Graph Exploration Methods: DFS vs. BFS, Strong Components, Acyclic Graphs

1. Graph -definition V : vertex set（頂点集合） A⊆V×V: arc set（枝集合） vertex （頂点） arc（枝） G=(V, A) discrete structure（離散構造）

2. 3.グラフ探索 to examine the vertices and arcs of a graph, systematically. • depth first search(DFS)　深さ優先探索 • breadth first search(BFS)　 幅優先探索

3. DFS（深さ優先探索） from a vertex v already treated, we proceed to any vertex w adjacent to v which was not yet treated and, after having examined w, continue to look for the next unlabeled vertex directly from w.

4. DFS example true v w Initialization a (= s) Av c d b e g f h

5. DFS property(1) Each arc in the connected component of s is used exactly once in each direction during DFS a (= s) c d b e g f h

6. DFS property(2) A’={(p(v), v) | v ∈V} When G is connected, T=(V, A’) is a spanning directed tree with root s. --- Depth first search tree a (= s) c d b e g f h

7. BFS（幅優先探索） Process all the neighbors of the starting vertex. Then process all the neighbors of neighbors of the starting vertex. And so on.

8. BFS example Q= [ ] a b c d e f g h v w Av a (= s) c d b e g f h

9. BFS property (1) Each arc in the connected component of s is used exactly once in each direction during BFS a (= s) c d b e g f h

10. BFS property (2) A’={(p(v), v) | v ∈V} When G is connected, T=(V, A’) is a spanning directed tree with root s. ---- Breadth first search tree a (= s) c d b e g f h

11. BFS property (3) If v and v’ are two vertices in the same connected component, there is a path of shortest length l between v and v’. Then v and v’ are said to have distance l = d(v, v’) At the end of BFS, d(s, t) = d(v), if d(v) is defined. a (= s) c d b e g f h

12. For digraph We can apply DFS and BFS to digraphs. (DFS) Av a (= s) c d b e g f h

13. Finding strongly connected components We can determine strongly connected components in a digraph G by executing the (modified) DFS both for G and for the directed graph having opposite orientation.

14. DFS+ nr(v) gives the order in which the vertex v is reached Nr(v) gives the order in which the examination of the vertex v is finished input Graph G and a vertex s begin set nr(v)←0, p(v)←0, Nr(v)←0 for all v∈V; set u(a)←false for all a∈A; set i ←1, v←s, nr(s)←1, j←0; repeat while there exists w∈Av with u(vw)=false do begin choose some w∈Av with u(vw)=false ; set u(vw)←true; ifnr(w) = 0then set p(w)←v, i←i+1, nr(w)←i, v←w; end set j←j+1, Nr(v)← j,v←p(v); untilv=s and u(sw)=true for all w∈As; end.

15. DFS+ example 6 5 2 3 4 1

16. DFS+ property in the same tree, but v is not descendant nor ancestor of x and x is accessible from v in different DFS trees v is descendant of x v x x x v v If Nr(v) <Nr(x) for two vertices v, x v is accessible from x

17. Algorithm for strongly connected component SCC input digraph G begin apply DFS+ to G and obtain Nr; construct the digraph H by reversing the orientation of all arcs of G; set k←0; repeat choose the vertex r in H for which Nr(r) is maximal; apply DFS+ to H where the starting vertex s=r. set k←k+1, Ck←{v ∈V | nr(v) ≠0}; set H←H\Ck; until the vertex set of H is empty; end. C1, C2, …,Ck are vertex sets of strongly connected components

18. SCC example 6 3 2 5 1 4

19. SCC correctness two vertices v and w are in the same set Ci two vertices v and w are in the same strong component ⇔ (⇒) v and w are in the same strong component ⇒ there are directed paths from v to w and from w to v in G ⇒ there are directed paths from v to w and from w to v in H ⇒ w.l.o.g. v is reached before w in DFS on H, v∈Ci , then w∈Ci

20. SCC correctness two vertices v and w are in the same set Ci two vertices v and w are in the same strong component ⇔ (⇒) v and w are in the same set Ci ⇒ v and w are in the same DFS tree Ti Let x be the root of Ti ⇒ Nr(x)>Nr(v) and v is accessible from x Nr(x)>Nr(w) and w is accessible from x ⇒ x is accessible from v x is accessible from w x v w

21. Acyclic（非閉路）digraph not contain a directed cycle

22. Acyclic graph example 1 2 4 3 5 topological order :o 7 6 for any arc(v, w), o(v) < o(w) 8 socks undershorts shoes shirt pants tie belts jacket

23. Topological sort Let G be a directed acyclic graph. Then G contains at least one vertex with din (v) = 0, where din (v) denotes the number of arcs having head v. 4 3 5 1 8 6 2 7 9