lecture 15: depth first search shang-hua teng. graphs g= (v,e) b e c f d a b e c f d a directed...

Lecture 15:Depth First Search

Shang-Hua Teng

Graphs• G= (V,E)

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E

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FD

A B

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• Directed Graph (digraph)– Degree: in/out

• Undirected Graph – Adjacency is symmetric

Graph Basics• The size of a graph is the number of its vertices• If two vertices are connected by an edge, they

are neighbors • The degree of a node is the number of edges it

has• For directed graphs,

– The in-degree of a node is the number of in-edges it has

– The out-degree of a node is the number of out-edges it has

Paths, Cycles• Path:

– simple: all vertices distinct• Cycle:

– Directed Graph: • <v0,v1,...,vk > forms cycle if v0=vk

and k>=1• simple cycle: v1,v2..,vk also distinct

– Undirected Graph: • <v0,v1,...,vk > forms (simple) cycle

if v0=vk and k>=3• simple cycle: v1,v2..,vk also distinct

B

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A path <A,B,F>path <A,B,F>

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A simple cycle simple cycle <E,B,F,E><E,B,F,E>

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Asimple cycle <A,B,C,A>= <B,C,A,B>simple cycle <A,B,C,A>= <B,C,A,B>

Connectivity

• Undirected Graph: connected– every pair of vertices is connected by a

path– one connected component– connected components:

• equivalence classes under “is reachable from” relation

• Directed Graph: strongly connected– every pair of vertices is reachable from

each other– one strongly connected component– strongly connected components:

• equivalence classes under “mutually reachable” relation

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Aconnectedconnected

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A2 connected 2 connected componentscomponents

strongly strongly connected connected componentcomponent

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not not strongly strongly connectedconnected

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Subgraphs• A subgraph S of a graph G

is a graph such that – The edges of S are a

subset of the edges of G– The edges of S are a

subset of the edges of G• A spanning subgraph of G

is a subgraph that contains all the vertices of G

Subgraph

Spanning subgraph

Trees and Forests• A (free) tree is an undirected

graph T such that

– T is connected

– T has no cycles

This definition of tree is different from the one of a rooted tree

• A forest is an undirected graph without cycles

• The connected components of a forest are trees

Tree

Forest

Spanning Trees and Forests• A spanning tree of a connected

graph is a spanning subgraph that is a tree

• A spanning tree is not unique unless the graph is a tree

• Spanning trees have applications to the design of communication networks

• A spanning forest of a graph is a spanning subgraph that is a forest

Graph

Spanning tree

Complete Graphs

A graph which has edge between all pair of vertex pair is complete.

A complete digraph has edges between any two vertices.

Graph Search (traversal)• How do we search a graph?

– At a particular vertices, where shall we go next?

• Two common framework: the breadth-first search (BFS) and the depth-first search (DFS)– In BFS, one explore a graph level by level away

(explore all neighbors first and then move on) – In DFS, go as far as possible along a single path until

reach a dead end (a vertex with no edge out or no neighbor unexplored) then backtrack

Depth-First Search• The basic idea behind this algorithm is

that it traverses the graph using recursion– Go as far as possible until you reach a

deadend– Backtrack to the previous path and try the

next branch– The algorithm in the book is overly

complicated, we will stick with the one on the right

– The graph below, started at node a, would be visited in the following order: a, b, c, g, h, i, e, d, f, j

dfs(g, current){ mark current as visited x = set of nodes adjacent

to current for(i gets next in x*) if(i has not been

visited yet) dfs(g, i)}

* - we will assume that adjacentnodes will be processed in numeric or alphabetical order

a b

h i

c

g

e

j

d

f

Color Scheme

• Vertices initially colored white

• Then colored gray when discovered

• Then black when finished

Time Stamps

• Discover time d[u]: when u is first discovered

• Finish time f[u]: when backtrack from u

• d[u] < f[u]

DFS Examplesourcevertex

DFS Example

1 | | |

| | 3 |

2 | |

sourcevertex

d f

DFS Example

1 | | |

| | 3 | 4

2 | |

sourcevertex

d f

DFS Example

1 | | |

| 5 | 3 | 4

2 | |

sourcevertex

d f

DFS Example

1 | | |

| 5 | 63 | 4

2 | |

sourcevertex

d f

DFS Example

1 | 8 | |

| 5 | 63 | 4

2 | 7 |

sourcevertex

d f

DFS Example

1 | 8 | |

| 5 | 63 | 4

2 | 7 9 |

sourcevertex

d f

DFS Example

1 | 8 | |

| 5 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 | 8 |11 |

| 5 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 |12 8 |11 |

| 5 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 |12 8 |11 13|

| 5 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 |12 8 |11 13|

14| 5 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 |12 8 |11 13|

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

PseudocodeDFS(G)

1. For each v in V,

2. color[v]=white; [u]=NIL

3. time=0;

4. For each u in V

5. If (color[u]=white)

6. DFS-VISIT(u)

1. DFS-VISIT(u)

2. color[u]=gray;

3. time = time + 1; d[u] = time;

4. For each v in Adj(u) do

5. If (color[v] = white)

6. [v] = u;

7. DFS-VISIT(v);

8. color[u] = black;

9. time = time + 1; f[u]= time;

Complexity Analysis

There is only one DFS-VISIT(u) for each vertex u.

Ignoring the recursion calls the complexity is O(deg(u)+1)

The recursive call on v is charged to DFS-VISIT(v)

Initialization complexity is O(V)

Overall complexity is O(V + E)

DFS Forest

1. [v] = u then (u,v) is an edge

2. All descendant of u are reachable from u

3. defines a spanning forest

Parenthesis Lemma Relation between timestamps and ancestry

• u is an ancestor of v if and only if [d[u], f[u]] [d[v], f[v]]

• u is a descendent of v if and only if [d[u], f[u]][d[v], f[v]]

• u and v are not related if and only if [d[u], f[u]] and [d[v], f[v]] are disjoint

DFS: Tree (Forest) Edges

• DFS introduces an important distinction among edges in the original graph:– Tree edge: encounter new (white) vertex

• The tree edges form a spanning forest

DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

Tree edges

DFS: Back Edges

• DFS introduces an important distinction among edges in the original graph:– Tree edge: encounter new (white) vertex – Back edge: from descendent to ancestor

• Encounter a grey vertex (grey to grey)

DFS Example

Tree edges Back edges

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS: Forward Edges

• DFS introduces an important distinction among edges in the original graph:– Tree edge: encounter new (white) vertex – Back edge: from descendent to ancestor– Forward edge: from ancestor to descendent

• Not a tree edge, though

• From grey node to black node

DFS Example

Tree edges Back edges Forward edges

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS: Cross Edges

• DFS introduces an important distinction among edges in the original graph:– Tree edge: encounter new (white) vertex – Back edge: from descendent to ancestor– Forward edge: from ancestor to descendent– Cross edge: between a tree or subtrees

• From a grey node to a black node

DFS Example

Tree edges Back edges Forward edges Cross edges

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

DFS: Types of edges

• DFS introduces an important distinction among edges in the original graph:– Tree edge: encounter new (white) vertex

– Back edge: from descendent to ancestor

– Forward edge: from ancestor to descendent

– Cross edge: between a tree or subtrees

• Note: tree & back edges are important; most algorithms don’t distinguish forward & cross

DFS: Undirected Graph

• Theorem: If G is undirected, a DFS produces only tree and back edges

• Proof by contradiction:– Assume there’s a forward edge

• But F? edge must actually be a back edge (why?)

sourceF?

DFS: Undirected Graph

• Theorem: If G is undirected, a DFS produces only tree and back edges

• Proof by contradiction:– Assume there’s a cross edge

• But C? edge cannot be cross:• must be explored from one of the

vertices it connects, becoming a treevertex, before other vertex is explored

• So in fact the picture is wrong…bothlower tree edges cannot in fact betree edges

source

C?

DFS Undirected Graph

• Theorem: An undirected graph is acyclic iff a DFS yields no back edges– If acyclic, no back edges (because a back edge implies

a cycle– If no back edges, acyclic

• No back edges implies only tree edges (Why?)• Only tree edges implies we have a tree or a forest• Which by definition is acyclic

• Thus, can run DFS to find whether a graph has a cycle

Example: DFS of Undirected Graph

AA

BB

CC

DD

EEFF

GG

G=(V,E)G=(V,E)Adjacency ListAdjacency List::A: B,C,D,FA: B,C,D,FB: A,C,DB: A,C,DC: A,B,D,E,FC: A,B,D,E,FD: A,B,C,GD: A,B,C,GE: C,GE: C,GF: A,CF: A,CG: D,EG: D,E


AA

BB

CC

DD

EEFF

GG

Adjacency ListAdjacency List::A: B,C,D,FA: B,C,D,FB: A,C,DB: A,C,DC: A,B,D,E,FC: A,B,D,E,FD: A,B,C,GD: A,B,C,GE: C,GE: C,GF: A,CF: A,CG: D,EG: D,E


AA

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EEFF

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TT


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AA

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Example: (continued)

DFS of Undirected Graph

DFS TreeDFS Tree

AA

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EEFF

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TTBBTT

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Using DFS & BFS

A directed graph G is acyclic if and only if a Depth-A directed graph G is acyclic if and only if a Depth-First Search of G yields no back edges.First Search of G yields no back edges.

• Using DFS to Detect Cycles:

• Using BFS for Shortest Paths:A Breadth-First Search of G yields shortest path information: A Breadth-First Search of G yields shortest path information:

For each Breadth-First Search tree, the path from its root u For each Breadth-First Search tree, the path from its root u to a vertex v yields the shortest path from u to v in Gto a vertex v yields the shortest path from u to v in G..

Directed Acyclic Graphs

• A directed acyclic graph or DAG is a directed graph with no directed cycles:

Directed Acyclic Graphs (DAGs)

• DAG - digraph with no cycles

• compare: tree, DAG, digraph with cycle

D E

CB

A

D E

CB

A

D E

CB

A

Where DAGs are used

• Syntactic structure of arithmetic expressions with common sub-expressions

e.g. ((a+b)*c + ((a+b)+e)*(e+f)) * ((a+b)*c)

*+

**

+ ++

a b

c

e f

Where DAGs are used?

• To represent partial orders

• A partial order R on a set S is a binary relation such that– for all a in S a R a is false (irreflexive)

– for all a, b, c in S if a R b and b R c then a R c (transitive)

• examples: “less than” (<) and proper containment on sets

• S ={1, 2, 3}

• P(S) - power set of S

(set of all subsets)

{1, 2, 3}

{1, 2} {1, 3} {2, 3}

{1} {2} {3}

{ }

DAGs in use

• To model course prerequisites or dependent tasks

Year 1 Year 2 Year 3 Year 4

Compilerconstruction

Prog.Languages

DS&APUMA

Data &Prog.

DiscreteMath

DataComm 1

DataComm 2

OpSystems

Real timesystems

DistributedSystems

DFS and DAGs

• Theorem: a directed graph G is acyclic iff a DFS of G yields no back edges:– => if G is acyclic, will be no back edges

• Trivial: a back edge implies a cycle

– <= if no back edges, G is acyclic• Proof by contradiction: G has a cycle a back edge

– Let v be the vertex on the cycle first discovered, and u be the predecessor of v on the cycle

– When v discovered, whole cycle is white– Must visit everything reachable from v before returning from

DFS-Visit()– So path from uv is greygrey, thus (u, v) is a back edge

lecture 15: depth first search shang-hua teng. graphs g= (v,e) b e c f d a b e c f d a directed...

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