An infinite family of Steiner triple systems without parallel classes
Daniel Horsley (Monash University)
Joint work with Darryn Bryant (University of Queensland)
Part 1:
Steiner triple systems and parallel classes
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Steiner triple systems
An STS(7)
Theorem(Kirkman 1847) An STS(v ) exists if and only if v ≥ 1 and v ≡ 1 or 3 (mod 6).
Parallel classes
An STS(9) An STS(9) with a PC
Parallel classes
An STS(9) An STS(9) with a PC
Parallel classes
An STS(9)
An STS(9) with a PC
Parallel classes
An STS(9)
An STS(9) with a PC
Parallel classes
An STS(9)
An STS(9) with a PC
Almost parallel classes
An STS(13)
An STS(13) with an APC
Almost parallel classes
An STS(13)
An STS(13) with an APC
Almost parallel classes
An STS(13)
An STS(13) with an APC
Almost parallel classes
An STS(13)
An STS(13) with an APC
A question
Question What can we say about when an STS(v ) has a PC/APC?
If v ≡ 3 (mod 6), the STS(v ) might have a PC. If v ≡ 1 (mod 6), the STS(v ) might have an APC.
A question
Question What can we say about when an STS(v ) has a PC/APC?
If v ≡ 3 (mod 6), the STS(v ) might have a PC.
If v ≡ 1 (mod 6), the STS(v ) might have an APC.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Small orders
I The unique STS(7) has no APC.
I The unique STS(9) has a PC.
I Both STS(13)s have an APC.
I 70 of the 80 STS(15)s have a PC.
I All but 2 of the 11, 084, 874, 829 STS(19)s have an APC.(Colbourn et al.)
I 12 STS(21)s are known to have no PC.(Mathon, Rosa)
STSs without PCs/APCs seem rare.
Conjectures
Conjecture(Mathon, Rosa) There is an STS(v ) with no PC for all v ≡ 3 (mod 6) except v = 3, 9.
Conjecture(Rosa, Colbourn) There is an STS(v ) with no APC for all v ≡ 1 (mod 6) except v = 13.
Conjectures
Conjecture(Mathon, Rosa) There is an STS(v ) with no PC for all v ≡ 3 (mod 6) except v = 3, 9.
Conjecture(Rosa, Colbourn) There is an STS(v ) with no APC for all v ≡ 1 (mod 6) except v = 13.
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Progress on these conjectures
v ≡ 1 (mod 6)
Theorem(Wilson, 1992) For each odd n there is an STS(2n− 1) with no APC.
Wilson’s examples are projective triple systems.
Theorem(Bryant, Horsley, 2013) For each n ≥ 1 there is an STS(2(3n) + 1) with no APC.
v ≡ 3 (mod 6)
Up until recently, the only known STSs of order 3 (mod 6) without PCs had order 15 or 21.
Theorem(Bryant, Horsley, 201?) For each v ≡ 27 (mod 30) such that ordp(−2) ≡ 0 (mod 4) for every prime divisor p of v − 2, there is an STS(v ) with no PC. There are infinitely many such values of v .
Part 2:
Our result
Construction
I Let v = 5n + 2 and G = Z5× Zn(remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction
I Let v = 5n + 2 and G = Z5× Zn (remember v ≡ 27 (mod 30)). Note n ≡ 5 (mod 6).
I Let theweightof a subset of G be the sum of its elements.
I Think of the 3-subsets of G with weight(0, 0).
I Every 2-subset of G {x , y } is in exactly one such 3-subset, namely {x, y , −x − y }, unless y = −2x or x = −2y .
I So the 3-subsets of G with weight(0, 0)give a PSTS(v − 2) on G with unused edges {{g , −2g } : g ∈ G \ {(0, 0)}}.
-x 2x
-4x
8x
-16x
x -2x
4x
-8x
-x -16x
I Every point in G \ {(0, 0)} is in one such cycle.
I Consider the weights of these edges.
I For each g ∈ G \ {(0, 0)} there is exactly one unused edge of weightg.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0)
(0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2.
I Add an STS(7) not containing {(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2.
I Add an STS(7) not containing {(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2.
I Add an STS(7) not containing {(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour. I Add triples made from the blue edges
and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour.
I Add triples made from the blue edges and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour.
I Add triples made from the blue edges and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour.
I Add triples made from the blue edges and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2
(4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
I Remove {(0, 0), (2, 0), (3, 0)} and {(0, 0), (1, 0), (4, 0)}.
I Add vertices(0, 0)1and(0, 0)2. I Add an STS(7) not containing
{(0, 0),(0, 0)1,(0, 0)2} on the specified vertices.
I Properly 2-edge-colour the remaining cycles so that(∗, b)and(∗, −b) always receive the same colour.
I Add triples made from the blue edges and(0, 0)1and from the green edges and(0, 0)2.
I The result is an STS(87) which I claim has no PC.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2 (4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
Suppose the STS(87) contains a PC. I The sum of the weights of the triples
in the PC is the sum of the vertex labels which is(0, 0).
I If(0, 0)1 and(0, 0)2are in the same triple, then the rest have weight (0, 0). Contradiction.
I LetT1andT2be the triples in the PC containing(0, 0)1and(0, 0)2. I IfT1is in the STS(7), then the
triples exceptT2have weight(∗, 0). Contradiction.
I SoT1andT2are not in the STS(7). By the properties of the edge colouring, their weights cannot add to(∗, 0). But the rest have weight (∗, 0). Contradiction.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2 (4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
Suppose the STS(87) contains a PC. I The sum of the weights of the triples
in the PC is the sum of the vertex labels which is(0, 0).
I If(0, 0)1 and(0, 0)2are in the same triple, then the rest have weight (0, 0). Contradiction.
I LetT1andT2be the triples in the PC containing(0, 0)1and(0, 0)2. I IfT1is in the STS(7), then the
triples exceptT2 have weight(∗, 0). Contradiction.
I SoT1andT2are not in the STS(7). By the properties of the edge colouring, their weights cannot add to(∗, 0). But the rest have weight (∗, 0). Contradiction.
Construction example: v = 87, G = Z5 × Z17
(0, 0) (0, 0)1
(0, 0)2 (4, 0)
(2, 0) (1, 0) (3, 0)
(a, 1) (3a, 15)
(4a, 4)
(2a, 9) (a, 16) (3a, 2) (4a, 13)
(2a, 8)
(a, 3) (3a, 11)
(4a, 12)
(2a, 10) (a, 14) (3a, 6) (4a, 5)
(2a, 7)
| {z }
for each a ∈ Z5
Suppose the STS(87) contains a PC.
I The sum of the weights of the triples in the PC is the sum of the vertex labels which is(0, 0).
I If(0, 0)1 and(0, 0)2are in the same triple, then the rest have weight (0, 0). Contradiction.
I LetT1andT2be the triples in the PC containing(0, 0)1and(0, 0)2. I IfT1is in the STS(7), then the
triples exceptT2 have weight(∗, 0). Contradiction.
I SoT1andT2are not in the STS(7). By the properties of the edge colouring, their weights cannot add to(∗, 0). But the rest have weight (∗, 0). Contradiction.
