CS170 Discussion Section 4: 9/18

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1 CS170 Discussion Section 4: 9/18 1. Alien alphabet. Suppose you have a dictionary of an alien language which lists words in some sorted lexicographical ordering. For example, given the following list of words: [ baaa abcd abca cabb cadd ] Looking at just the first letter, we can see that words that start with a b come before words that start with an a, which come before words that start with a c. In this way, we can see that b < a < c. You can then notice that cabb comes before cadd in the list, so b < d. You can now see that the ordering of the alphabet is: b < d < a < c Assuming all words will be of the same constant length, give an efficient algorithm to determine the lexicographical ordering for any input list. Begin by setting up a graph with one node for each symbol in the alphabet. Then walk through the list of words linearly, from the start of the list. Consider each adjacent pair of words, say (w 1, w 2 ). Look through the characters in (w 1, w 2 ) one by one until you find the first index i in which they differ. This indicates that w 1 [i] comes before w 2 [i] in the alien language. Thus, you should then draw an edge from symbol w 1 [i] to w 2 [i], indicating that w 1 [i] is before w 2 [i] in this ordering. Then, continue onto the next pair of words. At the end of this process, you should have a DAG. After drawing in the edges, linearize the DAG and to get a topological ordering. If we define the number of words given as W, and the size of the alphabet as A, then the overall runtime is linear in W and A : Θ( W + A ). 2. True Source. Design an efficient algorithm that given a directed graph G determines whether there is a some vertex v from which every other vertex can be reached. This vertex need not be unique. Hint: first solve this for directed acyclic graphs. Note that running DFS from every single vertex is not efficient. In directed acyclic graphs, this is easy to check. We just need to see if the number of source nodes (zero indegree) is 1 or more than 1. Certainly if it is more than 1, there is no true source, because one cannot reach either source from the other. But if there is only 1, that source can reach every other vertex, because if v is any other vertex, if we keep taking one of the incoming edges, starting at v, we have to either reach the source, or see a repeat vertex. But the fact that the graph is acyclic means that we can t see a repeat vertex, so we have to reach the source. This means that the source can reach any vertex in the graph. 1

2 Now for general graphs, we first form the SCCs, and the metagraph. Now if there is only one source SCC, any vertex from it can reach any other vertex in the graph, but if there are more than one source SCCs, there is no single vertex that can reach all vertices. 2

3 3. Shortest Cycle. Give an algorithm that takes as input an undirected, unweighted graph, and returns the length of the shortest cycle in the graph (if the graph is acyclic, it should say so). Your algorithm should take time at most O( V 2 + V E ). Consider the simpler problem of finding the shortest cycle through vertex s in a graph. We can solve it with a version of BFS starting from s and modified as follows: when exploring a vertex v and checking edges, if the edge (v, w) is encountered and w has been explored, we have found a cycle. We can calculate the length of the cycle through depth(v) + depth(w) + 1. At the end of the BFS exploration for s, return the shortest cycle length found. We can be sure this returns the length of the shortest cycle containing s because all cycles from s take the form of some s v path, plus some w s path (where w can be equal to s in which case this path has length 0), plus the edge (v, w), so it considers all the cycles. We know our depth(v) and depth(w) values will be correct, so the length it computes is correct. The complete algorithm is to run this modified BFS from each vertex in the graph, and return the minimum of all the results (the shortest cycle must go through some vertex, and that vertex was one of those we checked). This runs in Θ( V ( V + E )) = Θ( V 2 + V E ) time. Note: This algorithm wouldn t work if you used DFS. Why do you think that is? Try to construct a counterexample for which DFS would return the wrong cycle length. As another extension question, think of a counterexample to show why the BFS algorithm can t be modified to work on weighted graphs. Note: adding k dummy nodes to represent a length k edge doesn t work in this case, since the runtime of your algorithm then depends on the weight of the largest edge. This violates our runtime requirement. What algorithm would you use to find shortest cycles in a weighted graph? 4. Dijkstra and negative edge weights. In general, Dijkstra s algorithm doesn t work on graphs with negative edge weights. Here is one attempt to fix it: (a) Add a large number M to every edge so that there are no negative weights left. (b) Run Dijkstra to find the shortest path in the new graph. (c) Return the path Dijkstra s algorithm found, but with the old edge weights (subtract M from the weight of each edge). Show that this algorithm doesn t work even when the graph is guaranteed not to contain a negative weight cycle by finding such a graph for which the algorithm must give the wrong answer. What is your intuition for why this algorithm does not work? The intuition here is that this scheme increases the length of a path containing more edges by a larger amount. Thus, in our counterexample, we would like to have a shortest 3

4 path that has many more edges than some other path of larger weight. Building on this, we come up with the following counterexample: A 2-3 S 1 T The shortest S T path is S A T (weight 1), but after we add M = 3 to each edge, the shortest path changes to S T (weight 4, while that of the original one becomes 5), so we end up returning the wrong path. 5. Dijsktra and negative edge weights, part 2 The previous problem notwithstanding, for certain problems in which we have strong guarantees about where negative edges can occur, we can still use Dijkstra s algorithm. Consider a directed graph in which only the vertex s can have negative edges from it, and all other edges must be nonnegative. There are guaranteed to be no negative weight cycles. (a) Argue that in this graph, standard Dijkstra s algorithm correctly finds distances from s to all other vertices. Yes, if there is no negative cycle, then Dijkstra s algorithm would work. Proof: Consider the proof of Dijkstra s algorithm. The proof depended on the fact that if we know the shortest paths for a subset S V of vertices, and if (u, v) is an edge going out of S such that v has the minimum estimate of distance from s among the vertices in V \S, then the shortest path to v consists of the (known) path to u and the edge (u, v). We can argue that this still holds even if the edges going out of the vertex s are allowed to be negative. Let (u, v) be the edge out of S as described above. For the sake of contradiction, assume that the path claimed above is not the shortest path to v. Then there must be some other path from s to v which is shorter. Since s S and v S, there must be some edge (i, j) in this path such that i S and j S. But then, the distance from s to j along this path must be greater than that the estimate of v, since v had the minimum estimate. Also, the edges on the path between j and v must all have non-negative weights since the only negative edges are the ones out of s. Hence, the distance along this path from s to v must be greater than the estimate of v, which leads to a contradiction. (b) What if we want to find distances from a different vertex v to all other vertices? Suggest an algorithm for this task. A shortest path from v to any other vertex t may or may not pass through s. To find a shortest path from v to t that does not pass through s, we delete s from the graph and run Dijkstra s algorithm from v. Let d v,t be the length of the shortest path found. For the other case we need to 4

5 Find a shortest path from v to s. This can be done by running Dijkstra s algorithm from v. Since all the negative edges leave s and there is no negative cycle, Dijkstra s algorithm will correctly find a shortest path from v to s. Let this distance be d v,s. Find a shortest path from s to t. From Part (a) we know that this can be done by running Dijkstra s algorithm from s. Let this distance be d s,t. The shortest distance is then given by d v,t = min(d v,t, d v,s + d s,t ). A path corresponding to the shortest distance can be found in a similar way as Dijkstra s algorithm. The algorithm has same asymptotic running time as Dijkstra s algorithm since it runs Dijkstra s algorithm three times and deleting the vertex s from the graph can be done in linear time. Also, we ve been discussing the distance from v to a vertex t, but the same algorithm and analysis will give the distance to all vertices in the graph. 5

(Dijkstra s Algorithm) Consider the following positively weighted undirected graph for the problems below: 8 7 b c d h g f

(Dijkstra s Algorithm) Consider the following positively weighted undirected graph for the problems below: 8 7 b c d h g f CS6: Algorithm Design and Analysis Recitation Section 8 Stanford University Week of 5 March, 08 Problem 8-. (Graph Representation) (a) Discuss the advantages and disadvantages of using an adjacency matrix