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Outline of the lecture

RefreshingDFS traversal

Refreshing the problem of articulation points

Relationship betweenarticulation points and DFS traversal

Key insight and ideas

Recursive nature ofDFS traversal(for efficient implementation of the ideas)

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Depth First Search (DFS) traversal

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A recursivealgorithm for traversing graphs

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DFS traversal ofG

DFS(v){ Visited(v)true;

For each neighbor wofv

{ if (Visited(w) = false)

{ DFS(w);

}

}

}

DFS-traversal(G)

{

For each vertex vV { Visited(v) false }

For each vertex v V { If (Visited(v ) = false) DFS(v) }

} 4

..;

..;You will need to add only a few extra

statements here to obtain efficient

algorithms for a variety of problems.

DFN[v]dfn++;

dfn0;

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Insight into DFS through an example

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DFS(v) computes a tree rooted at v

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A DFStree rooted at v

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Question:

Is there anything unique about a DFStree ?

Answer:yes.

Every non-tree edge is between an ancestor

and descendant in the DFStree.

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u

xy

It can never happen

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Every non tree edge is between ancestorand a descendant

of DFS tree

A short proof:Let (x,y) be a non-tree edge.

Letxget visited before y.

Question:

If we remove all vertices visited prior tox, does y still lie in the connected

component ofx?

Answer:yes.

DFSpursued fromx will have a path to yin DFStree.Hencexmust be ancestor of y in the DFStree.

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Remember the following picture for DFS

traversal

non-tree edgeback edge

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A novel application of DFS traversal

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Computing all articulation points in a graph G

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No.

The removal of any of {v,f,u} can destroy

connectivity.

v,f,u are called the articulation points of G.

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Does the graph remain

connected even after removal of

any single vertex ?

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A formal definition of articulaton point

Definition:A vertexxis said to be articulation point if there exist two distinct vertices

uand vsuch that every path between uand vpasses throughx.

Observation:A graph is biconnected if none of its vertices is an articulation point.

AIM:

Design an algorithmto compute all articulation points in a given graph.

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vu x

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Articulation points and DFS traversal

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Key insight and ideas

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Some observations

Question:Cana leaf node be ana.p. ?

Answer: Never

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Some observations

Question: Necessary condition for the rootto be an a.p. ?

Answer: root must have more than one child in the DFS tree.

Question: Is this condition sufficient as well ?

Answer: Yes.

The only case left : internal nodes (other than the root node) in the DFS tree ?

AIM:

To find necessary and sufficient conditions for an internal node to be articulation point.

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vu x

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How will an internal node xlook like in DFS tree ?

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x

root

T1

At least one of uand v

must be descendant ofx.

Where will uand v

be relative tox?

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Case 1:Exactly one of uand vis a descendant of x in DFS tree

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x

u

v

root

w

y

No back edge from

T1to ancestor ofx

T1

uwyv:

a u-vpath not passing throughx

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Case 2:both uand vare descendants of x in DFS tree

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x

rootroot

uv

Can u and vbelong to

T1simultaneously?

T1

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Case 2:both uand vare descendants of x in DFS tree

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x

root

u v

w q

y

z

T1 T2

At least one of T1andT2have no back edge

to ancestor ofx

uwzyqv:

a u-vpath not passing throughx

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Necessary condition for x to be articulation point

Necessary condition:

xhas at least one child y s.t. there is noback

edge from subtree(y) to ancestorofx.

Question:Is this condition sufficient also?

Answer: yes.

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x

root

yu

v

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Articulation points and DFS

Let G=(V,E) be a connected graph.

Perform DFS traversal from any graph and get a DFS tree T.

No leaf of T is an articulation point.

root of T is an articulation pointif root has more than one child.

For any internal node ??

Theorem1: An internal nodexis articulation pointif and only if it has achild ysuch that there is no back edge from subtree(y) to any ancestor ofx.

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Efficient algorithm for Articulation points

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Exploitingrecursive nature of DFS

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Starting point

Theorem1: An internal nodexis articulation pointif and only if it has a child ysuchthat there is no back edge from subtree(y) to any ancestor ofx.

A new function

High_pt(v):

DFNof the highest ancestor of vto which there is a back edge from subtree(v).

Question: Can we express Theorem 1in terms of High_pt?

Answer: Yes

A nodexis articulation pointiffit has a child, say y, such that High_pt(y) dfn(x).

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Towards an O(m+n) time algorithm

In order to compute all articulation points of a graph G=(V,E) , it suffices to

know High_pt(v) for each vin V.

So our aim is to compute High_pt(v) for each vin Vefficiently so that overall

time complexity is O(m+n).

For this we take a closer look at High_pt(v).

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Acloser look atHigh_pt(v)

Question: Can we express High_pt(v) in terms ofits childrenand proper ancestors?

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v

root

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Acloser look atHigh_pt(v)

Question: Can we express High_pt(v) in terms ofits childrenand proper ancestors?

High_pt(v) =

min ? ?

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v

root

If w=child(v)

If w = proper

ancestor ofv(v,w) E

High_pt(w)

DFN(w)

Fully internalize the above equation

and then only proceed.

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Thenovelalgorithm

The algorithm will output an array AP[]such that

AP[v]= trueif and only if vis an articulation point.

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Algorithm for articulation points in a graph G

DFS(v)

{ Visited(v)true; DFN[v]dfn++;

For each neighbor wofv

{ if (Visited(w) = false)

{ DFS(w);

}

}

}

DFS-traversal(G)

{ dfn0;

For each vertex vV { Visited(v) false; }

For each vertex v V { If (Visited(v ) = false) DFS(v) }

} 29

..;

..;

Parent(w)v;

AP[v]false

What statements should we write here so

that at the end of DFS(v),

We have computed High_pt(v)

Determined if vis articulation point and

computed AP[v] accordingly.

High_pt[v] ;

..;

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Algorithm for articulation points in a graph G

DFS(v)

{ Visited(v)true; DFN[v]dfn++;

For each neighbor wofv

{ if (Visited(w) = false)

{ DFS(w);

}

}

}

DFS-traversal(G)

{ dfn0;

For each vertex vV { Visited(v) false; }

For each vertex v V { If (Visited(v ) = false) DFS(v) }

} 30

..;

..;

Else if (parent(v) w)

High_pt(v) min(DFN(w),High_pt(v))

High_pt(v) min(High_pt(v),High_pt(w));

Parent(w)v;

AP[v]false

High_pt[v] ;

..;If High_pt(w) DFN[v] AP[v] true

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Conclusion

Theorem2: For a given graph G=(V,E), all articulation pointscan becomputed in O(m+n) time.

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