CSC 505, Fall 2000: Week 8

Transcription

1 Objecties: CSC 505, Fall 2000: Week 8 learn abot the basic depth-first search algorithm learn how properties of a graph can be inferred from the strctre of a DFS tree learn abot one nontriial application of DFS to obtain a linear-time graph algorithm Page 1 of 13

2 Depth-first search: a simple example Page 2 of 13

3 DFS: the algorithm procedre DFS(G) is isits all ertices of the graph for each ertex V [G] do color[] white; π[] nil time 0 for each ertex V [G] do if color[] = white then DFS-Visit() end DFS procedre DFS-Visit() is color[] gray; d[] time time + 1 Insert any beginning of isit code here. for each ertex adj[] do Explore edge Insert code to process edge here if color[] = white then π[] ; DFS-Visit() color[] black; f [] time time + 1 Insert any end of isit code here. end DFS-Visit Page 3 of 13

4 Edge classification by DFS Dring DFS-Visit( ), is white gray black d[] < d[] < f [] < f [] tree edge d[] < d[] < f [] < f [] back edge (edge to an ancestor in the tree) d[] < f [] < f [] cross edge if d[] < d[] forward edge if d[] > d[] In general, is an ancestor of iff d[] < d[] < f [] < f []; is nrelated to if d[] < f [] < d[] < f [] or ice-ersa. Page 4 of 13

5 Cycles and back edges Lemma 8.1. A graph G has a cycle if and only if eery DFS of G has a back edge. Proof. (back edge => cycle) In order for to be a back edge, there mst be a path of tree edges from to (since is an ancestor of ). (cycle => back edge) Let f be the finish times of an arbitrary DFS and let C be a cycle of G. Let be the ertex with minimm f in C and let be the ertex after in C. Since f [] < f [], mst be a back edge (all other kinds hae f [] > f []). Page 5 of 13

6 Topological sort A directed acyclic graph (dag) is a directed graph with no cycles. A topological sort of a dag is a permtation of the ertices in which eery edge of the dag goes from an earlier ertex to a later ertex (a nmbering for which eery edge goes from a lower to a higher nmber). Obseration: Any permtation deried from a DFS sorting ertices by decreasing finish time is a topological sort. The conerse is also tre: Yo can always access the ertices in the oter loop of DFS in reerse order of any topological sort and get that same permtation back. Page 6 of 13

7 Strong components Two ertices and are in the same strong component of a directed graph if there is a path from to and a path from to. A E C A directed graph whose ertices are all in the same strong component is strongly connected. B D F G X W Y Z The dag of strong components. Page 7 of 13

8 Some qestions 1. Can a DFS forest of a strongly connected graph hae more than one tree? 2. If there is only one tree in a DFS forest, is the graph strongly connected? 3. If two ertices are in the same strong component, are they in the same tree of any DFS? Page 8 of 13

9 Forefathers The forefather of ertex with respect to a DFS tree, denoted ff() is a ertex sch that (a) there is a path from to, and (b) f[] is maximm among all that are reachable from by a path. B G A C D E F Page 9 of 13

10 Forefather => ancestor Lemma 8.2. If = ff() then is an ancestor of in the DFS tree. Proof. If = ff() then f [] f [] (since has a path to itself). There are three possibilities. x X X different trees different branches ancestor Page 10 of 13

11 Same forefather <=> same strong component Theorem 8.3. Vertices x and y hae the same forefather in a DFS tree iff they are in the same strong component. Proof. Part 1. ff(x) =ff(y) = x and y are in the same strong component. Sppose ff(x) =ff(y) =. Then x and y by definition of ff. Also, x and y by Lemma 8.2. Ths x y and y x as desired. Part 2. x and y in the same strong component = ff(x) =ff(y). Let x = ff(x) and y = ff(y). If x y assme f [ x ] < f [ y ]. Bt x y and y y, meaning x y, which contradicts the fact that x = ff(x). Page 11 of 13

12 How to compte forefathers Consider the ertex with latest finish time (largest f[]) after a DFS. Q: Who can reach? A: Exactly the ertices whose forefather is? Q: How do we find these ertices? A: Do a DFS-Visit from on G R, the graph with all edges reersed. After eliminating these ertices, we hae redced to an instance hae one fewer strong component (and at least one less ertex). Repeat the aboe process starting with the remaining ertex of latest finishing time. The algorithm takes Θ(m + n), as does any algorithm based on a constant nmber of applications of DFS. Page 12 of 13

13 The algorithm on an example B G A C D E F Page 13 of 13