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Radio-kk-Labeling of Cycles for Large kkthanks: Research is partially supported by the National Science Foundation grant DMS-1600778 and NASA NASA MIRO grant NX15AQ06A.

Colin Bloomfield, Daphne Der-Fen Liu   and Jeannette Ramirez California State University Los Angeles, currently Vanderbilt University; email:
[email protected] author.California State University Los Angeles, Los Angeles, USA; emails: [email protected] and [email protected].
Abstract

Let GG be a simple connected graph. For any two vertices uu and vv, let d(u,v)d(u,v) denote the distance between uu and vv in GG. A radio-kk-labeling of GG for a fixed positive integer kk is a function ff which assigns to each vertex a non-negative integer label such that for every two vertices uu and vv in GG, |f(u)f(v)|kd(u,v)+1|f(u)-f(v)|\geqslant k-d(u,v)+1. The span of ff is the difference between the largest and smallest labels of f(V)f(V). The radio-kk-number of a graph GG, denoted by rnk(G)rn_{k}(G), is the smallest span among all radio-kk-labelings admitted by GG. A cycle CnC_{n} has diameter d=n/2d=\lfloor n/2\rfloor. In this paper, we combine a lower bound approach with cyclic group structure to determine the value of rnk(Cn)\operatorname{\mathnormal{rn_{k}}}(C_{n}) for kn3k\geqslant n-3. For dk<n3d\leqslant k<n-3, we obtain the values of rnk(Cn)\operatorname{\mathnormal{rn_{k}}}(C_{n}) when nn and kk have the same parity, and prove partial results when nn and kk have different parities. Our results extend the known values of rnd(Cn)rn_{d}(C_{n}) and rnd+1(Cn)rn_{d+1}(C_{n}) shown by Liu and Zhu [15], and by Karst, Langowitz, Oehrlein and Troxell [10], respectively.

Keywords: Frequency assignment problem; radio labelling; radio-kk-labeling, radio-kk-number; cyclic groups.

1 Introduction

Radio-kk-labeling is motivated by the channel assignment problem which was first introduced by Hale [8]. In the channel assignment problem, we must both avoid signal interference between radio stations and minimize the range of frequencies used. Radio-kk-labeling models this problem with a graph where each station or transmitter is a vertex and neighboring stations are connected by an edge. The distance between two stations is then the usual distance between vertices on a graph, and a station’s assigned frequency is represented by a non-negative integer label of the corresponding vertex. If two stations are close enough for their signals to interfere, they must be assigned frequencies whose difference is determined by their proximity: The closer the stations are geographically, the larger their separation in assigned frequencies must be. The goal is to assign frequencies to all stations in a way that minimizes the range of the frequencies used while still avoiding interference.

Let GG be a graph. The distance between two vertices uu and vv, denoted by d(u,v)\operatorname{d}(u,v), is the length of a shortest path between uu and vv. The diameter of GG is the largest distance between any two vertices in V(G)V(G), which we denote by diam(G){\rm diam}(G) or sometimes dd when the graph is clear in the context.

For a positive integer kk, a function f:V(G){0,1,2,3,}f\colon V(G)\to\{0,1,2,3,\cdots\} is a radio-kk-labeling of a graph GG, if for all u,vV(G)u,v\in V(G), uvu\neq v,

|f(u)f(v)|kd(u,v)+1.|f(u)-f(v)|\geqslant k-\operatorname{d}(u,v)+1.

Sometimes we say ff is a labeling instead of radio-kk-labeling or omit mentioning the labeled graph GG if it is clear from the context. The span of a radio-kk-labeling ff of GG is defined as span(f)max{|f(u)f(v)|:u,vV(G)}.\operatorname{span}(f)\coloneqq\max\{|f(u)-f(v)|:u,v\in V(G)\}. We follow the convention of labeling at least one vertex 0, so the span of any radio-kk-labeling is the largest label assigned to a vertex vV(G)v\in V(G).

The radio-kk-number of GG is the minimum span of a radio-kk-labeling admitted by GG. We call a radio-kk-labeling ff of a graph GG optimal if the span is equal to the radio-kk-number. A special case for the radio-kk-labeling of a graph GG is when k=diam(G)=dk={\rm diam}(G)=d, in which a radio-dd-labeling is also known as radio labeling and the radio number of GG is denoted by rn(G)rn(G), where rn(G)=rnd(G)rn(G)=rn_{d}(G).

The notion of radio labeling was introduced by Chartrand, Erwin and Zhang [4]. The radio number for special families of graphs has been studied widely in the literature. Liu and Zhu [15] determined the radio number of cycles and paths. A general lower bound of the radio number of trees was given by Liu [13], and then applied by several authors to special families of trees (cf. [12, 7, 2, 14]). Khennoufa and Togni [11] studied the radio number for hypercubes by using generalized binary Gray codes. Zhou [20, 21, 22] and Martinez et al. [16] investigated the radio number for Cayley graphs and generalized prism graphs, respectively. Niedzialomski [18] studied radio graceful graphs (where GG admits a surjective radio labeling from V(G)V(G) to {1,2,,|V(G)|}\{1,2,\cdots,|V(G)|\}) and showed that the Cartesian product of tt copies of a complete graph is radio graceful for certain tt, providing infinitely many examples of radio graceful graphs of arbitrary diameter. For two positive integers m,n3m,n\geqslant 3, the toroidal grid Gm,nG_{m,n} is the Cartesian product of cycles CmC_{m} and CnC_{n}, CmCnC_{m}\Box C_{n}. Morris et al. [17] determined rn(Gn,n)rn(G_{n,n}), and Saha and Panigrahi [19] determined rn(Gm,n)rn(G_{m,n}) when mn0(mod2)mn\equiv 0\pmod{2}. Bantva and Liu [1] considered the radio number for block graphs and line graphs of trees.

When k=1k=1, radio-kk-labeling is equivalent to vertex coloring. Radio-kk-labelings of graphs for other values of kk have have also been studied extensively. There are many known results when k=2k=2 (cf. [6, 3]), this case is called L(2,1)L(2,1)-labeling, which can be generalized to L(h,k)L(h,k)-labeling. Recently, Chavez et al. studied optimal radio-kk-labeling for trees [5].

For cycles CnC_{n} with n3n\geqslant 3, the value of rnd(Cn)rn_{d}(C_{n}) was completely determined in [15]. Also, rnd1(Cn)rn_{d-1}(C_{n}) when n0(mod4)n\not\equiv 0\pmod{4} has been determined in [9], where bounds are given for the case when n0(mod4)n\equiv 0\pmod{4}. Karst, Langowitz, Oehrlein, and Troxell in [10] studied radio-kk-labeling when kdiam(Cn)k\geqslant{\rm diam}(C_{n}) and determined the value of rnk(Cn)rn_{k}(C_{n}) when k=diam(Cn)+1k={\rm diam}(C_{n})+1.

In this paper we investigate the values of rnk(Cn)rn_{k}(C_{n}) for all kdiam(Cn)k\geqslant{\rm diam}(C_{n}). In particular, we determine the values of rnk(Cn)rn_{k}(C_{n}) for kn3k\geqslant n-3, and for k<n3k<n-3 with nkn\equiv k (mod 2). For k<n3k<n-3 and knk\not\equiv n (mod 2), we establish lower and upper bounds and show these bounds are sharp for some values of kk and nn. A major tool we use is a combination of a lower bound approach (see 3) with some basic properties of finite cyclic groups.

2 Preliminary and Notations

In what follows, we let \mathbb{Z} denote the set of integers, \mathbb{N} the set of natural numbers and 0{0}\mathbb{N}_{0}\coloneqq\mathbb{N}\cup\{0\}. For n,mn,m\in\mathbb{Z}, such that nmn\leqslant m, we define [n,m]{n,n+1,,m}[n,m]\coloneqq\{n,n+1,\ldots,m\}. For nn\in\mathbb{N}, we let n\mathbb{Z}_{n} denote the cyclic group of order nn. For mnm\in\mathbb{Z}_{n}, let m\langle m\rangle be the subgroup of n\mathbb{Z}_{n} generated by mm. That is, m={tm:t}\langle m\rangle=\{tm:t\in\mathbb{Z}\} (mod nn). Let n/m\mathbb{Z}_{n}/\langle m\rangle denote the cyclic quotient group of n\mathbb{Z}_{n} modulo m\langle m\rangle, which is defined on the cosets i/m{jn:ijm}i/\langle m\rangle\coloneqq\{j\in\mathbb{Z}_{n}:i-j\in\langle m\rangle\} for ini\in\mathbb{Z}_{n}.

We regard CnC_{n} as a Cayley graph with vertex set n\mathbb{Z}_{n} and edges generated by ±1\pm 1. That is, V(Cn)={0,1,2,,n1}V(C_{n})=\{0,1,2,\cdots,n-1\} where uu and vv are adjacent if uv±1u\equiv v\pm 1 (mod nn). In the proofs we will often make use of an additive structure on V(Cn)=nV(C_{n})=\mathbb{Z}_{n} which is addition modular nn. The modest amount of group theory mentioned in this paper can be found in any introductory textbook on abstract algebra.

By definition, when kdiam(G)k\geqslant{\rm diam}(G), every radio-kk-labeling of GG is injective. So we can order the vertices of GG such that the labeling is strictly increasing. Thus, given a radio-kk-labeling ff of CnC_{n} with kn/2k\geqslant\lfloor n/2\rfloor, we denote the vertices of V(Cn)V(C_{n}), xix_{i} for i[0,n1]i\in[0,n-1] such that if i<ji<j, then f(xi)<f(xj)f(x_{i})<f(x_{j}) for i,j[0,n1].i,j\in[0,n-1]. For i[0,n2]i\in[0,n-2], we define the label gap and the distance gap respectively by

fif(xi+1)f(xi)anddid(xi+1,xi).f_{i}\coloneqq f(x_{i+1})-f(x_{i})\ \ \mbox{and}\ \ d_{i}\coloneqq d(x_{i+1},x_{i}).

These notations are used throughout the paper. By definition, fik+1dif_{i}\geqslant k+1-d_{i}.

Definition 1.

When defining a labeling, we first specify an ordering of the vertices x0,x1,,xn1x_{0},x_{1},\ldots,x_{n-1} of CnC_{n} by defining a jumping (or jump) sequence jij_{i} with 1jin21\leqslant j_{i}\leqslant\frac{n}{2}, i[0,n1]i\in[0,n-1]. This jumping sequence gives the “clockwise distance" of xi+1x_{i+1} from xix_{i}, namely, if xi=tnx_{i}=t\in\mathbb{Z}_{n}, then xi+1=t+jix_{i+1}=t+j_{i} (mod nn).

Definition 2.

Let 𝐣=(j0,,jt1)\mathbf{j}=(j_{0},\ldots,j_{t-1}), ji[1,n/2]j_{i}\in[1,\lfloor n/2\rfloor], i[0,t1]i\in[0,t-1]. For nn\in\mathbb{N}, we define the jump sequence generated by 𝐣\mathbf{j}, genn(𝐣)=(ai)i=0n1\operatorname{gen}_{n}(\mathbf{j})=(a_{i})_{i=0}^{n-1}, with a0=0a_{0}=0 and am=i=0m1jia_{m}=\sum_{i=0}^{m-1}j_{i}, where the indices of the jumps are taken modulo tt and the sums modulo nn. We call 𝐣\mathbf{j} a periodic jump sequence with tt jumps. We call the range of genn(𝐣)\operatorname{gen}_{n}(\mathbf{j}) the set generated by 𝐣\mathbf{j}, and denote it as Sn(𝐣)S_{n}(\mathbf{j}).

Here is an example of Definition 2.

Example 1.

Let n=12n=12 and 𝐣=(6,4)\mathbf{j}=(6,4), so t=2t=2. Then genn(𝐣)=gen12((6,4))=(a0,a1,\operatorname{gen}_{n}(\mathbf{j})=\operatorname{gen}_{12}((6,4))=(a_{0},a_{1},\dots, a11)=(0,6,10,4,8,2,6,0,4,10,2,8).a_{11})=(0,6,10,4,8,2,6,0,4,10,2,8). And Sn(𝐣)=S12((6,4))={0,2,4,6,8,10}.S_{n}(\mathbf{j})=S_{12}((6,4))=\{0,2,4,6,8,10\}.

Notice that in 1, the jump sequence does not determine a permutation and so does not define a labeling for CnC_{n}. In practice, we will define a jump sequence and then prove that it determines a permutation of the vertices of CnC_{n}.

Lemma 1.

Fix n3n\geqslant 3. Let 𝐣=(j0,j1)\mathbf{j}=(j_{0},j_{1}), h=n(j0+j1)h=n-(j_{0}+j_{1}), and H=hH=\langle h\rangle. Then

|Sn(𝐣)|{|H|if j0H,2|H|if j0H.|\operatorname{S}_{n}(\mathbf{j})|\leqslant\begin{cases}|H|&\text{if $j_{0}\in H$},\\ 2|H|&\text{if $j_{0}\not\in H$}.\end{cases}
Proof.

Note that in n\mathbb{Z}_{n}, H=h=h=j0+j1H=\langle h\rangle=\langle-h\rangle=\langle j_{0}+j_{1}\rangle. Hence j0+j1Hj_{0}+j_{1}\in H. If j0Hj_{0}\in H, then j1Hj_{1}\in H. Thus, Sn(𝐣)H\operatorname{S}_{n}(\mathbf{j})\subseteq H, so |Sn(𝐣)||H||\operatorname{S}_{n}(\mathbf{j})|\leqslant|H|.

If j0Hj_{0}\not\in H, we look at the first 2|H|2|H| vertices obtained from the jump sequence. The set of even and odd indexed terms, respectively, are

E={0,j0+j1,2(j0+j1),,(|H|1)(j0+j1)},O=j0+E.E=\{0,j_{0}+j_{1},2(j_{0}+j_{1}),\ldots,(|H|-1)(j_{0}+j_{1})\},\ \ O=j_{0}+E.

Since EHE\subseteq H and j0Hj_{0}\not\in H, we have that EE and OO belong to different cosets of n/H\mathbb{Z}_{n}/H. Hence, |O||H||O|\leqslant|H| and |E||H||E|\leqslant|H|. Thus, |Sn(𝐣)|=|OE|2|H||\operatorname{S}_{n}(\mathbf{j})|=|O\cup E|\leqslant 2|H|. ∎

We will frequently use the following lemma in our proofs.

Lemma 2.

Let nn be even and h<nh^{*}<n. Let h\langle h^{*}\rangle be the cyclic subgroup of n\mathbb{Z}_{n} generated by hh^{*}. Then

gcd(n,h)={gcd(n2,h)if n2h,2gcd(n2,h)otherwise.\gcd(n,h^{*})=\begin{cases}\gcd(\frac{n}{2},h^{*})&\text{if $\frac{n}{2}\in\langle h^{*}\rangle$},\\ 2\gcd(\frac{n}{2},h^{*})&\text{otherwise}.\end{cases}
Proof.

Let n=2lmn=2^{l}m, h=2abh^{*}=2^{a}b, where l,m,bl,m,b\in\mathbb{N}, a0a\in\mathbb{N}_{0}, and both mm and bb are odd. If n2h\frac{n}{2}\in\langle h^{*}\rangle, then gcd(n,h)n2\gcd(n,h^{*})\mid\frac{n}{2}. It follows that a<la<l and so gcd(n,h)=gcd(n2,h)\gcd(n,h^{*})=\gcd(\frac{n}{2},h^{*}).

If n2h\frac{n}{2}\notin\langle h^{*}\rangle, then gcd(n,h)n2\gcd(n,h^{*})\nmid\frac{n}{2} and so lal\leqslant a. It follows that gcd(n,h)=2gcd(n2,h)\gcd(n,h^{*})=2\gcd(\frac{n}{2},h^{*}). ∎

For any n2n\geqslant 2 and kn/2k\geqslant\lfloor n/2\rfloor, the following definition and notation will be used throughout the paper:

Φ(n,k)3k+3n2.\varPhi(n,k)\coloneqq\left\lceil\frac{3k+3-n}{2}\right\rceil. (1)

The next result of Karst, Langowitz, Oehrlein, and Troxell [10] plays a critical role in our proofs. For completeness we include a proof.

Proposition 3.

[10] For n>2n>2 and kdk\geqslant d where d=n2d=\lfloor\frac{n}{2}\rfloor and Φ(n,k)3k+3n2\varPhi(n,k)\coloneqq\left\lceil\frac{3k+3-n}{2}\right\rceil, we have the following lower bounds:

rnk(Cn){Φ(n,k)(n22)+kn2+1if n is even;Φ(n,k)(n12)if n is odd.\operatorname{\mathnormal{rn_{k}}}(C_{n})\geqslant\begin{cases}\varPhi(n,k)(\frac{n-2}{2})+k-\frac{n}{2}+1&\mbox{\rm{if $n$ is even;}}\\ \varPhi(n,k)(\frac{n-1}{2})&\mbox{\rm{if $n$ is odd.}}\end{cases}
Proof.

Assume ff is a radio-kk-labeling of CnC_{n}. For i[0,n3]i\in[0,n-3] the following hold:

  1. 1.

    f(xi+1)f(xi)k+1dif(x_{i+1})-f(x_{i})\geqslant k+1-d_{i},

  2. 2.

    f(xi+2)f(xi+1)k+1di+1f(x_{i+2})-f(x_{i+1})\geqslant k+1-d_{i+1},

  3. 3.

    f(xi+2)f(xi)k+1d(xi,xi+2).f(x_{i+2})-f(x_{i})\geqslant k+1-d(x_{i},x_{i+2}).

Adding these three inequalities we get

2(f(xi+2)f(xi))3k+3(d(xi,xi+2)+di+1+di).2(f(x_{i+2})-f(x_{i}))\geqslant 3k+3-(d(x_{i},x_{i+2})+d_{i+1}+d_{i}).

Since nd(xi,xi+2)+di+1+din\geqslant d(x_{i},x_{i+2})+d_{i+1}+d_{i}, we have

fi+1+fi=f(xi+2)f(xi)3k+3n2=Φ(n,k).f_{i+1}+f_{i}=f(x_{i+2})-f(x_{i})\geqslant\left\lceil\frac{3k+3-n}{2}\right\rceil=\varPhi(n,k).

Hence, we obtain the lower bounds for the radio-kk-number of CnC_{n}:

rnk(Cn){Φ(n,k)(n22)+kd+1if n is even;Φ(n,k)(n12)if n is odd.\operatorname{\mathnormal{rn_{k}}}(C_{n})\geqslant\begin{cases}\varPhi(n,k)(\frac{n-2}{2})+k-d+1&\text{if $n$ is even;}\\ \varPhi(n,k)(\frac{n-1}{2})&\text{if $n$ is odd.}\end{cases}

The result follows. ∎

Denote the lower bounds of rnk(Cn)rn_{k}(C_{n}) from 3 by LB(n,k){\rm LB}(n,k):

rnk(Cn)LB(n,k):={Φ(n,k)(n22)+kn2+1if n is even;Φ(n,k)(n12)if n is odd.rn_{k}(C_{n})\geqslant{\rm LB}(n,k):=\begin{cases}\varPhi(n,k)(\frac{n-2}{2})+k-\frac{n}{2}+1&\text{if $n$ is even;}\\ \varPhi(n,k)(\frac{n-1}{2})&\text{if $n$ is odd.}\end{cases} (2)

The next result from [10] greatly reduces the number of inequalities to verify when determining if a map ff is a radio-kk-labeling.

Proposition 4.

[10] If kdk\geqslant d, then f:V(Cn)0f\colon V(C_{n})\to\mathbb{N}_{0} is a radio-kk-labeling of CnC_{n} if and only if the following hold:

  1. 1.

    fik+1di,for i[0,n2]f_{i}\geqslant k+1-d_{i},\quad\text{for }i\in[0,n-2]

  2. 2.

    fi+fi+1k+1d(xi,xi+2),for i[0,n3]f_{i}+f_{i+1}\geqslant k+1-d(x_{i},x_{i+2}),\quad\text{for }i\in[0,n-3].

We do not include a proof of this result, however, the proof of 5 in the next section follows similarly.

3 Exact Values for rnk(Cn)rn_{k}(C_{n})

In this section, we completely determine the radio-kk-number for CnC_{n} when kn3k\geqslant n-3 (Theorem 6) and when k<n3k<n-3 wile nn and kk have the same parity (Theorem 8).

For kn3k\geqslant n-3, we make use of the following lemma, which is a stronger version of Proposition 4 for the special case when kn3k\geqslant n-3.

Lemma 5.

For kn3k\geqslant n-3, a function f:V(Cn)0f\colon V(C_{n})\to\mathbb{N}_{0} is a radio-kk-labeling of CnC_{n} if and only if fikdi+1f_{i}\geqslant k-d_{i}+1 for all i[0,n2]i\in[0,n-2].

Proof.

The forward direction follows from the definition of radio-kk-labeling. Let kn3k\geqslant n-3 and ff satisfy fikdi+1f_{i}\geqslant k-d_{i}+1, for all i[0,n2]i\in[0,n-2]. By Proposition 4 we only need to show that fi+fi+1k+1d(xi,xi+2)f_{i}+f_{i+1}\geqslant k+1-d(x_{i},x_{i+2}) is satisfied. Let i[0,n3]i\in[0,n-3]. Then f(xi+2)f(xi+1)+k+1di+1f(x_{i+2})\geqslant f(x_{i+1})+k+1-d_{i+1} and f(xi+1)f(xi)+k+1di.f(x_{i+1})\geqslant f(x_{i})+k+1-d_{i}. Combining these two inequalities,

f(xi+2)f(xi)+2k+2(di+1+di).f(x_{i+2})\geqslant f(x_{i})+2k+2-(d_{i+1}+d_{i}). (3)

Since (di+1+di)n1(d_{i+1}+d_{i})\leqslant n-1 (\because di,di+1n/2d_{i},d_{i+1}\leqslant n/2 and di+1+dind_{i+1}+d_{i}\neq n) and kn3k\geqslant n-3, we have

f(xi+2)f(xi)+2k+2(n1)f(xi)+kf(xi)+k+1d(xi,xi+2).f(x_{i+2})\geqslant f(x_{i})+2k+2-(n-1)\geqslant f(x_{i})+k\geqslant f(x_{i})+k+1-d(x_{i},x_{i+2}).

Hence the result follows. ∎

Theorem 6.

Let kn3k\geqslant n-3. Then

rnk(Cn)={(n22)(2k+3n)+kn2+1if n is even;(n12)(2k+3n)if n is odd.rn_{k}(C_{n})=\begin{cases}(\frac{n-2}{2})(2k+3-n)+k-\frac{n}{2}+1&\text{if $n$ is even};\\ (\frac{n-1}{2})(2k+3-n)&\text{if $n$ is odd}.\end{cases}
Proof.

Suppose ff is a radio-kk-labeling for CnC_{n}. From Equation (3):

f(xi+2)f(xi)2k+2(di+di+1)2k+3n.f({x_{i+2}})-f(x_{i})\geqslant 2k+2-(d_{i}+d_{i+1})\geqslant 2k+3-n.

Let Φ(n,k)=2k+3n\Phi^{\prime}(n,k)=2k+3-n. If nn is odd, then

span(f)i=0(n3)/2[f(x2i+2)f(x2i)](n12)Φ(n,k).\displaystyle\operatorname{span}(f)\geqslant\sum_{i=0}^{(n-3)/2}[f(x_{2i+2})-f(x_{2i})]\geqslant\left(\frac{n-1}{2}\right)\Phi^{\prime}(n,k).

If nn is even, the last vertex cannot be paired with others and so,

span(f)i=0(n4)/2[f(x2i+2)f(x2i)]+fn2(n22)Φ(n,k)+k+1n/2.\operatorname{span}(f)\geqslant\sum_{i=0}^{(n-4)/2}[f(x_{2i+2})-f(x_{2i})]+f_{n-2}\geqslant\left(\frac{n-2}{2}\right)\Phi^{\prime}(n,k)+k+1-n/2.

Therefore, the lower bounds are established.

Next we find a labeling which meets the lower bounds. Suppose n=2q+1n=2q+1 for qq\in\mathbb{N}. Let the distance gap be di=qd_{i}=q for all i[0,n2]i\in[0,n-2]. Since d=qd=q and gcd(2q+1,q)=1\gcd(2q+1,q)=1, if we let ji=dij_{i}=d_{i}, then jij_{i} determines a permutation of n\mathbb{Z}_{n}. Define the label gap by fi=k+1df_{i}=k+1-d. Then fik+1dif_{i}\geqslant k+1-d_{i} for all i[0,n2]i\in[0,n-2]. Therefore, ff is a valid radio-kk-labeling of CnC_{n} and span(f)\operatorname{span}(f) meets the desired lower bound.

Now consider the case when nn is even. That is, n=2qn=2q for qq\in\mathbb{N}, and d=qd=q. For i[0,n2]i\in[0,n-2] we define the jumping and labeling sequences respectively by:

ji={di is even;d1i is odd.fi={k+1di is even;k+2di is odd.j_{i}=\begin{cases}d&\text{$i$ is even};\\ d-1&\text{$i$ is odd}.\end{cases}\hskip 21.68121ptf_{i}=\begin{cases}k+1-d&\text{$i$ is even};\\ k+2-d&\text{$i$ is odd}.\end{cases}\hskip 14.45377pt

It is easy to see that the jump sequence gives a permutation of V(Cn)V(C_{n}), and it holds that fi=k+1dif_{i}=k+1-d_{i} for all ii. By 5, ff is a radio-kk-labeling of CnC_{n} with the desired span. ∎

Now we turn to the case when dk<n3d\leqslant k<n-3 and nkn\equiv k (mod 22). Recall Φ(n,k)\varPhi(n,k) and LB(n,k){\rm LB}(n,k) from Eq. 1 and Eq. 2, respectively.

Lemma 7.

Let ff be a radio-kk-labeling of CnC_{n}, where nn and kk have the same parity. For any i[0,n3]i\in[0,n-3], if fi+fi+1=Φ(n,k)f_{i}+f_{i+1}=\varPhi(n,k), then d(xi,xi+2){nk2,nk21}d(x_{i},x_{i+2})\in\{\frac{n-k}{2},\frac{n-k}{2}-1\}.

Proof.

Suppose fi+fi+1=Φ(n,k)f_{i}+f_{i+1}=\varPhi(n,k). Because nn and kk have the same parity, by the definition of radio-kk-labeling, we have

(3kn+4)/2=Φ(n,k)=fi+fi+12(k+1)(di+di+1).(3k-n+4)/2=\varPhi(n,k)=f_{i}+f_{i+1}\geqslant 2(k+1)-(d_{i}+d_{i+1}).

Hence, di+di+1n+k2d_{i}+d_{i+1}\geqslant\frac{n+k}{2}. Combining this with the fact that d(xi,xi+2)n(di+di+1)d(x_{i},x_{i+2})\leqslant n-(d_{i}+d_{i+1}) we get d(xi,xi+2)(nk)/2d(x_{i},x_{i+2})\leqslant(n-k)/2. Moreover, because f(xi+2)f(xi)=fi+fi+1=Φ(n,k)k+1d(xi,xi+2)f(x_{i+2})-f(x_{i})=f_{i}+f_{i+1}=\varPhi(n,k)\geqslant k+1-d(x_{i},x_{i+2}), we obtain d(xi,xi+2)(nk2)/2d(x_{i},x_{i+2})\geqslant(n-k-2)/2. Therefore, the result follows. ∎

7 shows that when nn and kk have the same parity, in order to keep the Φ\varPhi-function value for any three consecutive vertices, xi,xi+1,xi+2x_{i},x_{i+1},x_{i+2}, the distance between xix_{i} and xi+2x_{i+2} must be either nk2\frac{n-k}{2} or nk21\frac{n-k}{2}-1.

Theorem 8.

If nn and kk have the same parity and n2kn3\lfloor\frac{n}{2}\rfloor\leqslant k\leqslant n-3, then rnk(Cn)=LB(n,k)rn_{k}(C_{n})={\rm LB}(n,k).

Proof.

Throughout the proof (in fact, throughout the paper), in every defined labeling, d<di+di+1nd<d_{i}+d_{i+1}\leqslant n holds for all ii considered, which implies that n(di+di+1)=d(xi,xi+2).n-(d_{i}+d_{i+1})=d(x_{i},x_{i+2}). Also, in every defined jump sequence, no jump jij_{i} will exceed the diameter of the graph, so that ji=di.j_{i}=d_{i}.

Case 1. Both nn and kk are even. Denote n=4q+tn=4q+t and k=2q+2m+tk=2q+2m+t, where t{0,2}t\in\{0,2\} and 0m<q10\leqslant m<q-1. Let h=qm1h=q-m-1 and p=gcd(d,h)p=\gcd(d,h). Define the jump sequence (or distance gap) by:

ji=di={di is even,dhi is odd and inxp1,x,dh1i is odd and i=nxp1for somex.j_{i}=d_{i}=\begin{cases}d&i\text{ is even,}\\ d-h&i\text{ is odd and }i\neq\frac{nx}{p}-1,\ x\in\mathbb{N},\\ d-h-1&i\text{ is odd and }i=\frac{nx}{p}-1\ \mbox{for some}\ x\in\mathbb{N}.\end{cases}

Claim: The jump sequence gives a permutation.

Proof of Claim) Recall from Section 1 that V(Cn=nV(C_{n}=\mathbb{Z}_{n}. If p=1p=1, then our jumping strategy goes as follows: Start at the vertex x0x_{0}, then jump dd to the next vertex x1x_{1}. The vertices x0x_{0} and x1x_{1} form a coset of n/d.\mathbb{Z}_{n}/\langle d\rangle. (Note, each coset in n/d\mathbb{Z}_{n}/\langle d\rangle consists of two elements, {i,i+d}\{i,i+d\} (mod nn).) After x1x_{1} we then jump dhd-h to x2x_{2}, which is in another coset of n/d\mathbb{Z}_{n}/\langle d\rangle. After x2x_{2} we then jump dd to the vertex x3x_{3}. The vertices x2x_{2} and x3x_{3} form another coset of n/d\mathbb{Z}_{n}/\langle d\rangle.

We repeat this process until each vertex is landed exactly once. This can be done since gcd(d,h)=gcd(d,dh)=1\gcd(d,h)=\gcd(d,d-h)=1. So |Sn(𝐣)|=n|S_{n}(\mathbf{j})|=n with 𝐣=(d,dh)\mathbf{j}=(d,d-h). Therefore, 𝐣=(d,dh)\mathbf{j}=(d,d-h) is a permutation on n.\mathbb{Z}_{n}.

If p1p\neq 1, by Lemma 2, gcd(n,h)=gcd(d,h)=p\gcd(n,h)=\gcd(d,h)=p if dhd\in\langle h\rangle; otherwise gcd(n,h)=2gcd(d,h)=2p\gcd(n,h)=2\gcd(d,h)=2p. If dhd\in\langle h\rangle, then the order of each coset in n/h\mathbb{Z}_{n}/\langle h\rangle is n/pn/p. Also, Sn(𝐣)S_{n}(\mathbf{j}) with 𝐣=(d,dh)\mathbf{j}=(d,d-h) covers one coset of n/h\mathbb{Z}_{n}/\langle h\rangle. So, |Sn(𝐣)|=n/p|S_{n}(\mathbf{j})|=n/p. Similarly, if dhd\not\in\langle h\rangle, by 1, Sn(𝐣)S_{n}(\mathbf{j}) with 𝐣=(d,dh)\mathbf{j}=(d,d-h) covers two cosets in n/h\mathbb{Z}_{n}/\langle h\rangle. Thus, |Sn(𝐣)|=n/p|S_{n}(\mathbf{j})|=n/p.

After making np1\frac{n}{p}-1 jumps, we jump dh1d-h-1 to the next vertex, which is 1+h-1+\langle h\rangle in n/h\mathbb{Z}_{n}/\langle h\rangle. Then we continue jumping 𝐣=(d,dh)\mathbf{j}=(d,d-h), which will cover one or two new cosets, depending on whether dhd\in\langle h\rangle, as shown above. We continue this process until all vertices are covered exactly once. Thus the jump sequence determines a permutation. \blacksquare

Next we define the labeling sequence as follows:

fi={2m+1+(t/2)if i is even,q+m+1+(t/2)if i is odd.f_{i}=\begin{cases}2m+1+(t/2)&\quad\text{if $i$ is even},\\ q+m+1+(t/2)&\quad\text{if $i$ is odd}.\end{cases}

To complete the proof for Case 1, it suffices to show:

Claim: fif_{i} is a valid radio-kk-labeling with the desired span.

Proof of Claim) By Proposition 4, we need to show that fik+1dif_{i}\geqslant k+1-d_{i} and fi+fi+1k+1d(xi,xi+2).f_{i}+f_{i+1}\geqslant k+1-d(x_{i},x_{i+2}). We call these the first and the second inequality, respectively. We begin by verifying the first inequality. Suppose ii is even. Then ji=di=d=2q+(t/2)j_{i}=d_{i}=d=2q+(t/2) and

k+1di=2q+2m+t+12q(t/2)=2m+1+(t/2)=fi.k+1-d_{i}=2q+2m+t+1-2q-(t/2)=2m+1+(t/2)=f_{i}.

Suppose ii is odd. Then ji=di{q+m+1+(t/2),q+m+(t/2)}j_{i}=d_{i}\in\{q+m+1+(t/2),q+m+(t/2)\} and

k+1di2q+2m+t+1(q+m+t/2)=q+m+1+(t/2)=fi.k+1-d_{i}\leqslant 2q+2m+t+1-(q+m+t/2)=q+m+1+(t/2)=f_{i}.

Now we verify the second inequality. By the facts that d(xi,xi+2)=n(di+di+1)d(x_{i},x_{i+2})=n-(d_{i}+d_{i+1}) and di+di+1{3q+m+t,3q+m+t+1}d_{i}+d_{i+1}\in\{3q+m+t,3q+m+t+1\}, we obtain

fi+fi+1=q+3m+t+2\displaystyle f_{i}+f_{i+1}=q+3m+t+2 =3q+m+t+12q+2m+1\displaystyle=3q+m+t+1-2q+2m+1
di+di+12q+2m+1\displaystyle\geqslant d_{i}+d_{i+1}-2q+2m+1
=k+1d(xi,xi+2).\displaystyle=k+1-d(x_{i},x_{i+2}).

Therefore, ff is a radio-kk-labeling with the desired span. \blacksquare

Case 2. Both nn and kk are odd. Let n=4q+3n=4q+3 and k=2q+2m+1k=2q+2m+1 for 0mq10\leqslant m\leqslant q-1, or n=4q+1n=4q+1 and k=2q+2m+1k=2q+2m+1 for 0mq20\leqslant m\leqslant q-2. Then Φ(n,k)=(3kn+4)/2\varPhi(n,k)=(3k-n+4)/2. Let

z=d+q+m+12ands=ngcd(n,z).z=\left\lceil\frac{d+q+m+1}{2}\right\rceil\hskip 14.45377pt\mbox{and}\hskip 14.45377pts=\frac{n}{\gcd(n,z)}.\hskip 36.135pt

Case 2.1. Suppose Φ(n,k)=(3kn+4)/2\varPhi(n,k)=(3k-n+4)/2 is even. Define the jump sequence and labeling sequence respectively by:

ji={z+1if i=sx1,xzotherwise;andfi=Φ(n,k)2for all i.j_{i}=\begin{cases}z+1&\text{if $i=sx-1$},\ x\in\mathbb{N}\\ z&\text{otherwise};\end{cases}\hskip 14.45377pt\mbox{and}\hskip 14.45377ptf_{i}=\frac{\varPhi(n,k)}{2}\quad\text{for all $i$}.

We claim that our jump sequence gives a permutation. Let V(Cn)=nV(C_{n})=\mathbb{Z}_{n}. If gcd(n,z)=1(n,z)=1, then by the second formula of jij_{i} in the above, z=n\langle z\rangle=\mathbb{Z}_{n} gives a permutation. If gcd(n,z)>1(n,z)>1, since s=|z|s=|\langle z\rangle|, the jump sequence labels each coset of n/z\mathbb{Z}_{n}/\langle z\rangle, and then transitions into the next unlabeled coset by jumping z+1z+1 (the first formula in jij_{i} above). Continue this process until all vertices are labeled. Hence, the jump sequence is a permutation.

It remains to show that fif_{i} is a valid radio-kk-labeling. Consider the case that n=4q+3n=4q+3 and k=2q+2m+1k=2q+2m+1. Then Φ(n,k)=q+3m+2\varPhi(n,k)=q+3m+2. By the assumption that Φ(n,k)\varPhi(n,k) is even, qq and mm must have the same parity, and

ji={3q+m+42if i=sx1,x3q+m+22otherwise;andfi=q+3m+22for all i.j_{i}=\begin{cases}\frac{3q+m+4}{2}&\text{if $i=sx-1$},\ x\in\mathbb{N}\\ \frac{3q+m+2}{2}&\text{otherwise};\end{cases}\hskip 14.45377pt\mbox{and}\hskip 14.45377ptf_{i}=\frac{q+3m+2}{2}\quad\text{for all $i$}.

To verify the first inequality, as di=ji(3q+m+2)/2d_{i}=j_{i}\geqslant(3q+m+2)/2, we have

k+1di2q+2m+23q+m+22=q+3m+22=fi.\displaystyle k+1-d_{i}\leqslant 2q+2m+2-\frac{3q+m+2}{2}=\frac{q+3m+2}{2}=f_{i}.

To verify the second inequality, using the facts that di+di+1=nd(xi,xi+2)d_{i}+d_{i+1}=n-d(x_{i},x_{i+2}) and di+di+1{3q+m+2,3q+m+3}d_{i}+d_{i+1}\in\{3q+m+2,3q+m+3\}, we obtain

fi+fi+1=q+3m+2\displaystyle f_{i}+f_{i+1}=q+3m+2 =3q+m+3(2q2m+1)\displaystyle=3q+m+3-(2q-2m+1)
di+di+1(2q2m+1)\displaystyle\geqslant d_{i}+d_{i+1}-(2q-2m+1)
=kd(xi,xi+2)+1.\displaystyle=k-d(x_{i},x_{i+2})+1.

Therefore, ff is a radio-kk-labeling with the desired span. The case when n=4q+1n=4q+1 follows similarly.

Case 2.2. Suppose Φ(n,k)=(3kn+4)/2\varPhi(n,k)=(3k-n+4)/2 is odd. If n=4q+3n=4q+3, Φ(n,k)=q+3m+2\varPhi(n,k)=q+3m+2, so qq and mm must have different parities. It follows that d+q+m+1=3q+m+2d+q+m+1=3q+m+2 is odd. Otherwise, n=4q+1n=4q+1 and Φ(n,k)=q+3m+3\Phi(n,k)=q+3m+3. Then qq and mm have the same parity and again d+q+m+1d+q+m+1 is odd. Hence z=(d+q+m+2)/2z=(d+q+m+2)/2 always holds. Let h=n2zh=n-2z.

Case 2.2.1. Suppose gcd(n,h)=1.\gcd(n,h)=1. Define the jump sequence and the labeling sequence respectively by:

ji=zfor all iandfi={Φ(n,k)2if i is even,Φ(n,k)2if i is odd.j_{i}=z\quad\text{for all $i$}\hskip 28.90755pt\mbox{and}\hskip 28.90755ptf_{i}=\begin{cases}\lfloor\frac{\varPhi(n,k)}{2}\rfloor&\text{if $i$ is even},\\ \lceil\frac{\varPhi(n,k)}{2}\rceil&\text{if $i$ is odd}.\end{cases}

To show that the jump sequence is a permutation, we need to verify that z=n\langle z\rangle=\mathbb{Z}_{n}. Since gcd(n,h)=1\gcd(n,h)=1, we have that h=n=h=2z\langle h\rangle=\mathbb{Z}_{n}=\langle-h\rangle=\langle 2z\rangle. This implies that gcd(n,2z)=1\gcd(n,2z)=1. Hence gcd(n,z)=1\gcd(n,z)=1 and so z=n\langle z\rangle=\mathbb{Z}_{n}, implying the jump sequence is a permutation.

Now we show fif_{i} is a valid radio-kk-labeling. We prove the case when n=4q+1n=4q+1. (The case when n=4q+3n=4q+3 follows similarly.) Then Φ(n,k)=q+3m+3\varPhi(n,k)=q+3m+3. For the first inequality, one can see that k+1di=Φ(n,k)2fik+1-d_{i}=\left\lfloor\frac{\varPhi(n,k)}{2}\right\rfloor\leqslant f_{i}. For the second inequality, since di+di+1=nd(xi,xi+2)d_{i}+d_{i+1}=n-d(x_{i},x_{i+2}) and di+di+1=3q+m+2d_{i}+d_{i+1}=3q+m+2, one can verify that fi+fi+1kd(xi,xi+2)+1f_{i}+f_{i+1}\geqslant k-d(x_{i},x_{i+2})+1.

Therefore, we have a valid radio-kk-labeling with the desired span, which is exactly the lower bound in Proposition 3.

Case 2.2.2. Suppose c=gcd(n,h)>1c=\gcd(n,h)>1 for some odd cc. Then c=h\langle c\rangle=\langle h\rangle, and there are cc cosets in n/c\mathbb{Z}_{n}/\langle c\rangle. Recall that

h={qmif n=4q+3,qm1if n=4q+1.h=\begin{cases}q-m&\text{if $n=4q+3$},\\ q-m-1&\text{if $n=4q+1$}.\end{cases}

Define the jumping sequence and labeling sequence respectively by:

ji={dif i is even, i[0,n2nc],dhif i is odd, i=2ncx1,xi[0,n1nc],dh+1if i is odd, i2ncx1,x, and i[0,n1nc](nh)/2if i[nnc,n2];j_{i}=\begin{cases}d&\quad\text{if $i$ is even, $i\in[0,n-2-\frac{n}{c}]$},\\ d-h&\quad\text{if $i$ is odd, $i=\frac{2n}{c}x-1,x\in\mathbb{N}$, $i\in[0,n-1-\frac{n}{c}]$},\\ d-h+1&\quad\text{if $i$ is odd, $i\not=\frac{2n}{c}x-1,\ x\in\mathbb{N}$, and $i\in[0,n-1-\frac{n}{c}]$, }\\ (n-h)/2&\quad\text{if $i\in[n-\frac{n}{c},n-2]$};\end{cases}
fi={k+1dif i is even and i[0,n2nc],Φ(n,k)(k+1d)if i is odd and i[0,n1nc],Φ(n,k)2if i is even and i(n1nc,n2],Φ(n,k)2if i is odd and i(nnc,n2].f_{i}=\begin{cases}k+1-d&\quad\text{if $i$ is even and $i\in[0,n-2-\frac{n}{c}]$},\\ \varPhi(n,k)-(k+1-d)&\quad\text{if $i$ is odd and $i\in[0,n-1-\frac{n}{c}]$},\\ \lfloor\frac{\varPhi(n,k)}{2}\rfloor&\quad\text{if $i$ is even and $i\in(n-1-\frac{n}{c},n-2]$},\\ \lceil\frac{\varPhi(n,k)}{2}\rceil&\quad\text{if $i$ is odd and $i\in(n-\frac{n}{c},n-2]$}.\end{cases}

Consider the case that n=4q+1n=4q+1 and k=2q+2m+1k=2q+2m+1. (The case for n=4q+3n=4q+3 and k=2q+2m+1k=2q+2m+1 follows similarly.) Note that d=2qd=2q, Φ(n,k)=q+3m+3\varPhi(n,k)=q+3m+3, and h=qm1h=q-m-1.

Claim: Our jumping sequence is a permutation.

Proof of Claim) We show that we label two cosets of n/c\mathbb{Z}_{n}/\langle c\rangle at a time by alternately jumping 2q2q and q+m+2q+m+2 (see the first and third formulas in jij_{i} above). That is, 𝐣=(2q,q+m+2)\mathbf{j}=(2q,q+m+2).

Note that h=n(3q+m+2)h=n-(3q+m+2). This implies h=3q+m+2=c\langle h\rangle=\langle 3q+m+2\rangle=\langle c\rangle. Hence, it is enough to show that 2qc2q\not\in\langle c\rangle. Because gcd(4q+1,2q)=1\gcd(4q+1,2q)=1, we have 2q=n\langle 2q\rangle=\mathbb{Z}_{n}. If 2qc2q\in\langle c\rangle, then c=n\langle c\rangle=\mathbb{Z}_{n}, contradicting to gcd(n,h)=c>1(n,h)=c>1. Thus, 2qc2q\notin\langle c\rangle, and our jump sequence begins by labeling two cosets in n/c\mathbb{Z}_{n}/\langle c\rangle.

Next we show that the first two cosets labeled are c\langle c\rangle and c+c2\langle c\rangle+\lfloor\frac{c}{2}\rfloor, by proving that 2qc2+c2q\in\lfloor\frac{c}{2}\rfloor+\langle c\rangle. Let n=4q+1=tcn=4q+1=tc for some odd tt. Then

2q=tc12=c(t1)2+c12.2q=\frac{tc-1}{2}=\frac{c(t-1)}{2}+\frac{c-1}{2}.

Thus, 2qc2+c2q\in\lfloor\frac{c}{2}\rfloor+\langle c\rangle, and so 2q+c=c2+c2q+\langle c\rangle=\lfloor\frac{c}{2}\rfloor+\langle c\rangle. Hence, the first two labeled cosets are c\langle c\rangle and c+c2\langle c\rangle+\lfloor\frac{c}{2}\rfloor.

Afterwards, we jump q+m+1q+m+1 (the second formula in jij_{i} given above) and land in the coset c1\langle c\rangle-1 in n/c\mathbb{Z}_{n}/\langle c\rangle. Continue the same process repeatedly as above, we label the first c1c-1 cosets in pairs in the following order:

(c,c+c2)(c1,c+c21)(c+1c2,c+1).\left(\langle c\rangle,\langle c\rangle+\left\lfloor\frac{c}{2}\right\rfloor\right)\rightarrow\left(\langle c\rangle-1,\langle c\rangle+\left\lfloor\frac{c}{2}\right\rfloor-1\right)\rightarrow\cdots\rightarrow\left(\langle c\rangle+1-\left\lfloor\frac{c}{2}\right\rfloor,\langle c\rangle+1\right).

After the last pair of cosets above, the only coset that was left unlabelled is c+c2\langle c\rangle+\lceil\frac{c}{2}\rceil. Again, we jump q+m+1q+m+1 and land in that coset.

It remains to show that jumping nh2\frac{n-h}{2} (the last formula in the jij_{i} given above) is a permutation of c+c2\langle c\rangle+\lceil\frac{c}{2}\rceil. This is true because nn is odd, so gcd(n,nh2)=gcd(n,nh)=gcd(n,h)=c\gcd(n,\frac{n-h}{2})=\gcd(n,n-h)=\gcd(n,h)=c. \blacksquare

Finally we show that fif_{i} is a radio-kk-labeling. The first inequality can be obtained directly from our definitions that for all ii, k+1difi.k+1-d_{i}\leqslant f_{i}. Next we verify the second inequality by listing all possible cases in Table 1. Note, for the last case (last row), we need to show 3q+5m+42q+7m+62\frac{3q+5m+4}{2}\geqslant\frac{q+7m+6}{2}. Recall 0mq20\leqslant m\leqslant q-2. Then qm+1q\geqslant m+1, which implies 3q+5m+42q+7m+62\frac{3q+5m+4}{2}\geqslant\frac{q+7m+6}{2}. For an example of this case, see Example 3.

Therefore, we have a valid radio-kk-labeling with the desired span. The proof is complete. ∎

ii\in i+1i+1\in di+di+1d_{i}+d_{i+1}\leqslant fi+fi+1f_{i}+f_{i+1} kd(xi+xi+2)+1k-d(x_{i}+x_{i+2})+1\leqslant
[0,n1nc][0,n-1-\frac{n}{c}] [0,n1nc][0,n-1-\frac{n}{c}] 3q+m+23q+m+2 q+3m+3q+3m+3 q+3m+3q+3m+3
[nnc,n2][n-\frac{n}{c},n-2] [nnc,n2][n-\frac{n}{c},n-2] 3q+m+23q+m+2 q+3m+3q+3m+3 q+3m+3q+3m+3
{n1nc}\{n-1-\frac{n}{c}\} {nnc}\{n-\frac{n}{c}\} 5q+3m+42\frac{5q+3m+4}{2} 3q+5m+42\frac{3q+5m+4}{2} q+7m+62\frac{q+7m+6}{2}
ii is odd ii is even
Table 1: All possibilities for checking the second inequality for Case 2.2.2.
Example 2.

In Figure 1 we let q=4q=4 and m=1m=1, so n=16n=16 and k=10k=10. We get Φ(16,10)=9\varPhi(16,10)=9, h=2h=2, and p=gcd(d,h)=2p=\gcd(d,h)=2. In this example we label one coset of n/h\mathbb{Z}_{n}/\langle h\rangle at a time. With 𝐣=(8,6)\mathbf{j}=(8,6), S16((8,6))S_{16}((8,6)) covers one coset (circular black vertices) of 16/2\mathbb{Z}_{16}/\langle 2\rangle. We then jump 55 into a new coset (gray diamond vertices) of 16/2\mathbb{Z}_{16}/\langle 2\rangle to complete the labeling.

Refer to caption
Refer to caption
Figure 1: Steps of finding an optimal radio-1010-labeling for C16C_{16} in Example 2.
Example 3.

In Figure 2 we let n=27n=27 and k=19k=19. Then d=13d=13 and h=3h=3. Notice that gcd(n,h)=gcd(27,3)=3=h\gcd(n,h)=\gcd(27,3)=3=h and Φ(27,19)=17\varPhi(27,19)=17. We label two cosets of n/h\mathbb{Z}_{n}/\langle h\rangle at a time. With 𝐣=(13,11)\mathbf{j}=(13,11), S27((13,11))S_{27}((13,11)) covers two cosets (solid black and white circles) of 27/3\mathbb{Z}_{27}/\langle 3\rangle. We then jump 1010 into a new coset (gray diamonds) of 27/3\mathbb{Z}_{27}/\langle 3\rangle. We then jump 1212 consecutively to finish labeling the last coset.

Refer to caption
Figure 2: Example 3: Steps of finding an optimal radio-1919-labeling for C27.C_{27}.

4 Results for k<n3k<n-3 and nkn\not\equiv k (mod 22)

The case when nn and kk have different parities is more complicated, and so more interesting! Following the work in the previous section when nn and kk have the same parity, we will use the lower bound proved in 3. On the other hand, there are properties we use in the previous section that are not valid anymore for the case when nn and kk have different parities. The following lemma reveals the fact that in searching for an optimal radio-kk-labeling that achieves the lower bound for CnC_{n} we would lose some flexibility we had in 7. Consequently, we show that the lower bound in 3 is not always achieved.

Lemma 9.

Let ff be a radio-kk-labeling for CnC_{n}, where nn and kk have different parities. For any i[0,n3]i\in[0,n-3], if fi+fi+1=Φ(n,k)=(3kn+3)/2f_{i}+f_{i+1}=\varPhi(n,k)=(3k-n+3)/2, then d(xi,xi+2)=k+1Φ(n,k)=(nk1)/2d(x_{i},x_{i+2})=k+1-\varPhi(n,k)=(n-k-1)/2.

Proof.

This lemma essentially follows from Lemma 2.1 in [10]. One may also prove this lemma with the same approach taken in the proof of Lemma 7. ∎

9 shows that when nn and kk have different parities, in order to keep the Φ\varPhi-function value for any consecutive three vertices, xi,xi+1,xi+2x_{i},x_{i+1},x_{i+2}, the distance between xix_{i} and xi+2x_{i+2} must be fixed as nk12\frac{n-k-1}{2}. Throughout the section we denote this fixed value by hh. That is,

hk+1Φ(n,k)=nk12.h\coloneqq k+1-\varPhi(n,k)=\frac{n-k-1}{2}. (4)

This observation gives us the following lower bound. Recall the definition of LB(n,k){\rm LB}(n,k) from Eq. 2.

Proposition 10.

Suppose k<n3k<n-3 and nk(mod2)n\not\equiv k\pmod{2}, and let p=gcd(n,h)p=\gcd(n,h), where hh is defined in Eq. 4. Then

LB(n,k)+p21rnk(Cn).{\rm LB}(n,k)+\left\lceil\frac{p}{2}\right\rceil-1\leqslant rn_{k}(C_{n}). (5)
Proof.

The proof follows from 1 and 9 which collectively imply that one can label at most (2n)/p(2n)/p vertices while obtaining the Φ\Phi-function value. ∎

When nn is even and kk is odd, we establish lower and upper bounds as follows and show the bounds are sharp for some kk and nn.

Theorem 11.

Suppose nn is even and kk is odd with n2k<n3\frac{n}{2}\leqslant k<n-3. Let p=gcd(n,h)p=\gcd(n,h). Then

LB(n,k)+p21rnk(Cn)LB(n,k)+p1.{\rm LB}(n,k)+\left\lceil\frac{p}{2}\right\rceil-1\ \leqslant rn_{k}(C_{n})\ \leqslant{\rm LB}(n,k)+p-1.

Moreover, if dhd\notin\langle h\rangle then the lower bound in the above is sharp.

Proof.

The lower bound follows from Proposition 10. As d=n2d=\frac{n}{2}, to prove the upper bound and the “moreover” part, by Lemma 2, it suffices to define a radio-kk-labeling ff which satisfies all the following:

  • If dhd\in\langle h\rangle, then ff labels one coset of n/h\mathbb{Z}_{n}/\langle h\rangle at a time and increases the span by 1 each time it switches to an unlabeled coset.

  • If dhd\not\in\langle h\rangle, then ff labels two cosets of n/h\mathbb{Z}_{n}/\langle h\rangle at a time and increases the span by 1 each time it switches to a new pair of cosets.

Let p=gcd(d,h)p^{*}=\gcd(d,h). As nn is even and kk is odd, we let n=4qn=4q and n=4q+2n=4q+2; and in either case, let k=2q+2m+1k=2q+2m+1. Define the jumping sequence and labeling sequence (for both cases) respectively by:

ji={dif i is even,dh1if i=npx1,0<x<p,dhif i is odd and inpx1,0<x<p;j_{i}=\begin{cases}d&\quad\text{if $i$ is even},\\ d-h-1&\quad\text{if $i=\frac{n}{p^{*}}x-1,\quad 0<x<p^{*}$},\\ d-h&\quad\text{if $i$ is odd and $i\not=\frac{n}{p^{*}}x-1,\quad 0<x<p^{*}$};\end{cases}
fi={k+1dif i is even,dh+1if i=npx1,0<x<p,dhif i is odd and inpx1,0<x<p.f_{i}=\begin{cases}k+1-d&\quad\text{if $i$ is even},\\ d-h+1&\quad\text{if $i=\frac{n}{p^{*}}x-1,\quad 0<x<p^{*}$},\\ d-h&\quad\text{if $i$ is odd and $i\not=\frac{n}{p^{*}}x-1,\quad 0<x<p^{*}$}.\end{cases}

Then ff defines a radio-kk-labeling; the proof follows similarly to the proofs in Theorem 8. Whenever ji=dh1=q+mj_{i}=d-h-1=q+m, we have fi1+fi=Φ(n,k)+1f_{i-1}+f_{i}=\varPhi(n,k)+1. This happens p1p^{*}-1 times while labeling. Thus, span(f)=LB(n,k)+p1(f)={\rm LB}(n,k)+p^{*}-1.

If dhd\in\langle h\rangle, then gcd(n,h)=gcd(d,h)=p=p\gcd(n,h)=\gcd(d,h)=p=p^{*}, so the upper bound holds. If dhd\not\in\langle h\rangle, then gcd(n,h)=2gcd(d,h)\gcd(n,h)=2\gcd(d,h), or p=2pp=2p^{*}; hence the lower bound is sharp. ∎

Corollary 12.

Suppose nn is even and kk is odd with n2k<n3\frac{n}{2}\leqslant k<n-3. Recall that h=nk12h=\frac{n-k-1}{2}. If gcd(n2,h)=1\gcd(\frac{n}{2},h)=1, then rnk(Cn)=LB(n,k)rn_{k}(C_{n})={\rm LB}(n,k).

Proof.

Assume d=n2hd=\frac{n}{2}\in\langle h\rangle. Then by our assumption, p=gcd(n,h)=gcd(n2,h)=1p=\gcd(n,h)=\gcd(\frac{n}{2},h)=1. From Theorem 11 we get that the upper and lower bounds meet and rnk(Cn)=LB(n,k).rn_{k}(C_{n})={\rm LB}(n,k). If dhd\not\in\langle h\rangle, we have p=gcd(n,h)=2gcd(d,h)=2p=\gcd(n,h)=2\gcd(d,h)=2. The moreover part in Theorem 11 implies that rnk(Cn)=LB(n,k)+p21=LB(n,k)rn_{k}(C_{n})={\rm LB}(n,k)+\lceil\frac{p}{2}\rceil-1={\rm LB}(n,k). ∎

We conjecture that the upper bound in Theorem 11 is tight when dhd\in\langle h\rangle (see Conjecture 1).

Now we turn to the case that nn is odd and kk is even. The remaining two results of this section are devoted to this case. Recall Φ(n,k)\varPhi(n,k), LB(Cn){\rm LB}(C_{n}), and hh defined in Eq. 1, Eq. 2, and Eq. 4, respectively.

Theorem 13.

Let nn be odd and kk even. If gcd(n,h)=1\gcd(n,h)=1 and hh is odd, then rnk(Cn)=LB(n,k).rn_{k}(C_{n})={\rm LB}(n,k).

Proof.

We define a jump sequence and labeling as follows, for all ii:

ji=nh2 and fi=Φ(n,k)2.j_{i}=\frac{n-h}{2}\text{ \quad and \quad}f_{i}=\frac{\varPhi(n,k)}{2}.

It is easy to see that our jumping sequence gives a permutation, since gcd(n,nh2)=gcd(n,h)=1\gcd(n,\frac{n-h}{2})=\gcd(n,h)=1, thus nh2=n.\langle\frac{n-h}{2}\rangle=\mathbb{Z}_{n}.

Now we show that ff is a valid radio-kk-labeling by verifying 1 and 2 in 4. The first inequality is easy to see. For the second inequality, since d(xi,xi+2)=n(di+di+1)=hd(x_{i},x_{i+2})=n-(d_{i}+d_{i+1})=h, we obtain k+1d(xi,xi+2)=k+1h=Φ(n,k)=fi+fi+1.k+1-d(x_{i},x_{i+2})=k+1-h=\varPhi(n,k)=f_{i}+f_{i+1}. Therefore, ff is a radio-kk-labeling with the desired span as in 3. ∎

Theorem 14.

Suppose n=4q+in=4q+i and k=2q+2m+i1k=2q+2m+i-1 where i{1,3}i\in\{1,3\} and 0mq20\leqslant m\leqslant q-2. If hnh\mid n, then

rnk(Cn)=LB(n,k)+h12.{rn}_{k}(C_{n})={\rm LB}(n,k)+\frac{h-1}{2}.
Proof.

Recall from Eq. 4, h=(nk1)/2h=(n-k-1)/2. Since n=4q+in=4q+i and k=2q+2m+i1k=2q+2m+i-1 we get h=qmh=q-m. That LB(n,k)+h12rnk(Cn){\rm LB}(n,k)+\frac{h-1}{2}\leqslant rn_{k}(C_{n}) holds follows from 10.

To show that rnk(Cn)LB(n,k)+h12rn_{k}(C_{n})\leqslant{\rm LB}(n,k)+\frac{h-1}{2}, we define a jumping and a labeling sequence in the following:

ji={dif i is even and i[0,n2nh],dhif i is odd, i=2nhx1,xi[0,nnh],dh+1if i is odd and i2nhx1,xi[0,n2nh]nh2if i[nnh,n2];j_{i}=\begin{cases}d&\text{if $i$ is even and $i\in[0,n-2-\frac{n}{h}]$},\\ d-h&\text{if $i$ is odd, $i=\frac{2n}{h}x-1,x\in\mathbb{N}$, $i\in[0,n-\frac{n}{h}]$},\\ d-h+1&\text{if $i$ is odd and $i\not=\frac{2n}{h}x-1,x\in\mathbb{N}$, $i\in[0,n-2-\frac{n}{h}]$, }\\ \frac{n-h}{2}&\text{if $i\in[n-\frac{n}{h},n-2]$};\end{cases}
fi={k+1dif i is even and i[0,n2nh],Φ(n,k)(k+1d)+1if i is odd, i=2nhx1,xi[0,nnh],Φ(n,k)(k+1d)if i is odd and i2nhx1,xi[0,n2nh]Φ(n,k)2if i[nnh,n2].f_{i}=\begin{cases}k+1-d&\text{if $i$ is even and $i\in[0,n-2-\frac{n}{h}]$},\\ \varPhi(n,k)-(k+1-d)+1&\text{if $i$ is odd, $i=\frac{2n}{h}x-1,x\in\mathbb{N}$, $i\in[0,n-\frac{n}{h}]$},\\ \varPhi(n,k)-(k+1-d)&\text{if $i$ is odd and $i\not=\frac{2n}{h}x-1,x\in\mathbb{N}$, $i\in[0,n-2-\frac{n}{h}]$, }\\ \frac{\varPhi(n,k)}{2}&\text{if $i\in[n-\frac{n}{h},n-2]$}.\end{cases}

Finishing the proof by showing our jump sequence is a permutation and our labeling sequence is valid follows similarly to the proof of Theorem 8. ∎

Below is an example of Theorem 14.

Example 4.

In Figure 3 we let n=27n=27 and k=20k=20. Then d=13d=13 and h=3h=3. Notice that gcd(n,h)=gcd(27,3)=3=h\gcd(n,h)=\gcd(27,3)=3=h and Φ(27,20)=18\varPhi(27,20)=18. We label two cosets of n/h\mathbb{Z}_{n}/\langle h\rangle at a time. With 𝐣=(13,11)\mathbf{j}=(13,11), S27((13,11))S_{27}((13,11)) covers two cosets (solid black circles and gray diamonds) of 27/3\mathbb{Z}_{27}/\langle 3\rangle. We then jump 1010 into a new coset (larger gray circles) of 27/3\mathbb{Z}_{27}/\langle 3\rangle. Notice that the difference of f(x18)f(x16)=163144=19=Φ(27,20)+1f(x_{18})-f(x_{16})=163-144=19=\varPhi(27,20)+1.

Refer to caption
Figure 3: Example 4: Steps of finding an optimal radio 2020-labeling for C27.C_{27}.

5 Closing Remarks

Most results in this paper are extensions of results by Karst, Langowitz, Oehrlein, and Troxell [10], and by Liu and Zhu [15]. The method of using the Φ\varPhi-function to prove a lower bound for the radio kk-number for CnC_{n} was introduced in [15] when k=n2k=\left\lfloor\frac{n}{2}\right\rfloor, the diameter of CnC_{n}. Precisely, it was shown that rnn/2(Cn)=LB(n,d)rn_{\lfloor n/2\rfloor}(C_{n})={\rm LB}(n,d) for all nn. This method was applied and extended in [10], where the authors completely determined the radio kk-number for CnC_{n} when k=n2+1k=\lfloor\frac{n}{2}\rfloor+1, the diameter of CnC_{n} plus one. The lower bound is shown again to be close to the exact value.

Theorem 15.

[10] Let n=4q+rn=4q+r with q0q\geqslant 0 and 0r3.0\leqslant r\leqslant 3. For k=n2+1k=\lfloor\frac{n}{2}\rfloor+1,

(i)r=0:rnk(Cn)={LB(n,k)if q is even;LB(n,k)+1if q is odd.\displaystyle\text{{\rm(i)}}\ r=0:\ rn_{k}(C_{n})=\begin{cases}{\rm LB}(n,k)&\text{if $q$ is even};\\ {\rm LB}(n,k)+1&\text{if $q$ is odd.}\end{cases}
(ii)r=1,2:rnk(Cn)=LB(n,k).\displaystyle\text{{\rm(ii)}}\ r=1,2:\ rn_{k}(C_{n})={\rm LB}(n,k).
(iii)r=3:rnk(Cn)={LB(n,k)if q2 and it not a multiple of 3;LB(n,k)+1otherwise.\displaystyle\text{{\rm(iii)}}\ r=3:\ rn_{k}(C_{n})=\begin{cases}{\rm LB}(n,k)&\text{if $q\neq 2$ and it not a multiple of 3};\\ {\rm LB}(n,k)+1&\text{otherwise.}\end{cases}

Extending this lower bound approach with properties of cyclic groups, we were able to determine the values of rnk(Cn)rn_{k}(C_{n}) for most kn/2k\geqslant\lfloor n/2\rfloor. Many results in [15, 10] can be immediately obtained by this paper. For instance, when nn and d=n2d=\lfloor\frac{n}{2}\rfloor have different parities we have n=4q+2n=4q+2 with k=2q+1k=2q+1, or n=4q+1n=4q+1 with k=2qk=2q. When n=4q+2n=4q+2 and k=2q+1,k=2q+1, we have h=qh=q. Thus, gcd(d,h)=gcd(2q+1,q)=1.\gcd(d,h)=\gcd(2q+1,q)=1. Using Corollary 12, we have rnd(Cn)=LB(n,d)rn_{d}(C_{n})={\rm LB}(n,d). When n=4q+1n=4q+1 and k=2qk=2q we have h=qh=q. Thus gcd(n,h)=gcd(4q+1,q)=1\gcd(n,h)=\gcd(4q+1,q)=1. If hh is odd, by Proposition 13, rnd(Cn)=LB(n,d)rn_{d}(C_{n})={\rm LB}(n,d).

For the case that k=d+1k=d+1, when nn and d+1=n2+1d+1=\lfloor\frac{n}{2}\rfloor+1 have different parities we have n=4qn=4q with k=2q+1k=2q+1, or n=4q+3n=4q+3 with k=2q+2k=2q+2. When n=4qn=4q and k=2q+1k=2q+1, we have h=q1h=q-1. Consider the case when qq is even. Then, gcd(d,h)=gcd(2q,q1)=gcd(q+1,q1)=1\gcd(d,h)=\gcd(2q,q-1)=\gcd(q+1,q-1)=1. Using Corollary 12, we have rnd+1(Cn)=LB(n,d+1)rn_{d+1}(C_{n})={\rm LB}(n,d+1). When n=4q+3n=4q+3 and k=2q+2k=2q+2 we have h=qh=q. Thus, gcd(n,h)=gcd(4q+3,q)=gcd(3,q)=1\gcd(n,h)=\gcd(4q+3,q)=\gcd(3,q)=1 if qq is not a multiple of 33. If h=qh=q is odd, by Proposition 13 the radio kk-number is LB(n,k).{\rm LB}(n,k).

Although the values of rnk(Cn)rn_{k}(C_{n}) for most kk and nn with nn/2n\geqslant\lfloor n/2\rfloor have been determined, it remains open for other cases, especially when kk and nn have different parities. So far the evidence we obtained indicates that the upper bound in Theorem 11 might be always sharp for n/2hn/2\in\langle h\rangle.

Conjecture 1.

Suppose nn is even and kk is odd with n2k<n3\frac{n}{2}\leqslant k<n-3. Let h=nk12h=\frac{n-k-1}{2} and p=gcd(n,h)p=\gcd(n,h). If n/2hn/2\in\langle h\rangle, then

rnk(Cn)=LB(n,k)+p1.rn_{k}(C_{n})={\rm LB}(n,k)+p-1.

Acknowledgement. The authors are grateful to the two anonymous referees for their careful reading of the manuscript and for their constructive suggestions which led to a better presentation of the article.

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