CSE 101. Algorithm Design and Analysis Miles Jones Office 4208 CSE Building Lecture 5: SCC/BFS

Transcription

1 CSE 101 Algorithm Design and Analysis Miles Jones Office 4208 CSE Building Lecture 5: SCC/BFS

2 DECOMPOSITION There is a linear time algorithm that decomposes a directed graph into its strongly connected components. If explore is performed on a vertex u, then it will visit only the vertices that are reachable by u. What vertices will be visited when explore is performed on u if it is in a sink SCC?

3 SINK SCCS If explore is performed on a vertex that is in a sink SCC, then only the vertices from that SCC will be visited. This suggests a way to look for SCCs. Start explore on a vertex in a sink and visit its SCC. Remove the sink SCC from the graph and repeat.

4 SOURCE SCCS Ideally we would like to find a vertex in a sink SCC. There is not a direct way to do this. However, there is a way to find a vertex in a source SCC. The vertex with the greatest post number in any dfs output tree belongs to a source SCC. The vertex with the least post number in a dfs output does not necessarily belong to a sink SCC.

5 EXAMPLE OF LOW POST NUMBER NOT IN A SINK. C A D B E G F

6 VERTICES IN SOURCE SCCS The vertex with the greatest post number in any dfs output tree belongs to a source SCC. To prove this, we will state a more general property: If C and C are strongly connected components and there is an edge from a vertex in C to a vertex in C then the highest post number in C is greater than the highest post number in C

7 PROOF Case 1: DFS searches C before C : Then at some point dfs will cross into C and visit every edge in C then it will retrace its steps until it gets back to the first node in C it started with and assign it the highest post number C C

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9 PROOF Case 2: DFS searches C before C : Then DFS will visit all vertices of C before getting stuck and assign a post number to all vertices of C. Then it will visit some vertex of C later and assign post numbers to those vertices. C C

10 COROLLARY The strongly connected components can be linearized by arranging them in decreasing order of their highest post numbers.

11 A B C D E H F I G J K L M

12 HOW TO FIND SINK SCCS Given a graph G, let G R be the reverse graph of G. Then the sources of G R are the sinks of G. So if we perform DFS on G R then the vertex with the highest post number is in a source. This means that this vertex will be in a sink of G. So start with this vertex and explore the SCC. Then the vertex with the next greatest post number in G R is in the next SCC in linear order so start with that one next.

13 HOW TO DECOMPOSE A GRAPH INTO ITS SCCS: Run DFS on G R and keep track of the post numbers. Run DFS on G and order the vertices in decreasing order of the postnumbers from the previous step.

14 A B C D E H F I G J K L M

15 A B C D E H F I G J K L M

16 HOW TO DECOMPOSE A GRAPH INTO ITS SCCS: Run DFS on G R and keep track of the post numbers. Run DFS on G and order the vertices in decreasing order of the postnumbers from the previous step. How long does this take? I claim it is linear time for each step and so it is linear time in general

17 THINGS TO WORK ON How to make a linear time algorithm that inputs a directed graph and outputs its reverse. How to reorder the vertices in linear time by post-numbers of dfs on the reverse.

18 DFS IS GOOD FOR

19 DFS IS GOOD FOR Find what vertices can be reached by a given vertex Divide an undirected graph into connected components find cycles in graphs (directed or undirected.) Find sinks and sources in DAGs Topologically sort a DAG Make a directed graph into a DAG of its SCCs

20 DFS IS NOT GOOD FOR

21 DFS IS NOT GOOD FOR Finding shortest distances between vertices.

22 DISTANCE Definition: In a graph G with at least two vertices u and v, the distance from u to v is the length of the shortest path from u to v. If there is no path from u to v then we say that the distance is infinite. The distance from a vertex to itself is 0.

23 BALLS WITH STRINGS. F F C A B A A B E C E B F C E D D D

24 SHORTEST PATH TREE F F C A B A A B E C E B F C E D D D

25 BREADTH FIRST SEARCH (BFS) Given a graph G and a starting vertex s, BFS computes distances from s to every other node. It keeps this information in an array dist. To do this, BFS computes distances layer by layer. It sets dist(s)=0 and the next layer is all the vertices adjacent to s. If v is adjacent to s then it sets dist(v)=1. Once it has assigned distance values 0,1,2, d. The nodes at distance d+1 are the undiscovered nodes adjacent to the d nodes.

26 STACK VERSUS QUEUE The main difference between DFS and BFS is that DFS uses a stack and BFS uses a queue. The queue gives us some extra information. The queue starts with just the node s, the only one that has distance 0. Then for each subsequent distance, there is a point when the queue contains all the nodes at distance d and nothing else. As these nodes are ejected, their undiscovered neighbors are the next nodes injected into the end of the queue.

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28 BFS procedure BFS(G, s) Input: Graph G = (V,E), (directed or undirected) and a vertex s in V. Output: For all vertices u reachable from s, dist(u) is the distance from s to u. and for all vertices u not reachable from s, dist(u) = for each vertex u in V: dist(u)= dist(s) = 0 Q = [s] (queue just containing s) while Q is not empty u = eject(q) for all edges (u,v) in E if dist(v)= then inject(q,v) dist(v)=dist(u) + 1

29 EXAMPLE C A F B E D

30 CORRECTNESS OF BFS We want to show that BFS assigns dist() correctly to all vertices reachable from s. Proof by induction on distance. Base Case: dist(s) = 0. Inductive hypothesis: For any vertex v that is distance k from s, dist(v) = k. Inductive step: Suppose u is distance k + 1 (WTS: BFS assigns dist(u) = k + 1.)

31 CORRECTNESS OF BFS (PROOF CONT.) Inductive Step: Suppose u is distance k + 1 from s. dist(u) starts at infinity There exists a vertex w such that w is distance k from s and ( w,u) is in E. In the last for loop when w is ejected, we encounter the edge ( w,u) and dist(u) is infinity so we inject u into Q and set dist(u) = dist(w) + 1 = k + 1.

32 RUNTIME Each vertex is initially set to infinity. Then each vertex reachable by s is put into Q and taken out of Q so there are V initializations and at most 2 V queue operations. In the while loop, it eventually visits all edges reachable from s (once in directed graphs and twice in undirected graphs.) so there are at most E, (2 E ) times it visits an edge. So, worst case it takes 3 V + 2 E = O V + E operations. procedure BFS(G,s) for each vertex u in V: distance(u)=i nfinity dist(s) = 0 Q = [s] (queue just containing s) while Q is not empty u = eject(q) for all edges (u,v) in E if dist(v)=infi nity then inject(q,v) dist(v)=dist(u) + 1

33 SHORTEST PATH TREE Is there a way to output the dist values and the shortest path tree? procedure BFS(G,s) for each vertex u in V: distance(u)=i nfinity dist(s) = 0 Q = [s] (queue just containing s) while Q is not empty u = eject(q) for all edges (u,v) in E if dist(v)=infi nity then inject(q,v) dist(v)=dis t(u) + 1

34 BFS A B C D E H F I G J K L M

35 DFS VS BFS DFS gives info about the whole graph BFS gives info related to a given vertex within the graph BFS uses a queue DFS uses a stack BFS does not restart at other connected components since all vertices not connected to your starting vertex are distance infinity away.