• Aucun résultat trouvé

Classifying walks in the quarter plane

N/A
N/A
Protected

Academic year: 2022

Partager "Classifying walks in the quarter plane"

Copied!
73
0
0

Texte intégral

(1)

Marni Mishna

Classifying walks in the

quarter plane

(2)

Two walks in the quarter plane

Two step sets: A={



} B={



}

(3)

Two theorems

Theorem 1. [M.-Rechnitzer 07]

The generating function

for walks w in the quarter plane with steps from A = {} is not holonomic.

Theorem 2. [M.-Rechnitzer 07]

The generating function

for the walks in the quarter plane with steps from

(4)

Main Problem

Characterize (combinatorially) lattice walks with a holonomic generating function.

- By step set S

{



}

- By region (quarter plane, general wedge, ...) Develop general techniques and algorithms to automate this decision.

(5)

Why study lattice paths?

(6)

Why study lattice paths?

Classic combinatorial structure

 Arising in probability, statistical physics, queuing theory

(7)

Why study lattice paths?

Classic combinatorial structure

 Arising in probability, statistical physics, queuing theory

We understand much of the story already

(1/2-plane, slit plane, specific examples)

 Many are algebraic

 Several standard techniques

(8)

Why study lattice paths?

Classic combinatorial structure

 Arising in probability, statistical physics, queuing theory

We understand much of the story already

(1/2-plane, slit plane, specific examples)

 Many are algebraic

 Several standard techniques

Proving non-holonomy is still hard

 So far: very few examples proved, many suspected.

(9)

Walks in the 1/4 plane

(10)

Walks in the 1/4 plane

Because two conditions is harder than 1

(11)

Walks in the 1/4 plane

Because two conditions is harder than 1

“Kreweras” walks: S={}

 Kreweras, Gessel : algebraic

 Bousquet-Mélou: constructive approach

 Bernardi: combinatorial bijection

(12)

Walks in the 1/4 plane

Because two conditions is harder than 1

“Kreweras” walks: S={}

 Kreweras, Gessel : algebraic

 Bousquet-Mélou: constructive approach

 Bernardi: combinatorial bijection

Knights walks: S={(2,1), (1,2)}

 Bousquet-Mélou+Petkovšek: non-holonomic

(13)

Walks in the 1/4 plane

Because two conditions is harder than 1

“Kreweras” walks: S={}

 Kreweras, Gessel : algebraic

 Bousquet-Mélou: constructive approach

 Bernardi: combinatorial bijection

Knights walks: S={(2,1), (1,2)}

 Bousquet-Mélou+Petkovšek: non-holonomic

“Gessel” walks: S={}

(14)

Walks in the 1/4 plane

Because two conditions is harder than 1

“Kreweras” walks: S={}

 Kreweras, Gessel : algebraic

 Bousquet-Mélou: constructive approach

 Bernardi: combinatorial bijection

Knights walks: S={(2,1), (1,2)}

 Bousquet-Mélou+Petkovšek: non-holonomic

“Gessel” walks: S={}

 conjectured holonomic

|S|=3: classified (M.)

(15)

Why is holonomy interesting?

(16)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

(17)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

Efficient coefficient generation

(18)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

Efficient coefficient generation

Hope for solving the system

(19)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

Efficient coefficient generation

Hope for solving the system

Asymptotic template for coefficients

(20)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

Efficient coefficient generation

Hope for solving the system

Asymptotic template for coefficients

Indicator of structure

“Everything is non-holonomic unless it is

holonomic by design.” - Flajolet, Gerhold, Salvy

(21)

Why is holonomy interesting?

Holonomic functions satisfy a nice ODE

Efficient coefficient generation

Hope for solving the system

Asymptotic template for coefficients

Indicator of structure

“Everything is non-holonomic unless it is

holonomic by design.” - Flajolet, Gerhold, Salvy

“Guessability” (gfun, rate, …)

(22)

Examples and Non-Examples

✓ Algebraic: Dyck paths, Motzkin paths, ...

✓ Shuffles of Dyck paths

✓ Walks in 1/4 plane with steps {}

✓ k-regular graphs (Gessel) x Regular graphs

x Partitions

x Knights walks {(2, -1), (-1, 2)} (B.-M.+P) x (Flajolet+Gerhold+Salvy)

? Pattern avoiding permutations

? Walks in 1/4 plane with steps {}

(23)

Proving (non)-holonomicity

(24)

Proving (non)-holonomicity

✓ Find the explicit differential equation

(25)

Proving (non)-holonomicity

✓ Find the explicit differential equation

✓ Find a recurrence for coefficients

(26)

Proving (non)-holonomicity

✓ Find the explicit differential equation

✓ Find a recurrence for coefficients

✓ Using closure properties and known holonomic functions

(27)

Proving (non)-holonomicity

✓ Find the explicit differential equation

✓ Find a recurrence for coefficients

✓ Using closure properties and known holonomic functions

x Infinite number of singularities

(28)

Proving (non)-holonomicity

✓ Find the explicit differential equation

✓ Find a recurrence for coefficients

✓ Using closure properties and known holonomic functions

x Infinite number of singularities

x Poles of dense in bounded region then

is not holonomic.

(29)

Proving (non)-holonomicity

✓ Find the explicit differential equation

✓ Find a recurrence for coefficients

✓ Using closure properties and known holonomic functions

x Infinite number of singularities

x Poles of dense in bounded region then

is not holonomic.

x Q(0,0;t) not holonomic ⇒ Q(x,y;t) not holonomic

(30)

Ultimate Goal

(31)

Ultimate Goal

“The generating function of walks with steps from S is holonomic if(f) the set S has

combinatorial property _______.”

(fill in the blank...)

(32)

Ultimate Goal

“The generating function of walks with steps from S is holonomic if(f) the set S has

combinatorial property _______.”

(fill in the blank...)

Example:

...W(t) is holonomic if S has small height

variations and is symmetric across the y-axis.

(Bousquet-Mélou+ Petkovšek)

(33)

Ultimate Goal

“The generating function of walks with steps from S is holonomic if(f) the set S has

combinatorial property _______.”

(fill in the blank...)

Example:

...W(t) is holonomic if S has small height

variations and is symmetric across the y-axis.

(Bousquet-Mélou+ Petkovšek)

A coherent, complete, combinatorial theory of

(34)

Criteria for holonomicity?

(35)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

(36)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition

(37)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

(38)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

3. S = reverse(S) reverse()=

(39)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

3. S = reverse(S) reverse()=

4. S = reflect(reverse(S)) reflect()=

(40)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

3. S = reverse(S) reverse()=

4. S = reflect(reverse(S)) reflect()=

5. S={} (Kreweras) or S=reverse(Kreweras)

How are Kreweras and reverse(Kreweras) “combinatorial”?

(41)

Criteria for holonomicity?

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

3. S = reverse(S) reverse()=

4. S = reflect(reverse(S)) reflect()=

5. S={} (Kreweras) or S=reverse(Kreweras)

(42)

Evidence (part 1)

All known + conjectured “nice” cases

 square lattice

 diamond lattice

 triangular lattice

 step sets of cardinality 3

 ...

(43)

Evidence (part 2)

rat alg

Isomorphism classes for cardinality 3 step sets in the quarter plane

(44)

Theorem 1. (M.+Rechnitzer ) The generating function

for walks w in the quarter plane with steps from A = {} is not holonomic.

Theorem 2. (M.+Rechnitzer) The generating function

for the walks in the quarter plane with steps from B ={} is not holonomic.

Is it difficult to prove these theorems?

Recall our two theorems

(45)

Proving Thm 1: S= {  }

(46)

Proving Thm 1: S= {  }

(47)

Proving Thm 1: S= {  }

A walk is a shorter walk and a step.

(48)

Proving Thm 1: S= {  }

A walk is a shorter walk and a step.

Kernel version:

(49)

Proving Thm 1: S= {  }

A walk is a shorter walk and a step.

Kernel version:

Roots of K(x,y) are special:

(50)

Proving Thm 1: S= {  }

A walk is a shorter walk and a step.

Kernel version:

Roots of K(x,y) are special:

1. Satisfy a nice recurrence

(51)

Proving Thm 1: S= {  }

A walk is a shorter walk and a step.

Kernel version:

Roots of K(x,y) are special:

1. Satisfy a nice recurrence

(52)

The iterated kernel method

Using and the above equation, we show that

We substitute x=1, and rearrange (1) to get

(53)

Infinitely many singularities

(54)

Infinitely many singularities

(55)

Infinitely many singularities

(56)

Infinitely many singularities

(57)

Infinitely many singularities

1. Simple pole at 1/3 (dominant) 2. Singularities from Yn

These singularities are distinct.

3. Divergence of the sum

Aside from poles, no divergence.

The radius of convergence of Q(1,1) is bounded.

(58)

Infinitely many singularities

1. Simple pole at 1/3 (dominant) 2. Singularities from Yn

These singularities are distinct.

3. Divergence of the sum

Aside from poles, no divergence.

The radius of convergence of Q(1,1) is bounded.

(59)

Infinitely many singularities

1. Simple pole at 1/3 (dominant) 2. Singularities from Yn

These singularities are distinct.

3. Divergence of the sum

Aside from poles, no divergence.

The radius of convergence of Q(1,1) is bounded.

The singularities of W(t) are 1/3 and the singularities

(60)

Where are the singularities?

The singularities of Yn(1;t) are the solutions to inside the unit circle.

(61)

Theorem (M.+Rechnitzer): The number of lattice paths of length n with steps from {} confined to the quarter plane is asymptotic to

where

Asymptotic result

(62)

Other non-holonomic class?

Conjecture (M.+ Laferrière) Walks with steps {}

in wedge bounded by centered on x-axis has non-holonomic generating function.

Evidence

Combinatorial similarity to {} in ¼ plane.

Bi-variate GF counting walks ending on x=k:

• •

is rational and as k increases, poles fill unit circle.

(63)

Hints of non-holonomicity

counts walks in the wedge bounded by . shows evidence of

being non-holonomic, as does

(64)

Combinatorial criteria

Conjecture (M.) The generating function W(t) for walks in the quarter plane with steps from S {0, ±1}x{0, ±1}

is holonomic iff one of the following holds:

1. The condition reduces to a half plane condition 2. S has x- or y- axis symmetry

3. S = reverse(S) reverse()=

4. S = reflect(reverse(S)) reflect()=

5. S={} (Kreweras) or S=reverse(Kreweras) Any indications on how to prove this?

(65)

The group of the walk

To each step set S we define G(S), a group of transformations which fix the polynomial

Motivation: Fayolle+Iasnogordski+Malyshev e.g. S={} = {(-1,0), (1,0), (0,1), (0,-1)}

(66)

Finite group = holonomy?

(67)

Finite group = holonomy?

Key Observation: Only (+ all) known or hypothesized holonomic classes have a finite group.

(68)

Finite group = holonomy?

Key Observation: Only (+ all) known or hypothesized holonomic classes have a finite group.

Idea: Perhaps this is important.

(69)

Finite group = holonomy?

Key Observation: Only (+ all) known or hypothesized holonomic classes have a finite group.

Idea: Perhaps this is important.

(70)

Finite group = holonomy?

Key Observation: Only (+ all) known or hypothesized holonomic classes have a finite group.

Idea: Perhaps this is important.

F+I+M give reasons why the group is finite.

(71)

Finite group = holonomy?

Key Observation: Only (+ all) known or hypothesized holonomic classes have a finite group.

Idea: Perhaps this is important.

F+I+M give reasons why the group is finite.

Can they be interpreted in generating function language or combinatorial terms?

(72)

Projects and goals

Prove the holonomy of the Gessel walks

Prove the conjecture and “combinatorialize” F+I+M approach to quarter plane

Classify walks in other wedges (with Laferrière):

 Strategies for “OR” constraints

Consider other infinite Cayley graphs

Prove that miracles don’t exist- rather understand precisely when they do.

(73)

merci beaucoup!

Références

Documents relatifs

Keywords: Random walk in the quarter plane, generating function, singularity analysis, boundary value problem3. AMS 2000 Subject Classification: primary 60G50; secondary

At first sight, this cut increases even more the level of difficulty: we have now two functional equations and more unknowns to deal with, but in the particular case of

This approach concerns several types of problems pertaining to various mathematical areas, and allows one to obtain many quantities: the stationary distribution of ergodic RWQP

We shall give in Part 2 three examples to illustrate our approach (in particular, how to work out and simplify the expressions given in Theorem 1), namely Kreweras’ walk, the

Walks in the quarter plane; counting generating function; holonomy; group of the walk; Riemann surface; elliptic functions; uniformization; universal covering AMS 2000

A detailed exposition of the compensation approach can be found in [1, 3, 4], in which it has been shown to work for two-dimensional random walks on the lattice of the first

In this article, we prove that for the whole class of walks with small steps and zero drift killed at the boundary of the quadrant Z 2 + , there exists exactly one non-zero

Asymptotic expansion of stationary distribution for reflected Brownian motion in the quarter plane via analytic approach... ASYMPTOTIC EXPANSION OF STATIONARY DISTRIBUTION FOR