Download counting linear extensions of

HELSINGIN YLIOPISTO
HELSINGFORS UNIVERSITET
UNIVERSITY OF HELSINKI
MATEMAATTIS-LUONNONTIETEELLINEN TIEDEKUNTA
MATEMATISK-NATURVETENSKAPLIGA FAKULTETEN
FACULTY OF SCIENCE
COUNTING LINEAR EXTENSIONS OF
SPARSE POSETS
Kustaa Kangas, Teemu Hankala, Teppo Niinimäki, Mikko Koivisto
Department of Computer Science
Counting the linear extensions of a partially ordered set (poset) is a
fundamental problem with several applications. We present two exact
algorithms that target sparse posets in particular. The first algorithm
PROBLEM
breaks the counting task into subproblems recursively. The second
algorithm uses variable elimination via inclusion–exclusion and runs
in polynomial time for posets with a cover graph of bounded treewidth.
RECURSIVE COUNTING
A poset P is a set of n elements ordered by a
transitive asymmetric relation < P .
For all x, y ∈ P either
Denoting minimal elements by min( P),
`( P) =
I
∑
`( P \ x )
x ∈ min( P)
1. x < P y
If P can be partitioned into pairwise
disconnected sets A and B,
2. y < P x
3. x and y incomparable
The cover relation ≺ P is the
transitive reduction of < P .
Represented visually by
the cover graph.
A linear extension is a bijection σ : P → [n]
such that
x ≺ P y ⇒ σx < σy
| P|
| A|
Implementations [1]:
• R1: Applies rule I, runs in O(2n )
time in the worst case
• R14-a: Applies rules I and II, often
greatly expedites R1
• R24: Applies rule II together with
other techniques [2]
103
103
Small average degree
101
103
Small maximum indegree
0
101
Bipartite
101
100
10−1
100 0
20
40
60
80
Percentage of posets solved
The inverse may lead to
fewer subproblems.
102
R14-a
100
R1
R24
100
10−1
Rules I and II break the poset
into subproblems recursively.
102
Time (s)
Time (s)
102
Problem: Compute `( P), the number of
linear extensions. (#P-complete)
Time (s)
An element x with no y < P x is minimal.
`( P) = `( A) · `( B) ·
II
10−1
100 0
20
40
60
80
Percentage of posets solved
20
40
60
80
Percentage of posets solved
100
VARIABLE ELIMINATION VIA INCLUSION–EXCLUSION
For every mapping σ : P → [n] let
∏
With the elimination order (d, b, e, a, c) the sum factorizes:


!
∑ ∑ [b<d]  ∑ [d<e] ∑ [a<d] ∑ [a<c][c<e] 
[σx < σy ] .
x ≺P y
d
Therefore,
`( P) =
∑
Φ(σ) .
σ : P → [n]
bijection
The inclusion–exclusion principle removes
the bijectivity constraint:
n n
`( P) = ∑
(−1)n−k ∑ Φ(σ)
k
k=0
σ : P → [k]
{z
}
|
→ can use variable elimination (e.g. [3,4])
{z
λb ( d )
a
c
|
}
|
∑
{z
λc ( a,e)
}
{z
λa (d,e)
The summation task is
|
}
{z
}
λe ( d )
[ a<c] [ a<d] [b<d] [c<e] [d<e]
a,b,c,d,e
• Good order → the λ are small and fast to compute
where a = σa , b = σb , . . .
• A cover graph of low treewidth has good orders
103
IMPLEMENTATION
The implementation VEIE
runs in O(nt+4 ) time for
n elements and a cover
graph of treewidth t.
101
101
VEIE
100
R1
R14-a
100
10−1
t =3
102
30
40
[1] Implementation available at www.cs.helsinki.fi/u/jwkangas/lecount/
[2] M. Peczarski. New results in minimum-comparison sorting. Algorithmica, 40(2):133–145, 2004.
[3] R. Dechter. Bucket elimination: A unifying framework for reasoning. Artificial Intelligence, 113(12):41–85, 1999.
[4] U. Bertelè and F. Brioschi. Nonserial Dynamic Programming. Academic Press, 1972.
103
t =2
102
Time (s)
Then, if σ is a bijection, we have
(
1, if σ is a linear extension,
Φ(σ) =
0, otherwise.
e
b
|
Time (s)
Φ(σ) =
EXAMPLE
50
60 70 80
Poset size (n)
90
10−1
100 30
40
50
60 70 80
Poset size (n)
90
100