Problem Set 1 - Remarks and Solutions - Graph Theory | 1016 467 | Study notes Mathematics

Graph Theory - 1016 - 467

Spring 2005

Set 1- Remarks and Solutions

1.4 The graph is

€

. This can be shown by the brute force method of simply checking all of the vertex

connections directly. However, to reduce the work you might consider making use of the symmetries in

this problem. S itself is symmetric about 0, that is,

€

−S=S

. This, together with the given rules for joining

vertices, allows us to prove that

€

v,w

{ }

is an edge if and only if

€

−v, −w

{ }

is an edge. Some direct

checking is still needed, but it is much less than if no notice is taken of the symmetries.

1.22 Let G be a disconnected graph. Then

€

V G

( )

≥2

. If

€

V G

( )

, then

€

G=E2

, so

€

G=K2

, which has

diameter 1. Therefore suppose

€

V G

( )

≥3

Claim: If v and w are vertices in

€

, then

€

d v,w

( )

≤2

Proof: If

€

v,w

{ }

∈E G

( )

, then

€

d v,w

( )

, and our claim is satisfied. Suppose therefore that

€

v,w

{ }

∉E G

( )

. Then

€

v,w

{ }

∈E G

( )

. As G is disconnected, there must be a vertex z not joined to v by

any path in G . In particular,

€

v,z

{ }

∉E G

( )

, and so

€

v,z

{ }

∈E G

( )

. Also, were

€

z,w

{ }

∈E G

( )

, then zwv

would be a path joining z and v in G, which is forbidden. Therefore

€

z,w

{ }

∉E G

( )

, and so

€

z,w

{ }

∈E G

( )

. With this, vzw is now a path in

€

joining v and w. That is,

€

d v,w

( )

, and the claim is

proved.

€

d v,w

( )

≤2

for all vertices v and w in

€

, we have

€

diam G

( )

≤2

1.24 i. The graph

€

is bipartite, with bipartition

€

A=r1,u1,w1,t1,y1,q1

{ }

and

€

B=x1,v1 ,s1,z1

{ }

ii. The graph

€

is not bipartite. Suppose it were, with bipartition A, B, and

€

v2∈A

. Now

€

v2w2r2x2u2v2

is a cycle of length 5 in

€

. So we have

€

v2, r2,u2∈A

and

€

w2, x2,v2∈B

. As

€

v2∈A∩B

, we have a

contradiction, and so no bipartition of

€

is possible.

1.26 For a given nonempty set of integers V, we construct a graph G as follows. Put

€

V G

( )

, and join vertices

v and w with an edge if and only if

€

v+w≡1 mod 2

Claim: G is bipartite.

Proof: Put

€

A=v v ≡0 mod 2

{ }

, and

€

B=w w ≡1 mod 2

{ }

€

v, w∈A

, then

€

v+w≡0≠1 mod 2

. If

€

v, w∈B

, then

€

v+w≡2≠1 mod 2

. If

€

v∈A

and

€

w∈B

, then

€

v+w≡1 mod 2

. Therefore a pair of vertices are joined by an edge in G if and only if one vertex is in A

and the other in B.

If either

€

A=∅

€

B=∅

, then G is edgeless. If both

€

A≠ ∅

and

€

B≠ ∅

, then G is the complete bipartite

graph

€

KA , B

1.28 In this problem, recall the usual labeling of

€

, where adjacent bit strings differ in exactly one place. Parts

a and b can then be regarded as asking for special subgraphs of

€

Note: A union of graphs is taken as a disjoint union. In particular, the expression

€

G∪G

denotes the graph

made up of two disjoint copies of G.

a. i.

€

R2=Q2=K2∪K2

ii. Adjacent vertices in

€

differ in 2 or 3 places. Removing from each vertex the edge that joins it to

its binary complement gives

€

. When you’ve done this, you see that

€

R3=K4∪K4

, shown

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Graph Theory - 1016 - 467 Spring 2005 Set 1- Remarks and Solutions 1.4 The graph is € K 5. This can be shown by the brute force method of simply checking all of the vertex connections directly. However, to reduce the work you might consider making use of the symmetries in this problem. S itself is symmetric about 0, that is, € −S = S. This, together with the given rules for joining vertices, allows us to prove that €

{ v,w} is an edge if and only if

{ − v, − w} is an edge. Some direct

checking is still needed, but it is much less than if no notice is taken of the symmetries. 1.22 Let G be a disconnected graph. Then €

V (G ) ≥ 2. If

V (G ) = 2 , then

G = E 2 , so € G = K 2 , which has diameter 1. Therefore suppose €

V (G ) ≥ 3.

Claim : If v and w are vertices in € G , then €

d (v,w ) ≤ 2.

Proof : If €

{ v,w} ∈ E ( G), then

d (v,w ) = 1 , and our claim is satisfied. Suppose therefore that

{ v,w} ∉ E ( G). Then

{ v,w} ∈ E (G ). As G is disconnected, there must be a vertex z not joined to v by

any path in G. In particular, €

{ v,z} ∉ E (G ), and so

{ v,z} ∈ E ( G). Also, were

{ z,w} ∈ E (G ), then zwv

would be a path joining z and v in G, which is forbidden. Therefore €

{ z,w} ∉ E (G ), and so

{ z,w} ∈ E ( G). With this, vzw is now a path in

G joining v and w. That is, €

d (v,w ) = 2 , and the claim is

proved. As €

d (v,w ) ≤ 2 for all vertices v and w in

G , we have €

diam ( G) ≤ 2.

1.24 i. The graph € G 1 is bipartite, with bipartition €

A = { r 1 ,u 1 ,w 1 ,t 1 ,y 1 ,q 1 } and

B = { x 1 ,v 1 ,s 1 ,z 1 }.

ii. The graph € G 2 is not bipartite. Suppose it were, with bipartition A, B, and € v 2 ∈ A. Now € v 2 w 2 r 2 x 2 u 2 v 2 is a cycle of length 5 in € G 2. So we have € v 2 , r 2 ,u 2 ∈ A and € w 2 , x 2 ,v 2 ∈ B. As € v 2 ∈ A ∩ B, we have a contradiction, and so no bipartition of € G 2 is possible. 1.26 For a given nonempty set of integers V, we construct a graph G as follows. Put €

V (G ) = V , and join vertices

v and w with an edge if and only if € v + w ≡ 1 mod 2. Claim : G is bipartite. Proof : Put €

A = { v v ≡ 0 mod 2 } , and

B = { w w ≡ 1 mod 2 }.

If € v, w ∈ A , then € v + w ≡ 0 ≠ 1 mod 2. If € v, w ∈ B, then € v + w ≡ 2 ≠ 1 mod 2. If € v ∈ A and € w ∈ B, then € v + w ≡ 1 mod 2. Therefore a pair of vertices are joined by an edge in G if and only if one vertex is in A and the other in B. If either € A = ∅ or € B = ∅, then G is edgeless. If both € A ≠ ∅ and € B ≠ ∅, then G is the complete bipartite graph €

K A , B.

1.28 In this problem, recall the usual labeling of € Q (^) n , where adjacent bit strings differ in exactly one place. Parts a and b can then be regarded as asking for special subgraphs of € Q (^) n. Note: A union of graphs is taken as a disjoint union. In particular, the expression € G ∪ G denotes the graph made up of two disjoint copies of G. a. i. €

R 2 = Q 2 = K 2 ∪ K 2.

ii. Adjacent vertices in € Q 3 differ in 2 or 3 places. Removing from each vertex the edge that joins it to its binary complement gives R 3. When you’ve done this, you see that R 3 = K 4 ∪ K 4 , shown

below. 000 011 101

b. If two 3-bit binary strings differ in exactly 3 places, then each is the binary complement of the other. So in € S 3 each vertex is joined only to its binary complement. Therefore €

S 3 = K 2 ∪ K 2 ∪ K 2 ∪ K 2.

2.2 a. In the graphs € Cn , for € n ≥ 3 , every vertex has degree 2, and so no vertex has odd degree. b. The graph € K 2 × C 3 has six vertices, all of degree 3. It is not a complete graph, since it has just 9 edges, while € K 6 has 15. c. Think of the given condition this way: Every edge joins vertices of different degrees. 1 1 1

d. The easiest way is to take the complement of your answer in part c! Vertices that are not adjacent will then have different degrees. This complement looks a bit busy if you simply draw it using the above configuration. It looks nicer if you move things about, so that there are no unsightly crossings. 1 3 3

2.4 Let G be a graph with 6 vertices and 10 edges. If € δ (^) (G ) = 3 and € Δ (^) (G ) = 4 , then every vertex has degree 3 or

Let x denote the number of degree 3 vertices, y the number of degree 4 vertices. Then € x + y = 6 and € 3 x + 4 y = 2 ⋅ 10 = 20. Solving this system we get € x = 4 and € y = 2. To get the desired example, start with € K 2 × C 3 , a graph with six vertices, all of degree 3, and add a single edge joining some pair of vertices that are not already adjacent.

Problem Set 1 - Remarks and Solutions - Graph Theory | 1016 467, Study notes of Mathematics

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Download Problem Set 1 - Remarks and Solutions - Graph Theory | 1016 467 and more Study notes Mathematics in PDF only on Docsity!

{ v,w} is an edge if and only if

{ − v, − w} is an edge. Some direct

V (G ) ≥ 2. If

V (G ) = 2 , then

V (G ) ≥ 3.

d (v,w ) ≤ 2.

{ v,w} ∈ E ( G), then

d (v,w ) = 1 , and our claim is satisfied. Suppose therefore that

{ v,w} ∉ E ( G). Then

{ v,w} ∈ E (G ). As G is disconnected, there must be a vertex z not joined to v by

{ v,z} ∉ E (G ), and so

{ v,z} ∈ E ( G). Also, were

{ z,w} ∈ E (G ), then zwv

{ z,w} ∉ E (G ), and so

{ z,w} ∈ E ( G). With this, vzw is now a path in

d (v,w ) = 2 , and the claim is

d (v,w ) ≤ 2 for all vertices v and w in

diam ( G) ≤ 2.

A = { r 1 ,u 1 ,w 1 ,t 1 ,y 1 ,q 1 } and

B = { x 1 ,v 1 ,s 1 ,z 1 }.

V (G ) = V , and join vertices

A = { v v ≡ 0 mod 2 } , and

B = { w w ≡ 1 mod 2 }.

K A , B.

R 2 = Q 2 = K 2 ∪ K 2.

S 3 = K 2 ∪ K 2 ∪ K 2 ∪ K 2.