-1}\right)\frac{n^2}{2} }[/math].
- Let [math]\displaystyle{ G(V,E) }[/math] be a graph with [math]\displaystyle{ |V|=n }[/math]. If [math]\displaystyle{ G }[/math] has no [math]\displaystyle{ k }[/math]-clique, [math]\displaystyle{ k\ge 2 }[/math], then
First proof. (induction) - [math]\displaystyle{ \square }[/math]
Second proof. (weight shifting) - [math]\displaystyle{ \square }[/math]
Third proof. (the probabilistic method) - [math]\displaystyle{ \square }[/math]
Fourth proof. Let [math]\displaystyle{ G(V,E) }[/math] be a [math]\displaystyle{ k }[/math]-clique-free graph on [math]\displaystyle{ n }[/math] vertices with a maximum number of edges.
- Claim: [math]\displaystyle{ G }[/math] does not contain three vertices [math]\displaystyle{ u,v,w }[/math] such that [math]\displaystyle{ uv\in E }[/math] but [math]\displaystyle{ uw\not\in E, vw\not\in E }[/math].
Suppose otherwise. There are two cases.
- Case.1: [math]\displaystyle{ d(w)\lt d(u) }[/math] or [math]\displaystyle{ d(w)\lt d(v) }[/math]. Without loss of generality, suppose that [math]\displaystyle{ d(w)\lt d(u) }[/math]. We duplicate [math]\displaystyle{ u }[/math] by creating a new vertex [math]\displaystyle{ u' }[/math] which has exactly the same neighbors as [math]\displaystyle{ u }[/math] (but [math]\displaystyle{ uu' }[/math] is not an edge). Such duplication will not increase the clique size. We then remove [math]\displaystyle{ w }[/math]. The resulting graph [math]\displaystyle{ G' }[/math] is still [math]\displaystyle{ k }[/math]-clique-free, and has [math]\displaystyle{ n }[/math] vertices. The number of edges in [math]\displaystyle{ G' }[/math] is
- [math]\displaystyle{ |E(G')|=|E(G)|+d(u)-d(w)\gt |E(G)|\, }[/math],
- which contradicts the assumption that [math]\displaystyle{ |E(G)| }[/math] is maximal.
- Case.2: [math]\displaystyle{ d(w)\ge d(u) }[/math] and [math]\displaystyle{ d(w)\ge d(v) }[/math]. Duplicate [math]\displaystyle{ w }[/math] twice and delete [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math]. The new graph [math]\displaystyle{ G' }[/math] has no [math]\displaystyle{ k }[/math]-clique, and the number of edges is
- [math]\displaystyle{ |E(G')|=|E(G)|+2d(w)-(d(u)+d(v)+1)\gt |E(G)|\, }[/math].
- Contradiction again.
The claim implies that [math]\displaystyle{ uv\not\in E }[/math] defines an equivalence relation on vertices (to be more precise, it guarantees the transitivity of the relation, while the reflexivity and symmetry hold directly). Graph [math]\displaystyle{ G }[/math] must be a complete multipartite graph [math]\displaystyle{ K_{n_1,n_2,\ldots,n_{k-1}} }[/math] with [math]\displaystyle{ n_1+n_2+\cdots +n_{k-1}=n }[/math]. Optimize the edge number, we have the Turán graph.
- [math]\displaystyle{ \square }[/math]