Designs and i -Block-Intersection Graphs David Pike Memorial - - PowerPoint PPT Presentation

Designs and i-Block-Intersection Graphs David Pike

Memorial University of Newfoundland

Definition: A combinatorial design D consists of a set V of elements (called points), together with a set B of subsets (called blocks)

• f V .

A balanced incomplete balanced design, BIBD(v, k, λ), is a design in which:

• |V | = v,
• for each block B ∈ B, |B| = k, and
• each 2-subset of V occurs in precisely λ blocks of B.

A BIBD(v, 3, 1) is a Steiner triple system, STS(v). A BIBD(v, 3, 2) is a twofold triple system, TTS(v).

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Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1

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Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6

Slide 3 of 32

Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

Slide 3 of 32

Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

1 2 3 4 5 6 7 8 9 A B C

Slide 3 of 32

Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

1 2 3 4 5 6 7 8 9 A B C {0,2,7} {8,9,C}

Slide 3 of 32

Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

1 2 3 4 5 6 7 8 9 A B C {0,2,7} {8,9,C} {1,3,8} {9,A,0}

Slide 3 of 32

Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

1 2 3 4 5 6 7 8 9 A B C {0,2,7} {8,9,C} {1,3,8} {9,A,0} {2,4,9} {A,B,1}

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Example: a BIBD(13,3,1) ... i.e., a STS(13):

λ(v

2)

(k

2)

v = 13 k = 3 λ = 1 r = λ(v−1)

k−1

= 6 b =

λ(v

2)

(k

2) = 26

1 2 3 4 5 6 7 8 9 A B C

This is equivalent to a K3-decomposition of K13

{0,2,7} {8,9,C} {1,3,8} {9,A,0} {2,4,9} {A,B,1} {3,5,A} {B,C,2} {4,6,B} {C,0,3} {5,7,C} {0,1,4} {6,8,0} {1,2,5} {7,9,1} {2,3,6} {8,A,2} {3,4,7} {9,B,3} {4,5,8} {A,C,4} {5,6,9} {B,0,5} {6,7,A} {C,1,6} {7,8,B}

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Example: a TTS(6) ... i.e., a BIBD(6,3,2): V = {1, 2, 3, 4, 5, 6} Blocks:

{1,2,3} {2,3,5} {1,2,4} {2,4,6} {1,3,6} {2,5,6} {1,4,5} {3,4,5} {1,5,6} {3,4,6}

This is equivalent to a K3-decomposition of 2K6

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Example: a TTS(6) ... i.e., a BIBD(6,3,2): V = {1, 2, 3, 4, 5, 6} Blocks:

{1,2,3} {2,3,5} {1,2,4} {2,4,6} {1,3,6} {2,5,6} {1,4,5} {3,4,5} {1,5,6} {3,4,6}

This is equivalent to a K3-decomposition of 2K6 Theorem A STS(v) exists if and only if v ≡ 1 or 3 (mod 6). A TTS(v) exists if and only if v ≡ 0 or 1 (mod 3).

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Applications of Designs: Design of experiments: Suppose we need to test how v varieties of something interact in pairs (e.g., crop varieties, strains of bacteria, electronic components, etc.).

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Applications of Designs: Design of experiments: Suppose we need to test how v varieties of something interact in pairs (e.g., crop varieties, strains of bacteria, electronic components, etc.). Suppose also that we are able to test all k

2

• pairs of any

k-subset of varieties in parallel.

Slide 5 of 32

Applications of Designs: Design of experiments: Suppose we need to test how v varieties of something interact in pairs (e.g., crop varieties, strains of bacteria, electronic components, etc.). Suppose also that we are able to test all k

2

• pairs of any

k-subset of varieties in parallel. Testing a k-subset is costly (e.g., each test corresponds to a crop to plant and grow, a petri dish to prepare and incubate,

• r a prototypic circuit board to be built).

So we wish to minimise the number of tests that are necessary.

Slide 5 of 32

Applications of Designs: Design of experiments: Suppose we need to test how v varieties of something interact in pairs (e.g., crop varieties, strains of bacteria, electronic components, etc.). Suppose also that we are able to test all k

2

• pairs of any

k-subset of varieties in parallel. Testing a k-subset is costly (e.g., each test corresponds to a crop to plant and grow, a petri dish to prepare and incubate,

• r a prototypic circuit board to be built).

So we wish to minimise the number of tests that are necessary. Solution: use a BIBD(v, k, 1) to determine which k elements will comprise each test.

Slide 5 of 32

Applications of Designs: Team Building: Suppose a professor has v graduate students who are to be sent to work at a field station, k at a time, and so that each pair

• f students works together λ times.

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Applications of Designs: Team Building: Suppose a professor has v graduate students who are to be sent to work at a field station, k at a time, and so that each pair

• f students works together λ times.

How can the groups of k students to be formed?

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Applications of Designs: Team Building: Suppose a professor has v graduate students who are to be sent to work at a field station, k at a time, and so that each pair

• f students works together λ times.

How can the groups of k students to be formed? Solution: use a BIBD(v, k, λ).

Slide 6 of 32

Applications of Designs: Team Building: Suppose a professor has v graduate students who are to be sent to work at a field station, k at a time, and so that each pair

• f students works together λ times.

How can the groups of k students to be formed? Solution: use a BIBD(v, k, λ). What about Scheduling? How can travel costs to/from the field station be minimised? Are some orderings of the crops/bacteria/electronics to be tested better than other orderings? How can we nicely order the blocks of a design?

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Definition: Given a combinatorial design D with block set B, the block-intersection graph of D is the graph having B as its vertex set, and in which two vertices B1 and B2 are adjacent if and only if B1 ∩ B2 = ∅.

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Definition: Given a combinatorial design D with block set B, the block-intersection graph of D is the graph having B as its vertex set, and in which two vertices B1 and B2 are adjacent if and only if B1 ∩ B2 = ∅. Example: A TTS(4): V = {1, 2, 3, 4} B =

• {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
• 134

124 234 123

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Question (Graham, 1987) Is the block-intersection graph of a STS(v) Hamiltonian?

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Some Subsequent Discoveries:

• BIBD(v, k, λ) ⇒ Hamiltonian

(Horák and Rosa, 1988)

• PBD(v, K, 1) with max K 2 min K ⇒ Hamiltonian

(Alspach, Heinrich and Mohar, 1990)

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Some Subsequent Discoveries:

• BIBD(v, k, λ) ⇒ Hamiltonian

(Horák and Rosa, 1988)

• PBD(v, K, 1) with max K 2 min K ⇒ Hamiltonian

(Alspach, Heinrich and Mohar, 1990)

• BIBD(v, k, 1) with k 3 ⇒ edge-pancyclic

(Alspach and Hare, 1991)

• PBD(v, K, 1) with min K 3 ⇒ edge-pancyclic

(Hare, 1995)

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Some Subsequent Discoveries:

• BIBD(v, k, λ) ⇒ Hamiltonian

(Horák and Rosa, 1988)

• PBD(v, K, 1) with max K 2 min K ⇒ Hamiltonian

(Alspach, Heinrich and Mohar, 1990)

• BIBD(v, k, 1) with k 3 ⇒ edge-pancyclic

(Alspach and Hare, 1991)

• PBD(v, K, 1) with min K 3 ⇒ edge-pancyclic

(Hare, 1995)

• BIBD(v, k, λ) ⇒ pancyclic

(Mamut, Pike and Raines, 2004)

• PBD(v, K, λ) with max K λ min K ⇒ pancyclic

(Case and Pike, 2008)

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Some Subsequent Discoveries:

• BIBD(v, k, λ) ⇒ Hamiltonian

(Horák and Rosa, 1988)

• PBD(v, K, 1) with max K 2 min K ⇒ Hamiltonian

(Alspach, Heinrich and Mohar, 1990)

• BIBD(v, k, 1) with k 3 ⇒ edge-pancyclic

(Alspach and Hare, 1991)

• PBD(v, K, 1) with min K 3 ⇒ edge-pancyclic

(Hare, 1995)

• BIBD(v, k, λ) ⇒ pancyclic

(Mamut, Pike and Raines, 2004)

• PBD(v, K, λ) with max K λ min K ⇒ pancyclic

(Case and Pike, 2008)

• BIBD(v, k, λ) with k 3 ⇒ cycle extendible (Abueida and Pike, 2013)
• PBD(v, K, λ) with λ 2 and max K λ(min K − 1)

⇒ cycle extendible

(Luther and Pike, 201x)

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Definition: Given a combinatorial design D with block set B, the i-block-intersection graph of D is the graph having B as its vertex set, and in which two vertices B1 and B2 are adjacent if and only if |B1 ∩ B2| = i.

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Definition: Given a combinatorial design D with block set B, the i-block-intersection graph of D is the graph having B as its vertex set, and in which two vertices B1 and B2 are adjacent if and only if |B1 ∩ B2| = i. Observations: When i 1, the i-block-intersection graph of D is a subgraph

• f the traditional block-intersection graph.

However, when i = 0, the 0-block-intersection graph of D is the graph complement of the traditional block-intersection graph.

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Example: Our previous TTS(6) The graph

136 124 123 156 145 346 246 235 256 345

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Example: Our previous TTS(6) The traditional block-intersection graph:

136 124 123 156 145 346 246 235 256 345

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Example: Our previous TTS(6) The 1-block-intersection graph:

136 124 123 156 145 346 246 235 256 345

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Example: Our previous TTS(6) The 1-block-intersection graph:

136 124 123 156 145 346 246 235 256 345

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Example: Our previous TTS(6) The 2-block-intersection graph:

136 124 123 156 145 346 246 235 256 345

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Definition: A κ-intersecting Gray code for a design consists of an ordering

• f the blocks, (B1, B2, B3, . . . , Bb), such that consecutive blocks

intersect in exactly κ points (i.e., |Bi ∩ Bi+1| = κ). If, additionally, |Bb ∩ B1| = κ, then we have a κ-intersecting cyclic Gray code. Observations: A κ-intersecting Gray code for a design is equivalent to a Hamilton path in the design’s κ-block-intersection graph. A κ-intersecting cyclic Gray code for a design is equivalent to a Hamilton cycle in the design’s κ-block-intersection graph.

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Some results for κ = 0 The following designs have Hamiltonian 0-BIGs:

• every BIBD(v, 3, λ) with v 17

(Dewar, 2007)

• every BIBD(v, k, 1) with v 2k2 + 1

(Dewar, 2007)

• every BIBD(v, k, 1) with v >
• 1+

√ 5 2

• k2 + k

(LeGrow, Pike and Poulin, 201x)

• every BIBD(v, k, λ) with v > 2k2 + 1

(LeGrow, Pike and Poulin, 201x)

and hence they have 0-intersecting cyclic Gray codes. In many cases, Hamilton cycles can be found in polynomial time.

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Some results for κ = 1 The following designs have Hamiltonian 1-BIGs:

• every BIBD(v, k, 1)

(Horák and Rosa, 1988)

• every BIBD(v, 3, λ) with v 12

(Horák, Pike and Raines, 1999)

• every BIBD(v, 4, λ) with v 136

(Jesso, Pike and Shalaby, 2011)

• every BIBD(v, 5, λ) with v 305

(Jesso, 2011)

and hence they have 1-intersecting cyclic Gray codes. The computational complexity of finding Hamilton cycles in 1-block-intersection graphs is not yet known.

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What about κ = 2? Which BIBD(v, k, λ) have 2-intersecting cyclic Gray codes? For k = 3 and λ = 2, we equivalently ask which TTS have Hamiltonian 2-BIGs? Some TTS with non-Hamiltonian 2-BIGs:

• the TTS(6) that we have already seen
• a TTS(19) found in 1983 by M. Colbourn and Johnstone

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Theorem (Dewar, 2007) There exists a TTS(v) with a Hamiltonian 2-BIG if

• v ≡ 1 or 4 (mod 12) where v ≡ 0 (mod 5), or
• v ≡ 3 or 7 (mod 12) where v 7.

Recall: a TTS(v) exists if and only if v ≡ 0 or 1 (mod 3). Open Cases: Determine whether there exist TTS(v) with Hamiltonian 2-BIG when:

• v ≡ 25 or 40 (mod 60), or
• v ≡ 0, 6, 9 or 10 (mod 12).

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Observations: For the 2-BIG of a TTS to be connected, it is necessary that the TTS be simple (i.e., no blocks are repeated). Proof: if {x, y, z} and {x, y, z} both appear as blocks of a TTS, Proof: then they both have degree 0 in the 2-BIG. The 2-BIG of every simple TTS is cubic. Proof: Each block {x, y, z} has three distinct neighbours: Proof: {x, y, α}, {x, z, β} and {y, z, γ}. Theorem (Garey, Johnson and Tarjan, 1976) It is NP-complete to decide whether a general cubic graph is Hamiltonian.

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Observation: When v ≡ 1 or 3 (mod 6), an easy way to build a simple TTS(v) is to take the union of two STS(v) that share no blocks. Question: What does the 2-BIG of a TTS of this form look like?

Slide 18 of 32

Observation: When v ≡ 1 or 3 (mod 6), an easy way to build a simple TTS(v) is to take the union of two STS(v) that share no blocks. Question: What does the 2-BIG of a TTS of this form look like?

{x, y, z}

Slide 18 of 32

Observation: When v ≡ 1 or 3 (mod 6), an easy way to build a simple TTS(v) is to take the union of two STS(v) that share no blocks. Question: What does the 2-BIG of a TTS of this form look like?

{x, y, z} {x, y, α} {x, z, β} {y, z, γ}

Slide 18 of 32

Observation: When v ≡ 1 or 3 (mod 6), an easy way to build a simple TTS(v) is to take the union of two STS(v) that share no blocks. Question: What does the 2-BIG of a TTS of this form look like?

{x, y, z} {x, y, α} {x, z, β} {y, z, γ} {x, α, δ} {y, α, ǫ}

Slide 18 of 32

Observation: When v ≡ 1 or 3 (mod 6), an easy way to build a simple TTS(v) is to take the union of two STS(v) that share no blocks. Question: What does the 2-BIG of a TTS of this form look like?

{x, y, z} {x, y, α} {x, z, β} {y, z, γ} {x, α, δ} {y, α, ǫ} {x, δ, ζ} {α, δ, η}

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Lemma: The 2-BIG of a TTS(v) is bipartite if and only if its blocks can be partitioned into the blocks of two STS(v).

Slide 19 of 32

Example: Universal Cycle for a TTS(13)

1 2 3 8 12 4 9 10 3 13 4 7 10 5 13 8 6 10 11 2 13 9 7 3 11 12 5 9 8 1 10 12 2 7 8 11 4 5 2 6 9 11 1 13 12 6 7 1 5 3 6 4

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Example: Universal Cycle for a TTS(13)

1 2 3 8 12 4 9 10 3 13 4 7 10 5 13 8 6 10 11 2 13 9 7 3 11 12 5 9 8 1 10 12 2 7 8 11 4 5 2 6 9 11 1 13 12 6 7 1 5 3 6 4

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Example: Universal Cycle, Hamiltonian 2-BIG

1 2 3 8 12 4 9 10 3 13 4 7 10 5 13 8 6 10 11 2 13 9 7 3 11 12 5 9 8 1 10 12 2 7 8 11 4 5 2 6 9 11 1 13 12 6 7 1 5 3 6 4

Slide 20 of 32

Example: Universal Cycle, Hamiltonian 2-BIG, cubic

1 2 3 8 12 4 9 10 3 13 4 7 10 5 13 8 6 10 11 2 13 9 7 3 11 12 5 9 8 1 10 12 2 7 8 11 4 5 2 6 9 11 1 13 12 6 7 1 5 3 6 4

Slide 20 of 32

Example: Universal Cycle, Hamiltonian 2-BIG, cubic, bipartite

1 2 3 8 12 4 9 10 3 13 4 7 10 5 13 8 6 10 11 2 13 9 7 3 11 12 5 9 8 1 10 12 2 7 8 11 4 5 2 6 9 11 1 13 12 6 7 1 5 3 6 4

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Lemma: The 2-BIG of a TTS(v) is bipartite if and only if its blocks can be partitioned into the blocks of two STS(v). Corollary: Suppose D is a simple TTS whose blocks can be partitioned into the blocks of two STS. Then the 2-BIG of D is Class 1 (i.e., it has chromatic index χ′ = ∆). Theorem (Vizing, 1964) If G is a simple graph, then ∆(G) χ′(G) ∆(G) + 1. Theorem (Holyer, 1981) Determining whether the chromatic index of a cubic graph is 3 versus 4 is an NP-complete problem.

Slide 21 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

1 2 3 4 5 6

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Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

1 2 3 4 5 6

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

1 2 3 4 5 6

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

1 2 3 4 5 6

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

Pasch Configuration in Kv 1 2 3 4 5 6

Slide 22 of 32

Question: When does a TTS have a 2-BIG that is connected? Lemma: A bipartite 2-BIG is connected if and only if there are no trades

• f proper configurations between the two underlying STS.

STS #1 STS #2 . . . . . . . . . . . . {1,2,3} {1,4,5} {2,4,6} {3,5,6} {1,2,4} {1,3,5} {2,3,6} {4,5,6}

Pasch Configuration in Kv 1 2 3 4 5 6

Slide 22 of 32

Example: A TTS(9) with a disconnected 2-BIG

157 179 189 168 156 257 279 289 268 256 358 359 369 367 378 458 459 469 467 478 123 234 124 134

Slide 23 of 32

Definition: A snark is a graph that is cubic, 2-edge-connected and Class 2 (i.e., 3-regular, 2 κ′ ∆ = 3 and χ′ = ∆ + 1 = 4). Often the additional requirements of having girth at least 5 and having no non-trivial 3-edge-cuts are included in the definition. Observations: Snarks are not bipartite.

(bipartite ⇒ Class 1)

Snarks are not Hamiltonian.

(cubic Hamiltonian ⇒ Class 1)

Slide 24 of 32

Definition: A snark is a graph that is cubic, 2-edge-connected and Class 2 (i.e., 3-regular, 2 κ′ ∆ = 3 and χ′ = ∆ + 1 = 4). Often the additional requirements of having girth at least 5 and having no non-trivial 3-edge-cuts are included in the definition. Observations: Snarks are not bipartite.

(bipartite ⇒ Class 1)

Snarks are not Hamiltonian.

(cubic Hamiltonian ⇒ Class 1)

The 2-BIG of a simple TTS is cubic and (if not disconnected) 3-connected.

(M. Colbourn and Johnstone, 1983)

Slide 24 of 32

Definition: A snark is a graph that is cubic, 2-edge-connected and Class 2 (i.e., 3-regular, 2 κ′ ∆ = 3 and χ′ = ∆ + 1 = 4). Often the additional requirements of having girth at least 5 and having no non-trivial 3-edge-cuts are included in the definition. Observations: Snarks are not bipartite.

(bipartite ⇒ Class 1)

Snarks are not Hamiltonian.

(cubic Hamiltonian ⇒ Class 1)

The 2-BIG of a simple TTS is cubic and (if not disconnected) 3-connected.

(M. Colbourn and Johnstone, 1983)

Question: Do any simple TTS have 2-BIGs that are snarks?

Slide 24 of 32

Answer: Yes ... our previous TTS(6)

136 124 123 156 145 346 246 235 256 345

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Answer: Yes ... our previous TTS(6)

136 124 123 156 145 346 246 235 256 345

Question: Are there any others?

Slide 25 of 32

Some Empirical Results:

• 1 simple TTS(4)
• Its 2-BIG is K4 (Hamiltonian)
• 1 simple TTS(6)
• Its 2-BIG is the Petersen graph

(not Hamiltonian, but has a Hamilton path)

• 1 simple TTS(7)
• Its 2-BIG is Hamiltonian and bipartite
• 13 simple TTS(9)
• 2 have disconnected non-bipartite 2-BIGs
• 2 have Hamiltonian bipartite 2-BIGs
• 9 have Hamiltonian non-bipartite 2-BIGs

Slide 26 of 32

Some More Empirical Results:

• 394 simple TTS(10)
• 11 have disconnected 2-BIGs
• 383 have Hamiltonian 2-BIGs
• 88616310 simple TTS(12)
• 1213658 have disconnected 2-BIGs
• 87400079 have Hamiltonian 2-BIGs
• 2573 have 2-BIGs that are not Hamiltonian but have

Hamilton paths

• 16 have chromatic index 3 and so are Class 1
• 2557 are Class 2 and thus are “weak” snarks

(all 2557 have girth 3)

Slide 27 of 32

Empirical Results for TTS(13):

• estimated number of simple TTS(13) is about 195 billion
• 943 million actually constructed (in 2014)
• 37 years is the estimated time to construct all of them
• n a cluster of 70 cores
• simple TTS(13) with girth 5 or more:
• about 6 billion estimated total
• 648 million for which 2-BIGs have been analysed so far:
• 57 disconnected
• The vast majority are Hamiltonian and non-bipartite
• 115 with universal cycles (1 of which is bipartite)
• 7097 others that are Hamiltonian and bipartite
• 15933 that are Class 2, all with non-trivial 3-edge-cuts

(so no “strong” snarks, yet)

Slide 28 of 32

An Isomorphism Question: Colbourn and Johnstone established that there exist non-isomorphic TTS with isomorphic disconnected 2-BIG. But no example of two non-isomorphic TTS with isomorphic connected 2-BIG was known. Some New Facts: Of the 88616310 simple TTS(12), 185372 share their 2-BIG (92550 pairs, 32 trios, 44 quartets) 1504 of these 185372 TTS have disconnected 2-BIGs

Slide 29 of 32

Example: A quartet of non-isomorphic TTS(12):

#11234

1 2 4 1 2 3 1 3 4 1 10 12 6 10 12 1 11 12 5 6 10 1 9 11 1 5 6 1 7 9 1 5 7 2 4 6 2 3 5 2 6 7 2 5 9 2 8 11 2 8 12 2 10 11 2 7 10 2 9 12 3 4 8 4 6 9 4 5 12 4 7 12 4 8 11 4 5 11 4 9 10 4 7 10 3 6 11 3 6 7 3 5 12 3 7 8 3 10 11 3 9 10 3 9 12 1 8 10 6 8 12 6 9 11 1 6 8 5 8 10 5 8 9 7 11 12 7 8 9 5 7 11

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Example: A quartet of non-isomorphic TTS(12):

#21234

1 2 3 1 2 4 1 3 4 1 5 6 1 5 7 1 6 8 1 7 9 1 8 10 1 9 11 1 10 12 1 11 12 2 3 5 2 4 6 2 5 12 2 6 10 2 7 8 2 7 11 2 8 9 2 9 12 2 10 11 3 4 7 3 5 10 3 6 11 3 6 12 3 7 8 3 8 11 3 9 10 3 9 12 4 5 8 4 5 12 4 6 11 4 7 12 4 8 9 4 9 10 4 10 11 5 6 9 5 7 11 5 8 10 5 9 11 6 7 9 6 7 10 6 8 12 7 10 12 8 11 12

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Example: A quartet of non-isomorphic TTS(12):

#31234

1 2 3 1 3 4 1 2 4 1 5 7 5 7 8 1 7 9 5 8 10 1 9 11 1 8 10 1 11 12 1 10 12 2 3 5 3 4 10 3 5 12 3 10 11 3 6 9 3 6 7 3 8 9 3 8 12 3 7 11 2 4 6 2 5 11 2 7 10 2 7 12 2 6 9 2 9 10 2 8 11 2 8 12 4 5 9 4 5 12 4 7 10 4 6 12 4 8 9 4 8 11 4 7 11 1 5 6 6 7 8 5 9 11 1 6 8 5 6 10 6 10 11 7 9 12 6 11 12 9 10 12

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Example: A quartet of non-isomorphic TTS(12):

#41234

1 2 3 1 3 4 1 2 4 1 8 10 1 6 8 1 10 12 1 5 6 1 11 12 1 5 7 1 9 11 1 7 9 2 3 5 3 4 10 3 5 9 3 10 11 3 6 12 3 6 7 3 8 12 3 8 9 3 7 11 2 4 6 2 5 11 2 7 10 2 9 10 2 6 12 2 7 12 2 8 11 2 8 9 4 5 12 4 5 9 4 7 10 4 6 9 4 8 12 4 8 11 4 7 11 5 8 10 6 7 8 5 11 12 5 7 8 5 6 10 6 10 11 9 10 12 6 9 11 7 9 12

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Further Research:

• Generate and analyse the simple TTS(13) with girth at least 5.
• For each unsettled order v, find a TTS(v) with a Hamiltonian

2-BIG.

• For each v ≡ 0 or 1 (mod 3), find a TTS(v) with a

non-Hamiltonian 2-BIG.

• Find a characterisation for disconnected non-bipartite 2-BIGs.
• Find an efficient way of determining whether a given cubic

graph is the 2-BIG of some TTS.

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Thank-you. Acknowledgements:

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