Generation of oriented matroids using satisfiability solvers Lars - - PowerPoint PPT Presentation

Generation of oriented matroids using satisfiability solvers

Lars Schewe

TU Darmstadt

ICMS 2006

Application

Triangulations of surfaces

[ . . . ] It is well known [ . . . ] that every closed,

• rientable, topological 2-manifold M without

boundary is imbeddable in R3. In analogy to Steinitz’s theorem, one may ask whether every abstract 2-complex C homeomorphic to such a 2-manifold M is realizable by a 2-complex in R3, or in any Euclidean space. [ . . . ] If C is simplicial, it is trivially realizable by a 2-complex in R5; however, no example is known to contradict the conjecture that each simplicial C is realizable by a 2-complex in R3. (Grünbaum, 1967; Exercise 13.2.3)

A counterexample

Genus 6, 12 vertices Bokowski and Guedes de Oliveira, 2000 (Model: J. Bokowski, Photograph: N. Hähn)

Results

• 1. No triangulation of a surface of genus 6 using only

12 vertices admits a polyhedral embedding in R3.

• 2. There exist at least three triangulations of a surface
• f genus 5 using only 12 vertices that do not admit

a polyhedral embedding.

• 3. For every g ≥ 5 we can construct an infinite family
• f triangulations of a surface of genus g such that

none of these admit a polyhedral embedding in R3.

How to prove non-embeddability

Oriented matroid approach

7 3 4 7 2 1 7 4 3 7 1 2 5 6

How to prove non-embeddability

Oriented matroid approach

7 3 4 7 2 1 7 4 3 7 1 2 5 6

−→ Oriented Matroids −→

How to prove non-embeddability

Other approaches

◮ T

• pological obstructions (refined for the PL-case,

e.g. van-Kampen-Flores theorem)

◮ Geometric obstructions (further refinement of the

criteria above, see Novik)

◮ Geometric arguments (e.g. linking number applied

by Brehm) A hybrid strategy is described by Timmreck (to appear).

However . . .

None of the methods above was sucessfully applied to decide a genus 6, 12 vertex example.

Triangulations

7 3 4 7 2 1 7 4 3 7 1 2 5 6

Triangulations

7 3 4 7 2 1 7 4 3 7 1 2 5 6

◮ every edge is contained in

two triangles

Triangulations

7 3 4 7 2 1 7 4 3 7 1 2 5 6

◮ every edge is contained in

two triangles

◮ every link of a vertex is a

cycle

Bounds for combinatorial triangulations

Theorem (Jungerman and Ringel, 1980)

Let S be a surface of genus g (g = 2). Then there exists a triangulation of S with n vertices, if and only if: n − 3 2

• ≥ 6g

That means that with n vertices one can triangulate a surface of genus O(n2).

Number of combinatorial triangulations

g nmin # 4 1 1 7 1 2 10 865

(Lutz, 2003)

3 10 20

(Lutz, 2003)

4 11 821

(Lutz, 2005)

5 12 751 593

(Sulanke, 2005)

6 12 59

(Altshuler et al., 1996)

Polyhedral Embeddings

General construction

Theorem (McMullen, Schulz, Wills, 1983)

There exist triangulations using n vertices of surfaces of genus O (n logn), which admit a polyhedral embedding.

Polyhedral Embeddings

Current knowledge for g ≤ 5

g = 0 all triangulations are realizable (Steinitz) 1 ≤ g ≤ 4 all minimal vertex triangulations are realizable (genus 1, Cszásár, 1949; genus 2, Lutz and Bokowski, 2005; genus 3, 4: Hougardy, Lutz, Zelke, 2005) g = 5 Some minimal vertex triangulations are realizable (Hougardy, Lutz, Zelke, 2005)

Polyhedral Embeddings

Methods

Methods for small cases

◮ direct construction (Bokowski, Brehm) ◮ random coordinates (Lutz, 2005) ◮ small coordinates (Hougardy, Lutz, Zelke, 2005) ◮ local search (Hougardy, Lutz, Zelke, 2005)

Oriented matroids

Definition

Let p1, · · · , pn ∈ Rd and pi = (1, pi) ∈ Rd+1. Then we call χ(i1, · · · , id+1) := sgndet( pi1, · · · , pid+1) the chirotope of the point set. A chirotope is called uniform, if χ(i1, · · · , id+1) = 0, where the ij are pairwise disjoint.

Oriented matroids

1 2 3 4 5

Example: d = 2

◮ pi, pj, pk collinear

⇔ sgndet( pi, pj, pk) = 0

◮ pi, pj, pk ccw

⇔ sgndet( pi, pj, pk) = +

Oriented matroids

1 2 3 4 5

not part of the convex hull

Oriented matroids

1 2 3 4 5

part of the convex hull

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations :

r+1

• j=1

(−1)j det(x1, · · · , xj−1, xj+1, · · · , xr+1) det(xj, y1, · · · , yr−1) = 0

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

det(x2, x3) det(x1, y1) − det(x1, x3) det(x2, y1) + det(x1, x2) det(x3, y1) = 0

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

χ(b, c) χ(a, d) −χ(a, c) χ(b, d) +χ(a, b) χ(c, d)

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

χ(b, c) χ(a, d) = 0 −χ(a, c) χ(b, d) = 0 +χ(a, b) χ(c, d) = 0

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

χ(b, c) χ(a, d) = + −χ(a, c) χ(b, d) = − +χ(a, b) χ(c, d)

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

χ(b, c) χ(a, d) −χ(a, c) χ(b, d) = + +χ(a, b) χ(c, d) = −

Oriented matroids

For a combinatorial model we use a combinatorial version of the Grassmann-Plücker relations (r = 2):

χ(b, c) χ(a, d) −χ(a, c) χ(b, d) +χ(a, b) χ(c, d)

The higher-dimensional case can be reduced to the linear situation via contraction

Definition

Let E be a finite set, r ∈ N and χ : Er → {+, 0, −}. We call M = (E, χ) an oriented matroid, if: (B1) The mapping χ is alternating.

Definition

Let E be a finite set, r ∈ N and χ : Er → {+, 0, −}. We call M = (E, χ) an oriented matroid, if: (B1) The mapping χ is alternating. (B2) The set B = {{x1, · · · , xr} | χ(x1, · · · , xr) = 0} is the set of bases of a matroid.

Definition

Let E be a finite set, r ∈ N and χ : Er → {+, 0, −}. We call M = (E, χ) an oriented matroid, if: (B1) The mapping χ is alternating. (B2) The set B = {{x1, · · · , xr} | χ(x1, · · · , xr) = 0} is the set of bases of a matroid. (B3) For all σ ∈ E

r−2

and all subsets {x1, . . . , x4} ⊆ E \ σ

• ne of the two following conditions holds:

◮ s1 = s2 = s3 = 0 ◮ {s1, s2, s3} ⊇ {−, +}

Here we set: s1 = χ(σ, x1, x2)χ(σ, x3, x4) s2 = −χ(σ, x1, x3)χ(σ, x2, x4) s3 = χ(σ, x1, x4)χ(σ, x2, x3)

Generating oriented matroids

◮ Hyperline sequences (Bokowski, Guedes de

Oliveira, 2000)

◮ Cocircuit graphs (Finschi and Fukuda, 2002) ◮ Chirotope (Bremner, 2004)

Satisfiability: CNF

◮ Variables

x1 x2 x3

Satisfiability: CNF

◮ Literals

x1 x2 x3 ¬ x1 ¬ x2 ¬ x3

Satisfiability: CNF

◮ Clauses

x1 ∨ x2 ∨ x3 ¬ x1 ∨ ¬ x2 ∨ ¬ x3 ¬ x1 ∨ x2 x2 ∨ ¬ x3 x1 ∨ ¬ x2 ∨ ¬ x3 x1 ∨ ¬ x2 ∨ x3

Satisfiability: CNF

◮ CNF

x1 ∨ x2 ∨ x3

• ∧

¬ x1 ∨ ¬ x2 ∨ ¬ x3

• ∧

¬ x1 ∨ x2

• ∧

x2 ∨ ¬ x3

• ∧

x1 ∨ ¬ x2 ∨ ¬ x3

• ∧

x1 ∨ ¬ x2 ∨ x3

First steps towards the model

Theorem (Peirce)

Given: f : {0, 1}n → {0, 1}. Then: f(x) =

• v∈{z|f(z)=0}

  

• i∈{j|vj=0}

xi ∨

• i∈{j|vj=1}

¬xi   

First steps towards the model

χ(a)χ(b) = + ⇒ χ(c)χ(d) = −

First steps towards the model

χ(a)χ(b) = + ⇒ χ(c)χ(d) = − ¬xa ¬xb ¬xc ¬xd

First steps towards the model

χ(a)χ(b) = + ⇒ χ(c)χ(d) = − ¬xa ¬xb ¬xc ¬xd ¬xa ¬xb xc xd

Size of the model

In general (and before preprocessing)

◮ 1 variable per basis ◮ 16 clauses per Grassmann-Plücker relation ◮ 2 clauses per intersection condition

Genus 6 (after preprocessing)

◮ 494 variables ◮ between 225021 and 225148 clauses

SAT solvers

Solvers

◮ ZChaff (Fu, Mahajan, Malik, 2004) ◮ MiniSat (Eén and Sörensson, 2003–)

Preprocessor

◮ SatELite (Eèn and Biere, 2005)

Running times

Oriented matroids do not exist

Brehm’s triangulation of the Möbius’ strip ≈ 2s Genus 6, 12 vertices between 1h and 2h Genus 5, 12 vertices between 1h and 2h

Oriented matroids exist

Genus 1, 7 vertices ≈ 18s (2772 oriented matroids) Genus 5, 12 vertices ≈ 86s (until the first solution); ≈ 40min (for all)

Results

• 1. No triangulation of a surface of genus 6 using only

12 vertices admits a polyhedral embedding in R3.

• 2. There exist at least three triangulations of a surface
• f genus 5 using only 12 vertices that do not admit

a polyhedral embedding.

• 3. For every g ≥ 5 we can construct an infinite family
• f triangulations of a surface of genus g such that

none of these admit a polyhedral embedding in R3.

Open Questions

◮ Is there a geometric proof for non-embeddability? ◮ How can these results be transformed in a succinct

formal proof?

Thank you!