MCA 202: Discrete Mathematics Instructor Neelima Gupta [email protected]

MCA 202: Discrete Mathematics

InstructorNeelima Gupta

[email protected]

Thanks: Meghna 22 and Mrityunjaya 23 (MCA 202)

Introduction to Graphs


Map of Highways connecting Cities

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Two way Highways --- Undirected Graph


Map of One Way Highways connecting Cities

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One way Highways --- Directed Graph


Some of the Qs we often want to answer

• Is there a path from city ‘a’ to city ‘b’?• Is there a path to ‘c’ from ‘a’ via ’b’?• How can I visit all the cities and come back to

my original city without traveling any sector twice. (Euler Tour)

• Can I visit all the cities and come back to my original city without traveling any city twice? (Hamiltonian Circuit)


Map of Highways with cost of traveling

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Qs we may want to answer

• What is the cheapest path from city ‘a’ to city ‘b’? (Shortest Path Problem)

• Can I visit all the cities and come back to my original city without traveling any city twice? Can I get the cheapest such tour? (Traveling Salesman Problem)


Connecting the Cities in a Cheapest Way (Minimum Spanning Tree)

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Form a network of Computers in a cheapest way (MST)

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Graph

• A graph is a data structure to capture pairwise relationships amongst the objects.

• It is a collection of vertices and edges.• In the example of “map of highways

connecting cities”, cities can be represented by vertices and highways are binary relationships between cities which can be represented by edges.


Directed Graphs

V={v1, v2,…}

E V x V⊂E: V V

E {(u, v): u, v V }⊆ ∈E is a set of ordered pairs (u,v) of vertices from V representing an edge from u to v. v is stb the head of the arrow and u the tail.


Map of One Way Highways connecting Cities

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One way Highways --- Directed Graph


Undirected Graphs

V={v1, v2,…}

E is a collection of subsets of V of size 2. • However, with a slight abuse of notation we’ll continue

to write (u,v) for an undirected edge instead of {u,v} and whether (u,v) is an ordered pair or not will be clear from the context.

• To capture self-loops, subsets are allowed to be multisets.

• The edge (u, v) is stb incident on u and v. • u and v are stb adjacent to each other.



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Two way Highways --- Undirected Graph

Representation of Graphs

Thanks, Neeharika Chaudhary,Roll No.:26(MCA 2012)

2D Array Representation

• G=(V,E) v1 v2 ………………. vn

v1 v2

. . vn

G(Vi,Vj)=1 iff (Vi,Vj) Є E(G)


Undirected Graph

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MATRIX IS SYMMETRIC FOR AN UNDIRECTED GRAPH


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

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Directed Graph

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MATRIX IS NOT SYMMETRIC FOR A DIRECTED GRAPHThanks, Neeharika Chaudhary,Roll No.:26(MCA 2012)

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


Weighted Graphs

V={v1, v2,…}

W: E R

• Weights could be– cost of traveling – cost of connecting cities– cost of connecting computers

Thanks Neeraj Roll no-28(MCA 2012)

Adjacency List Representation


For Undirected graphs

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Size =2 X No. of Edges

=2|E|


• Size of Adjacency Matrix is theta (v2).• We know that |E| < O (v2). How?• When E<< v2 /2 i.e. when the graph is sparse,

then adjacency list representtaion saves a lot of space.


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For Directed Graphs

Χ

Size =|E|

∑ indeg(v)= ∑out deg(v)= |E|


Map of Highways with cost of traveling

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Multi - Graphs

• A graph in which there could be multiple edges between a pair of vertices.

• For directed graph, a multi-graph would be the one allowing multiple edges from a vertex u to another vertex v.


Map of Highways with more than one way to reach directly

• May be more than one highway.• One or more roadway, one or more railway,

one or more airway: each may have different cost in general


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Two way Highways --- Undirected Multi-Graph


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Two way Highways --- Undirected Multi-Graph

Paths and CircuitsPaths

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Thanks: Meghna 22, Mrityunjaya 23, Vibha Yadav 47 (MCA 202)

A path can be represented by a sequence of vertices or by a sequence of edges


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Thanks: Meghna 22 ,Mrityunjaya 23, Vibha Yadav 47 (MCA 202)



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Thanks: Meghna 22, Mrityunjaya 23, Vibha Yadav 47 (MCA 202)


Paths and CircuitsCircuits

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Paths and Circuits in a Directed GraphPaths

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A path can be represented by a sequence of vertices or by a sequence of directed edges


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Paths and Circuits in a Directed GraphCircuits

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Connected Graphs

• An undirected graph G is said to be connected if for every u and v V there is a path ∈between u and v.

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Example of a Disconnected Graph

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Directed Connected Graphs

• A directed graph G is said to be weakly connected if after removing the direction the graph is connected.

• A directed graph G is said to be strongly connected if for every u, v V(G) a directed ∈ ∃path from u to v.

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Example of a strongly connected graph


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Example of a graph that is weakly connected but not strongly connected


Degree of a Vertex

• In undirected graph G, degree(u) is defined as the number of edges incident on u.

• In a directed graph G, in-degree (u) is the number of edges with u as the arrow head.

• In a directed graph G, out-degree (v) is the number of edges with u as the tail of the arrow.

Vv

v)deg(For an undirected graph,

= 2|E|

where, V is the vertex set , E denotes the set of edges of the graph and deg(v) denotes the number of edges incident on vertex v.

Proof :Base case:For a graph with only one vertex,

deg( v1 )=0

which is true as the number of edges arezero and therefore, 2|E| =0.

Thanks Neelakshi Dang Roll No. 27 (MCA 2012)

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For a graph with two vertex,

Case(i) :when both vertex are disconnected,

= deg(v1)+ deg(v2)=0

which is true as the number of edges for the graph is zero and therefore, 2|E| =0.Case(ii):when both vertex are connected,

= deg(v1)+ deg(v2)= 1+1 =2

which is true as the number of edges for the graph is one therefore, 2|E| = 2(1) = 2 .


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v)deg(

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v)deg(

Induction Hypothesis: Let the hypothesis hold for every graph of vertex size n,

Let G be a graph with (n+1) vertices. Remove a vertex v* and all the edges incident on it, i.e. remove deg(v*) edges incident on v*, resulting in a graph G’ with n vertices.

→ = 2|E(G’)| by I.H. -


)'(

)(degGVv

vG'

1

Let deg(v*) = l , and let v1 , v2, … , vl be the neighbors of v* in G, thendegG’(vi) = degG(vi) – 1 , i = 1,2, … , l _and degG’(vi) = degG(vi) , v≠vi

→ |E(G’)| = |E(G)| - l -

From and , + = 2|E(G’)|

→ = 2|E(G’)|+l -


2

3

1 2

)'(

)(degGVv

vG

liivv

vG,...,2,1,

)(deg

l

i

iv1

]1)[deg(

4

Adding deg(v*) to both sides of eq. , we get = + degG(v*)

= 2|E(G’)|+ l + l

= 2|E(G’)|+ 2l = 2|E(G’)+ l | = 2|E(G)| (using )

Hence Proved.


4

)(

)(degGVv

G v )'(

)(degGVv

vG

3

Acyclic Graph: A Graph with no cycles


A graph with no cycles

Thanks Nakul kumar Roll No.24,Navneet kumar Roll No.25(MCA 2012)

•A connected graph with no cycles is called a tree.


Three equivalent statements about trees

• T is a tree i.e. T is a connected acyclic graph.• T is a connected graph with n-1 edges.• There is a unique path between every pair of

vertices.


Claim- A tree with n vertices has exactly (n-1) edges.Proof-: Base Case: for n=1. no. of edges =0.For n=2 No. of edges=1.Therefore, it holds true for n=1 and n=2.Induction Hypothesis:Let it be true for every tree with ≤ n vertices.


Let T be a tree with n+1 vertices.


After removing this edge we get two separate trees T1 and T2, such that E(T1) U E(T2) = E(T) – e and

|V(T1)| ≤ n and |V(T2)| ≤ n

|E(T1)| = |V(T1)|-1 (by induction hypothesis)

|E(T2)| = |V(T2)|-1 ( by induction hypothesis)

So, |E(T)| = |E(T1)|+|E(T2)|+1

|E(T)| = (|V(T1)|-1) + (|V(T2)|-1) + 1

= |V(T1)| + |V(T2)| - 1

= | V(T)| - 1 = n+1-1 = n Hence proved.Ex: Prove the converse i..e a connected graph with n-1 edges is acyclic.


Exercise:Prove that T is a tree iff there exists a unique simple path between every pair of vertices.

Proof:Using the definition of tree, If T is a tree, it is connected, i.e there exists some path between every pair of vertices.Suppose if possible, there exists two distinct paths between some pair of vertices ‘u’ and ‘v’.


Without loss of generality, we assume that there is no common vertex between them.

So, this shows there exists a cycle, which contradicts the definition of tree.Hence, a contradiction.

Assignment: Prove the converse.


Subgraphs

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Let G=(V,E) be a graph, then H=(V’,E’) is a subgraph of G if E’ is a subset of E and V’ is a subset of V, such that u, v is in V’ , for all (u,v) in E’.


Spanning Trees

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Given a graph G, a subgraph T of G is said to be a Spanning Tree ifI. T is a treeII. T spans all the vertices of G.

Documents

MCA 202: Discrete Mathematics Instructor Neelima Gupta [email protected]