Math Summer 2012

Transcription

1 Math Summer 2012 Final Exam You have one hour and fifty minutes to complete this exam. You are not allowed to use any electronic device. Be sure to give reasonable justification to all your answers. 1. Show that a connected graph G contains an Eulerian trail if and only if there are at most two vertices of odd degree. ( ) If the trail is a circuit, then there are zero vertices of odd degree. If the trail is not a circuit, then it has end vertices, say u and v. If w u, v is on the trail, then each time the trail comes to w through one edge, it must leave through a different edge. Therefore deg(w) is divisible by two. However, the trail passes through the end vertex and never leaves, and it begins at the start vertex without arriving there first. Therefore, these two vertices have odd degree. ( ) If there are zero vertices of odd degree, then G is Eulerian and the Eulerian circuit is also an Eulerian trail. If there are two vertices of odd degree, say u and v, then there are two cases. Case 1: u and v are adjacent. Form G from G by removing uv. Then every vertex of G has even degree, so it contains an Eulerian circuit. If we reorder the vertices on the circuit so that it starts at v and add uv at the beginning, then we have an Eulerian trail. Case 2: If u and v are not adjacent, form G from G by adding uv. Then G has an Eulerian circuit T. We re-order the circuit so that uv is last, then when uv is removed, we have an Eulerian trail in G. 2. If u and v are adjacent vertices in a graph, prove that their eccentricities differ by at most one. Suppose that u 1 and v 1 are vertices such that ecc(u) = d(u, u 1 ) and ecc(v) = d(v, v 1 ). We know that d(u, u 1 ) d(v, u 1 ) + 1, otherwise there would be a shorter route from u to u 1 by going through v. Similarly, Note that so or d(v, v 1 ) d(u, v 1 ) + 1. ecc(u) = d(u, u 1 ) d(v, u 1 ) + 1 ecc(v) + 1, ecc(v) = d(v, v 1 ) d(u, v 1 ) + 1 ecc(u) + 1. ecc(u) ecc(v) + 1 and ecc(v) ecc(u) + 1, 1 ecc(u) ecc(v) 1, which means that the difference between ecc(u) and ecc(v) is at most one. 3. Find a closed form for the generating function of the sequence {k 2 }. As seen in class, we know that kx k x = (1 x) 2.

2 Let a k = k 2. This sequence satisfies the recurrence relation a k = a k 1 + 2k 1, a 0 = 0. Let G(x) = a k x k. Then G(x) = a k x k = a 0 + a k x k = (a k 1 + 2k 1)x k = x a k x k + x (2k + 1)x k k 1 k 1 This implies that x = xg(x) + 2x (1 x) 2 + x 1 1 x. (1 x)g(x) = 2x2 (1 x) 2 + x 1 x G(x) = 2x2 (1 x) 3 + x (1 x) 2 = 2x2 + x(1 x) (1 x) 3 = x2 + x (1 x) Suppose the integers from 1 to n are arranged in some order around a circle, and let k be an integer with 1 k n. Show that there must exist a sequence of k adjacent numbers in the arrangement whose sum is at least k(n + 1)/2. Let the integers around the circle be labeled as a 1,..., a n. Let s j be the sum of the k adjacent integers starting at a j. Then n n(n + 1) s j = k, 2 i=1 since each number from 1 to n shows up k times in the sum. Then there are kn(n + 1)/2 pigeons that must be distributed between n holes (the s j ). By the Generalized Pigeonhole principle, there must be one s j whose value is at least k(n + 1)/2, and since the s j are integers, we can say it is at least k(n + 1)/2. 5. Assume that a vowel is one of the five letters A, E, I, O, or U. How many eleven letter sequences from the alphabet contain exactly three vowels? [You can assume each letter is only used once in the sequences. Also be aware that order matters here.] There are five vowels, so that leaves us with 21 consonants. Therefore, the number of unordered sequences with three vowels is ( )( ) The number of ways to order the 11 letters in the sequences is 11!. Therefore, the total number of 11 letter sequences with three vowels is ( )( ) ! = 81, 226, 696, 320, Give the definition of each of the following:

3 a) Graph A graph consists of two finite sets, V and E. Each element of V is called a vertex. The elements of E, called edges, are unordered pairs of vertices. b) Walk A walk in which the edges are distinct. c) Circuit A circuit is a trail that begins and ends at the same vertex. d) Eulerian circuit An Eulerian circuit is one that includes every edge of the graph. e) Hamiltonian cycle A Hamiltonian cycle is one that includes every vertex of the graph. f) Spanning tree Given a graph G and a subgraph T, we say that T is a spanning tree of G if T is a tree that contains every vertex of G. g) Distance between vertices In a connected graph G, the distance from vertex u to vertex v is the length (number of edges) of a shortest uv path in G. h) Eccentricity For a given vertex v of a connected graph, the eccentricity of v, denoted ecc(v), is defined to be the greatest distance from v to any other vertex in the graph. i) Adjacency matrix Let G be a graph with vertices v 1, v 2,..., v n. The adjacency matrix of G is the n n matrix A whose (i, j) entry, denoted by [A] i,j, is defined by [A] i,j = { 1 if vi v j E(G) 0 otherwise. j) Digraph A digraph is a graph where the edge set E consists of ordered pairs of vertices. k) Proper coloring We say that a function K : V (G) {1, 2,..., k} is a proper k-coloring of G if for every pair u, v of adjacent vertices, K(u) K(v) that is, if adjacent vertices are colored differently. l) Planar graph A graph G is said to be planar if it can be drawn in the plane in such a way that pairs of edges intersect only at vertices, if at all.

4 m) Characteristic path length The characteristic path length of a graph G, denoted L G, is the average distance between vertices, where the average is taken over all pairs of distinct vertices. n) The pigeonhole principle Let n be a positive integer. If more than n objects are distributed among n containers, then some container must contain more than one object. o) x k We define the falling factorial power x k as a product of k terms beginning with x, with each successive term one less than its predecessor: p) Pascal s triangle k 1 x k = x(x 1) (x k + 1) = (x i). Pascal s triangle is the triangle where the nth row is given by ( ) ( ) ( ) n n n,,..., 0 1 n for n 0. q) Generating function Given a sequence {a k } with k 0, its generating function G(x) is defined by G(x) = a k x k. i=0 r) The set o(g(n)) o(g(n)) = {f(n) : for every ε > 0 there exists n 0 (ε) with f(n) ε g(n) for n n 0 (ε)}. s) The set O(g(n)). O(g(n)) = {f(n) : there exists a constant C > 0 such that f(n) C g(n) for n n 0 }. t) The notation f(n) g(n. We say that f(n) g(n) if and only if lim n f(n) g(n) = 1.

5 7. Write the word True next to each statement that is always true. Write False next to each statement that is not true or is only sometimes true. For a graph G, κ(g) δ(g). False. The reverse inequality was proved in the homework. If v V (G), then N[v] is a connected graph. True. Between any two vertices, there is a path that goes through v. The graph K 2,6 has E 4 as a subgraph. True. Any graph having at least four vertices has E 4 as a subgraph. If G and H are graphs with 3 vertices and E(G) = E(H), then G and H are isomorphic. True. There number of edges completely determines the graph. of k. Let G k be a graph on n vertices that has k edges. Then diam(g k ) is an increasing function False. As more edges are added, the distance between vertices decreases, which means eccentricities decrease, so the diameter decreases. The number of labeled trees on n vertices is the same as the number of distinct orderings of n objects where two of the n objects are identical. False. The first number is n n 2, and the second is n!/2!. If G and H are isomorphic, applying the greedy algorithm to G and H will produce colorings that use the same number of colors. False. The greedy algorithm can use different numbers of colors depending on how the vertices are labeled. If G is a graph generated by a polyhedron, then it is four colorable. True. Such a graph is planar, and therefore 4-colorable. A sequence of positive integers of length 15 must have an increasing subsequence of length 4 or a decreasing subsequence of length 6. False. For the theorem from 2.4 to hold, we would have to more than 15 integers. Thus it is not always true. The number of 5 card hands that can be chosen from m decks is a strictly increasing function of m. False. As we saw, this number is the same whenever m 5.