3510: CS Design and Analysis of Algorithms
Problem Set 5 Solutions
Problem 1
Suppose that a communication network is modeled as a directed graph G = (V, E) in which each edge (u, v) E has an associated value 0 r(u, v) 1 that represents the probability that the channel from u to v will not fail. Assume that the probabilities are independent. Give an efficient algorithm that finds the most reliable path between two given vertices of a given network. Solution: For each edge (u, v) E, assign weight w(u, v) = - log r(u, v) if r(u, v) > 0, and w(u, v) = 0 otherwise. Run Dijkstra's algorithm on the G to find a shortest path from s to t, with respect to the new weights. We now show that this algorithm is correct. Because the probablities are independent, the probability that a path s = v0 v1 vk = t does not fail is k r(vi-1 , vi ). Thus findi=1 ing a most reliable path from s to t, i.e. maximizing k r(vi-1 , vi ), is equivalent to maximizing i=1 log
k i=1 r(vi-1 , vi )
=
= For 0 < r(u, v) < 1, log r(u, v) < 0. w(u, v) 0 for each (u, v) E, thus Dijkstra's algorithm applies. Alternatively, one can modify Dijkstra's algorithm to find the most reliable path.
k i=1 (- log r(vi-1 , vi ))
k i=1 log r(vi-1 , vi ). k i=1 w(vi-1 , vi ).
The latter is equivalent to minimizing -
k i=1 log r(vi-1 , vi )
=
Therefore
Problem 2
Let G = (V, E) be a weighted directed graph without negative-weight cycles, and let d be the maximum, over all (ordered) pairs of vertices u, v V , of the minimum number of edges in a shortest path from u to v in G. Give a simple modification to the Bellman-Ford algorithm so that the algorithm terminates in min (d + 1, n - 1) passes (as opposed to n - 1 passes), where n = |V |. Solution: By the path relaxation property, if the distance from s to a vertex v is , then a shortest path from s to v can be found in iterations of Bellman-Ford. Therefore, in d iterations, to every vertex a shortest path from s is found. Note that one cannot have the algorithm terminate in d iterations, since d was not known ahead of time. However, one can slightly modify the algorithm to terminate in d + 1 iterations: In each iteration, when relaxing each edge (u, v), check if Relax(u, v) yields any change. If there is no change for all edges, then to every vertex a shortest path from s has been found, so exit loop. Therefore d + 1 iterations suffice.
Problem 3
Prove that Dijkstra's algorithm correctly computes single-source short paths in weighted directed graphs where edges leaving the source may have negative weights, but all other edge weights are nonnegative and there are no negative-weight cycles. Solution: Exactly the same proof for the case where all edge weights are nonnegative, applies to this case.
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Problem 4
Let G = (V, E) be a weighted directed graph in which the edge are weights integers between 0 and w for some constant w. Give an implementation of Dijkstra's algorithm that computes shortest paths from a given source vertex of G in O(n + m) time, where n = |V | and m = |E|. Solution: First observe that for each vertex v V , at any time if d[v] < , then d[v] (n - 1)w. In order for d[v] < , an edge (u, v), with d[u] < , must be relaxed. It follows, by induction, that d[v] is at most the length of some simple path from s to v times the maximum edge weight, which is at most (n - 1)w. Also observe that in Dijkstra's algorithm, the values d[u] of vertices return by Extract-Min are monotonically increasing over time. The only way that such a value can change is by a Decrease-Key operation. Since the edge weights are nonnegative, when a vertex u is returned by Extract-Min and an edge (u, v) relaxed, we have d[u] d[v]. Thus every vertice extracted later has a value at least d[u]. We implement a priority queue so that every sequence of n Insert operations, n Extract-Min and m Decrease-Key operations take time O(n + m + k), where k = (n - 1)w Use an array A[0, ..., k], where A[j] consists of a doubly linked list of all elements whose value is j. Also have a separate list for vertices whose values are . The operations are implemented as follows: 1. Insert. To insert an element of value j, simply insert it into the list at A[j]. Time: O(1) per Insert. 2. Extract-Min. Maintain an index min of the smallest value extracted. To find an element of the smallest value, look at A[min]. If its list is nonempty, then remove an element from the list and return it. If the list is empty, then because of the monotonicity, increment min until a nonempty list is found (in which one removes an element from the list and return it, as before), or min > k (in which case the queue is empty). We increment min at most k times, and remove and return n elements. Thus the total time for all the Extract-Min operations is O(n + k). 3. Decrease-Key. To decrease the value of an elment from j to i, simply remove the element from the list A[j] and insert it into the list A[i]. Time: O(1) per Decrease-Key. Therefore, with this implementation, Dijkstra's algorithm runs in time O(n + m + k) = O(n + m + (n - 1)w) = O(n + m) since w = O(1).
Problem 5
Let G = (V, E) be a weighted connected undirected graph where all edge weights are distinct. How many minimum spanning trees can G have? Justify your answer. (Note: I intended this to be Part (3) of Problem 3 in the last problem set (Problem Set 4), but forgot to include it.) Solution: G has a unique MST if all its edge weights are unique. By Part (2) of Problem 3 in Problem Set 4, if the lightest edge crossing every cut of G is unique, then G has a unique MST. This condition is certainly satisfied if all the edge weights in G are unique.
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