Discrete Geometry and Extremal Graph Theory - IMSA

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and the proof is pretty much analogous to the one above. By the formula in the previous problem, it suffices to prove | ∑ m−1 k=1 (1 + zk ) n | ≤ (2 cos π m )n where z is a primitive m-th root of unity. Note that all the points 1 + z k must lie within the circle centered at the origin with radius 2 cos π . Thus, |(1 + m zk ) n | = |1 + z k | n ≤ 2 cos π m )n , and the result follows as before from the triangle inequality. 5. A graph G is bipartite if and only if it contains no odd cycle. Proof: If G = G 1 ∪ G 2 is bipartite, where G 1 and G 2 are disjoint subgraphs of G, then it clearly contains no odd cycle. This is because two vertices are neighbors if and only if they lie in different subsets, so a path must alternate between vertices in G 1 and G 2 . If G has no odd cycles, let us pick a vertex v and color it red. We color all neighbors of v blue, and all uncolored neighbors of these blue vertices red, and so on. We might as well assume that G is connected, so all vertices have a color. We claim that setting G 1 and G 2 to contain all red and blue vertices, respectively, shows that G is bipartite. Assume otherwise, and say that v 1 , v 2 are neighbors of the same color. This means that there exist paths from v to v 1 and v 2 with lengths of the same parity. In other words, the path from v 1 to v 2 through v must have an even length. (If the paths from v 1 to v and v 2 to v first meet at v 3 , then we may replace v with v 3 and the proof continues.) Now, this path from v 1 to v 2 with the edge joining v 1 and v 2 creates an odd cycle, contradiction. Thus, G is bipartite if and only if it contains no odd cycle. 6. Let G be a graph on n vertices without K k , a complete graph on k vertices. Then, G (k − 2)n2 has at most edges. Moreover, there exists a graph with this number of edges. 2(k − 1) Proof: I do admit to having seen this before, although I only vaguely remembered the ideas behind this proof of Turán’s Theorem. Let E(G) denote the number of edges of G. The proof proceeds by induction on n. Our base cases, for instance n = 1, 2, 3 are trivially true. Now, let G be a graph on n vertices without K k containing the maximal number of edges. Note that G must contain a K k−1 or else we could add edges, contradiction the maximality of E(G). Let H 1 be a K k−1 and let H 2 be G without H 1 . Note that E(H 1 ) = ( ) k−1 2 . By the induction hypothesis, we also have E(H 2 ) ≤ (k − 2)(n − k + 1)2 . 2(k − 1) If v ∈ H 2 , then v is neighbors with at most k − 2 vertices in H 1 or else we have a K k . This implies that the number of edges with one vertex in each H 1 and H 2 is at most (n − k + 1)(k − 2), so ( ) k − 1 E(G) ≤ + 2 (k − 2)(n − k + 1)2 2(k − 1) + (n − k + 1)(k − 2) = (k − 2)n2 2(k − 1) . Now, let n = (k −1)q +r where 0 ≤ r < k −1. Consider a partition G = G 1 ∪· · ·∪G k−1 where r of the subgraphs have q + 1 vertices and the rest have q vertices. We join two 6
vertices if and only if they lie in different G i . After a bit of computation, this shows that the bound above is essentially sharp. 7. We prove a few results regarding even and odd permutations. Lemma 1: Every permutation σ ∈ S n can be written as a product of transpositions. Proof: Write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles. It suffices to show that any cycle can be written as a product of transpositions. Let c = (i 1 i 2 · · · i k ) be a cycle. We show that c can be written as a product of transpositions by induction on k. For k = 1, 2, the result is trivial. Now, note that c = (i 1 i 2 · · · i k ) = (i 1 i 2 · · · i k−1 )(i k−1 i k ) and (i 1 i 2 · · · i k−1 ) can be written as a product of transpositions by the induction hypothesis. The result follows. Lemma 2: Let τ 1 , τ 2 , . . . , τ m ∈ S n be transpositions. Then, the number of cycles in τ 1 τ 2 · · · τ m has the same parity as m + n. Proof: We first prove the following: if σ ∈ S n is a permutation and τ ∈ S n is a transposition, then the number of disjoint cycles in σ and στ have different parity. As before, write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles and suppose τ = (ij). First we deal with the case where i and j are in the same cycle, say c t . If σ(i) = j (or vice versa), then c t = (ij) or (iji 1 · · · i s ). If c t = (ij), then we may replace c t with (i)(j) in στ. If c t = (iji 1 · · · i s ), then we replace c t with (ii 1 · · · i s )(j) In either case, the parity of the number of disjoint cycles changes. If i and j are in different cycles, say c t and c u respectively, we argue as follows. First, if c t = (i) and c u = (j), then στ is the product of all cycles c i and (ij) except for c t and c u . If c t = (i) and c u = (jj 1 · · · j v )) (or vice versa), then in στ, replace c u with (jij 1 · · · j v ) and remove (ij). If c t = (ii 1 · · · i s ) and c u = (jj 1 · · · j v ), then στ can be rewriten with c ′ t,u = (ij 1 · · · j v ji 1 · · · i s ) and (ij), c t , c u removed. Again, in either case, the parity changes. Now we prove Lemma 2 by induction on m, and our base cases are easily verified. We have τ 1 τ 2 · · · τ m+1 = στ m+1 , where we set σ = τ 1 τ 2 · · · τ m . The number of disjoint cycles in σ and στ m+1 have different parity, so this completes our induction. Comment: It immediately follows that a permutation σ can be written as a product of an even number of transpositions or an odd number of transpositions, but not both. 7
Page 1 and 2: Discrete Geometry and Extremal Grap
Page 3 and 4: We prove the following result using
Page 5: 3. Let R m (n) be the number of sub

Discrete Geometry and Extremal Graph Theory - IMSA

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