Equitable Coloring of Mycielskian of Some Graphs Equitable Coloring of Mycielskian Ofj

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Journal of Mathematical Extension Vol. 11, No. 3, (2017), 1-18 JournalISSN: 1735-8299 of Mathematical Extension URL: http://www.ijmex.com Vol. 11, No. 3, (2017), 1-18 ISSN: 1735-8299 URL: http://www.ijmex.com

Equitable Coloring of Mycielskian of Some Graphs Equitable Coloring of Mycielskian ofJ. Some Vernold Graphs Vivin∗ University College of Engineering Nagercoil

(Anna UniversityJ. Vernold Constituent Vivin∗ College) University College of Engineering Nagercoil K. Kaliraj Ramanujan(Anna Institute University for Advanced Constituent Study College) in Mathematics UniversityK. Kaliraj of Madras Ramanujan Institute for Advanced Study in Mathematics University of Madras Abstract. In a search for triangle-free graphs with arbitrarily large chromatic numbers, Mycielski developed a graph transformation that transforms a graph G into a new graph µ(G), we now call the Myciel- Abstract. In a search for triangle-free graphs with arbitrarily large skian of G, which has the same clique number as G and whose chro- chromatic numbers, Mycielski developed a graph transformation that matic number equals χ(G) + 1. This paper presents exact values of the transforms a graph G into a new graph µ(G), we now call the Myciel- equitable chromatic number χ for the Mycielski’s graph of complete skian of G, which has the same= clique number as G and whose chro- graphs µ(K ), the Mycielski’s graph of cycles µ(C ), the Mycielski’s matic numbern equals χ(G) + 1. This paper presentsn exact values of the graph of paths µ(Pn), the Mycielski’s graph of Helm graphs µ(Hn) and equitable chromatic number χ= for the Mycielski’s graph of complete the Mycielski’s graph of Gear graphs µ(Gn). graphs µ(Kn), the Mycielski’s graph of cycles µ(Cn), the Mycielski’s AMSgraph of Subject paths µ Classification:(Pn), the Mycielski’s05C15; graph 05C75 of Helm graphs µ(Hn) and Keywordsthe Mycielski’s and graph Phrases: of GearEquitable graphs µ coloring,(Gn). Mycielski’s graph, Helm graph, Gear graph AMS Subject Classification: 05C15; 05C75 Keywords and Phrases: Equitable coloring, Mycielski’s graph, Helm 1.graph, Introduction Gear graph The1. notion Introduction of equitable coloring was introduced by Meyer in 1973. This model of graph coloring has many applications. Everytime when we have toThe divide notion a system of equitable with binarycoloring conflicting was introduced relations by Meyer into inequal 1973. or This almost model of graph coloring has many applications. Everytime when we have Received: July 2016; Accepted: October 2016 to∗ divideCorresponding a system author with binary conflicting relations into equal or almost

Received: July 2016; Accepted: October 2016 ∗Corresponding author 1

1 2 J. VERNOLD VIVIN AND K. KALIRAJ equal conflict-free subsystems we can model this situation by means of equitable graph coloring. This subject is widely discussed in literature [1, 3, 4, 6, 7, 11, 12, 16, 18]. One motivation for equitable coloring suggested by Meyer [13] concerns scheduling problems. In this application, the vertices of a graph represent a collection of tasks to be performed, and an edge connects two tasks that should not be performed at the same time. A coloring of this graph represents a partition of tasks into subsets that may be performed simul- taneously. Due to load balancing considerations, it is desirable to per- form equal or nearly-equal numbers of tasks in each time step, and this balancing is exactly what an equitable coloring achieves. Furma´nczyk [7] mentions a specific application of this type of scheduling problem, namely assigning university courses to time slots in a way that spreads the courses evenly among the available time slots and avoids scheduling incompatible pairs of courses at the same time as each other, since then the usage of additional resources (e.g.rooms) is maximal. The notion of equitable colorability was introduced by Meyer [13]. How- ever, an earlier work of Hajnal and Szemer´edi[9] showed that a graph G with maximal degree ∆ is equitably k-colorable if k  ∆(G) + 1. Re- cently, Kierstead et al. [10] have given an O(∆ V (G) 2)-time algorithm | | for equitable (∆ + 1)-coloring of a graph G. In 1973, Meyer [13] formu- lated the following conjecture: Conjecture 1.1. [Equitable Coloring Conjecture (ECC)] For any connected graph G, other than complete graph or odd cycle, χ=(G)  ∆(G). This conjecture has been verified for all graphs on six or fewer vertices. Lih and Wu [11] proved that the Equitable Coloring Conjecture is true for all bipartite graphs. Wang and Zhang [16] considered a broader class of graphs, namely r-partite graphs. They proved that Meyer’s conjecture is true for complete graphs from this class. Also, the conjecture was confirmed for outerplanar graphs [17] and planar graphs with maximum degree at least 13 [18]. There are very few papers on the complexity of equitable coloring. First of all, a straightforward reduction from graph coloring to equitable coloring by adding sufficiently many isolated vertices to a graph, proves that it is NP-complete to test whether a graph ��

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�� EQUITABLE COLORING OF MYCIELSKIAN ... 5

In this paper, by G we mean a connected graph. From a graph G, by Mycielski’s construction [2, 14, 15], we get the Mycielskian µ (G) of G with V (µ (G)) = V U z , where ∪ ∪ { } V = V (G) = x , . . . , x ,U = y , . . . , y , and { 1 n} { 1 n} E (µ (G)) = E (G) y x : x N (x ) z , i = 1, . . . , n . ∪ { i ∈ G i ∪ { } } The paper is organized as follows. We start the next section with a theorem concerning the equitable coloring of Mycielskian of complete graphs. In the subsequent Sections 4, 5, 6 and 7 results concerning the equitable colorability of Mycielskian of cycles, paths, Helm graphs and Gear graphs are discussed. In this way we establish a new class of graphs that can be colored optimally in polynomial time and confirm the ECC conjecture.

3. Equitable Coloring of Mycielskian of Com- plete Graphs

Theorem 3.1. The equitable chromatic number of Mycielskian of complete graph χ=(µ(Kn)) = n + 1.

Proof. Let V (K ) = x : 1 i n . By Mycielski’s construction, n { i   } V (µ (K )) = V (K ) y : 1 i n z and n n ∪ { i   } ∪ { } E (µ (K )) = E (K ) y x : x N (x ) z , i = 1, 2, . . . , n . n n ∪ { i ∈ Kn i ∪ { } } Since z is adjacent with each vertex of y : 1 i n , in which each { i   } vertex is adjacent with z. Also µ (Kn) contains a n-clique, χ (µ (Kn))  n and hence χ= (µ (Kn))  n.

Assume that χ= (µ (Kn)) = n. (i.e), There exists a partition P = V ,V ,...,V , since V (µ (K )) = 2n+1, each V (1 i n) contains { 1 2 n} | n | i   either 2 or 3 vertices. Since each xi is adjacent with each yj (1  j  n) where i = j, x , y V (1 i n). Since z is adjacent to each y (1  i i ∈ i   i  i n), z / V for every i, (1 i n)  ∈ i   6 J. VERNOLD VIVIN AND K. KALIRAJ

i.e., Vi = 2 for all 1  i  n n | | V = 2n = 2n + 1 ⇒ | i|  i=1 This contradiction shows that χ= (µ (Kn))  n + 1. Now we partition the vertex set V (µ(Kn)) as follows, V = x , y : 1 i n , i { i i   } V = z . n+1 { } Clearly V (1 i n) and V are independent sets, since V = 2(1 i   n+1 | i|  i n) and V = 1 satisfying the condition V V 1, for any  | n+1| || i| − | j||  i = j, χ (µ(K )) n + 1. Hence χ (µ(K )) = n + 1.  = n  = n  4. Equitable Coloring of Mycielskian of Cycles

Theorem 4.1. The equitable chromatic number of Mycielskian of cycle 3 n = 4, 6, 8 χ=(µ(Cn)) = 4 if n  10, n is even 4 if n is odd.

 Proof. Let V (C ) = x : 1 i n be the set of vertices of C taken n { i   } n in cyclic order. By Mycielski’s construction, V (µ (C )) = V (C ) y : 1 i n z and n n ∪ { i   } ∪ { } E (µ (C )) = E (C ) y x : x N (x ) z , i = 1, 2, . . . , n . n n ∪ { i ∈ Cn i ∪ { } } Now we partition the vertex set of V (µ(Cn)) as follows: Case (i)a For n = 4:

V = x , x , z , 1 { 1 3 } V = x , x , y , 2 { 2 4 4} V = y , y , y . 3 { 1 2 3} EQUITABLE COLORING OF MYCIELSKIAN ... 7

Case (i)b For n = 6: V = x , x , x , z , 1 { 1 3 5 } V = x , x , x , y , y , 2 { 2 4 6 4 6} V = y , y , y , y . 3 { 1 2 3 5} Case (i)c For n = 8: V = x , x , x , x , z , 1 { 1 3 5 7 } V = x , x , x , x , y , y , 2 { 2 4 6 8 4 6} V = y , y , y , y , y , y . 3 { 1 2 3 5 7 8} It is clear that V V = for i = j. Since there exists a 5-cycle i ∩ j ∅  x1x2y3zy2x1, χ (µ (Cn))  3 and hence χ= (µ (Cn)) = 3.

Case (ii) If n  10, n is even and n is odd. Since there exists a 5-cycle as shown in the previous case, it is clear that χ= (µ (Cn))  3.

We know that V (µ (Cn)) = 2n + 1. Out of these 2n + 1, the vertices of | | n x : 1 i n z can be assigned with one color. i.e., + 1 2i   2 ∪ { } 2 vertices can be colored with one color. Hence remaining number  of n 3n vertices to be colored is (2n + 1) + 1 = . − 2 2      3n If χ (µ(C )) = 3, these vertices should be divided into two sets, = n 2   3n 3n each containing and + 1 vertices. 4 4     3n n 3n n Since = +1 and +1 = +1 for even n 10 and odd 4  2 4  2      n, we conclude that the partition of V (µ (Cn)) into three sets satisfying V V 1 for i = j is not possible and hence χ (µ(C )) 4, if || i| − | j||   = n  n  10 and n is odd.

Now the following partition gives the equitable coloring of µ (Cn) for even n  10 and odd n. 8 J. VERNOLD VIVIN AND K. KALIRAJ

If n  10, n is even: n V1 = x2i 1 : 1  i  z , − 2 ∪ { }  n  V = x : 1 i , 2 2i   2  n V3 = y2i 1 : 1  i  , − 2  n  V = y : 1 i . 4 2i   2   If n is odd: n 1 V1 = x2i 1 : 1  i  − z , − 2 ∪ { }   n 1 V2 = x2i : 1  i  − yn 1 , 2 ∪ { − }   n + 1 V3 = y2i 1 : 2  i  xn , − 2 ∪ { }   n 3 V = y : 1 i − y . 4 2i   2 ∪ { 1}   This implies χ= (µ (Cn)) = 4. 

5. Equitable Coloring of Mycielskian of Paths

Theorem 5.1. The equitable chromatic number of Mycielskian of path

2 n = 1,

χ=(µ(Pn)) = 3 if n 11, n = 10,    4 if n 10, n = 11.   Proof. Let V (P ) = x : 1 i n be the set of vertices of P . By n { i   } n Mycielski’s construction,

V (µ (P )) = V (P ) y : 1 i n z and n n ∪ { i   } ∪ { } E (µ (P )) = E (P ) y x : x N (x ) z , i = 1, 2, . . . , n . n n ∪ { i ∈ Pn i ∪ { } }

Now we partition the vertex set of V (µ(Pn)) as follows: EQUITABLE COLORING OF MYCIELSKIAN ... 9

Case (i) For n = 1 V = x , z , 1 { 1 } V = y . 2 { 1} It is obvious.

Case (ii) By this partitions χ= (µ (Pn)) = 3 1. For n = 2: V = x , z , 1 { 1 } V = x , y , 2 { 2 2} V = y . 3 { 1} 2. For n = 3: V = x , x , z , 1 { 1 3 } V = x , y , 2 { 2 2} V = y , y . 3 { 1 3} 3. For n = 4: V = x , x , z , 1 { 1 3 } V = x , x , y , 2 { 2 4 2} V = y , y , y . 3 { 1 3 4} 4. For n = 5: V = x , x , x , z , 1 { 1 3 5 } V = x , x , y , y , 2 { 2 4 2 4} V = y , y , y . 3 { 1 3 5} 5. For n = 6: V = x , x , x , z , 1 { 1 3 5 } V = x , x , x , y , y , 2 { 2 4 6 2 4} V = y , y , y , y . 3 { 1 3 5 6} 10 J. VERNOLD VIVIN AND K. KALIRAJ

6. For n = 7:

V = x , x , x , x , z , 1 { 1 3 5 7 } V = x , x , x , y , y , 2 { 2 4 6 2 4} V = y , y , y , y , y . 3 { 1 3 5 6 7} 7. For n = 8:

V = x , x , x , x , z , 1 { 1 3 5 7 } V = x , x , x , x , y , y , 2 { 2 4 6 8 2 4} V = y , y , y , y , y , y . 3 { 1 3 5 6 7 8} 8. For n = 9:

V = x , x , x , x , x , z , 1 { 1 3 5 7 9 } V = x , x , x , x , y , y , 2 { 2 4 6 8 2 4} V = y , y , y , y , y , y , y . 3 { 1 3 5 6 7 8 9} 9. For n = 11: n V1 = x2i 1 : 1  i  z , − 2 ∪ { }  n  V = x : 1 i y , y , y , 2 2i   2 ∪ { 2 4 6} V = y : 7 i n y , y , y . 3 { i   } ∪ { 1 3 5} It is clear that V V = for i = j. Since there exists a 5-cycle i ∩ j ∅  x1x2y3zy2x1, χ (µ (Pn))  3 and hence χ= (µ (Pn)) = 3. Case (iii) By this partition χ (µ (P )) = 4 for n 10, n = 11. = n   Since there exists a 5-cycle x1x2y3zy2x1, it is clear that χ= (µ (Pn))  3.

We know that V (µ (Pn)) = 2n + 1. Out of these 2n + 1, the vertices of | | n x : 1 i n z can be assigned with one color. i.e., + 1 2i   2 ∪ { } 2 vertices can be colored with one color. Hence remaining number  of n 3n vertices to be colored is (2n + 1) + 1 = . − 2 2      EQUITABLE COLORING OF MYCIELSKIAN ... 11

3n If χ (µ(P )) = 3, these vertices should be divided into two sets, = n 2   3n 3n each containing and + 1 vertices. 4 4     3n n 3n n Since = + 1 and + 1 = + 1 for n 10, n = 11, 4  2 4  2       we conclude that  the partition of V (µ (Pn)) into three sets satisfying V V 1 for i = j is not possible and hence χ (µ(P )) 4, for || i| − | j||   = n  n 10, n = 11.   Now the following partition gives the equitable coloring of µ (Pn) for n 10, n = 11.   n V = x : 1 i z , 1 2i   2 ∪ { }   n V2 = x2i 1 : 1  i  , − 2  n  V3 = y2i 1 : 1  i  , − 2  n  V = y : 1 i . 4 2i   2    This implies χ= (µ (Pn)) = 4. 

6. Equitable Coloring of Mycielskian of Helm Graphs

Theorem 6.1. The equitable chromatic number of Mycielski’s graph of Helm graph 4 if n is even χ=(µ(Hn)) = 5 if n is odd.

Proof. Let

V (H ) = x x : 1 i n x : 1 i n n { 0} ∪ { i   } ∪ i   and E (H ) = x x : 1 i n x x : 1 i n 1 n { 0 i   } ∪ { i i+1   − } x x x x : 1 i n . ∪ { n 1} ∪ i i     12 J. VERNOLD VIVIN AND K. KALIRAJ

By Mycielski’s construction,

V (µ (H )) = V (H ) y : 0 i n y : 1 i n z . n n ∪ { i   } ∪ i   ∪ { }   In µ (H ), y is adjacent with each of x : 1 i n , each y is ad- n 0 { i   } i jacent with the neighbour of xi (0  i  n), z is adjacent with each of y : 0 i n and y : 1 i n . Also each y is adjacent with { i   } { i   } i xi (1  i  n) Since x0 and y0 are non adjacent vertices, we use same color to color x0 and y let it be 1. Since y is adjacent with each vertex of x : 1 i 0 0 { i  n , the color 1 cannot be assigned to any vertex x : 1 i n .  } { i   } Case (i) If n is even: Since the set of vertices x : 1 i n induce a cycle of length n { i   } in µ (Hn), we use two colors 2 and 3 to color all the vertices of this cycle. Also same color cannot be assigned to color the vertices of y : 1 i n , since z is adjacent with each vertex of y : 0 i n , { i   } { i   } z cannot be assigned with any of the three colors and let it be assigned with color 4. Hence χ=(µ(Hn))  4.

We use the following partition to color µ (Hn) equitably with 4 given colors.

V1 = x0, y0, yi : 3  i  n , n V2 = x2i 1, y2i 1 : 1  i  y2 , − − 2 ∪  n  V = x , y : 1 i y ,  3 2i 2i   2 ∪ 1   V4 = z, xi : 1  i  n .     Clearly V V 1 for i = j. || i| − | j||   Hence χ=(µ(Hn))  4 and so χ= (µ (Hn)) = 4. Case (ii) If n is odd: Since the vertices x : 1 i n induce a cycle of odd length n, in { i   } µ (Hn), we use three colors (2,3 and 4) to color the vertices of this cycle. Now z is colored with a new color 5. Hence χ=(µ(Hn))  5. EQUITABLE COLORING OF MYCIELSKIAN ... 13

We use the following partition to color the vertices of µ(Hn) equitable in the following cases, Case (ii)a

1. For n = 3:

V1 = x0, y0, x3 , V2 = x1, y1, y3 , V3 = x2, y2, x1  , V4 = x3, y3, y2 , V5 = x1 , x2 , z .   2. For n = 6k 3, k 2: −  2n + 6 n 3 V = x , y x : i n y : − i n , 1 { 0 0} ∪ i 3   ∪ i 3       n n 3 V2 = x3i 2, y3i 2 : 1  i  y3 i : 1  i  − , − − 3 ∪ 3    n n 3 V3 = x3i 1, y3i 1 : 1  i  y3 i 2 : 1  i  − , − − 3 ∪ − 3    n  n 3 V4 = x3i, y3i : 1  i  y3 i 1 : 1  i  − , 3 ∪ − 3    2n + 3 V = x : 1 i z . 5 i   3 ∪ { }   Case (ii)b

1. For n = 5:

V1 = x0, y0, x4 , x5 , V2 = x1, y1, x4, y4, y3 , V3 = x2, y2, x5, y5, y1  , V4 = x3, y3, y2 , y4 , y5 , V5 = x1 , x2 , x3 , z .    14 J. VERNOLD VIVIN AND K. KALIRAJ

2. For n = 6k 1, k 2: −  2n + 2 n + 7 V = x , y x : i n y : i n , 1 { 0 0} ∪ i 3   ∪ i 2       n + 1 n + 1 V2 = x3i 2, y3i 2 : 1  i  y3 i : 1  i  , − − 3 ∪ 6     n + 1 n + 1 V3 = x3i 1, y3i 1 : 1  i  y3 i 2 : 1  i  , − − 3 ∪ − 6     n 2 n + 1 V4 = x3i, y3i : 1  i  − y3 i 1 : 1  i  , 3 ∪ − 6     2n 1 V = x : 1 i − z . 5 i   3 ∪ { }   Case (ii)c 1. For n = 7:

V1 = x0, y0, x6 , x7 , y1 , y4 , V2 = x1, y1, x4, y4, y3 , y6 , y7 , V = x , y , x , y , x , y , 3 { 2 2 5 5 7 7}  V4 = x3, y3, x6, y6, y2 , y5 , V5 = x1 , x2 , x3 , x4 , x5 , z . 2. For n = 6k + 1, k  2:   5n + 1 n + 11 V = x , y x : i n y : i n , 1 { 0 0} ∪ i 6   ∪ i 3       n + 2 n 7 V2 = x3i 2, y3i 2 : 1  i  y3 i+2 : 1  i  − , − − 3 ∪ 6     n + 2 n 7 V3 = x3i 1, y3i 1 : 1  i  y3 i+3 : 1  i  − , − − 3 ∪ 6     y , y ∪ 1 3 n 1 n 7 V = x , y : 1 i − y : 1 i − , 4 3i 3i   3 ∪ 3i+4   6     y , y ∪ 2 4 5n 5 V = x : 1 i − z . 5 i   6 ∪ { }   EQUITABLE COLORING OF MYCIELSKIAN ... 15

Clearly in all the cases V V 1 for i = j. || i| − | j||   Hence χ=(µ(Hn))  5 and so χ= (µ (Hn)) = 5 for n is odd. 

7. Equitable Coloring of Mycielskian of Gear Graphs

Theorem 7.1. The equitable chromatic number of Mycielski’s graph of Gear graph

3 if 3  n  5 χ=(µ(Gn)) = 4 if n  6

Proof. Let

V (G ) = x x : 1 i n x : 1 i n n { 0} ∪ { i   } ∪ i   and E (G ) = x x : 1 i n x x : 1 i n n { 0 i   } ∪ i i    x x : 1 i n 1 x x . ∪ i i+1   − ∪ n 1      By Mycielski’s construction,

V (µ (G )) = V (G ) y : 1 i n y : 1 i n z . n n ∪ { i   } ∪ i   ∪ { }   In µ (H ), y is adjacent with each vertex of x : 1 i n , each y n 0 { i   } i is adjacent with the neighbour of xi (1  i  n) and each yi is adjacent with the neighbours of xi (1  i  n), z is adjacent with each vertex of y : 1 i n y : 1 i n y . { i   } ∪ { i   } ∪ { 0} Case (i) For n = 3, 4, 5: Since the set of vertices x : 1 i n x : 1 i n induce an even { i   }∪{ i   } cycle in µ (Gn), it requires atleast two colors to color the vertices of this cycle (say 1 and 2). If we assign the same color 1 and 2 to color the vertices of y : 1 i n y : 1 i n then z will receive a new { i   } ∪ { i   } color other than 1 and 2. Hence χ=(µ(Gn))  3. 16 J. VERNOLD VIVIN AND K. KALIRAJ

1. For n = 3:

V1 = x0, x1 , x2 , x3 , z , V = x , x , x , y , y , 2 { 1 2 3 1 3} V3 = y0, y2, y1 , y2 , y3 .   2. For n = 4:

V1 = x0, x1 , x2 , x3 , x4 , z , V = x , x , x , x , y , y , 2 { 1 2 3 4 1 2} V3 = y0, y3, y4, y1 , y2 , y3 , y4 .   3. For n = 5:

V1 = x0, x1 , x2 , x3 , x4 , x5 , z , V = x , x , x , x , x , y , y , y , 2 { 1 2 3 4 5 1 2 3} V3 = y0, y4, y5, y1 , y2 , y3 , y4 , y5 .   Clearly V V 1 for i = j. || i| − | j||   Hence χ=(µ(Gn))  3 and hence χ= (µ (Gn)) = 3. Case (ii) For n  6: We know that V (µ (G )) = 4n + 3. Out of these 4n + 3, the vertices | n | of x : 1 i n x z can be assigned with one color. i.e., { i   } ∪ { 0} ∪ { } n + 2 vertices can be colored with one color. Hence remaining number of vertices to be colored is (4n + 3) (n + 2) = 3n + 1. − If χ=(µ(Gn)) = 3, these 3n + 1 vertices should be divided into two sets, 3n 3n + 1 each containing and vertices. 2 2 3n 3n + 1 Since = n + 2 and = n + 2 for n 6, we conclude that 2  2   the partition of V (µ (G )) into three sets satisfying V V 1 for n || i| − | j||  EQUITABLE COLORING OF MYCIELSKIAN ... 17 i = j is not possible and hence χ (µ(G )) 4.  = n  V = x : 1 i n x , 1 i   ∪ { 0} V = x : 1 i n z , 2 { i   } ∪ { } V = y : 1 i n y , 3 i   ∪ { 0} V = y : 1 i n . 4 { i   } Clearly V V 1 for i = j. || i| − | j||   Hence χ=(µ(Gn))  4 and so χ= (µ (Gn)) = 4. 

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Joseph Vernold Vivin Department of Mathematics Assistant Professor of Mathematics University College of Engineering Nagercoil (Anna University Constituent College) Nagercoil 629 004, Tamil Nadu, India E-mail: [email protected]

Kalimuthu Kaliraj Department of Mathematics Assistant Professor of Mathematics Ramanujan Institute for Advanced Study in Mathematics, University of Madras Chepauk, Chennai 600 005, Tamil Nadu, India E-mail: [email protected]