Download Week 6 math 61 Week 6 math 61 and more Exams Mathematics in PDF only on Docsity! Week 6 math 61 Equation for the degree of a vertex - δ(c) = 2|{l ε E(G) l is a loop at v}| + {e ε E(G) e is an edge incident to v}| If G is a graph then what does ∑δ(v) for all of v ε V(G) equal to - 2 * |E(G)| proof that ∑δ(v) for all of v ε V(G) = 2 * |E(G)| - define L(G) = {l ε E(g) | l is a loop} and S(G) = {e ε E(G) | e is not a loop } E(G) the total edges of G = L(G) U S(G) because the total number of edges is equal to the number of loop edges and non loop edges thus |E(G)| = |L(G)| + |S(G)| this also means that 2|E(G)| = 2|L(G)| + 2|S(G)| Now we define for a vertex v ε V(G), L(v) = {l ε L(G) | l is included to v and S(v) = {s ε S(G) | S is incident to v} By definition δ(v) = 2|L(v)| + |S(v)| thus ∑δ(v) = ∑(2|l(v)| + |s(v)|) ∑δ(v) = 2∑|l(v)| + ∑|s(v)| Also |L(G)| = ∑|L(v)| because we have that L(G) = the union of all L(v) since the sets {L(v)| v ε V(G)} are pairwise disjoint Also ∑|S(v)| = 2|S(G)| because very edge e ε S(G) is incident to two vertices plugging into the equation ∑δ(v) = 2∑|l(v)| + ∑|s(v)| that |L(G)| = ∑|L(v)| and that ∑|S(v)| = 2|S(G)| you get that ∑δ(v) = 2|L(G)| + 2|S(G)| proving the theorem does there exist a graph G with degree sequence (3,7,1,6) - No because ∑δ(v) for all of v ε V(G) = 2 * |E(G)| so it must be an even number Theorem about the number of vertices with an odd degree in a graph G - The number of vertices of odd degree in a graph G is even if a1, a2, ..., an εZ are positive odd integers then if n is odd what is true - the sum ∑ai is odd (adding up all the integers are odd) proof that a1, a2, ..., an εZ are positive odd integers then if n is odd then the sum ∑ai is odd - since each value ai is odd we can write that ai = 2*mi + 1 for some integer mi ε z Therefore ∑ai = ∑2 * mi + 1 = ∑2* mi + ∑1 = 2∑mi + ∑1 If we set that N = 2∑mi we see that N is even because multiply by two makes things even and if we set n = ∑1 this is odd because adding 1 an odd number of times is odd. So we have ∑ai N + m which is an odd + an even number which is odd proof of theorem that the number of odd degree vertices in a graph G is even - if we let N>=0 denote the number of odd vertices we want to show that n is even. We also have that ∑δ(v) is even because of the fact that it is equal to 2 * |E(G)|
