ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.07 / 477–488
https://doi.org/10.26493/1855-3974.2341.af2
(Also available at http://amc-journal.eu)
Avoidance in bowtie systems
Mike J. Grannell , Terry S. Griggs
The Open University, School of Mathematics and Statistics,
Walton Hall, Milton Keynes MK7 6AA, United Kingdom
Giovanni Lo Faro , Antoinette Tripodi *
Università di Messina, Dipartimento di Scienze Matematiche e Informatiche,
Scienze Fisiche e Scienze della Terra, Viale Ferdinando Stagno d’Alcontres 31,
98166 Messina, Italy
Received 20 May 2020, accepted 5 November 2021, published online 9 June 2022
Abstract
There are ten configurations of two bowties that can arise in a bowtie system. The
avoidance spectrum for three of these was determined in a previous paper (Aequat. Math.
85 (2013), 347–358). In this paper the avoidance spectrum for a further five configurations
is determined.
Keywords: Bowtie system, configuration, avoidance, Steiner triple system.
Math. Subj. Class. (2020): 05B30, 05B05, 05B070
1 Introduction
Let X = (V,E) be the graph with vertex set V = {x, a, b, c, d} and edge set
E = {xa, xb, xc, xd, ab, cd}. Such a graph is called a bowtie and will be represented
throughout this paper by the notation a, b − x − c, d. The vertex x is called the centre of
the bowtie and the other vertices are called endpoints. A decomposition of the complete
graph Kn into subgraphs isomorphic to X is called a bowtie system of order n and denoted
by BTS(n). An elementary counting argument shows that a necessary condition for the
existence of a BTS(n) is n ≡ 1 or 9 (mod 12). In a BTS(n), if every vertex of the com-
plete graph Kn occurs the same number of times as the centre of a bowtie, then the bowtie
*Corresponding author. G. Lo Faro and A. Tripodi were supported by INDAM (GNSAGA) and A. Tripodi
was supported by FFABR Unime 2019.
E-mail addresses: m.j.grannell@open.ac.uk (Mike J. Grannell), t.s.griggs@open.ac.uk (Terry S. Griggs),
lofaro@unime.it (Giovanni Lo Faro), atripodi@unime.it (Antoinette Tripodi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
478 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
system is said to be balanced, otherwise the system is said to be unbalanced. A necessary
condition for the existence of a balanced BTS(n) is n ≡ 1 (mod 12).
It is easy to see that, given a BTS(n), by regarding each of the two triangles of every
bowtie as separate entities, we have a Steiner triple system STS(n). We call this the asso-
ciated Steiner triple system of the bowtie system. Conversely, if n ≡ 1 or 9 (mod 12), it
is also true that the triangles of every STS(n) can be amalgamated to form bowties. This
is a consequence of the fact that the block intersection graph of every Steiner triple system
is Hamiltonian, see for example [2, Section 13.6]. If n ≡ 1 (mod 12), there exists a cyclic
STS(n), see also [2, Section 7.2], and this system will have an even number of full orbits.
It is then immediate that we can amalgamate triangles from pairs of orbits to form a bal-
anced BTS(n). Hence the necessary conditions for both BTS(n) and balanced BTS(n)
given above are also sufficient.
A configuration in a bowtie system (resp. Steiner triple system) is a small collection of
bowties (resp. triangles) which may occur in the system. The study of configurations in
STS(n) is now well established and the whole of Chapter 13 of [2] is devoted to various
results about them and in particular includes formulae for the number of occurrences of
all possible configurations of four or fewer triangles. Those for configurations of one, two
or three triangles are functions of n. Such configurations are called constant because the
number of occurrences is independent of the structure of the STS(n). Other configura-
tions are variable. There are 16 non-isomorphic configurations of four triangles of which
5 are constant and 11 are variable. An important concept is that of avoidance; given any
particular configuration in a bowtie system (resp. Steiner triple system), to determine the
spectrum of n for which there exists a BTS(n) (resp. STS(n)) which does not contain that
configuration. Avoidance sets for all configurations of four or fewer triangles in Steiner
triple systems are known. Most, particularly those for constant configurations, are easy
to determine but that for the so-called Pasch configuration (four triangles isomorphic to
{a, b, c}, {a, y, z}, {x, b, z}, {x, y, c}) was more challenging. It is n ≡ 1 or 3 (mod 6),
n ̸= 7, 13 and a complete solution appears in the two papers [7] and [6].
In this paper we will be concerned with the avoidance sets of configurations of two
bowties in a BTS(n). There are ten such configurations which were determined in [3] and
are illustrated in Figure 1. In this figure each triangle of a bowtie is represented by a path
on three vertices and, in each case, one bowtie is represented by solid lines and the second
by dashed lines. The intersection of two solid lines or two dashed lines is the centre of the
bowtie and the other four points are the endpoints. The ten configurations are each labelled
Ĉi for some value of i, 1 ≤ i ≤ 16, to reflect the fact that the bowtie configuration with that
label gives the configuration Ci in the standard listing of configurations of four triangles in
Steiner triple systems as given in [5] or [2, Section 13.1]. Indeed it was by examining all
16 possible configurations of four triangles in a Steiner triple system and identifying which
could be obtained from two bowties that the ten possible configurations of two bowties
were obtained.
There are four equations which connect the number of occurrences of the various con-
figurations of two bowties and these were proved in [3]. Denoting the number of occur-
rences of the configuration Ĉi by ci, the equations are the following.
4c7 + c8 + c11 + c15 = n(n− 1)(n− 5)/24. (1.1)
c11 + c12 + 2c14 + 3c15 + 4c16 = n(n− 1)/3. (1.2)
M. J. Grannell et al.: Avoidance in bowtie systems 479
Ĉ3 Ĉ7 Ĉ8 Ĉ9
Ĉ10 Ĉ11 Ĉ12 Ĉ14
Ĉ15 Ĉ16
Figure 1: Configurations of two bowties.
480 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
c8 + c9 + 2c10 + c11 + c12 + c14 = n(n− 1)(n− 7)/12. (1.3)
4c3 + c8 + 2c9 + c12 = n(n− 1)(n− 7)(n− 9)/72. (1.4)
If the bowtie system is balanced, there is a further equation.
c7 = n(n− 1)(n− 13)/288. (1.5)
All configurations are variable except that Ĉ7 is constant in balanced bowtie systems.
Avoidance sets for the three most compact configurations, Ĉ14, Ĉ15 and Ĉ16 have al-
ready been determined in [3]. The following theorem was proved.
Theorem 1.1. For each n ≡ 1 (mod 12) there exists both a balanced and an unbalanced
BTS(n) simultaneously avoiding Ĉ14, Ĉ15 and Ĉ16. For each n ≡ 9 (mod 12), n ̸=
9 there exists a (necessarily unbalanced) BTS(n) simultaneously avoiding Ĉ14, Ĉ15 and
Ĉ16.
Thus not only can each of these three configurations be avoided for all values of n for
which both balanced BTS(n) and unbalanced BTS(n) exist except for n = 9, they can all
be avoided simultaneously. There are precisely 12 non-isomorphic BTS(9)s which were
enumerated in [4]. All avoid Ĉ16, none avoid Ĉ15 and just one avoids Ĉ14. The details are
in [3].
In this paper, we consider five further configurations. In particular we show that BTS(n)
avoiding three of the least compact configurations Ĉ3, Ĉ7 and Ĉ8 do not exist if n > 13.
Our main results are that for each of the configurations Ĉ11 and Ĉ12, and for all admissible
values of n, there exists a BTS(n) avoiding that configuration, with the single exception of
Ĉ11 when n = 13. The situation for the two configurations Ĉ9 and Ĉ10 remains unresolved.
2 Avoiding Ĉ3, Ĉ7 and Ĉ8
We begin with Ĉ7. The number of bowties in a BTS(n) is n(n− 1)/12. Hence if n > 13,
there will be two bowties with a common centre. So the only possible systems which may
avoid Ĉ7 are balanced BTS(13)s, and indeed all such systems do avoid Ĉ7, and BTS(9)s.
Checking the data of the 12 non-isomorphic BTS(9)s from [3] shows that six of these do
avoid Ĉ7 and the other six do not. We state this formally as a theorem.
Theorem 2.1. The only bowtie systems to avoid Ĉ7 are six of the twelve non-isomorphic
BTS(9)s and all balanced BTS(13)s.
Next we consider Ĉ8 and begin with some observations. First, if a, b − x − c, d is a
bowtie in a BTS(n) which has no Ĉ8 configurations, then there are at most two bowties
whose centre is a. This is because any such bowtie must intersect the bowtie a, b−x− c, d
in a further point which can only be c or d. Similarly, there are at most two bowties whose
centre is b, c or d.
Secondly, in any BTS(n), a point x can be the centre of at most (n − 1)/4 bowties.
Thus if the BTS(n) has no Ĉ8 configurations and x is the centre of less then (n − 1)/4
bowties, then it is an endpoint of at least one other bowtie and so, by the above, there are
at most two bowties whose centre is x. As a consequence, in a BTS(n) which has no Ĉ8
configurations, each point x is the centre of 0, 1, 2 or (n− 1)/4 bowties. Furthermore, if a
point is the centre of (n− 1)/4 bowties, then all remaining points are the centre of at most
two bowties. We can now prove the following theorem.
M. J. Grannell et al.: Avoidance in bowtie systems 481
Theorem 2.2. A BTS(n) avoiding Ĉ8 can only exist if n ≤ 13.
Proof. Subtracting equation 1.2 from equation 1.1 and re-arranging terms gives
c8 = n(n− 1)(n− 13)/24− 4c7 + c12 + 2c14 + 2c15 + 4c16.
Hence c8 ≥ n(n− 1)(n− 13)/24− 4c7.
Now let ax be the number of bowties in a BTS(n) whose centre is x. Then c7 =∑
x∈V
(
ax
2
)
where V denotes the set of n points in the design. Suppose that n > 13 and
that the BTS(n) has no Ĉ8 configurations. Let m be the maximum number of bowties
centred on any point in the BTS(n). Then from the argument above either m = 2 or
m = (n− 1)/4 and all but one point is the centre of at most two bowties.
In either case
c7 =
∑
x∈V
(
ax
2
)
≤
(
(n− 1)/4
2
)
+ (n− 1) = (n− 1)(n+ 27)/32.
Hence
c8 ≥ n(n− 1)(n− 13)/24− (n− 1)(n+ 27)/8 = (n− 1)(n2 − 16n− 81)/24.
The right hand side of this expression is strictly positive for n ≥ 21, and the result follows.
In order to complete the avoidance spectrum for the configuration Ĉ8, we have the
following result.
Theorem 2.3. All BTS(9)s avoid Ĉ8 but no balanced BTS(13) avoids Ĉ8. There exist
unbalanced BTS(13)s which avoid Ĉ8.
Proof. Checking the data of the 12 non-isomorphic BTS(9)s from [3] shows that all avoid
Ĉ8. The fact that no balanced BTS(13) avoids Ĉ8 follows from an exhaustive com-
puter search of all 1, 411, 422 non-isomorphic systems identified in [4]. Two unbalanced
BTS(13)s on the point set {0, 1, 2, . . . , 12} which avoid Ĉ8 are given below. In the first
case the associated STS(13) is cyclic and in the second case it is non-cyclic.
(1) 0, 4− 1− 2, 5; 0, 7− 2− 3, 6; 2, 9− 4− 3, 7;
0, 6− 8− 1, 3; 4, 5− 8− 9, 12; 1, 7− 9− 5, 6;
0, 9− 10− 6, 7; 2, 8− 10− 3, 5; 0, 5− 11− 1, 10;
2, 12− 11− 4, 6; 3, 9− 11− 7, 8; 0, 3− 12− 4, 10;
1, 6− 12− 5, 7.
(2) 1, 4− 0− 2, 7; 6, 8− 0− 9, 10; 0, 12− 3− 1, 8;
2, 6− 3− 4, 7; 2, 9− 4− 5, 8; 1, 2− 5− 3, 10;
1, 7− 9− 5, 6; 2, 8− 10− 6, 7; 0, 5− 11− 1, 6;
2, 12− 11− 4, 10; 3, 9− 11− 7, 8; 1, 10− 12− 5, 7;
4, 6− 12− 8, 9.
Finally in this section we consider Ĉ3. We have a parallel result to Theorem 2.2 for the
configuration Ĉ8.
482 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
Theorem 2.4. A BTS(n) avoiding Ĉ3 can only exist if n ≤ 13.
Proof. Assume that n > 13, so that from Theorem 2.2, c8 > 0. From equation 1.3,
c9 < n(n − 1)(n − 7)/12 and c8 + c9 + c12 ≤ n(n − 1)(n − 7)/12. So by addition
c8 + 2c9 + c12 < n(n− 1)(n− 7)/6. From equation 1.4,
4c3 = n(n− 1)(n− 7)(n− 9)/72− (c8 + 2c9 + c12).
Therefore 4c3 > n(n − 1)(n − 7)(n − 21)/72. Throughout this proof all inequalities are
strict and since n > 13, i.e. n ≥ 21, we have that c3 > 0.
Again, to complete the avoidance spectrum for the configuration Ĉ3, we have the fol-
lowing result.
Theorem 2.5. The avoidance spectrum of the configuration Ĉ3 is the set {9, 13}.
Proof. The configuration Ĉ3 has ten points so all BTS(9)s avoid Ĉ3. A balanced BTS(13)
on the set Z13 which avoid Ĉ3 is the set of bowties (i+ 1), (i+ 4)− i− (i+ 2), (i+ 7),
0 ≤ i ≤ 12, with arithmetic modulo 13. An unbalanced BTS(13) can be obtained by
replacing the bowties 1, 4 − 0 − 2, 7 and 7, 10 − 6 − 8, 0 with the bowties 1, 4 − 0 − 6, 8
and 6, 10− 7− 0, 2.
3 Avoiding Ĉ11 and Ĉ12
The method we use to construct bowtie systems which avoid the configurations Ĉ11 and
Ĉ12 is similar to how we proved Theorem 1.1 on avoiding Ĉ14, Ĉ15 and Ĉ16 and uses stan-
dard techniques involving group divisible designs. It is however more intricate. We note
that all GDDs used in this paper exist (see [1, Section IV 4.1]). An essential component of
the construction is the following BTS(9) which is System (a)(I) in [4] and avoids both Ĉ11
and Ĉ12.
1, 2− 0− 3, 6; 4, 8− 0− 5, 7; 3, 5− 4− 1, 7;
6, 7− 8− 2, 5; 5, 6− 1− 3, 8; 3, 7− 2− 4, 6.
We begin by proving the following result.
Theorem 3.1. For each n ≡ 1, 9 (mod 24), there exists a BTS(n) avoiding Ĉ12.
Proof. Take a 3-GDD of type 4t, where t = 3s or 3s+ 1 and s ≥ 1. Denote the points of
the ith group, 1 ≤ i ≤ t, by (i, 1), (i, 2), (i, 3) and (i, 4). Inflate each point to two points,
i.e. a point (i, j) becomes two points (i, j) and (i, j′). Add a single new point ∞. On each
inflated group of 8 points augmented with the ∞ point place a copy of the BTS(9) above,
identifying the points as follows.
∞ = 0, (i, 1) = 1, (i, 1′) = 3, (i, 2) = 2, (i, 2′) = 6,
(i, 3) = 4, (i, 3′) = 5, (i, 4) = 8, (i, 4′) = 7.
On each of the original blocks of the GDD, say {(i1, j1), (i2, j2), (i3, j3)}, where i1 ̸=
i2 ̸= i3 ̸= i1, place the two bowties (i2, j2), (i3, j3) − (i1, j1) − (i2, j′2), (i3, j′3) and
(i2, j2), (i3, j
′
3)− (i1, j′1)− (i2, j′2), (i3, j3). The bowties in the resulting BTS(8t+1) can
be thought of as being of two types; (i) those resulting from a BTS(9) which we will call
M. J. Grannell et al.: Avoidance in bowtie systems 483
BTS bowties and (ii) those resulting from the blocks of the GDD which we will call GDD
bowties. We need to consider pairs of bowties which arise from all possibilities. There are
five cases to consider.
(1) Two GDD bowties which come from the same block of the GDD. By the construc-
tion these form a configuration Ĉ16.
(2) Two GDD bowties which come from different blocks of the GDD. There are four
possible scenarios.
(a) If the two bowties are disjoint then they form a configuration Ĉ3.
(b) If the centres of the two bowties are the same, then they have no further points
in common and we have a configuration Ĉ7.
(c) If the centre of one of the bowties is an endpoint of the other bowtie, then again
they have no further points in common and we have a configuration Ĉ8.
(d) If the two bowties have an endpoint in common, then they also have a further
endpoint in common and they form a configuration Ĉ10.
(3) Two BTS bowties which come from the same BTS(9). The configuration they form
is completely determined by the structure of the BTS(9) and so avoids Ĉ12 (and
Ĉ11).
(4) Two BTS bowties which come from different BTS(9)s. If the two bowties are dis-
joint then they form a configuration Ĉ3. Otherwise they can only intersect in the
point ∞ which will be the centre of both bowties and we have a configuration Ĉ7.
(5) A BTS bowtie and a GDD bowtie. If the two bowties are disjoint then they form a
configuration Ĉ3. If they have just one point in common then they also avoid Ĉ12.
Otherwise they have two points in common and these points will both be endpoints
of the GDD bowtie. Further, the two points will be (i, j) and (i, j′) for some i, j
such that 1 ≤ i ≤ t and 1 ≤ j ≤ 4. If either of these points is the centre of the
BTS bowtie, then the other point is an endpoint and we have a configuration Ĉ11.
Otherwise both points are endpoints of the BTS bowtie and, because of the way in
which the points of the BTS(9) were assigned to the points ∞, (i, j) and (i, j′), they
are in different triangles. Hence we have a configuration Ĉ10.
We now prove a parallel result for the configuration Ĉ11.
Theorem 3.2. For each n ≡ 1, 9 (mod 24), there exists a BTS(n) avoiding Ĉ11.
Proof. This follows the same steps as the previous theorem. However the way in which
each inflated group of 8 points augmented with the ∞ point is identified with the points of
the BTS(9) is different. In this case it is as follows.
∞ = 0, (i, 1) = 1, (i, 1′) = 2, (i, 2) = 3, (i, 2′) = 6,
(i, 3) = 4, (i, 3′) = 8, (i, 4) = 5, (i, 4′) = 7.
The construction of the GDD bowties is the same. Also, in the analysis of pairs of bowties,
the first four cases are the same. So we only need to consider case (5) of a BTS bowtie and
a GDD bowtie. Again, if the two bowties are disjoint then they form a configuration Ĉ3.
484 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
If they have just one point in common then they also avoid Ĉ11. Otherwise they have two
points in common and they are (i, j) and (i, j′) as before. Because of the way in which the
points of the BTS(9) were assigned to the points ∞, (i, j) and (i, j′), no BTS bowtie has
its centre at a point (i, j) (resp. (i, j′)) and an endpoint at the point (i, j′) (resp. (i, j)). So
both points are endpoints of the BTS bowtie. If they are in the same triangle then we have
a configuration Ĉ12. If they are in different triangles then we have a configuration Ĉ10.
We next consider the cases n ≡ 13, 21 (mod 24). In order to deal with bowtie systems
in these residue classes avoiding Ĉ12, the following further BTS(13) is used.
1, 4− 0− 9, 10; 2, 7− 0− 6, 8; 3, 12− 0− 5, 11;
1, 5− 2− 3, 6; 1, 8− 3− 5, 10; 2, 10− 8− 4, 5;
7, 9− 1− 10, 11; 1, 12− 6− 7, 10; 2, 4− 9− 5, 6;
4, 7− 3− 9, 11; 2, 12− 11− 4, 6; 4, 10− 12− 8, 9;
5, 12− 7− 8, 11.
This system avoids the configuration Ĉ12 and has the property that one point, namely 0,
is at the centre of three bowties and never appears as an endpoint. We can now prove the
following result.
Theorem 3.3. For each n ≡ 13, 21 (mod 24), there exists a BTS(n) avoiding Ĉ12.
Proof. Take a 3-GDD of type 4t61, where t = 3s or 3s + 1 and s ≥ 1. Proceed as
in Theorem 3.1 where in addition the points of the long group are denoted by (t + 1, j),
1 ≤ j ≤ 6. On this inflated group of 12 points augmented with the ∞ point place a copy
of the BTS(13) above, identifying the points as follows.
∞ = 0,
(t+ 1, 1) = 1, (t+ 1, 1′) = 10, (t+ 1, 2) = 4, (t+ 1, 2′) = 9,
(t+ 1, 3) = 2, (t+ 1, 3′) = 6, (t+ 1, 4) = 7, (t+ 1, 4′) = 8,
(t+ 1, 5) = 3, (t+ 1, 5′) = 5, (t+ 1, 6) = 12, (t+ 1, 6′) = 11.
The proof now follows that of Theorem 3.1 . This proves the result for all stated values of
n except n = 21. A solution for this value is the following.
15, 9− 3− 11, 17; 17, 9− 5− 11, 15; 18, 10− 3− 14, 19;
19, 10− 5− 14, 18; 16, 12− 3− 13, 20; 20, 12− 5− 13, 16;
18, 9− 4− 11, 19; 19, 9− 8− 11, 18; 16, 10− 4− 14, 20;
20, 10− 8− 14, 16; 15, 12− 4− 13, 17; 17, 12− 8− 13, 15;
16, 9− 6− 11, 20; 20, 9− 7− 11, 16; 15, 10− 6− 14, 17;
17, 10− 7− 14, 15; 18, 12− 6− 13, 19; 19, 12− 7− 13, 18;
0, 7− 3− 1, 5; 2, 3− 6− 5, 7; 2, 7− 8− 3, 4;
6, 8− 1− 4, 7; 0, 6− 4− 2, 5; 2, 9− 12− 11, 13;
2, 13− 14− 9, 10; 12, 14− 1− 10, 13; 0, 14− 11− 1, 9;
0, 12− 10− 2, 11; 0, 19− 15− 1, 17; 2, 15− 18− 17, 19;
2, 19− 20− 15, 16; 18, 20− 1− 16, 19; 0, 18− 16− 2, 17;
1, 2− 0− 9, 13; 8, 5− 0− 20, 17.
M. J. Grannell et al.: Avoidance in bowtie systems 485
Turning our attention to avoiding Ĉ11, we have shown by an exhaustive computer
search that there is no BTS(13) that avoids this configuration. So for the residue classes
13 and 21 (mod 24) we use the modified constructions given in Theorems 3.4 and 3.5.
For balanced BTS(13)s the minimum number of Ĉ11 configurations is 10 for both associ-
ated cyclic and non-cyclic STS(13)s. For unbalanced systems with the associated cyclic
STS(13), we find that the minimum is 5, but for unbalanced systems with the associated
non-cyclic STS(13), we find that the minimum is 4 and an example is given below.
0, 12− 3− 1, 8; 2, 6− 3− 4, 7; 3, 5− 10− 6, 7;
2, 4− 9− 3, 11; 2, 7− 0− 5, 11; 0, 10− 9− 8, 12;
0, 8− 6− 4, 12; 1, 7− 9− 5, 6; 0, 4− 1− 10, 12;
2, 5− 1− 6, 11; 2, 11− 12− 5, 7; 2, 8− 10− 4, 11;
4, 5− 8− 7, 11.
Theorem 3.4. For each n ≡ 21 (mod 24), there exists a BTS(n) avoiding Ĉ11.
Proof. Take a 3-GDD of type 3t, where t = 4s+3 and s ≥ 0. Denote the points of the ith
group, 1 ≤ i ≤ t, by (i, 1), (i, 2) and (i, 3). As before inflate each point to two points, i.e.
a point (i, j) becomes two points (i, j) and (i, j′). Add three new points ∞0, ∞1 and ∞2.
On each inflated group of 6 points augmented with the three ∞ points first place a copy of
the BTS(9) at the beginning of this Section, identifying the points as follows.
∞0 = 0, ∞1 = 1, ∞2 = 2,
(i, 1) = 3, (i, 1′) = 6, (i, 2) = 4, (i, 2′) = 8,
(i, 3) = 5, (i, 3′) = 7.
The triangle {∞0,∞1,∞2} now occurs 4s+ 3 times. Remove the bowties
∞1,∞2 −∞0 − (i, 1), (i, 1′)
for all i such that 2 ≤ i ≤ 4s+ 3 and replace them by the bowties
(2i, 1), (2i, 1′)−∞0 − (2i+ 1, 1), (2i+ 1, 1′), 1 ≤ i ≤ 2s+ 1.
We call these BTS⋆ bowties. The construction of the GDD bowties is as in the previous
three theorems.
We need to prove that a bowtie system constructed in this way avoids configuration
Ĉ11. The proof for the five cases involving just BTS bowties and GDD bowties is as
in Theorem 3.2. So any putative configuration Ĉ11 must contain a BTS⋆ bowtie. We
show that this is impossible. A configuration Ĉ11 consists of two bowties isomorphic to
c, y − x − b, z and a, z − y − d, e. The centre of every BTS⋆ bowtie is ∞0; however this
point never occurs as the endpoint of any bowtie. So y ̸= ∞0. Now suppose that x = ∞0
and that c, y − x− b, z is a BTS⋆ bowtie. Then without loss of generality y = (2i, 1) and
z = (2i + 1, 1) for some i such that 1 ≤ i ≤ 2s + 1, say i = q. Therefore the bowtie
a, z − y − d, e is a GDD bowtie and either d or e = (2q + 1, 1′) = b which means that we
do not have a configuration Ĉ11.
We note that, by using a 3-GDD of type 3t where t = 4s+1, s ≥ 1, the above theorem
can also be used to provide an alternative proof of the existence of a BTS(n) avoiding Ĉ11
for the residue class 9 (mod 24).
486 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
A BTS(21) avoiding Ĉ11 from the above theorem is given below. This will be needed
in the proof of the final theorem. It has the crucial property that one point, again namely 0,
is at the centre of five bowties and never appears as an endpoint.
1, 2− 0− 3, 6; 4, 8− 0− 5, 7; 10, 14− 0− 11, 13;
16, 20− 0− 17, 19; 9, 12− 0− 15, 18;
3, 5− 4− 1, 7; 9, 11− 10− 1, 13; 15, 17− 16− 1, 19;
6, 7− 8− 2, 5; 12, 13− 14− 2, 11; 18, 19− 20− 2, 17;
5, 6− 1− 3, 8; 11, 12− 1− 9, 14; 17, 18− 1− 15, 20;
3, 7− 2− 4, 6; 9, 13− 2− 10, 12; 15, 19− 2− 16, 18;
9, 15− 3− 12, 18; 10, 17− 3− 14, 19; 11, 16− 3− 13, 20;
9, 18− 6− 12, 15; 10, 19− 6− 14, 17; 11, 20− 6− 13, 16;
10, 16− 4− 14, 20; 11, 15− 4− 13, 18; 9, 17− 4− 12, 19;
10, 20− 8− 14, 16; 11, 18− 8− 13, 15; 9, 19− 8− 12, 17;
11, 17− 5− 13, 19; 9, 16− 5− 12, 20; 10, 15− 5− 14, 18;
11, 19− 7− 13, 17; 9, 20− 7− 12, 16; 10, 18− 7− 14, 15.
Theorem 3.5. For each n ≡ 13 (mod 24), except for n = 13, there exists a BTS(n)
avoiding Ĉ11.
Proof. Take a 3-GDD of type 4t101, where t = 3s+ 2, s ≥ 1. Proceed as in Theorem 3.2
where the points of the long group are denoted by (t + 1, j), 1 ≤ j ≤ 10. On this
inflated group of 20 points augmented with the ∞ point place a copy of the BTS(21)
above, identifying the points as follows.
∞ = 0,
(t+ 1, 1) = 1, (t+ 1, 1′) = 2, (t+ 1, 2) = 3, (t+ 1, 2′) = 6,
(t+ 1, 3) = 4, (t+ 1, 3′) = 8, (t+ 1, 4) = 5, (t+ 1, 4′) = 7,
(t+ 1, 5) = 10, (t+ 1, 5′) = 14, (t+ 1, 6) = 11, (t+ 1, 6′) = 13,
(t+ 1, 7) = 16, (t+ 1, 7′) = 20, (t+ 1, 8) = 17, (t+ 1, 8′) = 19,
(t+ 1, 9) = 9, (t+ 1, 9′) = 12, (t+ 1, 10) = 15, (t+ 1, 10′) = 18.
The proof now follows that of Theorem 3.2. This proves the result for all stated values of
n except n = 37. A solution for this value is given in Table 1 below.
Finally, we again note that, by using a 3-GDD of type 4t101 where t = 3s, s ≥ 1,
the above theorem can also be used to provide an alternative proof of the existence of a
BTS(n) avoiding Ĉ11 for the residue class 21 (mod 24).
M. J. Grannell et al.: Avoidance in bowtie systems 487
Table 1: A BTS(37) avoiding Ĉ11.
16, 34− 0− 17, 35; 18, 36− 0− 1, 19; 8, 26− 0− 9, 27;
10, 28− 0− 11, 29; 12, 30− 0− 13, 31; 14, 32− 0− 15, 33;
2, 20− 0− 7, 25; 3, 21− 0− 4, 22; 5, 23− 0− 6, 24;
7, 18− 6− 36, 1; 25, 36− 24− 18, 19; 24, 1− 7− 19, 36;
6, 19− 25− 1, 18; 9, 8− 1− 26, 27; 27, 8− 19− 26, 9;
7, 5− 2− 23, 25; 25, 5− 20− 23, 7; 14, 13− 3− 31, 32;
32, 13− 21− 31, 14; 15, 11− 4− 29, 33; 33, 11− 22− 29, 15;
12, 10− 6− 28, 30; 30, 10− 24− 28, 12; 11, 10− 1− 28, 29;
29, 10− 19− 28, 11; 15, 13− 2− 31, 33; 33, 13− 20− 31, 15;
6, 5− 3− 23, 24; 24, 5− 21− 23, 6; 12, 8− 4− 26, 30;
30, 8− 22− 26, 12; 14, 9− 7− 27, 32; 32, 9− 25− 27, 14;
13, 12− 1− 30, 31; 31, 12− 19− 30, 13; 11, 9− 2− 27, 29;
29, 9− 20− 27, 11; 7, 4− 3− 22, 25; 25, 4− 21− 22, 7;
15, 10− 5− 28, 33; 33, 10− 23− 28, 15; 14, 8− 6− 26, 32;
32, 8− 24− 26, 14; 15, 14− 1− 32, 33; 33, 14− 19− 32, 15;
6, 4− 2− 22, 24; 24, 4− 20− 22, 6; 11, 8− 3− 26, 29;
29, 8− 21− 26, 11; 12, 9− 5− 27, 30; 30, 9− 23− 27, 12;
13, 10− 7− 28, 31; 31, 10− 25− 28, 13; 18, 17− 16− 35, 36;
36, 17− 34− 35, 18; 2, 3− 1− 21, 16; 20, 21− 19− 3, 34;
19, 16− 2− 34, 21; 1, 34− 20− 16, 3; 10, 14− 4− 32, 16;
28, 32− 22− 14, 34; 22, 16− 10− 34, 32; 4, 34− 28− 16, 14;
8, 13− 5− 31, 16; 26, 31− 23− 13, 34; 23, 16− 8− 34, 31;
5, 34− 26− 16, 13; 9, 15− 6− 33, 16; 27, 33− 24− 15, 34;
24, 16− 9− 34, 33; 6, 34− 27− 16, 15; 11, 12− 7− 30, 16;
29, 30− 25− 12, 34; 25, 16− 11− 34, 30; 7, 34− 29− 16, 12;
4, 5− 1− 23, 17; 22, 23− 19− 5, 35; 19, 17− 4− 35, 23;
1, 35− 22− 17, 5; 12, 14− 2− 32, 17; 30, 32− 20− 14, 35;
20, 17− 12− 35, 32; 2, 35− 30− 17, 14; 9, 10− 3− 28, 17;
27, 28− 21− 10, 35; 21, 17− 9− 35, 28; 3, 35− 27− 17, 10;
11, 13− 6− 31, 17; 29, 31− 24− 13, 35; 24, 17− 11− 35, 31;
6, 35− 29− 17, 13; 8, 15− 7− 33, 17; 26, 33− 25− 15, 35;
25, 17− 8− 35, 33; 7, 35− 26− 17, 15; 8, 10− 2− 28, 18;
26, 28− 20− 10, 36; 20, 18− 8− 36, 28; 2, 36− 26− 18, 10;
12, 15− 3− 33, 18; 30, 33− 21− 15, 36; 21, 18− 12− 36, 33;
3, 36− 30− 18, 15; 9, 13− 4− 31, 18; 27, 31− 22− 13, 36;
22, 18− 9− 36, 31; 4, 36− 27− 18, 13; 11, 14− 5− 32, 18;
29, 32− 23− 14, 36; 23, 18− 11− 36, 32; 5, 36− 29− 18, 14.
488 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
ORCID iDs
Mike Grannell https://orcid.org/0000-0002-0429-0493
Giovanni Lo Faro https://orcid.org/0000-0001-6174-8627
Antoinette Tripodi https://orcid.org/0000-0003-0767-4457
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