imported>Etone |
imported>Etone |
| Line 1: |
Line 1: |
| == Forbidden Cliques ==
| |
| Extremal graph theory studies the problems like "how many edges that a graph <math>G</math> can have, if <math>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>K_3</math>.
| |
|
| |
|
| {{Theorem|Theorem (Mantel 1907)|
| | == Problem 1== |
| :Suppose <math>G(V,E)</math> is graph on <math>n</math> vertice without triangles. Then <math>|E|\le\frac{n^2}{4}</math>. | | Consider the following optimization problem. |
| }}
| | :'''Instance''': <math>n</math> positive integers <math>x_1<x_2<\cdots <x_n</math>. |
| | :Find two ''disjoint'' nonempty subsets <math>A,B\subset\{1,2,\ldots,n\}</math> with <math>\sum_{i\in A}x_i\ge \sum_{i\in B}x_i</math>, such that the ratio <math>\frac{\sum_{i\in A}x_i}{\sum_{i\in B}x_i}</math> is minimized. |
| | Give a pseudo-polynomial time algorithm for the problem, and then give an FPTAS for the problem based on the pseudo-polynomial time algorithm. |
|
| |
|
| 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.
| | == Problem 2== |
| | | In the ''maximum directed cut'' (MAX-DICUT) problem, we are given as input a directed graph <math>G(V,E)</math>. The goal is to partition <math>V</math> into disjoint <math>S</math> and <math>T</math> so that the number of edges in <math>E(S,T)=\{(u,v)\in E\mid u\in S, v\in T\}</math> is maximized. The following is the integer program for MAX-DICUT: |
| {{Prooftitle|First proof. (pigeonhole principle)|
| | :::<math> |
| We prove an equivalent theorem: Any <math>G(V,E)</math> with <math>|V|=n</math> and <math>|E|>\frac{n^2}{4}</math> must have a triangle.
| | \begin{align} |
| | | \text{maximize} &&& \sum_{(u,v)\in E}y_{u,v}\\ |
| Use induction on <math>n</math>. The theorem holds trivially for <math>n\le 3</math>.
| | \text{subject to} && y_{u,v} &\le x_u, & \forall (u,v)&\in E,\\ |
| | | && y_{u,v} &\le 1-x_v, & \forall (u,v)&\in E,\\ |
| Induction hypothesis: assume the theorem hold for <math>|V|\le n-1</math>.
| | && x_v &\in\{0,1\}, & \forall v&\in V,\\ |
| | | && y_{u,v} &\in\{0,1\}, & \forall (u,v)&\in E. |
| For <math>G</math> with <math>n</math> vertices, without loss of generality, assume that <math>|E|=\frac{n^2}{4}+1</math>, we will show that <math>G</math> must contain a triangle. Take a <math>uv\in E</math>, and let <math>H</math> be the subgraph of <math>G</math> induced by <math>V\setminus \{u,v\}</math>. Clearly, <math>H</math> has <math>n-2</math> vertices.
| | \end{align} |
| :'''Case.1:''' If <math>H</math> has <math>>\frac{(n-2)^2}{4}</math> edges, then by the induction hypothesis, <math>H</math> has a triangle. | |
| :'''Case.2:''' If <math>H</math> has <math>\le\frac{(n-2)^2}{4}</math> edges, then at least <math>\left(\frac{n^2}{4}+1\right)-\frac{(n-2)^2}{4}-1=n-1</math> edges are between <math>H</math> and <math>\{u,v\}</math>. By pigeonhole principle, there must be a vertex in <math>H</math> that is adjacent to both <math>u</math> and <math>v</math>. Thus, <math>G</math> has a triangle.
| |
| }}
| |
| | |
| {{Prooftitle|Second proof. (Cauchy-Schwarz inequality)|(Mantel's original proof)
| |
| 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_{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> | | </math> |
| where the last equation is due to Euler's equality <math>\sum_{v\in V}d_v=2|E|</math>. The theorem follows.
| | Let <math>x_v^*,y_{u,v}^*</math> denote the optimal solution to the '''LP-relaxation''' of the above integer program. |
| }}
| | * Apply the randomized rounding such that for every <math>v\in V</math>, <math>\hat{x}_v=1</math> independently with probability <math>x_v^*</math>. Analyze the approximation ratio (between the expected size of the random cut and OPT). |
| | | * Apply another randomized rounding such that for every <math>v\in V</math>, <math>\hat{x}_v=1</math> independently with probability <math>1/4+x_v^*/2</math>. Analyze the approximation ratio for this algorithm. |
| {{Prooftitle|Third proof. (inequality of the arithmetic and geometric mean)|
| |
| Assume that <math>G(V,E)</math> has <math>|V|=n</math> vertices and is triangle-free.
| |
| | |
| Let <math>A</math> be the largest independent set in <math>G</math> and let <math>\alpha=|A|</math>.
| |
| Since <math>G</math> is triangle-free, for very vertex <math>v</math>, all its neighbors must form an independent set, thus <math>d(v)\le \alpha</math> for all <math>v\in V</math>.
| |
| | |
| Take <math>B=V\setminus A</math> and let <math>\beta=|B|</math>.
| |
| Since <math>A</math> is an independent set, all edges in <math>E</math> must have at least one endpoint in <math>B</math>. Counting the edges in <math>E</math> according to their endpoints in <math>B</math>, we obtain <math>|E|\le\sum_{v\in B}d_v</math>. By the inequality of the arithmetic and geometric mean,
| |
| :<math>|E|\le\sum_{v\in B}d_v\le\alpha\beta\le\left(\frac{\alpha+\beta}{2}\right)^2=\frac{n^2}{4}</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|Theorem (Turán 1941)|
| |
| :Let <math>G(V,E)</math> be a graph with <math>|V|=n</math>. If <math>G</math> has no <math>r</math>-clique, <math>r\ge 2</math>, then
| |
| ::<math>|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>K_r</math>.
| |
| | |
| Partition <math>V</math> into <math>r-1</math> disjoint classes <math>V=V_1\cup V_2\cup\cdots\cup V_{r-1}</math>, <math>n_i=|V_i|</math>, <math>n_1+n_2+\cdots+n_{r-1}=n</math>. For every two vertice <math>u,v</math>, <math>uv\in E</math> if and only if <math>u\in V_i</math> and <math>v\in V_j</math> for distinct <math>V_i</math> and <math>V_j</math>. The resulting graph is a '''complete <math>(r-1)</math>-partite graph''', denoted <math>K_{n_1,n_2,\ldots,n_{r-1}}</math>. It is obvious that any <math>(r-1)</math>-partite graph contains no <math>r</math>-clique since only those vertices from different classes can be adjacent.
| |
| | |
| A <math>K_{n_1,n_2,\ldots,n_{r-1}}</math> has <math>\sum_{i<j}n_i n_j\,</math> edges, which is maximized when the numbers <math>n_i</math> are divided as evenly as possible, that is, if <math>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>1\le i\le r-1</math>.
| |
| | |
| {{Theorem|Definition|
| |
| :We call a complete multipartite graph <math>K_{n_1,n_2,\ldots,n_{r-1}}</math> with <math>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>i</math> a ''' Turán graph''', denoted <math>T(n,r-1)</math>.
| |
| }}
| |
| ;Example:Turán graph <math>T(13,4)</math>
| |
| [[File:Turan 13-4.svg|center|260px|Turán graph <math>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.
| |
| | |
| {{Prooftitle|First proof. (induction)|(Turán's original proof)
| |
|
| |
|
| Induction on <math>n</math>. It is easy to verify that the theorem holds for <math>n<r</math>.
| | == Problem 3== |
| | | Recall the MAX-SAT problem and its integer program: |
| Let <math>G</math> be a graph on <math>n</math> vertices without <math>r</math>-cliques where <math>n\ge r</math>. Suppose that <math>G</math> has a maximum number of edges among such graphs. <math>G</math> certainly has <math>(r-1)</math>-cliques, since otherwise we could add edges to <math>G</math>. Let <math>A</math> be an <math>(r-1)</math>-clique and let <math>B=V\setminus A</math>. Clearly <math>|A|=r-1</math> and <math>|B|=n-r+1</math>.
| | :::<math> |
| | | \begin{align} |
| By the induction hypothesis, since <math>B</math> has no <math>r</math>-cliques, <math>|E(B)|\le\frac{r-2}{2(r-1)}(n-r+1)^2</math>. And <math>E(A)={r-1\choose 2}</math>. Since <math>G</math> has no <math>r</math>-clique, every <math>v\in B</math> is adjacent to at most <math>r-2</math> vertices in <math>A</math>, since otherwise <math>A</math> and <math>v</math> would form an <math>r</math>-clique. We obtain that the number edges crossing between <math>A</math> and <math>B</math> is <math>|E(A,B)|\le (r-2)|B|=(r-2)(n-r+1)</math>. Combining everything together,
| | \text{maximize} &&& \sum_{j=1}^my_j\\ |
| :<math>|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>. | | \text{subject to} &&& \sum_{i\in S_j^+}x_i+\sum_{i\in S_j^-}(1-x_i)\ge y_j, && 1\le j\le m,\\ |
| }}
| | &&& x_i\in\{0,1\}, && 1\le i\le n,\\ |
| | | &&& y_j\in\{0,1\}, && 1\le j\le m. |
| {{Prooftitle|Second proof. (weight shifting)|(due to Motzkin and Straus)
| | \end{align} |
| | |
| Assign each vertex <math>v\in V</math> a nonnegative weight <math>w_v\ge 0</math>, and assume that <math>\sum_{v\in V}w_v=1</math>. We try to maximize the quantity
| |
| :<math>S=\sum_{uv\in E}w_uw_v</math>.
| |
| Let <math>W_u=\sum_{v:v\sim u}w_v\,</math> be the sum of the weights of <math>u</math>'s neighbors.
| |
| Note that <math>S</math> can also be computed as <math>S=\frac{1}{2}\sum_{u\in V}w_uW_u</math>.
| |
| For any nonadjacent pair of vertices <math>u\not\sim v</math>, supposed that <math>W_u\ge W_v</math>, then for any <math>\epsilon\ge 0</math>,
| |
| :<math>(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>S</math> by shifting all of the weight of the vertex <math>v</math> to the vertex <math>u</math>. It follows that <math>S</math> is maximized when all of the weight is concentrated on a complete subgraph, i.e., a clique.
| |
| | |
| Now if <math>w_u>w_v>0</math>, then choose <math>\epsilon</math> with <math>0<\epsilon<w_u-w_v</math> and change <math>w_u'=w_u-\epsilon</math> and <math>w_v'=w_v+\epsilon</math>. This changes <math>S</math> to <math>S'=S+\epsilon(w_u-w_v)-\epsilon^2>S</math>. Thus, the maximal value of <math>S</math> is attained when all nonzero weights are equal and concentrated on a clique.
| |
| | |
| <math>G</math> has at most an <math>(r-1)</math>-clique, thus <math>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>\sum_{v\in V}w_v=1</math>. In particular, for the case that all <math>w_v=\frac{1}{n}</math>,
| |
| :<math>S=\sum_{uv\in E}w_uw_v=\frac{|E|}{n^2}</math>.
| |
| Thus,
| |
| :<math>\frac{|E|}{n^2}\le \frac{r-2}{2(r-1)}</math>,
| |
| which implies the theorem.
| |
| }}
| |
| | |
| {{Prooftitle|Third proof. (the probabilistic method)|(due to Alon and Spencer)
| |
| | |
| Write <math>\omega(G)</math> for the number of vertices in a largest clique, called the '''clique number''' of <math>G</math>.
| |
| :'''Claim:''' <math>\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>V</math>, say <math>v_1,v_2,\ldots,v_n</math>. We construct a clique as follows:
| |
| *for <math>i=1,2,\ldots, n</math>, add <math>v_i</math> to <math>S</math> iff all vertices in current <math>S</math> are adjacent to <math>v_i</math>.
| |
| It is obvious that an <math>S</math> constructed in this way is a clique. We now show that <math>\mathbf{E}[|S|]\ge\sum_{v\in V}\frac{1}{n-d_v}</math>.
| |
| | |
| Let <math>X_v</math> be the random variable that indicates whether <math>v\in S</math>, i.e.,
| |
| :<math>
| |
| X_v=\begin{cases}
| |
| 1 & v\in S,\\ | |
| 0 & \mbox{otherwise.}
| |
| \end{cases} | |
| </math> | | </math> |
| Note that a vertex <math>v\in S</math> if <math>v</math> is ranked before all its <math>n-d_v-1</math> non-neighbors in the random ordering. The probability that this event occurs is <math>\frac{1}{n-d_v}</math>. Thus,
| | Recall that <math>S_j^+,S_j^-\subseteq\{1,2,\ldots,n\}</math> are the respective sets of variables appearing positively and negatively in clause <math>j</math>. |
| :<math>\mathbf{E}[X_v]=\Pr[v\in S]\ge\frac{1}{n-d_v}.</math>
| |
| Observe that <math>|S|=\sum_{v\in V}X_v</math>. Due to linearity of expectation,
| |
| :<math>\mathbf{E}[|S|]=\sum_{v\in V}\mathbf{E}[X_v]\ge\sum_{v\in V}\frac{1}{n-d_v}</math>.
| |
| There must exists a clique of at least such size, so that <math>\omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v}</math>. The claim is proved.
| |
|
| |
|
| Apply the Cauchy-Schwarz inequality
| | Let <math>x_i^*,y_j^*</math> denote the optimal solution to the '''LP-relaxation''' of the above integer program. In our class we learnt that if <math>\hat{x}_i</math> is round to 1 independently with probability <math>x_i^*</math>, we have approximation ratio <math>1-1/\mathrm{e}</math>. |
| :<math>\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>a_v=\sqrt{n-d_v}</math> and <math>b_v=\frac{1}{\sqrt{n-d_v}}</math>, then <math>a_vb_v=1</math> and so
| |
| :<math>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>\omega(G)\le r-1</math>. Recall the handshaking lemma <math>2|E|=\sum_{v\in V}d_v</math>. The above inequality gives us
| |
| :<math>n^2\le (r-1)(n^2-2|E|)</math>,
| |
| which implies the theorem.
| |
| }}
| |
|
| |
|
| 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 <font color=red>only</font> possible extremal graphs.
| | We consider a generalized rounding scheme such that every <math>\hat{x}_i</math> is round to 1 independently with probability <math>f(x_i^*)</math> for some function <math>f:[0,1]\to[0,1]</math> to be specified. |
| {{Prooftitle|Fourth proof.|
| | * Suppose <math>f(x)</math> is an arbitrary function satisfying that <math>1-4^{-x}\le f(x)\le 4^{x-1}</math> for any <math>x\in[0,1]</math>. Show that with this rounding scheme, the approximation ratio (between the expected number of satisfied clauses and OPT) is at least <math>3/4</math>. |
| Let <math>G(V,E)</math> be a <math>r</math>-clique-free graph on <math>n</math> vertices with a maximum number of edges.
| | * Derandomize this algorithm through conditional expectation and give a deterministic polynomial time algorithm with approximation ratio <math>3/4</math>. |
| :'''Claim:''' <math>G</math> does not contain three vertices <math>u,v,w</math> such that <math>uv\in E</math> but <math>uw\not\in E, vw\not\in E</math>. | | * Is it possible that for some more clever <math>f</math> we can do better than this? Try to justify your argument. |
| Suppose otherwise. There are two cases. | |
| * '''Case.1:''' <math>d(w)<d(u)</math> or <math>d(w)<d(v)</math>. Without loss of generality, suppose that <math>d(w)<d(u)</math>. We duplicate <math>u</math> by creating a new vertex <math>u'</math> which has exactly the same neighbors as <math>u</math> (but <math>uu'</math> is not an edge). Such duplication will not increase the clique size. We then remove <math>w</math>. The resulting graph <math>G'</math> is still <math>r</math>-clique-free, and has <math>n</math> vertices. The number of edges in <math>G'</math> is
| |
| ::<math>|E(G')|=|E(G)|+d(u)-d(w)>|E(G)|\,</math>,
| |
| :which contradicts the assumption that <math>|E(G)|</math> is maximal.
| |
| * '''Case.2:''' <math>d(w)\ge d(u)</math> and <math>d(w)\ge d(v)</math>. Duplicate <math>w</math> twice and delete <math>u</math> and <math>v</math>. The new graph <math>G'</math> has no <math>r</math>-clique, and the number of edges is | |
| ::<math>|E(G')|=|E(G)|+2d(w)-(d(u)+d(v)+1)>|E(G)|\,</math>.
| |
| :Contradiction again.
| |
|
| |
|
| 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_{r-1}}</math> with <math>n_1+n_2+\cdots +n_{r-1}=n</math>. Optimize the edge number, we have the Turán graph. | | ==Problem 4 == |
| }}
| | The following is the weighted version of set cover problem: |
|
| |
|
| == Forbidden Cycles ==
| | Given <math>m</math> subsets <math>S_1,S_2,\ldots,S_m\subseteq U</math>, where <math>U</math> is a universe of size <math>n=|U|</math>, and each subset <math>S_i</math> is assigned a positive weight <math>w_i>0</math>, the goal is to find a <math>C\subseteq\{1,2,\ldots,m\}</math> such that <math>U=\bigcup_{i\in C}S_i</math> and the total weight <math>\sum_{I\in C}w_i</math> is minimized. |
| 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.
| | * Give an integer programming for the problem and its linear programming relaxation. |
| Recall that the '''girth''' of a graph <math>G</math> is the length of the shortest cycle in <math>G</math>. A graph is triangle-free if and only if its girth <math>g(G)\ge 4</math>.
| | * Consider the following idea of randomized rounding: independently round each fractional value to <math>\{0,1\}</math> with the probability of the fractional value itself; and repeatedly apply this process to the variables rounded to 0 in previous iterations until <math>U</math> is fully covered. Show that this can return a set cover with <math>O(\log n)</math> approximation ratio with probability at least <math>0.99</math>. |
| Matel's theorem can be seen as a bound on the edge number of graphs with girth <math>g(G)\ge 4</math>. The next theorem extends this bound to the graphs with <math>g(G)\ge 5</math>, i.e., graphs without triangles and quadrilaterals ("squares").
| |
| | |
| {{Theorem|Theorem|
| |
| :Let <math>G(V,E)</math> be a graph on <math>n</math> vertices. If girth <math>g(G)\ge 5</math> then <math>|E|\le\frac{1}{2}n\sqrt{n-1}</math>.
| |
| }}
| |
| {{Proof|
| |
| Suppose <math>g(G)\ge 5</math>. Let <math>v_1,v_2,\ldots,v_d</math> be the neighbors of a vertex <math>u</math>, where <math>d=d(u)</math>. Let <math>S_i=\{v\in V\mid v\sim v_i\wedge v\neq u\}</math> be the set of neighbors of <math>v_i</math> other than <math>u</math>.
| |
| | |
| * For any <math>v_i,v_j</math>, <math>v_iv_j\not\in E</math> since <math>G</math> has no triangle. Thus, <math>S_i\cap\{u,v_1,v_2,\ldots,v_d\}=\emptyset</math> for every <math>i</math>.
| |
| * No vertex other than <math>u</math> can be adjacent to more than one vertices in <math>v_1,v_2,\ldots,v_d</math> since there is no <math>C_4</math> in <math>G</math>. Thus, <math>S_i\cap S_j=\emptyset</math> for any distinct <math>i</math> and <math>j</math>.
| |
| | |
| Therefore, <math>\{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>(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>\sum_{v:v\sim u}d(v)\le n-1</math>.
| |
| | |
| By Cauchy-Schwarz inequality,
| |
| :<math>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>|E|\le\frac{1}{2}n\sqrt{n-1}</math>.
| |
| }}
| |
| | |
| == Erdős–Stone theorem ==
| |
| We introduce a notation for the number of edges in extremal graphs with a specific forbidden substructure.
| |
| {{Theorem|Definition|
| |
| :Let <math>\mathrm{ex}(n,H)</math> denote the largest number of edges that a graph <math>G\not\supseteq H</math> on <math>n</math> vertices can have.
| |
| }}
| |
| With this notation, Turán's theorem can be restated as
| |
| {{Theorem|Turán's theorem (restated)|
| |
| :<math>\mathrm{ex}(n,K_r)\le\frac{r-2}{2(r-1)}n^2</math>. | |
| }}
| |
| | |
| Let <math>K_s^r=K_{\underbrace{s,s,\cdots,s}_{r}}</math> be the complete <math>r</math>-partite graph with <math>s</math> vertices in each class, i.e., the Turán graph <math>T(rs,r)</math>.
| |
| The Erdős–Stone theorem (also referred as the '''fundamental theorem of extremal graph theory''') gives an asymptotic bound on <math>\mathrm{ex}(n,K_s^r)</math>, i.e., the largest number of edges that an <math>n</math>-vertex graph can have to not contain <math>K_s^r</math>.
| |
| | |
| {{Theorem|Fundamental theorem of extremal graph theory (Erdős–Stone 1946)|
| |
| :For any integers <math>r\ge 2</math> and <math>s\ge 1</math>, and any <math>\epsilon>0</math>, if <math>n</math> is sufficiently large then every graph on <math>n</math> vertices and with at least <math>\left(\frac{r-2}{2(r-1)}+\epsilon\right)n^2</math> edges contains <math>K_{r,s}</math> as a subgraph, i.e.,
| |
| :::<math>\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>H</math> to a very natural parameter of <math>H</math>, its chromatic number.
| |
| | |
| Recall that <math>\chi(G)</math> is the '''chromatic number''' of <math>G</math>, the smallest number of colors that one can use to color the vertices so that no adjacent vertices have the same color.
| |
| | |
| {{Theorem|Corollary|
| |
| :For every nonempty graph <math>H</math>,
| |
| ::<math>\lim_{n\rightarrow\infty}\frac{\mathrm{ex}(n,H)}{{n\choose 2}}=\frac{\chi(H)-2}{\chi(H)-1}</math>.
| |
| }}
| |
| {{Prooftitle|Proof of corollary|
| |
| Let <math>r=\chi(H)</math>.
| |
| | |
| Note that <math>T(n,r-1)</math> can be colored with <math>r-1</math> colors, one color for each part. Thus, <math>H\not\subseteq T(n,r-1)</math>, since otherwise <math>H</math> can also be colored with <math>r-1</math> colors, contradicting that <math>\chi(H)=1</math>. By definition, <math>\mathrm{ex}(n,H)</math> is the maximum number of edges that an <math>n</math>-vertex graph <math>G\not\supseteq H</math> can have. Thus,
| |
| :<math>|T(n,r-1)|\le\mathrm{ex}(n,H)</math>.
| |
| It is not hard to see that
| |
| :<math>|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>H</math> with chromatic number <math>r</math> has that <math>H\subseteq K_s^r</math> for all sufficiently large <math>s</math>. We just connect all pairs of vertices from different color classes. Thus,
| |
| :<math>\mathrm{ex}(n,H)\le\mathrm{ex}(n,K_s^r)</math>.
| |
| Due to Erdős–Stone theorem,
| |
| :<math>\mathrm{ex}(n,K_s^r)=\left(\frac{r-2}{2(r-1)}+o(1)\right)n^2</math>.
| |
| Altogether, we have
| |
| :<math>
| |
| \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.
| |
| }}
| |
|
| |
|
| == References == | | == Problem 5== |
| * van Lin and Wilson. ''A course in combinatorics.'' Cambridge Press. Chapter 4.
| | Recall that the instance of '''set cover''' problem is a collection of <math>m</math> subsets <math>S_1,S_2,\ldots,S_m\subseteq U</math>, where <math>U</math> is a universe of size <math>n=|U|</math>. The goal is to find the smallest <math>C\subseteq\{1,2,\ldots,m\}</math> such that <math>U=\bigcup_{i\in C}S_i</math>. The frequency <math>f</math> is defined to be <math>\max_{x\in U}|\{i\mid x\in S_i\}|</math>. |
| * Aigner and Ziegler. ''Proofs from THE BOOK, 4th Edition.'' Springer-Verlag. [[media:PFTB_chap36.pdf| Chapter 36]].
| | * Give the primal integer program for set cover, its LP-relaxation and the dual LP. |
| * Diestel. ''Graph Theory, 3rd Edition''. Springer-Verlag 2000. [[media:Diestel2ed_chap7.pdf|Chapter 7]]. | | * Describe the complementary slackness conditions for the problem. |
| | * Give a primal-dual algorithm for the problem. Present the algorithm in the language of primal-dual scheme (alternatively raising variables for the LPs). Analyze the approximation ratio in terms of the frequency <math>f</math>. |