hw8 - MAT 145: Homework Solutions #8 Prepared by Maya Ahmed...

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MAT 145: Homework Solutions #8 Prepared by Maya Ahmed and Motohico Mulase June 16, 2003 1. Brualdi 11.12 Determine which pairs of the multi-graphs in Figure 11.40 are isomorphic and if iso- morphic, find an isomorphism. Figure 1: Problem 11.12 Figure 11.40 (i) (ii) (iii) Answer: Figure 2: Problem 11.12 Check for degree differences. Note that the degree sequences of vertices in all the graphs are same. (i) (ii) (iii) (3) (3) (3) (3) (3) (3) (3 )( 3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) (3) Graph (i) and (ii) are isomorphic. A relabelling of vertices that shows the isomorphism is shown in the figure below. The graph (iii) is not isomorphic since it contains a 4-cycle a-b-c-d, and the graphs (i) and (ii) have no 4-cycles. 1
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Figure 3: Problem 11.12 Isomorphism of graphs (i) (ii) (iii) 1 2 3 4 5 6 7 8 9 10 1 6 98 52 43 7 10 a b c d 2. Brualdi 11.20 Prove that a graph of order n with at least ( n - 1)( n - 2) 2 + 1 edges must be connected. Give an example of a disconnected graph of order n with one fewer edge. Answer: [By M. Mulase] We note that the complete graph has the largest number of edges among the graphs with the same order. A graph of order n (i.e., a graph that has n vertices) is discon- nected if it has two subgraphs of orders k and n - k that are not connected, here k is some integer in the range 1 k n - 1. Among all such graphs, the largest number of edges is achieved if we have the complete graph of order k and the complete graph of order n - k . If you add just one more edge to this situation, then the added edge should connect the two disjoint components to make a connected graph of order n . Thus we have to establish an inequality that ± k 2 ² + ± n - k 2 ² < ( n - 1)( n - 2) 2 + 1 for every 1 k n - 1. [ Note: It is a very common mistake that you prove this inequality only for k = 1 and think you are done. Of course that is not the right answer!] The proof goes as follows. First we notice that the inequality we want to establish is equivalent to ± k 2 ² + ± n - k 2 ² ( n - 1)( n - 2) 2 , 2
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or k ( k - 1) + ( n - k )( n - k - 1) ( n - 1)( n - 2) . To prove this, we consider the function f ( k ) = ( n - 1)( n - 2) - k ( k - 1) - ( n - k )( n - k - 1) , where k is a variable and n is a fixed constant. By an algebraic computation, we obtain f ( k ) = 2( - k 2 + nk - n + 1) = 2 ( k ( n - k ) - ( n - 1) ) . - (n - 1) 1 n - 1 0 k y Figure 4: y = k ( n - k ) - ( n - 1). From Figure 4, we see that if 1 k n - 1, then f ( k ) 0. This implies that ± k 2 ² + ± n - k 2 ² < ( n - 1)( n - 2) 2 + 1 for the same range of k , which establishes the claim. In class I gave a slightly different explanation, by removing edges from a complete graph. That method leads to exactly the same quadratic inequality a lot faster. The proof I give here seems to be a little easier to understand, though. An example of a disconnected graph of order 5 with only 6 edges is shown in the figure below.
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hw8 - MAT 145: Homework Solutions #8 Prepared by Maya Ahmed...

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