Graph (III)

Trees, Articulation point, Bridge, SCC

GGuy19-3-2011

Review

Shortest pathDijkstraBellman FordFloyd Warshall

Minimum Spanning TreePrim’sKruskal’s

Tree

• The following four conditions are equivalent:– G is connected and acyclic– G is connected and |E| = |V| - 1 – G is acyclic and |E| = |V| - 1– Between any pair of vertices in G, there exists a

unique path• G is a tree if at least one of the above

conditions is satisfied

Tree

• |E| = |V| - 1• Between any pair of vertices, there is a unique path• Adding an edge between a pair of non-adjacent

vertices creates exactly one cycle• Removing an edge from the tree breaks the tree into

two smaller trees

Tree

root

siblings

descendants children

ancestors

parent

Tree Diameter

Find the farthest node from root, say it uFind the farthest node from u, say it v

(u, v) = Tree diameter

DFS TreeDFS (vertex u) {mark u as visited

time = time + 1birth[u] = time;for each vertex v directly reachable from u

if v is unvisitedparent[v] = uDFS (v)

time = time + 1death[u] = time;

}

A

F

B

C

D

E

GH

DFS forest (Demonstration)A B C D E F G H

birth

death

parent

unvisited

visited

visited (dead)

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B

C

F

E

D

G

1 2 3 13 10 4 14

12 9 8 16 11 5 15

H

6

7

- A B - A C D C

Classification of edges

• Tree edge• Forward edge• Back edge• Cross edge

• Question: which type of edges is always absent in an undirected graph?

A

B

C

F

E

D

G

H

Determination of edge types

• How to determine the type of an arbitrary edge (u, v) after DFS?

• Tree edge– parent [v] = u

• Forward edge– not a tree edge; and– birth [v] > birth [u]; and– death [v] < death [u]

• How about back edge and cross edge?

Determination of edge types

Tree edge Forward Edge Back Edge Cross Edge

parent [v] = u

not a tree edgebirth[v] > birth[u]death[v] < death[u]

birth[v] < birth[u]death[v] > death[u]

birth[v] < birth[u]death[v] < death[u]

Articulation Point

For a undirected graph G,Node u is an Articulation Point if remove u from G will create more components.

Articulation Point

Bridge

For a undirected graph G,Edge (u, v) is an bridge if remove (u, v) from G will create more components.

Bridge

Finding AP and Bridge

DFS(vertex u)time = time + 1low[u] = vis[u] = timefor each vertex v directly reachable from u

if v is not visitedDFS(v)low[u] = min(low[u], low[v])

if v is visited and parent of u is not v

low[u] = min(low[u], vis[v])

Finding AP

Find the DFS tree of G and compute low[]

Vertex u is an AP if

1. u is the root and u has >1 children2. u is not the root and there exists children v where low[v] >= vis[u]

Finding Bridge

Find the DFS tree of G and compute low[]

Edge (u, v) is a Bridge if

1. v is u children and low[v] > vis[u]

Bi-connected component

A bi-connected component (or 2-connected component) is a maximal bi-connected subgraph

A bi-connected subgraph is a graph that has no articulation point

Bi-connected component

Each Bi-connected component isconnected by bridge

Each articulation point belongs to morethan one Bi-connected component

The Bi-connected components form a Tree

Bi-connected component

Strongly Connected Component

In a directed graph G, a Strongly Connected Component is a subgraph that there is a path from each vertex to every other vertex in the same SCC.

Strongly Connected Component

Finding SCC

• Compute the DFS forest of the graph G to get the death time of the vertices

• Reverse all edges in G to form G’• Compute a DFS forest of G’, but always choose

the vertex with the latest death time when choosing the root for a new tree

• The SCCs of G are the DFS trees in the DFS forest of G’

A

F

B

C

D

GH

SCC (Demonstration)

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F

B

C

D

E

GH

A B C D E F G H

birth

death

parent

1 2 3 13 10 4 14

12 9 8 16 11 5 15

6

7

- A B - A C D C

D

G

A E B

F

C

H

SCC (Demonstration)

D

G

A E B

F

C

H

A

F

B

C

D

GH

E

Finding SCC

Assume (u, v) belongs to the same SCCWLOG, assume u is visited first. Since the exists a path from u to v, we havedeath[u] > death[v]

When we do DFS on G’, we will call DFS(u) first. Since there exists a path from v to u in G, then there exists path from u to v in G’, hence v and u will be in the same tree.

Iterative Depth Searching (IDS)

Bi – Directional Searching (BDS)