On Erdos-Faber-Lovasz Conjecture

TL;DR

This paper investigates the Erdős-Faber-Lovász conjecture, proposing a coloring method based on intersection matrices and clique degrees, and proves the conjecture for specific graph classes under certain conditions.

Contribution

It introduces a new coloring approach using intersection matrices and clique degrees, and provides proof for the conjecture in classes with bounded clique degree vertices.

Findings

Coloring method using intersection matrices and clique degrees.

Proof of the conjecture for graphs with limited high clique degree vertices.

Conditions under which the graph is proven to be n-colorable.

Abstract

In 1972, Erd\"{o}s - Faber - Lov\'{a}sz (EFL) conjectured that, if $\textbf{H}$ is a linear hypergraph consisting of $n$ edges of cardinality $n$, then it is possible to color the vertices with $n$ colors so that no two vertices with the same color are in the same edge. In 1978, Deza, Erd\"{o}s and Frankl had given an equivalent version of the same for graphs: Let $G= \bigcup_{i=1}^{n} A_i$ denote a graph with $n$ complete graphs $A_1, A_2,$ $ \dots , A_n$, each having exactly $n$ vertices and have the property that every pair of complete graphs has at most one common vertex, then the chromatic number of $G$ is $n$. The clique degree $d^K(v)$ of a vertex $v$ in $G$ is given by $d^K(v) = |\{A_i: v \in V(A_i), 1 \leq i \leq n\}|$. In this paper we give a method for assigning colors to the graphs satisfying the hypothesis of the Erd\"os - Faber - Lov\'asz conjecture using intersection matrix of the cliques $A_i$'s of $G$ and clique degrees of the vertices of $G$. Also, we give theoretical proof of the conjecture for some class of graphs. In particular we show that: 1. If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_i$ ($1 \leq i \leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1, then $G$ is $n$-colorable. 2. If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_i$ ($1 \leq i \leq n$) has at most $\left \lceil {\frac{n+d-1}{d}} \right \rceil$ vertices of clique degree greater than or equal to $d$ ($2\leq d \leq n$), then $G$ is $n$-colorable.

Figures

Equations

$$X_1$$

$$\left(\genfrac{}{}{0ex}{}{d^K(v_1)}{2}\right)+\left(\genfrac{}{}{0ex}{}{d^K(v_2)}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{d^K(v_l)}{2}\right)=\frac{n(n-1)}{2},$$

$$\frac{n(n-1)}{2}$$

$$\frac{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{\left(\genfrac{}{}{0ex}{}{m}{2}\right)}$$

$$p$$

Full text

The clique degree $d^{K}(v)$ of a vertex $v$ in $G$ is given by $d^{K}(v)=|\{A_{i}:v\in V(A_{i}),1\leq i\leq n\}|$. In this paper we give a method for assigning colors to the graphs satisfying the hypothesis of the Erdös - Faber - Lovász conjecture using intersection matrix of the cliques $A_{i}$’s of $G$ and clique degrees of the vertices of $G$. Also, we give theoretical proof of the conjecture for some class of graphs. In particular we show that:

If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1, then $G$ is $n$-colorable. 2. 2.

If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\left\lceil{\frac{n+d-1}{d}}\right\rceil$ vertices of clique degree greater than or equal to $d$ ($2\leq d\leq n$), then $G$ is $n$-colorable.

Keyword Chromatic number, Erdös - Faber - Lovász conjecture

MSC[2010]* 05A05, 05B15, 05C15*

1 Introduction

One of the famous conjectures in graph theory is Erdös - Faber - Lovász conjecture. It states that if H is a linear hypergraph consisting of $n$ edges of cardinality $n$, then it is possible to color the vertices of H with $n$ colors so that no two vertices with the same color are in the same edge [1]. Erdös, in 1975, offered 50 USD [4, 10] and in 1981, offered 500 USD [3, 7] for the proof or disproof of the conjecture. Kahn [8] showed that the chromatic number of H is at most $n+o(n)$. Jakson et al. [6] proved that the conjecture is true when the partial hypergraph $S$ of H determined by the edges of size at least three can be $\Delta_{S}$-edge-colored and satisfies $\Delta_{S}\leq 3$. In particular, the conjecture holds when $S$ is unimodular and $\Delta_{S}\leq 3$. Sanhez - Arrayo [10] proved the conjecture for dense hypergraphs. Faber [5] proves that for fixed degree, there can be only finitely many counterexamples to EFL on this class (both regular and uniform) of hypergraphs.

Conjecture 1.1

If H is a linear hypergraph consisting of $n$ edges of cardinality $n$, then it is possible to color the vertices of H with $n$ colors so that no two vertices with the same color are in the same edge.

We consider the equivalent version of the conjecture for graphs given by Deza, Erdös and Frankl in 1978 [2, 10, 7, 9].

Conjecture 1.2

Let $G=\bigcup_{i=1}^{n}A_{i}$ denote a graph with $n$ complete graphs $(A_{1},A_{2},$ $\dots,A_{n})$, each having exactly $n$ vertices and have the property that every pair of complete graphs has at most one common vertex, then the chromatic number of $G$ is $n$.

Definition 1.3

Let $G=\bigcup_{i=1}^{n}A_{i}$ denote a graph with $n$ complete graphs $A_{1},A_{2},$ $\dots,A_{n}$, each having exactly $n$ vertices and the property that every pair of complete graphs has at most one common vertex. The clique degree $d^{K}(G)$ of a vertex $v$ in $G$ is given by $d^{K}(v)=|\{A_{i}:v\in V(A_{i}),1\leq i\leq n\}|$. The maximum clique degree $\Delta^{K}(G)$ of the graph $G$ is given by $\Delta^{K}(G)=max_{v\in V(G)}d^{K}(v)$.

From the above definition, one can observe that degree of a vertex in hypergraph is same as the clique degree of a vertex in a graph.

2 Coloring of $G$

Let $G$ be the graph satisfying the hypothesis of Conjecture 1.2. Let $\hat{H}$ be the graph obtained by removing the vertices of clique degree one from graph $G$. i.e. $\hat{H}$ is the induced subgraph of $G$ having all the common vertices of $G$.

Theorem 2.1

If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1, then $G$ is $n$-colorable.

**Proof: **Let $G$ be a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1. Let $\hat{H}$ be the induced subgraph of $G$ consisting of the vertices of clique degree greater than one in $G$. Define $X=\{b_{i,j}:A_{i}\cap A_{j}=\emptyset\}$, $X_{i}=\{v\in G:d^{K}(v)=i\}$ for $i=1,2,\dots,n$.

From [10], it is true that the vertices of clique degree greater than or equal to $\sqrt{n}$ can be assigned with at most $n$ colors. Assign the colors to the vertices of clique degree in non increasing order. Assume we next color a vertex $v$ of clique degree $1<k<\sqrt{n}$. At this point only vertices of clique degree $\geq k$ have been assigned colors. By assumption every $A_{i}$ ($1\leq i\leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1 and clique degree of $v$ is $k$ ($k<\sqrt{n}$), then for these $k$ $A_{i}^{\prime}s$ there are at most $k\sqrt{n}<n$ vertices have been colored. Therefore, there is an unused color from the set of $n$ colors, then that color can be assigned to the vertex $v$.

Theorem 2.2

If $G$ is a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\left\lceil{\frac{n+d-1}{d}}\right\rceil$ vertices of clique degree greater than or equal to $d$ ($2\leq d\leq n$), then $G$ is $n$-colorable.

**Proof: **Let $G$ be a graph satisfying the hypothesis of the Conjecture 1.2 and every $A_{i}$ ($1\leq i\leq n$) has at most $\left\lceil{\frac{n}{d}}\right\rceil$ vertices of clique degree greater than or equal to $d$ ($2\leq d\leq n$). Let $\hat{H}$ be the induced subgraph of $G$ consisting of the vertices of clique degree greater than one in $G$. Define $X=\{b_{i,j}:A_{i}\cap A_{j}=\emptyset\}$, $X_{i}=\{v\in G:d^{K}(v)=i\}$ for $i=1,2,\dots,m$.

Assign the colors to the vertices of clique degree in non increasing order. Assume we next color a vertex $v$ of clique degree $k$. At this point only vertices of clique degree $\geq k$ have been assigned colors. By assumption every $A_{i}$ ($1\leq i\leq n$) has at most $\sqrt{n}$ vertices of clique degree greater than 1 and clique degree of $v$ is $k$ ($k<\sqrt{n}$), then for these $k$ $A_{i}$’s there are at most $k(\left\lceil{\frac{n}{k}}\right\rceil-1)<n$ vertices have been colored. Therefore, there is an unused color from the set of $n$ colors, then that color can be assigned to the vertex $v$.

Let $G$ be a graph satisfying the hypothesis of the Conjecture 1.2, the intersection matrix is a square $n\times n$ matrix $C$ such that its element $c_{i,j}$ is $c$ when $A_{i}\cap A_{j}\neq\emptyset$ otherwise zero for $i\neq j$ and the diagonal elements of the matrix are all zero.

Given below is a method to color graph $G$ satisfying the hypothesis of the Conjecture 1.2 using intersection matrix (color matrix) of the cliques $A_{i}$’s of $G$ and clique degrees of the vertices of $G$.

Method for assigning colors to graph $G$:

Let $G$ be a graph satisfying the hypothesis of Conjecture 1.2. Let $\hat{H}$ be the induced subgraph of $G$ consisting of the vertices of clique degree greater than $1$ in $G$. Relable the vertex $v$ of clique degree greater than $1$ in $G$ by $u_{x}$, where $x=k_{1},k_{2},\dots,k_{j}$; vertex $v$ is in $A_{k_{i}},1\leq i\leq j$. Define $X=\{b_{i,j}:A_{i}\cap A_{j}=\emptyset\}$, $X_{i}=\{v\in G:d^{K}(v)=i\}$ for $i=1,2,\dots,n$.

Let $C$ be the intersection matrix of the cliques $A_{i}$’s of $G$ where $c_{i,j}=0$ if $A_{i}\cap A_{j}=\emptyset$ otherwise $c$ for $i\neq j$ and $c_{i,i}$ is [math]. Let $1,2,\dots,n$ be the $n$-colors.

Let $T_{i}=X_{i}$, $P_{i}=\emptyset$ and $S=\{j:T_{j}\neq\emptyset,2\leq j\leq n\}$.

If $S=\emptyset$, then the graph $G$ has no vertex of clique degree greater than one, which implies $G$ has exactly $n^{2}$(maximum number) vertices. i.e., $G$ is $n$ components of $K_{n}$. Otherwise, consider the intersection matrix $C$ as the color matrix and follow the steps.

Step 1:

If $S=\emptyset$, stop the process. Otherwise, let $\max(S)=k$, for some $k,2\leq k\leq n$. Then consider the sets $T_{k}$ and $P_{k}$, go to step 2.

Step 2:

If $T_{k}=\emptyset$, go to step 1. Otherwise, choose a vertex $u_{i_{1},i_{2},\dots,i_{k}}$ from $T_{k}$, where $i_{1}<i_{2}<\dots<i_{k}$ and go to Step 3.

Step 3:

Let $Y_{i}=\{y:$ color $y$ appears atleast $k-1$ times in the $i^{th}$ row of the color matrix $\}$, $i=1,2,\dots,n$. If $|{\bigcup_{i=i_{1}}^{i_{k}}Y_{i}}|=n$, let $B_{T}=\bigcup_{i=2}^{n}P_{i}$, $B_{P}=\emptyset$ and go to Step 4. Otherwise, construct a new color matrix $C_{1}$ by putting least $x$ in $c_{i,j}$, where $x\in\{1,2,3,\dots n\}\setminus\bigcup_{i=i_{1}}^{i_{k}}Y_{i}$, $i\neq j$, $i_{1}\leq i,j\leq i_{k}$. Then add the vertex $u_{i_{1},i_{2},\dots,i_{k}}$ to $P_{k}$ and remove it from $T_{k}$, go to Step 2.

Step 4:

Choose a vertex $v$ from $B_{T}$ such that $v\in A_{i}$, for some $i$, $i_{1}\leq i\leq i_{k}$. Let $B=\{i:v\in A_{i},1\leq i\leq n\}$ and go to Step 5.

Step 5:

Let $Y_{i}=\{y:$ color $y$ appears atleast $k-1$ times in the $i^{th}$ row of the color matrix$\}$, for every $i\in B$. If $|{\bigcup_{i\in B}Y_{i}}|=n$ add the vertex $v$ to $B_{P}$ and remove it from $B_{T}$, go to Step 4. Otherwise construct a new color matrix $C_{2}$ by putting $x$ in $c_{i,j}$, where $x\in\{1,2,3,\dots n\}\setminus\bigcup_{i\in B}Y_{i}$, $i\neq j$, $i,j\in B$. Go to Step 3.

Thus, we get the modified color matrix $C_{M}$. Then, color the vertex $v$ of $\hat{H}$ by $c_{i,j}$ of $C_{M}$, whenever $v\in A_{i}\cap A_{j}$. Then, extend the coloring of $\hat{H}$ to $G$. Thus $G$ is $n$-colorable.

Following is an example illustrating the above method.

Example 2.3

Let $G$ be the graph shown in Figure 1a.

*Let $V(A_{1})=\{v_{1},v_{2},v_{3},v_{4},v_{5},v_{6}\}$, $V(A_{2})=\{v_{1},v_{7},v_{8},v_{9},v_{10},v_{11}\}$,

$V(A_{3})=\{v_{1},v_{12},v_{13},v_{14},v_{15},v_{16}\}$, $V(A_{4})=\{v_{1},v_{17},v_{18},v_{19},v_{20},v_{21}\}$,

$V(A_{5})=\{v_{6},v_{7},v_{16},v_{22},v_{23},v_{24}\}$, $V(A_{6})=\{v_{9},v_{16},v_{19},v_{25},v_{26},v_{27}\}$.*

Relabel the vertices of clique degree greater than one in $G$ by $u_{A}$ where $A=\{i:v\in A_{i}\text{ for }1\leq i\leq 6\}$. The labeled graph is shown in Figure 1b. Figure 2 is the graph $\hat{H}$, where $\hat{H}$ is obtained by removing the vertices of clique degree 1 from $G$.

Let $X=\{b_{i,j}:A_{i}\cap A_{j}=\emptyset\}=\{b_{1,6},b_{4,5}\}$,

[TABLE]

$X_{2}=\{v\in G:d^{K}(v)=2\}=\{v_{6},v_{7},v_{9},v_{19}\}=\{u_{1,5},u_{2,5},u_{2,6},u_{4,6}\}$*,

$X_{3}=\{v\in G:d^{K}(v)=3\}=\{v_{16}\}=\{u_{3,5,6}\}$,

$X_{4}=\{v\in G:d^{K}(v)=4\}=\{v_{1}\}=\{u_{1,2,3,4}\}$,

$X_{5}=\emptyset$ and $X_{6}=\emptyset$.*

Let 1, 2, $\dots$, 6 be the six colors and C=\left(\begin{array}[]{cccccc}0&c&c&c&c&0\\ c&0&c&c&c&c\\ c&c&0&c&c&c\\ c&c&c&0&0&c\\ c&c&c&0&0&c\\ 0&c&c&c&c&0\\ \end{array}\right)

be the intersection matrix of order $6\times 6$.

Consider the sets $T_{i}=X_{i}$, $P_{i}=\emptyset$ for $i=1,2,\dots 6$ and $S=\{j:T_{j}\neq\emptyset,2\leq j\leq n\}=\{2,3,4\}$. Then by applying the method given above, we get the following.

Step 1: Since $S\neq\emptyset$ and $\max(S)=4$, then choose the sets $T_{4}=\{u_{1,2,3,4}\}$ and $P_{4}=\emptyset$. Go to step 2.

Step 2: Since $T_{4}\neq\emptyset$, choose the vertex $u_{1,2,3,4}$ from $T_{4}$, go to step 3.

Step 3: Since $Y_{1}=\emptyset$, $Y_{2}=\emptyset$, $Y_{3}=\emptyset$, $Y_{4}=\emptyset$ and $|Y_{1}\cup Y_{2}\cup Y_{3}\cup Y_{4}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=1,2,3,4}Y_{i}$ and construct a new color matrix $C_{1}$ by putting $1$ in $c_{i,j}$, $i\neq j$, $i,j=1,2,3,4$. Add the vertex $u_{1,2,3,4}$ to $P_{4}$ and remove it from $T_{4}$. Then

C_{1}=\left(\begin{array}[]{cccccc}0&1&1&1&c&0\\ 1&0&1&1&c&c\\ 1&1&0&1&c&c\\ 1&1&1&0&0&c\\ c&c&c&0&0&c\\ 0&c&c&c&c&0\\ \end{array}\right)* ,*

$T_{4}=\emptyset$, $P_{4}=\{u_{1,2,3,4}\}$. Go to step 2.

Step 2: Since $T_{4}=\emptyset$, go to step 1.

Step 1: Since $S\neq\emptyset$ and $\max(S)=3$, then choose the sets $T_{3}=\{u_{3,5,6}\}$ and $P_{3}=\emptyset$. Go to step 2.

Step 2: Since $T_{3}\neq\emptyset$, choose the vertex $u_{3,5,6}$ from $T_{3}$, go to step 3.

Step 3: Since $Y_{3}=\{1\}$, $Y_{5}=\emptyset$, $Y_{6}=\emptyset$, and $|Y_{3}\cup Y_{5}\cup Y_{6}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=3,5,6}Y_{i}$ and construct a new color matrix $C_{2}$ by putting $2$ in $c_{i,j}$, $i\neq j$, $i,j=3,5,6$. Add the vertex $u_{3,5,6}$ to $P_{3}$ and remove it from $T_{3}$. Then

C_{2}=\left(\begin{array}[]{cccccc}0&1&1&1&c&0\\ 1&0&1&1&c&c\\ 1&1&0&1&2&2\\ 1&1&1&0&0&c\\ c&c&2&0&0&2\\ 0&c&2&c&2&0\\ \end{array}\right),

$T_{3}=\emptyset$, $P_{3}=\{u_{3,5,6}\}$. Go to step 2.

Step 2: Since $T_{3}=\emptyset$, go to step 1.

Step 1: Since $S\neq\emptyset$ and $\max(S)=2$, then choose the sets $T_{2}=\{u_{1,5},u_{2,5},u_{2,6},u_{4,6}\}$ and $P_{2}=\emptyset$. Go to step 2.

Step 2: Since $T_{2}\neq\emptyset$, choose the vertex $u_{1,5}$ from $T_{2}$, go to step 3.

Step 3: Since $Y_{1}=\{1\}$, $Y_{5}=\{2\}$ and $|Y_{1}\cup Y_{5}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=1,5}Y_{i}$ and construct a new color matrix $C_{3}$ by putting $3$ in $c_{i,j}$, $i\neq j$, $i,j=1,5$. Add the vertex $u_{1,5}$ to $P_{2}$ and remove it from $T_{2}$. Then

C_{3}=\left(\begin{array}[]{cccccc}0&1&1&1&3&0\\ 1&0&1&1&c&c\\ 1&1&0&1&2&2\\ 1&1&1&0&0&c\\ 3&c&2&0&0&2\\ 0&c&2&c&2&0\\ \end{array}\right),

$T_{2}=\{u_{2,5},u_{2,6},u_{4,6}\}$, $P_{2}=\{u_{1,5}\}$. Go to step 2.

Step 2: Since $T_{2}\neq\emptyset$, choose the vertex $u_{2,5}$ from $T_{2}$, go to step 3.

Step 3: Since $Y_{2}=\{1\}$, $Y_{5}=\{2,3\}$ and $|Y_{2}\cup Y_{5}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=2,5}Y_{i}$ and construct a new color matrix $C_{4}$ by putting $4$ in $c_{i,j}$, $i\neq j$, $i,j=2,5$. Add the vertex $u_{2,5}$ to $P_{2}$ and remove it from $T_{2}$. Then

C_{4}=\left(\begin{array}[]{cccccc}0&1&1&1&3&0\\ 1&0&1&1&4&c\\ 1&1&0&1&2&2\\ 1&1&1&0&0&c\\ 3&4&2&0&0&2\\ 0&c&2&c&2&0\\ \end{array}\right),

$T_{2}=\{u_{2,6},u_{4,6}\}$, $P_{2}=\{u_{1,5},u_{2,5}\}$. Go to step 2.

Step 2: Since $T_{2}\neq\emptyset$, choose the vertex $u_{2,6}$ from $T_{2}$, go to step 3.

Step 3: Since $Y_{2}=\{1,4\}$, $Y_{6}=\{2\}$ and $|Y_{2}\cup Y_{6}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=2,6}Y_{i}$ and construct a new color matrix $C_{5}$ by putting $3$ in $c_{i,j}$, $i\neq j$, $i,j=2,6$. Add the vertex $u_{2,6}$ to $P_{2}$ and remove it from $T_{2}$. Then

C_{5}=\left(\begin{array}[]{cccccc}0&1&1&1&3&0\\ 1&0&1&1&4&3\\ 1&1&0&1&2&2\\ 1&1&1&0&0&c\\ 3&4&2&0&0&2\\ 0&3&2&c&2&0\\ \end{array}\right),

$T_{2}=\{u_{4,6}\}$, $P_{2}=\{u_{1,5},u_{2,5},u_{2,6}\}$. Go to step 2.

Step 2: Since $T_{2}\neq\emptyset$, choose the vertex $u_{4,6}$ from $T_{2}$, go to step 3.

Step 3: Since $Y_{4}=\{1\}$, $Y_{6}=\{2,3\}$ and $|Y_{4}\cup Y_{6}|<6$, choose the minimum color from the set $\{1,2,\dots,6\}\setminus\cup_{i=4,6}Y_{i}$ and construct a new color matrix $C_{6}$ by putting $4$ in $c_{i,j}$, $i\neq j$, $i,j=4,6$. Add the vertex $u_{4,6}$ to $P_{2}$ and remove it from $T_{2}$. Then

C_{6}=\left(\begin{array}[]{cccccc}0&1&1&1&3&0\\ 1&0&1&1&4&3\\ 1&1&0&1&2&2\\ 1&1&1&0&0&4\\ 3&4&2&0&0&2\\ 0&3&2&4&2&0\\ \end{array}\right),

$T_{2}=\emptyset$, $P_{2}=\{u_{1,5},u_{2,5},u_{2,6},u_{4,6}\}$. Go to step 2.

Step 2: Since $T_{2}=\emptyset$, go to step 1.

Step 1: Since $S=\emptyset$, stop the process.

Assign the colors to the graph $\hat{H}$ using the matrix $C_{M}=C_{6}$, i.e., color the vertex $v$ by the $(i,j)$-th entry $c_{i,j}$ of the matrix $C_{M}$, whenever $A_{i}\cap A_{j}\neq\emptyset$ (see Figure 3a), where the numbers 1, 2, 3, 4, 5, 6 corresponds to the colors Maroon, Tan, Green, Red, Blue, Cyan respectively. Extend the coloring of $\hat{H}$ to $G$ by assigning the remaining colors which are not used for $A_{i}$ from the set of $6$-colors to the vertices of clique degree one in each $A_{i}$, $1\leq i\leq 6$. The colored graph $G$ is shown in Figure 3b.

The following results give a relation between the number of complete graphs and clique degrees of a graph.

Theorem 2.4

Let $G$ be a graph satisfying the hypothesis of Conjecture 1.2. Then if the intersection of any two $A_{i}$’s is non empty, then

[TABLE]

where $\{v_{1},v_{2},\dots,v_{l}\}$ is the set of all vertices of clique degree greater than $1$ in $G$.

**Proof: **If $G$ is isomorphic to the graph $H_{n}$ for some $n$, then the result is obvious. If not there exists at least one vertex $v$ of clique degree greater than $2$. Define $I_{v}=\{i:v\in A_{i}\}$ then $d^{K}(v)=|I_{v}|=p$. For every unordered pair of elements $(i,j)$ of $I_{v}$ there is a vertex $b_{ij}(\text{ where }i<j)$ in $H_{n}$. Therefore corresponding to the elements of $I_{v}$ there are $\binom{p}{2}$ vertices in $H_{n}$. Since $G$ satisfies the hypothesis of Conjecture 1.2, there is no vertex $v^{\prime}$ different from $v$ in $G$ such that $v^{\prime}\in A_{i}\cap A_{j}$ where $i,j\in I_{v}$. Therefore for every vertex $v$ of clique degree greater than $1$ in $G$, there are $\binom{d^{K}(v)}{2}$ vertices of clique degree greater than $1$ in $H_{n}$. As there are $\frac{n(n-1)}{2}$ vertices of clique degree greater than $1$ in $H_{n}$, $\frac{n(n-1)}{2}=\binom{d^{K}(v_{1})}{2}+\binom{d^{K}(v_{2})}{2}+\dots+\binom{d^{K}(v_{l})}{2}$ where $\{v_{1},v_{2},\dots,v_{l}\}$ is the set of all vertices of clique degree greater than $1$ in $G$.

Corollary 2.5

If $G$ is a graph satisfying the hypothesis of conjecture 1.2, then $G$ has at most $\frac{\binom{n}{2}}{\binom{m}{2}}$ vertices of clique degree $m$ where $m\geq 2$.

**Proof: **Let $A=\{v_{1},v_{2},\dots,v_{l}\}$ be the set of vertices of clique degree greater than $1$ in $G$ and $p=\frac{\binom{n}{2}}{\binom{m}{2}}$. We have to prove that $G$ has at most $p$ vertices of clique degree $m$. Suppose $G$ has $q>p$ vertices of clique degree $m$. Then by the definition of $A$, it follows that, $q$ vertices are in $A$. Let those vertices be $v_{1},v_{2},\dots,v_{q}$. By Theorem 2.4 we get,

[TABLE]

which is a contradiction. Hence there are at most $\frac{\binom{n}{2}}{\binom{m}{2}}$ vertices of clique degree $m$ in $G$, where $m\geq 2$.