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Equitable total coloring of complete r -partite p -balanced graphs
Discrete Applied Mathematics (
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Equitable total coloring of complete r-partite p-balanced graphs✩ A.G. da Silva a , S. Dantas a , D. Sasaki b, * a b
IME, Fluminense Federal University, Brazil IME, Rio de Janeiro State University, Brazil
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Article history: Received 9 January 2017 Received in revised form 25 January 2018 Accepted 7 March 2018 Available online xxxx Keywords: r-partite graphs p-balanced graphs Equitable total coloring
a b s t r a c t A total coloring is equitable if the number of elements colored by any two distinct colors differs by at most one, and the smallest positive integer for which a graph has such coloring is called its equitable total chromatic number. In 1974, Bermond determined the total chromatic number of all complete r-partite p-balanced graphs. In 1994, Fu proved that there exist equitable (∆ + 2)-total colorings for all complete r-partite p-balanced graphs of odd order. Later, Wang (2002) conjectured that the equitable total chromatic number of a graph is either ∆ + 1 or ∆ + 2. In this work, we investigate the complete r-partite pbalanced graphs. We improve the bound given by Fu, when r and p are odd, by exhibiting an equitable (∆ + 1)-total coloring for these graphs. Furthermore, for the open cases, that is, if the order is even, we exhibit equitable (∆ + 2)-total colorings for all these graphs, which completely verifies Wang’s conjecture for this class. © 2018 Elsevier B.V. All rights reserved.
1. Introduction Let G = (V , E) be a simple connected graph. A k-total coloring of G is an assignment of k colors to the vertices and edges of G so that adjacent or incident elements have different colors. The total chromatic number of G, denoted by χ ′′ , is the smallest k for which G has a k-total coloring. Clearly, χ ′′ ≥ ∆ + 1 and the Total Coloring Conjecture (TCC) states that the total chromatic number of any graph is at most ∆ + 2, where ∆ is the maximum degree of the graph [1,11]. In 1989, Sánchez-Arroyo [9] proved that the problem of determining the total chromatic number of an arbitrary graph is NP-hard, and it remains NP-hard even for cubic bipartite graphs. If the difference between the cardinalities of any two color classes is either 0 or 1, then the total coloring is said to be equitable. The equitable total chromatic number of G, denoted by χe′′ , is the smallest k for which G has an equitable k-total coloring. Similarly to the total coloring problem, it was conjectured in 2002 by Wang [12] that the equitable total chromatic number of any graph is at most ∆ + 2 (Equitable Total Coloring Conjecture (ETCC)). Ever since, many papers have been published in this subject [3,4,6,7]. In 2016, Dantas et al. [4] proved that the problem of determining the equitable total chromatic number of a cubic bipartite graph is NP-complete. A graph is said to be r-partite if there exists a partition of its vertex set X1 = {x11 , x12 , . . . , x1p1 }, X2 = {x21 , x22 , . . . , x2p2 }, . . . , Xr = {xr1 , xr2 , . . . , xrpr } such that Xi is an independent set, with 1 ≤ i ≤ r. In some cases, we replace the notation xij by xi,j for the benefit of the reader. ✩ Partially supported by CNPq and FAPERJ. Corresponding author. E-mail addresses: [email protected] (A.G. da Silva), [email protected] (S. Dantas), [email protected] (D. Sasaki).
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https://doi.org/10.1016/j.dam.2018.03.009 0166-218X/© 2018 Elsevier B.V. All rights reserved.
Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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A complete r-partite graph is an r-partite graph in which there is an edge between any two vertices of different parts of the partition. We denote a complete r-partite graph having pi , i = 1, . . . , r, vertices in each independent set by Kp1 ,p2 ,...,pr . If pi = p for all i = 1, . . . , r, then the r-partite graph is said to be p-balanced. In this case, we denote the graph by Kr ×p . In 1974, Bermond determined the total chromatic number of all complete r-partite p-balanced graphs [2]. In 1994, Fu [5] investigated the equitable total coloring of complete bipartite graphs and complete r-partite graphs of odd order. In fact, for complete bipartite graphs, Fu proved that χe′′ = χ ′′ . Furthermore, considering complete r-partite graphs of odd order, Fu proved that there exist equitable (∆ + 2)-total colorings for all of these graphs. In this work, we investigate the complete r-partite p-balanced graphs. We improve the bound given by Fu, when r and p are odd, by exhibiting an equitable (∆ + 1)-total coloring for these graphs. Furthermore, we exhibit equitable (∆ + 2)-total colorings for the open cases, that is, the complete r-partite p-balanced graphs of even order. From these results, we verify the ETCC for the following class of graphs: 1. 2. 3. 4.
Kr ×p Kr ×p Kr ×p Kr ×p
with r with r with r with r
and p odd, χe′′ = ∆ + 1; ≥ 4 even and p odd, χe′′ = ∆ + 2; ≥ 4 even and p even, χe′′ ≤ ∆ + 2; odd and p even, χe′′ ≤ ∆ + 2.
Using similar techniques to those applied to some of the above results, we exhibit an equitable (∆ + 2)-total coloring of K2×p and an equitable (∆ + 1)-total coloring of Kp1 ,p2 , with p1 ̸ = p2 . This paper is organized as follows. In Section 2, we present definitions and known results that are useful throughout the next sections. In Section 3, we provide an algorithm to show that χe′′ = ∆ + 1, for Kr ×p with r odd and p odd. In Section 4, we provide an algorithm to obtain an equitable (∆ + 2)-total coloring of K2×p and the technique applied to this case is used in the next sections. In Section 5, we prove that χe′′ = ∆ + 2 for Kr ×p with r ≥ 4 even and p odd. In Section 6, we use the same algorithm presented in Section 5 to show an equitable (∆ + 2)-total coloring for Kr ×p with r ≥ 4 even and p even, establishing that χe′′ ≤ ∆ + 2. Moreover, we present an example with χe′′ = ∆ + 1. In Section 7, we present an equitable (∆ + 2)-total coloring for Kr ×p with r odd and p even, establishing that χe′′ ≤ ∆ + 2. Also, we present an infinite class of graphs within this case for which χe′′ = ∆ + 1. Finally, in the last section, we exhibit an equitable (∆ + 1)-total coloring for complete bipartite nonbalanced graphs. 2. Preliminaries In this section, we introduce results and definitions that are used throughout the paper. A k-edge coloring of G is an assignment of k colors to the edges of G so that adjacent edges have different colors. It is known that the complete graph Kn has a ∆-edge coloring when n is even and has a (∆ + 1)-edge coloring when n is odd [10]. A |V | matching in G is a set M ⊆ E such that no two edges in M share a common vertex. A matching P is perfect if |P | = 2 . Claim 1 (Soifer, 2008 [10]). Let Kn be the complete graph with n ≥ 2 even vertices. Since this graph has a ∆-edge coloring, there exist n − 1 disjoint perfect matchings in Kn . The case n = 2 being trivial, let n ≥ 4 even. In order to obtain such n − 1 perfect matchings M1 , M2 , . . . , Mn−1 , draw a polygon with n − 1 vertices v1 , v2 , . . . , vn−1 and place the vertex vn inside the polygon. Display the vertices of Kn in clockwise ascending order. The edge vi vi+1 , for all 1 ≤ i ≤ n − 2 is an element of Mi and the edge v1 vn−1 is an element of Mn−1 . All parallel diagonals (of the polygon) to the edge in the matching Mi (1 ≤ i ≤ n − 1) are elements of the same matching. For the sake of clarity, assume that i < j when we refer to an element (edge or diagonal) vi vj of the polygon. Observe that an element of the polygon is parallel to another one if they have the form vi vj and vi−1 vj+1 , respectively (if i ̸ = 1). In the case i = 1, a parallel element to v1 vj is vj+1 vn−1 . Since the polygon has an odd number of vertices, after assigning parallel diagonals of vi vj to the matching Mi , one vertex v remains. Thus, the edge vvn must be assigned to the matching Mi as well. Observe the following example in Fig. 1. We have M1 = {v1 v2 , v3 v5 , v4 v6 }, M2 = {v2 v3 , v1 v4 , v5 v6 }, M3 = {v3 v4 , v2 v5 , v1 v6 }, M4 = {v4 v5 , v1 v3 , v2 v6 } and M5 = {v1 v5 , v2 v4 , v3 v6 }. Claim 2 (Soifer, 2008 [10]). Let Kn be the complete graph with n ≥ 3 odd vertices. Since this graph has a (∆ + 1)-edge coloring, there exist n disjoint matchings in Kn . The algorithm to obtain such matchings is similar to the previous one just by deleting the vertex vn inside the polygon (and its corresponding edges in the matchings). Fig. 2 shows an example of the matchings of K5 . We have M1 = {v1 v2 , v3 v5 }, M2 = {v2 v3 , v1 v4 }, M3 = {v3 v4 , v2 v5 }, M4 = {v4 v5 , v1 v3 } and M5 = {v1 v5 , v2 v4 }. The isolated vertices in Fig. 2 represent the remaining vertex in each matching. The graph Kr is the complete graph having r vertices, in which r represents the number of parts of Kr ×p . In this case, we denote its matchings by Ri . Similarly, the graph Kp is the complete graph having p vertices, in which p represents the number of vertices in each part of Kr ×p . In this case, we denote its matchings by Pi . Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 1. Disjoint perfect matchings of K6 .
Fig. 2. Disjoint matchings of K5 .
Let Kr ×p be a graph. We define matching of distance i as the set of edges {xja xj+i,b } linking vertices of rows a and b (1 ≤ a, b ≤ p, a ̸ = b), for all 1 ≤ j ≤ r and j + i is taken modulo r. Since j + i is taken modulo r, we have j + i ≤ r, which implies that i ≤ r − 1 (since j ≥ 1). Note that each matching of distance i has r edges. We say that a color is represented in a vertex if the color is assigned to the vertex itself or to an edge that has such vertex as one of its ends. Throughout this paper, when we refer to results taken modulo i, if such result is congruent 0 modulo i, then we use i instead of 0. 3. Equitable total coloring of Kr ×p with r and p odd In this section, we improve the bound given by Fu [5] by presenting an algorithm to color these graphs with ∆ + 1 colors. We also present an example for K3×13 . Theorem 1. The graph Kr ×p with r and p odd has χe′′ = ∆ + 1. Proof. If p = 1, then we have Kr ×1 = Kr , which is the complete graph. It is known that such graphs have equitable (∆ + 1)total colorings [8]. So, assume that p ≥ 3. In all equitable (∆ + 1)-total coloring of Kr ×p every color must be represented in all vertices. Suppose that q vertices rp−q receive the same color i. It implies that the number of edges colored with color i is 2 . The total number of(elements that )
receive color i is
rp+q . 2
Since rp is odd, q must be odd. Furthermore, the number of elements of the graph rp + rp+1
p−1
rp2 (r −1) 2
p−1
divided by the number of colors ((r − 1)p + 1) has quotient 2 and remainder 2 . It implies that 2 colors are used ( ) rp+3 p−1 rp+1 p−1 times and (r − 1)p + 1 − 2 colors are used 2 times. Therefore, q = 1 or q = 3. We conclude that 2 triples of 2 3(p−1)
vertices receive one color each and the other rp − 2 vertices receive one different color each. In order to form the triples, we use Claim 2 to obtain the matchings Pi of the graph Kp . Take only those with odd index p−2−1 p−1 from 1 to p − 2, that is, P1 , P3 , . . . , Pp−2 , that totalizes 2 + 1 = 2 matchings, the exact number of triples that we
need. In each Pi taken, we need the ends of the first edge and the remaining vertex, that is, vi , vi+1 and vi+ p−1 +1 (all indexes 2
taken modulo p). Since we use only 3 vertices, we will use the notation Pi′ to refer to them. Distribute by rows the Pi′ ’s in a Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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′ table in which the columns represent the parts X1 , X2 , . . . , Xr . The first row of the table is P1′ , P3′ , P5′ , . . . , P2r −1 , the second is ′ ′ ′ ′ ′ ′ P2r +1 , P2r +3 , P2r +5 , . . . , P4r −1 , and so on. We have P1 in the column of X1 and P1 = {v1 , v2 , v2+ p−1 } (that is, the first element 2
of P1 and the remaining vertex). This means that the vertices x11 , x12 and x1,2+ p−1 must be assigned to the same color. The 2
process must be repeated to every Pi′ , with i = 1, 3, 5, . . . , p − 2. Notice that it might happen that Pi′ and Pj′ in the same column of the table share a common vertex. In this case, shift the elements of the row that contains Pj′ one unity to the right (note that the element of that row in the column of Xr is now in the column of X1 ). It is easy to see that a vertex of Kp appears in the first element of a Pi (i odd) precisely once and is the remaining vertex of only one matching of Kp . This implies that any Pi′ has a common vertex with exactly one Pj′ . Thus, shifting the elements of the row that contains Pj′ solves the problem. Up to this point, all vertices are colored. To color the horizontal edges we need the matchings of Kr (Claim 2). If a vertex xij receives ( a certain )color β , then look the matching Rl of Kr in which vi is the remaining vertex. We have Rl = {va vb , . . . , vc vd } |Rl | = r −2 1 . Then the edges xaj xbj , . . . , xcj xdj must be assigned to the same above mentioned color β . By doing that, if a given color is used in a certain vertex, then the color is represented in all vertices of that row. Since Kr ×p (with r and p odd) has an equitable total coloring with ∆ + 1 colors, this means that the colors still need to be represented in the vertices of the other rows. To represent the colors that have been used in 3 vertices and in the rows that contain such vertices, we use the matchings Pi of Kp (Claim 2) as follows. As explained above, the vertices x11 , x12 and x1,2+ p−1 receive the same color k and such color 2
has been represented in the vertices of the rows 1, 2 and 2 + 2 . Assuming that P1 − {v1 v2 } = {va vb , . . . , vc vd }, where { } p−1 a, b, . . . , c , d ∈ {3, 4, . . . , p} − 2 + 2 , assign color k to the matching of distance 1 (see Section 2) linking vertices of p−1
rows a and b; linking vertices of the next pair of rows following the order defined by {va vb , . . . , vc vd }; and so on until rows c and d. Repeat the process for all colors used in the triples of vertices described previously. Notice that by the end of this step, since the colors of the triples were applied on matchings of distance 1 related to P1 −{v1 v2 }, P3 −{v3 v4 }, · · · , Pp−2 −{vp−2 vp−1 }, the edges of the matchings of distance 1 linking rows 1 and 2, 3 and 4, · · · , p − 2 and p − 1 were not colored. Now, we need to represent the colors used in only one vertex (and the rows that contain it) in the other rows. In order to do that, we go from the part X1 to Xr , searching from the vertex xi1 to xip in each part to take the ones that received a color not used in any other vertex. It is important to leave one vertex of the last row (xα p ) to be the last one to apply this step. Let x1a be the first vertex in X1 that satisfies this condition. Thus, the color used in such vertex has been represented in the vertices of row a (in the horizontal edges) and still needs to be represented in the other ones. Take the matching of Kp in which va was the remaining vertex and assign the same color of x1a in the matching of distance 1 if such distance has not been previously used for any of the triples or matching of distance 2 otherwise. In the described order, repeat the process to each color that was used in one vertex using: either the matching of distance 1, if it was not used in the previous steps; or the matching of distance 2, otherwise. Since there are r − 1 distances between any two rows and we apply one matching of distance i per color and that color is associated with one matching Pj (there are p of those matchings), this means we can repeat the process (r − 1)p times. In other words, this process is possible to be done for (r − 1)p colors. However, we are supposed to use (r − 1)p + 1 = ∆ + 1 colors. The last vertex (xα p ) received a color that was represented in the last row. So it still needs to be represented in the other ones. Recall that when using the colors of the triples of vertices in edges, the matching of distance 1 linking rows 1 and 2, rows 3 and 4, · · · , rows p − 2 and p − 1, were not colored. Therefore, they receive the same color as xα p . By construction, we used exactly ∆ + 1 colors and each color was represented in every vertex. Also by construction, no incident or adjacent elements of the graph received the same color. Each color used in a triple of vertices was also used in rp−3 rp+3 rp−1 edges. So, each of these colors was used in 2 elements. The colors used in only one vertex were also used in 2 edges 2 rp+1 rp+3 rp+1 each. So, each of these colors were used in 2 elements. Since 2 − 2 = 1, we conclude that the difference between the cardinalities of any two color classes is at most 1. Therefore, the described algorithm provides an equitable (∆ + 1)-total coloring of Kr ×p when r and p are odd, as desired. □ In the following example, we exhibit an equitable total coloring of K3×13 with 27 = ∆ + 1 colors. We have P1 = {v1 v2 , v3 v13 , v4 v12 , v5 v11 , v6 v10 , v7 v9 }, remaining vertex is v8 ; P2 = {v2 v3 , v1 v4 , v5 v13 , v6 v12 , v7 v11 , v8 v10 }, remaining vertex is v9 ; P3 = {v3 v4 , v2 v5 , v1 v6 , v7 v13 , v8 v12 , v9 v11 }, remaining vertex is v10 ; P4 = {v4 v5 , v3 v6 , v2 v7 , v1 v8 , v9 v13 , v10 v12 }, remaining vertex is v11 ; P5 = {v5 v6 , v4 v7 , v3 v8 , v2 v9 , v1 v10 , v11 v13 }, remaining vertex is v12 ; P6 = {v6 v7 , v5 v8 , v4 v9 , v3 v10 , v2 v11 , v1 v12 }, remaining vertex is v13 ; P7 = {v7 v8 , v6 v9 , v5 v10 , v4 v11 , v3 v12 , v2 v13 }, remaining vertex is v1 ; P8 = {v8 v9 , v7 v10 , v6 v11 , v5 v12 , v4 v13 , v1 v3 }, remaining vertex is v2 ; P9 = {v9 v10 , v8 v11 , v7 v12 , v6 v13 , v1 v5 , v2 v4 }, remaining vertex is v3 ; P10 = {v10 v11 , v9 v12 , v8 v13 , v1 v7 , v2 v6 , v3 v5 }, remaining vertex is v4 ; P11 = {v11 v12 , v10 v13 , v1 v9 , v2 v8 , v3 v7 , v4 v6 }, remaining vertex is v5 ; P12 = {v12 v13 , v1 v11 , v2 v10 , v3 v9 , v4 v8 , v5 v7 }, remaining vertex is v6 ; P13 = {v1 v13 , v2 v12 , v3 v11 , v4 v10 , v5 v9 , v6 v8 }, remaining vertex is v7 . ′ Also, P1′ = {v1 , v2 , v8 }, P3′ = {v3 , v4 , v10 }, P5′ = {v5 , v6 , v12 }, P7′ = {v7 , v8 , v1 }, P9′ = {v9 , v10 , v3 }, P11 = {v11 , v12 , v5 }. ′ Table 1 shows the distribution of Pi before the shifting. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
A.G. da Silva et al. / Discrete Applied Mathematics ( Table 1 Distribution of Pi′ (i shifting.
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∈ {1, 3, 5, 7, 9, 11}) among the parts Xj before the
X1
X2
X3
P1′
P3′
P5′
′
′
P7
′ P11
P9
Table 2 Distribution of Pi′ (i ∈ {1, 3, 5, 7, 9, 11}) among the parts Xj after the shifting. X1
X2
X3
P1′
P3′
P5′
′ P11
P7′
P9′
Table 3 Distribution of matchings of distance 1 and 2 among the colors. 1→ 2→ 3→ 4→ 5→ 6→ 7→ 8→ 9→
P1 − {v1 v2 } dist 1 P11 − {v11 v12 } dist 1 P3 − {v4 v5 } dist 1 P7 − {v7 v8 } dist 1 P5 − {v5 v6 } dist 1 P9 − {v9 v10 } dist 1 P9 dist 2 P10 dist 1 P12 dist 1
10 → 11 → 12 → 13 → 14 → 15 → 16 → 17 → 18 →
P13 dist 1 P2 dist 1 P3 dist 2 P6 dist 1 P8 dist 1 P11 dist 2 P12 dist 2 P2 dist 2 P4 dist 1
19 → P5 dist 2 20 → P6 dist 2 21 → P7 dist 2 22 → P8 dist 2 23 → P10 dist 2 24 → P13 dist 2 25 → P1 dist 2 26 → P4 dist 2 27 →∗
′ . Therefore, the shifting is necessary in Note that P1′ and P7′ share a common vertex, as well as P3′ and P9′ , and P5′ and P11 ′ this case. Table 2 shows the distribution of Pi after the shifting. This means that each one of the following triples of vertices (obtained from the sets Pi′ ) receive a different color, which we indicate between parenthesis: x11 , x12 and x18 (color 1); x1,11 , x1,12 and x15 (color 2); x23 , x24 and x2,10 (color 3); x27 , x28 and x21 (color 4); x35 , x36 and x3,12 (color 5); x39 , x3,10 and x33 (color 6). Now we color the horizontal edges. To do so, we need the matchings of K3 , which are R1 = {v1 v2 }, R2 = {v2 v3 } and R3 = {v1 v3 }. Considering row 2, vertex x12 received color 1. The matching of K3 in which v1 is not used is R2 = {v2 v3 }. Therefore, we must assign color 1 to the edge x22 x23 . Repeat the process to every vertex that is part of a triple. The vertices not mentioned so far are the ones that are not part of a triple. There is no specific order to color them, as long as each one receives a different color that has not been previously used. For example, x13 was not assigned to any color. So, it can receive color 7. The matching of K3 in which v1 does not appear is R2 = {v2 v3 }. So, we must assign color 7 to the edge x23 x33 . This process must be repeated to all the vertices that are not part of a triple. So far, color 1 was assigned to every vertex in rows 1, 2 and 8. We have P1 − {v1 v2 } = {v3 v13 , v4 v12 , v5 v11 , v6 v10 , v7 v9 }. This means we have to assign color 1 to the matching of distance 1 linking vertices of rows 3 and 13; 4 and 12; 5 and 11; 6 and 10; and 7 and 9. Similarly, distribute colors 2, 3, 4, 5 and 6, that were used in triples of vertices. Color 7 has been already represented in the vertices of row 3. Note that vertex v3 is the remaining one in the matching P9 = {v9 v10 , v8 v11 , v7 v12 , v6 v13 , v1 v5 , v2 v4 }. Thus, we represent color 7 in the vertices linking rows 9 and 10; 8 and 11; 7 and 12; 6 and 13; 1 and 5; and 2 and 4 using matchings of distance 2 (note that the matching of distance 1 associated with P9 − {v9 v10 } has already been used with color 6). We exhibit in Fig. 3 a scheme of coloring of vertices and horizontal edges of K3×13 . Table 3 shows the distribution of colors among the edges that are not horizontal. To understand the notation, check the distribution of colors 1 and 7 done previously among this kind of edge. Recall that color 27 is used in matchings of distance 1 linking vertices of rows 1 and 2; 3 and 4; 5 and 6; 7 and 8; 9 and 10; and 11 and 12 (exactly the pairs of rows not used in the matchings of distance 1 of the triples). We exhibit in Fig. 4, the matchings of distance 1 and 2 linking vertices of two rows for a better understanding of the expressions dist 1 and dist 2 in Table 3.
4. Equitable total coloring of K2×p The equitable total chromatic number of K2×p was determined to be equal to ∆ + 2 by Fu [5]. We provide an elegant algorithm in this section to color such graphs with a technique that is also used throughout the next sections. Edge-coloring matrices are matrices whose entries represent the colors assigned to the edges of a graph. Let AX1 X2 = [aij ] be a p × p edge-coloring matrix in which the entry aij represents the color assigned to the edge that has x1i and x2j as its ends. For i = 1, . . . , p, let ci = (i, i + 1, . . . , p, 1, 2, . . . , i − 1) be the ith circular permutation of (1, . . . , p). Then AX1 X2 is such that Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 3. Coloring of vertices and horizontal edges of K3×13 .
Fig. 4. The bold edges form a matching of distance 1 and the remaining edges form a matching of distance 2.
its row i is ci , i = 1, . . . , p (see the following example of the matrix AX1 X2 and Fig. 5 for p = 3). 1 2 3
[ AX1 X2 =
2 3 1
3 1 . 2
]
Theorem 2. The algorithm above describes an equitable (∆ + 2)-total coloring of K2×p . Proof. It can be easily seen that each row and each column of AX1 X2 has p distinct entries. Thus for each vertex of K2×p , the p incident edges receive p distinct colors. Now, for i = 1, 2, coloring all the vertices of Xi with the color p + i, we obtain a (∆ + 2)-total coloring of K2×p (see Fig. 5 for K2×3 ). □ 5. Equitable total coloring of Kr ×p with r ≥ 4 even and p odd We begin this section showing that for these graphs, χe′′ ̸ = ∆ + 1. After, we provide an algorithm to show that χe′′ = ∆ + 2, contributing to the ETCC. Claim 3. There exists no equitable total coloring with ∆ + 1 colors for Kr ×p with r ≥ 4 even and p odd. Proof. Suppose, in order to obtain a contradiction, that complete r-partite graphs having p elements in each part of the partition, with r even and p odd, have an equitable (∆ + 1)-total coloring. Thus, each color must be represented in each one of the vertices. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 5. Equitable total coloring of K2×3 .
The case in which p = 1 is special, because such graphs are complete and have an even number of vertices, and it is known that they do not have equitable (∆ + 1)-total coloring [8]. So, we will assume that p ≥ 3. We claim that an odd number of vertices cannot receive the same color in any equitable (∆ + 1)-total coloring. Indeed, suppose that an odd number y of vertices receive the same color i. In this case, an odd number (rp − y) of vertices would remain and color i would be represented in these vertices by the edges that have them as ends. Since these edges are not adjacent, rp − y is even. This contradicts the fact that each color must be represented in each one of the vertices. We claim that an even number of vertices cannot receive the same color in any equitable (∆ + 1)-total coloring. Indeed, take an arbitrary part of the partition of V . Such part has p (odd) elements. Suppose that there exist colors a1 , a2 , . . . , ak used b1 , b2 , . . . , bk times (bi is even), respectively, to color the vertices of that given part. Note that we consider only one part of the partition of V because once a color is used in a vertex of any part, say Xi , it cannot be used in the vertices of the other parts, for the vertices of these parts are adjacent to the vertices of Xi . Thus, we would have b1 + b2 + · · · + bk vertices colored in such part. Note that this sum is even, since bi is even for all i = 1, 2, . . . , k. We would conclude that p − (b1 + b2 + · · · + bk ) vertices were not colored. Since this quantity is odd and we previously proved that there exists no color used an odd number of times to color vertices, we have a contradiction. Therefore, the graph cannot have an equitable (∆ + 1)-total coloring. □ Theorem 3. The graph Kr ×p with r ≥ 4 even and p odd has χe′′ = ∆ + 2. Proof. Using Claim 1 it is possible to organize all edge-coloring matrices as follows. If Ri = {va vb , . . . , vc vd } is a perfect matching of Kr (there are r − 1 disjoint perfect matchings in Kr ), then let Si = {AXa Xb , . . . , AXc Xd } be a set of edge-coloring matrices, with |Si | = |Ri |. To each set of matrices Si , assign numbers from 1 to p(r − 1) as follows: the entries of the matrices in S1 are the elements of the set {1, 2, 3, . . . , p}; in S2 , elements of the set {p + 1, p + 2, . . . , 2p}, and so on. In general, the entries of the matrices in Si are elements of the set {(i − 1)p + 1, (i − 1)p + 2, . . . , ip}. Note that each one of the sets of edge-coloring matrices has p elements and that if they are taken modulo p, the sets become {1, 2, . . . , p − 1, 0}. Based on that, the distribution of elements in the edge-coloring matrices is done similarly to the distribution in the case of complete bipartite balanced graphs. Since |Si | = |Ri | = 2r , each color from 1 to p(r − 1) was rp used 2 times in edges. When we divide the matrices into r − 1 groups, in each of those groups, part Xi appears precisely once (see Claim 1). Therefore, when the same set of colors is assigned to one of the sets of edge-coloring matrices, the way the entries are displayed in the matrices guarantees that no adjacent edges receive the same color. We replace the secondary diagonal entries of r −2 2 matrices by the color (r −1)p+1 and r −2 2 matrices by the color (r −1)p+2, which have not been used yet. The replacement is done as follows. AX1 X2 , AX2 X3 , . . . , AXr −2 Xr −1 (note that these matrices are related to S1 , S2 , . . . , Sr −2 , respectively) have the entries in their secondary diagonals replaced alternately by (r − 1)p + 1 and (r − 2)p + 2. Note that by the algorithm to divide the matrices into sets, all the ones that have their entries changed are elements of S1 , S2 , . . . , Sr −2 , so their secondary diagonals received previously different colors. We illustrate this procedure in the end of this section, by presenting an example of an equitable (∆ + 2)-total coloring of K4,3 , where the entries of the secondary diagonals of the matrices AX1 X2 and AX2 X3 were originally 3 and 6, respectively, but were replaced by 10 and 11, respectively, according to this step. Note that an element aij of AXk Xl is in the secondary diagonal if it is such that i + j = p + 1. Therefore, all entries in the secondary diagonal of one matrix receive the same color since i + j − 1 = p + 1 − 1 = p. In each group of edge-coloring matrices, the secondary diagonal will be p, 2p, 3p, . . . , (r − 2)p by the coloring of edge-coloring matrices presented in the beginning of this proof. The colors of the vertices will be the colors (r − 1)p + 1, (r − 1)p + 2 and the colors that have been changed in some of the secondary diagonals, which are p, 2p, 3p, . . . , (r − 2)p since these colors must be represented in vertices that are ends of edges that are colored by AX1 X2 , AX2 X3 , . . . , AXr −2 Xr −1 . To the vertices of Xi , for all 1 ≤ i ≤ r − 2, assign the color ip, to the vertices of Xr −1 , assign the color (r − 1)p + 1 and to the vertices of Xr , assign the color (r − 1)p + 2. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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(r −2)p
rp
The color (r − 1)p + 1 is used 2 times in edges and p times in the coloring of vertices, totalizing 2 times. The result is analogous for the color (r − 1)p + 2. The colors that were originally in the secondary diagonal of some matrices and were replaced, were originally used p times in 2r matrices. Then, they were replaced by a new color in p entries of a matrix. On the other side, those colors are used rp rp to color the p vertices of one of the parts of the partition of V . So, those colors are used 2 − p + p = 2 times in total. Each rp color is used precisely 2 times. Hence, the difference between the cardinalities of two color classes is 0 and this concludes the proof of the fact that the described algorithm gives an equitable (∆ + 2)-total coloring of Kr ×p with r ≥ 4 even and p odd. □ Now, we show the edge-coloring matrices that generate the equitable (∆ + 2)-total coloring of K4×3 . 1 2 10
2 10 1
10 1 , AX1 X3 = 2
[
4 5 11
5 11 4
11 4 , AX2 X4 = 5
[
[ AX1 X2 =
[ AX2 X3 =
]
]
7 8 9
8 9 7
9 7 , AX1 X4 = 8
[
7 8 9
8 9 7
9 7 , AX3 X4 = 8
[
]
]
4 5 6
5 6 4
6 4 5
1 2 3
2 3 1
3 1 . 2
]
]
The vertices in X1 are assigned to 3 = p, the ones in X2 , to 6 = 2p, in X3 , to 10 = (r − 1)p + 1 and the ones in X4 , to 11 = (r − 1)p + 2. 6. Equitable total coloring of Kr ×p with r and p even By using the algorithm presented in Section 5, we get an equitable (∆ + 2)-total coloring for these graphs, contributing to the ETCC. Theorem 4. The graph Kr ×p with r and p even has χe′′ ≤ ∆ + 2. Proof. The equitable (∆ + 2)-total coloring is obtained directly from the proof of Theorem 3 by applying the same technique. The parity of p was relevant in the previous section to prove that Kr ×p , with r ≥ 4 even and p odd does not have equitable (∆ + 1)-total coloring. The algorithm to prove that those graphs have χe′′ = ∆ + 2, however, does not require p to be odd. That is why the same algorithm works here. □ We present an equitable (∆ + 2)-total coloring of the graph K6×2 obtained by the algorithm. We show below their edgecoloring matrices. AX1 X2 AX2 X3 AX3 X4 AX4 X5
2 1
]
[
4 3
]
[
6 5
]
[
8 7
1 2 , AX4 X6 = 2 1
]
[
4 3 , AX5 X6 = 4 3
]
[
6 5 , AX1 X6 = 5 6
]
[
8 7 , AX2 X6 = 7 8
]
[
1 11 , AX3 X5 = 2 1
[
3 = 12
12 3 , AX1 X4 = 3 4
[
5 = 11
11 5 , AX2 X5 = 5 6
[
7 = 12
12 7 , AX1 X3 = 7 8
[
10 9 , AX2 X4 = 9 10
9 10
[
[
1 = 11
AX1 X5 =
]
]
[
]
]
]
]
[
10 9 , AX3 X6 = 9 10
] ]
10 . 9
The vertices of X1 receive color 2, of X2 color 4, of X3 color 6, of X4 color 8, of X5 color 11, and of X6 color 12. By the algorithm, we could show that χe′′ ≤ ∆ + 2. However, we present an equitable (∆ + 1)-total coloring for the graph K4×4 , which is a graph of this class. We will display in Figs. 6, 7, 8, 9, 10, 11, and 12, the distribution of colors among the elements of the graph. The reader can easily check that colors 1, 2, . . . , 8 were used in 2 vertices and 7 edges each, that is, in 9 elements each. Also, colors 9, 10, . . . , 13 were used in 8 edges each. It is also easy to check that no adjacent or incident elements received the same color. We get that the difference between the cardinalities of two color classes is either 0 or 1 and ∆ + 1 = (4 − 1)4 + 1 = 13. Therefore, the equitable total chromatic number of K4×4 is ∆ + 1. 7. Equitable total coloring of Kr ×p with r odd and p even We begin this section with an algorithm to show that χe′′ ≤ ∆ + 2. After that, we provide an algorithm to show that χe = ∆ + 1 if r ≥ 2p . If r < 2p , the algorithm does not work because there are not enough available matchings of distance to ′′
complete the coloring. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 6. The numbers represent the colors of the vertices, the dashed edges receive color 1, and the continuous edges receive color 2.
Fig. 7. The continuous edges receive color 3 and the dashed edges receive color 4.
Fig. 8. The continuous edges receive color 5 and the dashed edges receive color 6.
Fig. 9. The dashed edges receive color 7 and the continuous edges receive color 8.
Theorem 5. The graph Kr ×p with r is odd and p is even has χe′′ ≤ ∆ + 2. Proof. If p = 2, get the r matchings Ri ’s of Kr (Claim 2) and associate two colors to each matching in the following way: Ri is associated to 2i − 1 and 2i, for 1 ≤ i ≤ r. We have Ri = {va vb , . . . , vc vd } and vf is the remaining vertex in this matching. Let AXa Xb , . . . , AXc Xd be edge-coloring matrices related to the matching Ri and part Xf related to the remaining vertex vf . The entries of such matrices must be filled as in the complete bipartite balanced case. Since Kr has an odd number of vertices, the vertices of Xf do not have colors 2i − 1 and 2i represented by the edge-coloring matrices. To the vertex xf 1 assign color Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 10. The dashed edges receive color 9 and the continuous edges receive color 11.
Fig. 11. The continuous edges receive color 10.
Fig. 12. The dashed edges receive color 12 and the continuous edges receive color 13.
2i − 1 and to the vertex xf 2 , color 2i. Observe that, in this case, vertex xf 1 does not have color 2i represented and vertex xf 2 does not have color 2i − 1 represented, but since this is an equitable (∆ + 2)-total coloring, the vertices do not need to have all colors represented. Since all matchings of Kr have a remaining vertex, all vertices of Kr ×2 received a different color by the previous procedure. The argument to justify that adjacent edges are assigned to different colors in the complete bipartite balanced graphs also hold here. An edge and its ends are assigned to different colors as well because the part Xf receives colors 2i − 1 and 2i, which have only been attributed to the edge-coloring matrices of Ri and vf is the remaining vertex in such matching. We used two colors to each matching of Kr , which totalizes 2r colors. Also, ∆ + 2 = (r − 1)p + 2 = (r − 1)2 + 2 = 2r. Each color is used in 1 vertex and 2 times in each one of the r −2 1 = |Ri | (for all i = 1, 2, . . . , r) edge-coloring matrices in which it is applied. So, each color is used 1 + 2 r −2 1 = r times. Hence the difference between the cardinalities of any two color classes is 0 and we conclude that the total coloring is, indeed, equitable. p If p ≥ 4, we repeat the same process used for p = 2, 2 times. For this, we use the matching P1 of Kp (Claim 1) to determine the pairs of rows in which we apply the steps of case p = 2: if P1 = {v1 v2 , va vb , . . . , vc vd }, use 2r colors to color the edges linking vertices of rows 1 and 2 in the same way we colored the edges of the case p = 2 (edge-coloring matrices) and color vertices xi1 , xi2 , with 1 ≤ i ≤ r (vertices of Kr ×p that are in the part Xf related to the remaining vertex vf of matching Ri ). After that, using the same 2r colors, color the edges linking rows a and b and the vertices xia , xib , with 1 ≤ i ≤ r, analogously. Repeat the process until coloring the edges linking vertices in rows c and d and also the vertices in such rows. As explained p in the case p = 2, each color is used r times. Now, since the process is repeated 2 = |Pi | (for all i = 1, 2, . . . , r) times, rp each color is used 2 times. So far, every edge linking vertices of the pairs determined by P1 has been colored as well as all the horizontal edges and all vertices of the graph. Now, use one different color to each matching of available distance between the pairs determined by P2 , P3 , . . . , Pp−1 . Note that each matching of distance is a perfect matching of the graph. So, there are p − 1 − 2 + 1 = p − 2 matchings and r − 1 matchings of available distances (see Section 2), which implies that (p − 2)(r − 1) colors are used in this last step. And each of these colors is used in a disjoint perfect matching of Kr ×p , Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 13. Scheme of colors to be applied in elements linking vertices of rows 1 and 2; 3 and 7; 4 and 6; 5 and 8.
rp
that is, they are used in 2 edges. We conclude that the difference between any two color classes is 0. Also, we use a total of 2r + (p − 2)(r − 1) = ∆ + 2 colors. It is clear, by construction, that no adjacent or incident elements receive the same color, since the first part of the algorithm is based on the repetition of the case p = 2 and in the second part we used one different color to each matching of available distance (determined in Section 2) between the pairs determined by P2 , P3 , . . . , Pp−1 , as stated above. □ Observe the equitable total coloring of K3×8 with 18 = ∆ + 2 colors obtained by Theorem 5. We have matching P1 = {v1 v2 , v3 v7 , v4 v6 , v5 v8 } of Kp=8 . Also, matchings R1 = {v1 v2 }, R2 = {v2 v3 } and R3 = {v1 v3 } of Kr =3 . In Fig. 13 we present the coloring of the elements that link vertices of rows 1 and 2; 3 and 7; 4 and 6; and 5 and 8 (see P1 ). We have P2 = {v2 v3 , v1 v4 , v5 v7 , v6 v8 }. So, we use color 7 in the matching of distance 1 linking vertices of rows 2 and 3; 1 and 4; 5 and 7, 6 and 8; and color 8 in the matching of distance 2 linking vertices of the same pairs of rows. We represent this fact by stating: color 7 → P2 dist 1 and 8 → P2 dist 2. We have P3 = {v3 v4 , v2 v5 , v1 v6 , v7 v8 }. So, color 9 → P3 dist 1 and 10 → P3 dist 2. We have P4 = {v4 v5 , v3 v6 , v2 v7 , v1 v8 }. So, color 11 → P4 dist 1 and 12 → P4 dist 2. We have P5 = {v5 v6 , v4 v7 , v1 v3 , v2 v8 }. So, color 13 → P5 dist 1 and 14 → P5 dist 2. We have P6 = {v6 v7 , v1 v5 , v2 v4 , v3 v8 }. So, color 15 → P6 dist 1 and 16 → P6 dist 2. We have P7 = {v1 v7 , v2 v6 , v3 v5 , v4 v8 }. So, color 17 → P7 dist 1 and 18 → P7 dist 2. Theorem 6. The graph Kr ×p with r odd and p even has χe′′ = ∆ + 1 if r ≥
p . 2
Proof. If p = 2, assign the color i to the vertices xi1 ∈ Xi and xi2 ∈ Xi , for all i : 1, . . . , r. So, no adjacent vertices receive the same color. If a vertex xij received color i, take the matching Rl of Kr (Claim 2) in which vi is the remaining vertex. Such matching has the form Rl = {va vb , . . . , vc vd }. So, assign color i to the horizontal edges xaj xbj , . . . , xcj . Since vi is not an end of edges of Rl and the edges xaj xbj , . . . , xcj are related to the matching Rl , this means that incident elements receive different colors. Each color i that has been used in vertices was used twice (xi1 and xi2 ). We have |Rl | = r −2 1 . So, the color i has been used in 2 r −2 1 edges, totalizing r + 1 elements. rp Note that, so far, we have colored the horizontal edges and the vertices with 2 colors. Now, using different colors than the ones that have been already used, take r − 1 colors to color matchings of distance of graph Kr ×2 . Since the cardinality of those matchings is r, each one of those r − 1 colors is used r times and the difference between the cardinalities of two color classes is at most 1, as desired. p Now, consider the case p ≥ 4 and r ≥ 2 . Since there are p − 1 disjoint perfect matchings in Kp , in order to know how many times each matching Pi will be used to color the r parts of vertices of Kr ×p , divide r by p − 1. By Euclidean division, there exist positive integers s and q such that q(p − 1) + s = r. This means that we should use s matchings of Kp to color q + 1 parts of the partition each and p − 1 − s matchings to color q parts each. The coloring of the vertices is done as follows: if P1 = {va vb , . . . , vc vd } and if we use it to color the vertices of X1 then each of the following pairs of vertices receives a different color: x1a and x1b , . . . , x1c and x1d . The coloring of the horizontal edges is done similarly to the coloring of the horizontal edges in the case Kr ×p with r and p being odd. After coloring the horizontal edges, we guarantee that if a color was used in a vertex of a certain row, then it was represented in all the vertices of that row. Now, this color needs to be represented in the vertices of the other rows. To color the edges that are not horizontal, we begin with the colors that were used in vertices. If P1 = {va vb , vc vd , . . . , vg vh } and we used color i in the vertices x1a and x1b (admitting that P1 was used to color the vertices in X1 ), then we represent color i in the complement of P1 with respect to va vb , which we are going to denote by P1 − {va vb } = {vc vd , . . . , vg vh }. Color i must be represented in the vertices of rows c and d, and so on, being represented in the rows determined by {vc vd , . . . , vg vh } above, until being used in the vertices of rows g and h. Color i is used in matchings of distance 1. We repeat the process to every color used in vertices always using the next available matching of distance. The other colors, that is, the ones not used in vertices, are assigned according to the available matchings of distances in each perfect matching Pi of Kp . The number of rp colors used only in edges of the graph is ∆ + 1 − 2 . Indeed, since there are (p − 1) Pi′ s and there are r − 1 matchings of distance for each Pi , the number of matchings of distance is (r − 1)(p − 1). From this number, we remove the matchings Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 14. Coloring of vertices and horizontal edges of K3×4 .
(p
)
of distance colored with the colors of the vertices [s(q + 1) + (p − 1 − s)q] 2 − 1 . As explained above, s matchings are used in the coloring of q + 1 parts and p − 1 − s matchings are used in the coloring of q parts. It is easy to check that each p when applying a color of a pair (of vertices time we use a matching Pi to color a part, we use 2 − 1 matchings (of distance ) ) p to the complement(of Pi : (r − 1)(p − 1) − [ s(q + 1) + (p − 1 − s)q ] − 1 = rp − r − p + 1 + (q(p − 1) + s) 2p − 1 = 2 ) p rp rp rp − r − p + 1 − r 2 − 1 = 2 − p + 1 = ∆ + 1 − 2 . p p Note that there are r − 1 matchings of distance j and |Pi | = 2 . So, the complement of Pi has size 2 − 1. In order to have p p available matchings of distance to complete the coloring, we must have − 1 ≤ r − 1, which implies that r ≥ 2 . 2
rp
By construction we get that no adjacent or incident elements receive the same color. Also, we have that 2 colors are used in 2 vertices and then, represented in the other vertices through edges that have them as ends. Since each edge has 2 rp−2 rp rp ends, we get that those } colors are used in 2 edges, totalizing 2 + 1 elements. The other 2 − p + 1 colors are used in { rp2 (r −1) 2
−
/ rp2 =
( rp−2 ) rp 2
2
rp 2
− p + 1 perfect matchings of the graph Kr ×p , that is, in
rp 2
elements. We conclude that the
difference between two different color classes is at most 1, as desired. □ Observe the equitable (∆ + 1)-total coloring of K3×4 exhibited below. We have matchings P1 = {v1 v2 , v3 v4 }, P2 = {v2 v3 , v1 v4 } and P3 = {v3 v4 , v1 v2 } of Kp=4 . The coloring of the vertices of X1 is related to P1 , the vertices of X2 to P2 , and the vertices of X3 to P3 . Observe the coloring of vertices and horizontal edges in Fig. 14. Colors 1–6 are also used in the following edges: color 1 in the matching of distance 1 linking vertices of rows 3 and 4; color 2 in the matching of distance 1 linking vertices of rows 1 and 2; color 3 in the matching of distance 1 linking vertices of rows 1 and 4; color 4 in the matching of distance 1 linking vertices of rows 2 and 3; color 5 in the matching of distance 1 linking vertices of rows 2 and 4; color 6 in the matching of distance 1 linking vertices of rows 1 and 3. Colors 7, 8 and 9 are used in the following edges: 7 → P1 dist 2, 8 → P2 dist 2 and 9 → P3 dist 2. 8. Final considerations In this paper we improved the existing bound given by Fu [5] for graphs of odd order showing that these graphs have
χe′′ = ∆ + 1. We also showed equitable (∆ + 2)-total colorings for the open cases. In the case in which r is even and p is odd, we proved that χe′′ = ∆ + 2. However, for the cases in which r and p are even, and r is odd and p is even, we showed an equitable (∆ + 2)-total coloring for all of those graphs and gave examples of equitable total colorings with ∆ + 1 colors for graphs within these cases. This may suggest that, in both cases, we have χe′′ = ∆ + 1. Finally, in order to conclude the paper, we present equitable (∆ + 1)-total colorings for complete bipartite nonbalanced graphs by applying the same techniques used in the bipartite complete balanced graphs. It was established by Fu in [5] that χe′′ = ∆ + 1 for Kp1 ,p2 (with p1 ̸= p2 ). Let Kp1 ,p2 with p1 ̸ = p2 be a bipartite complete nonbalanced graph. Assume, without loss of generality, that p1 < p2 and let A = [aij ] be a (p1 + 1) × p2 matrix, in which the entry aij is the color assigned to the edge that has x1i and x2j as its ends if 1 ≤ i ≤ p1 ; and the color assigned to the vertex x2j if i = p1 + 1. Similarly to the complete bipartite balanced case, to the element aij of A, assign the color i + j − 1 mod (p1 ) if i + j − 1 ̸ ≡ 0 mod (p1 ) and the color p1 , otherwise. To the vertices of X1 , assign the color p1 + 1. Observe the following example of the matrix A and Fig. 15.
⎡
1 ⎢2 A=⎣ 3 4
2 3 4 1
3 4 1 2
⎤
4 1⎥ . 2⎦ 3
Theorem 7. The previous algorithm gives an equitable (∆ + 1)-total coloring of Kp1 ,p2 , with p1 ̸ = p2 . Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 15. Equitable 5-total coloring of K3,4 .
Proof. Let aij and akl be two different entries of A. If i = k, then j ̸ = l and the corresponding elements in Kp1 ,p2 are adjacent edges if 1 ≤ i ≤ p1 or vertices of X2 (which are nonadjacent) if i = p1 + 1. The proof provided for Theorem 2 also works in this situation and we get that adjacent edges receive different colors. So, each color of {1, 2, . . . , p2 } appears exactly once in each row of A. Since A has p1 + 1 rows, each one of those colors is used p1 + 1 times. Now, if j = l and i ̸ = k, the entries aij and akj are either two adjacent edges (if 1 ≤ i, k ≤ p1 ) or an incident vertex with an edge (if i = p1 + 1 or k = p1 + 1). Again, the proof of Theorem 2 also works in this case. In other words, each color of the set {1, 2, . . . , p2 } appears exactly once in each row. We get that p2 colors are used p1 + 1 times each. Also, the color p2 + 1 is used to color all the p1 vertices in X1 . Hence, the difference between the cardinalities of two color classes is at most 1, as desired. This concludes the proof that the algorithm gives an equitable (∆ + 1)-total coloring of the graph Kp1 ,p2 , with p1 ̸ = p2 . □ References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
M. Behzad, Graphs and their chromatic numbers (Ph.D. thesis), Michigan State University, 1965. J.C. Bermond, Nombre chromatique total du graphe r-parti complet, J. Lond. Math. Soc. 9 (1974) 279–285. T. Chunling, L. Xiaohui, Y. Yuansheng, L. Zhihe, Equitable total coloring of Cm □Cn , Discrete Appl. Math. 157 (2009) 596–601. S. Dantas, C.M.H. de Figueiredo, G. Mazzuoccolo, M. Preissmann, V.F. dos Santos, D. Sasaki, On the equitable total chromatic number of cubic graphs, Discrete Appl. Math. 209 (2016) 84–91. H.L. Fu, Some results on equalized total coloring, Congr. Numer. 102 (1994) 111–119. M.A. Gang, M.A. Ming, The equitable total chromatic number of some join graphs, Open J. Appl. Sci. 2 (4B) (2012) 96–99. H. Gui, W. Wang, Y. Wang, Z. Zhang, Equitable total-coloring of subcubic graphs, Discrete Appl. Math. 184 (2015) 167–170. M. Kubale, Graph Colorings, Vol. 352, American Mathematical Soc., 2004. A. Sánchez-Arroyo, Determining the total colouring number is NP-hard, Discrete Math. 78 (1989) 315–319. A. Soifer, The Mathematical Coloring Book, Springer-Verlag, 2008. V.G. Vizing, Some unsolved problems in graph theory, Russian Math. Surveys 23 (6) (1968) 125–141. W.F. Wang, Equitable total coloring of graphs with maximum degree 3, Graphs Combin. 18 (2002) 677–685.
Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Equitable total coloring of complete r-partite p-balanced graphs✩ A.G. da Silva a , S. Dantas a , D. Sasaki b, * a b
IME, Fluminense Federal University, Brazil IME, Rio de Janeiro State University, Brazil
article
info
Article history: Received 9 January 2017 Received in revised form 25 January 2018 Accepted 7 March 2018 Available online xxxx Keywords: r-partite graphs p-balanced graphs Equitable total coloring
a b s t r a c t A total coloring is equitable if the number of elements colored by any two distinct colors differs by at most one, and the smallest positive integer for which a graph has such coloring is called its equitable total chromatic number. In 1974, Bermond determined the total chromatic number of all complete r-partite p-balanced graphs. In 1994, Fu proved that there exist equitable (∆ + 2)-total colorings for all complete r-partite p-balanced graphs of odd order. Later, Wang (2002) conjectured that the equitable total chromatic number of a graph is either ∆ + 1 or ∆ + 2. In this work, we investigate the complete r-partite pbalanced graphs. We improve the bound given by Fu, when r and p are odd, by exhibiting an equitable (∆ + 1)-total coloring for these graphs. Furthermore, for the open cases, that is, if the order is even, we exhibit equitable (∆ + 2)-total colorings for all these graphs, which completely verifies Wang’s conjecture for this class. © 2018 Elsevier B.V. All rights reserved.
1. Introduction Let G = (V , E) be a simple connected graph. A k-total coloring of G is an assignment of k colors to the vertices and edges of G so that adjacent or incident elements have different colors. The total chromatic number of G, denoted by χ ′′ , is the smallest k for which G has a k-total coloring. Clearly, χ ′′ ≥ ∆ + 1 and the Total Coloring Conjecture (TCC) states that the total chromatic number of any graph is at most ∆ + 2, where ∆ is the maximum degree of the graph [1,11]. In 1989, Sánchez-Arroyo [9] proved that the problem of determining the total chromatic number of an arbitrary graph is NP-hard, and it remains NP-hard even for cubic bipartite graphs. If the difference between the cardinalities of any two color classes is either 0 or 1, then the total coloring is said to be equitable. The equitable total chromatic number of G, denoted by χe′′ , is the smallest k for which G has an equitable k-total coloring. Similarly to the total coloring problem, it was conjectured in 2002 by Wang [12] that the equitable total chromatic number of any graph is at most ∆ + 2 (Equitable Total Coloring Conjecture (ETCC)). Ever since, many papers have been published in this subject [3,4,6,7]. In 2016, Dantas et al. [4] proved that the problem of determining the equitable total chromatic number of a cubic bipartite graph is NP-complete. A graph is said to be r-partite if there exists a partition of its vertex set X1 = {x11 , x12 , . . . , x1p1 }, X2 = {x21 , x22 , . . . , x2p2 }, . . . , Xr = {xr1 , xr2 , . . . , xrpr } such that Xi is an independent set, with 1 ≤ i ≤ r. In some cases, we replace the notation xij by xi,j for the benefit of the reader. ✩ Partially supported by CNPq and FAPERJ. Corresponding author. E-mail addresses: [email protected] (A.G. da Silva), [email protected] (S. Dantas), [email protected] (D. Sasaki).
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https://doi.org/10.1016/j.dam.2018.03.009 0166-218X/© 2018 Elsevier B.V. All rights reserved.
Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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A complete r-partite graph is an r-partite graph in which there is an edge between any two vertices of different parts of the partition. We denote a complete r-partite graph having pi , i = 1, . . . , r, vertices in each independent set by Kp1 ,p2 ,...,pr . If pi = p for all i = 1, . . . , r, then the r-partite graph is said to be p-balanced. In this case, we denote the graph by Kr ×p . In 1974, Bermond determined the total chromatic number of all complete r-partite p-balanced graphs [2]. In 1994, Fu [5] investigated the equitable total coloring of complete bipartite graphs and complete r-partite graphs of odd order. In fact, for complete bipartite graphs, Fu proved that χe′′ = χ ′′ . Furthermore, considering complete r-partite graphs of odd order, Fu proved that there exist equitable (∆ + 2)-total colorings for all of these graphs. In this work, we investigate the complete r-partite p-balanced graphs. We improve the bound given by Fu, when r and p are odd, by exhibiting an equitable (∆ + 1)-total coloring for these graphs. Furthermore, we exhibit equitable (∆ + 2)-total colorings for the open cases, that is, the complete r-partite p-balanced graphs of even order. From these results, we verify the ETCC for the following class of graphs: 1. 2. 3. 4.
Kr ×p Kr ×p Kr ×p Kr ×p
with r with r with r with r
and p odd, χe′′ = ∆ + 1; ≥ 4 even and p odd, χe′′ = ∆ + 2; ≥ 4 even and p even, χe′′ ≤ ∆ + 2; odd and p even, χe′′ ≤ ∆ + 2.
Using similar techniques to those applied to some of the above results, we exhibit an equitable (∆ + 2)-total coloring of K2×p and an equitable (∆ + 1)-total coloring of Kp1 ,p2 , with p1 ̸ = p2 . This paper is organized as follows. In Section 2, we present definitions and known results that are useful throughout the next sections. In Section 3, we provide an algorithm to show that χe′′ = ∆ + 1, for Kr ×p with r odd and p odd. In Section 4, we provide an algorithm to obtain an equitable (∆ + 2)-total coloring of K2×p and the technique applied to this case is used in the next sections. In Section 5, we prove that χe′′ = ∆ + 2 for Kr ×p with r ≥ 4 even and p odd. In Section 6, we use the same algorithm presented in Section 5 to show an equitable (∆ + 2)-total coloring for Kr ×p with r ≥ 4 even and p even, establishing that χe′′ ≤ ∆ + 2. Moreover, we present an example with χe′′ = ∆ + 1. In Section 7, we present an equitable (∆ + 2)-total coloring for Kr ×p with r odd and p even, establishing that χe′′ ≤ ∆ + 2. Also, we present an infinite class of graphs within this case for which χe′′ = ∆ + 1. Finally, in the last section, we exhibit an equitable (∆ + 1)-total coloring for complete bipartite nonbalanced graphs. 2. Preliminaries In this section, we introduce results and definitions that are used throughout the paper. A k-edge coloring of G is an assignment of k colors to the edges of G so that adjacent edges have different colors. It is known that the complete graph Kn has a ∆-edge coloring when n is even and has a (∆ + 1)-edge coloring when n is odd [10]. A |V | matching in G is a set M ⊆ E such that no two edges in M share a common vertex. A matching P is perfect if |P | = 2 . Claim 1 (Soifer, 2008 [10]). Let Kn be the complete graph with n ≥ 2 even vertices. Since this graph has a ∆-edge coloring, there exist n − 1 disjoint perfect matchings in Kn . The case n = 2 being trivial, let n ≥ 4 even. In order to obtain such n − 1 perfect matchings M1 , M2 , . . . , Mn−1 , draw a polygon with n − 1 vertices v1 , v2 , . . . , vn−1 and place the vertex vn inside the polygon. Display the vertices of Kn in clockwise ascending order. The edge vi vi+1 , for all 1 ≤ i ≤ n − 2 is an element of Mi and the edge v1 vn−1 is an element of Mn−1 . All parallel diagonals (of the polygon) to the edge in the matching Mi (1 ≤ i ≤ n − 1) are elements of the same matching. For the sake of clarity, assume that i < j when we refer to an element (edge or diagonal) vi vj of the polygon. Observe that an element of the polygon is parallel to another one if they have the form vi vj and vi−1 vj+1 , respectively (if i ̸ = 1). In the case i = 1, a parallel element to v1 vj is vj+1 vn−1 . Since the polygon has an odd number of vertices, after assigning parallel diagonals of vi vj to the matching Mi , one vertex v remains. Thus, the edge vvn must be assigned to the matching Mi as well. Observe the following example in Fig. 1. We have M1 = {v1 v2 , v3 v5 , v4 v6 }, M2 = {v2 v3 , v1 v4 , v5 v6 }, M3 = {v3 v4 , v2 v5 , v1 v6 }, M4 = {v4 v5 , v1 v3 , v2 v6 } and M5 = {v1 v5 , v2 v4 , v3 v6 }. Claim 2 (Soifer, 2008 [10]). Let Kn be the complete graph with n ≥ 3 odd vertices. Since this graph has a (∆ + 1)-edge coloring, there exist n disjoint matchings in Kn . The algorithm to obtain such matchings is similar to the previous one just by deleting the vertex vn inside the polygon (and its corresponding edges in the matchings). Fig. 2 shows an example of the matchings of K5 . We have M1 = {v1 v2 , v3 v5 }, M2 = {v2 v3 , v1 v4 }, M3 = {v3 v4 , v2 v5 }, M4 = {v4 v5 , v1 v3 } and M5 = {v1 v5 , v2 v4 }. The isolated vertices in Fig. 2 represent the remaining vertex in each matching. The graph Kr is the complete graph having r vertices, in which r represents the number of parts of Kr ×p . In this case, we denote its matchings by Ri . Similarly, the graph Kp is the complete graph having p vertices, in which p represents the number of vertices in each part of Kr ×p . In this case, we denote its matchings by Pi . Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 1. Disjoint perfect matchings of K6 .
Fig. 2. Disjoint matchings of K5 .
Let Kr ×p be a graph. We define matching of distance i as the set of edges {xja xj+i,b } linking vertices of rows a and b (1 ≤ a, b ≤ p, a ̸ = b), for all 1 ≤ j ≤ r and j + i is taken modulo r. Since j + i is taken modulo r, we have j + i ≤ r, which implies that i ≤ r − 1 (since j ≥ 1). Note that each matching of distance i has r edges. We say that a color is represented in a vertex if the color is assigned to the vertex itself or to an edge that has such vertex as one of its ends. Throughout this paper, when we refer to results taken modulo i, if such result is congruent 0 modulo i, then we use i instead of 0. 3. Equitable total coloring of Kr ×p with r and p odd In this section, we improve the bound given by Fu [5] by presenting an algorithm to color these graphs with ∆ + 1 colors. We also present an example for K3×13 . Theorem 1. The graph Kr ×p with r and p odd has χe′′ = ∆ + 1. Proof. If p = 1, then we have Kr ×1 = Kr , which is the complete graph. It is known that such graphs have equitable (∆ + 1)total colorings [8]. So, assume that p ≥ 3. In all equitable (∆ + 1)-total coloring of Kr ×p every color must be represented in all vertices. Suppose that q vertices rp−q receive the same color i. It implies that the number of edges colored with color i is 2 . The total number of(elements that )
receive color i is
rp+q . 2
Since rp is odd, q must be odd. Furthermore, the number of elements of the graph rp + rp+1
p−1
rp2 (r −1) 2
p−1
divided by the number of colors ((r − 1)p + 1) has quotient 2 and remainder 2 . It implies that 2 colors are used ( ) rp+3 p−1 rp+1 p−1 times and (r − 1)p + 1 − 2 colors are used 2 times. Therefore, q = 1 or q = 3. We conclude that 2 triples of 2 3(p−1)
vertices receive one color each and the other rp − 2 vertices receive one different color each. In order to form the triples, we use Claim 2 to obtain the matchings Pi of the graph Kp . Take only those with odd index p−2−1 p−1 from 1 to p − 2, that is, P1 , P3 , . . . , Pp−2 , that totalizes 2 + 1 = 2 matchings, the exact number of triples that we
need. In each Pi taken, we need the ends of the first edge and the remaining vertex, that is, vi , vi+1 and vi+ p−1 +1 (all indexes 2
taken modulo p). Since we use only 3 vertices, we will use the notation Pi′ to refer to them. Distribute by rows the Pi′ ’s in a Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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′ table in which the columns represent the parts X1 , X2 , . . . , Xr . The first row of the table is P1′ , P3′ , P5′ , . . . , P2r −1 , the second is ′ ′ ′ ′ ′ ′ P2r +1 , P2r +3 , P2r +5 , . . . , P4r −1 , and so on. We have P1 in the column of X1 and P1 = {v1 , v2 , v2+ p−1 } (that is, the first element 2
of P1 and the remaining vertex). This means that the vertices x11 , x12 and x1,2+ p−1 must be assigned to the same color. The 2
process must be repeated to every Pi′ , with i = 1, 3, 5, . . . , p − 2. Notice that it might happen that Pi′ and Pj′ in the same column of the table share a common vertex. In this case, shift the elements of the row that contains Pj′ one unity to the right (note that the element of that row in the column of Xr is now in the column of X1 ). It is easy to see that a vertex of Kp appears in the first element of a Pi (i odd) precisely once and is the remaining vertex of only one matching of Kp . This implies that any Pi′ has a common vertex with exactly one Pj′ . Thus, shifting the elements of the row that contains Pj′ solves the problem. Up to this point, all vertices are colored. To color the horizontal edges we need the matchings of Kr (Claim 2). If a vertex xij receives ( a certain )color β , then look the matching Rl of Kr in which vi is the remaining vertex. We have Rl = {va vb , . . . , vc vd } |Rl | = r −2 1 . Then the edges xaj xbj , . . . , xcj xdj must be assigned to the same above mentioned color β . By doing that, if a given color is used in a certain vertex, then the color is represented in all vertices of that row. Since Kr ×p (with r and p odd) has an equitable total coloring with ∆ + 1 colors, this means that the colors still need to be represented in the vertices of the other rows. To represent the colors that have been used in 3 vertices and in the rows that contain such vertices, we use the matchings Pi of Kp (Claim 2) as follows. As explained above, the vertices x11 , x12 and x1,2+ p−1 receive the same color k and such color 2
has been represented in the vertices of the rows 1, 2 and 2 + 2 . Assuming that P1 − {v1 v2 } = {va vb , . . . , vc vd }, where { } p−1 a, b, . . . , c , d ∈ {3, 4, . . . , p} − 2 + 2 , assign color k to the matching of distance 1 (see Section 2) linking vertices of p−1
rows a and b; linking vertices of the next pair of rows following the order defined by {va vb , . . . , vc vd }; and so on until rows c and d. Repeat the process for all colors used in the triples of vertices described previously. Notice that by the end of this step, since the colors of the triples were applied on matchings of distance 1 related to P1 −{v1 v2 }, P3 −{v3 v4 }, · · · , Pp−2 −{vp−2 vp−1 }, the edges of the matchings of distance 1 linking rows 1 and 2, 3 and 4, · · · , p − 2 and p − 1 were not colored. Now, we need to represent the colors used in only one vertex (and the rows that contain it) in the other rows. In order to do that, we go from the part X1 to Xr , searching from the vertex xi1 to xip in each part to take the ones that received a color not used in any other vertex. It is important to leave one vertex of the last row (xα p ) to be the last one to apply this step. Let x1a be the first vertex in X1 that satisfies this condition. Thus, the color used in such vertex has been represented in the vertices of row a (in the horizontal edges) and still needs to be represented in the other ones. Take the matching of Kp in which va was the remaining vertex and assign the same color of x1a in the matching of distance 1 if such distance has not been previously used for any of the triples or matching of distance 2 otherwise. In the described order, repeat the process to each color that was used in one vertex using: either the matching of distance 1, if it was not used in the previous steps; or the matching of distance 2, otherwise. Since there are r − 1 distances between any two rows and we apply one matching of distance i per color and that color is associated with one matching Pj (there are p of those matchings), this means we can repeat the process (r − 1)p times. In other words, this process is possible to be done for (r − 1)p colors. However, we are supposed to use (r − 1)p + 1 = ∆ + 1 colors. The last vertex (xα p ) received a color that was represented in the last row. So it still needs to be represented in the other ones. Recall that when using the colors of the triples of vertices in edges, the matching of distance 1 linking rows 1 and 2, rows 3 and 4, · · · , rows p − 2 and p − 1, were not colored. Therefore, they receive the same color as xα p . By construction, we used exactly ∆ + 1 colors and each color was represented in every vertex. Also by construction, no incident or adjacent elements of the graph received the same color. Each color used in a triple of vertices was also used in rp−3 rp+3 rp−1 edges. So, each of these colors was used in 2 elements. The colors used in only one vertex were also used in 2 edges 2 rp+1 rp+3 rp+1 each. So, each of these colors were used in 2 elements. Since 2 − 2 = 1, we conclude that the difference between the cardinalities of any two color classes is at most 1. Therefore, the described algorithm provides an equitable (∆ + 1)-total coloring of Kr ×p when r and p are odd, as desired. □ In the following example, we exhibit an equitable total coloring of K3×13 with 27 = ∆ + 1 colors. We have P1 = {v1 v2 , v3 v13 , v4 v12 , v5 v11 , v6 v10 , v7 v9 }, remaining vertex is v8 ; P2 = {v2 v3 , v1 v4 , v5 v13 , v6 v12 , v7 v11 , v8 v10 }, remaining vertex is v9 ; P3 = {v3 v4 , v2 v5 , v1 v6 , v7 v13 , v8 v12 , v9 v11 }, remaining vertex is v10 ; P4 = {v4 v5 , v3 v6 , v2 v7 , v1 v8 , v9 v13 , v10 v12 }, remaining vertex is v11 ; P5 = {v5 v6 , v4 v7 , v3 v8 , v2 v9 , v1 v10 , v11 v13 }, remaining vertex is v12 ; P6 = {v6 v7 , v5 v8 , v4 v9 , v3 v10 , v2 v11 , v1 v12 }, remaining vertex is v13 ; P7 = {v7 v8 , v6 v9 , v5 v10 , v4 v11 , v3 v12 , v2 v13 }, remaining vertex is v1 ; P8 = {v8 v9 , v7 v10 , v6 v11 , v5 v12 , v4 v13 , v1 v3 }, remaining vertex is v2 ; P9 = {v9 v10 , v8 v11 , v7 v12 , v6 v13 , v1 v5 , v2 v4 }, remaining vertex is v3 ; P10 = {v10 v11 , v9 v12 , v8 v13 , v1 v7 , v2 v6 , v3 v5 }, remaining vertex is v4 ; P11 = {v11 v12 , v10 v13 , v1 v9 , v2 v8 , v3 v7 , v4 v6 }, remaining vertex is v5 ; P12 = {v12 v13 , v1 v11 , v2 v10 , v3 v9 , v4 v8 , v5 v7 }, remaining vertex is v6 ; P13 = {v1 v13 , v2 v12 , v3 v11 , v4 v10 , v5 v9 , v6 v8 }, remaining vertex is v7 . ′ Also, P1′ = {v1 , v2 , v8 }, P3′ = {v3 , v4 , v10 }, P5′ = {v5 , v6 , v12 }, P7′ = {v7 , v8 , v1 }, P9′ = {v9 , v10 , v3 }, P11 = {v11 , v12 , v5 }. ′ Table 1 shows the distribution of Pi before the shifting. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
A.G. da Silva et al. / Discrete Applied Mathematics ( Table 1 Distribution of Pi′ (i shifting.
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∈ {1, 3, 5, 7, 9, 11}) among the parts Xj before the
X1
X2
X3
P1′
P3′
P5′
′
′
P7
′ P11
P9
Table 2 Distribution of Pi′ (i ∈ {1, 3, 5, 7, 9, 11}) among the parts Xj after the shifting. X1
X2
X3
P1′
P3′
P5′
′ P11
P7′
P9′
Table 3 Distribution of matchings of distance 1 and 2 among the colors. 1→ 2→ 3→ 4→ 5→ 6→ 7→ 8→ 9→
P1 − {v1 v2 } dist 1 P11 − {v11 v12 } dist 1 P3 − {v4 v5 } dist 1 P7 − {v7 v8 } dist 1 P5 − {v5 v6 } dist 1 P9 − {v9 v10 } dist 1 P9 dist 2 P10 dist 1 P12 dist 1
10 → 11 → 12 → 13 → 14 → 15 → 16 → 17 → 18 →
P13 dist 1 P2 dist 1 P3 dist 2 P6 dist 1 P8 dist 1 P11 dist 2 P12 dist 2 P2 dist 2 P4 dist 1
19 → P5 dist 2 20 → P6 dist 2 21 → P7 dist 2 22 → P8 dist 2 23 → P10 dist 2 24 → P13 dist 2 25 → P1 dist 2 26 → P4 dist 2 27 →∗
′ . Therefore, the shifting is necessary in Note that P1′ and P7′ share a common vertex, as well as P3′ and P9′ , and P5′ and P11 ′ this case. Table 2 shows the distribution of Pi after the shifting. This means that each one of the following triples of vertices (obtained from the sets Pi′ ) receive a different color, which we indicate between parenthesis: x11 , x12 and x18 (color 1); x1,11 , x1,12 and x15 (color 2); x23 , x24 and x2,10 (color 3); x27 , x28 and x21 (color 4); x35 , x36 and x3,12 (color 5); x39 , x3,10 and x33 (color 6). Now we color the horizontal edges. To do so, we need the matchings of K3 , which are R1 = {v1 v2 }, R2 = {v2 v3 } and R3 = {v1 v3 }. Considering row 2, vertex x12 received color 1. The matching of K3 in which v1 is not used is R2 = {v2 v3 }. Therefore, we must assign color 1 to the edge x22 x23 . Repeat the process to every vertex that is part of a triple. The vertices not mentioned so far are the ones that are not part of a triple. There is no specific order to color them, as long as each one receives a different color that has not been previously used. For example, x13 was not assigned to any color. So, it can receive color 7. The matching of K3 in which v1 does not appear is R2 = {v2 v3 }. So, we must assign color 7 to the edge x23 x33 . This process must be repeated to all the vertices that are not part of a triple. So far, color 1 was assigned to every vertex in rows 1, 2 and 8. We have P1 − {v1 v2 } = {v3 v13 , v4 v12 , v5 v11 , v6 v10 , v7 v9 }. This means we have to assign color 1 to the matching of distance 1 linking vertices of rows 3 and 13; 4 and 12; 5 and 11; 6 and 10; and 7 and 9. Similarly, distribute colors 2, 3, 4, 5 and 6, that were used in triples of vertices. Color 7 has been already represented in the vertices of row 3. Note that vertex v3 is the remaining one in the matching P9 = {v9 v10 , v8 v11 , v7 v12 , v6 v13 , v1 v5 , v2 v4 }. Thus, we represent color 7 in the vertices linking rows 9 and 10; 8 and 11; 7 and 12; 6 and 13; 1 and 5; and 2 and 4 using matchings of distance 2 (note that the matching of distance 1 associated with P9 − {v9 v10 } has already been used with color 6). We exhibit in Fig. 3 a scheme of coloring of vertices and horizontal edges of K3×13 . Table 3 shows the distribution of colors among the edges that are not horizontal. To understand the notation, check the distribution of colors 1 and 7 done previously among this kind of edge. Recall that color 27 is used in matchings of distance 1 linking vertices of rows 1 and 2; 3 and 4; 5 and 6; 7 and 8; 9 and 10; and 11 and 12 (exactly the pairs of rows not used in the matchings of distance 1 of the triples). We exhibit in Fig. 4, the matchings of distance 1 and 2 linking vertices of two rows for a better understanding of the expressions dist 1 and dist 2 in Table 3.
4. Equitable total coloring of K2×p The equitable total chromatic number of K2×p was determined to be equal to ∆ + 2 by Fu [5]. We provide an elegant algorithm in this section to color such graphs with a technique that is also used throughout the next sections. Edge-coloring matrices are matrices whose entries represent the colors assigned to the edges of a graph. Let AX1 X2 = [aij ] be a p × p edge-coloring matrix in which the entry aij represents the color assigned to the edge that has x1i and x2j as its ends. For i = 1, . . . , p, let ci = (i, i + 1, . . . , p, 1, 2, . . . , i − 1) be the ith circular permutation of (1, . . . , p). Then AX1 X2 is such that Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 3. Coloring of vertices and horizontal edges of K3×13 .
Fig. 4. The bold edges form a matching of distance 1 and the remaining edges form a matching of distance 2.
its row i is ci , i = 1, . . . , p (see the following example of the matrix AX1 X2 and Fig. 5 for p = 3). 1 2 3
[ AX1 X2 =
2 3 1
3 1 . 2
]
Theorem 2. The algorithm above describes an equitable (∆ + 2)-total coloring of K2×p . Proof. It can be easily seen that each row and each column of AX1 X2 has p distinct entries. Thus for each vertex of K2×p , the p incident edges receive p distinct colors. Now, for i = 1, 2, coloring all the vertices of Xi with the color p + i, we obtain a (∆ + 2)-total coloring of K2×p (see Fig. 5 for K2×3 ). □ 5. Equitable total coloring of Kr ×p with r ≥ 4 even and p odd We begin this section showing that for these graphs, χe′′ ̸ = ∆ + 1. After, we provide an algorithm to show that χe′′ = ∆ + 2, contributing to the ETCC. Claim 3. There exists no equitable total coloring with ∆ + 1 colors for Kr ×p with r ≥ 4 even and p odd. Proof. Suppose, in order to obtain a contradiction, that complete r-partite graphs having p elements in each part of the partition, with r even and p odd, have an equitable (∆ + 1)-total coloring. Thus, each color must be represented in each one of the vertices. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 5. Equitable total coloring of K2×3 .
The case in which p = 1 is special, because such graphs are complete and have an even number of vertices, and it is known that they do not have equitable (∆ + 1)-total coloring [8]. So, we will assume that p ≥ 3. We claim that an odd number of vertices cannot receive the same color in any equitable (∆ + 1)-total coloring. Indeed, suppose that an odd number y of vertices receive the same color i. In this case, an odd number (rp − y) of vertices would remain and color i would be represented in these vertices by the edges that have them as ends. Since these edges are not adjacent, rp − y is even. This contradicts the fact that each color must be represented in each one of the vertices. We claim that an even number of vertices cannot receive the same color in any equitable (∆ + 1)-total coloring. Indeed, take an arbitrary part of the partition of V . Such part has p (odd) elements. Suppose that there exist colors a1 , a2 , . . . , ak used b1 , b2 , . . . , bk times (bi is even), respectively, to color the vertices of that given part. Note that we consider only one part of the partition of V because once a color is used in a vertex of any part, say Xi , it cannot be used in the vertices of the other parts, for the vertices of these parts are adjacent to the vertices of Xi . Thus, we would have b1 + b2 + · · · + bk vertices colored in such part. Note that this sum is even, since bi is even for all i = 1, 2, . . . , k. We would conclude that p − (b1 + b2 + · · · + bk ) vertices were not colored. Since this quantity is odd and we previously proved that there exists no color used an odd number of times to color vertices, we have a contradiction. Therefore, the graph cannot have an equitable (∆ + 1)-total coloring. □ Theorem 3. The graph Kr ×p with r ≥ 4 even and p odd has χe′′ = ∆ + 2. Proof. Using Claim 1 it is possible to organize all edge-coloring matrices as follows. If Ri = {va vb , . . . , vc vd } is a perfect matching of Kr (there are r − 1 disjoint perfect matchings in Kr ), then let Si = {AXa Xb , . . . , AXc Xd } be a set of edge-coloring matrices, with |Si | = |Ri |. To each set of matrices Si , assign numbers from 1 to p(r − 1) as follows: the entries of the matrices in S1 are the elements of the set {1, 2, 3, . . . , p}; in S2 , elements of the set {p + 1, p + 2, . . . , 2p}, and so on. In general, the entries of the matrices in Si are elements of the set {(i − 1)p + 1, (i − 1)p + 2, . . . , ip}. Note that each one of the sets of edge-coloring matrices has p elements and that if they are taken modulo p, the sets become {1, 2, . . . , p − 1, 0}. Based on that, the distribution of elements in the edge-coloring matrices is done similarly to the distribution in the case of complete bipartite balanced graphs. Since |Si | = |Ri | = 2r , each color from 1 to p(r − 1) was rp used 2 times in edges. When we divide the matrices into r − 1 groups, in each of those groups, part Xi appears precisely once (see Claim 1). Therefore, when the same set of colors is assigned to one of the sets of edge-coloring matrices, the way the entries are displayed in the matrices guarantees that no adjacent edges receive the same color. We replace the secondary diagonal entries of r −2 2 matrices by the color (r −1)p+1 and r −2 2 matrices by the color (r −1)p+2, which have not been used yet. The replacement is done as follows. AX1 X2 , AX2 X3 , . . . , AXr −2 Xr −1 (note that these matrices are related to S1 , S2 , . . . , Sr −2 , respectively) have the entries in their secondary diagonals replaced alternately by (r − 1)p + 1 and (r − 2)p + 2. Note that by the algorithm to divide the matrices into sets, all the ones that have their entries changed are elements of S1 , S2 , . . . , Sr −2 , so their secondary diagonals received previously different colors. We illustrate this procedure in the end of this section, by presenting an example of an equitable (∆ + 2)-total coloring of K4,3 , where the entries of the secondary diagonals of the matrices AX1 X2 and AX2 X3 were originally 3 and 6, respectively, but were replaced by 10 and 11, respectively, according to this step. Note that an element aij of AXk Xl is in the secondary diagonal if it is such that i + j = p + 1. Therefore, all entries in the secondary diagonal of one matrix receive the same color since i + j − 1 = p + 1 − 1 = p. In each group of edge-coloring matrices, the secondary diagonal will be p, 2p, 3p, . . . , (r − 2)p by the coloring of edge-coloring matrices presented in the beginning of this proof. The colors of the vertices will be the colors (r − 1)p + 1, (r − 1)p + 2 and the colors that have been changed in some of the secondary diagonals, which are p, 2p, 3p, . . . , (r − 2)p since these colors must be represented in vertices that are ends of edges that are colored by AX1 X2 , AX2 X3 , . . . , AXr −2 Xr −1 . To the vertices of Xi , for all 1 ≤ i ≤ r − 2, assign the color ip, to the vertices of Xr −1 , assign the color (r − 1)p + 1 and to the vertices of Xr , assign the color (r − 1)p + 2. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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(r −2)p
rp
The color (r − 1)p + 1 is used 2 times in edges and p times in the coloring of vertices, totalizing 2 times. The result is analogous for the color (r − 1)p + 2. The colors that were originally in the secondary diagonal of some matrices and were replaced, were originally used p times in 2r matrices. Then, they were replaced by a new color in p entries of a matrix. On the other side, those colors are used rp rp to color the p vertices of one of the parts of the partition of V . So, those colors are used 2 − p + p = 2 times in total. Each rp color is used precisely 2 times. Hence, the difference between the cardinalities of two color classes is 0 and this concludes the proof of the fact that the described algorithm gives an equitable (∆ + 2)-total coloring of Kr ×p with r ≥ 4 even and p odd. □ Now, we show the edge-coloring matrices that generate the equitable (∆ + 2)-total coloring of K4×3 . 1 2 10
2 10 1
10 1 , AX1 X3 = 2
[
4 5 11
5 11 4
11 4 , AX2 X4 = 5
[
[ AX1 X2 =
[ AX2 X3 =
]
]
7 8 9
8 9 7
9 7 , AX1 X4 = 8
[
7 8 9
8 9 7
9 7 , AX3 X4 = 8
[
]
]
4 5 6
5 6 4
6 4 5
1 2 3
2 3 1
3 1 . 2
]
]
The vertices in X1 are assigned to 3 = p, the ones in X2 , to 6 = 2p, in X3 , to 10 = (r − 1)p + 1 and the ones in X4 , to 11 = (r − 1)p + 2. 6. Equitable total coloring of Kr ×p with r and p even By using the algorithm presented in Section 5, we get an equitable (∆ + 2)-total coloring for these graphs, contributing to the ETCC. Theorem 4. The graph Kr ×p with r and p even has χe′′ ≤ ∆ + 2. Proof. The equitable (∆ + 2)-total coloring is obtained directly from the proof of Theorem 3 by applying the same technique. The parity of p was relevant in the previous section to prove that Kr ×p , with r ≥ 4 even and p odd does not have equitable (∆ + 1)-total coloring. The algorithm to prove that those graphs have χe′′ = ∆ + 2, however, does not require p to be odd. That is why the same algorithm works here. □ We present an equitable (∆ + 2)-total coloring of the graph K6×2 obtained by the algorithm. We show below their edgecoloring matrices. AX1 X2 AX2 X3 AX3 X4 AX4 X5
2 1
]
[
4 3
]
[
6 5
]
[
8 7
1 2 , AX4 X6 = 2 1
]
[
4 3 , AX5 X6 = 4 3
]
[
6 5 , AX1 X6 = 5 6
]
[
8 7 , AX2 X6 = 7 8
]
[
1 11 , AX3 X5 = 2 1
[
3 = 12
12 3 , AX1 X4 = 3 4
[
5 = 11
11 5 , AX2 X5 = 5 6
[
7 = 12
12 7 , AX1 X3 = 7 8
[
10 9 , AX2 X4 = 9 10
9 10
[
[
1 = 11
AX1 X5 =
]
]
[
]
]
]
]
[
10 9 , AX3 X6 = 9 10
] ]
10 . 9
The vertices of X1 receive color 2, of X2 color 4, of X3 color 6, of X4 color 8, of X5 color 11, and of X6 color 12. By the algorithm, we could show that χe′′ ≤ ∆ + 2. However, we present an equitable (∆ + 1)-total coloring for the graph K4×4 , which is a graph of this class. We will display in Figs. 6, 7, 8, 9, 10, 11, and 12, the distribution of colors among the elements of the graph. The reader can easily check that colors 1, 2, . . . , 8 were used in 2 vertices and 7 edges each, that is, in 9 elements each. Also, colors 9, 10, . . . , 13 were used in 8 edges each. It is also easy to check that no adjacent or incident elements received the same color. We get that the difference between the cardinalities of two color classes is either 0 or 1 and ∆ + 1 = (4 − 1)4 + 1 = 13. Therefore, the equitable total chromatic number of K4×4 is ∆ + 1. 7. Equitable total coloring of Kr ×p with r odd and p even We begin this section with an algorithm to show that χe′′ ≤ ∆ + 2. After that, we provide an algorithm to show that χe = ∆ + 1 if r ≥ 2p . If r < 2p , the algorithm does not work because there are not enough available matchings of distance to ′′
complete the coloring. Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 6. The numbers represent the colors of the vertices, the dashed edges receive color 1, and the continuous edges receive color 2.
Fig. 7. The continuous edges receive color 3 and the dashed edges receive color 4.
Fig. 8. The continuous edges receive color 5 and the dashed edges receive color 6.
Fig. 9. The dashed edges receive color 7 and the continuous edges receive color 8.
Theorem 5. The graph Kr ×p with r is odd and p is even has χe′′ ≤ ∆ + 2. Proof. If p = 2, get the r matchings Ri ’s of Kr (Claim 2) and associate two colors to each matching in the following way: Ri is associated to 2i − 1 and 2i, for 1 ≤ i ≤ r. We have Ri = {va vb , . . . , vc vd } and vf is the remaining vertex in this matching. Let AXa Xb , . . . , AXc Xd be edge-coloring matrices related to the matching Ri and part Xf related to the remaining vertex vf . The entries of such matrices must be filled as in the complete bipartite balanced case. Since Kr has an odd number of vertices, the vertices of Xf do not have colors 2i − 1 and 2i represented by the edge-coloring matrices. To the vertex xf 1 assign color Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 10. The dashed edges receive color 9 and the continuous edges receive color 11.
Fig. 11. The continuous edges receive color 10.
Fig. 12. The dashed edges receive color 12 and the continuous edges receive color 13.
2i − 1 and to the vertex xf 2 , color 2i. Observe that, in this case, vertex xf 1 does not have color 2i represented and vertex xf 2 does not have color 2i − 1 represented, but since this is an equitable (∆ + 2)-total coloring, the vertices do not need to have all colors represented. Since all matchings of Kr have a remaining vertex, all vertices of Kr ×2 received a different color by the previous procedure. The argument to justify that adjacent edges are assigned to different colors in the complete bipartite balanced graphs also hold here. An edge and its ends are assigned to different colors as well because the part Xf receives colors 2i − 1 and 2i, which have only been attributed to the edge-coloring matrices of Ri and vf is the remaining vertex in such matching. We used two colors to each matching of Kr , which totalizes 2r colors. Also, ∆ + 2 = (r − 1)p + 2 = (r − 1)2 + 2 = 2r. Each color is used in 1 vertex and 2 times in each one of the r −2 1 = |Ri | (for all i = 1, 2, . . . , r) edge-coloring matrices in which it is applied. So, each color is used 1 + 2 r −2 1 = r times. Hence the difference between the cardinalities of any two color classes is 0 and we conclude that the total coloring is, indeed, equitable. p If p ≥ 4, we repeat the same process used for p = 2, 2 times. For this, we use the matching P1 of Kp (Claim 1) to determine the pairs of rows in which we apply the steps of case p = 2: if P1 = {v1 v2 , va vb , . . . , vc vd }, use 2r colors to color the edges linking vertices of rows 1 and 2 in the same way we colored the edges of the case p = 2 (edge-coloring matrices) and color vertices xi1 , xi2 , with 1 ≤ i ≤ r (vertices of Kr ×p that are in the part Xf related to the remaining vertex vf of matching Ri ). After that, using the same 2r colors, color the edges linking rows a and b and the vertices xia , xib , with 1 ≤ i ≤ r, analogously. Repeat the process until coloring the edges linking vertices in rows c and d and also the vertices in such rows. As explained p in the case p = 2, each color is used r times. Now, since the process is repeated 2 = |Pi | (for all i = 1, 2, . . . , r) times, rp each color is used 2 times. So far, every edge linking vertices of the pairs determined by P1 has been colored as well as all the horizontal edges and all vertices of the graph. Now, use one different color to each matching of available distance between the pairs determined by P2 , P3 , . . . , Pp−1 . Note that each matching of distance is a perfect matching of the graph. So, there are p − 1 − 2 + 1 = p − 2 matchings and r − 1 matchings of available distances (see Section 2), which implies that (p − 2)(r − 1) colors are used in this last step. And each of these colors is used in a disjoint perfect matching of Kr ×p , Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 13. Scheme of colors to be applied in elements linking vertices of rows 1 and 2; 3 and 7; 4 and 6; 5 and 8.
rp
that is, they are used in 2 edges. We conclude that the difference between any two color classes is 0. Also, we use a total of 2r + (p − 2)(r − 1) = ∆ + 2 colors. It is clear, by construction, that no adjacent or incident elements receive the same color, since the first part of the algorithm is based on the repetition of the case p = 2 and in the second part we used one different color to each matching of available distance (determined in Section 2) between the pairs determined by P2 , P3 , . . . , Pp−1 , as stated above. □ Observe the equitable total coloring of K3×8 with 18 = ∆ + 2 colors obtained by Theorem 5. We have matching P1 = {v1 v2 , v3 v7 , v4 v6 , v5 v8 } of Kp=8 . Also, matchings R1 = {v1 v2 }, R2 = {v2 v3 } and R3 = {v1 v3 } of Kr =3 . In Fig. 13 we present the coloring of the elements that link vertices of rows 1 and 2; 3 and 7; 4 and 6; and 5 and 8 (see P1 ). We have P2 = {v2 v3 , v1 v4 , v5 v7 , v6 v8 }. So, we use color 7 in the matching of distance 1 linking vertices of rows 2 and 3; 1 and 4; 5 and 7, 6 and 8; and color 8 in the matching of distance 2 linking vertices of the same pairs of rows. We represent this fact by stating: color 7 → P2 dist 1 and 8 → P2 dist 2. We have P3 = {v3 v4 , v2 v5 , v1 v6 , v7 v8 }. So, color 9 → P3 dist 1 and 10 → P3 dist 2. We have P4 = {v4 v5 , v3 v6 , v2 v7 , v1 v8 }. So, color 11 → P4 dist 1 and 12 → P4 dist 2. We have P5 = {v5 v6 , v4 v7 , v1 v3 , v2 v8 }. So, color 13 → P5 dist 1 and 14 → P5 dist 2. We have P6 = {v6 v7 , v1 v5 , v2 v4 , v3 v8 }. So, color 15 → P6 dist 1 and 16 → P6 dist 2. We have P7 = {v1 v7 , v2 v6 , v3 v5 , v4 v8 }. So, color 17 → P7 dist 1 and 18 → P7 dist 2. Theorem 6. The graph Kr ×p with r odd and p even has χe′′ = ∆ + 1 if r ≥
p . 2
Proof. If p = 2, assign the color i to the vertices xi1 ∈ Xi and xi2 ∈ Xi , for all i : 1, . . . , r. So, no adjacent vertices receive the same color. If a vertex xij received color i, take the matching Rl of Kr (Claim 2) in which vi is the remaining vertex. Such matching has the form Rl = {va vb , . . . , vc vd }. So, assign color i to the horizontal edges xaj xbj , . . . , xcj . Since vi is not an end of edges of Rl and the edges xaj xbj , . . . , xcj are related to the matching Rl , this means that incident elements receive different colors. Each color i that has been used in vertices was used twice (xi1 and xi2 ). We have |Rl | = r −2 1 . So, the color i has been used in 2 r −2 1 edges, totalizing r + 1 elements. rp Note that, so far, we have colored the horizontal edges and the vertices with 2 colors. Now, using different colors than the ones that have been already used, take r − 1 colors to color matchings of distance of graph Kr ×2 . Since the cardinality of those matchings is r, each one of those r − 1 colors is used r times and the difference between the cardinalities of two color classes is at most 1, as desired. p Now, consider the case p ≥ 4 and r ≥ 2 . Since there are p − 1 disjoint perfect matchings in Kp , in order to know how many times each matching Pi will be used to color the r parts of vertices of Kr ×p , divide r by p − 1. By Euclidean division, there exist positive integers s and q such that q(p − 1) + s = r. This means that we should use s matchings of Kp to color q + 1 parts of the partition each and p − 1 − s matchings to color q parts each. The coloring of the vertices is done as follows: if P1 = {va vb , . . . , vc vd } and if we use it to color the vertices of X1 then each of the following pairs of vertices receives a different color: x1a and x1b , . . . , x1c and x1d . The coloring of the horizontal edges is done similarly to the coloring of the horizontal edges in the case Kr ×p with r and p being odd. After coloring the horizontal edges, we guarantee that if a color was used in a vertex of a certain row, then it was represented in all the vertices of that row. Now, this color needs to be represented in the vertices of the other rows. To color the edges that are not horizontal, we begin with the colors that were used in vertices. If P1 = {va vb , vc vd , . . . , vg vh } and we used color i in the vertices x1a and x1b (admitting that P1 was used to color the vertices in X1 ), then we represent color i in the complement of P1 with respect to va vb , which we are going to denote by P1 − {va vb } = {vc vd , . . . , vg vh }. Color i must be represented in the vertices of rows c and d, and so on, being represented in the rows determined by {vc vd , . . . , vg vh } above, until being used in the vertices of rows g and h. Color i is used in matchings of distance 1. We repeat the process to every color used in vertices always using the next available matching of distance. The other colors, that is, the ones not used in vertices, are assigned according to the available matchings of distances in each perfect matching Pi of Kp . The number of rp colors used only in edges of the graph is ∆ + 1 − 2 . Indeed, since there are (p − 1) Pi′ s and there are r − 1 matchings of distance for each Pi , the number of matchings of distance is (r − 1)(p − 1). From this number, we remove the matchings Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 14. Coloring of vertices and horizontal edges of K3×4 .
(p
)
of distance colored with the colors of the vertices [s(q + 1) + (p − 1 − s)q] 2 − 1 . As explained above, s matchings are used in the coloring of q + 1 parts and p − 1 − s matchings are used in the coloring of q parts. It is easy to check that each p when applying a color of a pair (of vertices time we use a matching Pi to color a part, we use 2 − 1 matchings (of distance ) ) p to the complement(of Pi : (r − 1)(p − 1) − [ s(q + 1) + (p − 1 − s)q ] − 1 = rp − r − p + 1 + (q(p − 1) + s) 2p − 1 = 2 ) p rp rp rp − r − p + 1 − r 2 − 1 = 2 − p + 1 = ∆ + 1 − 2 . p p Note that there are r − 1 matchings of distance j and |Pi | = 2 . So, the complement of Pi has size 2 − 1. In order to have p p available matchings of distance to complete the coloring, we must have − 1 ≤ r − 1, which implies that r ≥ 2 . 2
rp
By construction we get that no adjacent or incident elements receive the same color. Also, we have that 2 colors are used in 2 vertices and then, represented in the other vertices through edges that have them as ends. Since each edge has 2 rp−2 rp rp ends, we get that those } colors are used in 2 edges, totalizing 2 + 1 elements. The other 2 − p + 1 colors are used in { rp2 (r −1) 2
−
/ rp2 =
( rp−2 ) rp 2
2
rp 2
− p + 1 perfect matchings of the graph Kr ×p , that is, in
rp 2
elements. We conclude that the
difference between two different color classes is at most 1, as desired. □ Observe the equitable (∆ + 1)-total coloring of K3×4 exhibited below. We have matchings P1 = {v1 v2 , v3 v4 }, P2 = {v2 v3 , v1 v4 } and P3 = {v3 v4 , v1 v2 } of Kp=4 . The coloring of the vertices of X1 is related to P1 , the vertices of X2 to P2 , and the vertices of X3 to P3 . Observe the coloring of vertices and horizontal edges in Fig. 14. Colors 1–6 are also used in the following edges: color 1 in the matching of distance 1 linking vertices of rows 3 and 4; color 2 in the matching of distance 1 linking vertices of rows 1 and 2; color 3 in the matching of distance 1 linking vertices of rows 1 and 4; color 4 in the matching of distance 1 linking vertices of rows 2 and 3; color 5 in the matching of distance 1 linking vertices of rows 2 and 4; color 6 in the matching of distance 1 linking vertices of rows 1 and 3. Colors 7, 8 and 9 are used in the following edges: 7 → P1 dist 2, 8 → P2 dist 2 and 9 → P3 dist 2. 8. Final considerations In this paper we improved the existing bound given by Fu [5] for graphs of odd order showing that these graphs have
χe′′ = ∆ + 1. We also showed equitable (∆ + 2)-total colorings for the open cases. In the case in which r is even and p is odd, we proved that χe′′ = ∆ + 2. However, for the cases in which r and p are even, and r is odd and p is even, we showed an equitable (∆ + 2)-total coloring for all of those graphs and gave examples of equitable total colorings with ∆ + 1 colors for graphs within these cases. This may suggest that, in both cases, we have χe′′ = ∆ + 1. Finally, in order to conclude the paper, we present equitable (∆ + 1)-total colorings for complete bipartite nonbalanced graphs by applying the same techniques used in the bipartite complete balanced graphs. It was established by Fu in [5] that χe′′ = ∆ + 1 for Kp1 ,p2 (with p1 ̸= p2 ). Let Kp1 ,p2 with p1 ̸ = p2 be a bipartite complete nonbalanced graph. Assume, without loss of generality, that p1 < p2 and let A = [aij ] be a (p1 + 1) × p2 matrix, in which the entry aij is the color assigned to the edge that has x1i and x2j as its ends if 1 ≤ i ≤ p1 ; and the color assigned to the vertex x2j if i = p1 + 1. Similarly to the complete bipartite balanced case, to the element aij of A, assign the color i + j − 1 mod (p1 ) if i + j − 1 ̸ ≡ 0 mod (p1 ) and the color p1 , otherwise. To the vertices of X1 , assign the color p1 + 1. Observe the following example of the matrix A and Fig. 15.
⎡
1 ⎢2 A=⎣ 3 4
2 3 4 1
3 4 1 2
⎤
4 1⎥ . 2⎦ 3
Theorem 7. The previous algorithm gives an equitable (∆ + 1)-total coloring of Kp1 ,p2 , with p1 ̸ = p2 . Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.
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Fig. 15. Equitable 5-total coloring of K3,4 .
Proof. Let aij and akl be two different entries of A. If i = k, then j ̸ = l and the corresponding elements in Kp1 ,p2 are adjacent edges if 1 ≤ i ≤ p1 or vertices of X2 (which are nonadjacent) if i = p1 + 1. The proof provided for Theorem 2 also works in this situation and we get that adjacent edges receive different colors. So, each color of {1, 2, . . . , p2 } appears exactly once in each row of A. Since A has p1 + 1 rows, each one of those colors is used p1 + 1 times. Now, if j = l and i ̸ = k, the entries aij and akj are either two adjacent edges (if 1 ≤ i, k ≤ p1 ) or an incident vertex with an edge (if i = p1 + 1 or k = p1 + 1). Again, the proof of Theorem 2 also works in this case. In other words, each color of the set {1, 2, . . . , p2 } appears exactly once in each row. We get that p2 colors are used p1 + 1 times each. Also, the color p2 + 1 is used to color all the p1 vertices in X1 . Hence, the difference between the cardinalities of two color classes is at most 1, as desired. This concludes the proof that the algorithm gives an equitable (∆ + 1)-total coloring of the graph Kp1 ,p2 , with p1 ̸ = p2 . □ References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
M. Behzad, Graphs and their chromatic numbers (Ph.D. thesis), Michigan State University, 1965. J.C. Bermond, Nombre chromatique total du graphe r-parti complet, J. Lond. Math. Soc. 9 (1974) 279–285. T. Chunling, L. Xiaohui, Y. Yuansheng, L. Zhihe, Equitable total coloring of Cm □Cn , Discrete Appl. Math. 157 (2009) 596–601. S. Dantas, C.M.H. de Figueiredo, G. Mazzuoccolo, M. Preissmann, V.F. dos Santos, D. Sasaki, On the equitable total chromatic number of cubic graphs, Discrete Appl. Math. 209 (2016) 84–91. H.L. Fu, Some results on equalized total coloring, Congr. Numer. 102 (1994) 111–119. M.A. Gang, M.A. Ming, The equitable total chromatic number of some join graphs, Open J. Appl. Sci. 2 (4B) (2012) 96–99. H. Gui, W. Wang, Y. Wang, Z. Zhang, Equitable total-coloring of subcubic graphs, Discrete Appl. Math. 184 (2015) 167–170. M. Kubale, Graph Colorings, Vol. 352, American Mathematical Soc., 2004. A. Sánchez-Arroyo, Determining the total colouring number is NP-hard, Discrete Math. 78 (1989) 315–319. A. Soifer, The Mathematical Coloring Book, Springer-Verlag, 2008. V.G. Vizing, Some unsolved problems in graph theory, Russian Math. Surveys 23 (6) (1968) 125–141. W.F. Wang, Equitable total coloring of graphs with maximum degree 3, Graphs Combin. 18 (2002) 677–685.
Please cite this article in press as: A.G. da Silva, et al., Equitable total coloring of complete r-partite p-balanced graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.03.009.