Applications of BDF and DFS

Size: px

Start display at page:

Download "Applications of BDF and DFS"

Stuart Andrews
5 years ago
Views:

1 January 14, 2016

2 1 Deciding whether a graph is bipartite using BFS. 2 Finding connected components of a graph (both BFS and DFS) works. 3 Deciding whether a digraph is strongly connected. 4 Finding cut points in a graph. 5 Finding a cycle with the shortest length in a digraph. 6 Finding whether a digraph is acyclic and if so finding a topological ordering 7 Deciding whether a digraph is balanced 8 Coloring a digraph using a directed cycle of length 3, 4,.. (similar to 3-coloring)

3 Acyclic Digraphs and Topological Ordering A digraph D is acyclic if it does not contain any directed cycle. D is called DAG (directed acyclic graph).

4 Acyclic Digraphs and Topological Ordering A digraph D is acyclic if it does not contain any directed cycle. D is called DAG (directed acyclic graph). DAG can be used to model the job scheduling with precedence constraint. Suppose we want to schedule a set of jobs {J 1, J 2,..., J n } where there are some dependencies between them (precedence constraint). For certain pair i, j, job J i must be executed before job J j. We want to find an ordering of the jobs respecting the precedence constraints.

5 Topological ordering Let D be a digraph. We say an ordering v 1, v 2,..., v n of the nodes in D is a topological ordering if whenever v i v j is an arc of D, i < j. In other words, all the arcs are forward and there is no backward arc.

6 Topological ordering Let D be a digraph. We say an ordering v 1, v 2,..., v n of the nodes in D is a topological ordering if whenever v i v j is an arc of D, i < j. In other words, all the arcs are forward and there is no backward arc. Lemma Let D be an acyclic digraph. Then D has a node without in-degree.

7 Topological ordering Theorem Let D be a digraph. D has a topological ordering if and only if D is acyclic. Proof. If D contains a directed cycle C then in every ordering of the vertices of C at least one arc is backward. Conversely if D is acyclic we show that there is a topological ordering. We follow AC-Order algorithm. AC-Order(D) 1.If D is empty then return. 2.Else Let v be a node without in neighbor in D. 3. Printout v. 4. Call AC-Order(D v)

8 Topological ordering algorithm using queue AC-Order(D) 1. Initial queue Q to be empty. 2. For every node v set the Indegre[v] to be the number of nodes having arc to v. 3. For every vertex v, If (Indegree[v] = 0 ) { Q.add(v); } 4. While Q is not empty 5. u = Q.delete(); 6. Printout(u); 7. For every arc uw A(D) 8. Indegree[w] = Indegree[w] 1; 9. If (Indeegree[w] = 0) 10 Q.add(w);

9 Topological ordering algorithm using queue AC-Order(D) 1. Initial queue Q to be empty. 2. For every node v set the Indegre[v] to be the number of nodes having arc to v. 3. For every vertex v, If (Indegree[v] = 0 ) { Q.add(v); } 4. While Q is not empty 5. u = Q.delete(); 6. Printout(u); 7. For every arc uw A(D) 8. Indegree[w] = Indegree[w] 1; 9. If (Indeegree[w] = 0) 10 Q.add(w); The AC-Order(D) Algorithms runs in time O( D + E ).

10 Balanced Digraphs Let C be a cycle (induced). We oriented each edge of C (give direction) and we obtain an oriented cycle C We say C is balanced if the number of forward arcs and the number of backward arcs in C are the same (in a clockwise direction). We say digraph D is balanced if each of its oriented cycle is balanced. Alternatively, digraph D is balanced if there is a partitioning V 0, V 1,..., V k 1 of its vertices such that each arc of D is from a vertex in some V i to a vertex in V i+1. Design an algorithm that decides whether a given digraph D is balanced or not.

11 Balanced Checking Balanced (v, level) 1. set l(v) = level; 2. for every unvisited arc vu 3. set visit vu = true. 4. if (l(u) exists && l(u) l(v) + 1) 5. print-out Not balanced 6. exit(1); 7. else 8. Balanced(u, level+1); 9. for every unvisited arc uv 10 set visit uv = true. 11. if (l(u) exists && l(u) l(v) 1) 12. print-out Not balanced 13. exit(1); 14. else 15. Balanced(u, level-1);

12 Directed cycle coloring Let C 3 be a directed cycle on three vertices. V ( C 3 ) = {0, 1, 2} and arcs set A( C 3 ) = {01, 12, 20} We say a digraph D can be colored by C 3 if for every arc uv D whenever u is colored by i (i=0,1,2) then v is colored by i + 1 (sum is mod 3).

13 Directed cycle coloring Let C 3 be a directed cycle on three vertices. V ( C 3 ) = {0, 1, 2} and arcs set A( C 3 ) = {01, 12, 20} We say a digraph D can be colored by C 3 if for every arc uv D whenever u is colored by i (i=0,1,2) then v is colored by i + 1 (sum is mod 3). Design an algorithm that decides whether a given digraph D has a C 3 coloring.

14 Directed cycle coloring Let C 3 be a directed cycle on three vertices. V ( C 3 ) = {0, 1, 2} and arcs set A( C 3 ) = {01, 12, 20} We say a digraph D can be colored by C 3 if for every arc uv D whenever u is colored by i (i=0,1,2) then v is colored by i + 1 (sum is mod 3). Design an algorithm that decides whether a given digraph D has a C 3 coloring. Design an algorithm that decides whether a given digraph D has a C k coloring.

15 Strong Digraphs A digraph D is called strong, if for every two vertices u, v of D there is a directed path from u to v and there is a directed path from v to u. A directed cycle is a strong digraph.

16 Strong Digraphs A digraph D is called strong, if for every two vertices u, v of D there is a directed path from u to v and there is a directed path from v to u. A directed cycle is a strong digraph. A directed cycle is a strong digraph.

17 Strong Digraphs A digraph D is called strong, if for every two vertices u, v of D there is a directed path from u to v and there is a directed path from v to u. A directed cycle is a strong digraph. A directed cycle is a strong digraph. A strong component of D is a maximal subset U of D which is strong.

18 Strong Digraphs A digraph D is called strong, if for every two vertices u, v of D there is a directed path from u to v and there is a directed path from v to u. A directed cycle is a strong digraph. A directed cycle is a strong digraph. A strong component of D is a maximal subset U of D which is strong. Strong components are : S 1 = {a, b, c, j}, S 2 = {d, e}, and S 3 = {f, g, h, i}.

19 Strong Components of a digraph Finding strong components in a digraph. Input: digraph G = (V, E) Output: set of strongly connected components (sets of vertices)

20 Strong Components of a digraph Finding strong components in a digraph. Input: digraph G = (V, E) Output: set of strongly connected components (sets of vertices) Explaining Tarjan s algorithm : 1) The nodes are placed on a stack in the order in which they are visited.

21 Strong Components of a digraph Finding strong components in a digraph. Input: digraph G = (V, E) Output: set of strongly connected components (sets of vertices) Explaining Tarjan s algorithm : 1) The nodes are placed on a stack in the order in which they are visited. 2) When the depth-first search recursively explores a node v and its descendants, we may not popped them from the stack. Because there maybe a node v descendant of u which may have a path to a node earlier on the stack.

22 3) Each node v is assigned a unique integer index(v), The time when v is visited for the first time.

23 3) Each node v is assigned a unique integer index(v), The time when v is visited for the first time. 4) We maintain a value lowlink(v) that represents (roughly speaking) the smallest index of any node known to be reachable from v, including v itself.

24 3) Each node v is assigned a unique integer index(v), The time when v is visited for the first time. 4) We maintain a value lowlink(v) that represents (roughly speaking) the smallest index of any node known to be reachable from v, including v itself. 5) Therefore v must be left on the stack if lowlink(v) < index(v)

25 6) v must be removed as the root of a strongly connected component if lowlink(v) = index(v).

26 6) v must be removed as the root of a strongly connected component if lowlink(v) = index(v). 7) The value lowlink(v) is computed during the depth-first search from v.

27 Strong Components-Tarjan s Algorithm(D) function strongconnect(v) 1. index(v) := index; lowlink(v) := index; 2. index:= index + 1; Stack.push(v) 3. for each arc vw E do 6. if (index(w) is undefined) 7. strongconnect(w) 8. lowlink(v) :=min(lowlink(v), lowlink(w)) 9. else if (w is in Stack) 10. lowlink(v) := min(lowlink(v), index(w)) // If v is a root node, pop the stack for new strong component 11. if (lowlink(v) = index(v)) 12. repeat 13. w := Stack.pop() add w to current strong component 14. until (w = v) 15. output the current strongly component

28 Strong Components-Tarjan s Algorithm(D) 1. index := 0 2. Stack := empty 3. for each v in V do 4. if (index(v) is undefined) 5. strongconnect(v)

29 Biconnected Components Let G = (V, E) be a loop-free connected undirected graph. A vertex v in G is called an articulation point if κ(g v) > κ(g). G v has more connected components than v. A loop-free connected undirected graph with no articulation points is called biconnected.

30 Finding Articulation Points 1) We traverse the graph in DFS (preorder) manner. 2) For vertex x of G define dfi(x) to be the index of x in DFS (time we visit x). If y is a descendant of x then dfi(x) < dfi(y).

31 Finding Articulation Points 1) We traverse the graph in DFS (preorder) manner. 2) For vertex x of G define dfi(x) to be the index of x in DFS (time we visit x). If y is a descendant of x then dfi(x) < dfi(y). 3) Define low(x) = min{dfi(y) y is adjacent in G to either x or a descendant of x}

32 Let z be the parent of x in T. Then : 1) low(x) = dfi(z) : In this case T x contains no vertex that is adjacent to an ancestor of z (by back edge). Hence z is an articulation point of G. If T x contains no articulation points, then T x together with edge zx is a biconnected component of G. Remove T x and the edge xz and apply on the remaining subtree of T.

33 Let z be the parent of x in T. Then : 1) low(x) = dfi(z) : In this case T x contains no vertex that is adjacent to an ancestor of z (by back edge). Hence z is an articulation point of G. If T x contains no articulation points, then T x together with edge zx is a biconnected component of G. Remove T x and the edge xz and apply on the remaining subtree of T. 2) If low(x) < dfi(z) : there is a descendant of z in T x that is joined (by a back edge in G) to an ancestor of z.

34 Algorithm for Articulation Points 1 ) Find a DFS ordering x 1, x 2,..., x n of the vertices of G. 2) Start from x n and continue to x n 1, x n 2,..., x 1 and determine low(x j ) as follows : a) low (x j ) = min{dfi(z) z is adjacent in G to x j } b) If c 1, c 2,..., c m are children of x j, then low(x j ) = min{low (x j ), low (c 1 ), low (c 2 ),..., low (c m )} 3) Let w j be the parent of x j in T. If low(x j ) = dfi(w j ) then w j is an articulation point of G, unless w j is the root of T and w j has no child in T other than x j. Moreover, in either situation the subtree rooted at x j together with the edge w j x j is part of a biconnected component of G.

35 a e d c b(2) d(1) e(4) g f b c(3) a(5) f(6) h g(7) h(8) d(2,1) d(1,1) d d b(1,1) c(1,1) e(1,1) a(4,1) f(1,1) g(6,6) c(3,1) b(2,1) e(4,1) a(5,1) f(6,1) g(7,6) c b f g f a e g h(7,7) h(8,7) h

Searching in Graphs (cut points)

Searching in Graphs (cut points) 0 November, 0 Breath First Search (BFS) in Graphs In BFS algorithm we visit the verices level by level. The BFS algorithm creates a tree with root s. Once a node v is discovered by BFS algorithm we put