5. Biconnected Components of A Graph

If one city’s airport is closed by bad weather, can you still fly between any other pair of cities? If one computer in a network goes down, can a message be sent between any other pair of computers?

If any one vertex (and edges incident with it) is removed from a connected graph, is the remaining subgraph still connected?

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(a) graph. (b) Its biconnected components

Def’n: Let G=(V, E) be a connected, undirected graph.

A vertex aV is said to be an articulation point

if there exist vertices v and w such that

(1) v, w and a are distinct

(2) Every path between v and w must contain a.

Alternatively, a is an articulation point of G

if removing a splits G into two or more parts. subgraphs

2

Def’n: A connected graph G =(V, E) is said to be biconnected

if and only if it has no articulation points.

G1 G2 G3

Which of the above are biconnected?

Def’n: Let G' = (V', E') be a biconnected subgraph of a graph G = (V, E).

G' is said to be a biconnected component of G if G' is maximal

i.e., not contained in any other biconnected subgraph of G.

Examples are shown in the previous page !!!

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G = (V, E)

Two edges e1 and e2 in E are said to be related, i.e., (e1, e2) ∈ R

if e1 = e2 or if there is a cycle containing both e1 and e2.

(e, e) R e E

(e1 , e2) R and (e2 , e3) R (e1 , e3) R

(e1 , e2) R (e2 , e1) R why ?

Observation

e1 e2 e3 Why transitivity?

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An equivalence relation on E partitions E into equivalence classes !!!

Each subgraph consisting of the edges in an equivalence class and

the incident vertices form a biconnected component !!!

Can you prove it ? Homework.

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Gi

Lemma: For 1 i k , Let Gi =(Vi , Ei) be the biconnected components of a

connected undirected graph G = (V, E).

Then

(1) Gi is biconnected for each 1 i k ,

(2) For all i j , Vi Vj contains at most one vertex. (3) a is an articulation point of G

if and only if a Vi Vj for some i j [ proof ]

(1) Trivial why ?

(2) suppose that two distinct vertices v and w are in Vi Vj , i j

v

w

C1

Gj C2

Why?

(3) x

a

y

() () x

a

y

v

w

Homework.

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A B

C

D

E

A B

C

D

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F

A

B

C

D

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A

B

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D

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F

A is not an articulation point A is an articulation point

Can you now characterize an articulation point when A is the root ?

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G H

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A

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Can you characterize D ?

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An articulation point v in a depth-first search tree.

Every path from root to w passes through v !!!

v

w

root

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Theorem : Let G = (V, E) be a connected, undirected graph and let T = (V, ET) be a depth first spanning tree for G. Vertex a is an articulation point of G if and only if one of the following is true: (1) a is the root and a has two or more sons. (2) a is not the root and for some son s of a, there is no back edge between any descendant of s ( including s itself ) and a proper ancestor of a.

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[proof] i) The root is an articulation point if and only if it has two or more sons.

()

()

x y

x y a

a

a

x y

a

x y

why? a

x y

s

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(ii) Vertex a is not the root : Vertex a is an articulation point if and only if there exists no back edge between any descendant of s to a proper ancestor of a, where s is a son of a.

a

s

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()

x y a

y x

Either x or y is a proper descendant of a !!!

a

why ?

WLOG, let x be a proper descendant of a a

x x y

s

a

s This is not possible.

Why ? Only this

Two cases: (1) (2)

x

a

x

a

s s back edge no back edge

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case (2) : back edge

Suppose that y is not a (proper) descendant of a.

Thus, y must be a (proper) descendant of a.

() Easy, Exercise

x

a

y

w'

w

a

x y

This is not possible. s s’

s

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v

w

z

An algorithm for finding all biconnected components of a graph G = (V, E)

biconnected component

z

x y

v x y

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How to detect an articulation point ?

How to report a biconnected component ?

Assumption

Depth First Search

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3

Depth First Search number

G =(V, E)

Any relation between DFS numbers and articulation points ?

B

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4 5

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Vertices DSF#

A 1

B 2

C 3

D 6

E 5

F 4

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A

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w

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(a,b) a b

Tree edge : (a,b) a < b

Back edge : (a,b) a > b

If there is a back edge from x

to a proper ancestor w of v,

then w is reachable from v.

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v

x

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BACK[v] = w

v can back to w by following

(1) tree edges (2) back edges BACK[v]:= min{DFSNUMBER(v), min{w| (x,w) is a back edge, x is a descendant of v,

and w is an ancestor of v.}}

How far can a vertex back to ancesters in the DFS tree ?

v

w

x

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Initially (when v is first visited),

BACK[v] = DFSNUMBER[v]

why?

Now, how to update BACK[v] ?

(i) v w (back edge)

(ii) v w (backing up to v)

case(i) BACK[v]:= min{BACK[v], w} : back edge

why ? case(ii) BACK[v]:= min{BACK[v], BACK[w]} : back up to v why ?

v

w

w

case (i)

case (ii)

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why ?

How to detect an articulation point ?

v BACK[w] when w v (backing up to v).

v

w i) backing up to v

ii) v BACK[w]

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How to report ?

Well, …….

A set of edges.

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v

w

z

W A

B

y

r

a b

(v, w)

(y, v)

(r, z)

(y, a)

(z, y)

W

A

Edge Stack

when v is reached, pop edges until (v, w) is reached

yꞌ

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A

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(a) The complete depth- first search tree.

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1/1

2/2

3/1

(b) Proceed forward ; initialize values of back ; Detect back edge FA ; update back[F] ;

E

C

4/4

5/5

6/4 (c) Continue forward ; Detect back edge CE ; update back[C]

B

C

5/4

6/4

4/4 B

5/4 7/7

8/5

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5/4

4/4

(d) back[C]

begin

mark v "old";

DFSNUMBER[v] COUNT;

COUNT COUNT + 1;

BACK[v] DFSNUMBER[v];

for each vertex w on L[v] do Stack edges, here

if w is marked "new" then

begin

add(v, w) to T;

FATHER[w] v

SEARCHB(w)

if BACK[w] DFSNUMBER[v] then a

biconnected component has been found ; Pop, here

BACK[v] MIN(BACK[v], BACK[w])

end

else if w is not FATHER[v] then

BACK[v] MIN(BACK[v], DFSNUMBER[w]) end

end

procedure SEARCHB(v);

Report

O(|E|)

Why ?

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T := Ø ;

for all v in V do

mark v as "new"

end

count := 1 ;

for all "new" vertex v in V do

SEARCHB(v) |V| + (|E1| + |E2| + …) = |V| + |E|

end

Theorem : Algorithm Bicomp correctly finds the biconnected

components of G = (V, E) in O(|E| + |V|) time.

[proof ]

See the next page.

Algorithm : Bicomp.

O(|V|)

G = (V, E) : not connected

G1 G2 G3

Spanning forest.

G1

G2

G3 G4

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(i) Time complexity O(|V| + |E|)

(ii) Correctness Root ?

Even if the root is not an articulation point,

it can be treated as an articulation point !!!

why ?

The biconnected component containing the root

is emiting from the root.

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By induction on the # of biconnected components in G, we prove that our algorithm correctly

detects all articulation points and reports their corresponding biconnected components.

How to detect:

(i) w v (SEARCHB(w) is completed)

(ii) v BACK[w]

How to report:

Popping up the edges above (v, w) on EDGE STACK

Popping up the edges above (v, w) on EDGE STACK

v

w

(b = 1) Trivial

why ?

Depth First Search !!!

In this case, v must be the root !!! why ?

(b = i) Assume the induction hypothesis.

(b = i+1) (v,w)

b-1=i

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K-connectivity

k = 1 connectivity

k = 2 bi-connectivity

k = 3 tri-connectivity

k = 4

no known efficient algorithms

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6. Strongly Connected Components of a Digraph

Def'n : Let G = (V, E) be a directed graph. We can partition V into equivalent classes Vi, 1 i r, such that vertices v and w are

equivalent if and only if there are paths from v to w and w to v.

Let Ei, 1 i r, be the set of edges connecting the vertices

in Vi. The graph Gi = (Vi, Ei) is called a strongly connected

component of G. A graph is said to be strongly connected

if it has only one strongly connected component.

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9 G=(V, E)

G1

G2

G3

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1

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5 6

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9 G=(V, E)

G1

G2

G3 R = {(v, w) | there exist paths from v to w and w to v}

R is an equivalence relation !

why? Homework

R partitions V into a set of equivalence classes.

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(a) A digraph (b) Its strong components.

The condensation of the digraph

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G = (V, E)

A Weakly Connected Component

(Network Flow)

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How to find the strongly connected components of a graph ?

Assumption : Depth First Search

tree edge

descendant edge

cross edge

back edge

first visited

backing up to v

backing up from v

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v1

v8 v7

v6

v4

v3

v2

v5

v10

v9 v12

v11

v15 v14

v13

v18 v17

v16

G = (V, E)

v5 v6

v11

v4

v1

v2

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v7

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T = (V, ET)

v18

Spanning Forest

v10

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How to characterize the root of the DSF subtree for each strongly connected component ?

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Lemma : Let Gi = (Vi, Ei) be a strongly connected component of a directed graph

G = (V, E). Let T = (V, ET) be a depth-first search spanning forest of G.

Then the vertices of Gi together with the edges which are common to both

Ei and ET form a subtree of the spanning forest T.

[ proof ]

Gi =(Vi, Ei)

Take any pair of vertices v,w Vi

Depth first number

WLOG, let v < w.

In Gi, v w

P

why ?

x = min{ y| y is in P }

Note : x may be v itself.

v w

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w is a descendant of x !!!

Now, all vertices between x and w are also descendants of x.

why ?

DFS !!!

Since x v < w, v is also a descendant of x.

Any pair of vertices v, w in Gi have a common ancestor x in T.

Why ?

r min{ y| y is a common ancestor of vertices in Gi}

In T,

p

x

Once P reaches a descendant of x in the DFS forest T,

it cannot leave the subtree of x !!! why?

v x w

In Gi,

v w

P

x

x ≤ v < w

y

y

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r min{ y| y is a common ancestor of vertices in Gi}

For each vertex u in Gi , r ≤ u and r → u and u → r . ∴ u is in the subtree of r .

For each vertex u Gi , every vertex u' on the path from r to u in the subtree of r

is also contained in Gi

why? r u

'u

∴ r must be the root of the spanning subtree for strongly connected component Gi

iGu

r

u'

u

iGu

u' is in

the subtree of r

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v4 v5

v7

v6 v9

v8

v1

v2

v3 G = (V, E)

v5

v4

v1

v2

v3

1

2

3

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5

v6

v7

v8

6

7

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v9 9

(r1, r2, r3, ····, rk) the sequence of roots for strongly connected components 4 6 1 such that the DFS of each of these roots is terminated.

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(r1, r2,····, rk)

Observation

i < j Either rj is the root of a right subtree of ri or a proper ancester of ri. why ?

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Lemma : For each i, 1 i k, Gi consists of those vertices

which are descendant of ri but are contained in none of G1, G2,····, Gi-1.

[proof] The root rj for j > i cannot be a descendant of ri,

since the call of SEARCH(rj) terminates after SEARCH(ri).

(r1, r2, r3, ···, ri-1, ri, ri+1, ···, rk)

already

reported j > i

rj is the root of

a right subtree or a proper ancestor.

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w

v

r

LOWLINK[v] = min { DFSNUMBER(v) min{ w| there is a cross or back edge (x, w)

from a descendant x of v to a vertex w

that is not a descendant of v ,

and the root of the strongly

connected component

containing w is an ancestor of v } }

LOWLINK[v] v

Why ?

w

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x

Lemma : Let G = (V, E) be a directed graph. A vertex v is the root of a strongly connected component if and only if LOWLINK[v] = v.

[proof]

( ) Suppose that v is the root of a strongly connected component of G.

By definition of LOWLINK, LOWLINK[v] v for all v V. Suppose that LOWLINK[v] < v. Then, there are vertices w and r such that [1] w is reached by a cross or back edge from a descendant of v. [2] r is the root of the strongly connected component containing w.

[3] r is an ancestor of v.

[4] w < v

r and v must be in the same strongly connected component !!!

why ? From [1], [2], and [3].

[2] r is an ancestor of w r w

[4] r w < v r < v

[3] r is a proper ancestor of v.

v is not the root of a strongly connected component. #

LOWLINK[v] = v.

r v

w v

r

x

w

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w

( ) Now, suppose that LOWLINK[v] = v. Assume that v is not the root of the strongly connected component. Then, there is some proper ancestor r of v, which is the root, i.e.,

The path P goes from v to w to r.

There also exists a path from r to v. Why?

v and w are in the same strongly connected component.

LOWLINK[v] w < v. #

v

r

w

Moreover, there is a path P from v to r . Why? Consider the first edge of P from a descendant of v to a vertex w that is not a descendant of v. Then, w < v why ?

P P

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w

How to compute LOWLINK[v]

LOWLINK[v] := v

LOWLINK[v] = min{ LOWLINK[v], w }

LOWLINK[v] = min{ LOWLINK[v],

LOWLINK[w] }

w

w

v

v v

w

first visited

a back edge or a cross edge

backing up to v v

w

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if the root of the strongly connected component containing w is an ancestor of v.

How to check this ?

v

s' s

backing up from v

How to report a strongly connected component

LOWLINK[v] = v

f

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y

z

v LOWLINK[v] = v

x

y z

g

x

g

v

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q

How to check that the root of the strongly connected component

containing w is an ancestor of v?

What if w is on the stack ?

v and w have a common ancestor !

Why ?

r

v t

w u

q

r

v t

w u

u

v

w

t

q

r

…

u

v

q

…

w ∈ stack w ∈ stack

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Procedure SEARCHC(v): begin mark v "old"; DFSNUMBER[v] COUNT; COUNT COUNT + 1; O(|V|) LOWLINK[v] DFSNUMBER[v]; push v on STACK; for each vertex w on L[v] do O(|E|) if w is marked "new" then begin SEARCHC(w); LOWLINK[v] MIN(LOWLINK[v], LOWLINK[w]) end back edge or cross edge else {not new} if DFSNUMBER[w] < DFSNUMBER[v] need a table and w is on STACK then LOWLINK[v] MIN(DFSNUMBER[w], LOWLINK[v]) end if LOWLINK[v] = DFSNUMBER[v] then begin repeat begin pop x from top of STACK; print x end until x=v; print "end of strongly connected component" end end

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Algorithm : (Strongly connected components of a directed graph)

procedure StrongCC

begin

count := 1;

for all v in V do

mark v "new" O(|V|)

end

STACK := Ø

while there exists a vertex v marked "new" do

SEARCHC(v) O(|E|)

end.

end

Optimal !!!

O( |V| + |E| )

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Theorem : "StrongCC" correctly finds the strongly connected

components of G in O(|V| + |E|) time.

[proof]

Time complexity : Already shown

Correctness : ( By induction on # of strongly connected components)

"The algorithm correctly detects all the roots of the

subtrees for the strongly connected components and

reports these components."

Exercise.

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• November 18 (Friday)

• 10:00 ~ 11:00 class A

14:00 ~ 15:00 class B

• Room #4443

(Oh Sang-su lecture room)

Lecture Schedule