Combinatorics (Fall 2010)/Extremal graphs: Difference between revisions
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The claim implies that <math>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). | The claim implies that <math>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>G</math> must be a complete multipartite graph <math>K_{n_1,n_2,\ldots,n_k}</math> with <math>n_1+n_2+\cdots +n_k=n</math>. Optimize the edge number, we can show that <math>(n_1,n_2,\ldots,n_k)</math> must be as balanced as possible. The theorem follows. | ||
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Revision as of 09:08, 13 October 2010
Extremal Graph Theory
Mantel's theorem
Theorem (Mantel 1907) - If a graph [math]\displaystyle{ G }[/math] on [math]\displaystyle{ 2n }[/math] vertices contains [math]\displaystyle{ n^2+1 }[/math] edges, then [math]\displaystyle{ G }[/math] contains a triangle.
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+1) }[/math]-clique, [math]\displaystyle{ k\ge 2 }[/math], then
- [math]\displaystyle{ |E|\le\left(1-\frac{1}{k}\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+1) }[/math]-clique, [math]\displaystyle{ k\ge 2 }[/math], then
First proof. (induction) - [math]\displaystyle{ \square }[/math]
Second proof. (construction) - [math]\displaystyle{ \square }[/math]
Third proof. (shifting) - [math]\displaystyle{ \square }[/math]
Fourth proof. (the probabilistic method) - [math]\displaystyle{ \square }[/math]
Fifth proof. Let [math]\displaystyle{ G(V,E) }[/math] be a [math]\displaystyle{ (k+1) }[/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+1) }[/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+1) }[/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} }[/math] with [math]\displaystyle{ n_1+n_2+\cdots +n_k=n }[/math]. Optimize the edge number, we can show that [math]\displaystyle{ (n_1,n_2,\ldots,n_k) }[/math] must be as balanced as possible. The theorem follows.
- [math]\displaystyle{ \square }[/math]