5. biconnected components of a graph

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 in the network?

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

GI

E

C

F

B

A D

H J F

A D

H J

E

F

E

CB

GI

B

(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 vertex 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

Def’n: A 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.

Example is shown in the previous page !!!

Observation

G = (V, E)

Two edge e1 and e2 in E is said to be related

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 ?

Each subgraph consisting of the edges in an equivalent class and

the incident vertex is a biconnected component !!!

Can you prove it ?

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

Gi

Gj

C2

Why?

(3) x

a

y

() ()x

a

y

v

w

A B

C

D

E

A B

C

D

E

F

A

B

C

D

E

A

B

C

D

E

F

A is not an articulation point A is an articulation point

Can you now characterize anarticulation point when A is the root ?

A

B

C

D

E

F

GH

I

A

B

C

D

EF

G

H

I

Can you characterize D ?

An articulation point v in a depth-first search tree.Every path from root to w passes through v.

v

w

One or morebicomponents

w

v'

v

Theorem : Let G = (V, E) be a connected, undirected graph and let S = (V, T) 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.

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

()

()

x y

x ya

yx

(ii)()

x ya

yx

Since a is not the root, either x or y is a proper descendant of a !!!

awhy ?

WLOG, let x be a proper descendant of a a

x x y

s

a

sThis is not possible.Why ?

Only this

Two cases: (1) (2)

x

a

x

ass

back edgeno back edge

case (2) back edge y is not a descendant of a.

y is a descendant of a.

() Easy, Exercise

x

a

y

w'

w

a

x y

This is not possible.s s’

v

w

z

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

biconnected component

z

How to detect an articulation point ?

When to report a biconnected component ?

Assumption

Depth First Search

Depth First Search number

B

D E

A C FG =(V, E)

B

D

E

A

C

F

Any relation between DFS numbers and articulation point ?

1

2

3

4 5

A B C D E F 1 2 3 6 5 4

6

w’

B

D

E

A

C

F

1

2

3

4 5

v

Assume that (a,b) a b Tree edge : (a,b) a < b Back edge : (a,b) a > b

If there is a back edge from xto a proper ancestor of v,then v is reachable from x.

6

w

x

How far back in the DFS tree from each vertex ?

v

w

BACK[v]

v can back to BACK[v] by following (1) tree edges (2) back edges Initially, BACK[v] = v, where v is a DFS number why ?

BACK[v]:= min{BACK(v), min{w| (x,w) is a back edge, x is a descendant of v, and w is an ancestor of v.}

x

What if v BACK(w), when w v (back up to v) ?

assuming that BACK(w) = min{BACK(x) | xT(w)}

v is an articulation point !!! 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 ?

w

v

v

w

x

w

How to detect and when ?

v

wwhen ?

? or ?

why ?

How ? v BACK(w) !!!

How to report ? Well, ……. A set of edges.

v

w

z

WAB

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

A

D

F

E

B

C

JH

G

I

1

2

345

68

7

9

10

(a) The complete depth- first search tree.

A

D

F

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

4/4B

5/4 7/7

8/5

G

I

B5/4 7/5

8/5

G

I

E

B

C

5/4

4/4

(d) back[C]<dfsNumber[B] so back up to B updating back[B].

(e) Forward to G and I ; detect back edge IB ; update back[I] ; back[I]<dfsNumber[G]

(f) Back up to G updating back[G]; back[G]=dfsNumber[B] ; remove bicomponent ;

(g) Back up to B ; back[B]=dfsNumber[E] ; remove bicomonent ;

The action of the bicomponent algorithm on the graph in Fig. 4.25(detecting the first two bicomponents)

B

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

procedure SEARCHB(v);

O(|Ei|)

Report

O(|Ei|)why ?

Algorithm : Bicomp.

T := Ø ;

for all v in V do

mark v as "new"

end

for all "new" vertex v in V do

count := 1 ;

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.

O(|V|)

(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 emitted from the root.

( By induction on the # of biconnected components in G ) Reporting a biconnected component (i) w v (SEARCHB(w) is completed) (ii) v BACK(w)

The edges above (v, w) on STACK is exactly thoseedges in the biconnected component containing (v, w).

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

K-connectivity

k = 1 connectivity

k = 2 bi-connectivity

k = 3 tri-connectivity

k = 4

no known efficient algorithm

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 is a path from v to w and w to v. Let Ei, 1 i r, be the set of edges connecting the pairs of vertices in Vi. The graph Gi = (Vi, Ei) are called the strongly connected components of G. A graph is said to be strongly connected if it has only one strongly connected component.

1

2 3

4

56

7

89

G=(V, E)

G1

G2

G3

A

D

F

E

B

C

J

H

G

I

K

A

D

F

E

B

C

J

H

G

I

K

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

The condensation of the digraph

G = (V, E)

A Weakly Connected Component

How to find the strongly connected components of a graph ?

Assumption : Depth First Search

tree edgedescendant edgecross edgeback edge

v1

v8v7

v6

v4

v3

v2

v5

v10

v9 v12

v11

v15v14

v13

v18v17

v16

G = (V, E)

v5

v6

v11

v4

v1

v2

v3

v7

v8

v16

v17

T = (V, ET)

v18

Spanning Forest

v10

v12

v9

v13

v14

v15

1

2

3

4

5 6

78

9

10

1112

13

14

1516

17 18

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 tree T.[ proof ]

Gi =(Vi, Ei)

v,w Vi

Depth first number

WLOG, let v < w.In Gi,

v wP

why ?

x = min{ y| y is in P }

Note : x may be v itself. In T,

p

xOnce P reaches a descendant of x in T,it cannot leave the subtree of x !!! why?

v w

w is a descendant of x !!! why ?

Now, all vertices between x and w are also descendants of x. why ? DFS !!! ( by the way of T is connected )

Since x v < w, v is also a descendant of x. Any pair of vertices in Gi have a common ancestor in T. This is also in Gi. why ? r min{ y| y is a common ancestor of the vertices in Gi}

u Gi

why?

iGu ' Turu in )~('

iGur, r u

'u

(r1, r2,····, rk)

Observation

i < j Either ri is to the left of rj or a descendant of rj. why ?

v4 v5

v7

v6 v9

v8

v1

v2

v3

G = (V, E)

v5

v4

v1

v2

v3

1

2

3

4

5

v6

v7

v8

6

7

8

v9 9

(r1, r2, r3, ····, rk) the sequence of roots in the order in which 4 6 1 the DFS of these vertices are terminated.

strongly connected components

Lemma : For each i, 1 i k, Gi consists of those vertices which are descendant of ri but are 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).

w

v

r

or

LOWLINK[v] = min { {v} { w| there is a cross or back edge from a descendant of v to w, and the root of the strongly connected component containing w is an ancestor of v } }

LOWLINK[v] v

Why ?

w

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. why ? 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 (2) r is an ancestor of w r w (2),(4) r w < v r < v

(3) r is a proper ancestor of v. r and v must be in the same strongly connected component !!! why ?

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

LOWLINK[v] = v.

r v

w v

rw

( ) Now, suppose that LOWLINK[v] = v. Assume that v is not the root of the strongly connected component containing v. 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. Why?There also exists a path from r to v. Why?

r and w are in the same strongly connected component.LOWLINK[v] w < v. #

v

r

w

Then , 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.w<vWhy?

w

PP

How to compute LOWLINK[v]

LOWLINK[v] := v

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

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

w

w

w

v

v

v

v

w

How to report Gi

w LOWLINK[w] = w

xyv

g

x

y

v

gw

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]) 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" endend

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| )

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 calls to SEARCHC)

"Whenever SEARCHC terminates, LOWLINK(v) is correctly computed."

Exercise.