A k-edge coloring of G is said to be equitable if the number of edges, at any vertex, colored with a certain color differ by at most one from the number of edges colored with a different color at the same vertex. An STS(v) is said to be polychromatic if the edges in each triple are colored with three different colors. In this paper, we show that every STS(v) admits a 3-edge coloring that is both polychromatic for the STS(v) and equitable for the underlying complete graph. Also, we show that, for v 1 or 3 (mod 6), there exists an equitable k-edge coloring of K which does not admit any polychromatic STS(v), for k = 3 and k = v - 2

Abueida, Atif A.

Lefevre, James

Waterhouse, Mary

University of Queensland eSpace

AUSTRALASIAN JOURNAL OF COMBINATORICS
Volume 50 (2011), Pages 155–163
Equitable edge colored Steiner triple systems
Atif A. Abueida
Department of Mathematics
University of Dayton
300 College Park, Dayton, OH 45469-2316
U.S.A.
Atif.Abueida@notes.udayton.edu
James Lefevre Mary Waterhouse
School of Mathematics and Physics
The University of Queensland
Brisbane, Qld. 4072
Australia
jgl@maths.uq.edu.au maw@maths.uq.edu.au
Abstract
A k-edge coloring of G is said to be equitable if the number of edges, at
any vertex, colored with a certain color diﬀer by at most one from the
number of edges colored with a diﬀerent color at the same vertex. An
STS(v) is said to be polychromatic if the edges in each triple are colored
with three diﬀerent colors. In this paper, we show that every STS(v)
admits a 3-edge coloring that is both polychromatic for the STS(v) and
equitable for the underlying complete graph. Also, we show that, for
v ≡ 1 or 3 (mod 6), there exists an equitable k-edge coloring of Kv which
does not admit any polychromatic STS(v), for k = 3 and k = v − 2.
1 Introduction
An edge coloring of a graph G is an assignment of colors to the edges of G. A k-edge
coloring of G is an edge coloring of G in which k distinct colors are used, c1, c2, . . . , ck
say. We let E(ci) denote the set of edges that are assigned color ci, for i = 1, 2, . . . , k.
Also, let d(v) denote the degree of a vertex v.
A Steiner triple system of order v, denoted STS(v), is an ordered pair (V, T ),
where V is a v-set of symbols and T is a set of 3-element subsets of V called triples
such that any pair of symbols in V occurs together in exactly one triple. It is known
156 A.A. ABUEIDA, J. LEFEVRE AND M. WATERHOUSE
that a Steiner triple system (STS) of order v exists if and only if v ≡ 1 or 3 (mod
6); see [11].
An STS(v) is said to be resolvable if its triples can be partitioned into sets,
called parallel classes, such that every element of V appears exactly once within
each parallel class.
Let G and H be graphs. An H-decomposition of G is a set H = {H1, H2, . . . , Hp}
such that Hi is isomorphic to H for 1 ≤ i ≤ p and H partitions the edge set of G.
A K3-decomposition of Kv is equivalent to an STS(v), and we will use the terms
interchangeably. Similarly, we will often refer to K3 as a triple.
We color the triple {x, y, z} with the color triple 〈ci, cj, ck〉 by assigning the colors
ci, cj and ck to the edges {x, y}, {y, z} and {x, z} respectively. If i = j = k and k = i,
then we say that {x, y, z} is polychromatic. An STS(v) or K3-decomposition of some
graph G is said to be polychromatic if every triple in the STS(v) or decomposition
is polychromatic.
In general, consider the two graphs H and G, and an edge coloring of G. The
graph H is said to be polychromatic subgraph of G, if G contains a subgraph isomor-
phic to H, where all of whose edges are assigned distinct colors. Researchers devoted
some eﬀort to the study of determining the least number of colors used to color the
E(Kv) that forces a speciﬁed polychromatic subgraph to occur. Those problems are
called Anti-Ramsey problems. For more details about Anti-Ramsey problems and
polychromatic subgraphs, the reader is referred to [3, 4, 9, 12].
In [1], Bate studied complete graphs without polychromatic Cn for n = 3 and
n = 4. A good survey about monochromatic and polychromatic subgraphs in edge
colored graphs can be found in [10].
For each v ∈ V (G) let nv(ci) denote the number of edges in G of color ci incident
with vertex v, for i = 1, 2, . . . , k. A k-edge coloring of G is said to be equitable if
|nv(ci)− nv(cj)| ≤ 1 for all v ∈ V (G) and 1 ≤ i < j ≤ k.
If there exists an equitable k-edge coloring of G, then G is said to be equitably
k-edge colorable. The idea is that the number of edges of each color is as close as
possible at any vertex of the graph G.
Hilton and de Werra [8] found a suﬃcient condition for a simple graph to admit
an equitable edge coloring:
Theorem 1.1. [8] Let G be a simple graph and let k ≥ 2. If k  d(v) (for all
v ∈ V (G)) then G has an equitable edge coloring with k colors.
Specialized colorings of cycle systems (C3 and C4), in which some condition on
the coloring is satisﬁed, have received recent attention. We refer the interested reader
to [2, 5, 6, 7].
There are two natural questions one can ask about polychromatic Steiner triple
systems and equitable edge colorings of the complete graph. One is: does every
STS(v) admit a polychromatic k-edge coloring which is also an equitable k-edge
coloring of the complete graph of order v, Kv? The second is: For v ≡ 1 or 3
EDGE COLORED STEINER TRIPLE SYSTEMS 157
(mod 6), does every equitable k-edge coloring of Kv admit a polychromatic STS(v)?
In the following section we settle the ﬁrst question for k = 3. In the ﬁnal section
we settle the second question for k = 3 and k = v − 2.
Before proceeding, we deﬁne some notation. We let Km,n denote the complete
bipartite graph with m and n vertices in each partite set. We let Kx \Ky denote the
complete graph on x vertices with a hole of size y. When no confusion is likely to
arise, we denote the edge {x, y} by xy.
2 Polychromatic Steiner Triple Systems yielding equitable
colorings
One relatively straightforward method for constructing a polychromatic STS(v)
which yields an equitable 3-edge coloring of the underlying graph is based on cyclic
systems. In fact this method may be generalized to certain higher numbers of colors.
Theorem 2.1. Let S be a cyclic STS(v) on vertex set V , and let k ≥ 3 and k |
(v − 1)/2. Then there exists a polychromatic coloring of S which is an equitable
k-edge coloring of the complete graph on V .
Proof. The number of base blocks is (v− 1)/6, and they contain a total of (v− 1)/2
edges. Given that k ≥ 3 and k | (v−1)/2, we can assign these edges in equal numbers
to k color classes so that every base block is polychromatic. When these base blocks
are developed cyclically, with each triple assigned the coloring of the base block from
which it was developed, we obtain the required coloring.
In the case k = 3, we can prove a more general result—that in fact the required
coloring is possible for any STS.
Theorem 2.2. Every STS admits a 3-edge coloring which is both polychromatic (with
respect to the triples) and equitable (with respect to the vertices).
Proof. Given any STS on a vertex set V , we can trivially color the edges with colors
c1, c2, c3 so that each triple is polychromatic (in general this coloring is not equitable).
Suppose we have an arbitrary such coloring, and let nx(c) denote the number of edges
of color c ∈ {c1, c2, c3} incident with vertex x ∈ V in this coloring. Suppose further
that for some vertex 1 ∈ V , we have n1(ca) ≥ n1(cb) + 2. We will show that we can
construct a second coloring with the following properties, where n′x(c) denotes the
number of edges of color c incident with vertex x in this new coloring:
(1) Every triple is polychromatic.
(2) |n′1(ca)− n′1(cb)| < |n1(ca)− n1(cb)|.
(3) For every vertex x and any two colors ci and cj, we have
|n′x(ci)− n′x(cj)| ≤ |nx(ci)− nx(cj)|.
158 A.A. ABUEIDA, J. LEFEVRE AND M. WATERHOUSE
From this, the result will follow inductively.
We prove the result for a = 1 and b = 2; the other cases follow similarly. Separate
to the STS, we construct a digraph on the vertex set V as follows: For each triple
{x, y, z} in the STS with {x, z} ∈ E(c1), {y, z} ∈ E(c2), and {x, y} ∈ E(c3), we
construct a directed edge (x, y) from vertex x to vertex y. Then for every w ∈ V we
have outdegree(w) − indegree(w) = nw(c1) − nw(c2); note that any triple {w, p, q}
with {w, p} ∈ E(c1) and {w, q} ∈ E(c2) does not aﬀect either side of this equation.
It follows that we have outdegree(1)−indegree(1) ≥ 2. We now wish to show that
the digraph contains a directed path from 1 to some vertex for which the indegree
exceeds the outdegree. Let V ′ ⊆ V be the set of all vertices v for which there exists
a directed path from 1 to v (including 1 itself). By deﬁnition, every edge directed
away from a vertex of V ′ must be directed to another vertex of V ′, and hence the
total indegree summed over the vertices of V ′ must be greater than or equal to the
total outdegree. Since outdegree(1) > indegree(1), it follows that there is a vertex
2 ∈ V ′ with outdegree(2) < indegree(2), and hence with n2(c1) < n2(c2). Since
2 ∈ V ′, there is a directed path from vertex 1 to 2.
To construct the second coloring, we reverse the orientation of each edge in the
directed path from 1 to 2, and make the corresponding change in the edge coloring
of the STS (all other edge colorings are equal to the original coloring). That is, for
each directed edge (x, y) in this path, we change the coloring of the triple {x, y, z} in
the STS. By deﬁnition this triple must have original coloring {x, z} ∈ E(c1), {y, z} ∈
E(c2), and {x, y} ∈ E(c3); we replace this with the coloring {x, z} ∈ E(c2), {y, z} ∈
E(c1), and {x, y} ∈ E(c3). Observe that for all w ∈ V \ {1, 2} and c ∈ {c1, c2, c3},
we have n′w(c) = nw(c), and also n
′
1(c3) = n1(c3) and n
′
2(c3) = n2(c3). Note also that
all triples are polychromatic in the new coloring. The diﬀerence between the two
colorings is thus that n′1(c1) = n1(c1) − 1, n′1(c2) = n1(c2) + 1, n′2(c1) = n2(c1) + 1
and n′2(c2) = n2(c2) − 1. Since n1(c1) ≥ n1(c2) + 2 and n2(c1) < n2(c2), we have
|n′1(c1) − n′1(c2)| < |n1(c1) − n1(c2)| and |n′2(c1) − n′2(c2)| ≤ |n2(c1) − n2(c2)|. The
result follows.
3 Equitable edge coloring with no polychromatic Steiner
triple system
We now address the question of whether every equitable k-edge coloring of Kv (for a
given v ≡ 1 or 3 (mod 6)) admits a polychromatic STS(v) (that is, a decomposition
of the edge colored Kv into polychromatic triples). An immediate condition for the
existence of such a polychromatic STS(v) is that k ≥ 3. When k = 3 it is also
necessary that there are equal numbers of edges in each color class; this is forced (by
the equitable condition) when v ≡ 1 (mod 6) but not when v ≡ 3 (mod 6).
We will show that even if this condition holds, in the cases k = 3 and k = v − 2
there exists an equitable k-edge coloring of Kv which does not admit a polychromatic
STS(v), for every v ≡ 1, 3 (mod 6).
EDGE COLORED STEINER TRIPLE SYSTEMS 159
Unless otherwise speciﬁed, we will let the vertex set of Kv be {1, 2, . . . , v}, and
when using three colors, we denote the colors by c1, c2 and c3.
Lemma 3.1. There exists an equitable 3-edge coloring of Kv, with color classes of
equal size, that does not admit any polychromatic STS(v), for v ∈ 7, 9.
Proof. The following is an equitable 3-edge coloring of K7:
E(c1) = {12, 17, 27, 35, 36, 45, 46},
E(c2) = {15, 16, 25, 26, 34, 37, 47},
E(c3) = {13, 14, 23, 24, 56, 57, 67}.
The following is an equitable 3-edge coloring of K9:
E(c1) = {14, 15, 24, 25, 36, 37, 39, 49, 58, 67, 68, 89},
E(c2) = {13, 18, 19, 23, 28, 29, 46, 47, 48, 56, 57, 79},
E(c3) = {12, 16, 17, 26, 27, 34, 35, 38, 45, 59, 69, 78}.
In either case, it is easy to check that the edge 12 cannot appear in any polychromatic
triple; see Figure 1.
1
3
4
56
7
8
9 2
1
27
6 3
45
Figure 1: Equitably 3-edge colored copy of Kv which does not admit a polychromatic
STS(v), for v ∈ {7, 9}.
Lemma 3.2. There exists an equitable 3-edge coloring of K9\K3, with 11 edges in
each color class, that does not admit a polychromatic K3-decomposition.
Proof. Let the vertex set of V (K9\K3) be {a, b, c, 1, 2, 3, 4, 5, 6}, where a, b and c
correspond to the vertices in the hole. The following is an equitable 3-edge coloring
of K9 \K3:
E(c1) = {a5, a6, b5, b6, c5, c6, 12, 13, 14, 23, 24},
E(c2) = {a3, a4, b3, b4, c1, c2, 15, 16, 25, 26, 34},
E(c3) = {a1, a2, b1, b2, c3, c4, 35, 36, 45, 46, 56}.
The edges 12, 34 and 56 cannot appear in any polychromatic triple; see Figure 2.
160 A.A. ABUEIDA, J. LEFEVRE AND M. WATERHOUSE
b
2
34
5
6
a c
1
Figure 2: Equitably 3-edge colored copy of K9 \ K3 which does not admit a poly-
chromatic K3-decomposition.
Note that in the proof of Lemma 3.2, the edges 12, 34 and 56 are colored with
c1, c2 and c3 respectively and they are not in any polychromatic triple.
Lemma 3.3. Let the vertex set of K6,6 be ∪i=1,2{1i, 2i, . . . , 6i}, with the obvious
vertex partition. There exists an equitable 3-edge coloring of K6,6, with 12 edges
of each color, such that the edges 11x and 21x are assigned the same color for all
x ∈ {12, 22, . . . , 62}.
Proof. The following is an equitable 3-edge colouring of K6,6 which satisﬁes the
speciﬁed constraint; see Figure 3.
E(c1) = {1112, 2112, 1162, 2162, 3142, 4142, 3152, 4152, 5122, 6122, 5132, 6132},
E(c2) = {1122, 2122, 1132, 2132, 3112, 4112, 3162, 4162, 5142, 6142, 5152, 6152},
E(c3) = {1142, 2142, 1152, 2152, 3122, 4122, 3132, 4132, 5112, 6112, 5162, 6162}.
121 1 13 14 15 16
21 22 23 24 25 26
Figure 3: Equitably 3-edge colored copy of K6,6.
Theorem 3.4. For all v ≡ 1 or 3 (mod 6), v ≥ 7, there exists an equitable 3-edge
coloring of Kv, with (v − 1)/6 edges in each color class, that does not admit any
polychromatic STS(v).
EDGE COLORED STEINER TRIPLE SYSTEMS 161
Proof. We consider the cases v ≡ 1 (mod 6) and v ≡ 3 (mod 6) separately.
Case 1: v ≡ 1 (mod 6)
Let v = 6n + 1, for some positive integer n, and let the vertex set of Kv be
{∞} ∪
n⋃
i=1
Vi, where Vi = {1i, 2i, . . . , 6i}, for i = 1, 2, . . . , n.
For each i = 1, 2, . . . , n, place the equitable 3-edge coloring of K7 given by Lemma
3.1 on {∞}∪ Vi, such that vertex x in Lemma 3.1 corresponds to vertex xi ∈ Vi, for
x ∈ {1, 2, . . . , 6}, and vertex 7 corresponds to ∞.
Place the equitable 3-edge coloring of K6,6 given by Lemma 3.3 on Vi ∪ Vj , for
1 ≤ i < j ≤ n, such that vertices x1 and x2 in Lemma 3.3 correspond to vertices
xi ∈ Vi and xj ∈ Vj , for x ∈ {1, 2, . . . , 6}.
Note that Kv is equitably 3-edge colored.
The edge 1121 cannot appear in any polychromatic triple. To prove this consider
the triple 1121x. For any x ∈ V (Kv) \ {11, 21}, the edges 11x and 21x are assigned
the same colour. The result follows.
Case 2: v ≡ 3 (mod 6)
Let v = 6n+3 for some positive integer n, and let the vertex set of Kv be {a, b, c, 0, 1,
. . . , 6n− 1}.
If n = 1, then the result follows from Lemma 3.1. Suppose then that n ≥ 2.
Consider the following edge coloring: apply Lemma 3.2 to each of the induced graphs
K9\K3[a, b, c; 6t, 6t+1, . . . , 6t+5] for 0 ≤ t ≤ n−1. Apply Lemma 3.3 to each of the
bipartite graphs K6,6[6i, 6i+1, . . . , 6i+5; 6j, 6j+1, . . . , 6j+5] for 0 ≤ i = j ≤ n−1.
Finally, color the edges ab, ac and bc with the colors c1, c2 and c3 respectively.
Then Kv is equitably 3-edge colored and the edges 12, 34 and 56 cannot be in
any polychromatic triple.
Theorem 3.5. For all v ≡ 1 or 3 (mod 6), v ≥ 7, there exists an equitable (v − 2)-
edge coloring of Kv that does not admit any polychromatic STS(v).
Proof. Let v ≡ 1 or 3 (mod 6). Let 1, 2, . . . , v − 2 be the v − 2 available colors. It
is well known that there exists an idempotent commutative quasigroup of any odd
order. Let (Q, ◦) be an idempotent commutative quasigroup of order v − 2 where
Q = {1, 2, . . . , v − 2}. Consider an edge coloring of the complete graph Kv−2 by
coloring the edge ij ∈ E(Kv−2) with color i ◦ j.
Hence vertex i ∈ V (Kv−2) is incident with exactly one edge of color j ∈ {1, 2, . . . ,
v − 2} \ {i}, and no edges of color i.
Now, add two new vertices x and y and color the edges xi and yi with the color
i. Finally, color the edge xy with any of the v − 2 colors, say color 1.
Obviously, the coloring is equitable at all of the vertices 1, 2, . . . , v− 2 since each
vertex z is incident with a single edge colored with one of the colors 1, 2, . . . , v − 2
except the color z that has been used twice to color the edges zx and zy. At vertex
162 A.A. ABUEIDA, J. LEFEVRE AND M. WATERHOUSE
x (and similarly vertex y) all of the v − 2 colors are used once except color 1 which
is used twice.
It is obvious that the edge xy cannot be in any polychromatic triple since for
every vertex z ∈ {x, y}, the edges xz and yz receive the same color.
Lemma 3.6. Let v ≡ 1 or 3 (mod 6), v ≥ 7. Any equitably (v− 1)-edge colored copy
of Kv will admit a polychromatic STS(v).
Proof. This follows immediately by noting that the degree of every vertex in Kv is
v − 1.
We conjecture the following:
Conjecture 1. Let v ≡ 1 or 3 (mod 6), v ≥ 7. There exists an equitable k-edge
coloring of Kv that does not admit any polychromatic STS(v) for 2 ≤ k ≤ v − 2.
Obviously, the case when k = 2 is trivial.
Acknowledgments
The authors thank the referees for suggestions that lead to improvements in the
reading of the paper.
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(Received 26 Aug 2010; revised 9 Dec 2010)


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