Counting Linear Extensions of Sparse Posets Kustaa Kangas , Teemu - - PowerPoint PPT Presentation

HELSINGIN YLIOPISTO HELSINGFORS UNIVERSITET UNIVERSITY OF HELSINKI

Counting Linear Extensions of Sparse Posets

Kustaa Kangas, Teemu Hankala, Teppo Niinimäki, Mikko Koivisto July 13, 2016

University of Helsinki Department of Computer Science

Partially ordered set (poset)

A set P with an antisymmetric, transitive relation

Partially ordered set (poset)

Cover graph

Linear extensions

A linear extension = order preserving permutation

Counting linear extensions

#P-complete (Brightwell & Winkler, ’91) Applications: sequence analysis, preference reasoning, sorting, learning probabilistic models, ...

Counting linear extensions

Currently we can do O(2nn) time for n elements. We give two algorithms, based on

• 1. recursion (exploiting low connectivity)
• 2. variable elimination (exploiting low treewidth)

Counting linear extensions

Currently we can do O(2nn) time for n elements. We give two algorithms, based on

• 1. recursion (exploiting low connectivity)
• 2. variable elimination (exploiting low treewidth)

Recursive counting

ℓ(P) = the number of linear extensions of poset P Rule 1 ℓ(P) =

• x ∈ min(P)

ℓ(P \ x) Rule 2 ℓ(P) =

• (D,U)

ℓ(D) · ℓ(U) Rule 3 ℓ(P) =

k

• i=1

ℓ(Si) Rule 4 ℓ(P) = ℓ(A) · ℓ(B) · |P| |A|

Recursive counting

P \ a P \ b P

Recursive counting

P \ a P \ b P

Recursive counting

A B P

Recursive counting

Experiments on sparse posets for n = 30, . . . , 100

20 40 60 80 100 Percentage of posets solved 10−1 100 101 102 103 Time (s)

R14-a R134 R1 R24

Counting linear extensions

Currently we can do O(2nn) time for n elements. We give two algorithms, based on

• 1. recursion (exploiting low connectivity)
• 2. variable elimination (exploiting low treewidth)

Variable elimination

• a,b,c,d,e,f

φ1(a, b, d) φ2(a, c) φ3(b, c, e) φ4(d, f)

Variable elimination

• a,b,c,d,e,f

φ1(a, b, d) φ2(a, c) φ3(b, c, e) φ4(d, f)

Variable elimination

• a,b,c,d,e,f

φ1(a, b, d) φ2(a, c) φ3(b, c, e) φ4(d, f)

Variable elimination

• a,b,c,d,e,f

φ1(a, b, d) φ2(a, c) φ3(b, c, e) φ4(d, f) Polynomial time for bounded treewidth

Variable elimination

For every permutation σ : P → [n] define Φ(σ) =

• x ≺ y

[σx < σy]

Variable elimination

For every permutation σ : P → [n] define Φ(σ) =

• x ≺ y

[σx < σy] Φ(σ) = [σa < σc] [σa < σd] [σb < σd] [σc < σe] [σd < σe]

Variable elimination

For every permutation σ : P → [n] define Φ(σ) =

• x ≺ y

[σx < σy] Then, Φ(σ) = 1, if σ is a linear extension, 0, otherwise.

Variable elimination

For every permutation σ : P → [n] define Φ(σ) =

• x ≺ y

[σx < σy] Then, Φ(σ) = 1, if σ is a linear extension, 0, otherwise. As a consequence ℓ(P) =

• σ : P → [n]

bijection

Φ(σ)

Variable elimination

ℓ(P) =

• σ : P → [n]

bijection

Φ(σ)

Variable elimination

ℓ(P) =

• σ : P → [n]

bijection

Φ(σ) Can’t apply variable elimination because of the bijectivity constraint.

Variable elimination

ℓ(P) =

• σ : P → [n]

bijection

Φ(σ) =

• X ⊆ [n]

(−1)n−|X|

• σ : P → X

Φ(σ) =

n

• k=0

n k

• (−1)n−k
• σ : P → [k]

Φ(σ) Inclusion–exclusion principle

Variable elimination

ℓ(P) =

• σ : P → [n]

bijection

Φ(σ) =

• X ⊆ [n]

(−1)n−|X|

• σ : P → X

Φ(σ) =

n

• k=0

n k

• (−1)n−k
• σ : P → [k]

Φ(σ) O(nt+4) time for treewidth t

Variable elimination

VEIE: Variable elimination via inclusion–exclusion

30 40 50 60 70 80 90 100 Poset size (n) 10−1 100 101 102 103 Time (s) t = 2

VEIE R1 R14-a

Summary

Recursion: often fast in practice Variable elimination: polynomial time for bounded treewidth Thank you!