On Erdos-Faber-Lovasz Conjecture S. M. Hegde, Suresh Dara Department of Mathematical and Computational Sciences, National Institute of Technology Karnataka, Surathkal, Mangalore-575025 E-mail: smhegde@nitk.ac.in, suresh.dara@gmail.com Abstract 7 In 1972, Erdös - Faber - Lovász (EFL) conjectured that, if H is a linear 1 hypergraph consisting of n edges of cardinality n, then it is possible to color 0 2 the vertices with n colors so that no two vertices with the same color are in the n same edge. In 1978, Deza, Erdös and Frankl had given an equivalent version of a J the same for graphs: Let G = n A denote a graph with n complete graphs i=1 i 7 1 A1,A2,...,An, each havingexaSctly nvertices andhavethepropertythatevery pair of complete graphs has at most one common vertex, then the chromatic ] O number of G is n. C The clique degree dK(v) of a vertex v in G is given by dK(v) = A : v . |{ i ∈ h V(A ),1 i n . In this paper we give a method for assigning colors to the t i a ≤ ≤ }| m graphs satisfying the hypothesis of the Erds - Faber - Lovsz conjecture using [ intersection matrix of the cliques A ’s of G and clique degrees of the vertices of i 1 G. Also, we give theoretical proof of the conjecture for some class of graphs. v In particular we show that: 0 5 5 1. If G is a graph satisfying the hypothesis of the Conjecture 1.2 and every 4 A (1 i n) has at most √n vertices of clique degree greater than 1, 0 i ≤ ≤ 1. then G is n-colorable. 0 7 2. If G is a graph satisfying the hypothesis of the Conjecture 1.2 and every 1 A (1 i n) has at most n+d−1 vertices of clique degree greater than v: i ≤ ≤ d i or equal to d (2 d n), t(cid:6)hen G i(cid:7)s n-colorable. X ≤ ≤ r a Keyword Chromatic number, Erdös - Faber - Lovász conjecture MSC[2010] 05A05, 05B15, 05C15 1 Introduction One ofthe famousconjectures in graphtheory isErdös - Faber -Lova´sz conjecture. It states that if H is a linear hypergraph consisting of n edges of cardinality n, then it is 1 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 ∆ -edge-colored and satisfies ∆ 3. In particular, the S S ≤ conjecture holds when S is unimodular and ∆ 3. Sanhez - Arrayo [10] proved the S ≤ 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 = n A denote a graph with n complete graphs (A ,A , i=1 i 1 2 ...,A ), each having exactlySn vertices and have the property that every pair of com- n plete graphs has at most one common vertex, then the chromatic number of G is n. Definition 1.3 Let G = n A denote a graph with n complete graphs A ,A , i=1 i 1 2 ...,A , each having exactlySn vertices and the property that every pair of complete n graphs has at most one common vertex. The clique degree dK(G) of a vertex v in G is given by dK(v) = A : v V(A ),1 i n . The maximum clique degree ∆K(G) i i |{ ∈ ≤ ≤ }| of the graph G is given by ∆K(G) = max dK(v). v∈V(G) 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 Hˆ be the graph obtained by removing the vertices of clique degree one from graph G. i.e. Hˆ is the induced subgraph of G having all the common vertices of G. 2 Theorem 2.1 If G is a graph satisfying the hypothesis of the Conjecture 1.2 and every A (1 i n) has at most √n vertices of clique degree greater than 1, then G i ≤ ≤ is n-colorable. Proof: Let G be a graph satisfying the hypothesis of the Conjecture 1.2 and every A (1 i n) has at most √n vertices of clique degree greater than 1. Let Hˆ be the i ≤ ≤ induced subgraph of G consisting of the vertices of clique degree greater than one in G. Define X = b : A A = , X = v G : dK(v) = i for i = 1,2,...,n. i,j i j i { ∩ ∅} { ∈ } From [10], it is true that the vertices of clique degree greater than or equal to √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 < √n. At this point only vertices of clique degree k have been assigned ≥ colors. By assumption every A (1 i n) has at most √n vertices of clique degree i ≤ ≤ greater than 1 and clique degree of v is k (k < √n), then for these k A′s there are at i most k√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 (1 i n) has at most n+d−1 vertices of clique degree greater than or i ≤ ≤ d equal to d (2 d n), then G is n-(cid:6)colorab(cid:7)le. ≤ ≤ Proof: Let G be a graph satisfying the hypothesis of the Conjecture 1.2 and every A (1 i n) has at most n vertices of clique degree greater than or equal to d i ≤ ≤ d (2 d n). Let Hˆ be the in(cid:6)du(cid:7)ced subgraph of G consisting of the vertices of clique ≤ ≤ degree greater than one in G. Define X = b : A A = , X = v G : dK(v) = i,j i j i { ∩ ∅} { ∈ i for i = 1,2,...,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 k have been assigned colors. By assumption every A (1 i n) has at most √n i ≥ ≤ ≤ vertices of clique degree greater than 1 and clique degree of v is k (k < √n), then for these k A ’s there are at most k( n 1) < n vertices have been colored. Therefore, i k − there is an unused color from the(cid:6)se(cid:7)t 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 n matrix C such that its element c is c when A A = i,j i j × ∩ 6 ∅ otherwise zero for i = j and the diagonal elements of the matrix are all zero. 6 3 Given below is a method to color graph G satisfying the hypothesis of the Con- jecture 1.2 using intersection matrix (color matrix) of the cliques A ’s of G and clique i 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 Hˆ 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 , where x = x k ,k ,...,k ; vertex v is in A ,1 i j. Define X = b : A A = , 1 2 j ki ≤ ≤ { i,j i ∩ j ∅} X = v G : dK(v) = i for i = 1,2,...,n. i { ∈ } Let C betheintersection matrixofthecliques A ’sofG where c = 0if A A = i i,j i j ∩ otherwise c for i = j and c is 0. Let 1,2,...,n be the n-colors. i,i ∅ 6 Let T = X , P = and S = j : T = ,2 j n . i i i j ∅ { 6 ∅ ≤ ≤ } If S = , then the graph G has no vertex of clique degree greater than one, ∅ which implies G has exactly n2(maximum number) vertices. i.e., G is n components of K . Otherwise, consider the intersection matrix C as the color matrix and follow n the steps. Step 1: If S = , stop the process. Otherwise, let max(S) = k, for some k,2 k ∅ ≤ ≤ n. Then consider the sets T and P , go to step 2. k k Step 2: If T = , go to step 1. Otherwise, choose a vertex u from T , where k ∅ i1,i2,...,ik k i < i < < i and go to Step 3. 1 2 k ··· Step 3: Let Y = y : color y appears atleast k 1 times in the ith row of the color i { − matrix , i = 1,2,...,n. If ik Y = n, let B = n P , B = and go to } | i=i1 i| T i=2 i P ∅ Step 4. Otherwise, constructSa new color matrix C Sby putting least x in c , 1 i,j where x 1,2,3,...n ik Y , i = j, i i,j i . Then add the vertex ∈ { } \ i=i1 i 6 1 ≤ ≤ k u to P and removeSit from T , go to Step 2. i1,i2,...,ik k k Step 4: Choose a vertex v from B such that v A , for some i, i i i . Let T i 1 k ∈ ≤ ≤ B = i : v A ,1 i n and go to Step 5. i { ∈ ≤ ≤ } Step 5: Let Y = y : color y appears atleast k 1 times in the ith row of the color i { − matrix , for every i B. If Y = n add the vertex v to B and remove it } ∈ | i∈B i| P from B , go to Step 4. OtheSrwise construct a new color matrix C by putting T 2 x in c , where x 1,2,3,...n Y , i = j, i,j B. Go to Step 3. i,j ∈ { }\ i∈B i 6 ∈ S 4 Thus, we get the modified color matrix C . Then, color the vertex v of Hˆ by M c of C , whenever v A A . Then, extend the coloring of Hˆ to G. Thus G is i,j M i j ∈ ∩ 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 ) = v ,v ,v ,v ,v ,v , V(A ) = v ,v ,v ,v ,v ,v , 1 1 2 3 4 5 6 2 1 7 8 9 10 11 { } { } V(A ) = v ,v ,v ,v ,v ,v , V(A ) = v ,v ,v ,v ,v ,v , 3 1 12 13 14 15 16 4 1 17 18 19 20 21 { } { } V(A ) = v ,v ,v ,v ,v ,v , V(A ) = v ,v ,v ,v ,v ,v . 5 6 7 16 22 23 24 6 9 16 19 25 26 27 { } { } Relabel the vertices of clique degree greater than one in G by u where A = i : A { v A for 1 i 6 . The labeled graph is shown in Figure 1b. Figure 2 is the graph i ∈ ≤ ≤ } Hˆ, where Hˆ is obtained by removing the vertices of clique degree 1 from G. Let X = b : A A = = b ,b , i,j i j 1,6 4,5 { ∩ ∅} { } X = v G : dK(v) = 1 = v ,v ,v ,v ,v ,v ,v ,v ,v ,v , 1 2 3 5 8 10 11 12 13 14 15 { ∈ } { v ,v ,v ,v ,v ,v ,v ,v ,v ,v , 17 18 20 21 22 23 24 25 26 27 } X = v G : dK(v) = 2 = v ,v ,v ,v = u ,u ,u ,u , 2 6 7 9 19 1,5 2,5 2,6 4,6 { ∈ } { } { } X = v G : dK(v) = 3 = v = u , 3 16 3,5,6 { ∈ } { } { } X = v G : dK(v) = 4 = v = u , 4 1 1,2,3,4 { ∈ } { } { } X = and X = . 5 6 ∅ ∅ 0 c c c c 0   c 0 c c c c  c c 0 c c c  Let 1, 2, ..., 6 be the six colors and C =      c c c 0 0 c     c c c 0 0 c       0 c c c c 0    be the intersection matrix of order 6 6. × Consider the sets T = X , P = for i = 1,2,...6 and S = j : T = ,2 j i i i j ∅ { 6 ∅ ≤ ≤ n = 2,3,4 . Then by applying the method given above, we get the following. } { } Step 1: Since S = and max(S) = 4, then choose the sets T = u and 4 1,2,3,4 6 ∅ { } P = . Go to step 2. 4 ∅ Step 2: Since T = , choose the vertex u from T , go to step 3. 4 1,2,3,4 4 6 ∅ Step 3: Since Y = , Y = , Y = , Y = and Y Y Y Y < 6, choose 1 2 3 4 1 2 3 4 ∅ ∅ ∅ ∅ | ∪ ∪ ∪ | the minimum color from the set 1,2,...,6 Y and construct a new color i=1,2,3,4 i { }\ ∪ 5 (a) Graph G (b) Graph G after relabeling the verticesof clique degree greater than one Figure 1: Graph G: before and after relabeling the vertices matrix C by putting 1 in c , i = j, i,j = 1,2,3,4. Add the vertex u to P and 1 i,j 1,2,3,4 4 6 remove it from T . Then 4 6 Figure 2: Graph Hˆ 0 1 1 1 c 0   1 0 1 1 c c  1 1 0 1 c c  C =   , 1    1 1 1 0 0 c     c c c 0 0 c       0 c c c c 0    T = , P = u . Go to step 2. 4 4 1,2,3,4 ∅ { } Step 2: Since T = , go to step 1. 4 ∅ Step 1: Since S = and max(S) = 3, then choose the sets T = u and 3 3,5,6 6 ∅ { } P = . Go to step 2. 3 ∅ Step 2: Since T = , choose the vertex u from T , go to step 3. 3 3,5,6 3 6 ∅ Step 3: Since Y = 1 , Y = , Y = , and Y Y Y < 6, choose the 3 5 6 3 5 6 { } ∅ ∅ | ∪ ∪ | minimum color from the set 1,2,...,6 Y and construct a new color matrix i=3,5,6 i { }\∪ C by putting 2 in c , i = j, i,j = 3,5,6. Add the vertex u to P and remove it 2 i,j 3,5,6 3 6 from T . Then 3 0 1 1 1 c 0   1 0 1 1 c c  1 1 0 1 2 2  C =  , 2    1 1 1 0 0 c     c c 2 0 0 2       0 c 2 c 2 0    T = , P = u . Go to step 2. 3 3 3,5,6 ∅ { } Step 2: Since T = , go to step 1. 3 ∅ 7 Step 1: SinceS = andmax(S) = 2, then choosethe sets T = u ,u ,u ,u 2 1,5 2,5 2,6 4,6 6 ∅ { } and P = . Go to step 2. 2 ∅ Step 2: Since T = , choose the vertex u from T , go to step 3. 2 1,5 2 6 ∅ Step 3: Since Y = 1 , Y = 2 and Y Y < 6, choose the minimum color 1 5 1 5 { } { } | ∪ | from the set 1,2,...,6 Y and construct a new color matrix C by putting 3 i=1,5 i 3 { }\∪ in c , i = j, i,j = 1,5. Add the vertex u to P and remove it from T . Then i,j 1,5 2 2 6 0 1 1 1 3 0   1 0 1 1 c c  1 1 0 1 2 2  C =  , 3    1 1 1 0 0 c     3 c 2 0 0 2       0 c 2 c 2 0    T = u ,u ,u , P = u . Go to step 2. 2 2,5 2,6 4,6 2 1,5 { } { } Step 2: Since T = , choose the vertex u from T , go to step 3. 2 2,5 2 6 ∅ Step 3: Since Y = 1 , Y = 2,3 and Y Y < 6, choose the minimum color 2 5 2 5 { } { } | ∪ | from the set 1,2,...,6 Y and construct a new color matrix C by putting 4 i=2,5 i 4 { }\∪ in c , i = j, i,j = 2,5. Add the vertex u to P and remove it from T . Then i,j 2,5 2 2 6 0 1 1 1 3 0   1 0 1 1 4 c  1 1 0 1 2 2  C =  , 4    1 1 1 0 0 c     3 4 2 0 0 2       0 c 2 c 2 0    T = u ,u , P = u ,u . Go to step 2. 2 2,6 4,6 2 1,5 2,5 { } { } Step 2: Since T = , choose the vertex u from T , go to step 3. 2 2,6 2 6 ∅ Step 3: Since Y = 1,4 , Y = 2 and Y Y < 6, choose the minimum color 2 6 2 6 { } { } | ∪ | from the set 1,2,...,6 Y and construct a new color matrix C by putting 3 i=2,6 i 5 { }\∪ in c , i = j, i,j = 2,6. Add the vertex u to P and remove it from T . Then i,j 2,6 2 2 6 0 1 1 1 3 0   1 0 1 1 4 3  1 1 0 1 2 2  C =  , 5    1 1 1 0 0 c     3 4 2 0 0 2       0 3 2 c 2 0    T = u , P = u ,u ,u . Go to step 2. 2 4,6 2 1,5 2,5 2,6 { } { } 8 Step 2: Since T = , choose the vertex u from T , go to step 3. 2 4,6 2 6 ∅ Step 3: Since Y = 1 , Y = 2,3 and Y Y < 6, choose the minimum color 4 6 4 6 { } { } | ∪ | from the set 1,2,...,6 Y and construct a new color matrix C by putting 4 i=4,6 i 6 { }\∪ in c , i = j, i,j = 4,6. Add the vertex u to P and remove it from T . Then i,j 4,6 2 2 6 0 1 1 1 3 0   1 0 1 1 4 3  1 1 0 1 2 2  C =  , 6    1 1 1 0 0 4     3 4 2 0 0 2       0 3 2 4 2 0    T = , P = u ,u ,u ,u . Go to step 2. 2 2 1,5 2,5 2,6 4,6 ∅ { } Step 2: Since T = , go to step 1. 2 ∅ Step 1: Since S = , stop the process. ∅ Assign the colors to the graph Hˆ using the matrix C = C , i.e., color the vertex M 6 v by the (i,j)-th entry c of the matrix C , whenever A A = (see Figure 3a), i,j M i j ∩ 6 ∅ where the numbers 1, 2, 3, 4, 5, 6 corresponds to the colors Maroon, Tan, Green, Red, Blue, Cyan respectively. Extend the coloring of Hˆ to G by assigning the remaining colors which are not used for A from the set of 6-colors to the vertices of clique degree i one in each A , 1 i 6. The colored graph G is shown in Figure 3b. i ≤ ≤ 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 ’s is non empty, then i dK(v ) dK(v ) dK(v ) n(n 1) 1 + 2 + + l = − , (cid:18) 2 (cid:19) (cid:18) 2 (cid:19) ··· (cid:18) 2 (cid:19) 2 where v ,v ,...,v is the set of all vertices of clique degree greater than 1 in G. 1 2 l { } Proof: If G is isomorphic to the graph H for some n, then the result is obvious. n If not there exists at least one vertex v of clique degree greater than 2. Define I = i : v A then dK(v) = I = p. For every unordered pair of elements v i v { ∈ } | | (i,j) of I there is a vertex b ( where i < j) in H . Therefore corresponding to v ij n the elements of I there are p vertices in H . Since G satisfies the hypothesis of v 2 n (cid:0) (cid:1) 9 (a) Graph Hˆ (b) A 6 coloring of graph G Figure 3: The graphs Hˆ and G, after colors have been assigned to their vertices. Conjecture 1.2, there is no vertex v′ different from v in G such that v′ A A where i j ∈ ∩ i,j I . Therefore for every vertex v of clique degree greater than 1 in G, there are v ∈ dK(v) vertices of clique degree greater than 1 in H . As there are n(n−1) vertices of 2 n 2 (cid:0)clique(cid:1)degree greater than 1 in H , n(n−1) = dK(v1) + dK(v2) + + dK(vl) where n 2 2 2 ··· 2 v ,v ,...,v is the set of all vertices of cliq(cid:0)ue deg(cid:1)ree (cid:0)greate(cid:1)r than 1 i(cid:0)n G. (cid:1) 1 2 l { } Corollary 2.5 If G is a graph satisfying the hypothesis of conjecture 1.2, then G has 10

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