2004 SDU

Lectrue4-Properties of DFS

Properties of DFSClassification of edgesTopological 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

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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.

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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.

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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].

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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?

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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.

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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.

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Another Proof of<=(by induction)

<=Suppose the white path is u, v1, v2, …, vk= v, we will prove that each vi will become a descendent of u.

Base: v1will become a descendent of u.

Because v1Adj[u]. When v1 is scanned in Adj[u], if v1 is white, then [v1]u; otherwise, v1 must be discovered when u is gray. In either case, v1 is a descendent of u.

Induction: suppose v1, …, vi-1 will all become descendents of u.

Consider vi. When vi is scanned in Adj[vi-1], if vi is white, then [vi]vi-1; otherwise, vi must be discovered when u is gray, since vi is white at d[u], and before f[vi-1], thus before f[u], it becomes non-white. In either case, vi is a descendent of u.

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Exercises

Page 162 22.2-5, 22.2-7

Page 171 22.3-7, 22.3-8, 22.3-10

Page 175 22.4-3

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)

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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.

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Classification of Edges (Example)

1/8a

b

f

e c g

d2/7

3/6

4/5

9/14

10/11 12/13

T

T

TF

B T T

C

C

T: tree edge F: forward edge

B: back edge C: cross edge

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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?

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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.

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Singly Connected Graph(Page 172 22.3-12)

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?

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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 P1 = <u = u1,u2,…,uk, x1,x2,…xl, w,…,v>, the other path is P2 = <u = u1, u2, …,uk, y1,y2,…,ym,w,…,v>. That is, u, u2, …, uk are the first common vertices of P1 and P2; x1 and y1 are the first different vertices of P1 and P2; w is the first common vertex of P1 and P2 after the different vertices x1,…,xl, and y1, …, ym. Notice that w may be v, and one of the sets {x1, …, xl} and {y1, …, ym} may be empty.

u u2 uk

x1 x2

y1y2

w v

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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 u2. u2 cannot be gray. If u2 is black, then this is a forward edge, the conclusion is drawn. Otherwise, u2 is white, and (u, u2) is a tree edge.

When scanning u2’ adjacency list we will find u3. u3 cannot be gray, since otherwise u3 is an ancestor of u2 and therefore an ancestor of u since u is the parent of u2. If u3 is black, then (u2, u3) is a forward edge or cross edge, the proof is end. If u3 is white, (u2, u3) is a tree edge.Proceeding in this way, we can prove that either a forward edge or cross edge is met along <u, u2, …, uk, x1, …, xl, w> and along <u, u2, ..., uk, y1, …, ym, w>, or all the edges in these two sub-paths of P1 and P2 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.

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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?

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The Modification

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Modification (continued)DFS(G)

1 time0;

2 for (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 }

8 Print “G is singly connected”

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The Modification (Continued)

8 else if color[v] = BLACK

9 then {Print “G is not singly connected”;10 Return } //algorithm ended

1112

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.

1

3

2

21 3 4

4

How about if G contains a cycle?

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Topological Sort Problem

The problem Input: A directed acyclic graph G = (V, E). Output: A topological order of all the vertices of G.

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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?

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Application--Dressing Up

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The Algorithm

• Time Complexity Analysis:

• Line 1: (V + E)

• Line 2: (V)

• Line 3: (1)

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Correctness of the Algorithm

Lemma 22.11 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.

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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 v2. (u, v) is a forward edge, u is ancestor of v3. (u, v) is a cross edge, when u is still being processed, v is

black

In each case, f[u]>f[v].

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Another Method for Topological Sort

See Textbook, page 175, Exercise 22.4-5

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.

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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.

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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: 22.4-2