Volumetry
Generating function and random sampling
Application to combinatorics
Génération aléatoire et fonctions génératrices
pour les langages temporisés et les permutations
Nicolas Basset
University of Oxford
2 Juillet
1/34
Volumetry
Generating function and random sampling
Application to combinatorics
The context of timed automata theory
Introduced by Alur and Dill in 1990 to model and verify real-time
properties of embedded systems (cars, phones, pacemakers...).
Since then many people works...
• on practical issues:
• successful implemented tools;
• used in the industry.
• and also on a common theoretical challenge: lifting results
from automata theory to timed automata theory.
2/34
Volumetry
Generating function and random sampling
Application to combinatorics
The context of timed automata theory
Introduced by Alur and Dill in 1990 to model and verify real-time
properties of embedded systems (cars, phones, pacemakers...).
Since then many people works...
• on practical issues:
• successful implemented tools;
• used in the industry.
• and also on a common theoretical challenge: lifting results
from automata theory to timed automata theory.
A research program since 2009 ⊂ theoretical challenge
(Asarin, Degorre, B., Perrin, Béal)
Leverage quantitative properties from
• automata theory (e.g. Kleene like system of equations for
generating functions);
• symbolic dynamics (e.g. entropy);
• coding theory.
2/34
Volumetry
Generating function and random sampling
Application to combinatorics
This talk
Volumetry of timed languages
Generating function and random sampling for timed languages
Application to combinatorics of permutations
3/34
Volumetry
Generating function and random sampling
Application to combinatorics
Outline
Volumetry of timed languages
Generating function and random sampling for timed languages
Application to combinatorics of permutations
4/34
Volumetry
Generating function and random sampling
Application to combinatorics
Timed Automata [Alur and Dill]
Timed automaton = Finite state automaton + clocks, guards,
reset
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
d; x a , x d ∈ (0; 1)/x a := 0
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
d; x a , x d ∈ (0; 1)/x a := 0
(0, 0)
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
(0.4, 0.4)
d; x a , x d ∈ (0; 1)/x a := 0
0.4 s elapsed
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
(0, 0.4)
d; x a , x d ∈ (0; 1)/x a := 0
x d := 0
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
(0.3, 0.7)
q
0.3 s elapsed
d; x a , x d ∈ (0; 1)/x a := 0
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
d; x a , x d ∈ (0; 1)/x a := 0
x a := 0
(0.3, 0)
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
(0.9, 0.6)
d; x a , x d ∈ (0; 1)/x a := 0
0.6 s elapsed
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
How does it work?
Recognizes timed words e.g. (0.4,a)(0.3,d)(0.6,a)
xa
a; x a , x d ∈ (0; 1)/x d := 0
p
1
q
(0, 0.6)
x d := 0
d; x a , x d ∈ (0; 1)/x a := 0
O
1
xd
5/34
Volumetry
Generating function and random sampling
Application to combinatorics
The timed language recognized
a; x a , x d
∈ (0; 1)/x d := 0
p
q
d; x a , x d ∈ (0; 1)/x a := 0
a
b
a
b
a b
a
<1
<1
Language: (t1 , a)(t2 , d)(t3 , a)(t4 , d) . . . such that ti + ti +1 ∈ (0; 1).
6/34
Volumetry
Generating function and random sampling
Application to combinatorics
Volume of the hypercube
a, x ≤ d/x := 0
Timed word : (t1 , a)(t2 , a) . . . (tn , a)
dimension 1
dimension 2
dimension 3
dimension n
?
t1 ≤ d
t1 , t2 ≤ d
t1 , t2 , t3 ≤ d
t1 , . . . , tn ≤ d
Volume d
Volume d 2
Volume d 3
Volume d n
7/34
Volumetry
Generating function and random sampling
Application to combinatorics
Volume of the hypercube
a, x ≤ d/x := 0
Volume generating function: f (z) =
dimension 1
dimension 2
P
n∈N
d nz n =
dimension 3
1
1−dz .
dimension n
?
t1 ≤ d
t1 , t2 ≤ d
t1 , t2 , t3 ≤ d
t1 , . . . , tn ≤ d
Volume d
Volume d 2
Volume d 3
Volume d n
7/34
Volumetry
Generating function and random sampling
Application to combinatorics
Volume of the simplex
a, x ≤ 1
Timed word : (t1 , a)(t2 , a) . . . (tn , a)
dimension 1
dimension 2
dimension 3
dimension n
?
t1 ≤ 1
t1 + t2 ≤ 1
t1 + t2 + t3 ≤ 1 t1 +· · ·+tn ≤ 1
Volume 1
Volume 1/2
Volume 1/6
Volume 1/n!
8/34
Volumetry
Generating function and random sampling
Application to combinatorics
Volume of the simplex
a, x ≤ 1
Volume generating function: f (z) =
dimension 1
dimension 2
P
1 n
n∈N n! z
dimension 3
= exp(z).
dimension n
?
t1 ≤ 1
t1 + t2 ≤ 1
t1 + t2 + t3 ≤ 1 t1 +· · ·+tn ≤ 1
Volume 1
Volume 1/2
Volume 1/6
Volume 1/n!
8/34
Volumetry
Generating function and random sampling
Application to combinatorics
Measuring timed languages
π1
t1
π3
t3
π2
t2
• Choosing a timed word (~t, π) ∈ Ln = Discrete choice for the
path π + Continuous choice for vector of delays ~t.
• Given π, Lπ = {~t | (~t, π) ∈ Ln } ⊆ Rn is a polytope (e.g.
hypercube, simplex...)
• Measure of Ln : Vn =
P
π∈E n Vol(Lπ )
• Volume generating function : f (z) =
(typically ≈ ρn )
P
n∈N
Vn z n
9/34
Volumetry
Generating function and random sampling
Application to combinatorics
Outline
Volumetry of timed languages
Generating function and random sampling for timed languages
Application to combinatorics of permutations
10/34
Volumetry
Generating function and random sampling
Application to combinatorics
Generating functions of timed languages
Volume generating function:
X
f (z) =
Vn z n
a; x a , x d ∈ (0; 1)/x d := 0
n∈N
For the thick twin:
f (z) = tan(z) + 1/ cos(z)
5 4
z + ...
= 1 + z + 21 z 2 + 13 z 3 + 24
d; x a , x d ∈ (0; 1)/x a := 0
11/34
Volumetry
Generating function and random sampling
Application to combinatorics
Generating functions of timed languages
Volume generating function:
X
f (z) =
Vn z n
a; x a , x d ∈ (0; 1)/x d := 0
n∈N
For the thick twin:
f (z) = tan(z) + 1/ cos(z)
5 4
z + ...
= 1 + z + 21 z 2 + 13 z 3 + 24
d; x a , x d ∈ (0; 1)/x a := 0
Main idea: consider starting state s = (q, ~x ) as a parameter.
• L(s): languages starting from s.
• vk (s) = Vol(Lk (s)) for k ∈ N.
• f (s, z) =
P
k∈N vk (s)z
k.
• Take value in the initial state s0 : f (z) = f (s0 , z).
11/34
Volumetry
Generating function and random sampling
Application to combinatorics
Languages and volume equations, operator Ψ [AD09]
•
•
•
•
States s = (q, ~x ) ∈ S
Timed letter α = (t, δ) ∈ A
α
Successor operation s −
→ sα (with sα ∈ S ∪ {⊥}).
R
RM
P
Integral over timed letter α∈A .dα = δ∈∆ 0 .dt
Recursive equations for languages and volumes
Languages equations
Volumes equations
L0 (s) = ε if s is final
v0 (s)R= 1F (s)
S
Lk+1 (s) = α∈A αLk (sα ) vk+1 (s) = α∈A vk (sα ) dα
with Lk (⊥) = ∅
with vk (⊥) = 0
Volumes and operator ΨZ
Ψ(g )(s) =def
g (sα ) dα
with g (⊥) = 0.
A
vn = Ψ(vn−1 ) = Ψn (1F ).
12/34
Volumetry
Generating function and random sampling
Application to combinatorics
From languages equations to generating functions
Recall
Languages equations
Volumes equations
L0 (s) = ε if s is final
v0 (s)R= 1F (s)
S
Lk+1 (s) = α∈A αLk (sα ) vk+1 (s) = α∈A vk (sα ) dα
with Lk (⊥) = ∅
with vk (⊥) = 0
P
Implies f (s, z) = k∈N vk (s)z k fixed point of:
Z
f (sα , z) dα + 1F (s).
f (s, z) = z
(1)
α∈A
More succinctly: f = zΨf + 1F ; this characterises f
13/34
Volumetry
Generating function and random sampling
Application to combinatorics
From languages equations to generating functions
Recall
Languages equations
Volumes equations
L0 (s) = ε if s is final
v0 (s)R= 1F (s)
S
Lk+1 (s) = α∈A αLk (sα ) vk+1 (s) = α∈A vk (sα ) dα
with Lk (⊥) = ∅
with vk (⊥) = 0
P
Implies f (s, z) = k∈N vk (s)z k fixed point of:
Z
f (sα , z) dα + 1F (s).
f (s, z) = z
(1)
α∈A
More succinctly: f = zΨf + 1F ; this characterises f
Remark
(1) can be obtained directly from:
[
L(s) =
αL(sα ) + (ε if s is final).
α∈A
13/34
Volumetry
Generating function and random sampling
Application to combinatorics
Characterization of the generating function.
Recall f (s, z) =
Theorem
P
k∈N vk (s)z
k
and f (z) = f (s0 , z)
For z within the disc of convergence, the function f is the unique
solution of the integral equation:
f = zΨf + 1F .
Can we compute a closed form formula of f from this equation?
14/34
Volumetry
Generating function and random sampling
Application to combinatorics
Characterization of the generating function.
Recall f (s, z) =
Theorem
P
k∈N vk (s)z
k
and f (z) = f (s0 , z)
For z within the disc of convergence, the function f is the unique
solution of the integral equation:
f = zΨf + 1F .
Can we compute a closed form formula of f from this equation?
• In general too difficult: involve variables ~x ∈ Rd with d > 1.
14/34
Volumetry
Generating function and random sampling
Application to combinatorics
Characterization of the generating function.
Recall f (s, z) =
Theorem
P
k∈N vk (s)z
k
and f (z) = f (s0 , z)
For z within the disc of convergence, the function f is the unique
solution of the integral equation:
f = zΨf + 1F .
Can we compute a closed form formula of f from this equation?
• In general too difficult: involve variables ~x ∈ Rd with d > 1.
• For some subclass of timed automata, yes!
14/34
Volumetry
Generating function and random sampling
Application to combinatorics
Random sampling
Problem statement
Given L and n > 0,
Generate time words t1 a1 · · · tn an ∈ Ln
probability 1/Vn .
with density of
15/34
Volumetry
Generating function and random sampling
Application to combinatorics
Random sampling
Problem statement...again parametrised wrt. starting state
Given L and n > 0, and s ∈ S
Generate time words t1 a1 · · · tn an ∈ Ln (s) with density of
probability 1/vn (s).
15/34
Volumetry
Generating function and random sampling
Application to combinatorics
The recursive method
q
a, x < 1, y < 1, x := 0
b, x < 1, y < 1, x := 0
p
r
Let vk (p, (x, y ), a) weight (volume) of timed words of length k
starting from state (p, (x, y )) and the transition a:
From volume equation...
vk (p, (x, y )) = vk (p, (x, y ), a) + vk (p, (x, y ), b)
Z
vk−1 (q, (0, y + t))dt
vk (p, (x, y ), a) =
t|x+t∧y +t≤1
...to recursive random sampling
Pk (a|p, (x, y )) =
vk (p, (x, y ), a)
,
vk (p, (x, y ))
Pk (t|p, (x, y ), a) =
vk−1 (q,(0,y +t))
vk (p,(x,y ),a)
Pk (b|p, (x, y )) =
vk (p, (x, y ), b)
vk (p, (x, y ))
16/34
Volumetry
Generating function and random sampling
Application to combinatorics
The Boltzmann method (1/2)
• Parameter: z < Rconv(f ) and s ∈ S.
• Boltzmann sampler: Pz (w ) =
z |w|
f (s,z) .
17/34
Volumetry
Generating function and random sampling
Application to combinatorics
The Boltzmann method (1/2)
• Parameter: z < Rconv(f ) and s ∈ S.
• Boltzmann sampler: Pz (w ) =
z |w|
f (s,z) .
Why do we want this?
17/34
Volumetry
Generating function and random sampling
Application to combinatorics
The Boltzmann method (1/2)
• Parameter: z < Rconv(f ) and s ∈ S.
• Boltzmann sampler: Pz (w ) =
z |w|
f (s,z) .
Why do we want this?
• Words of a fixed length have the same probability to occurs.
• Tuning z allows us to choose the expected size of output
timed words.
• Easy to construct knowing generating functions:
17/34
Volumetry
Generating function and random sampling
Application to combinatorics
The Boltzmann method (2/2)
q
a, x < 1, y < 1, x := 0
p
b, x < 1, y < 1, x := 0
r
Let f (p, (x, y ), a, z) generating function starting from (p, (x, y ))
and a:
From generating function equation...
f (p, (x, y ), z) = z ×
R f (p, (x, y ), a, z) + z × f (p, (x, y ), b, z) + 1
f (p, (x, y ), a, z) = t|x+t∧y +t≤1 f (q, (0, y + t), z)dt
...to Boltzmann sampling
Pz (ε) =
1
z × f (p, (x, y ), a, z)
, Pz (a|p, (x, y )) =
, ...
f (p, (x, y ), z)
f (p, (x, y ), z)
Pk (t|p, (x, y ), a) =
f (q,(0,y +t),z)
f (p,(x,y ),a,z)
18/34
Volumetry
Generating function and random sampling
Application to combinatorics
Pros and cons of the two methods
• Recursive methods: Space complexity:
O(|Q|n|horloges|+1 log(n)) for length n (in bits).
• Boltzmann sampling: needs to access the generating
functions. No closed form formulae in general.
19/34
Volumetry
Generating function and random sampling
Application to combinatorics
Pros and cons of the two methods
• Recursive methods: Space complexity:
O(|Q|n|horloges|+1 log(n)) for length n (in bits).
A solution (for quasi-uniform generation): fix m << n and at
each step k = 1..n use vm (sk )/vm+1 (sk−1 ) instead of
vn−k (sk )/vn−k+1 (sk−1 )
• Boltzmann sampling: needs to access the generating
functions. No closed form formulae in general.
19/34
Volumetry
Generating function and random sampling
Application to combinatorics
Pros and cons of the two methods
• Recursive methods: Space complexity:
O(|Q|n|horloges|+1 log(n)) for length n (in bits).
• Boltzmann sampling: needs to access the generating
functions. No closed form formulae in general.
We can compute closed form formulae for 1.5 clocks timed
automata:
f (p, (x, 0), z) = cos(xz)+sin((1−x)z)
cos(z)
a; x a , x d ∈ (0; 1)/x d := 0
p
q
d; x a , x d ∈ (0; 1)/x a := 0
19/34
Volumetry
Generating function and random sampling
Application to combinatorics
Outline
Volumetry of timed languages
Generating function and random sampling for timed languages
Application to combinatorics of permutations
20/34
Volumetry
Generating function and random sampling
Application to combinatorics
Permutations with signature in a regular language.
σ(n)
7
7
ascent
6
σ = 6724512
6
5
Signature:
sg(σ) = adaada
5
descent
ascent
4
4
ascent
3
2
3
descent
d
p
ascent
2
a
q
a
1
0
1
n
0
1
2
3
4
5
6
7
21/34
Volumetry
Generating function and random sampling
Application to combinatorics
First problem statement
Given L ⊆ {a, d}∗ , compute the exponential generating function:
FL (z) =
X
σ, sg(σ)∈L
X
zn
z |σ|
=
αn (L)
|σ|!
n!
n≥1
where αn (L) = |{σ ∈ Sn | sg(σ) ∈ L}|.
EGF for the Up-up-down-down permutations
F(aadd)∗ (aa+ǫ) =
sinh(z)−sin(z)+sin(z) cosh(z)+sinh(z) cos(z)
.
1+cos(z) cosh(z)
22/34
Volumetry
Generating function and random sampling
Application to combinatorics
Second problem statement
Given L ⊆ {a, d}∗ and n ≥ 1, construct a uniform random
sampler for {σ ∈ Sn | sg(σ) ∈ L}. That is
Prob(output = σ) =
1
αn (L)
where αn (L) = |{σ ∈ Sn | sg(σ) ∈ L}|.
23/34
Volumetry
Generating function and random sampling
Application to combinatorics
Second problem statement
Given L ⊆ {a, d}∗ and n ≥ 1, construct a uniform random
sampler for {σ ∈ Sn | sg(σ) ∈ L}. That is
Prob(output = σ) =
1
αn (L)
where αn (L) = |{σ ∈ Sn | sg(σ) ∈ L}|.
A permutation without 2 consecutive descents (n = 100)
[75, 76, 7, 72, 81, 64, 77, 55, 97, 15, 95, 18, 98, 32, 93, 17, 67, 12, 49, 85,
22, 50, 21, 68, 57, 87, 27, 41, 52, 61, 91, 26, 30, 59, 33, 73, 5, 54, 39, 43,
28, 44, 14, 62, 11, 80, 40, 47, 45, 66, 56, 69, 86, 19, 78, 90, 37, 71, 51, 99,
13, 48, 4, 34, 83, 100, 1, 6, 46, 82, 9, 35, 60, 29, 84, 20, 58, 79, 2, 38, 96,
10, 23, 88, 3, 53, 94, 36, 89, 16, 31, 24, 63, 8, 74, 42, 65, 70, 92, 25]
23/34
Volumetry
Generating function and random sampling
Application to combinatorics
A geometric interpretation
Order polytopes for permutation σ or signature w ∈ {a, d}n−1 .
• O(σ)= set of vectors of [0, 1]n with same order as σ.
• O(w )= set of vectors of [0, 1]n with signature w .
A vector in O(132) ⊂ O(ad)
1
d
a
0
Volume Generating Function
FL (z) =
X
σ, sg(σ)∈L
z |σ| X
=
Vol(O(u))z |u|+1 .
|σ|!
u∈L
24/34
Volumetry
Generating function and random sampling
Application to combinatorics
Two transitions and their timed semantic
We consider timed automata with two kinds of transition
a;x a ,x d ∈(0;1)/x d :=0
Ascent: −−−−−−−−−−−−→
d;x a ,x d ∈(0;1)/x a :=0
Descent: −−−−−−−−−−−−→
The timed polytope associated to w ∈ {a, d}∗ is
(~t,w )
Pw =def {~t | (0, 0) −−−→ y for some y ∈ [0, 1]2 }
Example
(0.4,a)
(0.3,d)
(0.6,a)
(0, 0) −−−−→ (0.4, 0) −−−−→ (0, 0.3) −−−−→ (0.6, 0)
Thus (0.4, 0.3, 0.6) ∈ Pada
25/34
Volumetry
Generating function and random sampling
Application to combinatorics
Chain polytopes
The chain polytope of u ∈ {a, d}n−1 is the set C(u) of vectors
~t ∈ [0, 1]n such that for all i < j ≤ n and l ∈ {a, d},
wi · · · wj−1 = l j−i ⇒ ti + . . . + tj < 1.
Links between the three kinds of polytopes
For l ∈ {a, d} and u ∈ {a, d}n−1 :
φ
vol. preserving
Pul = C(u) −−ul−−−−−−−−−→ O(u).
Example
A vector (t1 , t2 , t3 , t4 , t5 ) ∈ [0, 1]5 belongs to C(daadd) = Pdaadda
iff
t1 +
t2
t2
+t3 +
t4
t4
+t5 + t6
<1
<1
<1
d
aa
dd
iff 1 − t1 > t2 < t2 + t3 < 1 − t4 > 1 − t4 − t5 > t6
iff (1 − t1 , t2 , t2 + t3 , 1 − t4 , 1 − t4 − t5 , t6 ) ∈ O(daadd).
26/34
Volumetry
Generating function and random sampling
Application to combinatorics
Reducing the two problems
Reduction for Problem 1 (exponential generating function)
FL (z) =
X
u∈L
Vol(O(u))z |u|+1 =
X
Vol(L′′w )z |w | = VGF (L′′ )(z).
w ∈L′′
(Recall: volume preserving transformation φw : L′′w → O(u).)
Reduction for Problem 2 (uniform sampling)
1. Choose uniformly an n-length timed word (~t, w ) ∈ L′′n ;
2. compute ~ν = φw (~t) ∈ On (L);
3. return σ such that ~ν ∈ O(σ) (using a sort).
27/34
Volumetry
Generating function and random sampling
Application to combinatorics
Reducing the two problems
Reduction for Problem 1 (exponential generating function)
FL (z) =
X
u∈L
Vol(O(u))z |u|+1 =
X
Vol(L′′w )z |w | = VGF (L′′ )(z).
w ∈L′′
(Recall: volume preserving transformation φw : L′′w → O(u).)
Reduction for Problem 2 (uniform sampling)
1. Choose uniformly an n-length timed word (~t, w ) ∈ L′′n ;
2. compute ~ν = φw (~t) ∈ On (L);
3. return σ such that ~ν ∈ O(σ) (using a sort).
Now it suffices to solve the problems for timed automata over
{a, d}.
27/34
Volumetry
Generating function and random sampling
Application to combinatorics
Only one clock is needed
Two new kinds of transition
(t,S)
• Straight: x −−−→ x + t if x + t < 1
to replace
(.,a)
(t,a)
(.,d)
(t,d)
. −−→ (x a , 0) −−−→ (x a + t, 0) if x a + t < 1;
. −−→ (0, x d ) −−−→ (0, x d + t) if x d + t < 1;
(t,T )
• Turn: x −−−→ t if x + t < 1
to replace
(.,a)
(t,d)
(.,d)
(t,a)
. −−→ (x a , 0) −−−→ (0, t) if x a + t < 1;
. −−→ (0, x d ) −−−→ (t, 0) if x d + t < 1;
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Volumetry
Generating function and random sampling
y
7
Application to combinatorics
The straight-turn encoding of signatures
7
a
6Turn
6
Straight
5
d
Straight
p
5
q
a
4 Turn
4
Turn
3
3
S
T
2
Turn
2
1
0
p
q
T
1
x
0
1
2
3
4
5
6
7
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Volumetry
Generating function and random sampling
Application to combinatorics
Language and VGF equations
No two consecutive descents
S
T
p
q
T
parametrized language equations
Lp (x) = ∪t≤1−x (t, S)Lp (x + t)∪ ∪t≤1−x (t, T)Lq (t) ∪ ǫ
Lq (x) =
∪t≤1−x (t, T)Lp (t)
parametrized VGF
fp (x, z) = z
fq (x, z) =
R
t≤1−x fp (x
R
+ t, z)dt+ z t≤1−x fq (t, z)dt + 1
R
z t≤1−x fp (t, z)dt
30/34
Volumetry
Generating function and random sampling
Application to combinatorics
The matrix notation
~f (x, z) = zMS
Z
1
~f (s, z)ds + zMT
x
Z
1−x
~f (t, z)dt + F
~
0
No two consecutive descents
S
T
p
MS =
q
T
1 0
0 1
1
~
, MT =
,F =
.
0 0
1 0
0
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Volumetry
Generating function and random sampling
Application to combinatorics
Solving the equation
~f (x, z) = zMS
Z
1
~f (s, z)ds + zMT
x
Z
1−x
~f (t, z)dt + F
~
0
!
!
~f (x, z)
~f (x, z)
−MS −MT
=z
~f (1 − x, z)
~f (1 − x, z)
MT
MS
!
!
~f (1, z)
~f (0, z)
−MS −MT
~
= exp z
and ~f (1, z) = F
~f (0, z)
~f (1, z)
MT
MS
∂
∂x
An algorithm to compute FL (z)
A1 (z) A2 (z)
−MS −MT
1. Compute
=def exp z
;
A3 (z) A4 (z)
MT
MS
2. return FL (z) the component of
~f (0, z) = [A1 (z)]−1 [I − A2 (z)]F
~ = [I − A3 (z)]−1 A4 (z)F
~
corresponding to the initial state q0 .
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Volumetry
Generating function and random sampling
Application to combinatorics
A classical example: the alternating permutations
e.g. σ = 94738251
T
~ = MT = (1), MS = (0).
Here F
0 −z
cos z
−MS −MT
= exp
=
exp z
z 0
MT
MS
sin z
− sin z
cos z
~ = [I − A3 (z)]−1 A4 (z)F
~)
(Recall FL (z) = [A1 (z)]−1 [I − A2 (z)]F
FL (z) =
cos z
1 + sin z
=
cos z
1 − sin z
33/34
Volumetry
Generating function and random sampling
Application to combinatorics
A classical example: the alternating permutations
e.g. σ = 94738251
T
~ = MT = (1), MS = (0).
Here F
0 −z
cos z
−MS −MT
= exp
=
exp z
z 0
MT
MS
sin z
− sin z
cos z
~ = [I − A3 (z)]−1 A4 (z)F
~)
(Recall FL (z) = [A1 (z)]−1 [I − A2 (z)]F
FL (z) =
cos z
1 + sin z
=
cos z
1 − sin z
For Boltzmann sampling:
f (x, z) = z
with
f (x, z) =
Z
1−x
f (t, z)dt + 1
0
cos(xz) + sin(1 − x)z
cos z
33/34
Volumetry
Generating function and random sampling
Application to combinatorics
Conclusion
Some idea exposed here
• Link between permutations, poset polytopes and timed
languages.
• Kleene like equations for timed languages, their volumes and
generating functions.
Future work
• Analytic combinatorics with continuous parameters
(measurable union, poset polytopes...).
• Statistical model checking (simulation) using random
sampling of timed words.
• Further exploration of combinatorics of permutations.
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