applications of bdf and dfs - indiana state universitycs.indstate.edu/~arash/algo2lec2.pdf · arash...

Applications of BDF and DFS

Arash Rafiey

January 14, 2016

Arash Rafiey Applications of BDF and DFS

1 Deciding whether a graph is bipartite using BFS.

2 Finding connected components of a graph (both BFS andDFS) 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 atopological ordering

7 Deciding whether a digraph is balanced

8 Coloring a digraph using a directed cycle of length 3, 4, ..(similar to 3-coloring)

Acyclic Digraphs and Topological Ordering

A digraph D is acyclic if it does not contain any directed cycle. Dis called DAG (directed acyclic graph).

DAG can be used to model the job scheduling with precedenceconstraint.

Suppose we want to schedule a set of jobs {J1, J2, . . . , Jn} wherethere are some dependencies between them (precedenceconstraint). For certain pair i , j , job Ji must be executed before

job Jj .

We want to find an ordering of the jobs respecting the precedenceconstraints.

Acyclic Digraphs and Topological Ordering

A digraph D is acyclic if it does not contain any directed cycle. Dis called DAG (directed acyclic graph).

DAG can be used to model the job scheduling with precedenceconstraint.

Suppose we want to schedule a set of jobs {J1, J2, . . . , Jn} wherethere are some dependencies between them (precedenceconstraint). For certain pair i , j , job Ji must be executed before

job Jj .

We want to find an ordering of the jobs respecting the precedenceconstraints.

Topological ordering

Let D be a digraph. We say an ordering v1, v2, . . . , vn of the nodesin D is a topological ordering if whenever vivj is an arc of D,i < j . In other words, all the arcs are forward and there is no

backward arc.

Let D be an acyclic digraph. Then D has a node without in-degree.

Let D be a digraph. We say an ordering v1, v2, . . . , vn of the nodesin D is a topological ordering if whenever vivj is an arc of D,i < j . In other words, all the arcs are forward and there is no

backward arc.

Let D be an acyclic digraph. Then D has a node without in-degree.

Theorem

Let D be a digraph. D has a topological ordering if and only if Dis acyclic.

Proof.

If D contains a directed cycle C then in every ordering of thevertices of C at least one arc is backward. Conversely if D isacyclic we show that there is a topological ordering. We followAC-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)

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 nodeshaving 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 |).

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 nodeshaving 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 |).Arash Rafiey Applications of BDF and DFS

Balanced Digraphs

Let C be a cycle (induced). We oriented each edge of C (givedirection) and we obtain an oriented cycle C ′

We say C ′ is balanced if the number of forward arcs and thenumber of backward arcs in C ′ are the same (in a clockwisedirection).

We say digraph D is balanced if each of its oriented cycle isbalanced.

Alternatively, digraph D is balanced if there is a partitioningV0,V1, . . . ,Vk−1 of its vertices such that each arc of D is from avertex in some Vi to a vertex in Vi+1.

Design an algorithm that decides whether a given digraph D isbalanced or not.

Balanced Checking

Balanced (v, level)

1. set `(v) = level ;2. for every unvisited arc vu3. set visit vu = true.4. if (`(u) exists && `(u) 6= `(v) + 1)5. print-out “Not balanced”6. exit(1);7. else8. Balanced(u, level+1);9. for every unvisited arc uv10 set visit uv = true.11. if (`(u) exists && `(u) 6= `(v)− 1)12. print-out “Not balanced”13. exit(1);14. else15. Balanced(u, level-1);

Directed cycle coloring

Let−→C3 be a directed cycle on three vertices.

V (−→C3) = {0, 1, 2} and arcs set A(

−→C3) = {01, 12, 20}

We say a digraph D can be colored by−→C3 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−→C3 coloring.

Design an algorithm that decides whether a given digraph D has a−→Ck coloring.

V (−→C3) = {0, 1, 2} and arcs set A(

−→C3) = {01, 12, 20}

V (−→C3) = {0, 1, 2} and arcs set A(

−→C3) = {01, 12, 20}

Strong Digraphs

A digraph D is called strong, if for every two vertices u, v of Dthere is a directed path from u to v and there is a directed pathfrom 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 isstrong.

Strong components are : S1 = {a, b, c , j}, S2 = {d , e}, andS3 = {f , g , h, i}.

Strong Digraphs

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 arevisited.

2) When the depth-first search recursively explores a node v andits descendants, we may not popped them from the stack. Becausethere maybe a node v descendant of u which may have a path to anode earlier on the stack.

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

4) We maintain a value lowlink(v) that represents (roughlyspeaking) the smallest index of any node known to be reachablefrom v, including v itself.

5) Therefore v must be left on the stack if lowlink(v) < index(v)

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

7) The value lowlink(v) is computed during the depth-first searchfrom v .

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

7) The value lowlink(v) is computed during the depth-first searchfrom v .

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 component11. if (lowlink(v) = index(v))12. repeat

13. w := Stack.pop() add w to current strong component14. until (w = v)15. output the current strongly component

Strong Components-Tarjan’s Algorithm(D)

1. index := 02. Stack := empty3. for each v in V do4. if (index(v) is undefined)5. strongconnect(v)

Biconnected Components

Let G = (V ,E ) be a loop-free connected undirected graph. Avertex 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 pointsis called biconnected .

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}

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}

Let z be the parent of x in T . Then :

1) low(x) = dfi(z) : In this case Tx contains no vertex that isadjacent to an ancestor of z (by back edge).Hence z is an articulation point of G .If Tx contains no articulation points, then Tx together with edgezx is a biconnected component of G .Remove Tx and the edge xz and apply on the remaining subtree ofT .

2) If low(x) < dfi(z) : there is a descendant of z in Tx that isjoined (by a back edge in G) to an ancestor of z .

Let z be the parent of x in T . Then :

1) low(x) = dfi(z) : In this case Tx contains no vertex that isadjacent to an ancestor of z (by back edge).Hence z is an articulation point of G .If Tx contains no articulation points, then Tx together with edgezx is a biconnected component of G .Remove Tx and the edge xz and apply on the remaining subtree ofT .

2) If low(x) < dfi(z) : there is a descendant of z in Tx that isjoined (by a back edge in G) to an ancestor of z .

Algorithm for Articulation Points

1 ) Find a DFS ordering x1, x2, . . . , xn of the vertices of G .

2) Start from xn and continue to xn−1, xn−2, . . . , x1 and determinelow(xj) as follows :

a) low ′(xj) = min{dfi(z)| z is adjacent in G to xj}b) If c1, c2, . . . , cm are children of xj , then

low(xj) = min{low ′(xj), low′(c1), low

′(c2), . . . , low′(cm)}

3) Let wj be the parent of xj in T . If low(xj) = dfi(wj) then wj isan articulation point of G , unless wj is the root of T and wj hasno child in T other than xj .Moreover, in either situation the subtree rooted at xj together withthe edge wjxj is part of a biconnected component of G .

b(2) e(4)

h(7,7)

g(6,6)

f(1,1)

a(4,1)

e(1,1)

d(2,1)

b(1,1)

c(1,1)

h(8,7)

g(7,6)

f(6,1)

a(5,1)

e(4,1)

c(3,1)

b(2,1)

d(1,1)dd

applications of bdf and dfs - indiana state universitycs.indstate.edu/~arash/algo2lec2.pdf · arash...

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