Article No. eujc.1999.0284
Available online at http://www.idealibrary.com on
Circular Chromatic Numbers of Distance Graphs with Distance Sets
Missing Multiples
†LINGLINGHUANG ANDGERARDJ. CHANG
Given positive integers m, k, s with m > sk, let Dm,k,srepresent the set {1, 2, . . . , m}\{k, 2k, . . . , sk}. The distance graph G(Z , Dm,k,s) has as vertex set all integers Z and edges connecting i and
j whenever |i − j| ∈ Dm,k,s. This paper investigates chromatic numbers and circular chromatic
numbers of the distance graphs G(Z, Dm,k,s). Deuber and Zhu [8] and Liu [13] have shown that
dm+sk+1s+1 e ≤ χ(G(Z , Dm,k,s)) ≤ dm+sk+1s+1 e + 1 when m ≥ (s + 1)k. In this paper, by establishing
bounds for the circular chromatic numberχc(G(Z, Dm,k,s)) of G(Z, Dm,k,s), we determine the
values ofχ(G(Z, Dm,k,s)) for all positive integers m, k, s and χc(G(Z, Dm,k,s)) for some positive
integers m, k, s. c
2000 Academic Press
1. INTRODUCTION
Given a set D of positive integers, the distance graph G(Z, D) has all integers as vertices, and two vertices are adjacent if and only if their difference is in D; that is, the vertex set is Z and the edge set is {uv : |u − v| ∈ D}. We call D the distance set. This paper studies chromatic and circular chromatic numbers of some distance graphs with certain distance sets. The circular chromatic number of a graph is a natural generalization of the chromatic num-ber of a graph, introduced by Vince [15] as the name “star chromatic numnum-ber.” Suppose p and q are positive integers such that p ≥ 2q. Let G be a graph with at least one edge. A(p, q)-coloring of G = (V, E) is a mapping c from V to {0, 1, . . . , p − 1} such that q ≤ |c(x) − c(y)| ≤ p − q for any edge x y in E. The circular chromatic number χc(G) of G is the infimum of the ratios p/q for which there exists a (p, q)-coloring of G.
Note that for p ≥ 2, a ( p, 1)-coloring of a graph G is simply an ordinary p-coloring of G. Therefore,χc(G) ≤ χ(G) for any graph G. Let G be a graph which is not a null graph. On the other hand, it has been shown [15] that for all finite graphs G, we haveχ(G) − 1 < χc(G). Applying a result of de Bruijn and Erd˝os [6], this can be proved also for infinite graphs. Therefore,χ(G) = dχc(G)e if G 6= Nn. In particular, two graphs with the same circular chromatic number also have the same chromatic number. However, two graphs with the same chromatic number may have different circular chromatic numbers. Thusχc(G) is a refinement ofχ(G), and it contains more information about the structure of the graph. It is usually much more difficult to determine the circular chromatic number of a graph than to determine its chromatic number.
The fractional chromatic number of a graph is another well-known variation of the chro-matic number. A fractional coloring of a graph G is a mapping c fromI(G), the set of all independent sets of G, to the interval [0, 1] such that P
x∈I ∈I(G)
c(I ) ≥ 1 for all vertices x of G. The fractional chromatic numberχf(G) of G is the infimum of the value P
I ∈I(G)
c(I ) of a fractional coloring c of G.
For any graph G, it is well known that
max{ω(G), |G|/α(G)} ≤ χf(G) ≤ χc(G) ≤ dχc(G)e = χ(G), (∗) †Supported in part by the National Science Council under grant NSC87-2115-M009-007.
where ω(G) (respectively, α(G)) is the clique (respectively, independence) number of G which is the maximum size of a pairwise adjacent (respectively, non-adjacent) vertex sub-set of V(G).
For simplicity, letω(S, D), α(S, D), χf(S, D), χc(S, D) and χ(S, D) denote the clique number, the independence number, the fractional chromatic number, the circular chromatic number and the chromatic number of the distance graph G(S, D), respectively.
For different types of distance sets D, the problem of determiningχ(Z, D) has been studied extensively, see Refs [4, 5, 7, 9, 10, 12, 16–19]. For instance, the case that D contains at most three integers were studied by Eggleton, Erd˝os and Skilton [9], Chen, Chang and Huang [5], Voigt [16], Deuber and Zhu [7], Zhu [18], and at last completely determined by Zhu [19].
Given positive integers m, k, s with m > sk, let Dm,k,sdenote the distance set {1, 2, . . . , m}\ {k, 2k, . . . , sk}. For s = 1, the chromatic number of G(Z , Dm,k,1) was first studied in [9, 12, 13] and finally completely determined by Chang, Liu and Zhu [4]. They also determined the fractional chromatic number of G(Z, Dm,k,1). The circular chromatic number of G(Z, Dm,k,1) was then determined by Chang, Huang and Zhu [2]. Recently, Liu and Zhu [14] determined the fractional chromatic number of G(Z, Dm,k,s) for a general s, which gives a lower bound ofχ(Z, Dm,k,s). Liu and Zhu [14], Deuber and Zhu [8] also studied χ(Z, Dm,k,s) for s = 2, 3, prime s + 1, and obtained some results for general s. Moreover, Deuber and Zhu [8] showed that for any values of m, k, s with m ≥ (s + 1)k, dm+sk+1
s+1 e ≤ χ(Z , Dm,k,s) ≤ dm+sk+1s+1 e + 1. In this paper, by establishing bounds for χc(Z, Dm,k,s), we determine the values ofχ(Z, Dm,k,s) for all positive integers m, k, s, and χc(Z, Dm,k,s) for some positive integers m, k, s.
Note that it becomes an easy case if m < (s + 1)k. Define a coloring f of G(Z, Dm,k,s) by: f(x) = x mod k for any x ∈ Z. As Dm,k,s contains no multiples of k, it can be easily verified that f is a proper coloring. Thus,χ(Z, Dm,k,s) ≤ k. As any consecutive k vertices in G(Z, Dm,k,s) form a clique, k ≤ ω(Z, Dm,k,s). This implies that all values in (∗) are equal to k for G = G(Z , Dm,k,s) if m < (s + 1)k (see Ref. [14]). Therefore, throughout the article, we assume m ≥ (s + 1)k.
The following table shows all results concerning the distance graph G(Z, Dm,k,s). Note that the value ofχf(Z, Dm,k,s) is determined in Ref. [14] and some value of χ(Z, Dm,k,s) is determined in Refs [8, 14]. Also, all values ofχc(Z, Dm,k,s) are given in this paper, which also implies the results ofχ(Z, Dm,k,s). Let
d = gcd(k, m + sk + 1), a = (m + sk + 1) mod d(s + 1), b = (m + sk + 1) mod(s + 1). Therefore, a = 0 means d(s + 1) | (m + sk + 1), a 6= 0 means d(s + 1) 6 | (m + sk + 1), b = 0 means (s + 1) | (m + sk + 1), b 6= 0 means (s + 1) 6 | (m + sk + 1).
Note that when m ≥ (s + 1)k with b 6= 0 and d > 1, we only know that m+sk+1s+1 ≤ χc(Z, Dm,k,s) ≤ m+sk+2s+1 , but still do not know the exact value ofχc(Z, Dm,k,s).
2. MAIN RESULTS
In the study of the chromatic number of the distance graphs G(Z, Dm,k,s) with distance sets Dm,k,s = {1, 2, . . . , m}\{k, 2k, . . . , sk}, Liu and Zhu [14] obtained the following result on fractional chromatic numbers, which asserts a lower bound for the circular chromatic numbers and chromatic numbers (by(∗)).
Conditions of parameters χf(Z, Dm,k,s) χc(Z, Dm,k,s) χ(Z, Dm,k,s) m< (s+1)k k[14] k[14] k[14] a = 0 m+sk+1s+1 [14] m+sk+1 s+1 d = 1 m ≥ (s+1)k b 6= 0 m+sk+1s+1 [14]
d
m+sk+1 s+1e
d > 1 ≤ m+sk+2 s+1 a 6= 0, b = 0 m+sk+2s+1 m+sk+1s+1 +1THEOREM1 ([14]). For positive integers m, k and s with m ≥ (s + 1)k, χf(Z, Dm,k,s) = m + sk + 1
s + 1 .
Liu and Zhu [14] also gave an upper bound ofχ(Z, Dm,k,s) as follows.
LEMMA2 ([14]). For positive integers m, k, s with m ≥ (s + 1)k and d = gcd(k, m + sk + 1), χ(Z, Dm,k,s) ≤ d & m + sk + 1 d(s + 1) ' .
By Lemma 2, it is clear that if d(s +1) | (m +sk +1), then χ(Z, Dm,k,s) ≤ m+sk+1s+1 . Hence we have
THEOREM3 ([14]). For positive integers m, k, s with m ≥ (s + 1)k and d = gcd(k, m + sk + 1), if d(s + 1) | (m + sk + 1), then
χc(Z, Dm,k,s) = χ(Z, Dm,k,s) = m + sk + 1 s + 1 .
Note that Theorem 3 only gives the values ofχc(Z, Dm,k,s) and χ(Z, Dm,k,s) under the condition d(s + 1) | (m + sk + 1), although for m+sk+1s+1 to be an integer we only need (s + 1) | (m + sk + 1).
Next, we show that if s + 1 divides m + sk + 1 but d(s + 1) does not, then χ(Z , Dm,k,s) > m+sk+1
s+1 . Let G[i, j] denote the subgraph of G(Z , Dm,k,s) induced by V [i, j] = {i, i + 1, . . . , j} for any integers i ≤ j.
LEMMA4. For positive integers m, k, s with m ≥ (s + 1)k and d = gcd(k, m + sk + 1), if(s + 1) | (m + sk + 1) but d(s + 1) 6 | (m + sk + 1), then
χc(G[0, m + sk + k − 1]) >m + sk + 1
s + 1 andχ(Z, Dm,k,s) >
m + sk + 1 s + 1 . PROOF. Sinceχc(G[0, m + sk + k − 1]) > χ(G[0, m + sk + k − 1]) − 1 and m+sk+1s+1 is an integer, it suffices to show thatχ(G[0, m + sk + k − 1]) > m+sk+1
sk + k − 1]) ≤ m+sk+1s+1 ; that is, G[0, m + sk + k − 1] has an m+sk+1s+1 -coloring f . For any integer 0 ≤ i ≤ k −1, the subgraph G[i, m +sk +i] has m +sk +1 vertices and independence number s + 1. Since f is anm+sk+1s+1 -coloring, each color class of f consists of exactly s + 1 vertices of G[i, m + sk + i]. It follows that f (i) = f (m + sk + i + 1) for any integer 0 ≤ i ≤ k − 2. Now, consider the color classes of f for the graph G[0, m + sk]. For each color class C = {x1, x2, . . . , xs+1}, where x1< x2< . . . < xs+1, the difference xi +1− xi of two consecutive vertices in C is called a gap. Note that there is at most one gap greater than m and all other gaps are equal to k. Suppose there is a gap greater than m + 1 and all other s − 1 gaps are equal to k. Let the first vertex x1= i and the last vertex xs+1= j. Then i ≤ k −2 and j ≥ m + (s − 1)k + i + 2, which imply that f (i) = f ( j) = f (m + sk + i + 1), contradicting 1 ≤ (m + sk + i + 1) − j ≤ k − 1. Therefore, all gaps are equal to k or exactly one gap is equal to m + 1 with the others equal to k. Then C is of the form {i, i + k, i + 2k, . . . , i + sk} (where each number is calculated modulo m + sk + 1).
Let u = m+sk+1d . Divide the vertex set of G[0, m + sk] into d subsets of the form {i, i + k, i + 2k, . . . , i + (u − 1)k} (mod m + sk + 1), each of size u. Then each of these d subsets is the union of some color classes of size s + 1, so s + 1 divides u, i.e., d(s + 1) | (m + sk + 1), a contradiction. Henceχ(G[0, m + sk + k − 1]) > m+sk+1
s+1 . 2
We then show thatχc(Z, Dm,k,s) ≤ m+sk+2s+1 for any positive integers m, k, s with m ≥ (s + 1)k. It follows that χ(Z, Dm,k,s) ≤ dm+sk+2s+1 e by (∗). Hence, χ(Z , Dm,k,s) = dm+sk+1s+1 e when(s + 1) 6 | (m + sk + 1), and χ(Z, Dm,k,s) = m+sk+1s+1 + 1 when (s + 1) | (m + sk + 1) but d(s + 1) 6 | (m + sk + 1). These, together with Theorem 3, give all values of the chromatic numbersχ(Z, Dm,k,s).
To calculate the upper bound ofχc(Z, Dm,k,s), we first give an (m +sk +2, s +1)-coloring c of the subgraph G[0, m + sk] and then extend it to an (m + sk + 2, s + 1)-coloring of G(Z, Dm,k,s). Intuitively, the coloring is the mapping c from V [0, m + sk] to {0, 1, . . . , m + sk} given in the following algorithm, although we in fact define it directly in the proof of Lemma 5. Algorithm. begin for j := 0 to m + sk do c( j) := −1; i := d − 1; c(i) := 0; repeat j := (i + k) mod (m + sk + 1); if c( j) 6= −1 then j := j − 1; c( j) := c(i) + 1; i := j ; until c(i) = m + sk end
LEMMA5. For positive integers m, k, s with m ≥ (s +1)k, there exists an (m +sk +2, s + 1)-coloring c of G[0, m + sk] such that c(x) = c(x − k) + 1 for k ≤ x ≤ m + sk.
PROOF. Suppose k = dk0and m + sk + 1 = dm0, where d = gcd(k, m + sk + 1). Since gcd(k0, m0) = 1, there exists an integer n such that k0n ≡ 1 (mod m0). Let ai = (in) mod m0
for 0 ≤ i ≤ m0− 1. Consider the mapping c from V [0, m + sk] to {0, 1, . . . , m + sk}, where
m + sk = dm0− 1, defined by
c(x) = aix + (d − 1 − jx)m0 for x = ixd + jx,
where 0 ≤ ix ≤ m0− 1 and 0 ≤ jx ≤ d − 1. Note that ix = bx/dc and jx = x mod d. It is straightforward to check that c is a one-to-one and hence onto mapping.
First, note that for k ≤ x ≤ m + sk, ix = ix−k+ k0 and jx = jx−k. Therefore, aix = (ixn) mod m0= (ix−kn + k0n) mod m0= aix−k+ 1 as ix 6= 0, and so c(x) = c(x − k) + 1.
Next, we show that s + 1 ≤ |c(x) − c(y)| ≤ (m + sk + 2) − (s + 1) for any edge x y in G[0, m + sk]. Let x = ixd + jx and y = iyd + jy, where 0 ≤ ix, iy ≤ m0− 1 and 0 ≤ jx, jy≤ d − 1. Without loss of generality, we may assume that c(x) > c(y).
Suppose 0< c(x) − c(y) ≤ s. Since m ≥ (s + 1)k, we have m0 ≥ (s + 1)k0 > sk0 > s. It follows that either (1) jx = jy and 0< aix − aiy ≤ s, or (2) jy = jx + 1 and m0− s ≤ aiy− aix < m0. In case (1), we have 0< (ix− iy)n mod m0≤ s. Hence ix− iy ≡ k0, 2k0, . . ., or sk0 (mod m0). It follows that x − y ≡ k, 2k, . . ., or sk (mod m + sk + 1), contradicting |x − y| ∈ Dm,k,s. In case (2), we have 0 ≤ aix ≤ s − 1 and m0− s ≤ aiy < m0. It follows that ix = 0, k0, 2k0, . . ., or (s − 1)k0, and iy = m0− k0, m0− 2k0, . . ., or m0− sk0. Hence iy− ix = m0− k0, m0− 2k0, . . ., or m0− sk0by m0− s ≤ aiy− aix < m0, which implies that y − x = (m + sk + 2) − k, (m + sk + 2) − 2k, . . ., or (m + sk + 2) − sk, a contradiction to y − x ∈ Dm,k,s. Therefore s + 1 ≤ c(x) − c(y).
Suppose m + sk + 2 − s ≤ c(x) − c(y) ≤ m + sk (note that m + sk is the largest color, also s ≥ 2). Since m0> s and m +sk +1 = dm0, we have that c(x)−c(y) ≥ (d −1)m0+ 2 and so jy− jx = d − 1, i.e., jx= 0 and jy = d − 1. Then m0− (s − 1) ≤ aix− aiy ≤ m0− 1. Hence 0 ≤ aiy ≤ s −2 and m0−(s −1) ≤ aix ≤ m0−1. It follows that iy = 0, k0, 2k0, . . ., or (s−2)k0, and ix = m0− k0, m0− 2k0, . . ., or m0− (s − 1)k0. Hence ix− iy= m0− k0, m0− 2k0, . . ., or m0−(s−1)k0by m0−(s−1) ≤ aix−aiy ≤ m0−1, which implies x −y = (m+sk+1)−k−(d− 1), (m +sk +1)−2k −(d −1), . . ., or (m +sk +1)−(s −1)k −(d −1) that is an integer larger than m + 1, contradicting |x − y| ∈ Dm,k,s. Therefore, c(x) − c(y) ≤ (m + sk + 2) − (s + 1). Thus, c is an(m + sk + 2, s + 1)-coloring of G[0, m + sk]. 2
THEOREM6. For positive integers m, k, s with m ≥ (s + 1)k, χc(Z, Dm,k,s) ≤ m + sk + 2
s + 1 .
PROOF. Let c be the (m + sk + 2, s + 1)-coloring of G[0, m + sk] given in Lemma 5. Consider the mapping c0: Z → {0, 1, . . . , m + sk + 1} defined by
c0(x) = c(x), for 0 ≤ x ≤ m + sk, (c0(x − k) + 1) mod(m + sk + 2), for x ≥ m + sk + 1, (c0(x + k) − 1) mod(m + sk + 2), for x < 0.
We show that c0 is a proper(m + sk + 2, s + 1)-coloring of G(Z, Dm,k,s) by induction. According to Lemma 5, c0 is proper in the subgraph G[0, m + sk]. Suppose c0is proper in G[0, x − 1] for x ≥ m + sk + 1. Let x y be any edge of G[0, x], i.e., y = x − i for some i ∈ Dm,k,s. Since x − k is adjacent to y − k in G[0, x − 1], by the induction hypothesis, s + 1 ≤ |c0(x − k) − c0(y − k)| ≤ (m + sk + 2) − (s + 1). It follows that s + 1 ≤ |c0(x) − c0(y)| ≤ (m + sk + 2) − (s + 1). Hence c0is proper in G(Z+, Dm,k,s) by induction. A similar argument works for negative vertices. Therefore, c0is a proper(m + sk + 2, s +
1)-coloring of G(Z, Dm,k,s). 2
According to(∗), Lemma 4, Theorems 1 and 6, we have the following values of the chro-matic numbers of the graphs G(Z, Dm,k,s) when d(s + 1) 6 | (m + sk + 1).
THEOREM7. Suppose m, k, s are positive integers with m ≥ (s +1)k and d = gcd(k, m + sk + 1). If (s + 1) 6 | (m + sk + 1), then χ(Z, Dm,k,s) = & m + sk + 1 s + 1 ' . If(s + 1) | (m + sk + 1) and d(s + 1) 6 | (m + sk + 1), then χ(Z, Dm,k,s) = m + sk + 1 s + 1 + 1.
The following lemma is useful in determining the circular chromatic numbers of the dis-tance graphs G(Z, Dm,k,s).
LEMMA8 ([15]). Ifχc(G) = p/q for any graph G, where p and q are relatively prime, then p ≤ |V (G)| and any ( p, q)-coloring of G is an onto mapping.
THEOREM9. For positive integers m, k, s with m ≥ (s + 1)k and d = gcd(k, m + sk + 1), if(s + 1) | (m + sk + 1) and d(s + 1) 6 | (m + sk + 1), then
χc(Z, Dm,k,s) = m + sk + 2 s + 1 .
PROOF. Supposeχc(G[0, m + sk + k − 1]) = p/q, where p and q are relatively prime. By Lemma 4, qp ≥ m+sk+1s+1 + 1q since(s + 1) | (m + sk + 1); and, by Lemma 8, p ≤ |V [0, m + sk + k − 1]| = m + (s + 1)k. Ifqp < m+sk+2s+1 , thenq1 < 1
s+1. Therefore q> s + 1, which implies that p > s+1q (m + sk + 1) ≥ s+2s+1(m + sk + 1) > m + (s + 1)k since m ≥ (s + 1)k, a contradiction. Hence, χc(Z, Dm,k,s) ≥ qp ≥ m+sk+2s+1 . By Theorem 6, we
haveχc(Z, Dm,k,s) = m+sk+2s+1 . 2
The next theorem determines the circular chromatic number of the distance graph G(Z, Dm,k,s) when k is relatively prime to m + sk + 1.
THEOREM10. For positive integers m, k, s with m ≥ (s + 1)k, if k is relatively prime to m + sk + 1, then
χc(Z, Dm,k,s) = m + sk + 1 s + 1 .
PROOF. By Theorem 1 and (∗), it suffices to show that χc(Z, Dm,k,s) ≤ m+sk+1s+1 , that is, G(Z, Dm,k,s) has an (m + sk + 1, s + 1)-coloring. Since k is relatively prime to m + sk + 1, there exists an integer n such that nk ≡ 1 (mod m + sk + 1). Consider the map-ping c defined by c(i) = (in) mod(m + sk + 1) for all i ∈ Z. Choose any edge i j of G(Z, Dm,k,s). If 0 ≤ |c(i) − c( j)| ≤ s or (m + sk + 1) − s ≤ |c(i) − c( j)| ≤ m + sk, then c(i) − c( j) ≡ 0, 1, . . . , s, −1, −2, . . ., or −s (mod m + sk + 1). It implies that i − j ≡ 0, k, . . . , sk, −k, −2k, . . ., or −sk (mod m + sk + 1), contradicting |i − j| ∈ Dm,k,s. Thus, c is an(m + sk + 1, s + 1)-coloring of G(Z, Dm,k,s).2
We conclude that all valuesχc(Z, Dm,k,s) are determined except for the case when (s + 1) 6 |(m + sk + 1) and gcd(k, m + sk + 1) > 1.
ACKNOWLEDGEMENTS
The authors thank Xuding Zhu for proposing this problem and for pointing out a gap in the proof to Theorem 9 in an old version. They also thank the referee for many useful suggestions.
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Received 7 April 1998 in revised form 23 November 1998 LINGLINGHUANG
Department of Hospital and Health Care Administration, Chungtai Institute of Health Science and Technology, Taichung, Taiwan E-mail: lhuang@chtai.ctc.edu.tw
AND
GERARDJ. CHANG
Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan E-mail: gjchang@math.nctu.edu.tw
