BFSand DFS

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Elementary Graph Algorithm

Breadth First Search(BFS)

AND

Depth First Search(DFS)

By :-

Prithviraj MohantyLect.EAST,Bhubaneswar

[email protected] 10/11/2009

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Review: Graph searching


Simplest searching algorithm.

Application:-Prim`s Algorithm: for findingminimum spanning tree andDijkstra`s singlesource shortest path algorithm.It finds the source vertex s first and discovers

all vertex rechable from s.

It discovers all vertices at distance k from sbefore discovering vertices at distance k+1

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Review: Graph Searching

Given: a graph G = (V, E), directed or

undirected

Goal: methodically explore every vertex andevery edge

Ultimately: build a tree on the graph

Pick a vertex as the root

Choose certain edges to produce a tree

Note: might also build aforestif graph is not

connected


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Review: Breadth-First Search

Explore a graph, turning it into a tree

One vertex at a time

Expand frontier of explored vertices across the

breadth of the frontier

Builds a tree over the graph

Pick asource vertex to be the root

Find (discover) its children, then their children,

etc.


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Review: Breadth-First Search

Again will associate vertex colors to guide

the algorithm

White vertices have not been discovered

All vertices start out white

Gray vertices are discovered but not fully explored

They may be adjacent to white vertices

Black vertices are discovered and fully explored They are adjacent only to black and gray vertices

Explore vertices by scanning adjacency list of

gray [email protected] 10/11/2009

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BFS - algorithm


BFS(G, s) // G is the graph and s is the starting node

1 for each vertex u V [G] - {s}

2 do color[u] WHITE // color of vertex u

3 d[u] // distance from source s to vertex u

4 [u] NIL // predecessor of u

5 color[s] GRAY

6 d[s] 0

7 [s] NIL

8 Q // Q is a F IFO - queue

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BFS algorithm contd..


9 ENQUEUE(Q, s)

10 while Q // iterates as long as there are gray vertices. Lines 10-18

11 do u DEQUEUE(Q)

12 for each vAdj[u]

13 do if color[v] = WHITE // discover the undiscovered adjacentvertices

14 then color[v] GRAY // enqueued whenever painted

gray15 d[v] d[u] + 1

16 [v] u

17 ENQUEUE(Q, v)18 color[u] BLACK // painted black whenever dequeued

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Breadth-First Search: Example

g

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g

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r s t u

v w x y


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g

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0

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g

g

r s t u

v w x y

sQ:


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1

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r s t u

v w x y

wQ: r


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1

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r s t u

v w x y

rQ: t x


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1

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g

r s t u

v w x y

Q: t x v


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1

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r s t u

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Q: v u y


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1

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r s t u

v w x y

Q: u y


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1

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r s t u

v w x y

Q: y


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1

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v w x y

Q:


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Breadth first search - analysis


Enqueue and Dequeue happen only once

for each node. - O(V).

Sum of the lengths of adjacency lists (E) (for adirected graph) and (2E) (for an undirected graph)

Initialization overhead - O(V)

Total runtime -O(V+E)

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Breadth-First Search: Properties

BFS calculates theshortest-path distance to

the source node

Shortest-path distance

H

(s,v) = minimum numberof edges from s to v, org if v not reachable from s

Proof given in the book

BFS builds breadth-first tree, in which paths to

root represent shortest paths in G

Thus can use BFS to calculate shortest path from

one vertex to another in O(V+E) time


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Breadth-First Search: Properties

contd

Lemma:1.1-Let G=(V,E) be a directed or undirectedgraph and let s V be an arbitary vertex.Then for any

edge (u,v) E, H(s,v) H(s,u)+1 Lemma:1.2-Let G=(V,E) be a directed or undirected

and suppose that BFS is run on G from a given sourc

vertex s V.Then upon termination ,for each vertex

vV,then value d[v] computed by BFS satisfiesd[v] H(s,v)


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Depth-First Search

It searches deeper the graph when possible.

Starts at the selected node and explores as far aspossible along each branch before backtracking.

Vertices go through white, gray and black stages ofcolor.

White initially

Gray when discovered first

Black when finished i.e. the adjacency list of thevertex is completely examined.


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Depth-First Search contd.

Also records timestamps for each vertex

d[v] when the vertex is first discovered

f[v] when the vertex is finished

T hese timestamps are integers between 1 and2V,since there is one discovery event and onefinishing event for each vertex in V.So forevery vertx v ,d[v]

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Depth-First Search: algorithm

DFS(G)

{

for each vertex u G->V

{

u->color = WHITE;}

time = 0;


{

if (u->color == WHITE)

DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]

{

if (v->color == WHITE)

DFS_Visit(v);

}

u->color = BLACK;time = time+1;

u->f = time;

}


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DFS(G)

1 for each vertex u V [G]2 do color[u] WHITE // color all

vertices white, set their parents NIL

3 [u] NIL4 time 0 // zero out time5 for each vertex u V [G] // call only for

unexplored vertices

6 do if color[u] = WHITE // this may resultin multiple sources

7 then DFS-VISIT(u)


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contd..

DFS-VISIT(u)

1 color[u] GRAY White vertex u has just beendiscovered.

2 time time +13 d[u] time // record the discovery time4 for each v Adj[u] Explore edge(u, v).

5 do if color[v] = WHITE

6 then [v] u // set the parent value

7 DFS-VISIT(v) // recursive call

8 color[u] BLACK Blacken u; it is finished.

9 f [u]

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Depth-First Search: The Code

DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);

}


u->f = time;

}

What does u->f [email protected] 10/11/2009

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DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);

}


u->f = time;

}

Willall

vertic

es eventuall

y becolored b


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DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);

}


u->f = time;

}

What w

illbe t

he runn

ing t


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DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);

}


u->f = time;

}

Running time: O(n2) because callDFS_Visit on each vertex,

and the loop over Adj[]can run as many as |V| [email protected] 10/11/2009

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DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]{


DFS_Visit(v);

}


u->f = time;

}

BUT, there is actually a tighter bound.

How many times willDFS_Visit() actually be [email protected] 10/11/2009

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DFS(G)

{


{

u->color = WHITE;}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;

u->d = time;

for each v u->Adj[]{


DFS_Visit(v);

}


u->f = time;

}

So, running time of DFS = O(V+E)[email protected] 10/11/2009

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DFS Example

source

vertex


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DFS Example

1 | | |

|||

| |

source

vertexd f


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DFS Example

1 | | |

|||

2 | |

source

vertexd f


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DFS Example

1 | | |

||3 |

2 | |

source

vertexd f


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DFS Example

1 | | |

||3 | 4

2 | |

source

vertexd f


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DFS Example

1 | | |

|5 | 63 | 4

2 | |

source

vertexd f


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DFS Example

1 | 8 | |

|5 | 63 | 4

2 | 7 |

source

vertexd f


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DFS Example

1 | 8 | |

|5 | 63 | 4

2 | 7 |

source

vertexd f


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DFS Example

1 | 8 | |

|5 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 | 8 |11 |

|5 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 |12 8 |11 |

|5 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 |12 8 |11 13|

|5 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 |12 8 |11 13|

14|5 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 |12 8 |11 13|

14|155 | 63 | 4

2 | 7 9 |10

source

vertexd f


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DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

source

vertexd f


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Properties of DFS

DFS produces a forest.

Important property of DFS is that discovery and

finishing times haveparenthesis structure.

Parenthesis theorem:-

In any DFS of a(directed or undirected)graph

G=(V,E),for any two vertices u and v extactly one of

the following three conditions holds:

Case-1:The intervals [d[u],f[u]] and [d[v],f[v]] entirelydisjoint and neither u nor v is a descendentof the

other in DFS forest.

[email protected] y 10/11/2009

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Properties of DFS contd...

Case 2:d[u]

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DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

source

vertexd f

Tree edges


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DFS: Kinds 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

Not a tree edge, though

From grey node to black node


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DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

source

vertexd f

Tree edges Backedges Forward edges


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DFS: Kinds 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; mostalgorithms dont distinguish forward & cross

edges.


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DFS: Kinds Of Edges

Thm 23.9: If G is undirected, a DFS producesonly tree and back edges

Proof by contradiction:

Assume theres a forward edge

But F? edge must actually be a

back edge (why?)

source

F?


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DFS: Kinds Of Edges

Thm 23.9: If G is undirected, a DFS producesonly tree and back edges

Proof by contradiction:

Assume theres a cross edge

But C? edge cannot be cross:

must be explored from one of the

vertices it connects, becoming a tree

vertex, before other vertex is explored

So in fact the picture is wrongboth

lower tree edges cannot in fact be

tree edges

source

C?


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DFS And Graph Cycles

Thm: 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 [email protected] 10/11/2009

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DFS And Cycles

How would you modify the code to detect cycles?

DFS(G)

{


{

u->color = WHITE;

}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);}

u->color = BLACK;

time = time+1;

u->f = time;

}[email protected] 10/11/2009

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DFS And Cycles

What willbe the running time?

DFS(G)

{


{

u->color = WHITE;

}

time = 0;


{


DFS_Visit(u);

}

}

DFS_Visit(u)

{

u->color = GREY;

time = time+1;u->d = time;

for each v u->Adj[]

{


DFS_Visit(v);}

u->color = BLACK;

time = time+1;

u->f = time;

}[email protected] 10/11/2009

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DFS And Cycles

What willbe the running time?

A: O(V+E)

We can actually determine if cycles exist in

O(V) time:

In an undirected acyclic forest, |E| e |V| - 1 So count the edges: if ever see |V| distinct edges,

must have seen a back edge along the way

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BFSand DFS