组合数学 (Fall 2011)/Counting and existence

Forbidden Cliques

Extremal graph theory studies the problems like "how many edges that a graph [math]\displaystyle{ G }[/math] can have, if [math]\displaystyle{ G }[/math] has some property?"

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].

We give three different proofs of the theorem. The first one uses induction and an argument based on pigeonhole principle. The second proof uses the famous Cauchy-Schwarz inequality in analysis. And the third proof uses another famous inequality: the inequality of the arithmetic and geometric mean.

First proof. (pigeonhole principle) We prove an equivalent theorem: Any [math]\displaystyle{ G(V,E) }[/math] with [math]\displaystyle{ |V|=n }[/math] and [math]\displaystyle{ |E|\gt \frac{n^2}{4} }[/math] must have a triangle. Use induction on [math]\displaystyle{ n }[/math]. The theorem holds trivially for [math]\displaystyle{ n\le 3 }[/math]. Induction hypothesis: assume the theorem hold for [math]\displaystyle{ |V|\le n-1 }[/math]. For [math]\displaystyle{ G }[/math] with [math]\displaystyle{ n }[/math] vertices, without loss of generality, assume that [math]\displaystyle{ |E|=\frac{n^2}{4}+1 }[/math], we will show that [math]\displaystyle{ G }[/math] must contain a triangle. Take a [math]\displaystyle{ uv\in E }[/math], and let [math]\displaystyle{ H }[/math] be the subgraph of [math]\displaystyle{ G }[/math] induced by [math]\displaystyle{ V\setminus \{u,v\} }[/math]. Clearly, [math]\displaystyle{ H }[/math] has [math]\displaystyle{ n-2 }[/math] vertices. Case.1: If [math]\displaystyle{ H }[/math] has [math]\displaystyle{ \gt \frac{(n-2)^2}{4} }[/math] edges, then by the induction hypothesis, [math]\displaystyle{ H }[/math] has a triangle. Case.2: If [math]\displaystyle{ H }[/math] has [math]\displaystyle{ \le\frac{(n-2)^2}{4} }[/math] edges, then at least [math]\displaystyle{ \left(\frac{n^2}{4}+1\right)-\frac{(n-2)^2}{4}-1=n-1 }[/math] edges are between [math]\displaystyle{ H }[/math] and [math]\displaystyle{ \{u,v\} }[/math]. By pigeonhole principle, there must be a vertex in [math]\displaystyle{ H }[/math] that is adjacent to both [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math]. Thus, [math]\displaystyle{ G }[/math] has a triangle. [math]\displaystyle{ \square }[/math]

Second proof. (Cauchy-Schwarz inequality) (Mantel's original proof) 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_{uv\in E}(d_u+d_v)=\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

The famous Turán's theorem generalizes the Mantel's theorem for triangles to cliques of any specific size. This theorem is one of the most important results in extremal combinatorics, which initiates the studies of extremal graph theory.

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{ r }[/math]-clique, [math]\displaystyle{ r\ge 2 }[/math], then [math]\displaystyle{ |E|\le\frac{r-2}{2(r-1)}n^2 }[/math].

We give an example of graphs with many edges which does not contain [math]\displaystyle{ K_r }[/math].

Partition [math]\displaystyle{ V }[/math] into [math]\displaystyle{ r-1 }[/math] disjoint classes [math]\displaystyle{ V=V_1\cup V_2\cup\cdots\cup V_{r-1} }[/math], [math]\displaystyle{ n_i=|V_i| }[/math], [math]\displaystyle{ n_1+n_2+\cdots+n_{r-1}=n }[/math]. For every two vertice [math]\displaystyle{ u,v }[/math], [math]\displaystyle{ uv\in E }[/math] if and only if [math]\displaystyle{ u\in V_i }[/math] and [math]\displaystyle{ v\in V_j }[/math] for distinct [math]\displaystyle{ V_i }[/math] and [math]\displaystyle{ V_j }[/math]. The resulting graph is a complete [math]\displaystyle{ (r-1) }[/math]-partite graph, denoted [math]\displaystyle{ K_{n_1,n_2,\ldots,n_{r-1}} }[/math]. It is obvious that any [math]\displaystyle{ (r-1) }[/math]-partite graph contains no [math]\displaystyle{ r }[/math]-clique since only those vertices from different classes can be adjacent.

A [math]\displaystyle{ K_{n_1,n_2,\ldots,n_{r-1}} }[/math] has [math]\displaystyle{ \sum_{i\lt j}n_i n_j\, }[/math] edges, which is maximized when the numbers [math]\displaystyle{ n_i }[/math] are divided as evenly as possible, that is, if [math]\displaystyle{ n_i\in\left\{\left\lfloor\frac{n}{r-1}\right\rfloor,\left\lceil\frac{n}{r-1}\right\rceil\right\} }[/math] for every [math]\displaystyle{ 1\le i\le r-1 }[/math].

Definition

We call a complete multipartite graph [math]\displaystyle{ K_{n_1,n_2,\ldots,n_{r-1}} }[/math] with [math]\displaystyle{ n_i\in\left\{\left\lfloor\frac{n}{r-1}\right\rfloor,\left\lceil\frac{n}{r-1}\right\rceil\right\} }[/math] for every [math]\displaystyle{ i }[/math] a Turán graph, denoted [math]\displaystyle{ T(n,r-1) }[/math].

Example

Turán graph [math]\displaystyle{ T(13,4) }[/math]

Turán graph [math]\displaystyle{ T(13,4) }[/math]

Turán's theorem has been proved for many times by different mathematicians, with different tools. We show just a few.

The first proof uses induction; the second proof uses a technique called "weight shifting"; and the third proof uses the probabilistic method. All of them are very powerful and frequently used proof techniques.

First proof. (induction) (Turán's original proof) Induction on [math]\displaystyle{ n }[/math]. It is easy to verify that the theorem holds for [math]\displaystyle{ n\lt r }[/math]. Let [math]\displaystyle{ G }[/math] be a graph on [math]\displaystyle{ n }[/math] vertices without [math]\displaystyle{ r }[/math]-cliques where [math]\displaystyle{ n\ge r }[/math]. Suppose that [math]\displaystyle{ G }[/math] has a maximum number of edges among such graphs. [math]\displaystyle{ G }[/math] certainly has [math]\displaystyle{ (r-1) }[/math]-cliques, since otherwise we could add edges to [math]\displaystyle{ G }[/math]. Let [math]\displaystyle{ A }[/math] be an [math]\displaystyle{ (r-1) }[/math]-clique and let [math]\displaystyle{ B=V\setminus A }[/math]. Clearly [math]\displaystyle{ |A|=r-1 }[/math] and [math]\displaystyle{ |B|=n-r+1 }[/math]. By the induction hypothesis, since [math]\displaystyle{ B }[/math] has no [math]\displaystyle{ r }[/math]-cliques, [math]\displaystyle{ |E(B)|\le\frac{r-2}{2(r-1)}(n-r+1)^2 }[/math]. And [math]\displaystyle{ E(A)={r-1\choose 2} }[/math]. Since [math]\displaystyle{ G }[/math] has no [math]\displaystyle{ r }[/math]-clique, every [math]\displaystyle{ v\in B }[/math] is adjacent to at most [math]\displaystyle{ r-2 }[/math] vertices in [math]\displaystyle{ A }[/math], since otherwise [math]\displaystyle{ A }[/math] and [math]\displaystyle{ v }[/math] would form an [math]\displaystyle{ r }[/math]-clique. We obtain that the number edges crossing between [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] is [math]\displaystyle{ |E(A,B)|\le (r-2)|B|=(r-2)(n-r+1) }[/math]. Combining everything together, [math]\displaystyle{ |E|=|E(A)|+|E(B)|+|E(A,B)|\le {r-1\choose 2}+\frac{r-2}{2(r-1)}(n-r+1)^2+(r-2)(n-r+1)=\frac{r-2}{2(r-1)}n^2 }[/math]. [math]\displaystyle{ \square }[/math]

Second proof. (weight shifting) (due to Motzkin and Straus) Assign each vertex [math]\displaystyle{ v\in V }[/math] a nonnegative weight [math]\displaystyle{ w_v\ge 0 }[/math], and assume that [math]\displaystyle{ \sum_{v\in V}w_v=1 }[/math]. We try to maximize the quantity [math]\displaystyle{ S=\sum_{uv\in E}w_uw_v }[/math]. Let [math]\displaystyle{ W_u=\sum_{v:v\sim u}w_v\, }[/math] be the sum of the weights of [math]\displaystyle{ u }[/math]'s neighbors. Note that [math]\displaystyle{ S }[/math] can also be computed as [math]\displaystyle{ S=\frac{1}{2}\sum_{u\in V}w_uW_u }[/math]. For any nonadjacent pair of vertices [math]\displaystyle{ u\not\sim v }[/math], supposed that [math]\displaystyle{ W_u\ge W_v }[/math], then for any [math]\displaystyle{ \epsilon\ge 0 }[/math], [math]\displaystyle{ (w_u+\epsilon)W_u+(w_v-\epsilon)W_v\ge w_uW_u+w_vW_v }[/math]. This means that we do not decrease [math]\displaystyle{ S }[/math] by shifting all of the weight of the vertex [math]\displaystyle{ v }[/math] to the vertex [math]\displaystyle{ u }[/math]. It follows that [math]\displaystyle{ S }[/math] is maximized when all of the weight is concentrated on a complete subgraph, i.e., a clique. Now if [math]\displaystyle{ w_u\gt w_v\gt 0 }[/math], then choose [math]\displaystyle{ \epsilon }[/math] with [math]\displaystyle{ 0\lt \epsilon\lt w_u-w_v }[/math] and change [math]\displaystyle{ w_u'=w_u-\epsilon }[/math] and [math]\displaystyle{ w_v'=w_v+\epsilon }[/math]. This changes [math]\displaystyle{ S }[/math] to [math]\displaystyle{ S'=S+\epsilon(w_u-w_v)-\epsilon^2\gt S }[/math]. Thus, the maximal value of [math]\displaystyle{ S }[/math] is attained when all nonzero weights are equal and concentrated on a clique. [math]\displaystyle{ G }[/math] has at most an [math]\displaystyle{ (r-1) }[/math]-clique, thus [math]\displaystyle{ S\le{r-1\choose 2}\frac{1}{(r-1)^2}=\frac{r-2}{2(r-1)} }[/math]. As we argued above, this inequality hold for any nonnegative weight assignments with [math]\displaystyle{ \sum_{v\in V}w_v=1 }[/math]. In particular, for the case that all [math]\displaystyle{ w_v=\frac{1}{n} }[/math], [math]\displaystyle{ S=\sum_{uv\in E}w_uw_v=\frac{|E|}{n^2} }[/math]. Thus, [math]\displaystyle{ \frac{|E|}{n^2}\le \frac{r-2}{2(r-1)} }[/math], which implies the theorem. [math]\displaystyle{ \square }[/math]

Third proof. (the probabilistic method) (due to Alon and Spencer) Write [math]\displaystyle{ \omega(G) }[/math] for the number of vertices in a largest clique, called the clique number of [math]\displaystyle{ G }[/math]. Claim: [math]\displaystyle{ \omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v} }[/math]. We prove this by the probabilistic method. Fix a random ordering of vertices in [math]\displaystyle{ V }[/math], say [math]\displaystyle{ v_1,v_2,\ldots,v_n }[/math]. We construct a clique as follows: for [math]\displaystyle{ i=1,2,\ldots, n }[/math], add [math]\displaystyle{ v_i }[/math] to [math]\displaystyle{ S }[/math] iff all vertices in current [math]\displaystyle{ S }[/math] are adjacent to [math]\displaystyle{ v_i }[/math]. It is obvious that an [math]\displaystyle{ S }[/math] constructed in this way is a clique. We now show that [math]\displaystyle{ \mathbf{E}[|S|]=\sum_{v\in V}\frac{1}{n-d_v} }[/math]. Let [math]\displaystyle{ X_v }[/math] be the random variable that indicates whether [math]\displaystyle{ v\in S }[/math], i.e., [math]\displaystyle{ X_v=\begin{cases} 1 & v\in S,\\ 0 & \mbox{otherwise.} \end{cases} }[/math] Note that a vertex [math]\displaystyle{ v\in S }[/math] if and only if [math]\displaystyle{ v }[/math] is ranked before all its [math]\displaystyle{ n-d_v-1 }[/math] non-neighbors in the random ordering. The probability that this event occurs is [math]\displaystyle{ \frac{1}{n-d_v} }[/math]. Thus, [math]\displaystyle{ \mathbf{E}[X_v]=\Pr[v\in S]=\frac{1}{n-d_v}. }[/math] Observe that [math]\displaystyle{ |S|=\sum_{v\in V}X_v }[/math]. Due to linearity of expectation, [math]\displaystyle{ \mathbf{E}[|S|]=\sum_{v\in V}\mathbf{E}[X_v]=\sum_{v\in V}\frac{1}{n-d_v} }[/math]. There must exists a clique of at least such size, so that [math]\displaystyle{ \omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v} }[/math]. The claim is proved. Apply the Cauchy-Schwarz inequality [math]\displaystyle{ \left(\sum_{v\in V}a_vb_v\right)^2\le\left(\sum_{v\in V}^na_v^2\right)\left(\sum_{v\in V}^nb_v^2\right) }[/math]. Set [math]\displaystyle{ a_v=\sqrt{n-d_v} }[/math] and [math]\displaystyle{ b_v=\frac{1}{\sqrt{n-d_v}} }[/math], then [math]\displaystyle{ a_vb_v=1 }[/math] and so [math]\displaystyle{ n^2\le\sum_{v\in V}(n-d_v)\sum_{v\in V}\frac{1}{n-d_v}\le\omega(G)\sum_{v\in V}(n-d_v). }[/math] By the assumption of Turán's theorem, [math]\displaystyle{ \omega(G)\le r-1 }[/math]. Recall the handshaking lemma [math]\displaystyle{ 2|E|=\sum_{v\in V}d_v }[/math]. The above inequality gives us [math]\displaystyle{ n^2\le (r-1)(n^2-2|E|) }[/math], which implies the theorem. [math]\displaystyle{ \square }[/math]

Our last proof uses the idea of vertex duplication. It does not only prove the edge bound of Turán's theorem, but also shows that Turán graphs are the only possible extremal graphs.

Fourth proof. Let [math]\displaystyle{ G(V,E) }[/math] be a [math]\displaystyle{ r }[/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{ r }[/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{ r }[/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_{r-1}} }[/math] with [math]\displaystyle{ n_1+n_2+\cdots +n_{r-1}=n }[/math]. Optimize the edge number, we have the Turán graph. [math]\displaystyle{ \square }[/math]

Forbidden Cycles

Another direction to generalize Mantel's theorem other than Turán's theorem is to see a triangle as a 3-cycle rather than 3-clique. We then ask for the extremal bound for graphs without certain cycle structures.

Girth

Recall that the girth of a graph [math]\displaystyle{ G }[/math] is the length of the shortest cycle in [math]\displaystyle{ G }[/math]. A graph is triangle-free if and only if its girth [math]\displaystyle{ g(G)\ge 4 }[/math]. Matel's theorem can be seen as a bound on the edge number of graphs with girth [math]\displaystyle{ g(G)\ge 4 }[/math]. The next theorem extends this bound to the graphs with [math]\displaystyle{ g(G)\ge 5 }[/math], i.e., graphs without triangles and quadrilaterals ("squares").

Theorem

Let [math]\displaystyle{ G(V,E) }[/math] be a graph on [math]\displaystyle{ n }[/math] vertices. If girth [math]\displaystyle{ g(G)\ge 5 }[/math] then [math]\displaystyle{ |E|\le\frac{1}{2}n\sqrt{n-1} }[/math].

Proof. Suppose [math]\displaystyle{ g(G)\ge 5 }[/math]. Let [math]\displaystyle{ v_1,v_2,\ldots,v_d }[/math] be the neighbors of a vertex [math]\displaystyle{ u }[/math], where [math]\displaystyle{ d=d(u) }[/math]. Let [math]\displaystyle{ S_i=\{v\in V\mid v\sim v_i\wedge v\neq u\} }[/math] be the set of neighbors of [math]\displaystyle{ v_i }[/math] other than [math]\displaystyle{ u }[/math]. For any [math]\displaystyle{ v_i,v_j }[/math], [math]\displaystyle{ v_iv_j\not\in E }[/math] since [math]\displaystyle{ G }[/math] has no triangle. Thus, [math]\displaystyle{ S_i\cap\{u,v_1,v_2,\ldots,v_d\}=\emptyset }[/math] for every [math]\displaystyle{ i }[/math]. No vertex other than [math]\displaystyle{ u }[/math] can be adjacent to more than one vertices in [math]\displaystyle{ v_1,v_2,\ldots,v_d }[/math] since there is no [math]\displaystyle{ C_4 }[/math] in [math]\displaystyle{ G }[/math]. Thus, [math]\displaystyle{ S_i\cap S_j=\emptyset }[/math] for any distinct [math]\displaystyle{ i }[/math] and [math]\displaystyle{ j }[/math]. Therefore, [math]\displaystyle{ \{u,v_1,v_2,\ldots,v_d\}\cup S_1\cup S_2\cup\cdots\cup S_d\subseteq V }[/math] implies that [math]\displaystyle{ (d+1)+|S_1|+|S_2|+\cdots+|S_d|=(d+1)+(d(v_1)-1)+(d(v_2)-1)+\cdots+(d(v_d)-1)\le n }[/math], so that [math]\displaystyle{ \sum_{v:v\sim u}d(v)\le n-1 }[/math]. By Cauchy-Schwarz inequality, [math]\displaystyle{ n(n-1)\ge \sum_{u\in V}\sum_{v:v\sim u}d(v)=\sum_{v\in V}d(v)^2\ge\frac{\left(\sum_{v\in V}d(v)\right)}{n}=\frac{4|E|^2}{n} }[/math], which implies that [math]\displaystyle{ |E|\le\frac{1}{2}n\sqrt{n-1} }[/math]. [math]\displaystyle{ \square }[/math]

Hamiltonian cycle

We now look at graphs which does not have large cycles. In particular, we consider graphs without Hamiltonian cycles.

For a Hamiltonian graph, every vertex must has degree 2. And the graph satisfying this condition with maximum number of edges is the graph composed by a [math]\displaystyle{ (n-1) }[/math]-clique and the one remaining vertex is connected to the clique by one edge. This graph has [math]\displaystyle{ {n-1\choose 2}+1 }[/math] edges, and has no Hamiltonian cycle. It is not very hard to realize that this is the largest possible number of edges that a non-Hamiltonian graph can have.

Since it is not very interesting to bound the number of edges of non-Hamiltonian graphs, we consider a more informative graph invariant, its degree sequence.

Dirac's Theorem

A graph [math]\displaystyle{ G(V,E) }[/math] on [math]\displaystyle{ n }[/math] vertices has a Hamiltonian cycle if [math]\displaystyle{ d_v\ge\frac{n}{2} }[/math] for all [math]\displaystyle{ v\in V }[/math].

Proof. Suppose to the contrary, the theorem is not true and there exists a non-Hamiltonian graph with [math]\displaystyle{ d_v\ge\frac{n}{2} }[/math] for all [math]\displaystyle{ v\in V }[/math]. Let [math]\displaystyle{ G }[/math] be such a graph with a maximum number of edges. Then adding any edge to [math]\displaystyle{ G }[/math] creates a Hamiltonian cycle. Thus, [math]\displaystyle{ G }[/math] must have a Hamiltonian path, say [math]\displaystyle{ v_1v_2\cdots v_n }[/math]. Consider the sets, [math]\displaystyle{ S=\{i\mid v_iv_n\in E\} }[/math]; [math]\displaystyle{ T=\{i\mid v_{i+1}v_1\in E\} }[/math]. Therefore, [math]\displaystyle{ S\subseteq\{v_1,v_2,\ldots,v_{n-1}\} }[/math] contains the neighbors of [math]\displaystyle{ v_n }[/math]; and [math]\displaystyle{ T\subseteq\{v_1,v_2,\ldots,v_{n-1}\} }[/math] contains the predecessors (along the Hamiltonian path) of the neighbors of [math]\displaystyle{ v_1 }[/math]. It holds that [math]\displaystyle{ S,T\subseteq\{v_1,v_2,\ldots,v_{n-1}\} }[/math]. Since [math]\displaystyle{ d_v\ge\frac{n}{2} }[/math] for all [math]\displaystyle{ v\in V }[/math], [math]\displaystyle{ |S|,|T|\ge\frac{n}{2} }[/math]. By the pigeonhole principle, there exists some [math]\displaystyle{ v_i\in S\cap T }[/math]. We can construct the following Hamiltonian cycle: [math]\displaystyle{ v_1v_{i+1}v_{i+2}\cdots v_nv_{i}v_{i-1}\cdots v_1\cdots }[/math], which contradict to the assumption that [math]\displaystyle{ G }[/math] is non-Hamiltonian. [math]\displaystyle{ \square }[/math]

Erdős–Stone theorem

We introduce a notation for the number of edges in extremal graphs with a specific forbidden substructure.

Definition

Let [math]\displaystyle{ \mathrm{ex}(n,H) }[/math] denote the largest number of edges that a graph [math]\displaystyle{ G\not\supseteq H }[/math] on [math]\displaystyle{ n }[/math] vertices can have.

With this notation, Turán's theorem can be restated as

Turán's theorem (restated)

[math]\displaystyle{ \mathrm{ex}(n,K_r)\le\frac{r-2}{2(r-1)}n^2 }[/math].

Let [math]\displaystyle{ K_s^r=K_{\underbrace{s,s,\cdots,s}_{r}} }[/math] be the complete [math]\displaystyle{ r }[/math]-partite graph with [math]\displaystyle{ s }[/math] vertices in each class, i.e., the Turán graph [math]\displaystyle{ T(rs,r) }[/math]. The Erdős–Stone theorem (also referred as the fundamental theorem of extremal graph theory) gives an asymptotic bound on [math]\displaystyle{ \mathrm{ex}(n,K-s^r) }[/math], i.e., the largest number of edges that an [math]\displaystyle{ n }[/math]-vertex graph can have to not contain [math]\displaystyle{ K_s^r }[/math].

Fundamental theorem of extremal graph theory (Erdős–Stone 1946)

For any integers [math]\displaystyle{ r\ge 2 }[/math] and [math]\displaystyle{ s\ge 1 }[/math], and any [math]\displaystyle{ \epsilon\gt 0 }[/math], if [math]\displaystyle{ n }[/math] is sufficiently large then every graph on [math]\displaystyle{ n }[/math] vertices and with at least [math]\displaystyle{ \left(\frac{r-2}{2(r-1)}+\epsilon\right)n^2 }[/math] edges contains [math]\displaystyle{ K_{r,s} }[/math] as a subgraph, i.e., [math]\displaystyle{ \mathrm{ex}(n,K_s^r)= \left(\frac{r-2}{2(r-1)}+o(1)\right)n^2 }[/math].

The theorem is called fundamental because of its single most important corollary: it relate the extremal bound for an arbitrary subgraph [math]\displaystyle{ H }[/math] to a very natural parameter of [math]\displaystyle{ H }[/math], its chromatic number.

Recall that [math]\displaystyle{ \chi(G) }[/math] is the chromatic number of [math]\displaystyle{ G }[/math], the smallest number of colors that one can use to color the vertices so that no adjacent vertices have the same color.

Corollary

For every nonempty graph [math]\displaystyle{ H }[/math], [math]\displaystyle{ \lim_{n\rightarrow\infty}\frac{\mathrm{ex}(n,H)}{{n\choose 2}}=\frac{\chi(H)-2}{\chi(H)-1} }[/math].

Proof of corollary Let [math]\displaystyle{ r=\chi(H) }[/math]. Note that [math]\displaystyle{ T(n,r-1) }[/math] can be colored with [math]\displaystyle{ r-1 }[/math] colors, one color for each part. Thus, [math]\displaystyle{ H\not\subseteq T(n,r-1) }[/math], since otherwise [math]\displaystyle{ H }[/math] can also be colored with [math]\displaystyle{ r-1 }[/math] colors, contradicting that [math]\displaystyle{ \chi(H)=1 }[/math]. By definition, [math]\displaystyle{ \mathrm{ex}(n,H) }[/math] is the maximum number of edges that an [math]\displaystyle{ n }[/math]-vertex graph [math]\displaystyle{ G\not\supseteq H }[/math] can have. Thus, [math]\displaystyle{ |T(n,r-1)|\le\mathrm{ex}(n,H) }[/math]. It is not hard to see that [math]\displaystyle{ |T(n,r-1)|\ge {r-1\choose 2}\left\lfloor\frac{n}{r-1}\right\rfloor^2\ge{r-1\choose 2}\left(\frac{n}{r-1}-1\right)^2=\left(\frac{r-2}{2(r-1)}-o(1)\right)n^2 }[/math]. On the other hand, any finite graph [math]\displaystyle{ H }[/math] with chromatic number [math]\displaystyle{ r }[/math] has that [math]\displaystyle{ H\subseteq K_s^r }[/math] for all sufficiently large [math]\displaystyle{ s }[/math]. We just connect all pairs of vertices from different color classes. Thus, [math]\displaystyle{ \mathrm{ex}(n,H)\le\mathrm{ex}(n,K_s^r) }[/math]. Due to Erdős–Stone theorem, [math]\displaystyle{ \mathrm{ex}(n,K_s^r)=\left(\frac{r-2}{2(r-1)}+o(1)\right)n^2 }[/math]. Altogether, we have [math]\displaystyle{ \frac{r-2}{r-1}-o(1)\le\frac{|T(n,r-1)|}{{n\choose 2}}\le \frac{\mathrm{ex}(n,H)}{{n\choose 2}} \le \frac{\mathrm{ex}(n,K_s^r)}{{n\choose 2}}=\frac{r-2}{r-1}+o(1) }[/math] The theorem follows. [math]\displaystyle{ \square }[/math]

References

(声明: 资料受版权保护, 仅用于教学.)

(Disclaimer: The following copyrighted materials are meant for educational uses only.)

• van Lin and Wilson. A course in combinatorics. Cambridge Press. Chapter 4.
• Aigner and Ziegler. Proofs from THE BOOK, 4th Edition. Springer-Verlag. Chapter 36.
• Diestel. Graph Theory, 3rd Edition. Springer-Verlag 2000. Chapter 7.