 2004 SDU Lectrue4-Properties of DFS Properties of DFS Classification of edges Topological sort

 2004 SDU 2 Properties of DFS Definition: In a rooted forest (especially, depth-first forest), any vertex u on the simple path from the root to v is called an ancestor of v, and v is called a descendant of u. Note: u =  [v] if and only if DFS-VISIT (v) was called during a search of u’s adjacency list. u is called the parent of v. So: Vertex v is a descendant of vertex u in the depth-first forest if and only if v is discovered during the time in which u is gray (when v is discovered, DFS-VISIT(  [v]) is called, DFS-VISIT(   [v]), …DFS- VISIT(u)) is called).

 2004 SDU 3 Properties of DFS (Cont.) Parenthesis Theorem: In any DFS of a graph (directed or undirected), for each pair of vertices u, v, exactly one of the following conditions holds: 1. u is a descendant of v, and [d[u], f[u]] is a subinterval of [d[v], f[v]]. 2. u is an ancestor of v, and [d[v], f[v]] is a subinterval of [d[u], f[u]]. 3.Neither u is a descendant of v nor v is a descendant of u, and [d[u], f[u]] and [d[v], f[v]] are disjoint. Page:166

 2004 SDU 4 Proof It is obvious that each pair u, v satisfies exactly one of the following: 1.u is a descendant of v 2.v is a descendant of u 3.neither is descendant of the other It is enough to prove that 1 holds if and only if [d[u], f[u]] is a subinterval of [d[v], f[v]]; 2 holds if and only if [d[v], f[v]] is a subinterval of [d[u], f[u]]; 3 holds if and only if [d[u], f[u]] and [d[v], f[v]] are disjoint.

 2004 SDU 5 Proof (Continued) If 1 holds, then u must be discovered when v is gray, that means d[v]  d[u]<f[u]  f[v]. In the opposite, if d[v]  d[u]<f[u]  f[v], then u is discovered when v is gray, that is, u is a descendant of v. Then second case is similar to the above. If neither is descendant of the other. Without loss of generality, suppose d[u]<d[v], then, since v is not a descendant of u, v must be discovered after u finishes, that is, f[u]<d[v], hence, [d[u], f[u]] is disjoint with [d[v],f[v]]. In the opposite, if [d[u],f[u]] and [d[v],f[v]] are disjoint, then neither is discovered when the other is gray, so, neither is descendant of the other.

 2004 SDU 6 Nesting of descendants’ intervals Corollary: Vertex v is a proper descendant of vertex u in the depth-first forest for a graph G if and only if d[u]< d[v] < f[v] < f[u].

 2004 SDU 7 White-path theorem Theorem:  In a depth-first forest of a graph G = (V, E), vertex v is a descendant of u if and only if at time d[u], vertex v can be reached from u along a path consisting entirely of white vertices. Note! 1: All vertices along the path are white 2: At this time 3: Does BFS satisfy this theorem?

 2004 SDU 8 Proof of White-path theorem =>Assume v is a descendant of u, there is a unique path from u to v in the DFS forest. Suppose w is an arbitrary vertex on the path from u to v in the depth-first forest, then w is a descendant of u. By the nesting of descendants’ interval corollary, d[u]<d[w], and so at time d[u], w is white.

 2004 SDU 9 Proof of White-path theorem <=(By contradiction) Suppose at time d[u], there is a path from u to v consisting only of white vertices, but v does not become a descendant of u in the depth-first tree. Without loss of generality, assume that every other vertex along the path becomes a descendant of u (otherwise, let v be the closest vertex to u along the path that does not become a descendant of u). Let w be the predecessor of v in the path. Hence, w is a descendant of u, and by parenthesis theorem, d[u]  d[w]<f[w]  f[u]. Since v is not descendent of u, it can not be a son of w. Hence, when scanning v in Adj[w], v cannot be white, which means d[v]<f[w]  f[u]. By parenthesis theorem, [d[v], f[v]] must be disjoint with [d[u], f[u]], so, d[v]<d[u], which contradicts to that at time d[u], v is white.

 2004 SDU 10 Another Proof of<= (by induction) <=Suppose the white path is u, v 1, v 2, …, v k = v, we will prove that each v i will become a descendent of u. Base: v 1 will become a descendent of u. Because v 1  Adj[u]. When v 1 is scanned in Adj[u], if v 1 is white, then  [v 1 ]  u; otherwise, v 1 must be discovered when u is gray. In either case, v 1 is a descendent of u. Induction: suppose v 1, …, v i-1 will all become descendents of u. Consider v i. When v i is scanned in Adj[v i-1 ], if v i is white, then  [v i ]  v i-1 ; otherwise, v i must be discovered when u is gray, since v i is white at d[u], and before f[v i-1 ], thus before f[u], it becomes non-white. In either case, v i is a descendent of u.

 2004 SDU 11 Exercises Page , Page , , Page

 2004 SDU Applications of DFS 1.Classifying edges of a graph (for directed or undirected graph) 2.Topological sort (for directed acyclic graph) 3.Strongly connected components decomposing (for directed graph)

 2004 SDU 13 Classification of Edges Let G  = (V, E  ) be the depth-first forest produced by a depth first search on directed graph G = (V, E), then the edges of G can be classified into four categories (the non-tree edges are classified according to the relationship between the end points): 1.Tree edges: Edges in G  。 2.Back edges: Edges (u, v) connecting a vertex u to an ancestor v. Self- loops in directed graphs are considered to be back edges. 3.Forward edges: Those non-tree edges (u, v) connecting a vertex u to a descendant v. 4.Cross edges: All other edges. They can go between vertices in the same depth-first tree as long as neither vertex is ancestor of the other, or they go between vertices in different trees. Recalling that for each pair of vertices u, v, there are only three probabilities between them: u is descendant of v, opposite, neither.

 2004 SDU 14 Classification of Edges (Example) 1/8 a b f e cg d 2/7 3/6 4/5 9/14 10/11 12/13 T T T F B T T C C T: tree edge F: forward edge B: back edge C: cross edge

 2004 SDU 15 Observation In a depth-first search, an edge (u, v) can be classified according to the color of v when (u, v) is explored:  WHITE indicates a tree edge  GRAY indicates a back edge  BLACK indicates a forward or cross edge How to distinguish a forward edge and a cross edge? −d[u] < d[v] means a a forward edge −d[u] > d[v] means a cross edge −We can also classify the edges after the DFS finishes according to d[], f[],  [] values. How?

 2004 SDU 16 Classifying Edges in an Undirected Graph In an undirected graph, (u, v) and (v, u) are in fact the same edge. To avoid ambiguity, we classify the edge according to whichever of (u, v) or (v, u) is encountered first during the execution of the DFS algorithm. In a depth-first search of an undirected graph G, every edge of G is either a tree edge or a back edge. Proof: Let (u, v) be an arbitrary edge of G, and W.L.O.G, suppose d[u] <d[v]. According to the white-path theorem (uv is a white path at time d[u]), v will become a descendant of u. When we scan u’s adjacency list in DFS-VISIT(u), v will be found since (u, v) is an edge of G. –v is WHITE, which indicates (u, v) is explored first in the direction from u to v, (u, v) becomes a tree edge. –v is not WHITE, then DFS-VISIT(v) has been called before the scan comes to v when scanning u’s adjacency list in DFS-VISIT(u). Note that u is also in v’s adjacency list, so (u, v) must be first explored in the direction from v to u. Since u is an ancestor of v, (v, u) becomes a back edge.

 2004 SDU 17 Singly Connected Graph (Page ) A directed graph G = (V, E) is singly connected if u  v (v is reachable from u) implies that there is at most one simple path from u to v for all vertices u, v  V. How to determine whether or not a directed graph is singly connected?

 2004 SDU 18 A directed graph is not singly connected  There exists a vertex u such that during the call of DFS-VISIT(u), a forward edge or cross edge will be found. <=If a forward edge xy is found during the call of DFS-VISIT(u), then there are two paths from x to y: one is xy, the other is the path from x to y in the corresponding depth-first tree. If there is a cross edge xy, then there are two paths from s to y: one is the path from s to y in the depth-first tree, the other is the path from s to x plus (x, y). =>If G is not singly connected, then there exist two paths from some vertex u to some vertex v. Suppose one path is P 1 =, the other path is P 2 =. That is, u, u 2, …, u k are the first common vertices of P 1 and P 2 ; x 1 and y 1 are the first different vertices of P 1 and P 2 ; w is the first common vertex of P 1 and P 2 after the different vertices x 1,…,x l, and y 1, …, y m. Notice that w may be v, and one of the sets {x 1, …, x l } and {y 1, …, y m } may be empty. uu2u2 ukuk x1x1 x2x2 y1y1 y2y2 wv

 2004 SDU 19 Consider the call of DFS-VISIT(u). Notice that all the vertices in the two paths will be descendents of u according to white-path theorem. When scanning u’s adjacency list we will find u 2. u 2 cannot be gray. If u 2 is black, then this is a forward edge, the conclusion is drawn. Otherwise, u 2 is white, and (u, u 2 ) is a tree edge. When scanning u 2 ’ adjacency list we will find u 3. u 3 cannot be gray, since otherwise u 3 is an ancestor of u 2 and therefore an ancestor of u since u is the parent of u 2. If u 3 is black, then (u 2, u 3 ) is a forward edge or cross edge, the proof is end. If u 3 is white, (u 2, u 3 ) is a tree edge. Proceeding in this way, we can prove that either a forward edge or cross edge is met along and along, or all the edges in these two sub-paths of P 1 and P 2 are tree edges. In the former case, the proof is end. In the latter case, w will be proved to have two predecessors, which is a dilemma. Therefore, only the former case is possible, and the proof is end.

 2004 SDU 20 The Answer Consider all vertices that is reachable from x. A depth- first search on G starting only from x will generate a tree T rooted at x, each vertex of the tree is reachable from x in the graph G. Other vertices that are not in the tree are not reachable from x in the graph G. If an edge (u, v)  G, where u, v  T, (u, v)  T, then (u, v) is  A back edge, or  A forward edge (indicates not singly connected) or  A cross edge (indicates not singly connected) Make each vertex v as the starting vertex, do depth-first search. How to modify the DFS-Algorithm? And the time complexity?

 2004 SDU 21 The Modification

 2004 SDU 22 Modification (continued) DFS(G) 1time  0; 2for (each vertex u  V[G]) 2 do{for (each vertex v  V[G]) 3 do{ color[v]  WHITE; 4  [v]  NIL; 5 } 6 DFS-VISIT(u) 7 } 8Print “G is singly connected”

 2004 SDU 23 The Modification (Continued) 8 else if color[v] = BLACK 9 then {Print “G is not singly connected”; 10 Return } //algorithm ended Add Remove

 2004 SDU 24 Topological Sort A topological sort of a directed acyclic graph G = (V, E) is a linear ordering of all its vertices such that if G contains an edge (u, v), then u appears before v in the ordering How about if G contains a cycle?

 2004 SDU 25 Topological Sort Problem The problem  Input: A directed acyclic graph G = (V, E).  Output: A topological order of all the vertices of G.

 2004 SDU 26 Applications of Topological Sort Assembly lines in industries Courses arrangement in schools A general life-related application—dressing order Data Structure Java Language Advanced Mathematics Graph Theory Graph Algorithms Can you give the topological order?

 2004 SDU 27 Application--Dressing Up

 2004 SDU 28 The Algorithm Time Complexity Analysis: Line 1:  (V + E) Line 2:  (V) Line 3:  (1)

 2004 SDU 29 Correctness of the Algorithm Lemma  A directed graph G is acyclic if and only if a depth-first search of G yields no back edges.  Proof: =>: Proof by contradiction. Suppose that there is a back edge (u, v). Then, vertex v is an ancestor of vertex u in the depth-first search forest. There is thus a path from v to u in G, and the back edge (u, v) completes a cycle, a contradiction. <=: Proof by contradiction. Suppose that G contains a cycle c. Let v be the first vertex to be discovered in c, (u, v) is an edge in c. At time d[v], the vertices of c form a path of white vertices from v to u. By the white-path theorem, vertex u becomes a descendant of v in the depth-first forest. Therefore, (u, v) is a back edge, a contradiction.

 2004 SDU 30 Correctness of the Algorithm Theorem 22.12: TOPOLOGICAL-SORT(G) produces a topological sort of a directed acyclic graph G. Proof: Suppose that DFS is run on a given dag G = (V, E). It suffices to show that if (u, v) is an edge in G, then f[u] > f[v]. By lemma 22.11, (u, v) cannot be a back edge, therefore, there are only three cases: 1.(u, v) is a tree edge, u is ancestor of v 2.(u, v) is a forward edge, u is ancestor of v 3.(u, v) is a cross edge, when u is still being processed, v is black  In each case, f[u]>f[v].

 2004 SDU 31 Another Method for Topological Sort See Textbook, page 175, Exercise Repeat: 1.Select a vertex v of zero in-degree, append it to the end of a list 2.Delete v and all edges leaving it from G Return the list.

 2004 SDU 32 Correctness First we prove that a list of all vertices can be constructed in this way, i.e., at each time a zero-in-degree vertex must exist. This is easy to prove by noticing that otherwise a cycle will be found by tracing along the entering edges. Second, we prove that such a list is a topological list. By induction on V. 1.V=2, at most only one edge, trivial. 2.Suppose it is true for V<n. When V=n. Suppose the algorithm outputs a list L. In the construction of L, algorithm will first select a vertex S of in- degree zero and put it at the most left of L. Let G’ be the graph obtained from G by deleting S and the edges leaving S. Obviously, G’ is still a DAG and V’<n. By induction hypothesis, the list L’ constructed by the algorithm for G’ is a topological list of G’, i.e., all edges in G’ point from left to right in the list L’. Notice that L’ is exactly obtained from L by deleting S. In L, if an edge of G is in G’, then both of its end vertices are in L’, and thus points from left to right, otherwise, it must points from S to right.

 2004 SDU 33 Time complexity of the second topological sort algorithm. If we compute the in-degrees of all the vertices beforehand, and keep all the vertices of zero-in- degree into a list L, then we use constant time for each run of step1, and Adj[v] time for each run of step2. For executing step2, we just need to delete vertex v from the adjacency lists, and for each vertex x in Adj[v], we decline its in-degree by one, and insert it into L if its in-degree changes to zero. Hence, the total time is  (V+E) since v will be any vertex in the graph.

 2004 SDU 34 Exercises Page 175: