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Extremal Graph Theory

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Extremal Graph Theory 2/12/2016 Ma/CS 6b Class 17: Extremal Graph Theory Paul Turán By Adam Sheffer Extremal Graph Theory The subfield of extremal graph theory deals with questions of the form: ◦ What is the maximum number of edges that a graph with 푛 vertices can have without containing a given subgraph 퐻? ◦ Characterize the graphs that obtain this maximum number of edges. A graph without 퐾5 1 2/12/2016 푒푥 푛,퐻 We denote by 푒푥 푛, 퐻 the maximum number of edges in a graph with 푛 vertices and no subgraph 퐻. Example. Recall that 푃푖 is a simple path of length 푖. What is 푒푥 4, 푃3 ? 3 Graphs Without Triangles Problem. Find 푒푥 푛, 퐾3 . ◦ What graph contains a large number of edges but does not have 퐾3 as a subgraph? ◦ Any complete bipartite graph. 2 2/12/2016 Maximizing the Number of Edges What value of 푚 maximizes the number of edges in 퐾푚,푛−푚? ◦ The number of edges is 푚 푛 − 푚 . ◦ The maximum of 푛2/4 is obtained for 퐾푛/2,푛/2. ◦ Can we find a graph without triangles and with more edges? Mantel’s Theorem 2 Theorem. 푒푥 푛, 퐾3 = 푛 /4 . Proof. Let 퐺 = 푉, 퐸 be a triangle-free graph with 푉 = 푛. ◦ 푑푖 = deg 푣푖. ◦ If 푣푖, 푣푗 ∈ 퐸, then 푣푖 and 푣푗 do not have common neighbors, so 푑푖 + 푑푗 ≤ 푛. ◦ We thus have 푑 + 푑 ≤ 푛 퐸 . 푣푖,푣푗 ∈퐸 푖 푗 ◦ Since every 푑푖 appears in 푑푖 elements 2 푛 퐸 ≥ 푑푖 + 푑푗 = 푑푖 . 푑푖,푑푗 ∈퐸 푖 3 2/12/2016 The Cauchy–Schwarz Inequality The Cauchy–Schwarz inequality. For any 푎1, 푎2, … , 푎푛, 푏1, 푏2, … , 푏푛 ∈ ℝ, we have 푛 2 푛 푛 2 2 푎푖푏푖 ≤ 푎푖 푏푖 . 푖=1 푖=1 푖=1 Equality holds iff the vectors 푎1, … , 푎푛 and 푏1, … , 푏푛 are linearly dependent. Available online Completing the Proof We proved that 2 푛 퐸 ≥ 푑푖 + 푑푗 = 푑푖 . 푑푖,푑푗 ∈퐸 푖 Recall that 푖 푑푖 = 2|퐸|. By Cauchy-Schwarz with 푎푖 = 푑푖 and 푏푖 = 1: 2 2 2 푑푖 ≤ 푑푖 1 . 푖 푖 푖 ◦ We thus have 4 퐸 2 푛2 푛 퐸 ≥ 푑2 ≥ → 퐸 ≤ 푖 푛 4 푖 4 2/12/2016 The Maximum Graph is Unique Claim. The only graph that has the maximum number of edges 2 푒푥 푛, 퐾3 = 푛 /4 is 퐾 푛/2 , 푛/2 . Proof. We consider the case of an even 푛. 2 ◦ In our proof for 푒푥 푛, 퐾3 = 푛 /4, we used Cauchy-Schwarz to obtain 2 2 2 푖 푑푖 ≤ 푖 푑푖 푖 1 , which implied 4 퐸 2 푛 퐸 ≥ 푑2 ≥ . 푖 푖 푛 ◦ Equality holds if and only if for every 1 ≤ 푖 ≤ 푛, we have 푑푖 = 푛/2. Inequality of Arithmetic and Geometric Means AM-GM Inequality. For any 푎1, 푎2, … , 푎푛 ∈ ℝ, we have 푎 + ⋯ + 푎 1 푛 ≥ 푎 ⋯ 푎 1/푛. 푛 1 푛 No Shirt available. Marketing opportunity? 5 2/12/2016 Recall: Independent Sets Consider a graph 퐺 = 푉, 퐸 . An independent set in 퐺 is a subset 푉′ ⊂ 푉 such that there is no edge between any two vertices of 푉′. Finding a maximum independent set in a graph is a major problem in theoretical computer science. ◦ No polynomial-time algorithm is known. Mantel’s Theorem: Second Proof 퐺 = 푉, 퐸 – a triangle-free graph with 푉 = 푛. 훼 – size of the largest independent set 퐴 in 퐺. The set of neighbors of any 푣 ∈ 푉 is an independent set, so 훼 ≥ deg 푣. 퐵 = 푉 ∖ 퐴, so |퐵| = 푛 − 훼. By the AM-GM inequality, we have 훼 + 푛 − 훼 2 푛2 퐸 ≤ deg 푣 ≤ 훼 퐵 ≤ = . 2 4 푣∈퐵 6 2/12/2016 Generalizing the Problem We move from 푒푥 푛, 퐾3 to 푒푥 푛, 퐾푟 . ◦ For some 푟 > 3, what graph contains many edges but no 퐾푟? ◦ An 푟-partite graph is a graph with 푟 parts, and no edge between two vertices of the same part (generalizing bipartite graphs). ◦ Example. The figure presents the complete 4-partite graph 퐾3,3,3,4. Turán Graphs The Turán graph 푇 푛, 푟 is a complete 푟-partite graph with 푛 vertices, such that each part consists of either 푛/푟 or 푛/푟 vertices. ◦ The graph in the figure is 푇 13,4 . ◦ 푇 푛, 푟 − 1 contains no copy of 퐾푟 and has 푛2 1 about 1 − edges. 2 푟−1 7 2/12/2016 Turán’s Theorem Theorem. If 퐺 = 푉, 퐸 is a graph that contains no copy of 퐾푟 and 푉 = 푛, then 푛2 1 퐸 ≤ 1 − . 2 푟−1 Proof. By induction on 푛. ◦ Induction basis. Easily holds when 푛 < 푟. ◦ Induction step. 퐺 = 푉, 퐸 – a graph that contains no copy of 퐾푟, with 푉 = 푛 and 퐸 = 푒푥 푛, 퐾푟 . ◦ 퐺 contains a copy of 퐾푟−1. Otherwise, adding an edge to 퐺 would not yield a copy of 퐾푟, contradicting the maximality of 퐺. 퐺 = 푉, 퐸 – a graph that contains no copy of 퐾푟, with 푉 = 푛 and 퐸 = 푒푥 푛, 퐾푟 edges. 퐴 – a subgraph of 퐺 that is a 퐾푟−1. 푟 − 1 퐴 contains edges. 2 By the hypothesis, 퐺 ∖ 퐴 has at most (푛−푟+1)2 1 1 − edges. 2 푟−1 Since each vertex of 푉 ∖ 퐴 is adjacent to at most 푟 − 2 vertices of 퐴, there are 푛 − 푟 + 1 푟 − 2 edges between 푉\퐴 and 퐴. Combining the above, we have 푟 − 1 푟 − 2 푛 − 푟 + 1 2 1 퐸 ≤ + 1 − 2 2 푟 − 1 푛2 + 푛 − 푟 + 1 푟 − 2 ≤ 1 − 1/ 푟 − 1 . 2 8 2/12/2016 Which US President Proved a Mathematical Theorem? Cliques Given a graph 퐺 = 푉, 퐸 , a clique of 퐺 is any subgraph that is a complete graph. No polynomial-time algorithm for finding a maximum clique in a graph is known. ◦ The problem is equivalent to finding a maximum independent set in a graph. ◦ Define the graph 퐺푐 = (푉, 퐸푐) such that 푣, 푢 ∈ 퐸푐 iff 푣, 푢 ∉ 퐸. Finding a maximum clique in 퐺 is equivalent to finding a maximum independent set in 퐺푐. 9 2/12/2016 A Lower Bound on the Clique Size Lemma. Let 퐺 = 푉, 퐸 be a graph with 푉 = 푣1, … , 푣푛 , and let 푑푖 = deg 푣푖. Then 퐺 contains a clique of size at least 푛 1 푖=1 . 푛−푑푖 1 1 1 1 1 1 + + + + = 3 + 5 − 4 5 − 4 5 − 3 5 − 3 5 − 2 3 Lemma. Let 퐺 = 푉, 퐸 be a graph with 푉 = 푣1, … , 푣푛 , and let 푑푖 = deg 푣푖. Then 퐺 contains a clique of size at least 푛 1 푖=1 . 푛−푑푖 We saw a probabilistic proof for: ◦ Theorem. A graph 퐺 = (푉, 퐸) has an independent set of size at least 1 . 1 + deg 푣 푣∈푉 The lemma is obtained by applying the theorem on 퐺푐. 10 2/12/2016 Proof #2 of Turán’s Theorem 퐺 = 푉, 퐸 – a graph that contains no copy of 퐾푟, with 푉 = 푣1, … , 푣푛 . 1 푑푖 = deg 푣푖 , 푎푖 = 푛 − 푑푖, 푏푖 = . 푛 − 푑푖 By Cauchy-Schwarz 푛 2 푛 푛 2 2 푎푖푏푖 ≤ 푎푖 푏푖 . 푖=1 푖=1 푖=1 Since 푎푖푏푖 = 1, we have 푛 푛 2 2 2 푛 ≤ 푎푖 푏푖 . 푖=1 푖=1 1 푑푖 = deg 푣푖 , 푎푖 = 푛 − 푑푖, 푏푖 = . 푛 − 푑푖 We have 푛 푛 푛 푛 2 2 2 1 푛 ≤ 푎푖 푏푖 = 푛 − 푑푖 . 푛 − 푑푖 푖=1 푖=1 푖=1 푖=1 By the previous lemma, 퐺 has a clique of size at 푛 1 least 푖=1 . 푛−푑푖 푛 1 By the assumption, 푖=1 ≤ 푟 − 1. Thus, 푛−푑푖 푛 2 2 푛 ≤ 푟 − 1 푛 − 푑푖 = 푟 − 1 푛 − 2 퐸 . 푖=1 푛2 푛2 1 ≤ 푛2 − 2 퐸 → 퐸 ≤ 1 − 푟 − 1 2 푟 − 1 11 2/12/2016 Proof #3: Balancing Weights 퐺 = 푉, 퐸 – a graph that contains no copy of 퐾푟, with 푉 = 푣1, … , 푣푛 . 1 We assign a weight 푤 = to every 푣 . We set 푖 푛 푖 푊 = 푤 푤 . 푣푖,푣푗 ∈퐸 푖 푗 1 Initially we have 푊 = 퐸 . We wish to move 푛2 weights to increase 푊. 1/3 1/2 1/3 1/3 1/2 0 1 1 1 1 2 1 1 1 1 ⋅ + ⋅ = ⋅ + ⋅ 0 = 3 3 3 3 9 2 2 2 4 Balancing Weights (cont.) 푣푖, 푣푗 – two vertices of 푉 with positive weights, such that 푣푖, 푣푗 ∉ 퐸. 푁푖, 푁푗 – the sum of the weights of the neighbors of 푣푖 and 푣푗, respectively. WLOG, assume that 푁푖 ≤ 푁푗. By moving the weight of 푣푖 to 푣푗, we change 푊 by 푤푖푁푗 − 푤푖푁푖 ≥ 0. 0.1 푣푖 0.1 푣푖 0.1 0 0.1 0.1 0.1 0.1 0.1 0.2 푣푗 0.2 푣푗 0.2 12 2/12/2016 More Balancing of Weights. If there exist 푣푖, 푣푗 with positive weights and 푣푖, 푣푗 ∉ 퐸, we can move the weight vertex of one to the other without decreasing 푊. We repeatedly do this until the entire weight is on the vertices of a complete subgraph 퐾푚. Consider 푣푖, 푣푗 ∈ 퐾푚 with 푤푗 − 푤푖 = 2휀 > 0. Then moving 휀 from 푤푗 to 푤푖 changes 푊 by 2 휀 1 − 푤푖 − 휀 − 1 − 푤푗 = 휀 푤푗 − 푤푖 − 휀 ≥ 휀 . 0.2 0.25 0.25 0.25 0.3 0.25 0.25 0.25 Concluding the Proof We conclude that we can move the entire weight to a subgraph 퐾푚 where each vertex of 1 the subgraph has a weight of .
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