Article No. ej970199
Circular Chromatic Numbers and Fractional Chromatic Numbers of
Distance Graphs
GERARDJ. CHANG†, LINGLINGHUANG† ANDXUDINGZHU‡ This paper studies circular chromatic numbers and fractional chromatic numbers of distance graphs G(Z, D) for various distance sets D. In particular, we determine these numbers for those D sets of size two, for some special D sets of size three, for D= {1, 2, . . . , m, n} with 1 ≤ m < n, for D= {q, q + 1, . . . , p} with q ≤ p, and for D = {1, 2, . . . , m} − {k} with 1 ≤ k ≤ m.
c
1998 Academic Press
1. INTRODUCTION
Suppose S is a subset of a metric spaceM with a metric δ, and D a subset of positive real numbers. The distance graph G(S, D), with a distance set D, is the graph with vertex set
S in which two vertices x and y are adjacent iffδ(x, y) ∈ D. Distance graphs, first studied
by Eggleton et al. [7], were motivated by the well-known plane-coloring problem: What is the minimum number of colors needed to color all points of a euclidean plane so that points at unit distances are colored with different colors. This problem is equivalent to determining the chromatic number of the distance graph G(R2, {1}). It is well-known that the chromatic number of this distance graph is between 4 and 7 (see [12, 15]). However, the exact number of colors needed remains unknown.
For distance graphs on the real line R or the integer set Z , the problem of finding the chromatic numbers of G(R, D) or G(Z, D) for different D sets has been studied extensively (see [3, 10, 13, 14, 17, 18, 20, 22]). Two recent papers [3, 14] related distance graphs to the T -coloring problem. Chromatic numbers and fractional chromatic numbers of distance graphs were used to derive bounds for T -spans of the corresponding T -colorings, and vice versa. In this paper, we study circular chromatic numbers and fractional chromatic numbers of distance graphs G(Z, D) for various D sets.
The circular chromatic number of a graph is a natural generalization of the chromatic number of a graph, introduced by Vince [16] under the name the ‘star chromatic number’ of a graph. Suppose p and q are positive integers such that p ≥ 2q. A (p, q)-coloring of a graph
G = (V, E) is a mapping c from V to {0, 1, . . . , p − 1} such that kc(x) − c(y)kp ≥ q for
any edge x y in E, wherekakp= min{a, p − a}. The circular chromatic number χc(G) of G
is the infimum of the ratios p/q for which there exist (p, q)-colorings of G.
Note that 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. On the other hand, it has been shown [16] that for all graphs G, we haveχ(G) − 1 < χc(G). Therefore, χ(G) = dχc(G)e. 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. Thus χc(G) is a refinement of χ(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 main results of this article determine the circular chromatic numbers of various distance graphs. These results may be viewed as improvements
†Supported in part by the National Science Council under grant NSC86-2115-M009-002. ‡Supported in part by the National Science Council under grant NSC87-2115-M110-004.
on previous results concerning the chromatic numbers of these distance graphs presented in [3, 4, 7, 13, 17, 21].
The fractional chromatic number of a graph is another well-known variation of the chromatic 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 in 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). (∗)
For simplicity, letω(S, D), α(S, D), χf(S, D), χc(S, D) and χ(S, D) denote, respectively,
the clique number, the independence number, the fractional chromatic number, the circular chromatic number, and the chromatic number of a distance graph G(S, D).
Chromatic numbers of distance graphs with distance sets|D| ≤ 2 were determined by Chen
et al. [4] and Voigt [17]. Chromatic numbers of distance graphs with|D| = 3 were determined
by Zhu [21]. In Section 2, we use a ‘multiplier method’ to establish an upper bound for the circular chromatic number of a distance graph G(Z, D) with an arbitrary distance set D. This upper bound is then used to determine the circular chromatic numbers and the fractional chromatic numbers of those distance graphs with distance sets D for|D| = 2, for some special
D with|D| = 3, for D = {1, 2, . . . , m, n} with 1 ≤ m < n, and for D = {q, q + 1, . . . , p}
with q≤ p. The chromatic number for G(Z, D) with D = {q, q + 1, . . . , p} was determined in [7, 13].
Chromatic numbers of distance graphs with distance sets of the form Dm,k = {1, 2, . . . , m}−
{k}, with 1 ≤ k ≤ m, were studied in [3, 7, 13, 14]. Partial results concerning chromatic numbers of such distance graphs were obtained in [7, 13, 14], and a complete solution was recently obtained by Chang et al. [3]. The authors of [3] also obtained circular chromatic numbers of such distance graphs for some special values of m and k. In Section 3, we determine the circular chromatic numbersχc(Z, Dm,k) for all integer pairs m, k.
2. MULTIPLIERMETHOD FORχf(Z, D)ANDχc(Z, D)
In this section we use a ‘multiplier method’ to establish an upper bound onχc(Z, D) for an
arbitrary D set. We then use this upper bound to determine circular chromatic numbers for some D sets.
The multiplier method was used in [2] to study the density of D-sets, and was also used in [11] to study fractional chromatic numbers and circular chromatic numbers of circulant graphs. In taking distance graphs to be ‘infinite’ circulant graphs, Theorem 2.2 is parallel to a result in [11]. Half of the proof of Theorem 2.3 is parallel to an argument in [2].
LEMMA2.1. Suppose D is a set of positive integers, and that p and r are positive integers. Let
dD(p, r) = min{kri mod pkp: i ∈ D}. If dD(p, r) ≥ 1, then χc(Z, D) ≤ p/dD(p, r).
PROOF. It is straightforward to verify that the coloring defined as c(i) = (ri mod p) for
Let fD = inf{p/dD(p, r) : dD(p, r) ≥ 1}. The function is well defined since dD(p, r) is
always an integer between 0 andbp/2c. Theorem 2.2 follows from Lemma 2.1.
THEOREM2.2. For any set D of positive integers,χc(Z, D) ≤ fD.
It is known [4, 17] that if D contains exactly two relatively prime integers, thenχ(Z, D) = 2 when the two integers are odd andχ(Z, D) = 3 when the two integers have different parities. We first use fDto determineχc(Z, D) and χf(Z, D) for D with |D| = 2.
THEOREM2.3. If D= {a, b} and gcd(a, b) = 1, then
χf(Z, D) = χc(Z, D) = fD= (a + b)/b(a + b)/2c.
PROOF. Suppose both a and b are odd. Since 2≤ ω(Z, D) and dD(2, 1) = 1, the theorem
follows from(∗) and Theorem 2.2.
Suppose that a and b have different parities, i.e., a+ b is odd. Assume that a + b = p. Since gcd(p, b − a) = 1, there exists a positive integer r such that r(b − a) ≡ 1 (mod p). Since r(b + a) ≡ 0 (mod p), it follows that 2rb ≡ −2ra ≡ 1 (mod p). Hence, ra ≡ −rb ≡ (p − 1)/2 (mod p), which implies that dD(p, r) = (p − 1)/2. Hence, according to Theorem
2.2,χc(Z, D) ≤ fD ≤ 2p/(p − 1) = (a + b)/b(a + b)/2c. On the other hand, it is easy to
see that G(Z, D) contains the odd cycle Cp. Thus, 2 p/(p − 1) ≤ p/α(Cp) ≤ χf(Cp) ≤
χf(Z, D) ≤ χc(Z, D). This completes the proof of the theorem. 2
Note that precisely the same arguments in the first two lines of the proof above also give that χf(Z, D) = χc(Z, D) = fD = 2 if D contains only odd integers.
We now consider circular chromatic numbers and fractional chromatic numbers of distance graphs G(Z, D) with |D| = 3. Zhu [21] proved the following result for chromatic numbers, which provides a range for circular chromatic numbers.
THEOREM2.4 ([21]). If D = {a, b, c}, where a < b < c are positive integers with
gcd(a, b, c) = 1, then χ(Z, D) = 2, if a, b, c are odd,
4, if a= 1 and b = 2 and c ≡ 0 (mod 3),
4, if a+ b = c and a 6≡ b (mod 3),
3, otherwise.
THEOREM2.5. If D= {a, a + 1, c}, with a + 1 < c, where c + a = (2a + 1)k + r, with k≥ 1 and 0 ≤ r ≤ 2a, then
χf(Z, D) ≤ χc(Z, D) ≤ fD≤
(
(c + a)/(ak), if 0≤ r ≤ a,
(c + a + 1)/(ak + r − a), if a+ 1 ≤ r ≤ 2a.
PROOF. Note that ck≡ −ak (mod c + a) and c(k + 1) ≡ −(a + 1)(k + 1) (mod c + a + 1). Therefore, dD(c + a, k) = ak for all r, and dD(c + a + 1, k + 1) = ak + r − a when
a+ 1 ≤ r ≤ 2a. The theorem then follows. 2
THEOREM2.6. If D = {a, a + 1, c} with a + 1 < c and c + a ≡ 2a or 0 (mod 2a + 1), thenχf(Z, D) = χc(Z, D) = fD= 2 + 1/a.
PROOF. Since G(Z, D) contains the odd cycle C2a+1, according to(∗), 2 + 1/a = (2a +
1)/α(C2a+1) ≤ χf(C2a+1) ≤ χf(Z, D). On the other hand, since c +a ≡ 2a or 0 (mod 2a +
1), it follows from Theorem 2.5 that fD ≤ 2 + 1/a. 2
Denote the subgraph of G(Z, D) induced by Vi = {0, 1, · · · , i} as Gi.
THEOREM2.7. If D = {2, 3, c}, with 3 < c, where c + 2 = 5k + r, with k ≥ 1 and
0≤ r ≤ 4, then χf(Z, D) = χc(Z, D) = fD = (c + 2)/2k, if r= 1, 2, (c + 3)/(2k + 1), if r= 3, 5/2, if r= 4, 0.
PROOF. The case in which r = 4 or 0 follows from Theorem 2.6. For the other cases, Theorem 2.5 implies that fD ≤ (c + 2)/2k when r = 1, 2, and fD≤ (c + 3)/(2k + 1) when r = 3. Therefore it suffices to show that α(Gc+1) ≤ 2k when r = 1, 2 and α(Gc+2) ≤ 2k + 1
when r= 3.
Consider the graph Gc+2for r = 1, 2, 3. Decompose the vertex set {0, 1, · · · , c+2} into k+1
subsets Ii = {5i, 5i+1, . . . , 5i+4} for 0 ≤ i ≤ k−1, and J = {5k, . . . , 5k+r = c+2}. Then, J = {c+1, c+2} when r = 1, J = {c, c+1, c+2} when r = 2, and J = {c−1, c, c+1, c+2}
when r = 3. Suppose that Gc+2has an independent set S of size 2k+ 2. We may assume
that 0∈ S and then c /∈ S. Since every five consecutive vertices in Gc+2form a 5-cycle, we
conclude that|Ii∩ S| = |J ∩ S| = 2 for 0 ≤ i ≤ k − 1. Then c + 1 ∈ S, and hence, 1 /∈ S.
Since|I0∩ S| = 2, 2 and 3 are not in S. We therefore conclude that 4 ∈ S. In a general
step, using the fact that|Ii ∩ S| = 2 and 5(i − 1) − 1 ∈ S, it is straightforward to derive that
5i− 1 ∈ S. Therefore, 5k − 1 ∈ S. Since 5k − 1 = c when r = 1, and 5k − 1 is adjacent to
c+ 1 when r = 2 or 3, we have contradictions. Hence, α(Gc+2) ≤ 2k + 1 for r = 1, 2, 3.
Moreover, for the case in which r= 1 or 2, any independent set S0of Gc+2of size 2k+ 1 that
contains the vertex 0 does not contain the vertex c+ 1. Hence, c + 2 ∈ S0andα(Gc+1) ≤ 2k.
This completes the proof of the theorem. 2
THEOREM2.8. Suppose D= {a, b, a + b}, with 0 < a < b and gcd(a, b) = 1. If a ≡ b
(mod 3), then χf(Z, D) = χc(Z, D) = fD= 3.
PROOF. Since gcd(a, b) = 1 and a ≡ b (mod 3), we have a, b, c 6≡ 0 (mod 3) and so
dD(3, 1) = 1. The theorem then follows from (∗) and the fact that {0, a, a + b} is a clique. 2
THEOREM2.9. If D= {1, 2, · · · , m, n}, with 1 ≤ m < n, then χf(Z, D) = χc(Z, D) = fD=
(
m+ 1, if n6≡ 0 (mod m + 1), m+ 1 + 1/k, if n= k(m + 1).
PROOF. Suppose n 6≡ 0 (mod m + 1). Since m + 1 ≤ ω(G) and dD(m + 1, 1) = 1, the
theorem follows from(∗) and Theorem 2.2.
Suppose n= k(m + 1). Since every independent set of Gncontains at most one vertex from
any m+1 consecutive vertices, and at most one vertex from {0, n}, α(Gn) = k. Consequently, m+ 1 + 1/k = (n + 1)/α(Gn) ≤ χf(Gn) ≤ χf(Z, D). Also, fD ≤ (n + 1)/dD(n + 1, k) =
m+ 1 + 1/k. The theorem then follows. 2
COROLLARY2.10. If D = {1, 2, 3k}, where k ≥ 1, then χf(Z, D) = χc(Z, D) =
TABLE1. Conditions of a, b, c χf(Z, D), χc(Z, D), fD χ(Z, D) a, b, c are odd 2 2 a= 1, b = 2, c = 3k 3+1k (Corollary 2.10) 4 c= a + b, a 6≡ b (mod 3) ? 4 c= a + b, a ≡ b (mod 3) 3 (Theorem 2.8) b= a + 1, c ≡ a or a + 1 (mod 2a + 1) 2+1a (Theorem 2.6) r= 1, 2 c2k+2(Theorem 2.7) 3 a= 2, b = 3, c + 2 = 5k + r r = 3 2kc+3+1(Theorem 2.7) Otherwise ?
This is one of the two cases covered by Theorem 2.4 in which we haveχ(Z, D) = 4. The other is that in which D = {a, b, c}, a + b = c and a 6≡ b (mod 3). We note that, in this case, the chromatic number of G(Z, D) is easily determined. However, the circular chromatic numbers of G(Z, D) are still unknown, except for some special values of a and b.
We summarize the results for D= {a, b, c} with a < b < c and gcd(a, b, c) = 1 in Table 1.
THEOREM2.11. If D = {q, q + 1, . . . , p}, with q ≤ p, then χf(Z, D) = χc(Z, D) = fD= 1 + p/q.
PROOF. Since dD(p + q, 1) = q, we conclude that fD≤ (p + q)/q. On the other hand, it
is quite obvious thatα(Gp+q−1) = q. Hence, χf(Z, D) = χc(Z, D) = fD = 1 + p/q. 2
THEOREM2.12. If D= [1, r], where r is any real number greater than or equal to 1, then
χf(R, D) = χc(R, D) = 1 + r.
PROOF. We first consider the case in which r= p/q is rational. Let D0= {q, q+1, . . . , p}. It is then straightforward to verify that each connected component of G(R, D) is isomorphic to G(Z, D0). According to Theorem 2.11, χf(R, D) = χc(R, D) = 1 + r.
When r is irrational, then let(ri : i = 1, 2, . . . ) and (ri0 : i = 1, 2, . . . ) be sequences of
rational numbers such that ri0≤ r ≤ rifor each i and limi→∞ri0= limi→∞ri = r. The above
argument then shows that 1+ ri0 ≤ χf(R, D) ≤ χc(R, D) ≤ 1 + ri for each i . Therefore,
χf(R, D) = χc(R, D) = 1 + r. 2
It was shown by Eggleton et al. [10] (Theorem 2) that if a prime distance graph has a proper
k-coloring, then it has a periodic k-coloring. The proof in fact shows that any k-colorable
distance graph has a periodic k-coloring. We remark that an argument parallel to the proof of Theorem 2 of [10] shows that if a distance graph G(Z, D) has a (p, q)-coloring, then it has a periodic(p, q)-coloring. Also we note that a (p, q)-coloring derived by the multiplier method is always a periodic(p, q)-coloring.
3. CIRCULARCHROMATICNUMBERχc(Z, Dm,k)
As mentioned in the introduction, Chang et al. [3] determined the chromatic number and the fractional chromatic number of the distance graph G(Z, Dm,k), where Dm,k= {1, 2, . . . , m}−
TABLE2. Conditions of m, k, r, s χf(Z, Dm,k) χc(Z, Dm,k) χ(Z, Dm,k) 2k> m k k k r > s m+k+12 gcd(m + k + 1, k) = 1 m+k+1 2 2k≤ m r = 0 gcd(m + k + 1, k) 6= 1 m+k+1 2 m+k+2 2 1≤ r ≤ s ? m+k+32
{k} and 1 ≤ k ≤ m. They also determined the circular chromatic number of G(Z, Dm,k) for
some pairs of integers m and k.
Let m+ k + 1 = 2rm0and k = 2sk0, where m0and k0are both odd. Table 2 shows their results.
The circular chromatic numbersχc(Z, Dm,k) remain unknown for those pairs of integers m, k corresponding to the question mark in Table 2. In this section, we shall fill in the
unknown part of Table 2 by showing thatχc(Z, Dm,k) = m+k+22 when 2k≤ m and r ≤ s and
gcd(m + k + 1, k) 6= 1.
The following lemma was proven in [16] and is used frequently in our proofs.
LEMMA3.1 ([16]). If G has a circular chromatic number pq(where p and q are relatively prime), then p≤ |V (G)|, and any (p, q)-coloring c of G is an onto mapping from V (G) to
{0, 1, . . . , p − 1}.
As in the preceding section, we denote the subgraph of G(Z, Dm,k) induced by Vi =
{0, 1, . . . , i} as Gi. We shall first derive a lower bound forχc(Z, Dm,k).
LEMMA3.2. Suppose 2k≤ m. Let m +k +1 = 2rm0and k= 2sk0, where r and s are non-negative integers and m0and k0are odd integers. If 1≤ r ≤ s, then χc(Gm+2k−1) > m+k+12 .
PROOF. Since m+ k + 1 is even and χc(Gm+2k−1) > χ(Gm+2k−1) − 1, it suffices to show
thatχ(Gm+2k−1) > m+k+12 . Assume to the contrary thatχ(Gm+2k−1) ≤ m+k+12 , and that c
is am+k+12 -coloring of Gm+2k−1.
For each integer i with 0≤ i ≤ k − 2, consider the subgraph of Gm+2k−1induced by the m+ k + 1 vertices {i, i + 1, . . . , i + m + k}. This graph has an independence number 2.
Therefore, each of them+k+12 colors is used at most, and thus exactly, twice in this subgraph. Consequently, vertices i and i+ m + k + 1 have the same colors for all 0 ≤ i ≤ k − 2. Therefore, for each j ∈ S := {0, 1, . . . , m + k}, the only possible vertices in S having the same color as j are j+ k and j − k.
Consider the circulant graph C(m + k + 1, k), with vertex set S and in which vertex i is adjacent to vertex j iff j≡ i +k or i −k (mod m +k +1). It follows from the discussion in the preceding paragraph that two vertices x and y in S have the same color only if x y is an edge of the circulant graph C(m + k + 1, k). Since the intersection of each color class with S contains exactly two vertices, the coloring induces a perfect matching of C(m + k + 1, k). However,
Since C(m + k + 1, k) has a perfect matching, each cycle has an even length. This implies that r> s, contrary to the assumption r ≤ s. Hence, χ(Gm+2k−1) > m+k+12 . 2
LEMMA3.3. Suppose 2k ≤ m. If m + k + 1 is odd and gcd(m + k + 1, k) 6= 1, then
χc(Gm+k) > m+k+12 , and hence,χc(Gm+2k−1) > m+k+12 .
PROOF. First, it is clear thatχc(Gm+k) ≥ α(Gmm+k+1+k) = m+k+12 . Suppose χc(Gm+k) = m+k+1
2 . Since m+k +1 and 2 are relatively prime, every (m +k +1, 2)-coloring c of Gm+kis
onto and hence is one-to-one; i.e., there exists an ordering x0, x1, x2, . . . , xm+kof Vm+ksuch
that c(xi) = i for 0 ≤ i ≤ m + k. Therefore, X = (x0, x1, . . . , xm+k, x0) is a cycle in the
complement G0of Gm+k.
Let m= ak + b, where 0 ≤ b < k. Since all vertices of {k − 1, k, . . . , m + 1} are of degree two in G0, the following paths must be on the cycle X :
Pi : i, k + i, 2k + i, . . . , ak + i, (a + 1)k + i for 0≤ i ≤ b; Pj : j, k + j, 2k + j, . . . , ak + j for b+ 1 ≤ j ≤ k − 1.
For each vertex u, let N(u) = {v ∈ Vm+k: uv ∈ E(G0)}. Since N(k −1) = {2k −1, m +k}
and m+ k = (a + 1)k + b, we have that PbPk−1is a path of the cycle X . Since N(k − 2) =
{2k − 2, m + k − 1, m + k} and vertex m + k is on the path PbPk−1, we have that Pb−1Pk−2
is a path of the cycle X . Continuing this process, we have that Pt0 = Pb+1+tPt, where the
index b+ 1 + t is taken modulo k, is a path of the cycle X for 0 ≤ t ≤ k − 1. Since gcd(m + k + 1, k) 6= 1, we have gcd(b + 1, k) 6= 1. Therefore, these paths Pt0form at least 2 disjoint cycles, contrary to our assumption that X is a cycle. Thus, the coloring c does not exist andχc(Gm+k) > m+k+12 .
Since Gm+kis a subgraph of Gm+2k−1, we conclude thatχc(Gm+2k−1) > m+k+12 . 2
THEOREM3.4. Suppose 2k≤ m. Let m + k + 1 = 2rm0and k= 2sk0, where r and s are non-negative integers and m0and k0are odd integers. If r ≤ s and gcd(m + k + 1, k) 6= 1, thenχc(Z, Dm,k) ≥ m+k+22 .
PROOF. Supposeχc(Gm+2k−1) = qp, where p and q are relatively prime. Then, p ≤
|Vm+2k−1| = m + 2k and qp > m+k+12 according to Lemmas 3.2 and 3.3. If q ≥ 3, then p > q2(m + k + 1) ≥ 32(m + k + 1) > m + 2k, a contradiction. Hence, q ≤ 2 and so
χc(Z, Dm,k) ≥ qp ≥ m+k+22 . 2
Now we give an(m +k +2, 2)-coloring of G(Z, Dm,k) to show that χc(Z, Dm,k) ≤ m+k+22 .
We first give an(m + k + 2, 2)-coloring of Gm+k that is a variation of the coloring given
in Theorem 2.1 after a shift operation. It is then extended to an(m + k + 2, 2)-coloring of
G(Z, Dm,k).
LEMMA3.5. If 2k ≤ m, then Gm+khas an(m + k + 2, 2)-coloring c such that c(x) = c(x − k) + 1 for k ≤ x ≤ m + k.
PROOF. Suppose m+ k + 1 = dm0 and k = dk0, where gcd(m + k + 1, k) = d. Since gcd(m0, k0) = 1, there exists an integer n such that nk0 ≡ 1 (mod m0). Let ai = in (mod m0)
for 0≤ i ≤ m0−1. Consider the mapping c from Vm+kto{0, 1, . . . , dm0−1 = m+k} defined
by c(x) = ai + jm0, where x= id + (d − 1 − j), with 0 ≤ i ≤ m0− 1 and 0 ≤ j ≤ d − 1.
For any edge x y in Gm+k, we shall prove thatkc(x) − c(y)km+k+2 ≥ 2. Suppose to the
and y= i2d+(d −1− j2). For the case in which c(x) = c(y), we have ai1 = ai2and j1= j2,
which imply i1 = i2and x = y, a contradiction to xy being an edge. For the case in which c(x) = c(y) + 1, either (1) ai1 = ai2+ 1 and j1 = j2, or (2) ai1 = 0 and ai2 = m0− 1 and j1= j2+1. In subcase (1), we have i1≡ i2+k0(mod m0). Thus, x − y = k or y − x = m +1,
a contradiction. In subcase (2), we have i1= 0 and i2 = m0− k0. Thus, y− x = m + 2, a
contradiction. Similarly, it is impossible that c(x) + 1 = c(y). This completes the proof of
the lemma. 2
THEOREM3.6. If 2k≤ m, then χc(Z, Dm,k) ≤ m+k+22 .
PROOF. Let c be the coloring of Gm+k given in Lemma 3.5. Consider the mapping c0of G(Z, Dm,k) defined by c0(x) = c(x), for 0≤ x ≤ m + k, (c0(x − k) + 1) mod (m + k + 2), for m+ k + 1 ≤ x, (c0(x + k) − 1) mod (m + k + 2), for 0> x.
We now show that c0 is a proper (m + k + 2, 2)-coloring of G(Z, Dm,k) by induction.
According to Lemma 3.5, c0is proper in Gm+k. Suppose c0is proper in Gx−1for x ≥ m+k+1.
Let x y be an edge in Gx; i.e., y= x − i for some i ∈ Dm,k. First, c0(y) is not equal to c0(x)
mod(m +k +2) or (c0(x)−1) mod (m +k +2), since c0(y) ≡ c0(x)−2 (mod m +k +2) when
i= 2k, and y = x −i is adjacent to x −k in Gx−1when i6= 2k, where c0(x −k) = (c0(x)−1)
mod (m + k + 2). Also, c0(y − k) is not equal to c0(x) mod (m + k + 2), since x − k is adjacent to y− k in Gx−1and c0(x − k) = (c0(x) − 1) mod (m + k + 2). Hence, c0(y) is not
equal to(c0(x) + 1) mod (m + k + 2). By induction, c0is proper for non-negative vertices in
G(Z+, Dm,k). Similar arguments work for negative vertices. This completes the proof of the
theorem. 2
Combining Theorems 3.4 and 3.6 and results in [3], we have
THEOREM3.7. Suppose 2k≤ m. Let m + k + 1 = 2rm0and k= 2sk0, where r and s are non-negative integers and m0and k0are odd integers. If r ≤ s and gcd(m + k + 1, k) 6= 1, thenχc(Z, Dm,k) = m+k+22 ; otherwise,χc(Z, Dm,k) = m+k+12 .
ACKNOWLEDGEMENTS
The authors thank a referee for raising the problem of the existence of periodic (p, q)-colorings for distance graphs, which we answer at the end of Section 2.
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Received 8 May 1997 and accepted 21 November 1997 G. J. CHANG Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan E-mail: [email protected] LINGLINGHUANG† Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan E-mail: [email protected] XUDINGZHU Department of Applied Mathematics, National Sun Yat-sen University, Kaoshing 80424, Taiwan E-mail: [email protected]
