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November 2017

Edge Colorings of Complete Multipartite Graphs

Forbidding Rainbow Cycles

Peter Johnson

johnspd@auburn.edu

Andrew Owens

Auburn University, ado0004@tigermail.auburn.edu

Follow this and additional works at:https://digitalcommons.georgiasouthern.edu/tag Part of theDiscrete Mathematics and Combinatorics Commons

This article is brought to you for free and open access by the Journals at Digital Commons@Georgia Southern. It has been accepted for inclusion in Theory and Applications of Graphs by an authorized administrator of Digital Commons@Georgia Southern. For more information, please contact

digitalcommons@georgiasouthern.edu.

Recommended Citation

Johnson, Peter and Owens, Andrew (2017) "Edge Colorings of Complete Multipartite Graphs Forbidding Rainbow Cycles,"Theory and Applications of Graphs: Vol. 4 : Iss. 2 , Article 2.

DOI: 10.20429/tag.2017.040202

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Abstract

It is well known that if the edges of a finite simple connected graph onnvertices are colored so that no cycle is rainbow, then no more thann−1colors can appear on the edges. In previous work, it has been shown that the essentially different rainbow-cycle-forbidding edge colorings ofKnwithn−1colors appearing are in 1-1 correspondence with (can be encoded by) the (isomorphism classes of) full binary trees withnleafs. In the encoding, the natural Huffman labeling of each tree arising from the assignment of 1 to each leaf plays a role. Very recently, it has been shown that a similar encoding holds for rainbow-cycle-forbidding edge colorings ofKa,bwitha+b−1colors appearing. In this case the binary trees are given Huffman labelings arising from certain assignments of (0,1) or (1,0) to the leafs. (Sibling leafs are not allowed to be assigned the same label.) In this paper we prove the analogous result for completer-partite graphs, forr >2.

1

Introduction

SupposeHis a subgraph of a graphGand the edges ofGare colored. LetC[X]denote the set

of colors onG[X], the subgraph ofGinduced byX, for anyX ⊆ V(G). Let C = C[V(G)]

for short.

Definitions. H is said to be a rainbowsubgraph of G(with respect to the coloring of G) if

no two edges inH are the same color. We say that rainbow subgraphs from a certain class of

graphs areforbiddenby the coloring ofGif every subgraph ofGin that class of graphs is not

rainbow with respect to the coloring onG.

Ramsey problems in graphs are generally about coloring the edges of a graph with as few colors as necessary so that no subgraph of a specified class is monochromatic. In this paper we look at a particular anti-Ramsey problem: that is, a problem that involves coloring the edges of a graph with as many different colors as possible so that no member of a specified class of subgraphs is rainbow.

Reference [4] is an excellent survey of anti-Ramsey results, including a section onGallai

colorings, which are edge colorings of complete graphs that forbid rainbowK3’s. In [1]it is

proven that an edge coloring ofKnforbids rainbowK3’s if and only if it forbids rainbow cycles

of all lengths (this could be a well known folkloric result). So, one of the main results of[1], a

characterization of all edge colorings ofKnwithn−1colors appearing which forbid rainbow

cycles, is a theorem about certain kinds of Gallai colorings. [The result in [1] follows easily

from a characterization of Gallai colorings in general, discussed in [4], of which the authors of

[1]were unaware. However, the statement of the result in[1]does not arise in any obvious way

from the earlier result on Gallai colorings. Further, that statement and its proof, in[1], inspired

the main result in[2]which inspired the main result of this paper.]

We will prove in this paper a result for complete multipartite graphs with at least three parts

analogous to results already proven for complete graphs in[1]and complete bipartite graphs in

[2].

The following proposition is well known; the first author heard about it from a friend who heard about it from a friend, so it may be folkloric. We do not know of any particular reference for it, so we include its short proof.

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Proof. SupposeGis edge-colored with more than|V(G)| −1colors. Pick|V(G)|edges with

no two edges bearing the same color. Let H be the subgraph of G induced by these edges.

Because|E(H)| ≥ |V(H)|, H must necessarily contain a cycle and this cycle is necessarily

rainbow sinceH is rainbow.

Moreover, Proposition 1.2 shows that the maximum number of colors stated above can be

achieved in any connected graphG.

Proposition 1.2. If G is connected, there is an edge-coloring of G with |V(G)| − 1 colors appearing such that rainbow cycles are forbidden.

Proof. LetT be a spanning tree inG. Color the|V(G)| −1edges ofT so thatT is rainbow.

Pickv0 ∈ V(G) = V(T). Let v0 be the root ofT. We then sort the vertices ofG into levels

(distances within T from the root). DefineSj = {v ∈V(G)|distT(v, v0) = j}. SinceT is a

tree, eachv ∈Sj has exactly one neighbor inT inSj−1 forj >0. We saySj−1 is "above"Sj.

There are no edges ofT among the vertices ofSj; otherwiseT is not a tree.

We then orderV(G)\ {v0}(call the verticesv1, . . . , vn−1) where the first|S1|vertices are

an arbitrary ordering ofS1, the next|S2|vertices are an arbitrary ordering ofS2, and so on.

Ifvivj ∈ E(G)\E(T)and0≤ i < j ≤ n−1, colorvivj with the color of the edge ofT

which joinsvj with its unique neighbor on the level above it.

SupposeH is a cycle inG. Let j be the largest index such thatvj ∈ V(H). Then the two

edges incident tovj inHare the same color by our construction. SoHis not rainbow.

Corollary 1.3. For any graphG, the greatest number of colors that can appear in a rainbow-cycle-forbidding edge-coloring of G is |V(G)| −c where cis the number of components of

G.

Definitions. A connected graphGisJL-coloredif it is edge-colored with|V(G)| −1colors

appearing and rainbow cycles are forbidden. In an edge-colored graph, a color cis said to be

dedicatedto a vertexvif every edge coloredcis incident tov.

The last two propositions are necessary for the proof of the main result. We believe it is likely these two propositions will play a pivotal role in future work on JL-colorings.

Proposition 1.4. IfGis JL-colored, then every vertex inV(G)has a color dedicated to it.

Proof. Letn=|V(G)|. Ifv ∈ V(G)has no color dedicated to it, then alln−1colors appear

onG−v, which has onlyn−1vertices, and there are no rainbow cycles inG−v. But this is

impossible by Corollary 1.3.

Proposition 1.5. IfGis JL-colored, there are at least two vertices inGwith exactly one dedi-cated color each.

Proof. Let n = |V(G)|. We will count the number of ordered pairs in the set (v, c) : v ∈

V(G), c ∈ C andc is dedicated to v . For each v ∈ V(G), let dv be the number of colors

dedicated to v. Note that dv ≥ 1 for all v ∈ V(G) by Proposition 1.4. The number of such

ordered pairs(v, c)is

X

v∈V(G)

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The inequality holds since no color can be dedicated to more than two vertices. If each vertex

had two or more colors dedicated to it, then X

v∈V(G)

dv ≥ 2n.Similarly,

X

v∈V(G)

dv ≥ 2n−1if

exactly one vertex has one color dedicated to it. So we must have at least two vertices with exactly one color dedicated to each.

Definition. If a graph G is edge-colored, and all edges incident to v ∈ V(G) bear the same

color, thenv is said to beunicoloredin the coloring.

2

The Main Result

Theorem 2.1. LetG=Kn1,...,nmbe a complete multipartite graph with parts of sizen1, . . . , nm,

withm ≥3. An edge coloring of G is a JL-coloring if and only if there is a partition ofV(G)

into non-empty subsetsRandS which satisfy the following:

1. AllR−Sedges inGhave the same color (let us call it green).

2. The sets of colors on the complete multipartite subgraphsG[R]andG[S]induced byR

andS, respectively, are disjoint, and neither set contains green.

3. The induced colorings ofG[R]andG[S]are JL-colorings.

The main result of[2]is precisely Theorem 2.1 in the casem= 2.

Proof. Let|V(G)|=n. Suppose thatE(G)is colored, and thatV(G)is partitioned intoRand

Ssatisfying the stipulated requirements. Let|R|=rand|S|=s.

We verify that the coloring of G is a JL-coloring. Since the colorings of G[R] andG[S]

are JL-colorings, these colorings use r −1 and s −1 colors, respectively. Also, the set of

colors onG[R],G[S]are disjoint and neither contains green, so we see thatGis colored with

(r−1) + (s−1) + 1 =r+s−1 =n−1colors appearing. LetCbe any cycle inG. IfCis

contained in eitherG[R]orG[S], thenC is not rainbow since both subgraphs are JL-colored.

IfChas vertices in bothRandS, then, becauseCis a cycle,Cmust contain at least twoR−S

edges. Then two ofC’s edges are green, soCis not rainbow. Thus,Gis JL-colored.

Notice that the "if" claim of the theorem holds for any connected graphGif the requirement

thatG[R]andG[S]be connected is added.

The forward implication is more difficult. From here on assume thatGis JL-colored.

Note. If Ghas a JL-coloring and v ∈ V(G)is unicolored in this coloring, then the partition

R={v}andS =V(G)\ {v}satisfies the three conditions in the main theorem.

To see this, let green be the color incident tov. Then1), allR−Sedges are green.

For2), note thatG[R] =v =K1 has no edges. So certainly the sets of colors onG[R]and

G[S]are disjoint. Sincev is unicolored by the color green, green must be dedicated to v and

thus green cannot appear in the graphG[S].

Finally, for 3) we know that G[R] has a single vertex, so r = 1 and the number of

col-ors used is r − 1 = 1− 1 = 0. Also, G[S] has n −1 vertices so s = n − 1. The

JL-coloring ofGhasn−1colors and the color green is not used inG[S], soG[S]is colored with

n−2 = (n−1)−1 = s−1colors appearing. SinceGis JL-colored, we see that neitherG[R]

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From here, the proof proceeds by induction onn. At some points in the proof we may be

applying the induction hypothesis to a complete bipartite subgraph of G. The induction

hy-pothesis holds in such cases by the main result of[2].

We start with|V(G)|= 3. Since we assume the number of non-empty independent sets is

mwithm≥3, we havem= 3in our base case and there is one vertex in each set. ThusK3 is

the graph in the base case. For a JL-coloring ofK3, we must use two colors. Let green be the

color that appears on two edges and let v be the vertex incident to both of these green edges.

Thenv is unicolored and we are done.

Now we can assume|V(G)| ≥ 4. By Proposition 1.5, we can find a vertexv with exactly

one color dedicated to it. Let red be the color dedicated tov inG. SoG−v is colored with

n−2 = (n−1)−1 = |V(G−v)| −1colors appearing. SinceGhas no rainbow cycles,G−v

has no rainbow cycles, thusG−vis JL-colored. Then, by our induction hypothesis we have a

partitionR0 6=∅andS0 6=∅ofV(G)\ {v}that satisfies conditions 1), 2), and 3) of the main

theorem.

First, we take care of the special case where one of R0, S0 is a singleton. Without loss of

generality, letS0 = {u}. Then all edges touinG−v are green. We may assumeuis not in

v’s part (otherwise there is nouv edge and thusuis unicolored inG). For the same reason, we

may assume the edgeuv is not green. Letc6=green be the color onuv.

Case 1: Supposecis dedicated touin G.

Sincecis dedicated touinG,cdoes not appear inC[V(G)\{u}]. In particular, this implies

cis not inC[R0]. The only colors inC not inC[R0]are green and red. Thus cmust be red.

Since red is dedicated inGto bothv andu, then red appears only on the edgeuv.

Consider an edgevwwhere the vertexwis not inu’s part. SinceGis a complete

multipar-tite graph, there is a cycle inGwith edgesuv,uw, andvw. We knowuvis red anduwis green,

sovw is either red or green (sinceGhas no rainbow cycles). This impliesvw must be green

since red appears only onuv. Thus all edges fromv to parts other thanu’s part are green.

If all edges incident to v exceptuv are green, then the partition R = R0 and S = {u, v}

satisfies the desired conditions.

Now, we may assume for somew ∈ R0 inu’s part, the edge vwis not green. Letvw be

blue where blue ∈ C. If all edges incident to w are blue, then w is unicolored and we are

done. Suppose there is some edge wxnot colored blue. Note that x ∈ R0. Then wxcannot

be green sincew, x∈ R0. Also, it is not red since red only appears onuv. Letwxbe yellow.

This creates a rainbow 4-cycle whereuv is red,uxis green,wxis yellow andvwis blue. This

impossibility finishes Case 1 under the supposition thatmin (|S0|,|R0|) = 1.

Case 2: Supposecis not dedicated touin G.

Since green is the only other color incident tou, it follows that green is dedicated touinG

by Proposition 1.4. Therefore nov−R0 edge is green.

Subcase 1: Suppose c is red. Let’s consider the edgevw for anyw ∈ R0 not in u’s part.

Note thatG[{u, v, w}]is a three cycle. Also,uv is red anduwis green. This impliesvwis red

since green is dedicated tou. So everyvwis red for everyw∈R0wherewis not inu’s part.

Again, if v is unicolored we are done, so we can assume there is somex ∈ R0 inu’s part

wherevxis not colored red. Let’s sayvxis blue. As above, ifxis unicolored, we are done. So

for somey ∈R0 whereyis not inu’s part,xy /∈ {blue, green, red}. Then{uv, uy, xy, vx}is

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Subcase 2: Suppose c is not red. Soc∈C[R0]. Let’s saycis blue. As in subcase 1, since

green is dedicated inGtouwe know that all edgesvw, wherew∈R0andwis not inu’s part,

are blue.

Red is dedicated tov, so there must be somexinu’s part wherevxis red. Pick any vertex

y 6= v not inu’s part. Then{uv, uy, xy, vx}is a four cycle whereuv is blue,uyis green and

vxis red. Since green is dedicated touand red is dedicated to v, xymust be blue. We know

blue is not dedicated toxinG(since it appears onuv). Sinceywas arbitrary, it follows that red

is the only color other than blue incident toxand thus red is dedicated tox, inG. This means

that red appears only on the edgevx.

Let’s takeS ={v, x}andR = V(G)− {v, x}. It will suffice to show this choice satisfies

conditions 1), 2), and 3) of our main theorem to dispose of the special case min(|R0|,|S0|) = 1

We begin by showing that blue∈/ C[R]. First we note that all edges incident toxinG−v

are blue; this was part of the proof that vx is red. Recall that R0 = V(G−v)− {u}. By

assumptionG[R0]was JL-colored, and thusxhas a color dedicated to it in G[R0]. That color

must be blue. Letab ∈ E(G[R]). If eithera orbis the vertexu, then the edge abis green. If

neithera norb isu, then ab ∈ E(G[R0]). Since blue is dedicated toxinG[R0], the edgeab

cannot be blue.

Now we will show that either allR −S edges are blue, or there is a unicolored vertex in

G. We have already shown all edges fromxtoRare blue. We have also shown that all edges

fromv tow ∈ Rwherewis not inu’s part are blue. If all edges fromv to vertices inu’s part

are blue (other than the redvxedge), then we have shown what we needed to show.

So considerz ∈ R wherez is inu’s part andvz is not blue. We note thatvz is not green,

because if it were green then for any w in a part other than those of uand v, the three cycle

{vz, wz, vw}is rainbow sincevzis green,vwis blue as above, andwzis neither blue nor green

sincez, w ∈V(G[R0])and neither vertex isx.

Let yellow ∈ {/ blue, red, green} be the color on vz. By the argument in the paragraph

above, for anyw∈V(G)\ {v}which is in neitherv’s part nor inz’s (u’s) part (which implies

thatw∈R0),wzmust be colored yellow. Ifzis unicolored by yellow, then we are done. Ifzis

not unicolored, letzz0 be an edge not colored yellow. This color is inC[R0]sincez, z0 ∈ R0.

In particular, the color is neither blue nor green. Ifz0 is not inv’s part, then we have a rainbow

three cycle{vz, vz0, zz0}wherevz is yellow, vz0 is blue, and zz0 is neither blue nor yellow.

Alternatively, ifz0is inv’s part then we have a rainbow four cycle{uz0, uv, vz, zz0}whereuz0

is green,uv is blue,vz is yellow, andzz0 is not yellow, blue nor green. Thus we either have a

unicolored vertexz, or condition1)is satisfied and allR−Sedges are blue.

For2), we have red appearing as the only color inC[S]and since red is only onvx, red is

not inC[R]. Since blue is dedicated toxinG[R0], and all edges incident touare green except

uv, blue∈/C[R]. SoC[R]andC[S]are disjoint and neither set contains blue.

For 3), since G has no rainbow cycles, neither G[R] nor G[S] can have rainbow cycles.

SinceG[S]has only one edge and that edge is red,G[S]has the appropriate number of colors.

We need G[R] to have n −3colors. We note that G has n −1 colors appearing. We have

already shown red and blue are not inC[R]. All othern−3colors must appear somewhere and

since the oneG[S]edge is red and allR−Sedges are blue (the only other edges are inG[R]),

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We may now assume that |V(R0)|,|V(S0)| ≥ 2. If R0 is a subset of only one part of

G, then there are no edges in G[R0] and since |V(R0)| ≥ 2, it follows that G[R0] is not

JL-colored. Therefore, it follows that bothR0 andS0 have representatives in at least two different

parts. Also, we note that green is not dedicated to any vertex inG−v (and thus not inG) since

|V(R0)|,|V(S0)| ≥2.

Without loss of generality, letx ∈R0 be such thatvxis red. Consider any edgevwwhere

w∈ S0andwis not inx’s part. By assumptionvxis red. We knowwxis green sincew ∈S0

andx∈R0. Since there are no rainbow cycles, this impliesvwmust be red or green.

Now consider an edgevw wherew ∈ S0 andw is in x’s part. Pick a vertexy ∈ R0 and

z ∈ S0 where neithery norz are inx’s part. In the four cyclevxywwe have vxis red, xyis

yellow where yellow is inC[R0], andywis green. This impliesvwis red, green, or yellow. In

the four cyclevxzwwe havevxis red,xzis green, andzwis blue where blue is inC[S0]. This

impliesvwis red, green, or blue. Thusvwmust be red or green.

We now know that eachv−S0 edge is red or green. We have two final cases to consider to

complete the proof. Either allv−S0 edges are green, or at least onev−S0edge is red.

Assume all v −S0 edges are green. We aim to show allv −R0 edges are either red or a

color inC[R0].

Consider the edgevwwherew ∈ R0 andwis not inx’s part. Then we have a three cycle

with the edges{vx, wx, vw}wherevxis red andwxis a color inC[R0]sincew, x∈R0. This

impliesvwis either red or a color inC[R0]sinceGhas no rainbow cycles.

Now we consider the edgevwwherew ∈ R0, wis in x’s part and w 6= x. There is some

color dedicated towinG. Red is not dedicated towbecause red is onvx, green is not dedicated

tow, because green is not dedicated to anything inG, and clearly no color dedicated towcould

be a color in C[S0]. Let yellow be a color dedicated to w where yellow is in C[R0]and say

yellow appears on an edge uwwhere u ∈ R0. Then we have a four cycle {vx, ux, uw, vw}

that cannot be rainbow. We note thatuxis an edge inG[R0]and is thus colored by a color in

C[R0]. Moreover, it is a color inC[R0]other than yellow, since yellow is dedicated tow. This

impliesvwis either red or some color inC[R0].

We takeR = {v} ∪R0 and S = S0. Since we have assumed all v−S0 edges are green

we have that all R − S edges are green. We have just shown C[R] and C[S] are disjoint

and neither contains green. Again, neither G[R] nor G[S] has a rainbow cycle since G has

no rainbow cycles. Finally, |S| − 1 colors appear on G[S] = G[S0], since the latter was

JL-colored by the induction hypothesis. Also, the colors C[R0]∪ {red} appear on G[R], so

(|R0| −1) + 1 = |R0| = |R| −1colors appear onG[R]. Thus, both G[R] andG[S]are

JL-colored.

Now we assume there is some redv−S0edge. By the same argument as above, that showed

that allv−S0edges are either red or green, this implies that allv−R0edges are red or green.

We now show that this implies that all edges incident tov are red. Thusvis unicolored and we

are done.

To see that all edges incident tov are red, assume that there is some edgevythat is green.

There is av−S0 red edge by assumption and there is also av−R0 red edge (namelyvx). So,

without loss of generality, lety∈R0.

Ifxandyare not in the same part, then we note that{vx, xy, vy}is a rainbow three cycle

sincexymust be colored by some color inC[R0]. Let us assumexandyare in the same part.

Let blue be the color dedicated toxin the JL-coloring ofG−v. Since green is not dedicated to

any vertex inG−v, blue must be inC[R0]. Let blue be on an edgeuxwhereu∈R0. Then the

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not blue since blue is dedicated tox, anduyis green. Thus, the green edgevycannot exist and

so all edges incident tovmust be red.

3

Encoding JL-colorings of Complete Multipartite Graphs

Definitions. A full binary tree is a tree with exactly one vertex of degree two and all other

vertices of degrees 1 or 3. The vertex of degree 2 is theroot of the tree, and the vertices of

degree 1 areleafs. Furthermore, every non-leaf (a vertex of degree 3 or the root) has exactly two

children, which we callsiblings. The non-leaf vertex is called theparentof the two children.

The children of a vertex of degree3are its two neighbors other than the neighbor which is on

the path joining it to the root.

Figure 1: A full binary tree with7leafs

In[1]it is shown that the JL-colorings ofKn,n > 1are in1−1correspondence with the

(isomorphism classes of) the full binary trees withnleafs.

In the case of complete bipartite graphs the situation is a little more complicated: each

JL-coloring ofKm,n can be encoded by a certain vertex labeling of a full binary tree withm+n

leafs, and conversely, every labeling of the vertices of a full binary tree with ordered pairs of non-negative integers, satisfying certain requirements, encodes a JL-coloring.

Theorem 2.1 implies an analogous result in which full binary trees withn1+. . .+nrleafs,

equipped with certain labelings of the tree vertices withr-tuples of non-negative integers,

en-code JL-colorings ofKn1,...,nr.

In a Huffman labeling of a full binary tree, each leaf is given a certain label from a commu-tative semigroup. The label assigned to any parent vertex is the sum of the labels of the parent

vertex’s two children; by this rule, every vertex of the tree gets a label. See [3] for Huffman

labeling in coding theory.

In particular, a Huffman labeling of a full binary tree withr-tuples of non-negative integers

is a labeling such that the label of each parent is the coordinate-wise sum of the label of its

chil-dren. Such a labeling produces a JL-coloring of a completer-partite graph given the following

conditions:

1. Each leaf has weight 1 (a 1 appears in one coordinate and zeros appear elsewhere).

2. Sibling leafs are orthogonal.

3. For eachj ∈ {1, . . . , r}, at least one leaf has the label with a1in positionjof ther-tuple.

To see how Theorem 2.1 implies a correspondence between JL-colorings of complete

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RT

RT ST

(1;2;1) (1;0;2)

(1;1;0) (0;1;1) (0;0;1) (1;0;1)

(1;0;0) (0;0;1) (0;0;1)

(0;1;0) (0;1;0)

(1;0;0)

green

yellow blue

red purple orange

(2,2,3)

Figure 2: A Huffman labeling for a JL-coloring ofK2,2,3

Consider Figure 2 and letG=K2,2,3, with partsP1,P2, andP3 satisfying|P1|=|P2|= 2, and

|P3|=3. First we will see how the labeled tree in Figure 2 tells us how to color the edges ofG.

The label on the root is(|P1|,|P2|,|P3|). Call the children of the rootRT andST. The labels

onRT andST tell the colorist how to partitionV(G)into two sets R andS: the label on the

vertexRT is(|R∩P1|,|R∩P2|,|R∩P3|), soRhas one vertex fromP1, two fromP2, and one

fromP3. SinceRandSpartitionV(G), the sum of the labels onRT andST must be(2,2,3).

When labeling the vertices from the top down, as was done in this example, not every

par-tition ofV(G)is permitted. No label with only one non-zero entry, and that entry greater than

one, can appear. For instance,(2,0,0)is not permitted as a label on any vertex of the full

bi-nary tree in a labeling of the tree in Figure 2 representing a JL-coloring ofG=K2,2,3. This is

becauseG[R], a complete multipartite graph, must be JL-colorable by condition 3 of the main

theorem, and a complete1-partite graph with more than one vertex is not JL-colorable.

To return to our example: the line joiningRT andST with the word "green" above it is not

part of the tree. This line indicates to the reader that the colorist will color allR−Sedges in

Gwith the same color, call it green, that will never be used again.

Next, the JL-coloring of G[R] ' K1,2,1 and ofG[S] ' K1,0,2 ' K1,2 begins as the

JL-coloring of G began, and so on. Theorem 2.1 guarantees both that an edge-coloring of a

complete multipartite graph Kn1,...,nr for r ≥ 3, ni ≥ 0, and i = 1, . . . , r derived from a

properly labeled full binary tree with root label(n1, . . . , nr), as indicated in the example, will

be a JL-coloring of the graph, and that every JL-coloring ofKn1,...,nr is so derivable.

We end the paper with an observation: the same full binary tree with different labelings may produce different JL-colorings of the same graph, and even JL-colorings of different complete multipartite graphs.

Notice that the coloring of K2,2,3 produced by the labeled full binary tree in Figure 2 has

9green edges. The coloring ofK2,2,3 produced by the labeled full binary tree of Figure 3 has

10green edges. Since there are16edges inK2,2,3, it is clear that these two labelings produce

different colorings ofK2,2,3.

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(2;2;0) (1;0;2)

(1;1;0) (1;1;0) (0;0;1) (1;0;1)

(1;0;0) (0;0;1) (1;0;0)

(0;1;0) (0;1;0)

(1;0;0)

(3,2,2)

green

Figure 3: A Huffman labeling of the tree in Figure 2 for a different JL-coloring ofK2,2,3

(2;2;0) (2;0;1)

(1;1;0) (1;1;0) (1;0;0) (1;0;1)

(1;0;0) (0;0;1)

(1;0;0)

(0;1;0)

(0;1;0)

(1;0;0)

(4,2,1)

Figure 4: A Huffman labeling of the same full binary tree in Figures 2 and 3, for a JL-coloring ofK4,2,1

References

[1] A. Gouge, D. Hoffman, P. Johnson, L. Nunley, and L. Paben, Edge colorings ofKnwhich

forbid rainbow cycles,Utilitas Mathematica, 83 (2010), 219-232.

[2] P. Johnson and C. Zhang, Edge colorings of Km,n withm+n−1colors which forbid

rainbow cycles,Theory and Applications of Graphs, 4 (2017), issue 1, Article 1, 17 pp.

[3] D. Hankerson, G.A. Harris, and P.D. Johnson, Introduction to Information Theory, 2nd

edition, Chapman & Hall/CRC, Boca Raton, 2003

[4] S. Fujita, C. Magnant and K. Ozeki, Rainbow generalizations of Ramsey Theory—a

dynamic survey, Theory and Applications of Graphs, vol. 0, issue 1, article 1. DOI

Figure

Figure 1: A full binary tree with 7 leafs
Figure 2: A Huffman labeling for a JL-coloring of K2,2,3
Figure 3: A Huffman labeling of the tree in Figure 2 for a different JL-coloring of K2,2,3

References

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