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assignement1sol

# assignement1sol - CO453 Network Design Winter 2007...

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CO453: Network Design – Winter 2007 Instructor: Chaitanya Swamy Solutions to Assignment 1 Throughout G = ( V, E ) with | V | = n, | E | = m . Q1: The algorithm we will design will be very similar to Dijkstra’s algorithm. Let s denote the starting point, and d ( v ) denote the time taken to travel from s to v along the fastest path, i.e., d ( v ) = min s v paths P T ( P ), where T ( P ) is the time taken to travel along path P . Observe that analogous to shortest-path distances, we now have d ( v ) f e ( d ( u )) for every edge ( u, v ), and this will form the basis of our greedy algorithm. As in Dijkstra’s algorithm, we will maintain a set S of “explored” nodes, the ones for which we have correctly calculated d ( v ), and we will maintain a label ( v ) for the nodes not in S , which is our current estimate for the least time taken to reach v from s . We will update ( v ) based on the above rule but only consider edges ( u, v ) where u S , then pick the node w / S with smallest ( . ) value and add it to the set S , and repeat. 1) Initialize S ← { s } with d ( s ) = 0. Set ( v ) = for all v / S . Let v * = s . [ v * denotes the last node added to S .] 2) While S = V , (a) For every edge e = ( v * , v ) where v / S , update ( v ) min ( v ) , f e ( d ( v * )) . (b) Let w / S be such that ( w ) = min v / S ( v ). (c) Set S S ∪ { w } , d ( w ) = ( w ) , v * = w . Here we have presented the improved implementation done in class that yields a better running time. The proof of correctness and running-time analysis also proceed as in Dijsktra’s algorithm. Consider any node w , and the set S just before w is added to it (in step 2(c)). First, note that there is an s w path P w such that the time taken to travel along P w is d ( w ). Note that d ( w ) = ( w ) = min u S :( u ,w ) E f u ,w ( d ( u ) ) , so if u S is the node that attains the minimum in ( w ), then P w is obtained by adding the edge ( u , w ) to the path P u (which is obtained recursively). (We can easily keep track of these paths in the algorithm.) Now we show that d ( w ) is indeed the least time taken to reach w along any s w path. Consider any s w path P . Let v be the first node on P that is not in S , and u S be the node on P just before v . Let P be the portion of path P from s to v . Let t 1 and t 2 be the times at which we reach u and v respectively by traveling along path P . In Dijsktra’s algorithm, we used the fact that edge costs are non-negative to argue that the cost of P is at least the cost of P . Similarly, we will lower bound T ( P ) here using the given properties of the f e functions. Let e 1 = ( v, v 1 ) , e 2 , . . . , e k = ( v k - 1 , w ) be the portion of path P from v to w . Then the time taken to reach w by traveling along P is T ( P ) = f e k ( f e k - 1 ( . . . f e 2 ( f e 1 ( t 2 )) . . . )) t 2 since f e ( t ) t for every edge e . Also, t 2 = f u,v ( t 1 ) f u,v ( d ( u ) ) since t 1 d ( u ) and the f e functions are monotonic. Thus, we have T ( P ) f u,v ( d ( u ) ) ( v ) ( w ), where the last two inequalities follow from the definition of the ( . ) values, and since we choose w to add to the set S . This holds for any path P , which completes the correctness proof.

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