CSOR W4231.002 – Spring, 2019 hw4solns.pdf - CSOR W4231.002 – Spring 2019 hw4 solutions 1 Solution to problem 1 G has a cycle that contains e =(u v

# CSOR W4231.002 – Spring, 2019 hw4solns.pdf - CSOR...

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CSOR W4231.002 – Spring, 2019 hw4 solutions 1. Solution to problem 1 G has a cycle that contains e = ( u, v ) if and only if there is another path between v and u . Remove e = ( u, v ) from G . Run BFS( G , v ). If u appears in the BFS tree, then G has a cycle that contains e . Running time: O ( n + m ). 2. Solution to problem 2 For clarity, we’ll rename the special nodes v and w as s and t , respectively. Then the algorithm is a modified BFS, starting at node s : for each node u , we also store the number of shortest s - u paths in u.num . Initialize s.num = 1. Return t.num . . . . while Q 6 = do u = dequeue ( Q ) for each ( u, v ) E do if discovered [ v ] = 0 then . . . v.num = u.num else if dist [ v ] = dist [ u ] + 1 then v.num = v.num + u.num end if end for end while return t.num 3. Solution to problem 3 Run BFS ( G, s ). Then s appears in layer 0 and t in some layer L d with d > n/ 2. Note that distances are integers in undirected graphs, hence d ≥ d n/ 2 e . Then some layer between 1 and d - 1, say layer i , must contain exactly 1 node. Otherwise, layers 1 to d - 1 contain at least 2( d - 1) nodes, which is at least 2( d n/ 2 e - 1) = 2 · b n/ 2 c nodes. Note that s and t don’t appear in these layers but 2 · b n/ 2 c + 2 > n , the number of nodes in G ! Removing the node in layer i disconnects s from t since all graph edges that are not tree edges span at most 1 layer of the BFS tree. Running time: O ( m + n )

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4. Solution to problem 4 Modify procedure Update in Dijkstra’s algorithm to maintain in dist [ v ] the length of the longest edge on any s - v path. Update ( u, v ) if dist [ v ] > max { dist [ u ] , w uv } then dist [ v ] = max { dist [ u ] , w uv } prev [ v ] = u end if Now run Dijkstra’s algorithm from the origin s . Correctness and running time follow from Dijkstra’s algorithm. 5. Solution to problem 5 Note that the resulting graph is undirected, hence xy = yx . Run Dijkstra’s algorithm twice, once from s and once from t , thus computing the distances d s ( x ) between s and every other node x and the distances d t ( y ) between t and every other node y .
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