Graph Definitions and Applications

Graphs: Very Useful Abstractions

• Graphs apply to many diverse areas: social sciences, linguistics, physical sciences, communication engineering, and many more.

• Why? Because they are a useful abstraction.– Abstractions capture the essence of a real-world problem.– Abstractions discard irrelevant details and let us focus on

the essence of the problem to be solved.– Abstractions lend themselves to precise (mathematical)

specification.– Mathematically defined abstractions lead to correct

computer implementations.

Graphs in Programming

• Testing: paths in programs:

• Call graphs in programs• Software products consist of modules that invoke

each other• Deadlock in operating systems

…

T F T

FT

F

Sequence If-then-else While Repeat-until Case

Graphs in Transportation& Product Delivery

• Airline connections: getting from here to there

• Highways with mileage between cities• Traveling salesman problem: visit all cities

with minimal distance• Traffic networks studied by transportation

and city planners• Pipelines for delivering water or gas

Graphs in Networks• Social networks

– Friend of a friend– Co-collaborators

• Computer networks– Connect all, minimal number of connections– (Minimal) spanning trees, LANs, WANs

• Scheduling networks– Workflow from source to sink– Critical path, slack time– Earliest starting time, latest completion time

Problems that Don’t Initially Look Like Graph Problems

Have you ever had to work your way through a maze of interpreters at an international conference?

Swedish and Indonesian delegates wish to talk to each other, but among the sixteen interpreters, not one speaks both Swedish and Indonesian. A way to solve this problem is to form a chain of interpreters.

There are, however, only four booths; interpreters must be in booths, and each booth can contain only one interpreter.

Problem: Does a solution exist? If so, which four interpreters will you place in the booths?

Interpreters at the Great World Conference

Swedish—(

Booth #1 Booth #2 Booth #3 Booth #4

)—Indonesian

(1)PortugueseIndonesian

(2)Polish

EnglishGerman

(3)Italian

NorwegianEnglish

(4)KoreanTurkish

(5)EnglishPolish

(6)SwedishTurkish

(7)SpanishChinese

Japanese

(8)French

Portuguese

(9)Dutch

German

(10)Swedish

Norwegian

(11)Japanese

IndonesianRussian

(12)DutchFrench

ChineseRussian

Portuguese

(13)PolishItalian

Swedish

(14)GermanChineseKorean

(15)Dutch

SpanishIndonesian

(16)FrenchEnglish

Table of Spoken LanguagesA 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B

Chinese 1 1 1

Dutch 1 1 1 1

English 1 1 1 1

French 1 1 1

German 1 1 1

Indonesian 1 1 1 1

Italian 1 1

Japanese 1 1

Korean 1 1

Norwegian 1 1

Polish 1 1 1

Portuguese 1 1 1

Russian 1 1

Spanish 1 1

Swedish 1 1 1 1

Turkish 1 1

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B

A 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0

1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1

2 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0

3 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0

4 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0

5 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0

6 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0

7 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0

8 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0

9 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0

10 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0

11 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1

12 0 1 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1 0

13 1 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0

14 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 0 0 0

15 0 1 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 1

16 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0

B 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0

“Can Speak to” Matrix

Graph of “Can Speak to” Matrix

10 3Norwegian English

2

Eng

lish

English

13Polish

Swed

ish

6

Swedish

Sw

edis

h

5

Polish

Polis

h

English

Eng

lish

14

Ger

man

9German

Germ

an

4

Turkish

Korean

7

Dutch

Dutch

DutchChinese

French

Chinese Chinese

Span

ish

8

1

11

Port

ugue

se

Portuguese

Portuguese

French

Japa

nese

Indonesian

Indonesian

Rus

sian

Indo

nesi

an

Italian

Swedish

Indonesian15

12

A B

Indonesian

Indo

nesi

an

Swedish

Swedish

16

French

Graph Definition• Hyp: edges either all ordered or all unordered

G = (V, E), whereV is a set of vertices/nodes, andE is a relation on VxV

(E is a set of all-ordered or all-unordered pairs)• Example

1

2 3

V = {1, 2, 3}

E = {(1, 2), (2, 3), (3, 3)}

A Few Special Graphs• H is a subgraph of G if VH is a subset of VG and EH

is a subset of EG

• A graph G is complete if all nodes in G are connected to all other nodes in G (often denoted K1, …, Kn)

• A graph G is planar if it can be drawn (in 2D) with no crossing lines

K5

B3

A graph is planar if and only if it does not contain either K5 or B3 as a

subgraph.

Adjacency Matrix• Definition: Let G = (V, E) be a simple digraph. Let V = {v1,

v2,…vn} be the vertices (nodes). Order the vertices from v1 to vn. The n n matrix A whose elements are given by

aij = 1, if (vi, vj) E 0, otherwise

is the adjacency matrix of the graph G.

• Example:

{

1 2

3

4 1010v4

1000v3

1100v2

0010v1

v4v3v2v1

A =

Powers of Adjacency Matrices

• Powers of adjacency matrices: A2, A3, …• Can compute powers

– Using ordinary arithmetic:

– Using Boolean arithmetic:

aij = k=1

n

aikakj

aij = k=1

n

aikakj

1 2

3

4

Powers Using Ordinary Arithmetic

10104

10003

11002

00101

4321

A =

21104

10103

20102

11001

4321

A2 =

41204

21103

31202

20101

4321

A3 =

<2,3,4> <2,4,4>

<4,2,4> <4,4,4>

<1,2,3,4> <1,2,4,4>

<2,4,2,4> <2,4,4,4> <2,3,4,4>

<3,4,2,4> <3,4,4,4>

<4,2,3,4> <4,2,4,4> <4,4,2,4> <4,4,4,4>

<1,2,4>

<3,4,4>

Powers Using Ordinary Arithmetic (continued…)

The element in the ith row and the jth column of An is equal to the number of paths of length n from the ith node to the jth node.

72404

41203

62302

31201

4321

A4 =

<1,2,3,4,4> <1,2,4,2,4> <1,2,4,4,4>

<2,4,2,4,4> <2,4,4,4,4> <2,3,4,2,4> <2,3,4,4,4> <2,4,2,3,4> <2,4,4,2,4><3,4,4,4,4> <3,4,2,4,4> <3,4,4,2,4> <3,4,2,3,4>

<4,4,4,4,4> <4,4,4,2,4> <4,4,2,4,4> <4,2,4,4,4><4,2,4,2,4> <4,2,3,4,4> <4,4,2,3,4>

1 2

3

4

Powers Using Ordinary Arithmetic (continued…)

72404

41203

62302

31201

4321

10104

10003

11002

00101

4321

A =

21104

10103

20102

11001

4321

A2 =

41204

21103

31202

20101

4321

A3 =

A1 + A2 = # of ways in 2 or fewer

A1 + A2 + A3 = # of ways in 3 or fewer

A1 + A2 + A3 + A4 = # of ways in 4 or fewer

0 + 1 + 2 =

1 + 2 + 4 =

Powers Using Boolean Arithmetic• Using Boolean arithmetic, we compute reachability.• We can compute the paths of length 2 by A2, 3 by A3… and of

length n by An. Since the longest possible path we care about for reachability is n, we obtain all paths by

A A2 A3 … An.• The reachability matrix is the transitive closure of A.• Algorithm:

R = A;for (i = 2 to n){ compute Ai; R = R Ai;}O(n4)

(n-1)nnn

for (j = 1 to n){ for (k = 1 to n) { ai

jk = 0;

for (m = 1 to n) { ai

jk = aijk (ai-1

jm a1mk);

} }}

1 2

3

4

Reachability

10104

10003

11002

00101

4321

A =

11104

10103

11102

11101

4321

A A2 =

11104

11103

11102

11101

4321

A A2 A3 =

11104

11103

11102

11101

4321

A A2 A3 A4 =

1st iteration 2nd iteration 3rd iteration

Algorithms: Proof of Correctness• The algorithm terminates

– No loops: obvious– Loops: descending sequence on well-founded poset

(intuitively, fixed start and end place)

• The algorithm produces the correct result– No loops: Reason that steps lead to the desired conclusion– Loops:

• Find an appropriate loop invariant a T/F condition that stays the same as the loop executes, is based on the number of iterations, and leads to the conclusion we need.

• Prove by induction that the loop invariant holds as the loop executes.

Loop Invariant Example: A[i]i=1

n

sum = 0 for i = 1 to n sum = sum + A[i]Loop invariant: After k iterations, sum = A[1] + A[2] + … + A[k].

Induction proof:

Basis: k = 0 (initialization, before the loop begins), sum = 0, which is the sum after 0 iterations. (Or, we could start with k=1: After 1 iteration, sum = A[1].)

Induction: By the induction hypothesis, after k iterations, sum = A[1] + A[2] + … + A[k]. In the next iteration we add A[k+1]. Thus, after k+1 iterations, sum = A[1] + A[2] + … + A[k+1].

Loop Invariant Example: Selection Sort-- A is an array of n distinct integers for i = 1 to n

find location j of the smallest integer in A[i] … A[n] swap A[i] and A[j]

Loop invariant:

After k iterations, A[1], …, A[k] are sorted A[k] < all of A[k+1], …, A[n].

Induction proof:

Basis: k = 0 (initialization, before the loop begins), A is an array of n distinct integers which may or may not be sorted. (Or, we could start with k=1: After 1 iteration A[1] is sorted A[1] < all of A[1+1], …, A[n].)

Induction: By the induction hypothesis, after k iterations, A[1], …, A[k] are sorted A[k] < all of A[k+1], …, A[n]. Then since A[k] < all of A[k+1], …, A[n], and we find the smallest integer in A[k+1], …, A[n] to swap with A[k+1], A[1], …, A[k+1] are sorted A[k+1] < all of A[(k+1)+1], …, A[n].

1 2 3 9 6 5 8

Proof of Correctness for Reachability

• Loop invariant: After z outer loops, axy = 1 iff y is reachable from x along a path of length z+1 or less from x to y.

• Note: the loop terminates after n1 outer loops since i runs from 2 to n. Thus, when the algorithm terminates, if the loop invariant holds, axy = 1 iff y is reachable from x along a path of length (n1)+1 = n or less from x to y, and thus is reachable since n is the longest a path can be (that starts at a node N, visits all other nodes once, and comes back to N).

• Inductive proof– Basis: z=0 (i.e., before entering the outer loop), R = A and axy is 1 iff there is a path from x to y;

thus axy = 1 iff y is reachable from x along a path of length 0+1 = 1 or less from x to y.– Induction: By the induction hypothesis we have that after z outer loops, axy = 1 iff y is

reachable from x along a path of length z+1 or less from x to y. In the z+1st iteration we compute A(z+2), in which axy = 1 iff there is a path of length z+1 from x to some node q and an edge from q to y, i.e. a path of length z+2 from x to y, and we add these reachable possibilities for paths of length z+2 to the result R. Hence, after z+1 outer loops, axy = 1 iff y is reachable from x along a path of length (z+1)+1 = z+2 or less from x to y.

R = A;for (i = 2 to n){ compute Ai; R = R Ai;}

for (j = 1 to n){ for (k = 1 to n) { ai

jk = 0;

for (m = 1 to n) { ai

jk = aijk (ai-1

jm a1mk);

} }}