Using SAT Solvers in Combinatorics and Geometry Manfred Scheucher - - PowerPoint PPT Presentation

Using SAT Solvers

in Combinatorics and Geometry

Manfred Scheucher

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Boolean satisfiability problem

• Given Boolean formula, is there an assignment such

that the formula is true?

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Boolean satisfiability problem

• Given Boolean formula, is there an assignment such

that the formula is true?

• NP-complete, but quite good heuristics

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Why to use SAT Solvers?

• Prove/disprove existence of structure

usually faster

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Why to use SAT Solvers?

• Prove/disprove existence of structure
• Induction base
• Induction step
• Find (counter)examples
• Counting occurences

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Today’s Topics

• Orthogonal Symmetric Chain Decompositions

(with Karl D¨ aubel, Sven J¨ ager, and Torsten M¨ utze)

• (Disjoint) Holes in Point Sets
• L-shaped Point Set Embeddings of Trees

(with Torsten M¨ utze)

• Universal Point Sets for Planar Graphs

(with Hendrik Schrezenmaier and Raphael Steiner)

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L-Shaped Point Set Embeddings

joint work with Torsten M¨ utze arXiv:1807.11043

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L-Shaped Point Set Embeddings

• vertices of T drawn as points of P
• edges drawn as unions of two axis parallel line segments

T . . . tree on n vertices P . . . set of m points

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L-Shaped Point Set Embeddings

f(T) . . . minimum number m s.t. tree T admits a planar L-shaped embedding in any set of m points fd(n) := max

T : tree on n vertices

• max. deg. ∆(T ) ≤ d

f(T)

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L-Shaped Point Set Embeddings

f(T) . . . minimum number m s.t. tree T admits a planar L-shaped embedding in any set of m points fd(n) := max

T : tree on n vertices

• max. deg. ∆(T ) ≤ d

f(T)

• f4(n) ≤ O(n1.58)

[S.’15, Aichholzer-Hackl-S.’16]

• f3(n) ≤ O(n1.22), f4(n) ≤ O(n1.55)

[Biedl et al.’17]

• no non-trivial lower bound (≥ n is trivial)
• f4(n) ≤ n2

[Di Giacomo, Frati, Fulek, Grilli, Krug ’13]

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SAT Model

• T . . . tree on vertices {v1, . . . , vn}
• formulate Boolean satisfiability instance:

∃ solution iff. T admits an L-shaped embedding in P

• P . . . point set {P1, . . . , Pn}

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SAT Model: Variables

• Mi,j . . . vertex vi is mapped to point Pj
• Ha,b . . . edge ab is connected horizontally to a

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SAT Model: Clauses

• Injective mapping V to P

every vertex vi has to be mapped:

• j

Mi,j no two vertices vi1, vi2 are mapped to the same point: ¬Mi1,j ∨ ¬Mi2,j

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b) a b Ha,b ∨ Hb,a, ¬Ha,b ∨ ¬Hb,a

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b)

• No overlapping edges

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b)

• No overlapping edges
• rder of points is important

p q r

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b)

• No overlapping edges

If p left of q and r, and va → p, vb → q, vc → r, then not both edges ab and ac horizonally connected

• nly depends on point set
• nly depends on variables

p q r

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b)

• No overlapping edges

If p left of q and r, and va → p, vb → q, vc → r, then not both edges ab and ac horizonally connected For each three points p, q, r with p left of q and r: ¬Ma,p ∨ ¬Mb,q ∨ ¬Mc,r ∨ ¬Ha,b ∨ ¬Ha,c

• nly depends on point set
• nly depends on variables

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SAT Model: Clauses

• Injective mapping V to P
• L-shaped edges:

ab connects either vertically or horizontally to a (and b)

• No overlapping edges
• No crossing edges

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Results

T13 P13 Theorem: Every tree on n ≤ 12 vertices admits an L-shaped embedding in every set of n points. Theorem: T13 has no L-shaped embedding in P13.

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Results

• Further examples for n ∈ {13, 14, 16, 17, 18, 19, 20}
• If cyclic order fixed, infinite family

Theorem: Every tree on n ≤ 12 vertices admits an L-shaped embedding in every set of n points. Theorem: T13 has no L-shaped embedding in P13. . . .

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And now for something completely different

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Orthogonal Chain Decompositions

joint work with Karl D¨ aubel, Sven J¨ ager, and Torsten M¨ utze arXiv:1810.09847

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Orthogonal Chain Decompositions

• Boolean lattice Qn

can be partitioned into

• n

⌊n/2⌋

• chains [Dilworth ’50]

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Orthogonal Chain Decompositions

∅ {4} {3} {3,4} {2} {2,4} {2,3} {2,3,4} {1} {1,4} {1,3} {1,3,4} {1,2} {1,2,4} {1,2,3} {1,2,3,4}

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Orthogonal Chain Decompositions

• Boolean lattice Qn

can be partitioned into

• n

⌊n/2⌋

• chains [Dilworth ’50]
• orthogonal chain-decompositions:

any two chains from two decompositions have at most one element in common

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Orthogonal Chain Decompositions

• ∃ two orthogonal CDs in Qn≥2 [Shearer-Kleitman ’79]
• moreover, they conjectured ⌊n/2⌋ + 1

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Orthogonal Chain Decompositions

• ∃ two orthogonal CDs in Qn≥2 [Shearer-Kleitman ’79]
• ∃ three orthogonal CDs in Qn≥24 [Spink ’17]
• moreover, they conjectured ⌊n/2⌋ + 1

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Orthogonal Chain Decompositions

• ∃ two orthogonal CDs in Qn≥2 [Shearer-Kleitman ’79]
• ∃ three orthogonal CDs in Qn≥24 [Spink ’17]
• moreover, they conjectured ⌊n/2⌋ + 1

Proof Idea:

• 1. (specific) orthogonal CDs for Q5 and Q7
• 2. Lemma ”Qa, Qb ⇒ Qa+b”

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Orthogonal Chain Decompositions

• ∃ two orthogonal CDs in Qn≥2 [Shearer-Kleitman ’79]
• ∃ three orthogonal CDs in Qn≥24 [Spink ’17]
• moreover, they conjectured ⌊n/2⌋ + 1

n 1 2 3 4 5 6 7 8 9 10 11 almost-orth. 1 2 2 2 3 3* 4* 3* 3* 3 4* ⌊n/2⌋ + 1 1 2 2 3 3 4 4 5 5 6 6 found via SAT solvers!

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Orthogonal Chain Decompositions

• ∃ two orthogonal CDs in Qn≥2 [Shearer-Kleitman ’79]
• ∃ three orthogonal CDs in Qn≥24 [Spink ’17]
• moreover, they conjectured ⌊n/2⌋ + 1

n 1 2 3 4 5 6 7 8 9 10 11 almost-orth. 1 2 2 2 3 3* 4* 3* 3* 3 4* ⌊n/2⌋ + 1 1 2 2 3 3 4 4 5 5 6 6

Theorem: ∃ four orthogonal CDs in Qn≥60. [D¨ aubel-J¨ ager-M¨ utze-S. ’18]

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SAT Instance

• Problem: naive CNF formulation much too big

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SAT Instance

• use symmetries
• Problem: naive CNF formulation much too big

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SAT Instance

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

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SAT Instance

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

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SAT Instance

• iteratively add clauses to “correct errors”
• use symmetries

(incremental solving supported by minisat/glucose)

• Problem: naive CNF formulation much too big

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And now for something more completely different

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)?

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)? 7-gon

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)?

• Theorem. 2k−2 + 1 ≤ g(k) ≤

2k−4

k−2

• . [Erd˝
• s–Szekeres ’35]

equality conjectured by Szekeres, Erd˝

• s offered 500$ for a proof

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)?

• Theorem. 2k−2 + 1 ≤ g(k) ≤

2k−4

k−2

• . [Erd˝
• s–Szekeres ’35]
• Theorem. g(k) ≤ 2k+o(k). [Suk ’17]

. . .

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)?

• Theorem. 2k−2 + 1 ≤ g(k) ≤

2k−4

k−2

• . [Erd˝
• s–Szekeres ’35]
• Theorem. g(k) ≤ 2k+o(k). [Suk ’17]

Known: g(4) = 5, g(5) = 9, g(6) = 17 computer assisted proof, 1500 CPU hours [Szekeres–Peters ’06] . . .

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Classical Erd˝

• s–Szekeres
• Given n points in the plane in general position,

is there subset size k in convex position (”k-gon”)?

• Theorem. 2k−2 + 1 ≤ g(k) ≤

2k−4

k−2

• . [Erd˝
• s–Szekeres ’35]
• Theorem. g(k) ≤ 2k+o(k). [Suk ’17]

Known: g(4) = 5, g(5) = 9, g(6) = 17 computer assisted proof, 1500 CPU hours [Szekeres–Peters ’06] 1 hour using SAT solvers! . . .

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SAT Formulation

• Variables for triple-orientations: χabc ∈ {+, −}

+ –

a b c d

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SAT Formulation

• Variables for triple-orientations: χabc ∈ {+, −}
• Axiomatize ”point set”: chirotope/signotope axioms
• Alternating axioms:

χiπ(1),iπ(2),iπ(3) = sgn(π) · χi1,i2,i3

• Exchange axioms:

For any x1, . . . , xr, y1, . . . , yr: If χyi,x2,...,xr · χy1,...,yi−1,x1,yi+1,...,yr ≥ 0 for every i, then χx1,...,xr · χy1,...,yr ≥ 0

Θ(n6) many

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SAT Formulation

• Variables for triple-orientations: χabc ∈ {+, −}
• Axiomatize ”point set”: chirotope/signotope axioms
• Crossings via triple-orientations

χabc = χabd and χcda = χcdb a b c d

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SAT Formulation

• Variables for triple-orientations: χabc ∈ {+, −}
• Axiomatize ”point set”: chirotope/signotope axioms
• Crossings via triple-orientations
• k-gons via crossings

(every 4-tuple in k-gon is in convex positon)

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Variant: k-Holes

• k-hole . . . k-gon with no interior points

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Variant: k-Holes

• k-hole . . . k-gon with no interior points
• h(4) = 5, h(5) = 10, 30 ≤ h(6) ≤ 463, h(7) = ∞

Harborth ’78 Overmars ’02 Gerken ’08, Nicolas ’07, Koshelev ’09 Horton’83

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Variant: k-Holes

• k-hole . . . k-gon with no interior points
• h(4) = 5, h(5) = 10, 30 ≤ h(6) ≤ 463, h(7) = ∞
• h3(n) and h4(n) quadratic

number of 3- and 4-holes, resp.

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Variant: k-Holes

• k-hole . . . k-gon with no interior points
• h(4) = 5, h(5) = 10, 30 ≤ h(6) ≤ 463, h(7) = ∞
• h3(n) and h4(n) quadratic
• h5(n) = Ω(n log4/5 n) and h6(n) = Ω(n)

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Variant: k-Holes

• k-hole . . . k-gon with no interior points
• h(4) = 5, h(5) = 10, 30 ≤ h(6) ≤ 463, h(7) = ∞
• h3(n) and h4(n) quadratic
• h5(n) = Ω(n log4/5 n) and h6(n) = Ω(n)
• hk(n) determined for small values of n

n 9 10 11 12 13 14 15 16 17 18 19 h5(n) 1 2 3 3 6 9 11 ≤ 16 ≤ 21 ≤ 26

via SAT S.’13 Harborth ’78 & Dehnhardt ’87

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Couting via SAT

• k-holes via crossings and containments
• containment p ∈ △(a, b, c) via ”no crossings”
• count with variables Xa,b,c,...;ℓ :

a, b, c, . . . form the ℓ-th k-hole (in lexicographic order)

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And now for something more completely different

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Universal Point Sets for Planar Graphs

joint work with Hendrik Schrezenmaier and Raphael Steiner arXiv:1811.06482

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Universal Point Sets for Planar Graphs

Definition: n-universal point set S: ∀ planar n-vertex graph G can be drawn straight-line on S.

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Universal Point Sets for Planar Graphs

Definition: n-universal point set S: ∀ planar n-vertex graph G can be drawn straight-line on S.

• n × n grid is n-universal [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≤ n2

4 − O(n) [Bannister–Cheng–Devanny–Eppstein ’14]

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Universal Point Sets for Planar Graphs

Definition: n-universal point set S: ∀ planar n-vertex graph G can be drawn straight-line on S.

• n × n grid is n-universal [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≤ n2

4 − O(n) [Bannister–Cheng–Devanny–Eppstein ’14]

• |S| ≥ n + Ω(√n) [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≥ 1.235n(1 + o(1)) [Kurowski ’04]

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Universal Point Sets for Planar Graphs

Definition: n-universal point set S: ∀ planar n-vertex graph G can be drawn straight-line on S.

• ∃ n-universal sets for n ≤ 10 [Cardinal–Hoffmann–Kusters ’15]
• ∄ n-universal sets for n ≥ 15 [Cardinal–Hoffmann–Kusters ’15]
• n × n grid is n-universal [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≤ n2

4 − O(n) [Bannister–Cheng–Devanny–Eppstein ’14]

• |S| ≥ n + Ω(√n) [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≥ 1.235n(1 + o(1)) [Kurowski ’04]

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Universal Point Sets for Planar Graphs

Definition: n-universal point set S: ∀ planar n-vertex graph G can be drawn straight-line on S.

• ∃ n-universal sets for n ≤ 10 [Cardinal–Hoffmann–Kusters ’15]
• ∄ n-universal sets for n ≥ 15 [Cardinal–Hoffmann–Kusters ’15]
• n × n grid is n-universal [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≤ n2

4 − O(n) [Bannister–Cheng–Devanny–Eppstein ’14]

• |S| ≥ n + Ω(√n) [De Fraysseix–Pach–Pollack ’90]
• . . .
• |S| ≥ 1.235n(1 + o(1)) [Kurowski ’04]

∄ 11-universal set on 11 points |S| ≥ 1.293n(1 + o(1))

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Universal Point Sets for Planar Graphs

SAT model for a fixed set S and fixed graph G ∈ G:

• Injective mapping V (G) → S
• No two edges cross (planarity)

(similar idea as in L-shaped model)

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Universal Point Sets for Planar Graphs

SAT model for a fixed set S and fixed graph G ∈ G:

• Injective mapping V (G) → S
• No two edges cross (planarity)

(similar idea as in L-shaped model)

• all graphs can be handled simultaneously
• all point sets simultaneously (chirotope/signotope axioms)

. . . but solvers do not terminate . . . All in one SAT instance:

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Universal Point Sets for Planar Graphs

• Enumerate all triangulations on 11 vertices (plantri)

(1,249)

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Universal Point Sets for Planar Graphs

• Enumerate all triangulations on 11 vertices (plantri)
• Enumerate all chirotopes on 11 points

via chirotope axioms, 20 CPU hours, 100 GB storage (2,343,203,071) (1,249)

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Universal Point Sets for Planar Graphs

• Enumerate all triangulations on 11 vertices (plantri)
• Enumerate all chirotopes on 11 points
• Test each pair S and G (SAT)

priority queue (”hard” triangulations first)

• Start with G as 11-vertex triangulations with maximum degree 10

(2,343,203,071) (1,249)

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Universal Point Sets for Planar Graphs

• Enumerate all triangulations on 11 vertices (plantri)
• Enumerate all chirotopes on 11 points
• Test each pair S and G (SAT)
• Start with G as 11-vertex triangulations with maximum degree 10
• For remaining G-universal sets, use IP to find minimal set of

triangulations which need to be added (2,343,203,071) (1,249)

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Universal Point Sets for Planar Graphs

• Enumerate all triangulations on 11 vertices (plantri)
• Enumerate all chirotopes on 11 points
• Test each pair S and G (SAT)
• 500 CPU days later:

conflict collection of 23 stacked triangulations !

• Start with G as 11-vertex triangulations with maximum degree 10
• For remaining G-universal sets, use IP to find minimal set of

triangulations which need to be added previously: 7393 (2,343,203,071) (1,249)

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Thank you for your attention!

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