Subdivisions of Graphs

Subdivisions of Graphs.

Dissertation

submitted in partial fulfillment of the requirements

for the degree of

Master of Technology

by

Namrata P. Tholiya(Roll no. 06305031)

under the guidance of

Prof. Ajit A. Diwan

Department of Computer Science and Engineering

Indian Institute of Technology Bombay

2008

Dissertation Approval Sheet

This is to certify that the dissertation entitled

Subdivisions of Graphsby

Namrata P. Tholiya(Roll no. 06305031)

is approved for the degree of Master of Technology.

Prof. Ajit A. Diwan

(Supervisor)

Prof. Sundar Vishwanathan

(Internal Examiner)

Prof. Bharat Adsul

(Internal Examiner)

Prof. Murali Srinivasan

(Chairperson)

Date:

Place:

v

INDIAN INSTITUTE OF TECHNOLOGY BOMBAYCERTIFICATE OF COURSE WORK

This is to certify that Ms. Namrata P. Tholiya was admitted to the candidacy

of the M.Tech. Degree and has successfully completed all the courses required for the

M.Tech. Programme. The details of the course work done are given below.

Sr.No. Course No. Course Name Credits

Semester 1 (Jul – Nov 2006)

1. CS699 Software Laboratory 4

2. CS601 Algorithms & Complexity 6

3. CS615 Formal Specification and Verification of Programming 6

4. CS631 Relational Database Management Systems 6

5. IT608 Data Mining 6

6. CS694 Seminar 4

7. HS699 Communication and Presentation Skills (P/NP) 4

Semester 2 (Jan – Apr 2007)

8. CS640 Natural Computing 6

9. CS613 Functional Programming 6

10. CS616 Parallelizing Compilers 6

11. IT628 ITProject Management 6

12. CS686 Object Oriented Systems 6

Semester 3 (Jul – Nov 2007)

13. CS641 (Advanced) Computer Networks 6

14. CS691 R&D Project 6

M.Tech. Project

15. CS696 M.Tech. Project Stage - I (Jul 2006) 22

16. CS697 M.Tech. Project Stage - II (Jan 2007) 28

17. CS698 M.Tech. Project Stage - III (Jul 2008) 40

I.I.T. Bombay Dy. Registrar(Academic)

Dated:

ix

Acknowledgements

I take this opportunity to express my sincere gratitude towards Prof. Ajit A. Diwan

for his constant support and encouragement. His excellent guidance has been instrumental

in making this project work a success.

I would like to thank Prof. Sundar Vishwanathan for useful insights and guidance

towards the project. I would like to thank members of the Algorithms Group at CSE

for their valuable suggestions and helpful discussions.

I would also like to thank my family and friends, who have been a source of encour-

agement and inspiration throughout the duration of the project. I would like to thank

the entire CSE family for making my stay at IIT Bombay a memorable one.

Namrata P. TholiyaI. I. T. Bombay

June 25th, 2008

xi

Abstract

For certain graphs H, every graph with minimum degree at least |H| − 1 contains a

subdivision of graph H. For instance, this relation is true if H is a cycle or a tree.

Through this project, we aim to find the general set of graphs that satisfy this property.

We attempt to do this by proving this condition for different cases of graph H using the

general techniques which we have developed. The cases that we study are when H is P53,

P63, P7

3, wheel Wd i.e. Cl ∗ K1 where l ≥ 3 and caterpillars. Also, we show that every

graph with minimum degree d contains a subdivision of a graph H with d + 1 vertices

and 3d − 3 edges for d ≥ 2. In addition, the proofs give us a polynomial time algorithm

to find a subdivision of H in a graph satisfying the degree condition.

Contents

Abstract i

1 Introduction 1

2 Mader’s Theorem 3

2.1 Mader’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Generalization of Mader’s Theorem . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 External Configuration 13

3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.1.1 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.1.2 External Configuration . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.1.3 General Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2 General Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Different cases of graph H . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.1 Edge having d− 1 triangles . . . . . . . . . . . . . . . . . . . . . . 16

3.3.2 P53 (P3 ∗K2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.3 P63(P4 ∗K2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3.4 Subdivision of wheels Wd where d ≥ 3 . . . . . . . . . . . . . . . . 19

3.3.5 Caterpillars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Internal Configuration 23

4.1 Internal Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.2 General Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.3 Good-Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.3.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.3.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.4 Bad-Configuration Example . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5 List of good configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.6 General Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

iii

iv Contents

4.7 Applications of Good-Configurations . . . . . . . . . . . . . . . . . . . . . 35

4.7.1 P53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.7.2 P63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.7.3 P73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.7.4 Wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.7.5 2-Caterpillar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.7.6 Path through the vertices . . . . . . . . . . . . . . . . . . . . . . . 40

5 Conclusion and Future Work 43

5.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.2 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

References 45

Chapter 1

Introduction

The problem of testing if one graph contains a subdivision of another graph has many

applications in graph theory. For example, checking for planarity can be done by testing

if a given graph contains a subdivision of graph K5 or K3,3. One way to test if a graph

G contains a subdivision of a fixed graph H is to choose vertices in G, corresponding to

vertices in H and then find internally disjoint paths between specified pairs of vertices in

G such that paths to be found in G correspond to edges in H. Finding the paths itself

is a difficult problem and the number of choices for the vertices is O(nd) where n is the

number of vertices in the graph G and d is the number of vertices in the graph H. Hence,

this makes it as a hard problem.

For certain graphs H, every graph with minimum degree at least |H| − 1 contains a

subdivision of graph H. Our aim is to find for what kind of graphs H, every graph G

with minimum degree at least |H|−1 contains a subdivision of graph H. We also develop

a polynomial time algorithm to find a subdivision of H in a graph satisfying the degree

condition.

In the report, we study Mader’s theorem [1], proof of which helps us to get d internally

vertex disjoint paths between end vertices of an edge of a graph G with minimum degree

at least d. We also see how this helps us to solve one case of graph H where H is a graph

with d−1 triangles having a common edge. Two more cases that we study are when H is

a graph with r edges starting from the same vertex and the ith edge has di − 1 triangles,

for 1 ≤ i ≤ r such that∑r

i=1 di = d and the another case is when H is a graph that can

be obtained from a tree by adding edges joining one vertex to all the other d vertices [2].

We define the concept of reduction based on the proof of Mader’s theorem and two

types of configurations ‘External Configuration’ and ‘Internal Configuration’. Using these

two concepts we define two different techniques which help us to prove the property being

studied for the different cases of the graph H.

The cases that we study are when H is P53 i.e. P3 ∗K2 where P3 ∗K2 can be defined as

the graph obtained by connecting each vertex of P3 to each vertex of K2. Also, the cube

2

of a graph G is the graph obtained by adding edges to G between those pairs of vertices

that are at distance 2 or 3 in G. Thus the cube of P5 is the graph with vertices 0, 1, 2,

3, 4 and two vertices are adjacent iff the difference between them is ≤ 3. This is same

as K5 − e. The property has been proved by Pelikan J [3]. Another cases that we study

are when H is P63 (P4 ∗K2) and P7

3. Also, we see the proof for subdivision of wheels by

our technique. We also prove the property when H is a 2 caterpillar with d+ 1 vertices.

Finally, the general result: Every graph of minimum degree ≥ d contains a subdivision of

a graph with d+ 1 vertices and 3d− 3 edges for d ≥ 2.

Chapter 2

Mader’s Theorem

We have developed a technique to prove for certain cases of graph H, every graph G having

minimum degree at least |H|−1 contains a subdivision of graphH. This technique is based

on the concept of reduction and configuration. To understand the concept of reduction,

which is based on the proof of Mader’s theorem, we need to study the theorem. In this

chapter, we study this theorem and then generalize it.

2.1 Mader’s Theorem

Theorem 2.1.1. Let G be a graph with minimum degree at least d. There exists an edge

xy in G such that there are d internally vertex-disjoint paths between x and y in G [1].

In order to prove this, we prove a slightly stronger result.

Theorem 2.1.2. Let G be a graph and K be a clique in G such that G−K contains an

edge. Suppose that every vertex in G − K has degree at least d in G. Then there is an

edge xy in G −K such that there are d internally vertex-disjoint paths between x and y

in G.

Proof. Let L be a maximal clique of the graph G containing K. Let uv be an edge in

G−K.

The base case is when G−L does not contain an edge. Hence, either G−L is empty

or G− L contains vertices with no edge between them.

1. Consider the case when G − L is empty. In this case, both u and v ∈ L. We are

given that every vertex in G−K has degree at least d in G. Hence, L contains at

least d+ 1 vertices forming a clique of size at least d+ 1. Hence, the vertex u has at

least d− 1 neighbours in L other than v and all these neighbours belong to L only.

Also, v ∈ L. Hence, all these neighbours are adjacent to v, form d − 1 internally

vertex-disjoint paths. Together with an edge uv, we get d vertex-disjoint paths.

2. Consider the other case when G − L is not empty and contains no edge. Hence,

either u ∈ G−L and v ∈ L or u and v both ∈ L. In the former case, there is no edge

from u in G−L. Hence, all d neighbours of u including v are in L and hence, these

3

4 2.1. Mader’s Theorem

are adjacent to the vertex v forming d−1 internally vertex-disjoint paths. Together

with an edge uv, we get d internally vertex-disjoint paths. Consider the latter case.

Let y1, y2, · · · , yk be the common neighbours of u and v and let xk+1, · · · , xd−1 be

the neighbours of u that are not adjacent to v other than v. Since, L is a clique,

the vertices xk+1 · · · , xd−1 are not in L. Each of these vertices has d− 1 neighbours

in L. Hence, we can find vertices yk+1, · · · , yd−1 in L such that yi /∈ {u, v}, yi 6= yj

for i 6= j and xi is adjacent to yi for k + 1 ≤ i ≤ d − 1. Hence, uyiv for 1 ≤ i ≤ k

and uxiyiv for k + 1 ≤ i ≤ d− 1 are d− 1 internally vertex-disjoint paths between

u and v. Together with an edge uv, we get required d vertex-disjoint paths.

Consider the case when G − L contains an edge. Let the vertices in L be ordered

v1, v2, · · · , vl. Since, L is a maximal clique, for every vertex v ∈ G − L, there is at least

one vertex vi in L which is not adjacent to v. Let π(v) = vi where i is the smallest index

such that v is not adjacent to vi and vi ∈ L. Remove the vertex v1 from the clique K. Let

(G′, K ′) be the tuple obtained by removing v1. This may change the degree of vertices.

As it is given that the vertex present in G−K has degree at least d in G, we need to add

an edge for each vertex v ∈ G′ − K ′ and adjacent to v1. Hence, we add an edge vπ(v)

for each such v. Hence, we get G′ = (G − v1) ∪ {vπ(v)|v ∈ V (G) \ V (L), π(v) > 1} and

K ′ = L− v1. Hence, K ′ is a clique in G′ such that G−L = G′−K ′ contains an edge and

every vertex in G′−K ′ has degree at least d in G′ since we have added an edge vπ(v) for

every v ∈ G′ −K ′ and adjacent to v1.

By induction, there is an edge xy in G′ − L′ such that there are d internally vertex

disjoint paths between x and y in G′ . We need to show that we can modify these paths

to get d internally vertex disjoint paths from x to y in G. Any path from x to y that does

not contain any of the edges vπ(v) is a path from x to y in G. All such paths are left

unchanged.

Let P1, P2, · · · , Pk be the paths from x to y in G′ that contain a vertex in L′. Divide

each path into two paths where it intersects L′. We get 2k disjoint paths out of which

k paths start from x and end in L′ while remaining k paths start from y and end in L′.

Edges which lie in clique L′ do not belong to these paths. Consider an edge in a path

which is of the form vπ(v). we call it as a bad edge. The bad edges for this iteration will

have exactly one endpoint in L′ . We have got these edges because of the removal of the

vertex v1. In G, v was joined with v1. After removal of v1, we have added the edge vπ(v)

to satisfy the degree constraint. Hence, we can replace such edges with the original one

i.e. vv1.

In other terms, let si be the vertex in L′ that is nearest to x in Pi and let ti be the

furthest such vertex. We may assume that either si = ti or siti is an edge in Pi, since L′

is a clique. Let s−i and t+i be the predecessor and successor of si and ti respectively in the

path Pi from x to y.

Let S = {s1, s2, · · · , sk} and T = {t1, t2, · · · , tk}.Let S ′ = {si ∈ S|π(s−i ) = si} and T ′ = {ti ∈ T |π(t+i ) = ti}.

2.1. Mader’s Theorem 5

Let S ′ = {vi1 , vi2 , · · · , vip} for i1 < i2 · · · < ip.

Let T ′ = {vj1 , vj2 , · · · , vjp} for j1 < j2 · · · < jp.

If π(v) = vr, by definition of π(v), vvi is an edge in G for all i < r. For each vertex

S = vir ∈ S ′, replace the edge s−s in the path containing s by the edge s−vir−1 , where

vi0 = v1. Similarly, for each vertex t = vjr ∈ T ′ replace the edge tt+ in the path containing

t by the edge vjr−1t+, where vj0 = v1. Delete the edge siti if si 6= ti.

This gives 2k internally vertex disjoint paths in G from {x, y} to L, exactly k of which

starts from and end in k distinct vertices in L. If two of these paths end in a common

vertex in L, their union is a path from x to y in G. Remaining paths can be obtained by

pairing up the paths starting from x arbitrarily with the remaining paths starting from

y in L and adding in an edge in L joining their endpoints in L. This gives k internally

vertex disjoint paths. Hence, in this way, we get d disjoint paths between x and y in

G.

2.1.1 Example

Figure 2.1: Graph G

Consider the figure 2.1. Here, the minimum degree of graph G is d = 3. Hence, there

exists an edge xy with 2 disjoint paths including xy. We need to find such edge and

required disjoint paths between the edge. Follow the procedure which is used to prove

Mader’s theorem, to get the required pair of vertices and the vertex-disjoint paths between

them. Table 2.3 shows the step by step procedure.

For example, the first step in the table shows that K = {1}, L = {1, 2, 3}. Vertex 1

is removed. π(3) = 2 and π(5) = 2. Hence, edges 2 − 3 and 2 − 5 are added as shown

in second figure of figure 2.1. In this way, finally we get K = {4, 5}, L = {4, 5, 6, 7} and

G − L as empty. We stop at this step i.e. step 5. Table 2.2 gives 3 disjoint paths at

the current step. For example, step 5 indicates G− L empty. Hence, 3 disjoint paths are

6-7, 6-4-7, 6-5-7. Step 4 indicates transition from step 5 to step 4 by adding vertex 3. 3

disjoint paths in this step are obtained by replacing bad edge 7 − 5 by 7 − 3. Hence, 3

disjoint paths are 6-7, 6-4-7, 6-5-3-7. In this way, final paths are 6-7, 6-4-8-2-7, 6-5-3-7.

6 2.2. Generalization of Mader’s Theorem

Step no K L REM π Added edges Figure

1 {1} {1, 2, 8} {1} π(3) = 2,

π(5) = 2

2-3 and 2-5 2nd fig of

figure 2.1

2 {2, 8} {2, 8} {2} π(3) = 8,

π(5) = 8,

π(7) = 8

3-8, 5-8 and 7-8 1st fig

of figure

2.2(a)

3 {8} {8, 3, 4} {8} π(5) = 4,

π(7) = 4

5-4 and 7-4 2nd fig

of figure

2.2(a)

4 {3, 4} {3, 4, 5} {3} π(7) = 5 5-7 figure

2.2(b)

5 {4, 5} {4, 5, 6, 7} {φ} figure

2.2(b)

Table 2.1: Finding K, L and π at each step to get 3 disjoint paths in figure 2.1

(a) (b)

Figure 2.2:

2.2 Generalization of Mader’s Theorem

Mader’s Theorem can be generalized to find the internally vertex-disjoint paths from the

root of a rooted tree to remaining vertices of that tree.

If S is a subset of vertices and v a vertex in S then a v − S fan is a collection of

(|S| − 1)(v− (S \ {v})) paths such that any two paths have only the vertex v in common.

Theorem 2.2.1. Let G be the graph with minimum degree δ(G) = d and let T be any

tree with d+ 1 vertices, rooted at a vertex r. Then, there is a subtree T ′ of G isomorphic

to T , with root r′ , such that G contains a r′ − V (T ′) fans [2].

In order to prove this theorem, we prove slightly stronger result. First we look at some

definitions which helps us to prove the theorem.

Definition 2.2.1. A subtree T of G is consistent with the ordered clique K, for every ver-

tex vi ∈ V (T )∩V (K), vi has at most one neighbour in T not contained in {v1, v2, · · · , vi−1},for 1 ≤ i ≤ k.

2.2. Generalization of Mader’s Theorem 7

Present Step Current Edges vπ(v) Vi0 vvi0 Final Edges

5 6-7, 6-4-7, 6-5-7

4 6-7, 6-4-7, 6-5-7 7-5 3 7-3 6-7, 6-4-7, 6-5-3-7

3 6-7, 6-4-7, 6-5-7 7-4 8 7-8 6-7, 6-4-8-7, 6-5-3-7

2 6-7, 6-4-7, 6-5-7 7-8 2 7-2 6-7, 6-4-8-2-7, 6-5-3-7

1 6-7, 6-4-7, 6-5-7 1 6-7, 6-4-8-2-7, 6-5-3-7

Table 2.2: Finding 3 disjoint paths in figure 2.1

Theorem 2.2.2. Let G be a graph and T a rooted tree with d + 1 vertices. Let K be an

ordered clique in G such that every vertex in G−K has degree at least d in G. Let d(T )

be the degree of the root. Suppose, G−K contains a vertex of degree at least d(T ). Then

there is a subtree T ′ of G isomorphic to T that satisfies the following properties.

1. The root r′ of T ′ and the neighbours of r′ in T ′ are contained in G−K.

2. T ′ is consistent with the ordered clique K.

3. G contains a r′ − V (T ′) fan.

Proof. Let the vertices of T be labeled as t1, t2, · · · , td+1 in breadth first order with td+1 as

a root having degree d(T ). Hence, every ti has exactly one neighbour tj for i < j for 1 ≤i ≤ d+ 1 as shown in second figure of figure 2.3. We call this unique neighbour as parent

of ti. All other neighbours are treated as children of ti. Let d(T ) = t and let d1, · · · , dt

be the degrees of root’s children. Let d′ =∑t

i=1 di ≤ d. Let L′ = L − {v1, · · · , vd−d′}.Therefore, in second figure of figure 2.3 d′ = 3 + 1 = 4 and d = 6.

Here, the base case is when G− L does not contain a vertex with degree ≥ t. Hence,

we can have three possibilities. 1. G−L is empty. 2. G−L contains vertices with degree

at least t in G −K. 3. G − L does not contain a vertex of degree at least t in G −K.

Consider these cases one by one.

1. Let G− L be empty. As every vertex in G−K has degree at least d in G, G must

be a clique of size at least d+1. Also, G−K contains a vertex of degree at least t in

G−K. Hence, L−K contains at least t+ 1 vertices. Hence, we can always choose

root and children of the root from G − K. Vertex with largest index in L i.e. vl

corresponds to root. Children of the root can be selected from rightmost vertex in L

i.e. vl−1 to left. Hence, ti of T corresponds to vi+l−d−1. This gives an isomorphism

from T to a subtree T ′ of G that satisfies all the properties.

2. Let G− L contain vertices of degree at least t in G−K. Among these vertices, we

choose maximum degree vertex r in G−L as the root of T ′. Let c1, c2, · · · , cs be the

neighbours of r in G−L, where s ≤ t, are children of r in T ′. Since, the degree of r


is at least d in G, r has at least d− s neighbours in L. Also, degree of r is at least

t in G−K, we can choose cs+1, · · · , ct from G−K. Also, d′ − s ≤ t− s. Let, these

neighbours be in L′ and let S = {cs+1, · · · , ct} for the vertex ci where 1 ≤ i ≤ s,

each one has at least d′ − t neighbours in V (L′) \ S. As, V (L′) contains at least

d′ vertices. If it has less than d′ − t neighbours in V (L′) \ S then it must have at

least s + 1 neighbours in G − L and at least t + 1 neighbours in G − K, which is

not possible. Because, it contradicts the choice of the root. We might have selected

s+ 1 neighbours from G−L instead of S. These d′− t neighbours are smaller than

ct in V (L′) \ S.

So, for each ci, 1 ≤ i ≤ t, there will be at least d′− t neighbours in V (L′) \ S and if

ci ∈ L, it has at least d′− t neighbours which are smaller than ci in V (L′)\S. Here,

the order is maintained i.e., selection is not done randomly to avoid the problem.∑ti=1 di−1 = d1+d2 · · · dt−t = d′−t, so, we can always find disjoint subsets S1, · · ·St

of vertices in V (L′) \ S such that |Si| = di−1 and every vertex in Si is adjacent to

ci in T ′. The remaining d − d′ vertices are mapped to the vertices v1, v2 · · · vd−d′

respectively and add the edges joining these to the vertex in T ′ corresponding to

parent.

To get r − V (T ′) fan in G, r also has d − t neighbours in V (L) \ S. If a vertex v

in T ′ is adjacent to r in G, then rv forms a path from r to v in the fan. Else, we

can always choose a vertex w that is adjacent to r such that rwv forms a path in

the fan and w does not belong to any other path. So, in this way, we get r− V (T ′)

fans in G.

3. Let G − L be not empty but it does not contain a vertex of degree at least t in

G −K. In this case, we select a vertex from G −K of degree at least t in G −Kand proceed as above.

The inductive step is when G−L contains a vertex of degree at least t in G−L. Let

the vertices in L be ordered v1, v2, · · · , vl. Since, L is a maximal clique, for every vertex

v ∈ G− L, there is at least one vertex vi in L which is not adjacent to v. Let π(v) = vi

where i is the largest index in L such that v is not adjacent to vi. Remove the vertex vl

from the clique K. Let (G′, K ′) be the tuple obtained by removing vl. This may change

the degree of vertices. As it is given that the vertex present in G−K has degree at least

d in G, we need to add an edge for each vertex v ∈ G′ −K ′ and adjacent to vl. Hence,

we add an edge vπ(v) for each such v. Hence, we get

G′ = (G− vl) ∪ {vπ(v)|v ∈ V (G) \ V (L), π(v) 6= vl} and K ′ = L− vl.

Hence, K ′ is a clique in G′ such that G−L = G′−K ′ contains an edge and every vertex in

G′−K ′ has degree at least d in G′ since we have added an edge vπ(v) for every v ∈ G′−K ′

and adjacent to vl.


By induction hypothesis, there is a tree T ′ in G′ isomorphic to T , that is consistent

with the ordered clique L′ , such that the root r′ of T ′ and its children are in G′−L′ and

contains r′ − V (T ′) fans.

Remove the vertex vl in L and find the largest index vertex in L which is not adjacent

to the vertex v in G−L, which is adjacent to vl. This is required to maintain consistency

with the ordered clique. Because, removing the smallest index vertex may affect the

children position and the children may get higher position than its parent, which disobey

the property of consistency with the ordered clique referred in definition 2.2.1.

So, here we may assume that for the children of the root, the path in the fan is just

the edge in T ′ . Further, any path in the fan that intersects L′ contains at most one edge

in L′ .

The only edges of G′ that are missing in G are edges of the form vπ(v) for v ∈ G′−L′.If neither T ′ nor any of the paths in the r−V (T ′) fan contain any of these edges, then T ′

is the required tree in T . Note that T ′ is a consistent with any ordered sub-clique of L′

having the same ordering of vertices as L′, in particular K − vl. Suppose, in tree T ′ and

in path of the fans contains edges of the form vπ(v). We call such edges as bad edges.

Suppose there is an edge of the form vπ(v) in subtree T ′. Let π(v) = vi where vi is

the largest index vertex in L which is not adjacent to v. Hence, vi ∈ L′ . We remove the

largest index vertex vl in L. Hence, v must be parent of vi. In the base case, root of T

corresponds to vertex with the largest index in L and ordering is done from rightmost to

innermost. Hence, all children of vi must be in L′ and smaller than the index of vi. This

will get clearer in the example given after the proof. Once, it enters the clique L, all other

children of that vertex remain in the clique only, with leftmost direction in L. Therefore,

T ′ cannot contain any other edge of the form uπ(u) with π(u) = π(v) = vi. We call such

bad edge in T ′ as outgoing edge. Suppose a path in the r − V (T ′) fan contains an edge

of the form vπ(v) with π(v) = vi. If vi ∈ V (T ′) then this path must be terminating in

vi. We call this edges bad incoming edges. So, we get a vertex vi satisfies the following

properties.

• There are no bad edges incident with vi.

• There is exactly one bad edge incident with vi either bad incoming edge or bad

outgoing edge.

• There are exactly two bad edges incident with vi one of which is bad incoming edge

and other one is bad outgoing edge.

This completes the proof of theorem 2.2.2. The proof of the theorem 2.2.1 follows

from theorem 2.2.2. Hence, the proof.


Figure 2.3:

2.2.1 Example

Consider the graph G and tree T as shown in figure 2.3. We need to find the tree T and

r−V (T ) fans in G. Here d(t) = 2. The table shows the step by step procedure to remove

the vertex and adding corresponding edges to maintain the degree of vertices in G− L.

Figure 2.4:

Figure 2.5:

We stop at step 5 as G − L does not contain a vertex of degree ≥ t(2). Hence, now

we will find the subtree T ′ and r′ − V (T ′) fans where r′ is the root of subtree T ′ graph

G′ corresponding to that of tree T . As G − L is empty, a vertex ti of T corresponds to

vertex vi+l−d−1 of graph G′. Therefore, here root r′ is the vertex 11. Hence, we get the

tree T ′ as shown in figure 2.6.


Step no K L REM π Added edges Figure

1 {1} {1, 2, 5} {5} π(4) = 2,

π(6) = 2,

π(9) = 1,

π(10) = 2

2-4, 2-6, 1-9 and 2-10 1st fig of

figure 2.4

2 {1, 2} {1, 2, 7, 8, 9} {9} π(6) = 7,

π(3) = 8

3-8 and 6-7 2nd fig of

figure 2.4

3 {1, 2, 7, 8} {1, 2, 7, 8} {8} π(6) = 1,

π(3) = 7,

π(10) = 7

3-7, 1-6 and 7-10 1st fig of

figure 2.5

4 {1, 2, 7} {1, 2, 7, 6} {6} π(3) = 2,

π(4) = 1,

π(11) = 1

2-3, 1-4 and 1-11 2nd fig of

figure 2.5

5 {1, 2, 7} {1, 2, 7, 3,4, 10, 11}

{φ} figure 2.6

Table 2.3: Finding K, L and π at each step to get 3 disjoint paths in figure 2.3

Figure 2.6:

Going back to step 4 from step 5, there are two bad edges 11−1 and 2−3 and removed

vertex is 6. So, replacing bad edge with actual one, we will get the edges 11−6 and 6−3.

So, the final path is 11− 6− 3 and the another path is obtained by joining 1 and 2 to get

11− 2− 1− 3 shown in figure 2.7. Hence, going back to step 3,2 and 1, there are no bad

edges. Hence, the final paths are as shown in figure 2.7.

Let say if we remove the vertex in ascending order i.e., from the smallest index in K,

moving from the base case to the previous graph, if we find the bad edge in a tree as

shown in figure 2.8(a), then replacing bad edge with the original one, we get parent index

small than that of its children. This may result some inside portion of the tree in a clique

and their children outside the clique, in some later stage. This creates a problem and also

does not satisfy the constraint. Hence, the removal of a vertex is done from rightmost

i.e., with the largest index vertex in L i.e., vl.

If there are more than two incoming or outgoing edges, then also it creates problem in

a selection of tree. Consider the figure 2.8(b). There are two outgoing edges. Replacing


Figure 2.7:

(a) If smallest index vertexremoved

(b) More than two bad edgespresent

Figure 2.8:

them by actual edges, shift both to the same point which creates a problem in forming a

tree i.e., tree gets destroy. Hence, the removal of vertex is done with the largest index in

L i.e. the last vertex in L and also, while selecting edges, if an edge enters the clique, care

is taken so that remaining part of tree remain in a clique only. Till there are no vertex

outside the clique, a vertex is not selected from the clique. This gives the required result.

Corollary 1. Let G be a graph with minimum degree δ(G) ≥ d. Then, for all 1 ≤ r ≤ d,

there is a vertex v in G and a set S of r neighbours of v, such that v is d-linked to S in

G i.e., there are d internally vertex disjoint paths with exactly di of them end in vi.

This is the specific case of theorem 2.2.1.

Chapter 3

External Configuration

In this chapter, we extend the technique used to prove Mader’s theorem and define the

concepts of reduction and internal configuration. Using these concepts, we develop a

general technique which helps us to prove the property “Every graph G with minimum

degree at least |H| − 1 contains a subdivision of the graph H” for the different cases of

the graph H. The cases that we study are when H is P53, P6

3, wheel Wd i.e., Cd ∗K1 for

d ≥ 3 and caterpillars.

To understand the technique which we have developed, first we study the following

definitions and concepts and then see the technique.

3.1 Definitions

3.1.1 Reduction

LetG be a graph andK be an ordered clique of the graphG. Let v1, v2 · · · vk be the vertices

of the ordered clique K. Let L = v1, ..., vk, vk+1, ..., vl be a maximal clique containing K.

As we have seen in the proof of Mader’s theorem in section 2.1, π(v) = vi where i is the

smallest index in L such that vi is not adjacent to the vertex v and v ∈ G−L. Let G′ be

a new graph obtained by removing vertex v1 from the clique L of the tuple (G,K) and

K ′ = L − v1. This might change the degree of the vertices present in G′ − K ′. Hence,

an edge vπ(v) is added for each v ∈ G′ − K ′ and v ∈ adj(v1) so that the degree of the

vertices in G′ −K ′ remain unchanged. Hence, we can define reduction as the process of

adding vertices vk+1 · · · vl in the clique K of the graph G and then removing a vertex v1

from the clique K such that the degree of vertices in G− L remains unchanged.

Definition 3.1.1. (Reduction) If (G′, K ′) is a tuple obtained by applying reduction on

the tuple (G,K) so that either G′ = G and K ′ = K + v1 or G′ = (G− v1) ∪ {vπ(v) | v ∈V (G) \ V (L), π(v) > 1} and K ′ = K − v1, then we can say that (G′, K ′) is obtained by

applying reduction on the tuple (G,K).

13

14 3.1. Definitions

3.1.2 External Configuration

Definition 3.1.2. A configuration C is an arbitrary graph H(C) with non-negative integer

weights assigned to the vertices of H(C).

The non-negative weight assigned to the vertex is denoted by w(v). In this chapter,

we refer external configuration as ‘configuration’.

Definition 3.1.3. Let G be a graph and K be an ordered clique in G. We say that the

pair (G,K) contains a given configuration C if

1. G − K contains a subdivision H ′ of H(C). Let v′ denote the vertex in H ′ corre-

sponding to the vertex v in H(C).

2. For every vertex v of H(C), there are w(v) paths from v′ to K in G that pairwise

intersect in v′ and are internally disjoint from H ′ and K also. Denote this set of

paths by P (v).

3. For different vertices u and v of H(C), the set of paths P (u) ∪ P (v) are internally

disjoint.

(a) C1 (b) (G, K)

Figure 3.1: Configuration C1 and Tuple (G,K)

3.1.3 General Arguments

Let G be a graph and K be an ordered clique of the graph G. Let (Gi, Ki) be a tuple

obtained by applying successively i reductions on the tuple (G,K). Let (Gi+1, Ki+1) be a

tuple obtained by applying reduction on the tuple (Gi, Ki). Hence, either Gi+1 = Gi− v1

or Gi+1 = Gi and Ki = Ki+1 − vl.

Lemma 3.1.1. If (Gi+1, Ki+1) contains an arbitrary configuration C and if Gi is obtained

by adding a vertex to Gi+1, then (Gi, Ki) will also contain the same configuration.

Proof. Let P1, P2, .., Pk be the paths in P (v) for a vertex v in H(C). Let the endpoint of Pi

in Ki+1 be ti and let t−i be the vertex that precedes ti in Pi. Let T ′ = {ti ∈ T |π(t−i ) = ti}.Let T ′ = {vi1 , · · · vip} where i1 < · · · < ip.

3.2. General Technique 15

If π(v′) = vr, by definition of π(v′), v′vi is an edge in Gi for all i < r. For each vertex

vir ∈ T ′, replace the edge t−t in the path containing t by the edge t−vir−1 where vi0 = v1.

Hence, P (v) to the clique Ki from the vertex v ∈ Gi−Ki remains same as in Gi+1−Ki+1.

Hence, we get the same configuration. Hence, we can say that if Gi is obtained by adding

a vertex to Gi+1, then Gi will contain the same configuration as that of Gi+1.

Lemma 3.1.2. Let (Gi+1, Ki+1) contain an arbitrary configuration C. If Ki is obtained

by removing a vertex vl from Ki+1 and if S is the set of vertices {v| Some path in P (v)

terminates in vl}, then (Gi, Ki) will contain a configuration obtained by adding a vertex

v, corresponding to the vertex vl, and, joining it to vertices of S, reducing their weights

by 1 and assigning maximum of all these weights in C to the new vertex vl.

Proof. S is the set of vertices {v| Some path of P (v) terminates in vl}. Hence, on removing

vl, the number of paths entering the clique Ki starting from vertices in S will reduce by

1. Let w be the maximum of the weights assigned to the vertices in C. Then there will

be at least w vertices present in the clique Ki. Also, vl was in the clique Ki+1. Hence, vl

is connected to all the vertices present in the clique Ki. As there are at least w vertices,

vl will have weight at least w. Hence, the proof. Also, Gi −Ki will contain a subdivision

of the graph obtained by adding a new vertex v and joining it to vertices in S.

Corollary 2. Let (Gi+1, Ki+1) contain an arbitrary configuration C. If Ki is obtained by

removing a vertex v1 from Ki+1 and if vl is not the endpoint of any path in P (v) for any

v in C, then (Gi, Ki) will also contain the same configuration.

Proof. If S is the set of vertices {v| Some paths in P (v) terminates in vl}, then S = φ.

Hence, from lemma 4.2.2, weights of the vertices in the configuration C remain unchanged.

Therefore, we will get the same configuration.

Corollary 3. Let (Gi+1, Ki+1) contain an arbitrary configuration C where C contains at

least two vertices with maximum weight. If Ki is obtained by removing a vertex vl from

Ki+1 and vl is the endpoint of a path in P (v) for exactly one vertex v in C, then (Gi, Ki)

will also contain the same configuration.

Proof. If S is the set of vertices {v| Some paths of P (v) terminates in vl}, then |S| = 1.

Let w is the maximum weight. From lemma 4.2.2, weight of the vertex v in S will reduce

by 1 and vl will get assigned the maximum weight w. At least one vertex in Ki is not the

endpoint of any path in P (v). Continuing the path from vl to this vertex, gives the set of

paths P (v) in Gi. Hence, we will get the same configuration.

3.2 General Technique

We have to prove that for certain graphs H, every graph with minimum degree at least

|H| − 1 contains a subdivision of H. For proving this property for a particular case of H,

16 3.3. Different cases of graph H

let the number of vertices in the graph H be d + 1. Let G be a graph having minimum

degree at least d. Let K be an ordered clique of the graph G. The following steps describe

our technique which is based on the above definitions and arguments.

1. Choose a set Sc(H) of configurations that depend on H such that the only con-

figuration in Sc(H) that can be contained in (G, φ) is H with 0 weights to the

vertices.

2. Start with (G, φ).

3. Apply the reduction on (G, φ) till the reduced graph Gt is a clique.

4. Gt will be a clique of size at least d + 1. Hence, show that Gt contains one of the

configurations in Sc(H).

5. Prove if (Gi+1, Ki+1) contains one of the configurations in Sc(H) then (Gi, Ki) also

contains one of the configurations in Sc(H) where 1 ≤ i ≤ t− 1.

6. Hence, (G, φ) contains one of the configuration in Sc(H). The only possible config-

uration that can be contained in (G, φ) is H with 0 weights to the vertices i.e., the

subdivision of H, as K is φ.

3.3 Different cases of graph H

We now prove the property being studied for different cases of graph H using the technique

defined in the previous section.

3.3.1 Edge having d− 1 triangles

We have to prove that if H is a graph obtained by adding d−1 triangles having a common

edge, then every graph G having minimum degree at least d contains a subdivision of H.

There exists a proof given by Mader in section 2.1.

Configurations

For this case, the configurations are as shown in figure 3.2.

3.3.2 P53 (P3 ∗K2)

We have to prove that if H is the maximal planar graph with 5 vertices i.e. P53(P3 ∗K2),

then every graph G having minimum degree at least 4 contains a subdivision of H.

Configurations

Different configurations that are possible for this case are as shown in figure 3.3. We

denote the set of these configurations by Sc(H) where Sc = {C1, · · ·C7}.

3.3. Different cases of graph H 17

(a) C1 (b) C2

(c) C3 (d) C4

Figure 3.2: Configurations for H

(a) C1 (b) C2 (c) C3

(d) C4 (e) C5

(f) C6 (g) C7

Figure 3.3: Configurations for P53

Theorem 3.3.1. Let G be a graph with minimum degree d ≥ 4, then G contains a

subdivision of a graph P53.

Proof. 1. Let K be a clique of a graph G represented as a tuple (G,K). Initially,

assume K as empty set. Let L be a maximal ordered clique containing K. Let

(Gt, Kt) be a tuple obtained by applying successively t reductions on the tuple

(G,K) so that Gt is a clique.

2. Since, degree of every vertex in Gt −Kt is 4, Gt will be a clique of size at least 5.

3. If Gt − Kt contains one vertex v, then it will have 4 (v,Kt) paths. Hence, it will

contain C1 configuration. If Gt−Kt contains 2 vertices u and v, then there will be 3

(u,Kt) and 3 (v,Kt) paths. Hence, (Gt, Kt) will contain C2 configuration. Similarly


for other cases, we can prove that (Gt, Kt) will contain one of the configurations in

Sc(H).

4. Assuming that (Gi+1, Ki+1) contains any of the configurations in Sc(H), we need to

prove that (Gi, Ki) will also contain one of the configurations in Sc(H).

5. Let (Gi+1, Ki+1) contain a configuration C from Sc(H). If Gi is obtained by adding

a vertex to Gi+1, then from lemma 4.2.1, we can show that we will get the similar

configuration.

6. Consider the other possibility when Gi = Gi+1 and Ki = Ki+1−vl. Let S be a set of

vertices {v| Some path in P (v) terminates in vl}. Now, we will see the configuration

obtained for all possible cases of C and S.

(a) If S = φ, from corollary 2, we will get the same configuration.

(b) If |S| = 1 and C ∈ {C2 · · ·C7}, then from corollary 3, we will get the similar

configuration.

From lemma 4.2.2, we can see that following configurations are obtained for

remaining cases of C and S.

(c) C = C1, S = {x1} ⇒ C2.

(d) C = C2, S = {x1, x2} ⇒ C3.

(e) C = C3, S = {x1, x2} ∨ {x2, x3} ∨ {x1, x3} ⇒ C4.

(f) C = C3, S = {x1, x2, x3} ⇒ C5.

(g) C = C4, S = {x1, x2} ∨ {x2, x4} ⇒ C4. In this case, when x5 joins to x2 and

x4, from lemma 4.2.2, x2 will have 0 path entering the clique, x4 will have one

path while x5 will have 2 paths entering the clique Ki. We will get the similar

configuration by removing the edge x2 − x3 and contracting the edge x1 − x3.

(h) C = C4, S = {x1, x4} ⇒ C5.

(i) C = C4, S = {x1, x2, x4} ⇒ C5.

(j) C = C5, S = {x1, x2} ∨ {x2, x4} ⇒ C6.

(k) C = C5, S = {x1, x2, x4} ⇒ C7.

(l) C = C6, S = {x1, x5} ⇒ C7.

7. Hence, (G,K) will contain one of these configurations. As K is φ, the only possible

configuration is the final configuration C7 i.e. subdivision of H.


3.3.3 P63(P4 ∗K2)

We have to prove that if H is P63, then for every graph G having minimum degree at

least 5 contains a subdivision of H.

Configurations

(a) C1 (b) C2 (c) C3 (d) C4

(e) C5 (f) C6 (g) C7

(h) C8 (i) C9 (j) C10

Figure 3.4: Configurations for P63

The configurations for this case are as shown in the figure 3.4. We denote the set of

these configurations by Sc(H) where Sc(H) = {C1, · · ·C10}.

Theorem 3.3.2. If H is P63 (P4 ∗K2), then every graph G having minimum degree at

least 5 contains a subdivision of H.

Proof. The proof of this is similar to P53. Let (Gi+1, Ki+1) be a tuple obtained by applying

reduction on the tuple (Gi, Ki) where Ki is a clique of a graph Gi. If (Gi+1, Ki+1) contains

one of the configuration C in Sc(H) and S is the set of vertices {v| Some path in P (v)

terminates in vl} where Ki = Ki+1−vl, then possible configurations that can be contained

in (Gi, Ki) for all possible cases of C and S are as shown in table 3.1. We can see that

the remaining configurations in table 3.1 from 3 to 19 are obtained from lemma 4.2.2.

3.3.4 Subdivision of wheels Wd where d ≥ 3

We have to prove that if H is a wheel Wd i.e. Cd ∗K1 where d ≥ 3, then for every graph

G with minimum degree at least d contains a subdivision of H.


Sr.no. C S New Configuration

1 C ∈ {C1, · · ·C10} φ C from corollary 2

2 C ∈ {C2, · · ·C10} |S| = 1 C from corollary 3

3 C1 x1 C2

4 C2 {x1, x2} C3

5 C3 {x1, x2} ∨ {x2, x3} ∨ {x1, x3} C4

6 C3 {x1, x2, x3} C5

7 C4 {x1, x2} ∨ {x2, x4} C4

8 C4 {x1, x4} C5

9 C4 {x1, x2, x4} C5

10 C5 {x1, x2} ∨ {x2, x4} C6

11 C5 {x1, x2, x4} C7

12 C6 {x1, x5} C7

13 C6 {x4, x5} C8

14 C6 {x1, x4, x5} C9

15 C7 {x1, x4} ∨ {x1, x5} ∨ {x4, x5} C9

16 C7 {x1, x4, x5} C10

17 C8 {x1, x6} ∨ {x1, x5} C9

18 C8 {x1, x5, x6} C9

19 C9 {x1, x6} C10

Table 3.1: Configurations obtained for P63

Theorem 3.3.3. If G is a simple graph with d ≥ 3, then G has a subdivision of Wd i.e.,

Cd ∗K1.

There exists a proof by Galen E Turner [4]. The proof by our technique is given below.

Configurations

The configurations are as shown in figure 3.5. We denote the set of these configurations

by Sc(H).

Proof. The proof of this is similar to P3 ∗ K2. Let (Gi+1, Ki+1) be a tuple obtained by

applying reduction on the tuple (Gi, Ki) where Ki is a clique of a graph Gi. If (Gi+1, Ki+1)

contains one of the configurations C in Sc(H) and S is the set of vertices {v| Some path

in P (v) terminates in vl} where Ki = Ki+1 − vl, then possible configurations that can be

contained in (Gi, Ki) for all possible cases of C and S are as shown in table 3.2. For the

configurations C ′r and C ′′r , one of v1/vr has d−r−1 paths and the other has d−r. If w1 is the

weight assigned to the vertex v1 and wr to the vertex vr, then {w1, wr} = {d−r−1, d−r}.Hence, rather than having two different configurations, we can have a single configuration

with {w1, wr} = {d− r− 1, d− r}. We can see that the remaining configurations in table


(a) C1 (b) C2 (c) Cr (d) C ′r (e) C ′′r

(f) Cd−2 (g) C ′d−2 (h) C ′′d−2 (i) Cd−1 (j) C ′d−1

(k) Cf1 (l) C ′f1(m) Cf2 (n) Cf

Figure 3.5: Configurations for Wd where d ≥ 3

3.2 from 3 to 25 are obtained from lemma 4.2.2.

3.3.5 Caterpillars

A caterpillar graph is a tree such that if all leaf vertices and their incident edges are

removed, the remainder of the graph forms a path. We have to prove that if H is a graph

obtained by adding di−1 triangles on every edge i in the caterpillar graph, then for every

graph G having minimum degree at least d contains a subdivision of H. The proof of this

is similar to the above proof.

Proving above result automatically implies for a graph with d + 1 vertices such that

there exists a path between r + 1 vertices and ith edge has di − 1 triangles such that

1 ≤ i ≤ r and∑r

i=1 di = d.

Configurations

(a) C1 (b) C2 (c) C3 (d) C4

Figure 3.6: Configurations

The configurations are as shown in figure 3.6. We denote the set of these configurations


Sr.no. C S New Configuration

1 C ∈ {C1, · · ·Cf} φ C from corollary 2

2 C ∈ {C2, · · ·Cf} |S| = 1 C from corollary 3

3 C1 {v1} C2

4 C2 {v1, v2} Cr

5 Cr {v1, vr} C ′r6 Cr {v, v1} ∨ {v, vr} Cr+1

7 C ′r {v1, vr} C ′′r8 C ′r {v, v1} C ′r+1

9 C ′r {v, vr} Cr+1

10 C ′′r {v1, vr} C ′′r11 C ′′r {v, v1} ∨ {v, vr} C ′r+1

12 Cd−2 {v1, vd−2} C ′d−2

13 Cd−2 {v, v1} ∨ {v, vd−2} Cd−1

14 C ′d−2 {v1, vd−2} C ′′d−2

15 C ′d−2 {v, v1} C ′d−1

16 C ′d−2 {v, vd−2} Cd−1

17 C ′′d−2 {v1, vr} C ′′d−2

18 C ′′d−2 {v, v1} ∨ {v, vd−2} C ′d−1

19 Cd−1 {v1, vd−1} Cf2

20 Cd−1 {v, v1} ∨ {v, vd−1} Cf1

21 C ′d−1 {v1, vd−1} Cf2

22 C ′d−1 {v, v1} C ′f1

23 C ′d−1 {v, vd−1} Cf1

24 Cf1 ∨ C ′f1{v1, vd} Cf

25 Cf2 {v, vd} Cf

Table 3.2: Configurations obtained for Cd ∗K1 where d ≥ 3

by Sc(H).

Chapter 4

Internal Configuration

In this chapter, we develop an another technique which proves the property “Every graph

G with minimum degree |H| − 1 contains a subdivision of the graph H” for different

cases of the graph H. This technique is sometimes easier than the technique which we

developed in external configuration and gives us more result compared to it.

In this chapter, we define the concept of reduction and internal configuration which

helps to define the technique. Then we define the technique. The cases that we study are

when H is P53, P6

3, P73, wheel Wd i.e., Cd ∗ K1 for d ≥ 3 and 2-caterpillars. Also, we

show that every graph with minimum degree d contains a subdivision of a graph H with

d+ 1 vertices and 3d− 3 edges for d ≥ 2.

To understand the technique which we have developed, first we study the following

definitions and concepts and then see the technique.

4.1 Internal Configuration

Definition 4.1.1. (Configuration) An internal configuration is a set of graphs (or

multigraphs) having the same vertex set, with a subset of vertices assigned non-negative

weights. The vertices with weights assigned are called external vertices of the configuration,

the other vertices are said to be internal.

In this chapter, we refer an ‘Internal configuration’ as a configuration.

Let (G0, K0) be the initial tuple where K0 is a clique of the graph G0. Let (Gi, Ki)

be the tuple obtained by applying successively i reductions on the tuple (G0, K0). Let

X = {v1, v2, · · · , vk} be a subset of vertices in Gi −Ki. We say that X is (c1, c2, ..., ck)-

joined to Ki, if Gi contains c = c1 + c2 + · · · + ck internally disjoint X − V (Ki) paths

such that exactly ci paths have vi as an endpoint and no two paths have both endpoints

common. Denote these paths by P (v) where P (v) be the paths in this set that have vi as

an endpoint.

Let C be a configuration with k external vertices {u1, u2, .., uk} having weights a1, a2, · · · , ak

respectively.

Definition 4.1.2. (Good Configuration) We call C a good configuration if it satisfies

the following for all initial tuples (G0, K0) and all tuples (Gi, Ki) that are obtained from

23

24 4.2. General Argument

(G0, K0) by reduction:

If Gi−Ki contains a set X = {v1, v2, · · · , vk} of k vertices that is (a1, a2, ..., ak)-joined

to Ki, then G0 contains a subdivision of some graph in C such that vertex ui corresponds

to vertex vi.

4.2 General Argument

Let C be a configuration containing a graph with k external vertices {u1, u2, .., uk} having

weights c1, c2, · · · , ck respectively and some internal vertices.

Let (G0, K0) be an initial tuple. Let (Gi−1, Ki−1) be a reduced tuple obtained by

applying successively i − 1 reductions on the tuple (G0, K0). Let (Gi, Ki) be obtained

by applying reduction on (Gi−1, Ki−1). Gi−1 is obtained by adding a vertex to the clique

Ki in Gi and replacing the bad edges or Gi−1 is the same as Gi but Ki−1 is obtained by

removing a vertex from the clique Ki.

Lemma 4.2.1. If a set X = {v1, v2, · · · , vk} in Gi − Ki is {c1, c2, · · · , ck}-joined to Ki

and if Gi−1 is obtained by adding a vertex v0 to the clique Ki in Gi, then X is also

{c1, c2, · · · , ck}-joined to Ki−1 in Gi−1.

Proof. Let P1, P2, .., Pk be the paths in P (v) for a vertex v in Gi −Ki. Let the endpoint

of Pi in Ki be ti and let t−i be the vertex that precedes ti in Pi.

Let T ′ = {ti ∈ T |π(t−i ) = ti}. Let T ′ = {vi1 , · · · vip} where i1 < · · · < ip. If π(v′) = vr,

by definition of π(v′), v′vi is an edge in Gi−1 for all i < r. For each vertex vir ∈ T ′,

replace the edge t−t in the path containing t by the edge t−vir−1 where vi0 = v0. Hence,

the number of paths entering the clique Ki−1 from the vertex v ∈ Gi−1 − Ki−1 remains

the same. Hence, X = {v1, v2, · · · , vk} is {c1, c2, · · · , ck}-joined to Ki−1 in Gi−1.

Lemma 4.2.2. Let (Gi−1, Ki−1) be obtained by removing the vertex v from the clique Ki

to outside of Ki and let S be the set of vertices {u|some path in P (u) terminates in v}.If a set X = {v1, v2, · · · , vk} in Gi−Ki is {c1, c2, · · · , ck}-joined to Ki and S is empty or

|S| = 1 and at least two different ci have maximum value , then X is also {c1, c2, · · · , ck}-joined to Ki−1 in Gi−1.

Proof. Consider the case when S is empty. In that case, the set of paths P (u) starting from

the vertex u ∈ Gi−1−Ki−1 remains same as that in Gi−Ki. Hence, X = {v1, v2, · · · , vk}is {c1, c2, · · · , ck}-joined to Ki−1 in Gi−1.

Let |S| = 1. Hence, only a path in P (u) terminates at v. Hence, |P (u)| is reduced

by 1. As the vertex v was in the clique Ki previously, it joins to all the vertices present

in the clique Ki−1. Since, the set of paths P (u) from the vertex u reduces by 1 and as

there are at least two ci with maximum value, there exists at least one vertex in Ki−1 to

which u is not joined by a path in p(u) but v is joined. Let w be such a vertex. v has

4.3. Good-Configuration 25

a path from only one vertex u ∈ X terminating at it. Hence, we can get a path from u

to w as u to v and v to w and add this path to the set P (u) in Gi−1 −Ki−1. Therefore,

|p(u)| in Gi−1 −Ki−1 remains same as that of in Gi −Ki. Hence, X = {v1, v2, · · · , vk} is

{c1, c2, · · · , ck}-joined to Ki−1 in Gi−1.

4.3 Good-Configuration

4.3.1 Example 1

Let (G0, K0) be an initial tuple. Let (Gi, Ki) be a reduced tuple obtained by applying

successively i reductions on the tuple (G0, K0).

Let C be a configuration containing two graphs with 4 external vertices {u1, u2, u3, u4}having weights {1, 1, 2, 2} respectively and no internal vertices as shown in figure 4.1. One

graph contain edges u1u2 and u3u4 while other graph contains edges u1u3 and u2u4. We

prove that this configuration is good.

(a) G1 in C (b) G2 in C

Figure 4.1: Configuration C

Proof. We use induction on the number of steps in the reduction used to obtain (Gi, Ki)

from (G0, K0).

The base case is when the number of step is 0 i.e. i = 0. It is trivially true, since the

if condition is false. As K0 is empty, there is no set in G0 −K0 that is (1, 1, 2, 2)-joined

to K0.

Let the number of steps i be greater than 0. Let (Gi, Ki) be the tuple obtained

after i reductions and let X = {v1, v2, v3, v4} be the set of vertices in Gi − Ki that is

{1, 1, 2, 2}-joined to Ki in Gi as shown in figure 4.2(a).

Let (Gi−1, Ki−1) be the reduced tuple obtained by applying i − 1 reductions on the

tuple (G0, K0). We know that either Gi−1 is obtained by adding a vertex to the clique Ki

in Gi or Gi−1 is same as Gi and Ki−1 is obtained by removing a vertex from the clique

Ki in Gi.

If the first case holds, by lemma 4.2.1, X is {1, 1, 2, 2}-joined to Ki−1 in Gi−1. By

applying induction on the number of steps, (G0, K0) contains a subdivision of some graph

in C, with vertex u1 corresponding to vertex v1, u2 to v2, u3 to v3 and u4 to v4.

Consider the other case. Let S ⊂ X be the subset of vertices {u|some path in P (u)

terminates in v}. If |S| ≤ 1, here we know that at least 2 different ci have maximum

value when |S| = 1, hence by lemma 4.2.2, X is {1, 1, 2, 2}-joined to Ki−1 in Gi−1. By

26 4.3. Good-Configuration

(a) (Gi, Ki) (b) (Gi−1, Ki−1) (c) (G′i−1, Ki−1)

(d) C ′ (e) G′0 (f) G0

Figure 4.2: {v1, v2} ⊂ S

applying induction on the number of steps, (G0, K0) contains a subdivision of some graph

in C, with vertex u1 corresponding to vertex v1, u2 to v2, u3 to v3 and u4 to v4.

Assume the set |S| ≥ 2. Consider the following cases.

1. {v1, v2} ⊂ S i.e. v is joined to v1 and v2 by paths as shown in figure 4.2(b). Consider

it as a path from v1 to v2. Delete all the vertices present on this path from v1 to

v2 including v1 and v2. Let, G′i−1 = Gi−1 − Q where Q is a path from v1 to v2

as shown in figure 4.2(c). Let G′0 be obtained by deleting all the vertices in Q.

Let C ′ be a configuration containing a graph with 2 external vertices {u′1, u′2} of

weight {1, 1} and an edge u′1u′2 (figure 4.2(d)). We can show that C ′ is a good

configuration. Since, {v3, v4} is {1, 1}-joined to Ki−1 in G′i−1 and (G′i−1, Ki−1) can

be obtained by reducing (G′0, K0), (G′0, K0) contains the subdivision of the graph in

the configuration C ′ containing path v3v4 maps to the edge u′1u′2. Together with a

path Q, we get (G0, K0) contains the subdivision of the graph in the configuration

C.

(a) (Gi, Ki) (b) (Gi−1, Ki−1) (c) (G′i−1, Ki−1)

Figure 4.3: {v2, v3} ⊂ S

2. {v1, v3} ⊂ S but v2 6∈ S i.e. v is joined to v1 and v3 by paths. Consider it as

a path from v1 to v3. Delete all the vertices present on this path from v1 to v3

including v1 and v3. Let G′i−1 = Gi−1 −Q where Q is a path from v1 to v3. Let G′0be obtained by deleting all the vertices in Q. Let C ′ be a configuration containing

a graph with 2 external vertices {u′1, u′2} of weight {1, 1} and an edge u′1u′2. We

4.3. Good-Configuration 27

Figure 4.4: {v2, v3} ⊂ S and Graph G′0

Figure 4.5: {v2, v3} ⊂ S and Graph G0

can show that C ′ is a good configuration. Since, {v2, v4} is {1, 1}-joined to Ki−1 in

G′i−1 and (G′i−1, Ki−1) can be obtained by reducing (G′0, K0), (G′0, K0) contains the

subdivision of the graph in the configuration C ′ containing path v2v4 that maps to

an edge u′1u′2 in C ′. Together with a path Q, we get (G0, K0) contains the subdivision

of the graph in the configuration C.

3. {v2, v4} ⊂ S but v1 6∈ S. Same argument holds as that of previous. Here, (G0, K0)

contains the subdivision of the graph in the configuration containing subdivision of

edges u1u3 and u2u4.

4. {v3, v4} ⊂ S but v1, v2 6∈ S. Again same argument holds as that of previous.

Here, (G0, K0) contains the subdivision of the graph in the configuration containing

subdivision of edges u1u2 and u3u4.

5. {v2, v3} ⊂ S i.e. v is joined to v2 and v3 by paths (figure 4.3(a)). Delete all the ver-

tices present on this path v2 to v3 including v2 but not v and v3. Hence, (G′i−1, Ki−1)

contains 4 vertices {v1, v3, v4, v} that are {1, 1, 2, 2}-joined to Ki−1 (figure 4.3(c)).

Hence, by applying induction on the number of steps, (G′0, K0) contains the subdi-

vision of the graph in the configuration C containing subdivision of edges u1u2 and

u3u4 or subdivision of edges u1u3 and u2u4 i.e. paths v1v3 and vv4 or v1v and v3v4

(figure 4.4). Adding back the path vv2, (G0, K0) contains paths v1v3 and v2v4 or v1v2

and v3v4. Hence, (G0, K0) contains the subdivision of the graph in the configuration

(figure 4.5).

6. {v1, v4} ⊂ S. Same argument holds as that of above.

This proves that the configuration C is a good configuration.

28 4.3. Good-Configuration

4.3.2 Example 2


successively i reductions on the tuple (G0, K0).

Let C be a configuration containing a graph with 3 external vertices {u1, u2, u3} having

weights {2, 2, 2} respectively and internal vertices u4 and u5. It contains edges u1u4, u1u5,

u2u4, u2u5, u3u4 and u4u5 as shown in figure 4.6. we prove that this configuration is good.

Figure 4.6: Configuration C


from (G0, K0).

The base case is when the number of step is 0 i.e. i = 0. It is trivially true, since the

if condition is false. As K0 is empty, there is no set in G0 −K0 that is (2, 2, 2)-joined to

K0.

Let the number of steps i be greater than 0. Let (Gi, Ki) be the tuple obtained after i

reductions and let X = {v1, v2, v3} be the set of vertices in Gi−Ki that is {2, 2, 2}-joined

to Ki in Gi.




Ki in Gi.

We assume that Gi−1 is obtained by removing a vertex v from the clique Ki in Gi and

|S| ≥ 2 where S ⊂ X be the subset of vertices {u|some path in P (u) terminates in v}.We need to consider the following cases.

1. {v1, v2} ⊂ S i.e. v is joined to v1 and v2 by paths. Let Q be a path from v1 to

v2 excluding the vertex v. Let G′i−1 = Gi−1 − Q. Let G′0 be obtained by deleting

all the vertices in Q. Hence, (G′i−1 −Ki−1) contains vertices {v1, v2, v3, v} that are

{1, 1, 2, 2}-joined to Ki−1. Let C ′ be a configuration containing two graphs with

4 external vertices {u′1, u′2, u′3, u′4} with weights {1,1,2,2} and 1 internal vertex u′5(sixth figure in figure 4.9). One graph contains edges u′1u

′5, u

′2u′5, u

′3u′5 and u′4u

′5

while other contains edges u′1u′5, u

′2u′5, u


′4. We can show that C ′ is a

good configuration. Hence, (G′0, K0) contains the subdivision of the graph in the

configuration C ′ containing paths v1v5, v2v5, v3v5 and vv5 or paths v1v5, v2v5, v3v

and vv5. Together with a path Q i.e. paths v1v and v2v, (G0, K0) contains the

subdivision of the graph in the configuration C.

4.4. Bad-Configuration Example 29

2. {v1, v2, v3} ⊂ S. LetQ contain paths v1v, v2v, v3v including v3. LetG′i−1 = Gi−1−Q.

Let G′0 be the corresponding graph obtained by deleting all the vertices in Q. Hence,

(G′i−1−Ki−1) contains vertices {v1, v2, v} that are {1, 1, 1}-joined to Ki−1. Let C ′ be

a configuration containing a graph having 3 external vertices {u′1, u′2, u′3} of weights

{1,1,1} and an internal vertex u′4 with edges u′1u′4, u


′4 (first figure in

figure 4.8). We can show that C ′ is a good configuration. Hence, (G′0, K0) contains

the subdivision of the graph in the configuration C ′ containing paths v1v5, v2v5 and

vv5. Together with paths v1v, v2v and v3v, (G0, K0) contains the subdivision of the

graph in the configuration C.

3. {v2, v3} ⊂ S. Following the same procedure as above we can show that (G′0, K0)

contains the configuration C ′ where C ′ contains a graph with 3 external vertices

{u1, u2, u3} of weights {1, 2, 2} and an internal vertex u′4 with edges u′1u′4, u

′2u′4, u

′2u′3

and u′3u′4 (2 figure in figure 4.8). Together with paths v2v and v3v, we get (G0, K0)

contains the subdivision of the graph in the configuration C. Same argument follows

when {v1, v3} ⊂ S.

This proves that the configuration C is a good configuration.

4.4 Bad-Configuration Example

(a) Configuration C (b) (G0, K0)

Figure 4.7: {v2, v3} ⊂ S and Graph G0


successively i reductions on the tuple (G0, K0). Let a configuration C contain a graph

with 4 external vertices {u1, u2, u3, u4} having weights {1, 1, 2, 2} and no internal vertices

as shown in figure 4.7(a). It contains edges u1u2 and u3u4. We need to show that C is

not a good configuration.

Let G0 contain 6 vertices {v1, v2, v3, v4, v5, v6}. Let G0 contain edges v1v5, v3v5, v4v5

and v2v6, v3v6, v4v6, v5v6. The reduced tuple is (G0, {v5, v6}). The vertices {v1, v2, v3, v4}are (1, 1, 2, 2)-joined to {v5, v6} but G0 does not contain two disjoint paths from v1 to v2

and v3 to v4 (figure 4.7(b)).

30 4.5. List of good configurations

4.5 List of good configurations

Here we list some configurations that can be shown to be good. These are useful in

proving that some other configuration are good. These are as shown in figure 4.8 and 4.9.

1.

Configuration C which contains a graph with 3 external

vertices {u1, u2, u3} with weights {1, 1, 1} and an internal

vertex u4. It contains edges u1u4, u2u4 and u3u4.

2.

Configuration C (which contains a graph with 3 external

vertices {u1, u2, u3} with weights {1, 2, 2} and an internal

vertex u4. It contains edges u1u4, u2u4, u2u3 and u3u4.

3.


vertices {u1, u2, u3} of weights {3, 3, 3} and internal ver-

tices u4, u5 and u6. It contains edges u1u4, u1u5, u1u6,

u2u4, u2u5 and u3u4. Also, {u1, u2, u3} = {0, 1, 2}

4.

Configuration C which contains two graphs with 4 external

vertices {u1, u2, u3, u4} having weights {1, 1, 2, 2} and no

internal vertices. One graph contains edges u1u2 and u3u4

and the other contains u1u3 and u2u4.

5.



tices u4 and u5. It contains edges u1u4, u1u5, u2u4, u2u5,

u3u5 and u4u5.


4.6 General Result

Theorem 4.6.1. Every graph of minimum degree d ≥ 2 contain a subdivision of some

graph with d+ 1 vertices and 3d− 3 edges.

This is the generalization of Dirac’s theorem [4] and Pelikan’s theorem [3]. Dirac’s

theorem says that if the graph G has minimum degree d ≥ 3 then it contains subdivision

of K4. It contains 6 i.e., 3∗3−3 edges. According to Pelikan, if the graph G has minimum

degree d ≥ 4, then it contains subdivision of K5 − e. It contains 9 i.e., 3 ∗ 4 − 3 edges.

Also, we prove in the next section that if the graph G has d ≥ 5, then it contains the

4.6. General Result 31

6.

Configuration C which contains two graphs with 4 external

vertices {u1, u2, u3, u4} having weights {1, 1, 2, 2} and an

internal vertex u5. One graph contains edges u1u5, u2u5,

u3u5 and u4u5 while the other contains edges u1u5, u2u5,

u3u5 and u3u4.

7.



tices u4 and u5. It contains edges u1u2, u1u4, u1u5, u2u4,

u2u5, u3u5 and u4u5.

8.


vertices {u1, u2, u3} of weights {3, 2, 3} and an internal ver-

tex u4. It contains edges u1u2, u1u3, u1u4, u2u4 and u3u4.

9.



tices u4, u5 and u6. It contains edges u1u4, u1u5, u1u6,

u2u4, u2u5 and u3u4. Also, {u1, u2, u3} = {0, 1, 2}


subdivision of P63 (3*5-3 edges) and if G has d ≥ 6, then it contains the subdivision of

P73 (3*6-3 edges).

To prove the theorem 4.6.1 we show that following configurations are good.

1. This configuration has 3 external vertices u1, u2, u3 having weight d each where

d ≥ 0. The graphs in the configuration are all possible graphs with d internal

vertices u4, u5, ..., ud+3 such that every internal vertex i is adjacent to exactly 3

vertices with a number less than i.

2. This configuration has 3 external vertices u1, u2, u3 having weights d, d, d − 1

respectively where d ≥ 1. The graphs in this configuration contain the edge u1u2

and d− 1 internal vertices u4, u5,...,ud+2 such that every internal vertex is adjacent

to exactly 3 vertices with a smaller number.

3. This configuration has k ≥ 3 external vertices u1, u2, u3, .. , uk having weights d, d,

d−1, d−2, ..., d−k+2 respectively where d ≥ k−2. The graphs in this configuration

are all possible graphs in which vertex u1 is adjacent to all other external vertices

u2, u3, .., uk and there are d− k + 1 internal vertices uk+1, uk+2, ..., ud+1 such that

every internal vertex is adjacent to exactly 3 vertices with a smaller number.

32 4.6. General Result

We show that all these configurations are good. Let (G0, K0) be an initial tuple. Let

(Gi, Ki) be a reduced tuple obtained by applying successively i reductions on the tuple

(G0, K0). We need to prove that the above configurations are good configurations. Essen-

tially, we need to show that depending upon the subset of external vertices that the new

vertex joins, the problem reduces to finding one of these configurations in (Gi−1, Ki−1)

where (Gi−1, Ki−1) is a reduced tuple obtained by applying successively i− 1 reductions

on the tuple.

(a) Graphs in C1 when d = 2,{u1, u2, u3} = {0, 1, 2}

(b) Graphs in C2 when d = 3,{u1, u2, u3} = {0, 1, 2}

(c) Graphs in C3 when d = 3,{u1, u2, u3} = {0, 1, 2}

Figure 4.10: Configurations for General result


from (G0, K0). Consider the configurations when there is no internal vertex. These

(a) Base 1 (b) Base 2 (c) Base 3

Figure 4.11: Base cases for General result

configurations just consist of paths as shown in figure 4.15. We can prove that (G0, K0)

contains the subdivision of the graph in the configuration by Mader’s argument.

4.6. General Result 33

Consider the case when there is one internal vertex. Apply base case. The base case

is when the number of step is 0 i.e. i = 0. It is trivially true, since the if condition is

false. As K0 is empty, there is no set in G0 −K0 that is (1, 1, 1)-joined or (2, 2, 1)-joined

or (k, k, k − 1, · · · , 1)-joined to K0.

Let the number of steps i be greater than 0. Let (Gi, Ki) be the tuple obtained after

i reductions.


|S| ≥ 2 where S ∈ X be the subset of vertices {u|some path in P (u) terminates in v}.If v joins to more than two vertices then in that case we can consider it as a new vertex

joining to any three vertices and hence, (Gi−1 − Ki−1) contains the same vertices with

p(u) of each vertex u of Gi−Ki reduces by 1 in Gi−1−Ki−1. Hence, by induction on the

number of steps, (G0, K0) contains the subdivision of the graph in the same configurations.

Hence, we can ignore this case. We need to consider the following cases.

1. Let (Gi −Ki) contains 3 vertices {v1, v2, v3} that are {d, d, d}-joined to Ki.

(a) Let {v1, v2} = S i.e. v joins to v1 and v2 by paths. Hence, Gi−1−Ki−1 contains

4 vertices {v1, v2, v3, v} that are {d − 1, d − 1, d, d}-joined to Ki−1. Let Q be

a path from v1 to v2 through v. Let G′i−1 be obtained by deleting the path

Q and vertex v2. Let G′0 be the graph obtained by deleting all the vertices

present on the path Q and the vertex v2 from G0. Hence, G′i−1−Ki−1 contains

3 vertices {v, v3, v1} that are {d, d, d− 1}-joined to Ki−1. Hence, by induction

on the number of steps, (G′0, K0) contains the subdivision of the graph in the

second configuration. Adding back the path Q in this graph, we get (G0, K0)

contains the subdivision of the graph in the first configuration.

2. Let (Gi −Ki) contain 3 vertices {v1, v2, v3} that are {d, d, d− 1}-joined to Ki.

(a) Let {v1, v2} = S i.e. v joins to v1 and v2 by paths. Let Q be a path from v1 to

v2 including v. Let G′i−1 = Gi−1−Q. Let G′0 be the graph obtained by deleting

the path Q from G0. Hence, (G′i−1 −Ki−1) contains 3 vertices {v1, v2, v3} that

are {d − 1, d − 1, d − 1}-joined to Ki−1. Hence, by applying induction on the

number of steps, (G′0, K0) contains the subdivision of the graph in the first

configuration. Together with the path Q, (G0, K0) contains the subdivision of

the graph in the second configuration.

(b) Let {v2, v3} = S i.e. v joins to v2 and v3 by paths. Using the same technique

as above, we get (G′0, K0) contains the subdivision of the graph in the third

configuration. Hence, (G0, K0) contains the subdivision of the graph in the

second configuration.

(c) Similarly, we can prove this when {v1, v3} ⊂ S.

34 4.6. General Result

3. Let (Gi−Ki) contain k vertices {v1, v2, · · · , vk} that are {d, d, d− 1, · · · , d− k+ 2}-joined to Ki where k ≥ 3.

(a) Consider the case when k = 3.

i. Let {v1, v2} = S. Let Q be a path from v1 to v2 including v. Let G′i−1 =

Gi−1 −Q. Let G′0 be the initial graph of G′i−1 obtained by deleting all the

vertices on the path Q. Hence, G′i−1−Ki−1 contains 3 vertices {v1, v2, v3}that are {d − 1, d − 2, d − 1}-joined to Ki−1. Hence, by induction on the

number of steps, (G′0, K0) contains the subdivision of the graph in the

second configuration. Adding the path Q, we get (G0, K0) contains the

subdivision of the graph in the third configuration.

ii. Let {v1, v3} = S. Let Q be a path from v1 to v3 including v. Let G′i−1 =

Gi−1 −Q. Let G′0 be the initial graph of G′i−1 obtained by deleting all the

vertices on the path Q. Hence, G′i−1−Ki−1 contains 3 vertices {v1, v2, v3}that are {d − 1, d − 1, d − 2}-joined to Ki−1. Hence, by induction on the

number of steps, (G′0, K0) contains the subdivision of the graph in the

second configuration. Adding the path Q, we get (G0, K0) contains the

subdivision of the graph in the third configuration.

iii. Let {v2, v3} = S. Let Q be a path from v2 to v and v3 to v excluding

v. Let G′i−1 = Gi−1 − Q. Let G′0 be the initial graph of G′i−1 obtained

by deleting all the vertices on the path Q. Hence, G′i−1 −Ki−1 contains 4

vertices {v1, v, v2, v3} that are {d, d, d − 1, d − 2}-joined to Ki−1. Hence,

by induction on the number of steps, (G′0, K0) contains the subdivision of

the graph in the third configuration. Adding the path Q, we get (G0, K0)

contains the subdivision of the graph in the third configuration.

(b) Consider k > 3.

i. Let v join to v1 and some other vertex in S ′ = {v2, v3, ..., vk} say v′ where

v′ ∈ S ′. Let Q be a path from v1 to v and v to v′ where v′ ∈ S ′. Let G′i−1

be obtained by deleting the path Q including vertex v′ ∈ S ′ such that vi

is the vertex with the smallest number of paths in set S ′. Similarly we

can obtain G′0 by deleting the path Q from G0 and the vertex v′. Assume

v′ = vi. Hence, G′i−1 − Ki−1 contains vertices {v1, .., vi−1, vi+1..vk} that

are {d − 1, d − 1, d − 2, .., d − k + 3}-joined to Ki−1. Hence, by applying

induction on the number of steps, (G′0, K0) contains the subdivision of the

graph in the third configuration. Adding back the path Q, we get (G0, K0)

contains the subdivision of the graph in the third configuration.

ii. Let v join to any two vertices (say v2,v3) other than v1. So, v can be

considered as the new vertex, join to v2, v3 and we need one more path from

v to some vertex in Gi−1−Ki−1 i.e. vv1. For u ∈ {v2, · · · , vk}, reduce P (u)

4.7. Applications of Good-Configurations 35

of every vertex by 1. Let Q be a path from v to v2 and v to v3. Let G′i−1 =

Gi−1−Q. Let G′0 be the initial graph of G′i−1. Hence, G′i−1−Ki−1 contains

vertices {v1, v, v2, v3, · · · , vk} that are {d, d, d−1, d−2, · · · , d−k+3}-joined

to Ki−1. Hence, by applying induction on the number of steps, we get

(G′0, K0) contains the subdivision of the graph in the third configuration.

Together with the path Q, (G0, K0) contains the subdivision of the graph

in the third configurations.

Hence, the proof.

4.7 Applications of Good-Configurations

Lemma 4.7.1. If G0 has minimum degree at least d, then there is a reduction (Gi, Ki)

of (G0, K0) such that Gi −Ki contains a subdivision of K3, the vertices of K3 are {d −2, d− 2, d− 2}-joined to Ki.

Proof. Let (G0, K0) be an initial tuple. Let (Gj, Kj) be a reduced tuple obtained by

applying successively j reductions on the tuple (G0, K0) so that Gj is a clique. Since, the

degree of every vertex in Gj −Kj is at least d, Gj is a clique of size at least d+ 1.

Let p(u) be the set of paths starting from vertex u which enters to the clique Kj. We

assume that Gj−1 is obtained by removing a vertex v from the clique Kj in Gi and |S| ≥ 2

where S ∈ X be the subset of vertices {u|some path in P (u) terminates in v}.If Gj − Kj contains one vertex v1, then it has d paths entering to the clique Kj.

Hence, (Gj−1, Kj−1) is obtained by removing a vertex v2 from the clique Kj and there

exists a path from v1 to v2. Hence, Gj−1 − Kj−1 contains two vertices {v1, v2} that are

{d−1, d−1}-joined to Kj−1 and a subdivision of K2. Similarly, Gj−2−Kj−2 contains three

vertices {v1, v2, v3} that are {d−2, d−2, d−2}-joined to Kj−2 and it contains subdivision

of K3 as there exists a cycle containing paths v1v2, v2v3 and v1v3. Let i = j − 2. Hence,

we can apply i reductions on (G0, K0) so that (Gi −Ki) contains subdivision of K3 and

vertices are {d− 2, d− 2, d− 2}-joined to Ki.

If Gj−Kj contains two vertices v1, v2, then we can apply the same technique as above

and here i = j − 1. If Gj −Kj contains three vertices {v1, v2, v3}, then as Gj is a clique,

we get K3 outside and each vertex in Gj − Kj has d − 2 paths entering to the clique.

Hence, if Gj −Kj contains one or two vertices, we can always get (Gi, Ki) from (Gj, Kj)

such that Gi −Ki contains K3 outside.

If Gj −Kj contains more than two vertices, then it is a clique as Gj and Kj both are

clique. Hence, we can always apply r reductions by adding vertices from Gj −Kj to the

clique Kj till Gj −Kj contains a subdivision of K3. Here, i = j + r. Hence, the vertices

of Gi −Ki are {d− 2, d− 2, d− 2}-joined to Ki where Gi −Ki contains the subdivision

of Kr. Hence, the proof.

36 4.7. Applications of Good-Configurations

4.7.1 P53

Theorem 4.7.1. If a graph G0 has minimum degree d ≥ 4 then it contains the subdivision

of P53 i.e. P3 ∗K2.

Figure 4.12: P53

Proof. Let (G0, K0) be an initial tuple. Let (Gi, Ki) be a reduced tuple obtained by

applying successively i reductions on the tuple (G0, K0).

We know that we can get K3 in Gi−Ki from lemma 4.7.1. Let X = {v1, v2, v3} be the

set of vertices in Gi −Ki that are {2, 2, 2}-joined to Ki. Let G′i is obtained by deleting a

path v1v2, v1v3 and v2v3. Hence, vertices of G′i −Ki are also {2, 2, 2}-joined to Ki. Let

G′0 be the initial graph of reduced graph G′i, obtained by deleting the above paths.

Let C be a configuration containing a graph with 3 external vertices {u1, u2, u3} of

weights {2, 2, 2} and internal vertices u4 and u5 as shown in the fifth figure of figure 4.8.

We know that it is a good configuration. As vertices of G′i−Ki are {2, 2, 2}-joined to Ki,

(G′0, K0) contains the subdivision of the graph in the configuration C. Adding deleted

paths back, we get (G0, K0) contains the subdivision of P53.

4.7.2 P63


of P63 i.e. P4 ∗K2.

Figure 4.13: P63



We know that we can get K3 in Gi − Ki from lemma 4.7.1. Let X = {v1, v2, v3} be

the set of vertices in Gi −Ki that are {3, 3, 3}-joined to Ki.

Let C be a configuration which contains a graph with 3 external vertices {u1, u2, u3}of weights {3, 3, 3} and internal vertices u4, u5 and u6. It contains edges u1u4, u1u5, u1u6,

u2u4, u2u5 and u3u4. Also, {u1, u2, u3} = {0, 1, 2} (shown in figure 9 of figure 4.9). To


show that this is a good configuration we need to show that following three configurations

are good.

1. Configuration C1 containing a graph with 3 external vertices {u1, u2, u3} of weights

{2, 2, 2} and internal vertices u4 and u5 (fifth figure of figure 4.8).

2. Configuration C2 which contains a graph with 3 external vertices {u1, u2, u3} of

weights {3, 3, 2} and internal vertices u4 and u5. It contains edges u1u2, u1u4, u1u5,

u2u4, u2u5, u3u5 and u4u5 (figure 7 of figure 4.9).

3. Configuration C3 which contains a graph with 3 external vertices {u1, u2, u3} of

weights {3, 2, 3} and an internal vertex u4. It contains edges u1u2, u1u3, u1u4, u2u4

and u3u4 (shown in figure 8 of figure 4.9).

We know that all these configurations are good. Hence, the configuration C is the good

configuration.

Let G′i is obtained by deleting a path v1v2, v1v3 and v2v3. Hence, vertices of G′i −Ki

are also {3, 3, 3}-joined to Ki. Let G′0 be the initial graph of reduced graph G′i, obtained

by deleting the above paths. Hence, (G′0, K0) contains the subdivision of the graph in

the configuration C. Adding back these deleted paths, we get (G0, K0) contains the

subdivision of P63.

4.7.3 P73


of P73.



We know that we can get K3 in Gi − Ki from lemma 4.7.1. Let X = {v1, v2, v3} be

the set of vertices in Gi −Ki that are {4, 4, 4}-joined to Ki.




Ki in Gi.


|S| ≥ 2 where S ⊂ X be the subset of vertices {u|some path in P (u) terminates in v}.We need to consider the following cases.

1. Let {v1, v2, v3} ⊂ S. Then we get K4 outside. Hence, vertices {v1, v2, v3, v} of

Gi−1−Ki−1 are {3, 3, 3, 3}-joined to Ki−1. Let Q contain paths from the vertex v to

other vertices including vertex v. Let G′i−1 = Gi−1−Q. Let G′0 be the initial graph

of G′i−1 obtained by deleting all the vertices present on the path Q from G0. Hence,


Figure 4.14: (Gi−1, Ki−1)

vertices {v1, v2, v3} of G′i−1 − Ki−1 are {3, 3, 3}-joined to Ki−1. Hence, (G′0, K0)

contains the subdivision of the graph in the configuration C where C is as shown

in ninth figure of figure 4.9. Adding back the deleted paths, (G0, K0) contains the

subdivision of P73.

2. Let v join to any two vertices in Gi − Ki say {v2, v3}. Then vertices of {v1, v2, v}of Gi−1 −Ki−1 are {4, 3, 4}-joined to Ki−1. We can always maintain this structure

(refer proof of P4 ∗K2 of section 3.3.3 in chapter 3). Hence, if next vertex comes out

join to all these three vertices, we get subdivision of K4 outside else we can maintain

the {4, 3, 4} structure. Hence, similar to above arguments, we get (G0, K0) contains

the subdivision of P73.

4.7.4 Wheel

Let a configuration C contain a graph with 3 external vertices a, b and c with weights d,

d and d−1 and internal vertices u1, u2, · · · , ud. It contains a path from u1, u2, · · · , ud with

vertex a joined to u1, vertex b joined to all u1, u2, · · · , ud and vertex c joined to ud.

Theorem 4.7.4. If G0 be the graph having minimum degree at least d+2, then it contains

the subdivision of wheel wd+2 i.e. Cd+2∗K1 or subdivision of the graph in the configuration

C.

We need to prove that C is a good configuration.


from (G0, K0). The base case is when the number of step is 0 i.e. i = 0. It is trivially

(a) C (b) {2, 2, 1}

Figure 4.15: Wheel Configuration and Case d = 2

true, since the if condition is false. As K0 is empty, there is no set in G0 − K0 that is

(d, d, d− 1)-joined to K0.


Let the number of steps i be greater than 0. We know that we can get K3 in Gi −Ki

from lemma 4.7.1. Let X = {v1, v2, v3} be the set of vertices in Gi − Ki. From lemma

4.7.1, it is {d, d, d− 1}-joined to Ki in Gi.


|S| ≥ 2 where S ∈ X be the subset of vertices {u|some path in P (u) terminates in v}.We need to consider the following cases.

1. Let {v1, v2} ⊂ S i.e. v joins to v1 and v2 by paths. Gi−1 − Ki−1 contains vertices

{v1, v2, v3, v} that are {d-1,d-1,d-2,d-1}-joined to Ki−1. Let G′i−1 be obtained by

deleting the path v1v, vv2 and vertex v1. Call this as Q. Let G′0 be the initial graph

of the reduced graph G′i−1 obtained by deleting all the internal vertices on the path

Q. Hence, (G′i−1, Ki−1) contains vertices {v, v2, v3} that are {d-1,d-1,d-2}-joined

to Ki−1. Hence, by induction on the number of steps, we get (G′0, K0) contains

a subdivision of the graph in the configuration. Together with path Q, (G0, K0)

contains the subdivision of the graph in the configuration C.

2. Let {v2, v3} ⊂ S or {v1, v2, v3} ⊂ S. Same argument holds as that of previous.

3. Let {v1, v3} ⊂ S i.e. v joins to v1 and v3 by paths. Gi−1 − Ki−1 contains vertices

{v1, v2, v3, v} that are {d − 1, d, d − 2, d}-joined to Ki−1. Let G′i−1 be obtained by

deleting the path vv1 and vv3 including vertex v3. LetG′0 be the corresponding initial

graph of G′i−1. Hence, (G′i−1, Ki−1) contains vertices {v, v2, v1} that are {d, d, d−1}-joined to Ki−1. Hence, by induction on the number of steps, (G′0, K0) contains the

subdivision of the graph in the configuration C. Together with deleted path and

the vertex v3, (G0, K0) contains the subdivision of the graph in the configuration C.

Here, we need to handle case {2,2,1} separately. Because if we use above argument then if

the vertex v join to v1 and v2, then vertices of Gi−1−Ki− 1 are {1, 1, 0}-joined to Ki−1.

So, we don’t get the subdivision of wheel. Therefore, instead of reducing the path p(u) of

each vertex u by 1, we reduce p(v1) and (p(v2) by 1 so that the vertices of Gi−1−Ki−1 are

{1, 1, 1}-joined to Ki−1. We know that the configuration C ′ as shown in figure 1 of figure

4.8 is good. Hence, (G0, K0) contains the subdivision of the graph in the configuration.

Hence, the proof.

4.7.5 2-Caterpillar

For the 2-caterpillar, we can define the configuration as consisting of two non-adjacent

external vertices 0 and 1, and d internal vertices numbered 2, 3, ..., d+ 1 such that every

internal vertex i has exactly two smaller numbered neighbours, and for i ≥ 3, either i is

adjacent to i− 1 and one of the smaller numbered neighbours of i− 1, or it is adjacent to

both the smaller numbered neighbours of i− 1.


Let (Gi−Ki) contains two vertices {v1, v2} that are {d, d}-joined to Ki. Then we need

to prove that (G0, k0) contains the subdivision of the graph present in the configuration.

The proof is similar to above.

4.7.6 Path through the vertices

Let V be a set of vertices and E a set of disjoint edges with endpoints in V . Let |V | = n

and |E| = k.

Define the configuration C(V,E) as follows:

V is the set of external vertices in the configuration, and there are no internal vertices.

All vertices in V that are not incident with an edge in E have count n− k− 1. For every

edge in E, one endpoint has count n − k − 2 and the other has count n − k − 1. The

graphs in the configuration are all graphs G(V, F ) such that F ∩E = ∅ and G(V, F ∪E)

is a path of length n− 1 that spans V .

We need to prove that the configuration C(V,E) is good.

Proof. Suppose n− k ≤ 1. The possible cases are n = 1, k = 0, or n = 2, k = 1 in which

case the result is trivial, since the configuration contains a graph without any edges.

Suppose n−k ≥ 2. In this case there are at least two vertices with counts n−k−1 ≥ 1

in the configuration. Let S be the subset of vertices that the new vertex x joins.

1. Suppose there exist u, v ∈ S such that {u, v} 6∈ E.

We consider the paths from u to x and v to x to form a path from u to v with x as

an intermediate vertex. Let E ′ = E∪{u, v}. Now the new configuration is obtained

as follows.

(a) If the edges in E ′ are disjoint, the new configuration is C(V,E ′) with the

counts of all vertices reduced by 1. This configuration is good by induction on

the number of reduction steps. Note that in this case, n remains same but k

increases, so all counts can be decreased by 1. Also for the added edge {u, v}both u and v have counts n− k − 2 in the new configuration.

(b) Suppose G(V,E ′) contains a component that is a path v, u ,w of length 2.

Then the new configuration is C(V −{u}, (E−{{u,w}})∪{{v, w}}) with the

counts of all vertices reduced by 1. In this case, n reduces by 1 but k remains

the same. Also, for the added edge {v, w}, the vertex v has count n − k − 2

while w has count at least n− k − 3.

(c) Suppose G(V,E ′) contains a component that is a path p, u ,v, q of length 3.

If either p or q has count n − k − 1 in the original configuration, then take

the new configuration to be C(V −{u, v}, (E−{{u, p}, {v, q}})∪{p, q}). with

counts of all vertices reduced by 1. If both p and q had counts n − k − 2 in

the initial configuration, but either p or q, say p, is not in S, then reduce the


counts for all vertices except p. This ensures that in the new configuration,

the newly added edge {p, q} has at least one endpoint with count n − k − 2.

Finally, if both p and q had count n − k − 2 initially and both are in S, take

the new configuration to be C(V − {p, q}, (E − {{u, p}, {v, q}}) ∪ {u, v}) with

counts of all vertices reduced by 1. In this case both u and v will have count

n− k − 2 in the new configuration.

2. Suppose S contains only the endpoints of some edge {u, v} in E. Without loss of

generality, assume u had count n − k − 2 and v had count n − k − 1 in the initial

configuration. The new vertex x will have count n− k− 1. We will consider x to be

an intermediate vertex in the path that terminates at u. Then the new configuration

required can be described as C((V −{u})∪{x}, (E−{u, v})∪{v, x}), with v having

count n − k − 2, x having count n − k − 1 and all other vertices having the same

count as earlier.

Chapter 5

Conclusion and Future Work

5.1 Summary

This concludes our outline of some of the major results that we have obtained in subdi-

vision of graphs. We have seen two different techniques External configuration technique

in chapter 3 and Internal configuration technique in chapter 4, prove the property “Every

graph G has minimum degree at least |H| − 1, it contains subdivision of H” for different

cases of H.

The cases for which we have proved the property are when H is P53, P6

3 and a wheel

Wd i.e., Cd ∗ K1 for d ≥ 3. We have seen the proof when H is a caterpillar graph. We

have seen the proof for the case when H is P73 and a graph containing d+ 1 vertices and

3d− 3 edges using internal configuration technique.

5.2 Future Work

We have seen the following properties:

1. If a graph G has minimum degree at least 4, then it contains a subdivision of P53.

2. If the graph G has minimum degree at least 5 then it contains a subdivision of P63.

3. If the graph G has minimum degree at least 6 then it contains the subdivision of

P73.

Looking at these three results, one important area of further research can be to determine

if the property holds when H is P83. It can be further determined if the property holds

for the general case when H is Pn3 .

We have seen that the property being studied holds when H is P3 ∗K2 and P4 ∗K2.

It can be further studied if the property holds when H is P5 ∗K2 and in the general case

when H is Pn ∗K2. Also, it can be determined if the property holds when H is Cn ∗K ′2.We have also seen the result holds when H is a 2-caterpillar. It can be a good topic

to test if the property is true when H is any 2-tree.

43

References

[1] W. Mader, “Existenz konfigurationen in n-gesattigten graphen und in graphen

genugend grober kantendichte,” Math. Ann., vol. 194, pp. 295–312, 1974.

[2] A. A. Diwan, “Generalization of Mader’s theorem,” Indian Institute of Technology,

Bombay, Tech. Rep., 2007.

[3] J. Pelikan, “Valency conditions for the existence of certain subgraphs.” Theory of

Graphs, Proc. Colloq., Tihany, pp. 251–258, 1968.

[4] G. E. Turner III, “A generalization of Dirac’s theorem: Subdivision of wheels,” Dis-

crete Mathematics, pp. 202–205, 28 July 2005.

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