MCA 520: Graph Theory - PowerPoint Presentation

Slide1

MCA 520: Graph Theory

Instructor

Neelima Gupta

ngupta@cs.du.ac.inSlide2

Table of Contents

Directed GraphsSlide3

Simple Digraph

May contain one loop at each vertex.

Distance: we say that a vertex y is at a distance d from a vertex x, if d is the length of a shortest path from x to y.Slide4

Motivation

Finite State Machine

For Example: a bulb controlled by two switches often called “three-way switch”.

Prediction Graphs

Functional GraphsSlide5

Underlying Graph

Weekly and Strongly Connected Components.Slide6

Subgraph

, Isomorphism, decomposition and Union are same as that in undirected graphs

Incidence and Adjacency MatricesSlide7

KernelSlide8

Theorem: Richardson

Every graph having no odd cycle contains a kernel.

Proof: First prove for strongly connected di-graphs D. Let y be any vertex in D. Let S be the set of all vertices at even distance from y.

Show that there is no edge between the vertices in S…left as exercise.

Let x be in V \ S. Then x is at odd distance say d_1 from y. Since D is strongly connected there is a path from x to y. Then the length of this path must be odd for else we have an odd walk and hence an odd cycle. Let x’ be the first vertex after x on this path. Then x’ must be at even distance from y for if there exists a (shorter) path of odd length from y

to x’,

then again we get an odd cycle…y to x’ to

y..since

x to y was odd, x’ to y is even and thus

y to x’ to

y is odd.Slide9

An example to show that this is not true if there is an odd cycle.Slide10

Proof continued by induction

Let n(D) =1 , single vertex graph, this itself is a kernel.

n(D) > 1, IH: true for a graph with < n(D) number of vertices. We’ll prove it is true for n(D). If D is strongly connected, we have already proved. So let D is not strongly connected. Then D contains a strongly connected component that has no outgoing edge. For

eg

. Consider D that is a simple directed path: v1 -> v2 -> v3 -> v4. Then D has 4 SCC, each consisting of a singleton and {v4} has no out going edge.

Since D’ is SCC, it has a kernel say S’. Remove S’ and its predecessors from the graph. Let D” be the remaining graph. By IH, D” has a kernel say S”.

Claim: S’ union S” is a kernel for DSlide11

Vertex Degrees

In-

deg

, Out-deg.

In-neighborhood/predecessor set

Out-neighborhood/successor set

Degree-Sum Result: Sum of

indegrees

= Sum of

Outdegrees

= number of edgesSlide12

Degree Sequence

Is a list of degree pairs (d

+

(v

i

) , d

-

(v

i

)).

Proposition: A list of pairs of non-negative integers is realizable as degree pairs of a digraph

iff

Σ

i

d

+

(v

i

)

=

Σ

i

d

-

(v

i

)

This is true when multiple edges are allowed

Proof: let m = sum. Consider m points and label

d

+

(v

i

)

points with label I and

d

-

(v

i

)

points with –j. For each point receiving label

i

and –j put an edge from I to j. Note that the resulting di-graph may contain loops and

muti

-edges…

.so not a simple di-graphSlide13

Characterization of DS for simple graphs (recall loops are allowed)

Split

of a digraph:

A constructive algorithm similar to the one for undirected graphs can be given

to test whether

a list

of pairs of non-negative integers is realizable as degree pairs of a digraph

or not. The algorithm uses split of a digraph…

…assignment.Slide14

Eulerian Di-graphs

Definitions of

Trail, Walk, Circuit/Cycle

remain the same with directions on them.

A di-graph is

E

ulerian

if it has an

Eulerian

Cycle.

Lemma: If G is a di-graph with d

+

(G) > 1 (or d

-

(G) > 1

) , then it contains a cycle.

A digraph G is

Eulerian

iff

d

+

(

v)

=

d

-

(

v) for all v and the underlying graph has at most one non-trivial component. Slide15

Orientations and Tournaments

2

n^2

simple di-graphs with n vertices – n

2

ordered pairs including loops, each is present or absent…..(0,1) choices

Orientation: for each edge in

a simple undirected graph G (no loops),

choose the direction ….three choices (0,+1,-1)…so …

3

n

C

2

Tournament: Orientation of a complete simple

simple graph

…no loops, two choices (+1, -1

) …so

…2

nC2

Slide16

Representing n-teams league match by an orientation of Kn

Let (

u,v

) be an edge if u wins.

Score(u) = number of matches u wins =

outdeg

(u).

Thus out-degree sequence is also called score sequence in a tournament.

Indegree

can be determined from

outdegree

(how?)Slide17

How to define the winner?

Suppose we define winner to be the team that has not lost to any other team…..there may be no winner, we may have a cycle.

Def2: that has won against maximum number of teams. Slide18

King of a tournament

Claim

: A team(/vertex) with maximum score (/out-degree) beats every other team either directly or by a path of length 2.

Definition:

In a di-graph, a

king

is a vertex from which every vertex is reachable by a path of length at most 2.

Theorem: Every tournament has a king

.

Proof: here we’ll essentially prove our claim

.

Note:

There

may be several vertices with maximum out-degree i.e. maximum

score and there may still be no clear winner

.

Worst is that all the vertices may have same out-degree.

The only thing we can claim is that there is at least one winner

.