Chapter 13: Graphs

GRAPH ADT

TOPOLOGICAL SORT

CS 240 1

SHORTEST PATH

MAXIMUM FLOW

MINIMUM SPANNING TREE

DEPTH-FIRST SEARCH

P AND NP PROBLEMS

CS 240 2

Graph DefinitionA graph G = (V, E) consists of a set of vertices, V, and a set of edges, E, each of which is a pair of vertices. If the edges are ordered pairs of vertices, then the graph is directed.

Undirected, unweighted, unconnected, loopless graph

(length of longest simple

path: 2)

Directed, unweighted,

acyclic, weakly

connected, loopless graph

(length of longest simple

path: 6)

Directed, unweighted,

cyclic, strongly

connected, loopless graph

(length of longest simple

cycle: 7)

Undirected, unweighted, connected graph, with

loops(length of

longest simple path: 4)

Directed,weighted,

cyclic, weakly connected,

loopless graph(weight of

longest simple cycle: 27)

3 4

2

4

23

5

6

3

6 5

Graphs are relatively unstructured, but they provide a platform for solving some of the most sophisticated problems in computer science.

CS 240 3

Graph RepresentationsAdjacency Matrix:

A B

D

H

E

G

F

C

ABCDEFGH

A B C D E F G H11000010

10110000

01000000

01000000

00001010

00000010

10001101

00000010

A B

D

H

E

G

F

C

ABCDEFGH

A B C D E F G H00000100

10100000

00001000

11000000

00010001

00010000

00010100

00000010

A B

D

H

E

G

F

C3 4

2

4

23

5

6

3

6 5

ABCDEFGH

A B C D E F G H2

35

4

6

62

345

3

The Problem: Most graphs are sparse (i.e., most vertex pairs are not edges), so the memory requirement is excessive (i.e., V2).

CS 240 4

Graph RepresentationsAdjacency List:

ABCDEFGH

AABBEGAG

BC

G

E

GD

F H

A B

D

H

E

G

F

C

ABCDEFGH

B

B 3H

G

6

2

4

C

A

3

53

5

4

2

D

E

G

G

6E

A B

D

H

E

G

F

C3

4

2

4

2

3

5

6

3

65

ABCDEFGH

B

EB

GCA

E

D

F

G

D

H

A B

D

H

E

G

F

C

CS 240 5

Topological SortA topological sort of an acyclic directed graph orders the vertices so that if there is a path from vertex u to vertex v, then vertex v appears after vertex u in the ordering.

One topological sort of the course prerequisite graph at right:

MATH 120, CS 111, MATH 125, CS 140, MATH 150, CS 150, MATH 224, CS 240, CS 234, MATH 152, CS 312, ECE 282, STAT 380, CS 321, MATH 250, MATH 321, CS 314, MATH 423, CS 325, CS 340, CS 425, ECE 381, CS 434, ECE 482, CS 330, CS 382, CS 423, CS 438, CS 454, CS 447, CS 499, CS 482, CS 456, ECE 483

MATH 125 CS 140

CS 150MATH 152

CS 240 CS 234

MATH 224

CS 340

STAT 380

CS 438

CS 454

CS 456

CS 325

CS 499

CS 434

CS 312

CS 482

CS 330

CS 423CS 447

MATH 150

CS 314

CS 425

CS 321

ECE 483

ECE 282ECE 381

ECE 482

CS 111

MATH 120

MATH 250

MATH 321

MATH 423

CS 382

CS 240 6

Topological Sort Algorithm Place all indegree-zero vertices

in a list. While the list is non-empty:

– Output an element v in the indegree-zero list.

– For each vertex w with (v,w) in the edge set E:• Decrement the indegree of w

by one.• Place w in the indegree-zero

list if its new indegree is 0.

G

A

H I

E F D

CB

013114130

03104030

021

3020

020

201

10

201

0

101

0

1

0

1

0

Indegree of A:Indegree of B:Indegree of C:Indegree of D:Indegree of E:Indegree of F:

Indegree of G:Indegree of H:Indegree of I:

Output: A E I B D G C H

0

F

CS 240 7

Shortest Path ProblemHow do you find the shortest path from a specified vertex to every other vertex in the directed graph?

Game Pathfinding

Robotic Maneuvering Web-Based Mapping

NetworkRouting

CS 240 8

G

A

H I

E F

CB

D(0,-)

(0,-)

(1,A)

(1,A)

(1,A)

(2,B)

(2,E)

(2,E)

(3,F)

G

A

H I

E F

CB

D

(0,-)

(1,A)

(1,A)

(1,A)

G

A

H I

E F

CB

D (0,-)

(1,A)

(1,A)

(1,A)

(2,B)

(2,E)

(2,E)

G

A

H I

E F

CB

D

(0,-)

(1,A)

(1,A)

(1,A)

(2,B)

(2,E)

(2,E)

(3,F)

(4,D)

G

A

H I

E F

CB

D

Case 1: Unweighted GraphsUse a breadth-first search, i.e., starting at the specified vertex, mark each adjacent vertex with its distance and its predecessor, until all vertices are marked. Algorithm’s time complexity: O(E+V).

Shortest Path Algorithms

CS 240 9

Case 2: Weighted Graphs With No Negative WeightsUse Dijkstra’s Algorithm, i.e., starting at the specified vertex, finalize the unfinalized vertex whose current cost is minimal, and update each vertex adjacent to the finalized vertex with its (possibly revised) cost and predecessor, until all vertices are finalized. Algorithm’s time complexity: O(E+V2) = O(V2).

G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(,-)

(,-)

(,-)

(,-)

(,-) (,-

)

(,-)

(,-)

(0,-)

(7,A)

(26,B)

(6,A)

(19,E)

(8,A) (22,E

)

(,-)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(27,E)

(6,A)

(19,E)

(8,A) (22,E

)

(,-)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(26,B)

(6,A)

(18,H)

(8,A) (16,G)

(,-)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(,-)

(6,A)

(,-)

(8,A) (,-)

(,-)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(26,B)

(6,A)

(19,E)

(8,A) (16,G)

(,-)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

CS 240 10

Note that this algorithm would not work for graphs with negative weights, since a vertex cannot be finalized when there might be some negative weight in the graph which would reduce a particular path’s cost.

(0,-)

(7,A)

(25,D)

(6,A)

(18,H)

(8,A) (16,G)

(20,F)

(23,D)

G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(26,B)

(6,A)

(18,H)

(8,A) (16,G)

(20,F)

(,-)G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(25,D)

(6,A)

(18,H)

(8,A) (16,G)

(20,F)

(23,D)

G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

(0,-)

(7,A)

(25,D)

(6,A)

(18,H)

(8,A) (16,G)

(20,F)

(23,D)

G

A

H I

E F

CB

D13

7

6

8

19

21

168

2

25

3

4 3

CS 240 11

Case 3: Weighted Graphs With Negative WeightsUse a variation of Dijkstra’s Algorithm without using the concept of vertex finalization, i.e., starting at the specified vertex, update each vertex adjacent to the current vertex with its (possibly revised) cost and predecessor, placing each revised vertex in a queue. Continue until the queue is empty. Algorithm’s time complexity: O(EV).

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(,-)

(,-)

(,-)

(,-)

(,-) (,-

)

(,-)

(,-)

(0,-)

(13,A)

(28,B)

(25,A)

(20,B)

(26,A) (,-

)

(,-)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(,-)

(25,A)

(23,E)

(26,A) (,-

)

(,-)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(28,B)

(25,A)

(20,B)

(26,A) (31,F)

(26,F)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(,-)

(25,A)

(,-)

(26,A) (,-

)

(,-)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(28,B)

(25,A)

(20,B)

(26,A) (32,G

)

(,-)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

CS 240 12

Note that this algorithm would not work for graphs with negative-cost cycles. For example, if edge IH in the above example had cost -5 instead of cost -3, then the algorithm would loop indefinitely.

(0,-)

(13,A)

(28,B)

(25,A)

(20,B)

(26,A) (31,F)

(26,F)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(17,D)

(22,H)

(20,B)

(26,A) (31,F)

(26,F)

(34,D)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(17,D)

(22,H)

(20,B)

(26,A) (31,F)

(26,F)

(34,D)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(17,D)

(22,H)

(20,B)

(26,A) (31,F)

(26,F)

(34,D)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(28,B)

(22,H)

(20,B)

(26,A) (31,F)

(26,F)

(,-)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

(0,-)

(13,A)

(17,D)

(22,H)

(20,B)

(26,A) (31,F)

(26,F)

(34,D)

G

A

H I

E F

CB

D

1325

26

15

-27

-9

6

11

6-9

-3

-4 8

CS 240 13

Maximum Flow Problem

Airline Routing

Image Segmentation

When paths have limited capacity, how do you maximize the overall benefit of the routes that are taken?

Network Capacity Planning

CS 240 14

Maximum Flow AlgorithmAssume that the directed graph G = (V, E) has edge capacities assigned to each edge, and that two vertices s and t have been designated the source and sink nodes, respectively.We wish to maximize the “flow” from s to t by determining how much of each edge’s capacity can be used so that at each vertex, the total incoming flow equals the total outgoing flow.This problem relates to such practical applications as Internet routing and automobile traffic control.

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

14 30

12

21

In the graph at left, for instance, the total of the incoming capacities for node C is 40 while the total of the outgoing capacities for node C is 35. Obviously, the maximum flow for this graph will have to “waste” some of the incoming capacity at node C. Conversely, node B’s outgoing capacity exceeds its incoming capacity by 5, so some of its outgoing capacity will have to be wasted.

CS 240 15

To find a maximum flow, keep track of a flow graph and a residual graph, which keep track of which paths have been added to the flow.Keep choosing paths which yield maximal increases to the flow; add these paths to the flow graph, subtract them from the residual graph, and add their reverse to the residual graph.

Original Graph Flow Graph Residual Graph

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

0

0

0

0

0

0

0

0

0

0

0

0

00

0

0

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

21

0

0

0

0

21

0

0

0

21

0

0

00

0

0

G

A

I

E F

CB

D

H

14

27

18

19

17

0

16

8

12

2

35

15

1430

12

21

21

21

21

CS 240 16

Original Graph Flow Graph Residual Graph

G

A

I

E F

CB

D

H

12

18

12

14

10

18

19

0

0

16

8

2

15

1430

4

21

21

211717

17

17

G

A

I

E F

CB

D

H

4

1717

0

10

18

5

0

0

16

8

12

2

15

1430

12

4

3521

21

17

31

14

G

A

I

E F

CB

D

H

0

12

012

0

10

6

5

0

4

8

2

4

3

14 1

8

4

35 2

1

21

17

17

17

31

14

12

12

12

12

23

3527

18

19

17

21

16

8

35

15

14 3

0

21

12

12

G

A

I

E F

CB

D

H

21

17

0

0

17

21

0

0

0

21

17

0

00

0

17

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

35

17

12

14

17

21

12

0

0

21

31

12

012

12

17

G

A

I

E F

CB

D

H

3527

18

19

17

21

16

8

12

23

35

15

14 3

0

12

21

G

A

I

E F

CB

D

H

35

17

0

14

17

21

0

0

0

21

31

0

00

0

17

G

A

I

E F

CB

D

H

CS 240 17

Original Graph Flow Graph Residual Graph

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

35

25

12

14

17

21

12

8

8

21

31

12

620

4

17

G

A

I

E F

CB

D

H

4 4

44

25 4 8

0

2

6

5

00

0

2

3

610

8

35 2

1

21

17

17

31

14

12

12

12

20

8

8

G

A

I

E F

CB

D

H

35

27

18

19

17

21

16

8

12

23

35

15

1430

12

21

G

A

I

E F

CB

D

H

35

25

16

14

17

21

16

8

12

21

31

12

1024

8

17

G

A

I

E F

CB

D

H

04

17

8 12

0

2

2

5

00

0

0

2

4

3

26

4

35 2

1

21

17

25

31

14

16

8

12

24

8

12

Thus, the maximum flow for the graph is 76.

CS 240 18

Minimum Spanning TreesAn alternative to Kruskal’s Algorithm (which develops a minimum spanning tree via forests) is Prim’s Algorithm, which is just a variation of Dijkstra’s Algorithm. Starting with a minimum-cost edge, it builds the tree by adding minimum-cost edges as long as they don’t create cycles. Like Kruskal’s, Prim’s Algorithm is O(ElogV)

B

A D

C

E

F G

H

B

A D

C

E

F G

H3

B

A D

C

E

F G

H5

4

3

B

A D

C

E

F G

H5

4

6

4

3

3

B

A D

C

E

F G

H5

4

6

3

MINIMUM SPANNING TREE

B

A D

C

E

F G

H5

4

6

4

7

3

3

B

A D

C

E

F G

H4

3

B

A D

C

E

F G

H5

4

6

4

3

ORIGINAL GRAPH

B

A D

C

E

F G

H5

47

6

7 4

7

8

6 93

83

CS 240 19

Depth-First SearchA convenient means to traverse a graph is to use a depth-first search, which recursively visits and marks the vertices until all of them have been traversed.A B C

F G H

D E

Original Graph

A B C

F G H

D E

Depth-First Search(Solid lines are part of

depth-first spanning tree; dashed lines are visits to

previously marked vertices)

Such a traversal provides a means by which several significant features of a graph can be determined.

A

B

C

F

G

HD

E

Depth-First Spanning Tree

CS 240 20

Depth-First Search Application: Articulation PointsA vertex v in a graph G is called an articulation point if its removal from G would cause the graph to be disconnected.Such vertices would be considered “critical” in applications like networks, where articulation points are the only means of communication between different portions of the network.A depth-first search can be used to find all of a graph’s articulation points.The only articulation points are the root (if it has more than one child) and any other node v with a child whose “low” number is at least as large as v’s “visit” number. (In this example: nodes B, C, E, and G.)

Original Graph

B E H

C F

GA D J

I

Depth-First Search (with nodes numbered as they’re

visited)

2 6 8

4 5

71 3 10

9

Nodes also marked with lowest-numbered vertex reachable via

zero or more tree edges, followed by at most one back edge

2/1 6/6 8/6

4/1 5/5

7/61/1 3/1 10/7

9/7

CS 240 21

Depth-First Search Application: Euler CircuitsAn Euler circuit of a graph G is a cycle that visits every edge exactly once.Such a cycle could be useful in applications like networks, where it could provide an efficient means of testing whether each network link is up.A depth-first search can be used to find an Euler circuit of a graph, if one exists. (Note: An Euler circuit exists only if the graph is connected and every vertex has even degree.)

Original Graph

B C D

H I

FA G E

J

Splicing the first two cycles yields cycle A(BCDFB)GHA.Splicing this cycle with the third cycle yields ABCD(FHIF)BGHA.Splicing this cycle with the fourth cycle yields ABCD(FEJF)HIFBGHANote that this traversal takes O(E+V) time.

After Removing First DFS Cycle: BCDFB

B C D

H I

FA G E

JAfter Removing Second

DFS Cycle: ABGHA

B C D

H I

FA G E

J

After Removing Third DFS Cycle: FHIF

B C D

H I

FA G E

JAfter Removing Fourth

DFS Cycle: FEJF

B C D

H I

FA G E

J

CS 240 22

P and NP ProblemsA problem is said to be a P problem if it can be solved with a deterministic, polynomial-time algorithm. (Deterministic algorithms have each step clearly specified.)

Example: The Knapsack ProblemGiven a set of n valuable jewels J1, J2, …, Jn with respective weights w1, w2, …, wn, and respective prices p1, p2, …, pn, as well as a knapsack capable of supporting a total weight of M.

A nondeterministic polynomial-time solution:totalWorth = 0;totalWeight = 0;for (i=1; i<=n; i++){ b[i] = choice(TRUE,FALSE); if (b[i]==TRUE) { totalWorth += p[i]; totalWeight += w[i]; }}if ((totalWorth >= T) && (totalWeight <= M)) cout << “YAHOO! I’M RICH!”;else cout << “@#$&%!”;

A problem is said to be an NP problem if it can be solved with a nondeterministic, polynomial-time algorithm. In essence, at a critical point in the algorithm, a decision must be made, and it is assumed that a magical “choice” function always chooses correctly.

Problem: Is there a way to pack at least T dollars worth of jewels, without exceeding the weight capacity of the knapsack?(It’s not as easy as it sounds; three lightweight $1000 jewels might be preferable to one heavy $2500 jewel, for instance.)