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Unformatted text preview: Fall Semester 2008 CS300 Algorithms Solution #8 1. (a) i. n 1 ii. min { m,n } (b) i. True; By definition, a graph with connectivity 4 is 1connected, 2connected, 3connected, and 4connected, but not 5connected. ii. False; Think about K 5 , a complete graph with 5 nodes. This graph is 1 connected, 2connected, 3connected, and 4connected, but not 5connected. Therefore, its connectivity is 4. 2. (a) Suppose that the condensation graph G contains a cycle ( v 1 ,v 2 , ··· ,v m ,v 1 ), m ≥ 2. Let u 1 be a vertex in the component S 1 and u m be a vertex in S m . Since G contains both a v 1 v m path and a v m v 1 path, G has both a u 1 u m path and a u m u 1 path. Thus, u 1 and u m belong to the same strong component of G . This produces a contradiction. So, G has no cycle. (b) • Case 1: A newly added edge connects two vertices in the same component. Then G * , the condensation graph of G * , has no cycle. Thus, the number of strongly connected components of G and that of G *...
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This note was uploaded on 02/04/2010 for the course COMPUTER S cs300 taught by Professor Unkown during the Spring '08 term at Korea Advanced Institute of Science and Technology.
 Spring '08
 Unkown
 Algorithms

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