CC BY
4
2021 Прикладная теория графов № 54
УДК 519.174.7 DOI 10.17223/20710410/54/5
THE PALETTE INDEX OF SIERPINSKI TRIANGLE GRAPHS AND SIERPINSKI GRAPHS
A. Ghazaryan
Yerevan State University, Yerevan, Armenia
E-mail: [email protected]
The palette of a vertex v of a graph G in a proper edge coloring is the set of colors assigned to the edges which are incident to v. The palette index of G is the minimum number of palettes occurring among all proper edge colorings of G. In this paper, we consider the palette index of Sierpinski graphs S™ and Sierpinski triangle graphs S3. In particular, we determine the exact value of the palette index of Sierpinski triangle graphs. We also determine the palette index of Sierpinski graphs S™ where p is even, p = 3, or n = 2 and p = 41 + 3.
Keywords: palette index, Sierpinski triangle graph, Sierpinski graph.
ОБ ИНДЕКСЕ ПАЛИТРЫ ТРЕУГОЛЬНИКА СЕРПИНСКОГО
И ГРАФА СЕРПИНСКОГО
A. Газарян
Ереванский государственный университет, г. Ереван, Армения
Палитра вершины v графа G в правильной раскраске рёбер — это набор цветов, присвоенных рёбрам, инцидентным v. Индекс палитры G — это минимальное количество палитр из всех правильных рёберных раскрасок G. Рассматриваются индексы палитры графов Серпинского S3 и треугольников Серпинского S3. Доказано, что индекс палитры треугольника Серпинского S3 равен 3, если n чётное, и 4 иначе; индекс палитры графа Sp3 равен 2 для чётного p и равен 3 для p = 3 или n = 2 и p = 41 + 3.
Ключевые слова: индекс палитры графа, треугольник Серпинского, граф Сер-пинского.
1. Introduction
In this paper, we use the standard notations of graph theory [1]. Graph coloring problems are even more challenging when there are some constraints on them, such as proper edge or proper vertex colorings. Usually, such constraints are naturally motivated by different applications in scheduling theory.
In [2], a new chromatic parameter is called the palette index of a graph and is defined as follows: for a given proper edge coloring of a graph G, we define the palette of a vertex v E V(G) as the set of all colors appearing on edges incident to v. The palette index s(G) of G is the minimum number of distinct palettes occurring in a proper edge coloring of G.
Mainly, the palette index was studied for regular graphs. In [2], it is shown that the palette index of a regular graph is 1 if and only if the graph is of Class 1, and it is different from 2. The palette index of complete graphs is also determined in [2].
Theorem 1 [2]. For every positive integer n > 1, we have
Í1, if n is even,
3, if n = 4/ + 3,
4, if n = 4/ + 1.
In [2], the authors also studied the palette index of cubic graphs. More specifically,
1, if G is of Class 1, s(G) = { 3, if G is of Class 2 and has a perfect matching, 4, if G is of Class 2 and has no perfect matching.
As mentioned in [3], the palette index of a d-regular graph of Class 2 satisfies the inequality 3 ^ á(G) ^ d +1.
The paper [3] investigates 4-regular graphs and proves that s(G) E {3,4,5} if G is 4-regular and of Class 2, and that all these values are, in fact, attained.
Since the computing of the chromatic index of cubic graphs is NP-complete [4], determining the palette index of a given graph is also NP-complete, even for cubic graphs [2]. This means that even determining, if a given graph has a palette index 1, is an NP-complete problem.
Vizing's edge coloring theorem yields an upper bound for the palette index of a general graph G with the maximum degree A, namely á(G) ^ 2A+1 — 2. It is not hard to construct graphs whose palette index is quite smaller than 2A+1 - 2. In [5], an infinite family of multigraphs is described, whose palette index grows asymptotically as A2; however, it is an open question whether there are such examples without multiple edges. Furthermore, in [5] it is conjectured that there is a polynomial p(A) so that for any graph with the maximum degree A, it holds the bound s(G) ^ p(A).
There are few results about the palette index of non-regular graphs. In [6], M. Horñak and J. Hudák have completely determined the palette index of the complete bipartite graphs Ka,b with min{a, b} ^ 5.
In [7], C. Casselgren and P. Petrosyan studied the palette index of bipartite graphs. In particular, they have determined the exact value of the palette index of grids and characterized the class of graphs whose palette index equals the number of vertices.
In [8], the palette index of trees is considered. In particular, the authors have proved
A 1A1
that s(T) ^ — . Moreover, they also have showed the sharpness of the bound by
i=i L 1 \
constructing trees TA for which the palette index is is equal to this value.
One of the fascinating graph families is Sierpinski graphs S". It has been introduced in [9] as a result of the study of the topological properties of the Lipscomb space, and it is still being studied extensively. An interesting fact is that is isomorphic to the graph of the Tower of Hanoi with n disks [9].
Sierpinski triangle graphs S" are quite similar to Sierpiáski graphs S^1 and are obtained from Sn+1 by a finite number of steps. These graphs are usually studied in conjunction with Sierpinski graphs and have quite interesting properties. Sierpinski triangle graphs are fractals of dimension d = log3/log2 « 1.585 and have been introduced in [10].
For Sierpinski graphs and other Sierpinski-type graphs, [11] is a good survey.
In [12], S. Klavzar has introduced an explicit labeling of the vertices of S". Also, he has proved that S is uniquely 3-colorable and S" is uniquely 3-edge-colorable.
In [13], the authors have studied and summarized vertex, edge, and total colorings of the Sierpinski triangle graphs SS and Sierpinski graphs SS.
In [14], some properties of graphs SS are given, including their cycle structure, domination number, and pebbling number.
These graphs are beneficial to other theories such as probability theory [15], dynamical systems theory [16], topology [17], etc.
Let us now give explicit definitions of Sierpinski graphs Sn and Sierpinski triangle
graphs SS, where n ^ 0 and p > 0 are integers. The graphs are defined as follows. The vertex set of Sp! is the set of all n-tuples of integers 0,1,... ,p-1, namely, V(Sn) = {0,1,... ,p-1}n. Two different vertices u = (u^ ..., un) and v = (v^ ..., vn) are adjacent if and only if there exists an h G {1,..., n} such that:
1) ut = vt for t = 1,..., h — 1;
2) uh = vh;
3) ut = vh and vt = uh for t = h + 1,..., n.
We will denote a vertex (u1, u2,..., un) by (u1u2 ... un) or even u1u2 ... un. The vertices (i... i), i G {0,... ,p — 1}, are called the extreme vertices of Sp\ We will denote by iSp = = SS[{v : v = (i... i)}] the subgraph of SS, where i = 0,1,... ,p — 1. Obviously, iSS+1 is
isomorphic to SS. Consequently, SS contains pn 1 copies of the graph Sp = Kp. We will call
—..—^ —J^---- „„„ o-^" "p
link edges all the edges of Sn that do not belong to the above-mentioned Kp.
As a result of contracting all the link edges of SS+1, we will get the Sierpinski triangle graph S3 where n ^ 0. We label the vertices of S3 as follows. Let (u1... urij ... j) and (u1... ur ji... i) be the endvertices of a link edge of SS+1 that is contracted to a vertex x of S31. Then we label x with (u1... ur)[i, j] or u1... ur [i, j] where r ^ n — 2. SS+1 contains three isomorphic copies of SS, and we denote these copies with i-S;S+1, where iSS+1 is the subgraph which contains (i... i), 0 ^ i ^ 2; see Fig. 1.
\022
22 *
Fig. 1. Labeling of S; (a) and of S3 (b)
111 1[1,2J [1,2] 2[1,2]
b
2
In this paper, the palette indices of Sierpinski triangle graphs SS and Sierpinski graphs SS are determined. Next, the palette index of Sp1 is determined, where p is even, or n = 2 and p = 41 + 3.
2. The palette index of Sierpinski triangle graphs
Let SS be the Sierpinski triangle graph. When n = 0, then SJ is isomorphic to K3, whose palette index is determined (Theorem 1). Below we consider the case n > 0. The graph has two kinds of vertices in terms of vertex degree: three vertices with degree 2 and the remaining vertices with degree 4. To color SS, we need at least four colors as A = 4.
a
Let us define edge coloring functions for the graph Si as follows:
0n(e)
n mod 4 + 1, (n + 1) mod 4+1, k(n + 2) mod 4+ 1,
if e G {[0,1]00, [1, 2]11, [0, 2]22}, if e G {[0,1]11, [1, 2][0, 2]}, if e G {[1, 2]22, [0,1][0, 2]}, if e G {[0, 2]00, [0,1][1, 2]},
where n G {1, 2, 3, 4}. Thus, for all colorings we have 4 different sets of palettes and we call this group A:
1) {{1, 2, 3, 4}, {1, 2}, {1, 3}, {1,4}};
2) {{1, 2, 3, 4}, {2,1}, {2, 3}, {2,4}};
3) {{1, 2, 3, 4}, {3,1}, {3, 2}, {3,4}};
4) {{1, 2, 3, 4}, {1, 4}, {2, 4}, {3,4}}.
We will identify two edge colorings if one is obtained by rotating or reflecting the other. Clearly, if we color Si in a manner that all vertices with cardinality 4 have the palette {1, 2, 3, 4}, then we will get a coloring After this, let us color the graph Sf so that all 4 degree vertices are colored with the palette {1, 2, 3,4}. That means we must use colorings to color iSf, where i = 0,1, 2. By considering all possible cases, we have four different sets of palettes and we call that group B:
1) {{1, 2, 3, 4}, {1, 2}, {1, 3}, {2, 3}};
2) {{1, 2, 3, 4}, {1, 2}, {1, 4}, {2,4}};
3) {{1, 2, 3, 4}, {1, 3}, {1, 4}, {3,4}};
4) {{1, 2, 3, 4}, {2, 3}, {2, 4}, {3,4}}.
Proposition 1. For every positive integer n, we have
s(Sn) ^ 4.
Proof. If n is 1 or 2, then we can color Sf using a set of palettes from group A or B. If we try to color Sf so that each iSf is colored the same as we colored Sf, where i = 0,1, 2, and the vertices [0,1], [0, 2], [1, 2] have the palette {1, 2, 3, 4}, then we will have all the possible sets of palettes of group A. We now prove the following stronger lemma:
Lemma 1. For every positive integer n, we can color Sf using every set of palettes from group A, if n is odd, and every set of palettes from group B, if n is even. As we have seen, the lemma is true for Sf, where n = 1, 2, 3. Assume the lemma holds for n > 0. We wish to find a satisfying coloring of Sf+1. By the induction assumption, we can color Sn using every set of palettes from group A or B depending on whether n is odd or not. In general, if we have colorings for iSf", then the palettes of vertices of degree two of iSf are the only essentials for coloring Sf", where i = 0,1, 2, m > 1. As we saw above in the process of coloring Sf (Sf), if we can color iSf (iSf) using every set of palettes from group A (B), then we can color Sf(Sf) using every set of palettes from group B (A). So we can color Sn+1 by using Sf colorings as we did for Sf or Sf. ■
Proposition 2. For every positive integer n, we have
s(sn) > 3.
Proof. Clearly, s(Sf) ^ 2, since there are vertices with only two different degrees. If we try to color Sf with one palette with cardinality 4 ({1, 2, 3, 4}), then we will get a
coloring . In Proposition 1 for coloring , we used only colorings for any subgraph S3 and tried all possible cases. That means that if we want to have only one palette with cardinality 4, we must use a set of palettes from group A or B, where each set of palettes contains three palettes with cardinality 2. Hence, s(Sn) ^ 3. ■
Proposition 2 means that if we are looking for a coloring with three palettes, then we should have one palette with cardinality 2 and two palettes with cardinality 4.
Proposition 3. For every positive integer n, if ¿(SJ) = 3, then the palettes for vertices with degree 4 should differ in only one color.
Proof. All possible cases for two palettes with degree 4 are described below:
1) {1, 2, 3,4} and {1, 2, 3, 5};
2) {l, 2, 3,4} and {l, 2, 5, ô};
3) {l, 2, 3,4} and {l, 5, 6, 7};
4) {l, 2, 3,4} and {5, 6, 7, 8}.
In Fig. 2, we show an example of the coloring of Sf with three palettes, where the palettes of cardinality 4 correspond to the first case.
We now show that the last three cases are impossible. Suppose we have colored Sf with three palettes. If s(Sn) = 3, then we have only one palette with cardinality 2. Let's denote that palette by {a, b}, where 1 ^ a < b ^ 8. We will call the colors 5, 6, 7, 8 the extra colors. Now, we change each extra color c with the color 9 — c from the set {1, 2, 3,4}. The proper edge coloring of Sf might be broken only on edges adjacent to the vertices of degree 2 if b is an extra color and 9 — b = a. But in cases 2-4, we can change the color b with another color different from a. So for the cases 2-4, we can color Sf with two palettes, which is a contradiction to Proposition 2. ■
By Proposition 3 and its proof, we may assume that if we have a coloring with three palettes, then the palettes are {1, 2, 3,4}, {1, 2, 3, 5}, and {4, 5}. Moreover, it is possible to use the palette {1, 2, 3, 5} exactly three times.
Let us now expose all possible colorings of Sf, where the palettes {1, 2, 3,4}, {1, 2, 3, 5}, and {4, 5} are used. After this, we will have this group of sets of palettes, and we will call this group C:
1) {{1,2, 3,4}, {1, 2, 3, 5}, {4, 5}, {1, 2}, {3,4}};
2) {{1,2, 3,4}, {1, 2, 3, 5}, {4, 5}, {1, 3}, {2,4}};
3) {{1,2, 3,4}, {1, 2, 3, 5}, {4, 5}, {1,4}, {2, 3}}.
To find colorings with three palettes of Sf, as a set of palettes of Sf, it is enough for us to use only sets of palettes of groups A and C, as we can use the palette {1,2,3,5} three times.
There is no coloring of Si with three palettes; thus, s(Sl) = 4. As shown in Fig. 2, we can color Sf by coloring each iSl using a set of palettes from group C, where i = 0,1, 2.
Let us see what sets of palettes we can have for by coloring each iSf using a set of palettes from group A and C, where i = 0,1, 2. We have three cases shown in Fig. 3.
b
Fig. 3. Cases a, b, c
3 4 c
For the case a, the possible combination is the following: all iSl have the same set of palettes from group C, where i = 0,1, 2, and that gives us a coloring with three palettes. Let us notice that we can not use this coloring of Si for coloring a subgraph of Sf, where n > 2, because we do not have a palette with cardinality 4 that has colors 4 and 5. The case b does not give a coloring with three palettes, and we can not use it for coloring Sf, where n > 2, because there are two vertices with the palette {4,5}. The case c gives a new group of sets of palettes, and we will call this group D:
1) 2)
3)
4)
5)
6)
1, 2, 3,4}, {1, 2, 3, 5} 1, 2, 3,4}, {1, 2, 3, 5} 1, 2, 3,4}, {1, 2, 3, 5} 1, 2, 3,4}, {1, 2, 3, 5} 1, 2, 3,4}, {1, 2, 3, 5} 1, 2, 3,4}, {1, 2, 3, 5}
{4, 5} {4, 5} {4, 5} {4, 5} {4, 5} {4, 5}
{1,2}} {1,3}} {1,4}} {2,3}} {2,4}}
{3,4}}.
Let us now see what sets of palettes we can have for Sf by coloring each iSf using a set of palettes from group B and D, where i = 0,1, 2. We have three cases (Fig. 4).
b
Fig. 4. Cases a, b, c
2 4
c
For the case a, there is no possibility to construct a proper edge coloring. The case b does not give a coloring with three palettes, and we can not use it for coloring Sf, where n > 3, because there are two vertices with the palette {4, 5}. And finally, the case c gives the same sets of palettes of group C that close the chain. Thus, depending on whether the number n > 0 is even or odd, we can color Sf with the palettes of any set of groups C or D, respectively.
Theorem 2. For every non-negative integer n, the palette index of Sf is determined by the formula
s(£f) = i3, if n iseven
I 4, if n is odd.
a
a
Proof. Clearly, 8(S°) = 3, so we will consider the cases when n > 0. We have shown that we can color Sf using any set of palettes from groups A and C if n is odd, and from groups B and D if n is even. Also, we have shown that we can color Sf with three palettes if each iSf is colored with palettes of a set of group C, where i = 0,1, 2. Since we have considered all possible cases of coloring Sf with three palettes, and it is
3 is colored using a set oi palettes nom group c, then s(Sf
w' f
possible only if iSf is colored using a set of palettes from group C, then S^?^ > 3 for
odd n. Consequently, by Proposition 1, 8(sU) = 4 if n is odd.
3. The palette index of Sierpinski graphs
In this section, we will examine the palette index of Sierpinski graphs S^", where p > 1.
If n = 1, then Sp" is isomorphic to Kp, whose palette index was determined (Theorem 1). Now, let us consider the cases when n > 1. Here, we have two kinds of vertices in terms of vertex degree: extreme vertices with degree n — 1 and the remaining vertices with degree n. So 8^) ^ 2.
Theorem 3. For every even integer p > 1 and every integer n > 1, we have
s(SJU ) = 2.
Proof. Since s(SJU) ^ 2, we just need a coloring of Spf with two palettes. We color all pu-1 copies of Sp = Kp in Spf with the palette {1,... ,p — 1}. Then we color all link edges with the color p. In this way all extreme vertices have the palette {1,... ,p — 1}, and the other vertices have the palette {1,... ,p}. ■
Now, consider the case when p is odd.
Proposition 4. For every odd integer p > 1 and every integer n > 1, we have
8(Sf) ^ 3.
Proof. In [13, Claim in Theorem 4.1], it has been proved for any integer n ^ 1 and any odd integer p > 1, that x'(SjU) = p and the palettes of extreme vertices of Spf are pairwise different. If we try to color Spf with two palettes, then we must use a single palette for coloring vertices with degree p, which means that we use only p colors. So we will have p palettes for extreme vertices [13, Claim in Theorem 4.1]. Hence, 8 (Spf) ^ 3. ■
Proposition 5. For every odd integer p > 1 and every integer n > 1, we have
8(Spu) ^
i 8(Kp), if n = 2,
8 (Kp) + 1, if n > 2.
Proof. As shown in [2], if p is odd, then we can always color Kp with three or four palettes, and there is only one vertex s with a unique palette Ps. For a coloring Sp, we color each complete graph iSp, and we keep the vertex s as the extreme vertex (0 ^ i ^ p — 1). At this moment, we have these palettes: Ps and Pl,...,Pm, where m is 2 or 3. Then we color all link edges with a new color c. Thus, we have these 8 (Kp) palettes: Ps and P1 U {c},..., Pm U {c}. To color Sp, we color each iSp as mentioned above, and again, we color uncolored link edges with the same color c. Thus, we have 8 (Kp) + 1 palettes: Ps and Ps U {c}, P1U {c},..., PmU {c}. To color Sp4, where n > 3, we can do the same steps. In that case, we do not create a new palette. ■
Corollary 1. For every integer p, we have
,4p+3
8 (S4p+3) = 3.
Proof. This result follows from Proposition 4, Proposition 5, and Theorem 1. ■ Theorem 4. For every positive integer n, we have
) = 3.
Proof. By Proposition 4, we obtain s(S3^) ^ 3. So we just need a coloring with three palettes to prove the theorem. If n =1, then we have K3, whose palette index is 3. If n = 2, we can color Sf as shown in Fig. 5.
Fig. 5. A coloring of S2 with three palettes Now, let us consider the two colorings of S| from Fig. 6.
Fig. 6. Group A
Let us denote this group of sets of palettes by A:
1) {{1, 2, 3}, {1, 2,4}, {1, 2}, {3, 4}};
2) {{1, 2, 3}, {1, 2,4}, {2, 3}, {3, 4}}.
Next, let us consider the coloring of Sf in Fig. 7.
Fig. 7. Group B
Figure 7, a gives a coloring of S|. We denote this single element group of sets of palettes by B:
1) {{1,2, 3}, {1, 2}, {1, 3}, {2, 3}}.
b
a
Figure 7, b shows how we can color Sf using palettes from group B, where n > 1.
In Fig. 8, we can see that, for any integer n ^ 2, we can color Sf using each set of palettes from group A by coloring one of iS3f using a set of palettes from group A, two others using a set of palettes from group B, and link edges that connect these iSf with colors 1, 2 or 3, where i = 0,1, 2. We can also see that for Sf we can color it with three palettes by coloring all iS3f using a set of palettes from group A, and link edges that connect these iSf with colors 1 or 3, where i = 0,1, 2. The theorem is proved. ■
Fig. 8. The coloring of Sf with three palettes
Acknowledgement. The author thanks P. A. Petrosyan for many useful comments and suggestions.
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