hw1soln

# hw1soln - CME 305 Discrete Mathematics and Algorithms...

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Unformatted text preview: CME 305: Discrete Mathematics and Algorithms Instructor: Professor Amin Saberi ([email protected]) HW#1 Solutions 1. (Lovasz, Pelikan, and Vesztergombi 7.3.9) Prove that at least one of G and G is connected. Here, G is a graph on the vertices of G such that two vertices are adjacent in G if and only if they are not adjacent in G . Solution: Let G be a disconnected graph in which case we can decompose it into k connected components C 1 ,C 2 ,...,C k . We want to show that G is connected i.e. there is a path between any u and v in G . In the case that u and v are in different components we know that there exist an edge (a path of length one) between them in G . In the case that u and v are in the same component, say C i , we can construct a path of two edges between them in G as follows. Pick any vertex w from some other component C j for j 6 = i and note that edges { u,w } and { w,v } are in G . Thus u,w,v is a path in G and hence G is connected. 2. (Lovasz, Pelikan, and Vesztergombi 8.5.6) A double star is a tree that has exactly two nodes that are not leaves. How many unlabeled double stars are there on n nodes? Solution: We can represent each unlabeled double star by an unordered pair { a,b } where a is the number of leaves adjacent to one of the two non-leaf nodes and b is the number of leaves adjacent to the other. Therefore, to compute the number of double stars we need to find the number of sets { a,b } (note that { a,b } = { b,a } ) such that a,b > 0 and a + b + 2 = n . The first condition comes from the fact that the non-leaf nodes must be adjacent to at least one leaf and the second condition is simply the statement that, taken together, the leaves and the two non-leaf nodes make up the entire node set. First, assume n is even. Then a can take on values in A = { 1 , 2 ,..., ( n- 2) / 2 } . A smaller value contradicts a > 0 and a larger one implies that b ≤ n/ 2- 2 so b ∈ A and we have counted { a,b } twice. Now, assume n is odd. Then a can take on values A = { 1 , 2 ,..., ( n- 3) / 2 } . A smaller value again contradicts a > 0 and a larger one implies b ≤ ( n- 3) / 2 so again b ∈ A resulting in double counting. Thus the number of unlabeled double stars is given by ( n- 2) / 2 whenever n is even and ( n- 3) / 2 whenever n is odd. 3. (Lovasz, Pelikan, and Vesztergombi 8.5.10) If C is a cycle, and e is an edge connecting two nonadjacent nodes of C , then we call e a chord of C . Prove that if every node of a graph G has degree at least 3, then G contains a cycle with a chord. Solution: Suppose u u 1 ...u k is the longest path in G ( V,E ). Since d ( u ) ≥ 3, there exist two neighbors w,v of u not equal to u 1 . Since u u 1 ...u k is the longest path, we must have that w,v ∈ { u 2 ,u 3 ,...,u k } . Suppose w = u i ,v = u j with i ≤ j . Then u u 1 ...u j u is a circle with chord u u i . 4. (Kleinberg and Tardos 4.33) Suppose you are given a directed graph G = ( V,E ) in which each edge has a cost of either 0 or 1. Also suppose thatin which each edge has a cost of either 0 or 1....
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hw1soln - CME 305 Discrete Mathematics and Algorithms...

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