On the existence of rainbows in 1-factorizations of K-2n

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On the Existence of Rainbows in

1-Factorizations of

K

_n

David E. Woolbright,

1

Hung-Lin Fu

2,3

1_{Department of Computer Science, Columbus State University, Algonquin Drive,}

Columbus, GA 31907, USA

2_{Department of Discrete and Statistical Sciences, 120 Math Annex, Auburn}

Uni-versity, Auburn, AL 36849-5307, USA

3_{Department of Applied Mathematics, National Chiao Tung University, Hsin}

Chu, Taiwan, Republic of China

Received November 11, 1995; accepted March 15, 1996

Abstract: A 1-factor of a graph G = (V, E) is a collection of disjoint edges which contain all the vertices of V . Given a 2n − 1 edge coloring of K2n, n ≥ 3, we prove there exists a 1-factor

of K2nwhose edges have distinct colors. Such a 1-factor is called a‘‘Rainbow.’’ c 1998 John

Wiley & Sons, Inc. J Combin Designs 6: 1–20, 1998

Keywords:1-factor; 1-factorization; edge coloring; rainbow

1. INTRODUCTION

A 1-factor in a graph G = (V, E) is a set of pairwise disjoint edges in E which contain all the vertices in V . A 1-factorization of G is a partition of the edges in E into 1-factors. These notions can be generalized to hypergraphs: If V = {v1, v2, . . . , vn} is a finite set and

E = {Ei|i ∈ I} is a family of nonempty subsets of V such that ∪i∈IEi= V , then the pair

(V, E) is a hypergraph with vertex set V and edge set E. A 1-factor of the hypergraph is a collection of pairwise disjoint edges which contain all the vertices of V . A 1-factorization of the hypergraph is a partition of the edges of E into 1-factors. If V = {v1, v2, . . . , vn}

and E is the set of all k-element subsets of V then (V, E) is denoted by Kk

nand is called the

Correspondence to: D. E. Woolbright

e-mail: woolbright david@colstate.edu; hlfu@math.nctu.edu.tw c

1998 John Wiley & Sons, Inc. CCC 1063 8539/98/010001-20

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2 WOOLBRIGHT AND FU

complete k-uniform hypergraph on n vertices. Zs. Baranyai [1] has proven that there exist 1-factorizations of all Kk

kn, k ≥ 2. The reader will note that if k = 2 then K2n2 is

simply the complete graph on 2n vertices.

In 1977 Alexander Rosa suggested the following interesting problem: Given a 1-factorization, F, of Kk

kn, n ≥ 3, prove there exists a 1-factor in Kknk whose edges belong

to n different 1-factors of F. The first author [2] has investigated this problem and has shown that for any 1-factorization, F, of Kk

kn, k ≥ 2, there exists a 1-factor whose edges

belong to at least n − 1 1-factors of F. There is a colorful way to think of Rosa's problem. Imagine coloring the edges of Kk

knin such a way that any two edges have the same color if

and only if they belong to a common 1-factor of F. Then Rosa's conjecture states that there exists a 1-factor in Kk

knwith the property that no two of its edges have the same color—a

colorful 1-factor. We will call such a 1-factor a rainbow. This is also commonly called an orthogonal 1-factor. Formally, an edge coloring of a graph G = (V, E), is a function

φ : E → {1, 2, · · ·} such that adjacent edges have distinct images. A k-edge coloring is an

edge coloring whose image set is {1, 2, . . . , k}. The purpose of this article is to show that Rosa's conjecture is true for certain complete graphs:

Theorem 1.1. For any 2n−1 edge coloring of K2n, n ≥ 3, there exists a 1-factor whose

edges have exactly n colors.

It remains an open problem whether rainbows exist in all 1-factorizations of Kk kn for

k ≥ 3.

2. EXTENDING A PREVIOUS RESULT

We begin the proof of the theorem mentioned above. Let G = (V, E) = K2nbe a complete

graph on 2n vertices, n ≥ 3, and F a 1-factorization of G. Assume that φ is a 2n − 1 edge coloring of G. It has been shown [2] that there exists a 1-factor, F , in G whose edges are colored with at least n−1 colors. A moment's reflection shows that if G = K6, no 1-factor

in G can have edges of only two colors. This means that F must have edges of 3 different colors. We have dispatched the case where G = K6and now consider G = K2n, n ≥ 4.

For brevity of notation we will denote an edge {vi, vj} as vivj and define edge sets

F = {e1 = v1v2, e2 = v3v4, . . . , en = v2n−1v2n} and F0 = F \ {e1}. We can assume

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RAINBOWS IN 1-FACTORIZATIONS OFK_2N 3 that the n − 1 edges of F0_{have colors n + 1, n + 2, . . . , 2n − 1, and that edge e}₁_{has color}

n + 1. Let T = {e ∈ E|e is incident with v1or v2and φ(e) ≤ n}. Obviously |T | = 2n.

Two cases can occur:

1. one edge in F0_{is incident with exactly four edges in T , or}

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4 WOOLBRIGHT AND FU

FIG. 2. (continued)

2. each of at least two edges in F0_{is incident with exactly three edges in T .}

Otherwise at most one edge in F0_{is incident with exactly 3 edges in T , and the other n − 2}

edges in F0 _{are incident with at most two edges in T . This, however, would account for}

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RAINBOWS IN 1-FACTORIZATIONS OFK_2N 5

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6 WOOLBRIGHT AND FU

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only 3 + 2(n − 2) = 2n − 1 of the edges of T , a contradiction. The proof breaks into the two cases stated above:

Case 1. There exists an edge, say e2 = v3v4, in F0incident with exactly four edges in

T [Fig. 1(a)]. If φ(v1v3) 6= φ(v2v4), then by replacing v1v2and v3v4in F with v1v3and

v2v4we have constructed a rainbow in G. A similar argument holds if φ(v1v4) 6= φ(v2v3).

Therefore, without loss of generality, we assume φ(v1v3) = φ(v2v4) = 1 and φ(v1v4) =

φ(v2v3) = 2 and consider the two subcases below:

Subcase 1.1. φ(v3v4) = x > n + 1.

Let A = E1,2 ∪ E3,4 where E1,2 = {e ∈ E|e is incident with v1or v2, and φ(e) ∈

{3, 4, . . . , n, x}}, and E3,4 = {e ∈ E|e is incident with v3or v4, and 3 ≤ φ(e) ≤ n}.

Obviously, |A| = 4(n − 2) + 2. By the pigeon-hole principle there exists an edge, say

v5v6, in F00= F0\ {v3v4} which is incident with at least 5 edges in A. We continue the

proof by examining all distinct ways in which 5 edges in A can be incident with the edges in F . In each case we demonstrate how F can be modified (by adding and deleting edges) to form a rainbow in G. Since there will be many cases to consider, we handle each case by drawing a figure of the graph with some accompanying statements that indicate how a rainbow can be constructed. In every figure, each edge is labeled with its color.

For example, omitting symmetric cases, there are two distinct ways [Fig. 2(a) and 2(b)] that 5 edges in A can be incident with edges v1v2, v3v4, and v5v6where deg(v5) = 4,

and deg(v6) = 1. Consider Figure 2(a). The statement ‘‘d 6= e → 1, d, e’’ means if

φ(v4v5) 6= φ(v2v6), then we can add to F edges v1v3, v4v5, and v2v6(which are colored

1, d, and e, respectively) and delete edges v1v2, v3v4, v5v6 from F to obtain a rainbow.

On the other hand, ‘‘d = e → 2, c, e’’ means that if φ(v4v5) = φ(v2v6), then (F \

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8 WOOLBRIGHT AND FU

FIG. 6.

and deg(v6) = 4 are analogous to the cases represented by Figures 2(a) and 2(b) and as

a result are omitted. Figures 2(c)–2(n) represent distinct ways in which 5 edges in A can be incident with edges v1v2, v3v4, and v5v6 where deg(v5) = 3, and deg(v6) = 2. We

consider cases where v5is adjacent to v1, v2, and v3and where v5is adjacent to v1, v3, and

v4. We omit the symmetric cases where v5 is adjacent to v1, v2, and v4and where v5is

adjacent to v2, v3, and v4. The cases with deg(v5) = 2 and deg(v6) = 3 being analogous,

are also omitted.

Subcase 1.2. φ(v3v4) = n + 1.

In this case we let A = {e ∈ E|e is incident with one of {v1, v2, v3, v4} and 3 ≤ φ(e) ≤

n}. Clearly, |A| = 4(n−2). By the pigeon-hole principle, either one edge in F00_{is incident}

with at least 5 edges in A, or every edge in F00_{is incident with exactly 4 edges in A. In}

handling subcase 1.1, we never used the edge colored x in constructing any rainbow. As a result, our previous arguments in subcase 1.1 apply here as well when one edge in F00_is

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FIG. 7.

We are left to consider the case where every edge in F00_{is incident with 4 edges in A.}

Call an edge in F00_{‘‘type one’’, if one of its vertices is incident with exactly 1 edge in A}

(and the other vertex with exactly 3). Call the edge ‘‘type two’’ if both vertices are incident with exactly two edges in A, and call it ‘‘type three’’ if one of its vertices is incident with 4 edges in A. Every edge in F00_{belongs to one of the three types. Figures 3(a) to 3(c)}

demonstrate how to obtain a rainbow in the cases where F00_{contains a type one edge, and}

so we proceed with the assumption that F00_{contains only edges of types two and three.}

Suppose v5v6is a type 3 edge and that v5is incident with 4 edges of A. Consider the

n edges colored 1, 2, . . . , n which are incident with v6. None of these edges is incident

with v1v2or v3v4since v5v6is a type three edge. By the pigeon-hole principle, there is an

edge in F00_{, say v}₇_v₈_{, which is incident with two of the n edges described above. Figure}

4 addresses the case in which that edge is type three and Figures 5(a) to 5(e) handle the cases in which that edge is type two. Notice that the edges of A which contain vertices

v7or v8are colored 3 or 4. We can do this without loss of generality since using three or

more colors results in an immediate rainbow by appropriately swapping edges. Note that in Figures 5(a) to 5(c), the edges of A which contain vertices v7or v8are adjacent to exactly

two other vertices. In Figure 5(d) these edges contain exactly 3 other vertices, and in Figure 5(e) these edges contain 4 other vertices. Because of the symmetry of the subgraph induced by vertices in {v1, v2, v3, v4} and its edge coloring, many subgraphs similar to 5(d) and

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10 WOOLBRIGHT AND FU

FIG. 8.

As a result of the remarks above, we assume that all edges in F00_{are of type two. Let}

v5v6 be a type two edge. Figures 6(a) to 6(d) indicate how to obtain a rainbow in the

cases where there are at least 3 vertices in {v1, v2, v3, v4} which are incident with edges

of A containing v5or v6. Figures 7(a) to 7(c) illustrate the cases in which exactly two

vertices in {v1, v2, v3, v4} are incident with 4 edges of A containing v5 or v6. In each

of these graphs we assume a = d and b = c, otherwise a rainbow is easily found. The reader should note that these subgraphs are isomorphic if you are allowed to relabel the colors. We must show that a rainbow can be constructed in each of these cases. Let

Bi,j = {vxvy ∈ F00|vivx, vivy, vjvx, vjvy ∈ A}. Without loss of generality, we assume

that v5v6 ∈ B3,4. Obviously |B3,4| ≤ (n−2)₂ since there are n − 2 edges in A which are

incident with v3and each edge in B3,4is incident with two of these edges in A.

Suppose |B3,4| = (n−2)₂ , then it is easy to see that |B1,2| is also(n−2)₂ . Recall that each

edge in F00_{is of type 2 and as a result is incident with exactly 4 edges in A. If |B}

3,4| = (n−2)₂ ,

then the edges of B3,4are incident with exactly 2(n − 2) edges in A. This means there are

2(n − 2) edges of A not incident with edges in B3,4. These edges are all incident with v1

or v2. Also there are n − 2 edges in F00\ B3,4. Each of these edges is incident with exactly

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FIG. 9.

in B3,4. Without loss of generality we assume v3v5 = v4v6 = 3 and v3v6 = v4v5 = 4,

otherwise a rainbow is achieved by swapping edges. For instance, if v3v5= 3 and v4v6= 4,

then (F \ {v3v4, v5v6}) ∪ {v3v5, v4v6} is a rainbow. Consider the n − 3 edges colored

5, 6, . . . , n, n + 1 which are incident with v5. At most 2((n−2)₂ − 1) = n − 4 of these edges

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an edge in B1,2. Call that edge v7v8. Figure 8(a) demonstrates how to construct a rainbow

in this case.

On the other hand, suppose we assume that |B3,4| < (n−2)₂ . In this case we consider the

n − 4 edges incident with v5which are colored 5, 6, . . . , n. Now |B3,4| ≤ (n−2)₂ − 1 and at

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FIG. 10.

v5v6. Thus, at least one of the n − 4 edges, say t, is incident with an edge in F00\ B3,4.

This means t is incident with an element of either B1,2, B1,3, B1,4, B2,3, or B2,4. Each of

these situations are handled in Figures 8(a) to 8(d).

Case 2. There exist at least two edges, say v3v4and v5v6in F0each of which are incident

with exactly 3 edges in T . We examine two subcases [Figure 1(b)–(c)].

Subcase 2.1. Consider Figure 1(b). Without loss of generality we may assume φ(v1v3) =

φ(v2v4) = 1, φ(v1v5) = φ(v2v6) = 2, φ(v1v4) = 3, and φ(v2v5) = 3 or 4. We also let

φ(v3v4) = x and φ(v5v6) = y and note that x 6= y and at least one of x and y is not equal

to n + 1. In what follows we assume that y 6= n + 1 and the strategy will be to find a 1-factor that includes y (we omit this edge in the accompanying diagrams).

If φ(v2v5) = 3 let E1 = {e = v1vz|φ(e) = 4, 5, . . . , n}, E2 = {e = v2vz|φ(e) =

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4, 5, . . . , n}. Let A = E1 ∪ E2 ∪ E3 ∪ E4, then no edge in A has two vertices in

{v1, v2, v3, v4} otherwise we can easily find a rainbow. Since the edges described above

are distinct, we have |A| = 3(n − 3) + (n − 2) = 4(n − 3) + 1 and by the pigeon hole principle there exists an edge v7v8 in F00\ {v5v6} which is incident with 5 edges in A.

Without loss of generality we may assume that v7is incident with 3 or 4 of the edges of A.

Figures 9(a) to (d) handle the cases in which v7is incident 4 edges of A. Figures 9(e) to

(p) handle the cases in which v7is incident 3 edges of A and v8is incident 2 edges of A.

In Figures 9(e, f, h, i, k, l, n, and o) one of the edges incident with v8is omitted. Since the

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If φ(v2v5) = 4 we let E1= {e = v1vz|φ(e) = 4, 5, . . . , n}, E2 = {e = v2vz|φ(e) =

3, 5, . . . , n}, E3 = {e = v3vz|φ(e) = 2, 4, 5, . . . , n}, and E4 = {e = v4vz|φ(e) =

2, 5, . . . , n}. Note that none of the 1-factors produced in Figure 9 use an edge with color 3. As a result, an analogous argument to that of the previous paragraph together with the diagrams in Figure 9 apply in this case as well.

Subcase 2.2. Consider Figure 1(c). Without loss of generality we may assume φ(v1v4) =

3, φ(v1v6) = 4, φ(v1v3) = φ(v2v4) = 1, and φ(v1v5) = φ(v2v6) = 2. Again we let

φ(v3v4) = x and φ(v5v6) = y 6= n + 1. Let E1 = {e = v1vz|z ≥ 7 and φ(e) =

5, 6, . . . , n}, E2 = {e = v2vz|z ≥ 7 and φ(e) = 5, 6, . . . , n}, E3 = {e = v3vz|z ≥ 7

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if two vertices in {v1, v2, v3, v5} are joined by an edge colored 1, 4, 5, 6, . . . , n, then we

can immediately find a rainbow. As a result, If A = E1∪ E2∪ E3∪ E5, then we assume

|A| = 4n − 14. Let F000 _{= F}00_{\ {v}

5v6}. If there is an edge, say v7v8, in F000 which is

incident with 5 edges in A, then we are able to construct a rainbow using the graphs in Figure 10. Therefore, we assume that each edge in F000_{is incident with at most 4 edges}

in A.

Let v7v8be an edge which is incident with exactly 4 edges in A and let S be the set of

4 edges. If the edges of S are incident with exactly two vertices in {v1, v2, v3, v5} then by

exchanging edges we obtain a 1-factor of a type which was handled in Case 1. For example, in Figure 11, the edges of S are incident with only v1and v5. If we remove edges v1v2and

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v5v6and add edges v1v5and v2v6we obtain a 1-factor in which an edge with a repeated

color (edge v1v5) satisfies the conditions in Case 1.

On the other hand, if the edges of S are incident with 3 or 4 of the vertices in {v1, v2, v3,

v5}, it is easy to see that either v7or v8is incident with all 4 edges in S, or two of the edges

in S are disjoint and have different colors. In the latter case, rainbows are easily found. For example, in Figure 12, we can construct a rainbow by using edges colored a, b, 2, and 3. The reader is left to check the 5 other possibilities.

We next consider the former case:

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FIG. 11.

2. If an edge in F000_{, say v}

7v8, is incident with 4 edges in A, then all 4 edges are incident

with v7or all are incident with v8.

Most of the edges in F000_{are incident with 4 edges in A. To see this, call an edge in F}000

‘‘type one’’ if it is incident with at most 3 edges in A, otherwise call the edge ‘‘type two.’’ Note that if there are at least 3 type one edges then the largest number of edges in A which are incident with edges in F000_{is (3)(3) + 4(n − 6) = 4n − 15. Since A contains 4n − 14}

edges, this would not account for all the edges of A. There must be at most two type one edges in F000_{. If v}

xvyis a type two edge, we call vxa ‘‘degree 4’’ vertex if 4 edges of A

contain vx. Otherwise, call the vertex a ‘‘degree 0’’ vertex. We intend to show that there

is an edge e, containing two degree 0 vertices, such that ϕ(e) ∈ {1, 2, . . . , n}.

Suppose v7v8∈ F000and v7is a degree 4 vertex. Consider the set of edges B = {e =

v8vx|φ(e) = 1, 2, . . . , n}. At most 4 of these edges are incident with type one edges since

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FIG. 13.

there are at most two type 1 edges. It is straightforward to show that if any edge in B is incident with a vertex in {v1, v2, v3, v4, v5}, then F can be modified to produce a rainbow.

One edge in B might be incident with v6and so there are at least n − 5 edges in B which

are incident with at most n − 6 edges in F000_{other than v}₁_v₂_{, v}₃_v₄_{, v}₅_v₆_{, v}₇_v₈_{and any type}

one edges. We see there is an edge, say v9v10, which is incident with 2 edges in B, and so

one of the two edges is incident with a degree 0 vertex. This is the edge e(=v8v9) alluded

to in the paragraph above. Figure 13 shows how to construct a rainbow in this case. REFERENCES

[1] Z. Baranyai, On the factorization of the complete uniform hypergraph, Colloq. Math. Soc. J´anos Bolyai, (1973), 91–108.

[2] D. E. Woolbright, On the size of partial 1-factors of 1-factorizations of the complete k-uniform hypergraph on kn vertices, Ars Combin. 6 (1978), 185–192.