[PPT] - Extended boxed product and application to synchronized trees Olivier PowerPoint Presentation

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Extended boxed product and application to synchronized trees

Olivier Bodini1, Antoine Genitrini2 et Nicolas Rolin1

1Universit´

e Paris 13 – LIPN

2UPMC Paris – LIP6

GASCom 2016

N. ROLIN (Universit´

e Paris 13 – LIPN) Extended boxed product GASCom 2016 1 / 32

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1 Introduction 2 Adaptating boxed product 3 Specification of synchronized trees 4 Uniform sampling

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 2 / 32

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Introduction

Increasing tree

Plane tree labeled with n nodes: Each label between 1 and n is taken Each path from the root to a leaf is a strictly increasing path ⇒ two nodes are sharing a label.

Shape of the tree

1 4 2 5 6 3 Increasing tree

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Introduction

Synchronized tree

Plane tree labeled with n + 1 nodes Each label between 1 and n is taken Each path from the root to a leaf is a strictly increasing path ⇒ two nodes are sharing a label.

Shape of the tree

1 4 5 2 5 6 3 Synchronized tree

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Introduction

Definitions

A combinatorial class C is a set of objects, with a size function, denoted by | · | : C → N and such that for every integer n, the subset Cn of objects of size n, is finite with cardinality Cn. We define the exponential generating function of a combinatorial class C to be: C(z) =

n≥0

Cn zn n!

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Introduction

Definitions

A combinatorial class C is a set of objects, with a size function, denoted by | · | : C → N and such that for every integer n, the subset Cn of objects of size n, is finite with cardinality Cn. We define the exponential generating function of a combinatorial class C to be: C(z) =

n≥0

Cn zn n! Small dictionary C = A + B → C(z) = A(z) + B(z) C = A ⋆ B → C(z) = A(z) · B(z)

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 5 / 32

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Introduction

Boxed product

The boxed product on the labeled classes: C = (A ⋆ B), Means that C is the product of A and B, and that the element with the smallest label comes from the A component. Its translation to equation is: C = A ⋆ B → C(z) =

v=z

v=0

dA dv (v)B(v)dv.

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Introduction

Specification of increasing trees

The increasing trees class T verifies the following specification: T = Z ⋆ Seq(T ) Z . . . T T T T

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Introduction

Specification of increasing trees

The increasing trees class T verifies the following specification: T = Z ⋆ Seq(T ) Z . . . T T T T Hence its generating function verifies: T(z) =

v=z

v=0

dv 1 − T(v) or T ′(z) = 1 1 − T(z) and T(0) = 0. So: T(z) = 1 − √ 1 − 2z

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Adaptating boxed product

Increasing trees with a synchronization

We can regroup the two identical nodes:

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 8 / 32

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Adaptating boxed product

Increasing trees with a synchronization

We can regroup the two identical nodes:

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 8 / 32

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Adaptating boxed product

Extended boxed operator

Let A, B and C combinatorial classes with no element of size 0.

C A B E =

can be expressed with boxed operator: E = C ⋆ (A ⋆ B) E(z) =

t=z

t=0

C ′(t)A(t)B(t)dt. blanc blanc blanc blanc

N. ROLIN (Universit´

e Paris 13 – LIPN) Extended boxed product GASCom 2016 9 / 32

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Adaptating boxed product

Extended boxed operator

Let A, B and C combinatorial classes with no element of size 0.

C A B E =

can be expressed with boxed operator: E = C ⋆ (A ⋆ B) E(z) =

t=z

t=0

C ′(t)A(t)B(t)dt. blanc blanc blanc blanc

C A B D =

can be expressed with boxed operator: D = A ⋆ (B ⋆ C) + B ⋆ (A ⋆ C) D(z) =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

y=0

y

x=0

A′(y)B′(x)C(x)dxdy.

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Adaptating boxed product

A more complicated example

We now look at the class Pk,p. The previous method gives:

Pk,p = A ⋆ X

1 ⋆ (Y 1 ⋆ (. . . ) + X 2 ⋆ (. . . ))

+ A ⋆ Y

1 ⋆ (X 1 ⋆ (. . . ) + Y 2 ⋆ (. . . ))

The two branches interlace. ⇒ The number of terms in the sum is k+p

k

R

Xp Yk Xp−1 Yk−1 . . . . . . X1 Y1 A Pk,p =

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Adaptating boxed product

Stanley method

Stanley showed that computing the number of linear extensions of a partial order reduces to compute the volume of convex polytopes.

z a b t Graph portraying the partial order ≻ : z ≻ a, z ≻ b, a ≻ t, b ≻ t

#{linear extension of ≻} = 4!

1

z=0

z

t=0

z

x=t

z

y=t

dzdtdxdy = 2

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Adaptating boxed product

Simple example : factorization

Factorize the generating function:

D(z) =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

y=0

y

x=0

A′(x)B′(y)C(x)dxdy

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 12 / 32

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Adaptating boxed product

Simple example : factorization

Factorize the generating function:

D(z) =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

y=0

y

x=0

A′(x)B′(y)C(x)dxdy we swap the integration order of y and x on the right term =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

x=0

z

y=x

A′(x)B′(y)C(x)dxdy

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e Paris 13 – LIPN) Extended boxed product GASCom 2016 12 / 32

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Adaptating boxed product

Simple example : factorization

Factorize the generating function:

D(z) =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

y=0

y

x=0

A′(x)B′(y)C(x)dxdy we swap the integration order of y and x on the right term =

z

x=0

x

y=0

A′(x)B′(y)C(y)dxdy +

z

x=0

z

y=x

A′(x)B′(y)C(x)dxdy we replace y and x by min(x, y) in C =

z

x=0

z

y=0

A′(x)B′(y)C(min(x, y))dxdy

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Adaptating boxed product

A more complicated example: factorization

R Xp Yk Xp−1 Yk−1 · · · · · · X1 Y1 A

Pk,p(z) =

z

u=0

u

x1=0

x1

x2=0

. . .

xp−1

xp=0

u

y1=0

y1

y2=0

. . .

yk−1

yk=0

min(xp,yk)

t=0

A′(u)X ′

1(x1)X ′ 2(x2) . . . X ′ p(xp)Y ′ 1(y1)Y ′ 2(y2) . . . Y ′ k(yk)R′(t)dudtdx1 . . .

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Adaptating boxed product

Simple example: reordering

We change the integration order: D(z) =

z

x=0

z

y=0

min(x,y)

t=0

A′(x)B′(y)C′(t)dxdydt

N. ROLIN (Universit´

e Paris 13 – LIPN) Extended boxed product GASCom 2016 14 / 32

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Adaptating boxed product

Simple example: reordering

We change the integration order: D(z) =

z

x=0

z

y=0

min(x,y)

t=0

A′(x)B′(y)C′(t)dxdydt D(z) =

z

x=0

min(x,z)

t=0

z

y=t

A′(x)B′(y)C′(t)dxdydt

N. ROLIN (Universit´

e Paris 13 – LIPN) Extended boxed product GASCom 2016 14 / 32

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Adaptating boxed product

Simple example: reordering

We change the integration order: D(z) =

z

x=0

z

y=0

min(x,y)

t=0

A′(x)B′(y)C′(t)dxdydt D(z) =

z

x=0

min(x,z)

t=0

z

y=t

A′(x)B′(y)C′(t)dxdydt D(z) =

z

t=0

z

x=t

z

y=t

A′(x)B′(y)C′(t)dxdydt

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Adaptating boxed product

A more complicated example: reordering

We change the integration order:

Pk,p(z) =

z

u=0

u

x1=0

x1

x2=0

. . .

xp−1

xp=0

u

y1=0

y1

y2=0

. . .

yk−1

yk=0

min(xp,yk)

t=0

A′(u)X ′

1(x1)X ′ 2(x2) . . . X ′ p(xp)Y ′ 1(y1)Y ′ 2(y2) . . . Y ′ k(yk)R′(t)dudtdx1 . . .

P(z) =

z

u=0

u

t=0

u

x1=t

x1

x2=t

. . .

xp−1

xp=t

u

y1=t

y1

y2=t

. . .

yk−1

yk=t

A′(u)X ′

1(x1)X ′ 2(x2) . . . X ′ p(xp)Y ′ 1(y1)Y ′ 2(y2) . . . Y ′ k(yk)R′(t)dudtdx1 . . .

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Specification of synchronized trees

Construction of synchronized trees

Construction: Shape in ”pendulum”

.

. .

.

. . . . .

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Specification of synchronized trees

Construction of synchronized trees

Construction: Shape in ”pendulum” Increasing forests F grafted on it

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Specification of synchronized trees

Construction of synchronized trees

S = Z ⋆ (F ⋆ S ⋆ F) + P With F the class of increasing forests. P

S P

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Specification of synchronized trees

Construction of synchronized trees

S(z) =

z

t=0

√1 − 2t √1 − 2z P′(t)dt

P(z) =

z

u=0

F 3(u)

z

t=0

F 2(t) √1 − 2t √1 − 2z 2 dtdu = 4 z2 + (3 z − 1)√−2 z + 1 − 4 z + 1 3 (4 z2 − 4 z + 1)

S P

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Specification of synchronized trees

Application to synchronized trees

S(z) =

2z 1−2 z − log

1

1−2 z

(4 √1 − 2 z)

S(z) = 1 z2 2! + 11 z3 3! + 122 z4 4! + 1518 z5 5! + 21423 z6 6! + . . . Sn = 2n n!

n

π

1

2 − log(n) 4n

·
1 + O

1

√n

.
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Uniform sampling

1 Introduction 2 Adaptating boxed product 3 Specification of synchronized trees 4 Uniform sampling

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Uniform sampling

Simple example : binary tree

B(z) = z + zB(z)B(z) We put a weight on the nodes of the left tree:

•
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Uniform sampling

Simple example : binary tree

B(z) = z + zB(z)B(z) We put a weight on the nodes of the left tree: B(z, u) = z + zB(zu)B(z) Left tree of size k: Pk,n = [ukzn]B(z, u) [zn]B(z, 1)

•
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Uniform sampling

More complex example : plane trees

T(z) = 1 1 − T(z) We put a weight for each child of the root:

•
• •
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Uniform sampling

More complex example : plane trees

T(z) = 1 1 − T(z) We put a weight for each child of the root: T(z, u) = 1 1 − uT(z) Root with k children: Pk,n = [ukzn]T(z, u) [zn]T(z, 1)

•
• •
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Uniform sampling

More complex example : plane trees

T(z) = zT(z)5

•
• •
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Uniform sampling

More complex example : plane trees

T(z) = zT(z)5 We sample the size of each subtree:

T(z, u1, u2, u3, u4, u5) = zT(zu1)T(zu2)T(zu3)T(zu4)T(zu5). Children from left to right of size respectively p1, p2, p3, p4 and p5 : Pp = [u1p1u2p2u3p3u4p4u5p5zn]T(z, u1, u2, u3, u4, u5) [zn]T(z, 1, 1, . . . , 1)

•
• •
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Uniform sampling

Synchronized trees

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Uniform sampling

Synchronized trees

Add three variables: u for the size of the trunk b for the size of the left branch g for the size of the right branch

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Uniform sampling

Trunk length

S(z, u) = −((u − 3)z + 1)√1 − 2 z (1 − 2 z)

1 2 u − 4 z2 + 4 z − 1

(4z2 − 4z + 1) (u2 − 4 u + 3)(1 − 2 z)

1 2 u

We sample a given length of the trunk m with probability Pm = [umzn]S(z, u) [zn]S(z, 1) .

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Uniform sampling

Trunk length

m

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Uniform sampling

P(z, b, g) =

(b + g) − (b + g + 1)(1 − 2 z)− 1

2 + (1 − 2 z)− b+g+1 2

(b + g)(b + g + 1)

. We calculate S(m)(z, b, g) using the parameter m, and choose a left and right side length of respectively l and r with probability Pl,r = [blgrzn]S(m)(z, b, g) [zn]S(m)(z, 1, 1) .

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Uniform sampling

Shape sampling

2(m + l + r + 1) + 1 instances of F To sample the shape we: use a variable for each instance of F draw a size repartition sample each F independently

m l r

F

F

F

F

F

F . . .

F

F F

F

F

F

F

F

F . . . . . .

F

F

F

F

F

F

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Uniform sampling

Label sampling

1

2

1

2

1

2 1 3 4

1

3 1 2 4 5

1
1

1

1
2

1

1

1

1
1

1 2

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Uniform sampling

Label sampling

1

2

1

2

1

2 1 3 4

1

3 1 2 4 5

1
1

1

1
2

1

1

1

1
1

1 2

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Uniform sampling

Label sampling

mark each node whose label is greater than the synchronized label. 2(l + r + m) + 1 different variables, one for each string. A string of length n has for equation 1−un+1

k

1−uk zn

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Uniform sampling

Label sampling

22
22
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Uniform sampling

Label sampling

1 6

2

18 19 3

4

21 10 16 17

5
8
9

14 11 20

7

12

13

15 22

13
3

1 7 8 12

6

9

4
10

16

5
2
22

14 11 15

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Uniform sampling

Label sampling

1 6

2

18 19 3

4

21 10 16 17

5
8
9

14 11 20

7

12

13

15 22

35
25

23 29 30 34

28

31

26
32

38

29
24
22

36 33 37

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Uniform sampling

Label sampling

1 6 2 18 19 3 4 21 10 16 17 5 8 9 14 11 20 7 12 13 15 22 37 33 36 24 29 38 32 26 31 28 34 30 23 35 25 29

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Uniform sampling

Label sampling

1 6 2 18 19 3 4 21 10 16 17 5 8 9 14 11 20 7 12 13 15 22 37 33 36 24 29 38 32 26 31 28 34 30 23 35 25 29

1

2

1

2

1

2 1 3 4

1

3 1 2 4 5

1
1

1

1
2

1

1

1

1
1

1 2

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Uniform sampling

Label sampling

1 6 2 18 19 3 4 21 10 16 17 5 8 9 14 11 20 7 12 13 15 22 37 33 36 24 29 38 32 26 31 28 34 30 23 35 25 29 1 6 35 2 18 19 3 25 29 23 30 34 4 21 17 10 16 28 31 5 26 8 32 38 9 14 11 29 20 7 12 24 13 15 22 36 33 37

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Conclusion

We showed: a generalized boxed product that deals with any kind of partial order the exact generating function for synchronized trees a method to uniformly sample synchronized trees

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