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Broadcast Chromatic Numbers of Graphs

Wayne Goddard, Sandra M. Hedetniemi, Stephen T. Hedetniemi

Clemson University

{goddard,shedet,hedet}@cs.clemson.edu

John M. Harris, Douglas F. Rall

Furman University

{John.Harris,Doug.Rall}@furman.edu

Abstract

A function π:V→ {1,...,k}is a broadcast coloring of order k

if π(u) = π(v) implies that the distance between uand vis more

than π(u). The minimum order of a broadcast coloring is called the

broadcast chromatic number of G, and is denoted χb(G). In this pa-

per we introduce this coloring and study its properties. In particular,

we explore the relationship with the vertex cover and chromatic num-

bers. While there is a polynomial-time algorithm to determine whether

χb(G)≤3, we show that it is NP-hard to determine if χb(G)≤4. We

also determine the maximum broadcast chromatic number of a tree,

and show that the broadcast chromatic number of the inﬁnite grid is

ﬁnite.

1 Introduction

The United States Federal Communications Commission has established

numerous rules and regulations concerning the assignment of broadcast fre-

quencies to radio stations. In particular, two radio stations which are as-

signed the same broadcast frequency must be located suﬃciently far apart

so that neither broadcast interferes with the reception of the other. The

0Correspondence to: W. Goddard, Dept of Computer Science, Clemson University,

Clemson SC 29634-0974, USA

1

geographical distance between two stations which are assigned the same fre-

quency is, therefore, directly related to the power of their broadcast signals.

These frequency, or channel, assignment regulations have inspired a vari-

ety of graphical coloring problems. One of these is the well-studied L(2,1)-

coloring problem [2]. Let d(u, v) denote the distance between vertices u

and v, and let e(u) denote the eccentricity of u. Given a graph G= (V, E),

an L(2,1)-coloring is a function c:V→ {0,1,...}such that (i) d(u, v) = 1

implies |c(u)−c(v)| ≥ 2, and (ii) d(u, v) = 2 implies |c(u)−c(v)| ≥ 1. For

a survey of frequency assignment problems, see [4].

In a similar way, Dunbar et al. [1] deﬁne a function b:V→ {0,1,...}to

be a dominating broadcast if for every u∈V(i) b(u)≤e(u), and (ii) b(u) = 0

implies there exists a vertex v∈Vwith b(v)>0 and d(u, v)≤b(v). A

broadcast is called independent if b(u) = b(v) implies that d(u, v)> b(u);

that is, broadcast stations of the same power must be suﬃciently far apart

so that neither can hear each other’s broadcast.

In this paper we introduce a new type of graph coloring. A function

π:V→ {1,...,k}is called a broadcast coloring of order kif π(u) = π(v)

implies that d(u, v)> π(u). The minimum order of a broadcast coloring of a

graph Gis called the broadcast chromatic number, and is denoted by χb(G).

Equivalently, a broadcast coloring is a partition Pπ={V1, V2,...,Vk}of V

such that each color class Viis an i-packing (pairwise distance more than i

apart). Note that in particular, every broadcast coloring is a proper coloring.

Also, if His a subgraph of G, then χb(H)≤χb(G).

Throughout this article, we assume that graphs are simple: no loops or

multiple edges. For terms and concepts not deﬁned here, see [3]. In particu-

lar, we shall use the following notation: α0(G) for the vertex cover number,

β0(G) for the independence number, χ(G) for the chromatic number, ω(G)

for the clique number, and ρr(G) for the largest cardinality of an r-packing.

2

2 Basics

Every graph Gof order nhas a broadcast coloring of order n, since one can

assign a distinct integer between 1 and nto each vertex in V. There is a

better natural upper bound.

Proposition 2.1 For every graph G,

χb(G)≤α0(G) + 1,

with equality if Ghas diameter two.

Proof. For the upper bound, give color 1 to every vertex in a maximum

independent set in G. Then give every other vertex a distinct color. Since

n−β0(G) = α0(G) by Gallai’s theorem, the result follows.

If a graph has diameter two, then no two vertices can receive the same

color i, for any i≥2. On the other hand, since the vertices which receive the

color 1 form an independent set, there are at most β0(G) such vertices. 2

From this it follows that the complete, the complete multipartite graphs,

and the wheels have broadcast chromatic number one more than their ver-

tex cover number. It also follows that computing the broadcast chromatic

number is NP-hard (since vertex cover number is NP-hard for diameter 2).

If a graph is bipartite and has diameter three, then there is also near

equality in Proposition 2.1.

Proposition 2.2 If Gis a bipartite graph of diameter 3, then α0(G)≤

χb(G)≤α0(G) + 1.

Proof. By the diameter constraint, each color at least 3 appears at most

once. Since the graph is bipartite of diameter 3, color 2 cannot be used

twice on the same partite set, and thus can be used only twice overall. 2

For an example of equality in the upper bound, consider the 6-cycle;

for the lower bound, consider the 6-cycle where two antipodal vertices have

been duplicated.

3

We now present the broadcast chromatic numbers for paths and cycles.

Proposition 2.3 For 2≤n≤3,χb(Pn) = 2; and for n≥4,χb(Pn) = 3.

Proof. The ﬁrst nentries in the pattern below represent a broadcast col-

oring for Pn.

12131213...

This is clearly best possible. 2

Proposition 2.4 For n≥3, if nis 3or a multiple of 4, then χb(Cn) = 3;

otherwise χb(Cn) = 4.

Proof. Since the cycle contains either P4or K3,χb(Cn)≥3. Let v0, v1,

...,vn−1, v0be the vertices of Cnand suppose there is a broadcast coloring

of order 3 with n≥4.

Then there cannot be two consecutive vertices neither of which has

color 1. For suppose that v2has color 2 and v3has color 3, say. Then

neither v0nor v1can receive color 2 or 3, and only one can receive color 1,

a contradiction. Since vertices with color 1 cannot be consecutive, it follows

that the vertices with color 1 alternate. In particular, nis even.

Say the even-numbered vertices have color 1. But then no two consec-

utive odd-numbered vertices can receive the same color, and so they must

alternate between colors 2 and 3. In particular, nis a multiple of 4. It

follows that if nis not a multiple of 4, then χb(Cn)≥4.

We consider now optimal broadcast colorings. For cycles with order na

multiple of 4, the pattern

1,2,1,3,1,2,1,3,...,1,2,1,3

is a broadcast coloring. When nis not a multiple of 4, the pattern consists

of repeated blocks of “1,2,1,3” with an adjustment at the very end:

n= 4r+ 1 : 1,2,1,3,1,2,1,3,...,1,2,1,3,4

n= 4r+ 2 : 1,2,1,3,1,2,1,3,...,1,2,1,3,1,4

n= 4r+ 3 : 1,2,1,3,1,2,1,3,...,1,2,1,3,1,2,4

4

Hence, if nis a multiple of 4, then χb(Cn)≤3; otherwise χb(Cn)≤4. 2

Just as the natural upper bound involves the vertex cover number, the

natural lower bound involves the chromatic number.

Proposition 2.5 For every graph G,

ω(G)≤χ(G)≤χb(G).

It would be nice to characterize those graphs where the broadcast chro-

matic number is equal to the clique number. It is certainly necessary that

the neighbors of any maximum clique form an independent set, and at least

one vertex of such a clique has no neighbors outside the clique (so it can

receive color 1). If the graph Gis a split graph, then this necessary condition

is suﬃcient. (Recall that a split graph is a graph whose vertex set can be

partitioned into two sets, Aand B, where Ainduces a complete subgraph

and Bis an independent set.)

On the other hand, a necessary condition for χb(G) = χ(G) is that the

clique number be large.

Proposition 2.6 For every graph G, if χb(G) = χ(G)then ω(G)≥χ(G)−2.

Proof. For, assume χb(G) = χ(G) = m. Consider the broadcast coloring as

a proper coloring. If one can reduce the color of every vertex colored mwhile

still maintaining a proper coloring, then one has a contradiction. So there

exists a vertex vmthat has a neighbor of each smaller color: say v1,...,vm−1

with vihaving color i(viis unique for i≥2). Now, if one can reduce the

color of the vertex vm−1, this too will enable one to lower the color of vm.

It follows that vm−1has a neighbor of each smaller color. By the properties

of the broadcast coloring, that neighbor must be vifor i≥3. By repeated

argument, it follows that the vertices v3, v4,...,vmform a clique. 2

In another direction, we note that if one has a broadcast coloring, then

one can choose any one color class Vito be a maximal i-packing. (Recolor

vertices far away from Viwith color iif necessary.) However, one cannot

ensure that all color classes are maximal i-packings.

5

3 Graphs with Small Broadcast Chromatic Num-

ber

We show here that there is an easy algorithm to decide if a graph has broad-

cast chromatic number at most 3. In contrast, it is NP-hard to determine

if the broadcast chromatic number is at most 4. We start with a character-

ization of graphs with broadcast chromatic number 2.

Proposition 3.1 For any connected graph G,χb(G) = 2 if and only if G

is a star.

Proof. We know that the star K1,m has broadcast chromatic number 2.

So assume χb(G) = 2. Then Gdoes not contain P4and diam(G)≤2. By

Proposition 2.1, it follows that α0(G) = 1; that is, Gis a star. 2

As regards those graphs with χb(G) = 3, we start with a characterization

of the blocks with this property. If Gis a graph, then we denote by S(G)

the subdivision graph of G, which is obtained from Gby subdividing every

edge once. In S(G) the vertices of Gare called the original vertices; the

other vertices are called subdivision vertices.

Proposition 3.2 Let Gbe a 2-connected graph. Then χb(G) = 3 if and

only if Gis either S(H)for some bipartite multigraph Hor the join of K2

and an independent set.

Proof. Assume that χb(G) = 3. Let π:V(G)→ {1,2,3}be a broadcast

coloring and let Vi=π−1(i), for 1 ≤i≤3. It follows that Viis an i-packing

for each i.

Let v∈V1. The set V1is an independent set, so N(v)⊆V2∪V3. But v

has at most one neighbor in V2, since V2is a 2-packing. Similarly, vhas at

most one neighbor in V3. Since vhas at least two neighbors, it follows that

vhas degree 2 and is adjacent to exactly one vertex in each of V2and V3.

It follows from Proposition 2.4 that the length of any cycle in Gis either 3

or a multiple of 4. Assume Gcontains a triangle {x, y, z}. These vertices

6

receive diﬀerent colors; say x∈V2and y∈V3. Then for every neighbor t

of xapart from y, it follows that t∈V1and (since it is too close to yto

have another neighbor of color 3) that N(t) = {x, y}. Since ycannot be a

cut-vertex, it follows that this is the whole of G.

So assume that every cycle length is a multiple of 4. By the proof of

Proposition 2.4, in any broadcast coloring of order 3 of a cycle, every alter-

nate vertex receives color 1. It follows that V2∪V3is an independent set.

(Since Gis a block, every edge lies in a cycle.) In particular, Gis the subdi-

vision of some multigraph Hwhere every subdvision vertex receives color 1.

Furthermore, since V2and V3are 2-packings in G, they are independent sets

in H; that is, (V2, V3) is a bipartition of H.

Conversely, to broadcast color the subdivision of a bipartite multigraph,

take V1as the subdivision vertices, and (V2, V3) as the original bipartition. 2

Now, in order to characterize general graphs with broadcast chromatic

number 3, we deﬁne a T-add to a vertex vas introducing a vertex wvand

a set Xvof independent vertices, and adding the edge vwvand some of the

edges between {v, wv}and Xv. By extending the above result one can show:

Proposition 3.3 Let Gbe a graph. Then χb(G) = 3 if and only if Gcan

be formed by taking some bipartite multigraph Hwith bipartition (V2, V3),

subdividing every edge exactly once, adding leaves to some vertices in V2∪V3,

and then performing a single T-add to some vertices in V3.

Thus there is an algorithm for determining whether a graph has broad-

cast chromatic number at most 3. The key is that the colors 2 and 3 can

seldom be adjacent. In particular, if vertices uand vare adjacent with u

with color 2 and vwith color 3, then any neighbor aof uapart from vhas

N(a)⊆ {u, v}. Apart from that, V2∪V3must be an independent set, while

every vertex of V1has degree at most 2. In particular, if two vertices with

degree at least 3 are joined by a path of odd length, then at one of the ends

of this path there must be two consecutive V2∪V3vertices.

So a graph can be tested for having broadcast chromatic number 3 by

identifying the places where V2and V3must be adjacent, coloring and trim-

7

ming these appropriately, trimming leaves that are in V1, and then seeing

whether what remains with the partial coloring is a subdivision of a bipartite

graph. We omit the details.

4 Intractable Colorings

In contrast to the above, the problem of determining whether a graph has

a broadcast 4-coloring is intractable. We will need the following gener-

alization. For a sequence of positive integers s1≤s2≤... ≤sk, an

(s1, s2,...,sk)-coloring is a weak partition π= (V1, V2, . . . , Vk), where Vj

is an sj-packing for 1 ≤j≤k. Then we deﬁne the decision problem:

(s1, s2,...,sk)-Coloring

Instance: Graph G

Question: Does Ghave an (s1, s2,...,sk)-coloring?

For example, a 3-coloring is a (1,1,1)-coloring. The Broadcast 4-

Coloring problem is equivalent to the (1,2,3,4)-Coloring problem. We

will need the intractability of a special 3-coloring problem.

Proposition 4.1 (1,1,2)-Coloring is NP-hard.

Proof. The proof is by reduction from normal 3-coloring. The reduction

is to form G0from Gas follows. Replace each edge uv by the following:

add a pentagon Puv and join u, v to nonadjacent vertices of Puv; add a

pentagon Quv and join u, v to adjacent vertices of Quv and add an edge

joining two degree-two vertices of Quv. (See Figure 1.) The vertices u, v are

called original in G0.

We claim that G0has a (1,1,2)-coloring iff Gis 3-colorable.

Assume that Ghas a 3-coloring with colors red, blue and gold. We will

3-color G0such that the gold vertices form a 2-packing. Start by giving the

original vertices of G0their color in G. For each two adjacent vertices u, v

of G, color gold one vertex from each of Puv and Quv chosen as follows.

8

Quv

Puv

u v

Figure 1: Replacing an edge uv

For Quv it is the degree-2 vertex; for Puv it is the degree-2 vertex that is

distance-3 from whichever of uor vis gold if one of them is gold, and it is

the degree-2 vertex at distance 2 from both otherwise. Then it is easy to

color the remaining vertices in Puv and Quv with red and blue.

Conversely, suppose G0has a (1,1,2)-coloring πwhere the gold vertices

form a 2-packing. Then adjacent vertices uand vcannot have the same

color. For, if they are both red or both blue, then there is no possible

coloring of Quv (since one of the vertices in the triangle is gold); if they are

both gold, then there is no possible coloring of Puv. That is, restricted to

V(G), the coloring πis a 3-coloring of G.2

Theorem 4.2 Broadcast 4-Coloring is NP-hard.

Proof. We reduce from (1,1,2)-Coloring as follows. Given a connected

graph H, form graph H0by quadrupling each edge and then subdividing each

edge. Thus, all original vertices have degree at least 4 and all subdivision

vertices have degree 2.

The (1,1,2)-coloring of Hwith red, blue and gold, becomes a broadcast

4-coloring (V1, V2, V3, V4) of H0by making V1all the subdivision vertices, V2

all the red vertices, V3all the blue vertices and V4all the gold vertices. On

the other hand, in a broadcast 4-coloring of H0, none of the original vertices

9

can receive color 1. If we maximize the number of vertices receiving color 1,

it follows that all subdivision vertices receive color 1, and all original vertices

receive color 2, 3, or 4. So, the vertices colored 2 or 3 form an independent

set in Hand the vertices colored 4 form a 2-packing in H. Thus we have a

(1,1,2)-coloring of H.2

Comment: This proves that Broadcast 4-Coloring is also NP-hard

for planar graphs. It is an open question what the complexity is for cubic

or 4-regular graphs. In another direction, it is easy to determine whether a

graph is (2,2,2)-colorable (as only paths of any length and cycles of length

a multiple of 3 are). But what is the complexity of (1,2,2)-Coloring?

5 Trees

We are interested in trees with large broadcast chromatic numbers. In order

to prove the best possible general result, it is necessary to examine the small

cases.

A tree of diameter 2 (that is, a star) has broadcast chromatic number 2.

A tree of diameter 3 has broadcast chromatic number 3. The case of a

tree of diameter 4 is more complicated, but one can still write down an

explicit formula. We say that a vertex is large if it has degree 4 or more,

and small otherwise. The key to the formula is the numbers of large and

small neighbors of the central vertex.

Proposition 5.1 Let Tbe a tree of diameter 4with central vertex v. For

i= 1,2,3, let nidenote the number of neighbors of vof degree i, and let L

denote the number of large neighbors of v. If L= 0 then

χb(T) = (4if n3≥2and n1+n2+n3≥3

3otherwise,

and if L > 0then

χb(T) =

L+ 3 if n3≥1and n1+n2+n3≥2

L+ 1 if n1=n2=n3= 0

L+ 2 otherwise.

10

2 5 6

23

4

Figure 2: The tree T5: the unlabeled vertices have color 1

Proof. To show the upper bound we need to exhibit a broadcast coloring.

A simple coloring is to color the center and the leaves not adjacent to

it with color 1, and the remaining vertices with unique colors. This uses

L+n3+n2+n1+ 1 colors. It is optimal when n1=n2= 0 and either

2≤n3≤3 and L= 0 or 0 ≤n3≤2 and L > 0. (Note that L+n3+n2≥2

by the diameter condition.)

Another good coloring is as follows: put color 1 on the small neighbors

of the center vand on the children of large neighbors; put color 2 on one

large neighbor (if one exists) and on one child of each small neighbor; put

color 3 on the remaining children of small neighbors; and put unique colors

on the remaining vertices. If L= 0, then this uses 4 colors if n3>0 and 3

values otherwise. If L > 0, then this uses L+ 3 colors if n3>0 and L+ 2

colors otherwise. This coloring is illustrated in Figure 2.

The only case not covered by the above two colorings is when L= 0

and n3= 1. In this case 3 colors suﬃce: use color 3 on the central vertex,

color 2 on its degree-3 neighbor and the children of its degree-2 neighbors,

and color 1 on the remaining vertices.

For a lower bound, proceed as follows. If the center is colored 1, then

the coloring uses L+n3+n2+n1+ 1 colors, which is at least the above

bound. So we may assume that the center is not colored 1.

11

A8B8

Figure 3: The smallest trees with χb(T) = 4

If a large neighbor receives any color other than 2, then either it or one

of its children receives a unique color. Thus we may remove it and induct.

So we may assume that every large neighbor of the center receives color 2.

This means there is at most one large neighbor.

At least three colors are always needed. For the case that L= 0, it is

enough to argue that 4 colors are needed if n3= 2, n2= 0 and n1= 1 (as

any other case contains this as a subgraph). (This is tree A8in Figure 3.) If

any degree-3 vertex receives color 1, then three more colors are needed for

its neighbors. On the other hand, the degree-3 vertices induce a P3, and so

require three new colors if 1 is not used.

In fact, this observation also takes care of the case where there is only

one large neighbor. 2

Proposition 5.2 The minimum order of a tree with broadcast chromatic

number 2is 2. For 3it is 4and for 4it is 8. Furthermore, P4is the unique

tree on 4vertices that needs 3colors. The two trees on 8 vertices that need

4colors are (i) the diameter-4tree with n3= 2,n1= 1 and L=n2= 0,

called A8; and (ii) the diameter-6tree where the two central vertices have

degree-3and for each central vertex its three neighbors have degrees 1,2and

3respectively, called B8. (These are depicted in Figure 3.)

Proof. By the above result, A8is the unique smallest tree with diame-

ter 4 that needs four colors. So we need only examine the small trees with

diameter 5 or more, which is easily done. 2

In another direction there is an extension result.

12

Proposition 5.3 Let Tbe a graph but not P4. Suppose Tcontains a path

t, u, v, w where thas degree 1and uand vhave degree 2. Then χb(T) =

χb(T−t).

Proof. Take an optimal broadcast coloring of T−t. Since T6=P4,T−t

contains P4and hence uses at least three colors.

If ureceives any color except 1, then one can color twith 1. So assume

ureceives color 1. If vreceives any color except 2, then one can color twith

color 2. So assume vreceives color 2. If wreceives any color except 3, then

one can color twith color 3. So assume wreceives color 3. But then one

can recolor as follows: vgets color 1, ugets color 2, and tgets color 1. 2

For example, this shows that χb(Pn) = 3 for all n≥4.

We are now ready to determine the maximum broadcast chromatic num-

ber of a tree. An extremal tree Tdfor d≥2 is constructed as follows: it

has diameter 4; n1=n3= 1, n2= 0, L=d−2 and all large vertices have

degree exactly 4. The tree Tdhas 4d−3 vertices and χb(Td) = d+ 1. The

tree T5is shown in Figure 2.

Theorem 5.4 For all trees Tof order nit holds that χb(T)≤(n+ 7)/4,

except when n= 4 or 8, when the bound is 1/4more, and these bounds are

sharp.

Proof. We have shown sharpness above. The proof of the bound is by

induction on n.

If n≤8 the result follows from Proposition 5.2. If diam(T)≤3, then

χb(T)≤3. If diam(T) = 4, then the bound follows from Proposition 5.1.

So assume n≥9 and diam(T)≥5.

Deﬁne a penultimate vertex as one with exactly one non-leaf neighbor.

Suppose some penultimate uhas degree 4 or more. Deﬁne T0to be the

tree after the removal of uand all its leaf-neighbors. Then take an optimal

broadcast coloring of T0, and extend to a broadcast coloring of Tby giving u

a new unique color and its leaf neighbors color 1. By the inductive hypothesis

13

it follows that χb(T)≤χb(T0) + 1 ≤(n+ 3)/4 + 1 = (n+ 7)/4, unless T0is

an exceptional tree.

But the three exceptional trees can be broadcast colored with 4 colors

so that, for any speciﬁc vertex v, neither it nor any of its neighbors receives

color 2. So let vbe u’s other neighbor and color T0thus; then color uwith

color 2 and its leaf-neighbors with color 1. So, in this case χb(T)≤4, which

establishes the bound.

So we may assume that every penultimate vertex has degree 2 or 3.

Now, deﬁne a late vertex as one that is not a penultimate, but at most

one of its neighbors is not a penultimate or a leaf. (For example, the third-

to-last vertex on a diametrical path.) For a late vertex v, deﬁne Tvas the

subtree consisting of v, all its penultimate neighbors, and any leaf adjacent

to one of these. Since diam(T)≥5, Tvis not the whole of T; let wbe v’s

other neighbor. Then deﬁne T0=T−Tv.

If |Tv|= 3, then vhas degree 2 and its penultimate neighbor has degree 2.

So by the above result, χb(T) = χb(T0) and we are done. Therefore we may

assume that |Tv| ≥ 4.

Give T0an optimal broadcast coloring. Then, if wis not colored 3, one

can extend this to a broadcast coloring of Tby giving va new unique color,

all its neighbors in Tvcolor 1 and the remaining vertices of Tvcolors 2 or 3.

We are done by induction—one new color for at least four vertices—unless

|Tv|= 4 and T0is an exceptional tree. But in this case one can readily argue

that χb(T)≤4.

So assume the vertex wreceives color 3 in any coloring of T0. (In partic-

ular, this means that T0is not one of the exceptional trees.) Now, we can

aﬀord to recolor wwith a new color and proceed as above if |Tv| ≥ 8. So

assume that |Tv| ≤ 7.

If vhas only one penultimate neighbor of degree 3, then one can color Tv

with colors 1 and 2 except for v, and so are done. So we may assume that v

has two degree-3 neighbors. But then vhas exactly two neighbors in Tvand

these have degree 3. But then one can color Tvwith colors 1 and 2, except

for one neighbor of v, with vreceiving color 1. And hence we are done by

14

the inductive hypothesis. 2

6 Grids

We will now investigate broadcast colorings of grids Gr,c with rrows and c

columns. The exact values for r≤5 are given in the following result

Proposition 6.1 χb(G2,c) = 5 for c≥6;χb(G3,c) = 7 for c≥12;χb(G4,c) =

8for c≥10; and χb(G5,c) = 9 for c≥10. The values for smaller grids are

as follows:

m\n2 3 4 5 6 7 8 9 10 11 12

2 3 4 4 4 5 . . .

3 4556666667. . .

4 5777778...

5 7 7 7 8 8 9 . . .

Proof. Consider the following coloring pattern. For c≥2, the ﬁrst c

columns indicate that the values for χb(G2,c) stated are in fact upper bounds.

214131...

131215...

Consider the following coloring pattern. For c≥3, the ﬁrst ccolumns

indicate that the values for χb(G3,c) stated are in fact upper bounds.

213121312131...

141516141517...

312131213121...

Consider the following coloring pattern. For c≥4, the ﬁrst ccolumns

indicate that the values for χb(G4,c) stated are in fact upper bounds.

1213121316...

3151714121...

1412131518...

2131612131...

15

Consider the following coloring pattern. Let iand jdenote the row and

column of a vertex, with 1 ≤i≤rand 1 ≤j≤c. Then assign color 1 to

every vertex with i+jodd; assign color 2 to every vertex with iand jodd

and i+jnot a multiple of 4; and assign color 3 to every other vertex with

iand jodd. The picture looks as follows.

2 1 3 1 2 1 3 1 . ..

1 – 1 – 1 – 1 – . ..

3 1 2 1 3 1 2 1 . ..

1 – 1 – 1 – 1 – . ..

2 1 3 1 2 1 3 1 . ..

(The uncolored vertices induce a copy of the grid in the square of the graph.)

For G5,c the uncolored vertices should be colored as follows:

46584759...

57495648...

It is to be noted that the patterns for G5,8and G5,9are exceptions.

Lower bounds in general can be veriﬁed by computer. Some can be

veriﬁed by hand. One useful idea is the following. For the lower bound for

G2,c where c≥6, note that any copy of G2,3contains a color greater than 3.

If one considers the three columns after a column containing a 4, then that

G2,3has at least one of its vertices colored 5 or greater. 2

The following table provides some more upper bounds. An asterisk in-

dicates an exact value.

m\n6 7 8 9 10 11 12 13 14 15 16

6 8∗9∗9∗9∗9∗9∗10 10 10 11 11

7 9∗9∗10 10 11 11 11 11 12 12

Often, the greedy approach produces a bound close to optimal. By the

time the grids have around 20 rows (together with several hundred columns),

the greedy approach uses more than 25 colors. As the following theorem

implies, these bounds are not best possible.

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Theorem 6.2 For any grid Gm,n,χb(Gm,n)≤23.

Proof. There is a broadcast coloring of the inﬁnite grid that uses 23 colors.

The coloring is illustrated below. This provides a broadcast coloring of any

ﬁnite subgrid.

As in the coloring of G5,n above, we start by coloring with 1s, 2s and 3s

such that the uncolored vertices occur in every alternate row and column.

Then the following coloring is used to tile the plane.

4 5 8 4 5 9 4 5 8 4 5 9

10 6 11 7 12 6 10 7 11 6 13 7

5 4 9 5 4 8 5 4 9 5 4 8

14 7 15 6 13 7 16 6 17 7 12 6

4 5 18 4 5 11 4 5 19 4 5 11

20 6 21 7 10 6 14 7 15 6 10 7

5 4 8 5 4 9 5 4 8 5 4 9

13 7 11 6 22 7 12 6 11 7 23 6

4 5 9 4 5 8 4 5 9 4 5 8

12 6 10 7 15 6 13 7 10 6 14 7

5 4 17 5 4 11 5 4 18 5 4 11

16 7 19 6 14 7 20 6 21 7 15 6

The coloring was found by placing the colors 4 through 9 in a speciﬁc pattern,

and then using a computer to place the remaining colors. 2

Schwenk [5] has shown that the broadcast chromatic number of the in-

ﬁnite grid is at most 22.

7 Other Grid-like Graphs

While the broadcast chromatic number of the family of grids is bounded,

this is not the case for the cubes. We start with a simple result on the

cartesian product 2with K2.

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Proposition 7.1 If χb(G)≥diam(G)+x, then χb(G2K2)≥diam(G2K2)+

2x−1.

Proof.

The result is true for x≤0, since χb(G2K2)≥χb(G). So assume x≥1.

Suppose there exists a broadcast coloring πof G2K2that uses at most

diam(G2K2) + 2x−2 colors. It follows that the 2x−1 biggest colors—call

them Z—are used at most once in G2K2. It follows that one of the copies

of Gis broadcast-colored by the colors up to diam(G2K2)−1 = diam(G)

together with at most x−1 colors of Z. Thus χb(G)≤diam(G) + x−1, a

contradiction. 2

For example, this shows that the broadcast chromatic number of the

cube is at least a positive fraction of its order. Next is a result regarding

the ﬁrst few hypercubes Qd.

Proposition 7.2 χb(Q1) = 2,χb(Q2) = 3,χb(Q3) = 5,χb(Q4) = 7, and

χb(Q5) = 15.

Proof. The values for Q1and Q2follow from earlier results. Consider

Q3. The upper bound is from Proposition 2.1. To see that ﬁve colors are

required, note that since β0(Q3) = 4, at most four vertices can be colored

1. Further, at most two vertices can be colored 2; but, if four vertices are

colored 1 then no more than one vertex can be colored 2. Therefore, the

number of vertices colored 1 or 2 must be at most ﬁve. Since diam(Q3) = 3,

no color greater than 2 can be used more than once. As there are eight

vertices in Q3, this means that at least ﬁve colors are required. The lower

bound for Q4follows from the value for Q3by the above proposition.

For a suitable broadcast coloring in each case, use the greedy algorithm

as follows. Place color 1 on a maximum independent set; then color with

color 2 as many as possible, then color 3 and so on. 2

We look next at the asymptotics. Bounds for the packing numbers of the

hypercubes are well explored in coding theory. For our purposes it suﬃces

18

to note the bounds:

ρj(Qk)≤2k

Pbj/2c

i=0 k

i.

and ρ2(Qk)≥2k−1/k by, for example, the (computer) Hamming code.

Proposition 7.3 χb(Qk)∼(1

2−O(1

k))2k.

Proof. For the upper bound, color a maximum independent set with color 1.

Then color as many vertices with color 2 as possible. Clearly one can choose

at least half a maximum 2-packing. And then use unique colors from there

on.

The lower bound is from the packing bounds above, together with the

fact that β0(Qk) = 2k−1.2

For the sake of interest, the following table gives some bounds computed

by using these approaches.

n6 7 8 9 10 11

χb(Qn)≥15 28 63 132 285 610

χb(Qn)≤25 49 95 219 441 881

We consider next another relative of the grid. Deﬁne a path of thick-

ness was the lexicographic product Pn[Kw], that is, every vertex of the path

is replaced by a clique of wvertices. A broadcast coloring of a thick path

is equivalent to assigning wdistinct colors to each vertex of the path and

requiring a broadcast coloring.

Let g(w) denote the broadcast chromatic number of the inﬁnite path of

thickness w. A natural approach is to assign one color to every vertex, then

a second and so on. So the following parameter arises naturally. Let f(m)

denote the broadcast chromatic number of the inﬁnite path given that no

color smaller than mis used.

Proposition 7.4 For all msuﬃciently large, f(m)≤3m−1. Indeed, for

all m f (m)≤3m+ 2.

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Proof. This result is easy to verify for small m. For the cases up to m= 33

a computer search produced a suitable coloring.

The bound f(m)≤3m−1 is by induction on m. The base case is

m= 34, where it can be checked by computer that the greedy algorithm

eventually settles into a cycle of length 176400 moves and uses no color more

than 101.

In general, take the broadcast coloring for f(m). Then replace the ver-

tices of color mwith the three colors in succession 3m, 3m+ 1,3m+ 2. The

result is still a broadcast coloring. 2

So if we have the thick path, it follows that g(w)≤(1 + o(1))3w. The

idea is to ﬁll the ﬁrst level, then the second and so on.

As for lower bounds, it holds for the path that at most 1/(i+ 1) of the

vertices can receive color i. Hence for H(s, t) = Pt

s1/i it follows that we

need tsuch that H(m, t)≤1. By standard estimates for the harmonic series,

H(1, t)∼ln(t)−γ, where γis the Euler constant. Thus f(m)≥em −o(m).

As for g(w), similar considerations imply that we need H(1, t)≥wso

that g(w)≥Ω(ew). It is unclear if this is the correct order of magnitude.

We note that we quickly run into problems with achieving a proportion of

1/(i+ 1) for any beyond the ﬁrst wcolors.

8 Open Problems

1. Can the broadcast chromatic number of a tree be computed in poly-

nomial time?

2. What is the maximum broadcast chromatic number of a grid (and

when is it ﬁrst obtained)?

3. What about other “grids” such as the three-dimensional grid or the

hexagonal lattice?

4. What is the maximum broadcast chromatic number of a cubic graph

on nvertices?

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9 Acknowledgment

The authors express their appreciation to Charlie Shi of Clemson University,

and to James Knisley of Bob Jones University. Their insightful comments

and eﬀorts helped to improve our results on broadcast colorings of grids.

References

[1] J. Dunbar, D. Erwin, T. W. Haynes, S. M. Hedetniemi, and S. T. Hedet-

niemi. Submitted.

[2] J. R. Griggs and R. K. Yeh, The L(2,1)-labeling problem on graphs,

SIAM J. Discrete Math.,9(1996) 309–316.

[3] W. Imrich and S. Klavˇzar, Product Graphs, John Wiley & Sons, New

York, 2000.

[4] R.A. Murphey, P.M. Pardalos and M.G.C. Resende, Frequency assign-

ment problems, In: Handbook of combinatorial optimization, Supplement

Vol. A, Kluwer Acad. Publ., 1999, 295–377.

[5] A. Schwenk, personal communication, 2002.

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