• 沒有找到結果。

T-colorings and T-edge spans of graphs

N/A
N/A
Protected

Academic year: 2021

Share "T-colorings and T-edge spans of graphs"

Copied!
7
0
0

加載中.... (立即查看全文)

全文

(1)

Graphs and

Combinatorics

( Springer-Verlag 1999

T-Colorings and T-Edge Spans of Graphs*

Shin-Jie Hu, Su-Tzu Juan, and Gerard J. Chang

Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan. e-mail: gjchang@math.nctu.edu.tw

Abstract. Suppose G is a graph and T is a set of non-negative integers that contains 0. A T-coloring of G is an assignment of a non-negative integer f …x† to each vertex x of G such that j f …x† ÿ f …y†j B T whenever xy A E…G†. The edge span of a T-coloring f is the maximum value of j f …x† ÿ f …y†j over all edges xy, and the T-edge span of a graph G is the minimum value of the edge span of a T-coloring of G. This paper studies the T-edge span of the dth power Cd

n of the n-cycle Cn for T ˆ f0; 1; 2; . . . ; k ÿ 1g. In particular, we ®nd the

exact value of the T-edge span of Cd

n for n 1 0 or 1 (mod d ‡ 1), and lower and upper

bounds for other cases.

1. Introduction

T-colorings were introduced by Hale [3] in connection with the channel assignment problem in communications. In this problem, there are n transmitters x1; x2; . . . ; xn

situated in a region. We wish to assign to each transmitter x a frequency f …x†. Some of the transmitters interfere because of proximity, meteorological, or other reasons. To avoid interference, two interfering transmitters must be assigned fre-quencies such that the absolute di¨erence of their frefre-quencies does not belong to the forbidden set T of non-negative integers and T contains 0. The objective is to make a frequency assignment that is e½cient according to certain criteria, while satisfying the above constraint.

To formulate the channel assignment problem graph-theoretically, we con-struct a graph G in which V…G† ˆ fx1; x2; . . . ; xng, and there is an edge between

transmitters xiand xj if and only if they interfere. Given graph G and a set T of

non-negative integers and T contains 0, a T-coloring of G is a function f from V…G† to the set of non-negative integers such that

xy A E…G† implies j f …x† ÿ f …y†j B T:

For the case when T ˆ f0g, T-coloring is the ordinary vertex coloring.

In channel assignments, the objective is to allocate the channels e½ciently. From the T-coloring standpoint, three criteria are important for measuring the

(2)

e½ciency: ®rst, the order of a T-coloring, which is the number of di¨erent colors used in f; second, the span of f, which is the maximum of j f …x† ÿ f …y†j over all vertices x and y; and third, the edge span of f, which is the maximum of j f …x† ÿ f …y†j over all edges xy. Given T and G, the T-chromatic number wT…G† is the minimum order of a T-coloring of G, the T-span spT…G† is the minimum span of a T-coloring of G, and the T-edge span espT…G† is the minimum edge span of a T-coloring of G.

Cozzens and Roberts [1] showed that the T-chromatic number wT…G† is equal to the chromatic number w…G†, which is the minimum number of colors needed to color the vertices of G so that adjacent vertices have di¨erent colors. The param-eter T-span of graphs has been studied extensively; for a good survey, see [6]; for recent results, see [2, 5, 7]. However, comparing to T-spans, there are relatively fewer known results about T-edge spans of graphs, see [1, 4].

Cozzens and Roberts [1] raised the problem of computing T-edge spans of non-perfect graphs when T ˆ f0; 1; 2; . . . ; k ÿ 1g. Liu [4] studied this problem for odd cycles. In this article, we consider Cd

n, the dth power of the n-cycle Cn. The

graph Cd

n has the vertex set V…Cnd† ˆ fv0; v1; . . . ; vnÿ1g and the edge set

E…Cd

n† ˆ 6

0UiUnÿ1fvivj: j ˆ i ‡ 1; i ‡ 2; . . . ; i ‡ dg;

where the index j for vjis taken modulo n. We ®nd the exact value of espT…Cnd† for

n 1 0 or 1 (mod d ‡ 1), and lower and upper bounds for other cases. 2. Previous results

In this section, we quote some known results about T-spans and T-edge spans, some of which will be used in Section 3.

The clique number o…G† of G is the maximum order of a clique (complete graph), a set of pairwise adjacent vertices. A complete graph of order n is denoted by Kn. The n-cycle is the graph Cn with vertex set V…Cn† ˆ fv0; v1; . . . ; vnÿ1g and

edge set E…Cn† ˆ fv0v1; v1v2; . . . ; vnÿ2vnÿ1; vnÿ1v0g. Note that Cn1is Cn.

The following are some known results on T-spans and T-edge spans.

Theorem 1. (Cozzens and Roberts [1]) The following statements hold for all graphs G and sets T.

(1) w…G† ÿ 1 U espT…G† U spT…G†.

(2) spT…KoG†† U espT…G† U spT…G† U spT…Kw…G††.

(3) If T is …k ÿ 1†-initial, i.e., T ˆ f0; 1; . . . ; k ÿ 1g U S where S contains no multi-ple of k, then spT…G† ˆ spT…Kw…G†† ˆ k…w…G† ÿ 1†.

Theorem 2. (Liu [4]) For any odd cycle Cn and T ˆ f0; 1; . . . ; k ÿ 1g;

espT…Cn† ˆ …n ‡ 1†kn ÿ 1

 

.

Figure 1 shows an example of Cn with T ˆ f0; 1; 2g for which wT…C7† ˆ

(3)

3. Edge spans for powers of n-cycles

This section gives results for T-edge spans of Cd

n for the …k ÿ 1†-initial set

T ˆ f0; 1; 2; . . . ; k ÿ 1g. We note that Cd

n G Kn for d V n2

j k

and espT…Kn† ˆ spT…Kn† ˆ k…n ÿ 1†.

Therefore, throughout this article we consider Cd

n only for d U n2

j k

ÿ 1 and assume n ˆ m…d ‡ 1† ‡ r, where m V 2 and 0 U r U d. Our main results are as follows. First, we give an upper bound and a lower bound for espT…Cd

n† (Theorem

4), both of them imply the exact value of esp…Cd

n† when r ˆ 0 (Theorem 5). We

then give a better upper bound when gcd…n; d ‡ 1† ˆ 1 (Theorem 6) and a better lower bound when r V 1 (Theorem 7), both of them imply the exact value when r ˆ 1 (Theorem 8).

Lemma 3. If n ˆ m…d ‡ 1† ‡ r with m V 2 and 0 U r U d, then w…Cd

n† ˆ d ‡ 1 and

w…Cd n† ˆ mn

l m

ˆ d ‡ 1 ‡l mmr . Proof. It is easy to see that o…Cd

n† ˆ d ‡ 1 since d ‡ 1 U n2

j k

; and w…Cd n† V mn

l m since any independent set of Cd

n contains at most m vertices. Letting niˆ n ÿ im

  , we have n ˆXmÿ1 iˆ0 ni:

Color the n vertices of Cd

n as 1; 2; . . . ; n0; 1; 2; . . . ; n1; 1; 2; . . . ; n2; . . . ; 1; 2; . . . ; nmÿ1.

This coloring is a proper vertex coloring since eachn ÿ i m V n ÿ m ‡ 1 m ˆ d ‡ r ‡ 1 m and so n ÿ im   V d ‡ 1. Hence w…Cd n† U mn l m . r

Theorem 4. If n ˆ m…d ‡ 1† ‡ r with m V 2 and 0 U r U d, then dk U espT…Cd

n† U spT…Cnd† ˆ dk ‡ mr

l m k.

Proof. The theorem follows from Theorem 1 and Lemma 3. r

(4)

Theorem 5. If n ˆ m…d ‡ 1† with m V 2, then espT…Cd

n† ˆ spT…Cnd† ˆ dk.

Proof. The theorem follows from Theorem 4 as r ˆ 0. r Theorem 6. Suppose n ˆ m…d ‡ 1† ‡ r with m V 2 and 0 U r U d. If gcd…n; d ‡ 1† ˆ 1, then espT…Cd

n† U dk ‡ rkm

  .

Proof. Since gcd…n; d ‡ 1† ˆ 1; d ‡ 1 is a generator of Zn using modulo n

addition, i.e., ji1 i …d ‡ 1† (mod n) for 0 U i U n ÿ 1 generates each integer

in f0; 1; . . . ; n ÿ 1g exactly once. In other words, we can consider V…Cd n† as

fvj0; vj1; . . . ; vjnÿ1g. Note that any m circularly consecutive vertices vja‡1; vja‡2; . . . ;

vja‡m (with indices a ‡ p considered modulo n) form an independent set in

Cd

n. Consequently, vjavjb is not an edge when 0 U a < b U n ÿ 1 with 1 U

minfb ÿ a; n ‡ a ÿ bg U m ÿ 1.

Now, consider the function f on V…Cd

n† de®ned by f …vji† ˆ

ik m  

for 0 U i U n ÿ 1. We claim that f is a T-coloring. For any edge vjavjb with 0 U a <

b U n ÿ 1, according to the preceding discussion, minfb ÿ a; n ‡ a ÿ bg V m, i.e., m U b ÿ a U n ÿ m ˆ md ‡ r. Then j f …vja† ÿ f …vjb†j ˆ bk m   ÿ akm   Vbkmÿak ‡ m ÿ 1 m V k ÿ 1 ‡ 1 m; Ubk ‡ m ÿ 1m ÿakm U…md ‡ r†km ‡ 1 ÿm1; 8 > > > < > > > : or j f …vja† ÿ f …vjb†j V k; U dk ‡ rkm   : 8 > < > : Therefore, f is a T-coloring of Cd n and espT…Cnd† U dk ‡ rkm   . r

Theorem 7. If n ˆ m…d ‡ 1† ‡ r with m V 2 and 1 U r U d, then espT…Cd n† V

dk ‡ mk  

.

Proof. Suppose espT…Cd

n† U dk ‡ mk

 

ÿ 1. Let f be a T-coloring for which espT…Cd

n† ˆ maxfj f …vi† ÿ f …vj†j : vivj A E…Cnd†g. Note that the m ‡ 1 vertices

vi…d‡1†; 0 U i U m, are pairwise non-adjacent except for v0vm…d‡1†A E…Cnd†. Let

ei; jˆ f …vi…d‡1†† ÿ f …vj…d‡1†† for 0 U i U j U m. Then

k U je0; mj ˆ Xmÿ1 iˆ0 ei; i‡1 U X mÿ1 iˆ0 jei; i‡1j

and so there exists at least one i such that jei; i‡1j V mk

 

. In other words, the set U ˆ fi : jei; i‡1j V mk

 

(5)

For any i A U, the d ‡ 2 vertices vj; i…d ‡ 1† U j U …i ‡ 1†…d ‡ 1†; are

pair-wise adjacent except that vi…d‡1†is not adjacent to v…i‡1†…d‡1†. Sort the d ‡ 2 values

f …vj†; i…d ‡ 1† U j U …i ‡ 1†…d ‡ 1†; into b1U b2U    U bd‡2. If fb1; bd‡2g 0

f f …vi…d‡1††; f …v…i‡1†…d‡1††g, then

esp…Cd n† V bd‡2ÿ b1ˆ Xd‡1 jˆ1 …bj‡1ÿ bj† V dk ‡ mk   ; a contradiction. Hence, fb1; bd‡2g ˆ f f …vi…d‡1††; f …v…i‡1†…d‡1††g and

jei; i‡1j ˆ j f …vi…d‡1†† ÿ f …v…i‡1†…d‡1††j ˆ

Xd‡1 jˆ1 …bj‡1ÿ bj† V …d ‡ 1†k: Also, bd‡2ÿ b2U esp…Cnd† U dk ‡ mk   ÿ 1; bd‡1ÿ b1U esp…Cnd† U dk ‡ mk   ÿ 1;

bi‡1ÿ biV k for 2 U i U d and so; bd‡1ÿ b2V …d ÿ 1†k:

Then jei; i‡1j ˆ bd‡2ÿ b1U …d ‡ 1†k ‡ 2 mk   ÿ 2. In conclusion, …d ‡ 1†k U jei; i‡1j U …d ‡ 1†k ‡ 2 mk   ÿ 2 for all i A U: On the other hand, jei; i‡1j U mk

 

ÿ 1 for all i B U. Let U be the disjoint union of U1and U2such that jU1j V jU2j and all ei; i‡1in U1(or U2) are of the same sign.

For the case jU1j > jU2j, we have

espT…Cd n† V jeo; mj ˆ X mÿ1 iˆ0 ei; i‡1 V X i A U1 jei; i‡1j ÿ X i A U2 jei; i‡1j ÿ X i B U jei; i‡1j V jU1j…d ‡ 1†k ÿ jU2j …d ‡ 1†k ‡ 2 mk   ÿ 2   ÿ …m ÿ jUj† mk   ÿ 1   ˆ …jU1j ÿ jU2j†…d ‡ 1†k ‡ …jU1j ÿ jU2j ÿ m† mk   ÿ 1   V …d ‡ 1†k ‡ …1 ÿ m† mk   ÿ 1   > dk ‡ mk   ÿ 1 since k > m mk   ÿ 1     ; a contradiction.

(6)

For the case jU1j ˆ jU2j, say Uiˆ fi1; i2; . . . ; iag for i ˆ 1; 2. Then k  je0; mj ˆ Xmÿ1 iˆ0 ei; i‡1 U Xa jˆ1 …e1j;1j‡1‡ e2j;2j‡1† ‡ X i B U jei; i‡1j UXa jˆ1 …d ‡ 1†k ‡ 2 mk   ÿ 2 ÿ …d ‡ 1†k   ‡X i B U k m   ÿ 1   ˆ a 2 mk   ÿ 2   ‡ …m ÿ 2a† mk   ÿ 1   ˆ m mk   ÿ 1   < k; a contradiction. r

Theorem 8. If n ˆ m…d ‡ 1† ‡ 1 with m V 2, then espT…Cd

n† ˆ dk ‡ mk

  .

Proof. The theorem follows from Theorems 6 and 7 and the fact that

gcd…n; d ‡ 1† ˆ 1. r

Note that Theorem 2 is a special case to the above theorem when n is odd and d ˆ 1. For the case where n V 5 is odd and d ˆn ÿ 32 , we have r ˆ 1; m ˆ 2, and Cd

n is isomorphic to the complement Cnof Cn. Thus, we have the following result.

Corollary 9. If n V 5 is odd, then espT…Cn† ˆ …n ÿ 2†k2

 

.

Acknowledgments. The authors thank a referee and Daphne Der-Fen Liu for many useful suggestions on revising the paper.

References

1. Cozzens, N.B., and Roberts, F.S.: T-colorings of graphs and the channel assignment problem, Congr. Numerantium 35, 191±208 (1982)

2. Griggs, J.R., and Liu, D.D.-F.: Channel assignment problem for mutually adjacent sites, J. Comb. Theory, Ser. A 68, 169±183 (1994)

3. Hale, W.K.: Frequency assignment: theory and applications, Proc. IEEE 68, 1497± 1514 (1980)

4. Liu, D.D.-F.: Graph Homomorphisms and the Channel Assignment Problem, Ph.D. Thesis, Department of Mathematics, University of South Carolina, Columbia, SC, 1991

(7)

5. Liu, D.D.-F.: T-graphs and the channel assignment problem, Discrete Math. 161, 198± 205 (1996)

6. Roberts, F.S.: T-colorings of graphs: recent results and open problems, Discrete Math. 93, 229±245 (1991)

7. Tesman, B.A.: List T-colorings of graphs, Discrete Appl. Math. 45, 277±289 (1993)

Received: May 13, 1996 Revised: December 8, 1997

數據

Fig. 1. C 7 with T ˆ f0; 1; 2g

參考文獻

相關文件

In particular, if s = f(t) is the position function of a particle that moves along a straight line, then f ′(a) is the rate of change of the displacement s with respect to the

Particles near (x, y, z) in the fluid tend to rotate about the axis that points in the direction of curl F(x, y, z), and the length of this curl vector is a measure of how quickly

Figure 6 shows the relationship between the increment ∆y and the differential dy: ∆y represents the change in height of the curve y = f(x) and dy represents the change in height

The best way to picture a vector field is to draw the arrow representing the vector F(x, y) starting at the point (x, y).. Of course, it’s impossible to do this for all points (x,

1、曾擔任以國家、重要城市為名,至少以二個版面以上刊登國際 新聞,且發行對象以全國或全球讀者為目標之平面媒體或通訊 社(例如:《美國新聞與世界報導》(U.S. News

The function f (m, n) is introduced as the minimum number of lolis required in a loli field problem. We also obtained a detailed specific result of some numbers and the upper bound of

If x or F is a vector, then the condition number is defined in a similar way using norms and it measures the maximum relative change, which is attained for some, but not all

Study the following statements. Put a “T” in the box if the statement is true and a “F” if the statement is false. Only alcohol is used to fill the bulb of a thermometer. An