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| == Forbidden Cliques == | | {{Orphan|date=December 2010}} |
| 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?"
| | '''Graham's number''' is a very, very big [[natural number]] that was defined by a man named Ronald Graham. Graham was solving a problem in an area of mathematics called [[Ramsey theory]]. He proved that the answer to his problem was smaller than Graham's number. |
| === 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>.
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| {{Theorem|Theorem (Mantel 1907)|
| | Graham's number is one of the biggest numbers ever used in a [[mathematical proof]]. Even if every digit in Graham's number were written in the tiniest writing possible, it would still be too big to fit in the [[observable universe]]. |
| :Suppose <math>G(V,E)</math> is graph on <math>n</math> vertice without triangles. Then <math>|E|\le\frac{n^2}{4}</math>.
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| }}
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| 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.
| | ==Context== |
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| {{Prooftitle|First proof. (pigeonhole principle)|
| | Ramsey theory is an area of mathematics that asks questions like the following: |
| 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.
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| Use induction on <math>n</math>. The theorem holds trivially for <math>n\le 3</math>.
| | {{quote|<p>Suppose we draw some number of points, and connect every pair of points by a line. Some lines are blue and some are red. Can we always find 3 points for which the 3 lines connecting them are all the same color?}} |
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| Induction hypothesis: assume the theorem hold for <math>|V|\le n-1</math>.
| | It turns out that for this simple problem, the answer is "yes" when we have 6 or more points, no matter how the lines are colored. But when we have 5 points or fewer, we can color the lines so that the answer is "no". |
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| 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.
| | Graham's number comes from a variation on this question. |
| :'''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.
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| :'''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.
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| }}
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| {{Prooftitle|Second proof. (Cauchy-Schwarz inequality)|(Mantel's original proof) | | {{quote|<p>Once again, say we have some points, but now they are the corners of an ''n''-dimensional [[hypercube]]. They are still all connected by blue and red lines. For any 4 points, there are 6 lines connecting them. Can we find 4 points that all lie on one [[Plane (mathematics)|plane]], and the 6 lines connecting them are all the same color?}} |
| 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
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| :<math>\sum_{uv\in E}(d_u+d_v)\le n|E|</math>.
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| Note that <math>d(v)</math> appears exactly <math>d_v</math> times in the sum, so that
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| :<math>\sum_{uv\in E}(d_u+d_v)=\sum_{v\in V}d_v^2</math>.
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| Applying Chauchy-Schwarz inequality,
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| :<math>
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| 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},
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| </math>
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| where the last equation is due to Euler's equality <math>\sum_{v\in V}d_v=2|E|</math>. The theorem follows.
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| }} | |
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| {{Prooftitle|Third proof. (inequality of the arithmetic and geometric mean)|
| | By asking that the 4 points lie on a plane, we have made the problem much harder. We would like to know: for what values of ''n'' is the answer "no" (for some way of coloring the lines), and for what values of ''n'' is it "yes" (for all ways of coloring the lines)? But this problem has not been completely solved yet. |
| Assume that <math>G(V,E)</math> has <math>|V|=n</math> vertices and is triangle-free.
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| Let <math>A</math> be the largest independent set in <math>G</math> and let <math>\alpha=|A|</math>.
| | In 1971, Ronald Graham and B. L. Rothschild found a partial answer to this problem. They showed that for ''n''=6, the answer is "no". But when ''n'' is very large, as large as Graham's number or larger, the answer is "yes". |
| 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>.
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| Take <math>B=V\setminus A</math> and let <math>\beta=|B|</math>.
| | One of the reasons this partial answer is important is that it means that the answer is eventually "yes" for at least some large ''n''. Before 1971, we didn't know even that much. |
| 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,
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| :<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>.
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| }}
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| === Turán's theorem === | | ==Definition== |
| 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.
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| {{Theorem|Theorem (Turán 1941)|
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| :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
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| ::<math>|E|\le\frac{r-2}{2(r-1)}n^2</math>.
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| }}
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| We give an example of graphs with many edges which does not contain <math>K_r</math>.
| | Graham's number is not only too big to write down all of its digits, it is too big even to write in [[scientific notation]]. In order to be able to write it down, we have to use [[Knuth's up-arrow notation]]. |
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| 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.
| | We will write down a [[sequence]] of numbers that we will call '''g1''', '''g2''', '''g3''', and so on. Each one will be used in an equation to find the next. '''g64 is''' Graham's number. |
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| 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>.
| | First, here are some examples of up-arrows: |
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| {{Theorem|Definition|
| | * <math>3\uparrow3</math> is 3x3x3 which equals 27. An arrow between two numbers just means the first number multiplied by itself the second number of times. |
| :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>.
| | * You can think of <math>3 \uparrow \uparrow 3</math> as <math>3 \uparrow (3 \uparrow 3)</math> because two arrows between numbers A and B just means A written down a B number of times with an arrow in between each A. Because we know what single arrows are, <math>3\uparrow(3\uparrow3)</math> is 3 multiplied by itself <math>3\uparrow3</math> times and we know <math>3\uparrow3 |
| }}
| | </math> is twenty-seven. So <math>3\uparrow\uparrow3</math> is 3x3x3x3x....x3x3, in total 27 times. That equals 7625597484987. |
| ;Example:Turán graph <math>T(13,4)</math>
| | * <math>3 \uparrow \uparrow \uparrow 3</math> is <math>3 \uparrow \uparrow (3 \uparrow \uparrow 3)</math> and we know <math>3\uparrow\uparrow3</math> is 7625597484987. So <math>3\uparrow\uparrow(3\uparrow\uparrow3)</math> is <math>3\uparrow \uparrow 7625597484987</math>. That can also be written as <math>3\uparrow(3\uparrow(3\uparrow(3\uparrow . . .(3\uparrow(3\uparrow(3\uparrow3)</math> with a total of 7625597484987 3s. This number is so huge, its digits, even written very small, could fill up the observable universe and beyond. |
| [[File:Turan 13-4.svg|center|260px|Turán graph <math>T(13,4)</math>]]
| | ** Although this number may already be beyond comprehension, this is barely the start of this giant number. |
| | * The next step like this is <math>3 \uparrow \uparrow \uparrow \uparrow 3</math> or <math>3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3)</math>. This is the number we will call '''g1'''. |
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| Turán's theorem has been proved for many times by different mathematicians, with different tools. We show just a few.
| | After that, '''g2''' is equal to <math>3\uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow 3</math>; the number of arrows in this number is '''g1'''. |
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| 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.
| | '''g3''' is equal to <math>3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3</math>, where the number of arrows is '''g2'''. |
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| {{Prooftitle|First proof. (induction)|(Turán's original proof)
| | We keep going in this way. We stop when we define '''g64''' to be <math>3\uparrow \uparrow \uparrow \uparrow \uparrow \ldots \uparrow \uparrow \uparrow \uparrow \uparrow 3</math>, where the number of arrows is '''g63'''. |
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| Induction on <math>n</math>. It is easy to verify that the theorem holds for <math>n<r</math>.
| | This is Graham's number. |
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| 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>.
| | ==Related pages== |
| | * [[Knuth's up-arrow notation]] |
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| 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,
| | [[Category:Mathematics]] |
| :<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>.
| | [[Category:Hyperoperations]] |
| }}
| | [[Category:Integers]] |
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| {{Prooftitle|Second proof. (weight shifting)|(due to Motzkin and Straus)
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| 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
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| :<math>S=\sum_{uv\in E}w_uw_v</math>.
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| Let <math>W_u=\sum_{v:v\sim u}w_v\,</math> be the sum of the weights of <math>u</math>'s neighbors.
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| Note that <math>S</math> can also be computed as <math>S=\frac{1}{2}\sum_{u\in V}w_uW_u</math>.
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| 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>,
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| :<math>(w_u+\epsilon)W_u+(w_v-\epsilon)W_v\ge w_uW_u+w_vW_v</math>.
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| 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.
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| 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.
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| <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>.
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| 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>,
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| :<math>S=\sum_{uv\in E}w_uw_v=\frac{|E|}{n^2}</math>.
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| Thus,
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| :<math>\frac{|E|}{n^2}\le \frac{r-2}{2(r-1)}</math>,
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| which implies the theorem.
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| }}
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| {{Prooftitle|Third proof. (the probabilistic method)|(due to Alon and Spencer)
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| Write <math>\omega(G)</math> for the number of vertices in a largest clique, called the '''clique number''' of <math>G</math>.
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| :'''Claim:''' <math>\omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v}</math>.
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| 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:
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| *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>.
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| It is obvious that an <math>S</math> constructed in this way is a clique. We now show that <math>\mathbf{E}[|S|]=\sum_{v\in V}\frac{1}{n-d_v}</math>.
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| Let <math>X_v</math> be the random variable that indicates whether <math>v\in S</math>, i.e.,
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| :<math>
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| X_v=\begin{cases}
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| 1 & v\in S,\\
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| 0 & \mbox{otherwise.}
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| \end{cases}
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| </math>
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| Note that a vertex <math>v\in S</math> if and only 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,
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| :<math>\mathbf{E}[X_v]=\Pr[v\in S]=\frac{1}{n-d_v}.</math>
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| Observe that <math>|S|=\sum_{v\in V}X_v</math>. Due to linearity of expectation,
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| :<math>\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>\omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v}</math>. The claim is proved.
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| Apply the Cauchy-Schwarz inequality
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| :<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>.
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| 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
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| :<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>
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| 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
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| :<math>n^2\le (r-1)(n^2-2|E|)</math>,
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| which implies the theorem.
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| }}
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| 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.
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| {{Prooftitle|Fourth proof.|
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| 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.
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| :'''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>.
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| Suppose otherwise. There are two cases.
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| * '''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
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| ::<math>|E(G')|=|E(G)|+d(u)-d(w)>|E(G)|\,</math>,
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| :which contradicts the assumption that <math>|E(G)|</math> is maximal.
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| * '''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
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| ::<math>|E(G')|=|E(G)|+2d(w)-(d(u)+d(v)+1)>|E(G)|\,</math>.
| |
| :Contradiction again.
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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). 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.
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| }}
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| == Forbidden Cycles ==
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| 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.
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| === Girth ===
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| 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>.
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| 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").
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| {{Theorem|Theorem|
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| :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>.
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| }}
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| {{Proof|
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| 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>.
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| * 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>.
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| * 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>.
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| Therefore, <math>\{u,v_1,v_2,\ldots,v_d\}\cap S_1\cup S_2\cup\cdots\cup S_d\subseteq V</math> implies that
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| :<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>,
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| so that <math>\sum_{v:v\sim u}d(v)\le n-1</math>.
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| By Cauchy-Schwarz inequality,
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| :<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>,
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| which implies that <math>|E|\le\frac{1}{2}n\sqrt{n-1}</math>.
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| }}
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| === Hamiltonian cycle ===
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| We now look at graphs which does not have large cycles. In particular, we consider graphs without '''Hamiltonian cycles'''.
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| 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>(n-1)</math>-clique and the one remaining vertex is connected to the clique by one edge. This graph has <math>{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.
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| 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'''.
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| {{Theorem|Dirac's Theorem|
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| :A graph <math>G(V,E)</math> on <math>n</math> vertices has a Hamiltonian cycle if <math>d_v\ge\frac{n}{2}</math> for all <math>v\in V</math>.
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| }}
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| {{Proof|
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| Suppose to the contrary, the theorem is not true and there exists a non-Hamiltonian graph with <math>d_v\ge\frac{n}{2}</math> for all <math>v\in V</math>. Let <math>G</math> be such a graph with a maximum number of edges. Then adding any edge to <math>G</math> creates a Hamiltonian cycle. Thus, <math>G</math> must have a Hamiltonian path, say <math>v_1v_2\cdots v_n</math>.
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| Consider the sets,
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| *<math>S=\{i\mid v_iv_n\in E\}</math>;
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| *<math>T=\{i\mid v_{i+1}v_1\in E\}</math>.
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| Therefore, <math>S\subseteq\{v_1,v_2,\ldots,v_{n-1}\}</math> contains the neighbors of <math>v_n</math>; and <math>T\subseteq\{v_1,v_2,\ldots,v_{n-1}\}</math> contains the predecessors (along the Hamiltonian path) of the neighbors of <math>v_1</math>. It holds that <math>S,T\subseteq\{v_1,v_2,\ldots,v_{n-1}\}</math>.
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| Since <math>d_v\ge\frac{n}{2}</math> for all <math>v\in V</math>, <math>|S|,|T|\ge\frac{n}{2}</math>. By the pigeonhole principle, there exists some <math>v_i\in S\cap T</math>. We can construct the following Hamiltonian cycle:
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| :<math>v_1v_{i+1}v_{i+2}\cdots v_nv_{i}v_{i-1}\cdots v_1\cdots</math>,
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| which contradict to the assumption that <math>G</math> is non-Hamiltonian.
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| }}
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| == Erdős–Stone theorem ==
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| We introduce a notation for the number of edges in extremal graphs with a specific forbidden substructure.
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| {{Theorem|Definition|
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| :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.
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| }}
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| With this notation, Turán's theorem can be restated as
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| {{Theorem|Turán's theorem (restated)|
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| :<math>\mathrm{ex}(n,K_r)\le\frac{r-2}{2(r-1)}n^2</math>.
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| }}
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| 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>.
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| 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>.
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| {{Theorem|Fundamental theorem of extremal graph theory (Erdős–Stone 1946)|
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| :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.,
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| :::<math>\mathrm{ex}(n,K_s^r)= \left(\frac{r-2}{2(r-1)}+o(1)\right)n^2</math>.
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| }}
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| 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.
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| 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.
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| {{Theorem|Corollary|
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| :For every nonempty graph <math>H</math>,
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| ::<math>\lim_{n\rightarrow\infty}\frac{\mathrm{ex}(n,H)}{{n\choose 2}}=\frac{\chi(H)-2}{\chi(H)-1}</math>.
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| }}
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| {{Prooftitle|Proof of corollary|
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| Let <math>r=\chi(H)</math>.
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| 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,
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| :<math>|T(n,r-1)|\le\mathrm{ex}(n,H)</math>.
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| It is not hard to see that
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| :<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>.
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| 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,
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| :<math>\mathrm{ex}(n,H)\le\mathrm{ex}(n,K_s^r)</math>.
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| Due to Erdős–Stone theorem,
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| :<math>\mathrm{ex}(n,K_s^r)=\left(\frac{r-2}{2(r-1)}+o(1)\right)n^2</math>.
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| Altogether, we have
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| :<math>
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| \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)
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| </math>
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| The theorem follows.
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| }}
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| == References ==
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| :('''声明:''' 资料受版权保护, 仅用于教学.)
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| :('''Disclaimer:''' The following copyrighted materials are meant for educational uses only.)
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| * van Lin and Wilson. ''A course in combinatorics.'' Cambridge Press. Chapter 4.
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| * Aigner and Ziegler. ''Proofs from THE BOOK, 4th Edition.'' Springer-Verlag. [[media:PFTB_chap36.pdf| Chapter 36]].
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| * Diestel. ''Graph Theory, 3rd Edition''. Springer-Verlag 2000. [[media:Diestel2ed_chap7.pdf|Chapter 7]].
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