Math 5707 : Graph Theory , Spring 2017 Midterm 2
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چکیده
1.2 Solution Proof. Let n = |V |. Note first that since |E| ≥ |V |, n can be neither 1 nor 2. Define a bipartite graph (H;V,E), where H is the simple graph with vertex set V (H) = E ∪V and edge set E (H) = {{v, e} | e ∈ E, v ∈ V \e}. (We WLOG assume that E ∩ V = ∅.) Thus, the edges in H connect each vertex v of G to the edges of G that do not contain v. A V -complete matching M in H would consist of a subset of the edges {v ∈ V, e ∈ E} of H connecting each v ∈ V to a distinct e ∈ E. Because of the definition of E (H),
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Math 5707 : Graph Theory , Spring 2017
1.1 Problem Fix a loopless multidigraphD = (V,A, φ). Let f : V → N be a configuration. Let h = ∑ f . Let n = |V |. Assume that n > 0. Let ` = (`1, `2, . . . , `k) be a legal sequence for f having length k ≥ ( n+ h− 1 n− 1 ) . Prove the following: (a) There exist legal sequences (for f) of arbitrary length. (b) Let q be a vertex of D such that for each vertex u ∈ V , there exists a path from u t...
متن کاملMath 5707 : Graph Theory , Spring 2017 Homework 5
1.1 Problem Fix a loopless multidigraphD = (V,A, φ). Let f : V → N be a configuration. Let h = ∑ f . Let n = |V |. Assume that n > 0. Let ` = (`1, `2, . . . , `k) be a legal sequence for f having length k ≥ ( n+ h− 1 n− 1 ) . Prove the following: (a) There exist legal sequences (for f) of arbitrary length. (b) Let q be a vertex of D such that for each vertex u ∈ V , there exists a path from u t...
متن کاملMATH 5707 Midterm 2
Proof. Let n = |V |, and assume without loss of generality that |E| = n (else we could remove edges from G until |E| = n and apply the argument). Also assume WLOG that V ∩ E = ∅ (else, we can rename the vertices). Note that we must have n ≥ 3, since a simple graph with 2 or fewer vertices can have at most 1 edge. Now we seek a bijection f : V → E with each v ∈ V satisfying v / ∈ f(v). Construct...
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تاریخ انتشار 2017