Midterm1Prac

# Midterm1Prac - Solutions to Practice Problems for Midterm 1...

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Peng Du University of California, San Diego 1 Exercise 3.11 Suppose e = ( u,v ), we only need to delete e in G and check whether there is a path from u to v by DFS or BFS. The algorithm runs in time O ( | E | ). 2 Exercise 3.22 For each s V , we use DFS or BFS to check whether all vertices can be reachable from s . The algorithm takes O ( | V || E | ) time. 3 Exercise 4.5 See the solution for homework four. 4 Exercise 4.8 See the solution for homework four. 5 Exercise 4.19 Let l 0 = { l e + c u : e = ( u,v ) E } and dist = dijkstra ( G,l 0 ,s ). We have cost [ w ] = dist [ w ] + c w for each w V . The algorithm has the same time com- plexity as dijkstra algorithm. 6 Exercise 4.20 Let d 1 = dijkstra ( G,l,s ) and d 2 = dijkstra ( G,l,t ). For each e 0 = ( u,v ) E 0 , we deﬁne d e 0 = min( d 1 [ t ] ,d 1 [ u ] + d 2 [ v ] + l e 0 ,d 1 [ v ] + d 2 [ u ] + l e 0 ), which is the shortest driving distance from s to t after adding road e 0 . Therefore, the

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Midterm1Prac - Solutions to Practice Problems for Midterm 1...

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