csc 201: design and analysis of algorithms lecture # 18 graph algorithms mudasser naseer 1...

CSC 201: Design and Analysis of Algorithms

Lecture # 18

Graph Algorithms

Mudasser Naseer 1 04/21/23

Depth-First Search

● Depth-first search is another strategy for exploring a graph■ Explore “deeper” in the graph whenever possible■ Edges are explored out of the most recently

discovered vertex v that still has unexplored edges■ When all of v’s edges have been explored,

backtrack to the vertex from which v was discovered


Depth-First Search

● Vertices initially colored white● Then colored gray when discovered● Then black when finished


Depth-First Search: The Code


DFS(G)

{

1 for each vertex u G->V {

2 u->color = WHITE;

}

3 time = 0;


5 if (u->color == WHITE)

6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;

4 for each v u->Adj[] {

5 if (v->color == WHITE)

6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


What does u->d represent?Mudasser Naseer 5 04/21/23

DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


What does u->f represent?Mudasser Naseer 6 04/21/23

DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


Will all vertices eventually be colored black?Mudasser Naseer 7 04/21/23

DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


What will be the running time?Mudasser Naseer 8 04/21/23

DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


Running time: O(n2) because call DFS_Visit on each vertex, and the loop over Adj[] can run as many as |V| times


DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}


DFS(G)

{


2 u->color = WHITE;

}

3 time = 0;



6 DFS_Visit(u);

}

}

DFS_Visit(u)

{

1 u->color = GREY;

2 time = time+1;

3 u->d = time;



6 DFS_Visit(v);

}

7 u->color = BLACK;

8 time = time+1;

9 u->f = time;

}BUT, there is actually a tighter bound.

How many times will DFS_Visit() actually be called?Mudasser Naseer 10 04/21/23


DFS(G)

{

for each vertex u G->V {

u->color = WHITE;

}

time = 0;

for each vertex u G->V {

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;

}

So, running time of DFS = O(V+E)Mudasser Naseer 11 04/21/23

● What is the running time of DFS? The loops on lines 1–2 and lines 4–6 of DFS take time (V), exclusive of the time to execute the calls to DFS-VISIT.

● The procedure DFSVISIT is called exactly once for each vertex v V∈ , since DFS-VISIT is invoked only on white vertices and the first thing it does is paint the vertex gray. During an execution of DFS-VISIT(v), the loop on lines 4–6 is executed |Adj[v]| times.

Since

the total cost of executing lines 4–6 of DFS-VISIT is (E). The running time of DFS is therefore (V + E).

Vv

EvAdj )(][


DFS Example

sourcevertex


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

What is the structure of the grey vertices? What do they represent?


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


DFS: Kinds of 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○ Can tree edges form cycles? Why or why not?


DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

Tree edges


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

○ Encounter a grey vertex (grey to grey)


DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

Tree edges Back edges


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


DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

Tree edges Back edges Forward edges


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

○ From a grey node to a black node


DFS Example

1 |12 8 |11 13|16

14|155 | 63 | 4

2 | 7 9 |10

sourcevertex

d f

Tree edges Back edges Forward edges Cross edges


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; most algorithms don’t distinguish forward & cross


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 cycle


DFS And Cycles

● What will be the running time?● A: O(V+E)● We can actually determine if cycles exist in

O(V) time:■ In an undirected acyclic forest, |E| |V| - 1 ■ So count the edges: if ever see |V| distinct edges,

must have seen a back edge along the way


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

BB

CC

DD

EEFF

GG


TT


AA

BB

CC

DD

EEFF

GG


TT

TT


AA

BB

CC

DD

EEFF

GG


TT

TTBB


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB TT


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB TT

TT


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB TT

TTBB


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB TT

TTBBTT


AA

BB

CC

DD

EEFF

GG


TT

TTBBTT

BB

BB TT

TTBBTT

BB



AA

BB

CC

DD

EEFF

GG

TT

TTBBTT

BB

BB TT

TTBBTT

BB

Example: (continued)

DFS of Undirected Graph

DFS TreeDFS Tree

AA

BB

CC

DD

EEFF

GG

TT

TTBBTT

BB

BB TT

TTBBTT

BB

AA

CC

DD

EE

FF

GG

BB

BB

TT

TT

TT

TT

TT

TT

BB

BB

BB

BB

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

csc 201: design and analysis of algorithms lecture # 18 graph algorithms mudasser naseer 1...

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