A Proof of Erdös  Faber  Lovász Conjecture ^{1}^{1}1This paper is presented in the 23 International Conference of Forum for Interdisciplinary Mathematics (FIM) on “Interdisciplinary Mathematical, Statistical and Computational Techniques  2014” organized by the NITK, Surathkal, INDIA, from 18/12/2014 to 20/12/2014.
Abstract
In 1972, Erdös  Faber  Lovász (EFL) conjectured that, if H is a linear hypergraph consisting of edges of cardinality , then it is possible to color the vertices with colors so that no two vertices with the same color are in the same edge. In 1978, Deza, Erdös and Frankl had given an equivalent version of the same for graphs: Let denote a graph with complete graphs , each having exactly vertices and have the property that every pair of complete graphs has at most one common vertex, then the chromatic number of is .
The clique degree of a vertex in is given by . In this paper we give an algorithmic proof of the conjecture using the symmetric latin squares and clique degrees of the vertices of .
keywords:
Chromatic number, Erdös  Faber  Lovász conjecture, Latin squaresMsc:
[2010] 05A05, 05B15, 05C151 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 edges of cardinality , then it is possible to color the vertices of H with colors so that no two vertices with the same color are in the same edge berge1990onvizing (). Erdös, in 1975, offered 50 pounds erdHos1975problems (); arroyo2008dense () and in 1981, offered 500USD erdHos1981combinatorial (); jensen2011graph () for the proof or disproof of the conjecture. Kahn kahn1992coloring () showed that the chromatic number of H is at most . Jakson et al. jackson2007note () proved that the conjecture is true when the partial hypergraph of H determined by the edges of size at least three can be edgecolored and satisfies . In particular, the conjecture holds when is unimodular and . Viji Paul et al. vijipaul2012dmsaa () established the truth of the conjecture for all linear hypergraphs on vertices with . Sanhez  Arrayo arroyo2008dense () proved the conjecture for dense hypergraphs. Faber faber2010uniformregular () proves that for fixed degree, there can be only finitely many counterexamples to EFL on this class (both regular and uniform) of hypergraphs. We consider the equivalent version of the conjecture for graphs given by Deza, Erdös and Frankl in 1978 deza1978intersection (); arroyo2008dense (); jensen2011graph (); mitchem2010arscombin ().
Conjecture 1.1
Let denote a graph with complete graphs , each having exactly vertices and have the property that every pair of complete graphs has at most one common vertex, then the chromatic number of is .
Definition 1.2
Let denote a graph with complete graphs , each having exactly vertices and the property that every pair of complete graphs has at most one common vertex. The clique degree of a vertex in is given by . The maximum clique degree of the graph is given by .
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.
Definition 1.3
Let and be two vertex disjoint graphs, and let be two vertices of respectively. Then, the graph obtained by merging the vertices and into a single vertex is called the concatenation of and at the points and (see kundu1980reconstruction ()).
Definition 1.4
A latin square is an array containing different symbols such that each symbol appears exactly once in each row and once in each column. Moreover, a latin square of order is an matrix with entries from an set , where every row and every column is a permutation of (see laywine1998discrete ()). If the matrix is symmetric, then the latin square is called symmetric latin square.
2 Results
We know that a symmetric matrix is determined by scalars. Using symmetric latin squares we give an coloring of constructed below. Then using the coloring of we give an coloring of all the other graphs satisfying the hypothesis of Conjecture 1.1.

Construction of :
Let be a positive integer and be copies of . Let the vertex set , .
 Step 1.

Let .
 Step 2.

Consider the vertices of and of . Let be the vertex obtained by the concatenation of the vertices and . Let the resultant graph be .
 Step 3.

Consider the vertices , of and , of . Let be the vertex obtained by the concatenation of vertices , and let be the vertex obtained by the concatenation of vertices , . Let the resultant graph be .

Continuing in the similar way, at the th step we obtain the graph (for the sake of convenience we take as ).
By the construction of one can observe the following:

is a connected graph and satisfying the hypothesis of Conjecture 1.1.

has exactly verticies of clique degree one and vertices of clique degree (each has exactly vertices of clique degree and one vertex of clique degree one, ).

, where and , have exactly one common vertex for .

has exactly vertices.

One can observe that in a connected graph if clique degree increases the number of vertices also increases, from this it follows that, is the graph with minimum number of vertices satisfying the hypothesis of Conjecture 1.1. If all the vertices of are of clique degree one, then will have vertices. Thus, .
Lemma 2.1
If is a graph satisfying the hypothesis of Conjecture 1.1, then can be obtained from for some in .
Proof: Let be a graph satisfying the hypothesis of Conjecture 1.1. Let be the new labeling to the vertices of clique degree greater than one in , where : vertex is in . Define for . Then the graph is constructed from as given below:
Step 1: For every common vertex in which is not in , split the vertex into two vertices such that vertex is adjacent only to the vertices of and the vertex is adjacent only to the vertices of in .
Step 2: For every vertex in where , merge the vertices into a single vertex in where and for .
Let be the graph obtained in Step 2. Let , be the set of all clique degree 1 vertices of of , of respectively, . Thus by splitting all the common vertices of which are not in and merging the vertices of corresponding to the vertices in , we get the graph . One can observe that , . Define a function by
One can observe that is an isomorphism from to .
From Lemma 2.1, one can observe that in there are at most common vertices.
Let be the graph satisfying the hypothesis of Conjecture 1.1. Let be the graph obtained by removing the vertices of clique degree one from graph . i.e. is the induced subgraph of having all the common vertices of .
Lemma 2.2
The chromatic number of is .
Proof: Let be the graph defined as above. Let (given below) be an matrix in which an entry , is a vertex of , belonging to both for and is the vertex of which belongs to . i.e.,
M=
Clearly is a symmetric matrix. We know that, for every in there is a symmetric latin square (see ye2011number ()) of order . Bryant and Rodger bryant2004completion () gave a necessary and sufficient condition for the existence of an edge coloring of (n even), and edge coloring of (n odd) using symmetric latin squares. Let be the vertices of and is the edge joining the vertices and of , where , then arrange the edges of in the matrix form where , for and for , we have
and let is a matrix given by
. Then, define a matrix as
.
Let be a matrix where (), is the color of (i.e., ) and is the color of . We call the color matrix of . Then is the symmetric latin square (seebryant2004completion ()). As the elements of are the vertices of , one can assign the colors to the vertices of from the color matrix , by the color (where denotes the value at the th entry in the color matrix ), for and to the vertex in and the color (where denotes the value at the th entry in the color matrix ), for to the vertex in . Hence is colorable.
As is the graph satisfying the hypothesis of Conjecture 1.1. With using the coloring of which is the graph satisfying the hypothesis of Conjecture 1.1 we extend the coloring of all possible graphs satisfying the hypothesis of Conjecture 1.1.
Theorem 2.3
If is a graph satisfying the hypothesis of Conjecture 1.1, then is colorable.
Proof: Let be a graph satisfying the hypothesis of Conjecture 1.1. Let be the induced subgraph of consisting of the vertices of clique degree greater than one in . For every vertex of clique degree greater than one in , label the vertex by where . Define , for .
Let be the colors and be the color matrix( of size ) as defined in the proof of Lemma 2.2. The following construction applied on the color matrix , gives a modified color matrix , using which we assign the colors to the graph . Then this coloring can be extended to the graph . Construct a new color matrix by putting for every in . Also, let for each
Construction:
Let , , and .
 Step 1:

If , let be the color matrix obtained in Step 4 and go to Step 5. Otherwise, choose a vertex from , where , and then choose vertices , , , , , , from corresponding to the set . Take and . Let appears more than once in the row or column in and appears exactly once in the row and column in . If choose a vertex from , otherwise choose a vertex from . Then add the vertex to and remove it from . Go to Step 2.
 Step 2:

If go to Step 3. Otherwise, choose a vertex from . Let , . If then, construct a new color matrix , replacing , by , where . Then add the vertex to and remove it from . Go to Step 3. Otherwise choose a color which appears exactly once either in row or in column of the color matrix and construct a new color matrix replacing , by . Then add the vertex to and remove it from . Go to Step 3.
 Step 3:

If , then add the vertex to and remove it from , go to Step 1. Otherwise, if choose a vertex from , if not choose a vertex from . Go to Step 4.
 Step 4:

Let . If , then add the vertex to and remove it from . Go to Step 3. Otherwise, let , . Construct a new color matrix by putting for every in and for every in B. Then add the vertex to and remove it from . Go to Step 3.
 Step 5:

If consider stop the process. Otherwise, choose a vertex from and go to Step 6.
 Step 6:

If appears exactly once in both row and column of the color matrix , then add the vertex to and remove it from , go to Step 5. Otherwise let , . Construct a new color matrix by putting in , where . Then add the vertex to and remove it from , go to Step 5.
Thus, in step 6, we get the modified color matrix . Then, color the vertex of by of , whenever . Then, extend the coloring of to by assigning the remaining colors which are not used for from the set of colors, to the vertices of clique degree one in , . Thus is colorable.
Corollary 2.4
arroyo2008dense () Consider a linear hypergraph H consisting of edges each of size at most and . If is dense then .
References
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