- 87 Views
- Uploaded on
- Presentation posted in: General

Lecture 5.4: Paths and Connectivity

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Lecture 5.4: Paths and Connectivity

CS 250, Discrete Structures, Fall 2011

NiteshSaxena

*Adopted from previous lectures by ZephGrunschlag

- Due at 11am on Nov 30 (Wed)
- Covers the chapter on Graphs (lecture 5.*)
- Thanksgiving is in the way – please try to start early
- Has a 10-pointer bonus
problem too

Little work will not hurt!

Lecture 5.4 -- Paths and Connectivity

- Grading now
- Expect results in a couple of days
- Graded HWs will be distributed next Tuesday

- Solution posted

Lecture 5.4 -- Paths and Connectivity

- Thursday, December 8, 10:45am- 1:15pm, lecture room
- Heads up!
- Please mark the date/time/place
- Emphasis on post mid-term 2 material
- Coverage:
- 65% post mid-term 2 (lectures 4.*, 5.*, 6.*), and
- 35% pre mid-term 2 (lecture 1.*. 2.* and 3.*)

- Our last lecture will be on December 6
- We plan to do a final exam review then

Lecture 5.4 -- Paths and Connectivity

- Paths and Isomorphism
- Connectivity

Lecture 5.4 -- Paths and Connectivity

Lecture 5.4 -- Paths and Connectivity

Lecture 5.4 -- Paths and Connectivity

DEF: Suppose G1 = (V1,E1 ) and G2 = (V2,E2 ) are pseudographs. Let f :V1V2 be a function s.t.:

- f is bijective
- for all vertices u,v in V1, the number of edges between u and v in G1is the same as the number of edges between f (u) and f (v ) in G2; or e(u, v, G1) = e(f(u), f(v), G2)
Then f is called an isomorphismand G1is said to be isomorphic to G2.

Lecture 5.4 -- Paths and Connectivity

Two graphs are isomorphic to each other if they satisfy the following properties:

- same number of vertices
- same number of edges
- same degrees at corresponding vertices
- Any subgraph of one is isomorphic to some subgraph of the other

Lecture 5.4 -- Paths and Connectivity

Theorem: Isomorphism is an equivalence relation

Proof: We need to prove that the isomorphism relation is reflexive, symmetric, and transitive.

Let’s use the whiteboard.

Lecture 5.4 -- Paths and Connectivity

A path in a graph is a continuous way of getting from one vertex to another by using a sequence of edges.

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

A path in a graph is a continuous way of getting from one vertex to another by using a sequence of edges.

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

A path in a graph is a continuous way of getting from one vertex to another by using a sequence of edges.

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

EG: could get from 1 to 3 circuitously as follows:

1-e12-e11-e33-e42-e62-e52-e43

e6

e1

e2

1

2

e5

e3

e4

e7

3

4

- Linkedin paths (let us see in practice)
- Facebook paths
- Internet paths
- …

Lecture 5.4 -- Paths and Connectivity

- Paths and circuits can also be serve as a good criteria for determining whether two graphs are isomorphic
- Look for (simple) circuits of different lengths. Example below:

u1

u2

u6

u3

u5

u4

Lecture 5.4 -- Paths and Connectivity

- Adjacency matrix A for a graph depicts all paths of length 1
- The matrix A2depicts number of paths of length 2
- In general, the matrix Akdepicts number of paths of length k

a

b

c

d

Lecture 5.4 -- Paths and Connectivity

- Adjacency matrix A for a graph depicts all paths of length 1
- The matrix A2depicts number of paths of length 2
- In general, the matrix Akdepicts number of paths of length k

a

b

c

d

Lecture 5.4 -- Paths and Connectivity

DEF: Let G be a pseudograph. Let u and v be vertices. u and v are connected to each other if there is a path in G which starts at u and ends at v. G is said to be connected if all vertices are connected to each other.

Note: Any vertex is automatically connected to itself via the empty path.

Lecture 5.4 -- Paths and Connectivity

Q: Which of the following graphs are connected?

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

A: First and second are disconnected. Last is connected.

1

2

3

4

Lecture 5.4 -- Paths and Connectivity

Can define a puzzling graph G as follows:

V = {3-letter English words}

E : two words are connected if we can get one word from the other by changing a single letter.

One small subgraph of G is:

Q: Is “fun” connected to “car” ?

rob

job

jab

Lecture 5.4 -- Paths and Connectivity

A: Yes: funfanfarcar

Or: funfinbinbanbarcar

Lecture 5.4 -- Paths and Connectivity

Thm1: Every connected graph with n vertices has at least n-1 edges

Proof: Use proof by mathematical induction

Basis Step (n=1): No. of edges is 0, which is less >= 0 (n-1)

Induction Step: Assume to be true for n = k and show it to be true for n = k + 1

We assumed that a connected graph with k vertices will have at least k–1 edges. Now, we add a new vertex to this graph to obtain a new graph with k vertices.

For the new graph to remain connected, the new vertex should be incident with at least one edge which is also incident with one of the vertices in the old graph. This means that the new graph should have a total of at least (k – 1) + 1 = k edges.

This proves the induction step.

Finally, combining the basis and induction steps, we get that the theorem is true for all n

Thm 2: Vertex connectedness in a simple graph is an equivalence relation

Proof: We will show that the “connectedness” relation is reflexive, symmetric and transitive

It is reflexive since every vertex is connected to itself via a path of length 0

It is symmetric because if a vertex u is connected to another vertex v, then there exists a path between u and v – just traverse the reverse path

It is transitive because if u and v are connected (via path p) and v and w are connected (via path q), then u and w are connected via a path p|q

- Thm 3: If a connected simple graph G is the union of graphs G1 and G2, then G1 and G2 must have a common vertex
- Proof: (very simple) let’s use the board.

Lecture 5.4 -- Paths and Connectivity

DEF: A connected component (or just component) in a graph G is a set of vertices such that all vertices in the set are connected to each other and every possible connected vertex is included.

Q: What are the connected components of the following graph?

1

6

2

7

8

5

3

4

Lecture 5.4 -- Paths and Connectivity

1

8

5

3

A: The components are {1,3,5},{2,4,6},{7} and {8} as one can see visually by pulling components apart:

6

2

7

4

Lecture 5.4 -- Paths and Connectivity

1

5

3

A: The components are {1,3,5},{2,4,6},{7} and {8} as one can see visually by pulling components apart:

6

2

7

8

4

Lecture 5.4 -- Paths and Connectivity

1

5

3

A: The components are {1,3,5}, {2,4,6}, {7} and {8} as one can see visually by pulling components apart:

6

2

7

8

4

Lecture 5.4 -- Paths and Connectivity

Not all connected graphs are treated equal!

Q: Rate following graphs in terms of their design value for computer networks:

1)

2)

3)

4)

A: Want all computers to be connected, even if 1 computer goes down:

1) 2nd best. However, there

is a weak link— “cut vertex”

2) 3rd best. Connected

but any computer can disconnect

3) Worst!

Already disconnected

4) Best! Network dies

only with 2 bad computers

Lecture 5.4 -- Paths and Connectivity

The network is best because it can only become disconnected when 2 vertices are removed. In other words, it is 2-connected. Formally:

DEF: A connected simple graph with 3 or more vertices is 2-connected if it remains connected when any vertex is removed. When the graph is not 2-connected, we call the disconnecting vertex a cut vertex.

Lecture 5.4 -- Paths and Connectivity

There is also a notion of N-Connectivity where we require at least N vertices to be removed to disconnect the graph.

Lecture 5.4 -- Paths and Connectivity

In directed graphs may be able to find a path from a to b but not from b to a. However, Connectivity was a symmetric concept for undirected graphs. So how to define directed Connectivity is non-obvious:

- Should we ignore directions?
- Should we insist that we can get from a to b in actual digraph?
- Should we insist that we can get from a to b and that we can get from b to a?

Lecture 5.4 -- Paths and Connectivity

Resolution: Don’t bother choosing which definition is better. Just define to separate concepts:

- Weakly connected : can get from a to b in underlying undirected graph
- Semi-connected (my terminology): can get from a to b OR from b to a in digraph
- Strongly connected : can get from a to b AND from b to a in the digraph
DEF: A graph is strongly (resp. semi, resp. weakly) connectedif every pair of vertices is connected in the same sense.

Lecture 5.4 -- Paths and Connectivity

Q: Classify the connectivity of each graph.

Lecture 5.4 -- Paths and Connectivity

A:

semi weakstrong

Lecture 5.4 -- Paths and Connectivity

- Rosen 10.3 and 10.4

Little work will not hurt!

Lecture 5.4 -- Paths and Connectivity