Walks with large steps in the quadrant Mireille Bousquet-Mlou, - - PowerPoint PPT Presentation

Walks with large steps in the quadrant

Mireille Bousquet-Mélou, CNRS, Université de Bordeaux based on work with Alin Bostan, INRIA Saclay, Paris Steve Melczer, University of Waterloo and École normale supérieure de Lyon

Outline

• I. Motivation
• II. A general approach... that solves some cases
• III. What can go wrong?
• IV. Some cases that work

Counting quadrant walks

Let S be a finite subset of Z2 (set of steps) and p0 ∈ N2 (starting point).

• Example. S = {10, ¯

10, 1¯ 1, ¯ 11}, p0 = (0, 0)

Counting quadrant walks

Let S be a finite subset of Z2 (set of steps) and p0 ∈ N2 (starting point). A path (walk) of length n starting at p0 is a sequence (p0, p1, . . . , pn) such that pi+1 − pi ∈ S for all i.

• Example. S = {10, ¯

10, 1¯ 1, ¯ 11}, p0 = (0, 0)

Counting quadrant walks

Let S be a finite subset of Z2 (set of steps) and p0 ∈ N2 (starting point). A path (walk) of length n starting at p0 is a sequence (p0, p1, . . . , pn) such that pi+1 − pi ∈ S for all i. What is the number q(n) of n-step walks starting at p0 and contained in N2? For (i, j) ∈ N2, what is the number q(i, j; n) of such walks that end at (i, j) ?

• Example. S = {10, ¯

10, 1¯ 1, ¯ 11}, p0 = (0, 0)

• (i, j) = (5, 1)

Counting quadrant walks: higher dimension

Let S be a finite subset of Zd (set of steps) and p0 ∈ Nd (starting point). A path (walk) of length n starting at p0 is a sequence (p0, p1, . . . , pn) such that pi+1 − pi ∈ S for all i. What is the number q(n) of n-step walks starting at p0 and contained in Nd? For i = (i1, . . . , id) ∈ Nd, what is the number q(i; n) of such walks that end at i?

• Example. S = {10, ¯

10, 1¯ 1, ¯ 11}, p0 = (0, 0)

• (i, j) = (5, 1)

Counting quadrant walks: higher dimension

Let S be a finite subset of Zd (set of steps) and p0 ∈ Nd (starting point). A path (walk) of length n starting at p0 is a sequence (p0, p1, . . . , pn) such that pi+1 − pi ∈ S for all i. What is the number q(n) of n-step walks starting at p0 and contained in Nd? For i = (i1, . . . , id) ∈ Nd, what is the number q(i; n) of such walks that end at i? The associated generating function: Q(x1, . . . , xd; t) =

• n≥0
• (i1,...,id)∈Nd

q(i1, . . . , id; n)xi1

1 · · · xid d tn

What is the nature of this series?

A hierarchy of formal power series

• The formal power series A(t) is rational if it can be written

A(t) = P(t)/Q(t) where P(t) and Q(t) are polynomials in t.

A hierarchy of formal power series

• The formal power series A(t) is rational if it can be written

A(t) = P(t)/Q(t) where P(t) and Q(t) are polynomials in t.

• The formal power series A(t) is algebraic (over Q(t)) if it satisfies a

(non-trivial) polynomial equation: P(t, A(t)) = 0.

A hierarchy of formal power series

• The formal power series A(t) is rational if it can be written

A(t) = P(t)/Q(t) where P(t) and Q(t) are polynomials in t.

• The formal power series A(t) is algebraic (over Q(t)) if it satisfies a

(non-trivial) polynomial equation: P(t, A(t)) = 0.

• The formal power series A(t) is D-finite (holonomic) if it satisfies a

(non-trivial) linear differential equation with polynomial coefficients: Pk(t)A(k)(t) + · · · + P0(t)A(t) = 0.

A hierarchy of formal power series

• The formal power series A(t) is rational if it can be written

A(t) = P(t)/Q(t) where P(t) and Q(t) are polynomials in t.

• The formal power series A(t) is algebraic (over Q(t)) if it satisfies a

(non-trivial) polynomial equation: P(t, A(t)) = 0.

• The formal power series A(t) is D-finite (holonomic) if it satisfies a

(non-trivial) linear differential equation with polynomial coefficients: Pk(t)A(k)(t) + · · · + P0(t)A(t) = 0.

• Nice closure properties + asymptotics of the coefficients

A hierarchy of formal power series

• The formal power series A(t) is rational if it can be written

A(t) = P(t)/Q(t) where P(t) and Q(t) are polynomials in t.

• The formal power series A(t) is algebraic (over Q(t)) if it satisfies a

(non-trivial) polynomial equation: P(t, A(t)) = 0.

• The formal power series A(t) is D-finite (holonomic) if it satisfies a

(non-trivial) linear differential equation with polynomial coefficients: Pk(t)A(k)(t) + · · · + P0(t)A(t) = 0.

• Nice closure properties + asymptotics of the coefficients
• Extension to several variables (D-finite: one DE per variable)

Classification of quadrant walks with small steps

Theorem

Assume S ⊂ {¯ 1, 0, 1}2. The series Q(x, y; t) is D-finite iff a certain group G associated with S is finite. It is algebraic iff, in addition, the “orbit sum” is zero. [mbm-Mishna 10], [Bostan-Kauers 10] D-finite [Kurkova-Raschel 12] non-singular non-D-finite [Mishna-Rechnitzer 07], [Melczer-Mishna 13] singular non-D-finite quadrant models with small steps: 79 |G|<∞: 23 OS=0: 19 D-finite OS=0: 4 algebraic |G|=∞: 56 Not D-finite

Classification of quadrant walks with small steps

quadrant models with small steps: 79 |G|<∞: 23 OS=0: 19 D-finite OS=0: 4 algebraic |G|=∞: 56 Not D-finite in probability Random walks Formal power series algebra Complex analysis Computer algebra effective closure properties arithmetic properties G-functions asymptotics D-finite series

Quadrant walks with large steps

A mathematical challenge: the small step condition seems crucial in all approaches (apart from computer algebra) Is the nice classification of walks with small steps robust? Large steps occur in “real life”: bipolar orientations of regular maps [Kenyon, Miller, Sheffield, Wilson, 15(a)]

N S

From quadrant walks to bipolar regular maps [KMSW 15(a)]

Fix p ≥ 1, and take a quadrant walk with two kinds of steps: SE steps (1, −1) NW steps (−i, j) with i, j ≥ 0 and i + j = p The construction starts from a quadrant excursion and a bipolar map reduced to an edge, and yields a bipolar map with faces of degree p + 2. Ex: p = 2

N S

From quadrant walks to bipolar regular maps [KMSW 15(a)]

The construction starts from a quadrant excursion and a bipolar map reduced to an edge, and yields a bipolar map with faces of degree p + 2. every SE step (1, −1) creates an edge. every NW step (−i, j) creates a face of degree i + j + 2 and an edge.

i + 1 j + 1 (1, −1)

• r (−i, j)

From quadrant walks to bipolar regular maps [KMSW 15(a)]

Proposition [Kenyon et al. 15(a)]

This construction is a bijection from quadrant excursions to bipolar maps with faces of degree p + 2.

N S

• steps ⇔ edges in the orientation (minus 1)

• II. A general approach for

quadrant walks...

which solves some cases.

quadrant models with small steps: 79 |G|<∞: 23 OS=0: 19 D-finite OS=0: 4 algebraic |G|=∞: 56 Not D-finite

A four step approach

• 1. Write a functional equation for the tri-variate series Q(x, y; t). It will

involve bi-variate series Q(x, 0; t), Q(0, y; t), . . . (called sections)

• 2. Compute the “orbit” of (x, y)
• 3. Find a functional equation free from sections
• 4. Extract from it Q(x, y; t)

Step 1: Write a functional equation

Example: S = {01, ¯ 10, 1¯ 1} (bipolar triangulations) Q(x, y; t) ≡ Q(x, y) = 1+t(y + ¯ x + x¯ y)Q(x, y)−t¯ xQ(0, y)−tx¯ yQ(x, 0) with ¯ x = 1/x and ¯ y = 1/y.

Step 1: Write a functional equation

Example: S = {01, ¯ 10, 1¯ 1} (bipolar triangulations) Q(x, y; t) ≡ Q(x, y) = 1+t(y + ¯ x + x¯ y)Q(x, y)−t¯ xQ(0, y)−tx¯ yQ(x, 0)

• r
• 1 − t(y + ¯

x + x¯ y)

• Q(x, y) = 1 − t¯

xQ(0, y) − tx¯ yQ(x, 0),

Step 1: Write a functional equation

Example: S = {01, ¯ 10, 1¯ 1} (bipolar triangulations) Q(x, y; t) ≡ Q(x, y) = 1+t(y + ¯ x + x¯ y)Q(x, y)−t¯ xQ(0, y)−tx¯ yQ(x, 0)

• r
• 1 − t(y + ¯

x + x¯ y)

• Q(x, y) = 1 − t¯

xQ(0, y) − tx¯ yQ(x, 0),

• r
• 1 − t(y + ¯

x + x¯ y)

• xyQ(x, y) = xy − tyQ(0, y) − tx2Q(x, 0)

Step 1: Write a functional equation

Example: S = {01, ¯ 10, 1¯ 1} (bipolar triangulations) Q(x, y; t) ≡ Q(x, y) = 1+t(y + ¯ x + x¯ y)Q(x, y)−t¯ xQ(0, y)−tx¯ yQ(x, 0)

• r
• 1 − t(y + ¯

x + x¯ y)

• Q(x, y) = 1 − t¯

xQ(0, y) − tx¯ yQ(x, 0),

• r
• 1 − t(y + ¯

x + x¯ y)

• xyQ(x, y) = xy − tyQ(0, y) − tx2Q(x, 0)
• The polynomial 1 − t(y + ¯

x + x¯ y) is the kernel of this equation

• The equation is linear, with two catalytic variables x and y (tautological

at x = 0 or y = 0) [Zeilberger 00]

• The series Q(0, y) and Q(x, 0) are the sections.

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

• A reflexive and symmetric relation: for x1, x2, y1, y2 ∈ Q(x, y),

(x1, y1) ∼ (x2, y2) if S(x1, y1) = S(x2, y2) and

• x1 = x2, or

y1 = y2.

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

• A reflexive and symmetric relation: for x1, x2, y1, y2 ∈ Q(x, y),

(x1, y1) ∼ (x2, y2) if S(x1, y1) = S(x2, y2) and

• x1 = x2, or

y1 = y2. (x, y) ∼ (x, ) and (x, y) ∼ ( , y)

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

• A reflexive and symmetric relation: for x1, x2, y1, y2 ∈ Q(x, y),

(x1, y1) ∼ (x2, y2) if S(x1, y1) = S(x2, y2) and

• x1 = x2, or

y1 = y2. (x, y) ∼ (x, x¯ y) and (x, y) ∼ ( , y)

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

• A reflexive and symmetric relation: for x1, x2, y1, y2 ∈ Q(x, y),

(x1, y1) ∼ (x2, y2) if S(x1, y1) = S(x2, y2) and

• x1 = x2, or

y1 = y2. (x, y) ∼ (x, x¯ y) and (x, y) ∼ (¯ xy, y)

Step 2: Compute the “orbit” of (x, y)

• The step polynomial:

S(x, y) =

• (i,j)∈S

xiyj S(x, y) = ¯ x + y + x¯ y

• A reflexive and symmetric relation: for x1, x2, y1, y2 ∈ Q(x, y),

(x1, y1) ∼ (x2, y2) if S(x1, y1) = S(x2, y2) and

• x1 = x2, or

y1 = y2. (x, y) ∼ (x, x¯ y) and (x, y) ∼ (¯ xy, y)

• Let ∼ be the transitive closure of this relation. The orbit of (x, y) is its

equivalence class. ∼ ∼ ∼ ∼ ∼ ∼ (¯ y, ¯ x) (x, x¯ y) (x, y) (¯ xy, y) (¯ xy, ¯ x) (¯ y, x¯ y)

Step 3: Find a functional equation free from sections

• The equation reads (with K(x, y) = 1 − tS(x, y)):

K(x, y)xyQ(x, y) = xy − tx2Q(x, 0) − tyQ(0, y)

• The orbit of (x, y) is

(x, y) ∼ (¯ xy, y) ∼ (¯ xy, ¯ x) ∼ (¯ y, ¯ x) ∼ (¯ y, x¯ y) ∼ (x, x¯ y)

Step 3: Find a functional equation free from sections

• The equation reads (with K(x, y) = 1 − tS(x, y)):

K(x, y)xyQ(x, y) = xy − tx2Q(x, 0) − tyQ(0, y)

• The orbit of (x, y) is

(x, y) ∼ (¯ xy, y) ∼ (¯ xy, ¯ x) ∼ (¯ y, ¯ x) ∼ (¯ y, x¯ y) ∼ (x, x¯ y)

• The value of S(x, y) (and K(x, y)) is the same over the orbit. Hence

K(x, y) xyQ(x, y) = xy − tx2Q(x, 0) − tyQ(0, y) K(x, y) ¯ xy2Q(¯ xy, y) = ¯ xy2 − t¯ x2y2Q(¯ xy, 0) − tyQ(0, y) K(x, y) ¯ x2yQ(¯ xy, ¯ x) = ¯ x2y − t¯ x2y2Q(¯ xy, 0) − t¯ xQ(0, ¯ x) · · · = · · · K(x, y) x2¯ yQ(x, x¯ y) = x2¯ y − tx2Q(x, 0) − tx¯ yQ(0, x¯ y).

Step 3: Find a functional equation free from sections

⇒ Form the alternating sum of the equation over all elements of the orbit: K(x, y)

• xyQ(x, y) − ¯

xy2Q(¯ xy, y) + ¯ x2yQ(¯ xy, ¯ x) − ¯ x¯ yQ(¯ y, ¯ x) + x¯ y2Q(¯ y, x¯ y) − x2¯ yQ(x, x¯ y)

• =

xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y (the orbit sum).

Step 3: Find a functional equation free from sections

⇒ Form the alternating sum of the equation over all elements of the orbit: xyQ(x, y) − ¯ xy2Q(¯ xy, y) + ¯ x2yQ(¯ xy, ¯ x) − ¯ x¯ yQ(¯ y, ¯ x) + x¯ y2Q(¯ y, x¯ y) − x2¯ yQ(x, x¯ y) = xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y 1 − t(y + ¯ x + x¯ y)

Step 4: Extract Q(x, y)

⇒ Form the alternating sum of the equation over all elements of the orbit: xyQ(x, y) − ¯ xy2Q(¯ xy, y) + ¯ x2yQ(¯ xy, ¯ x) − ¯ x¯ yQ(¯ y, ¯ x) + x¯ y2Q(¯ y, x¯ y) − x2¯ yQ(x, x¯ y) = xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y 1 − t(y + ¯ x + x¯ y)

Step 4: Extract Q(x, y)

⇒ Form the alternating sum of the equation over all elements of the orbit: xyQ(x, y) − ¯ xy2Q(¯ xy, y) + ¯ x2yQ(¯ xy, ¯ x) − ¯ x¯ yQ(¯ y, ¯ x) + x¯ y2Q(¯ y, x¯ y) − x2¯ yQ(x, x¯ y) = xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y 1 − t(y + ¯ x + x¯ y)

• Both sides are power series in t, with coefficients in Q[x, ¯

x, y, ¯ y].

Step 4: Extract Q(x, y)

⇒ Form the alternating sum of the equation over all elements of the orbit: xyQ(x, y) − ¯ xy2Q(¯ xy, y) + ¯ x2yQ(¯ xy, ¯ x) − ¯ x¯ yQ(¯ y, ¯ x) + x¯ y2Q(¯ y, x¯ y) − x2¯ yQ(x, x¯ y) = xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y 1 − t(y + ¯ x + x¯ y)

• Both sides are power series in t, with coefficients in Q[x, ¯

x, y, ¯ y].

• Extract the part with positive powers of x and y:

xyQ(x, y) = [x>0y>0] xy − ¯ xy2 + ¯ x2y − ¯ x¯ y + x¯ y2 − x2¯ y 1 − t(y + ¯ x + x¯ y) is a D-finite series. [Lipshitz 88]

A four step approach: valid in d dimensions

• 1. Write a functional equation for the (d + 1)-variate series

Q(x1, . . . , xd; t)

• 2. Compute the “orbit” of (x1, . . . , xd)
• 3. Find a linear combination free from d-variate series

Q(0, x2, . . . , xd−1; t), . . . (called sections)

• 4. Extract from it Q(x1, . . . , xd; t)

In 3 dimensions [Bostan, mbm, Kauers, Melczer 14(a)], [Bacher, Kauers, Yatchak 16]

A four step approach: valid in d dimensions

• 1. Write a functional equation for the (d + 1)-variate series

Q(x1, . . . , xd; t)

• 2. Compute the “orbit” of (x1, . . . , xd)
• 3. Find a linear combination free from d-variate series

Q(0, x2, . . . , xd−1; t), . . . (called sections)

• 4. Extract from it Q(x1, . . . , xd; t)

In 3 dimensions [Bostan, mbm, Kauers, Melczer 14(a)], [Bacher, Kauers, Yatchak 16] But also when d = 1!

• III. What can go wrong?

quadrant models with small steps: 79 |G|<∞: 23 OS=0: 19 D-finite OS=0: 4 algebraic |G|=∞: 56 Not D-finite

• 1. Write a functional equation for the (d + 1)-variate series

Q(x1, . . . , xd; t)

• 2. Compute the “orbit” of (x1, . . . , xd)
• 3. Find a linear combination free from d-variate series

Q(0, x2, . . . , xd−1; t), . . . (called sections)

• 4. Extract from it Q(x1, . . . , xd; t)

Step 1: Write a functional equation

This can be done systematically. Examples 1D walks on N with S = {−1, 2}: Q(x) = 1 + t(¯ x + x2)Q(x) − t¯ xQ(0).

Step 1: Write a functional equation

This can be done systematically. Examples 1D walks on N with S = {−1, 2}: Q(x) = 1 + t(¯ x + x2)Q(x) − t¯ xQ(0). Bipolar quadrangulations: quadrant walks with S = Q(x, y) = 1 + t(x¯ y + ¯ x2 + ¯ xy + y2)Q(x, y) − tx¯ yQ(x, 0) −t¯ x2 (Q0(y) + xQ1(y)) − t¯ xyQ0(y), where Qi(y) counts quadrant walks ending at abscissa i.

Step 2: Compute the “orbit” of (x1, . . . , xd)

Can be infinite Example: if S = {0¯ 1, ¯ 1¯ 1, ¯ 10, 11}, then S(x, y) = ¯ x(1 + ¯ y) + ¯ y + xy, (x, y) ∼ (x, ¯ x¯ y(1 + ¯ x)) and (x, y) ∼ (¯ x¯ y(1 + ¯ y), y), and the orbit is infinite: (x, y) · · · · · · · · · · · · · · · · · · ∼ ∼ ∼ ∼ (x, ¯ x¯ y(1 + ¯ x)) (¯ x¯ y(1 + ¯ y), y)

Step 2: Compute the “orbit” of (x1, . . . , xd)

Can be infinite Always finite when d = 1 When d = 1, the orbit consist of all elements x′ such that S(x) = S(x′). Example: for S = {−1, 2}, we solve 1/x + x2 = 1/x′ + x′2 and the orbit

• f x consists of

x, x1 = −x2 +

• x (x3 + 4)

2x , x2 = −x2 −

• x (x3 + 4)

2x .

Step 2: Compute the “orbit” of (x1, . . . , xd)

Can be infinite Always finite when d = 1 Software that computes finite orbits

Step 2: Compute the “orbit” of (x1, . . . , xd)

Can be infinite Always finite when d = 1 Software that computes finite orbits Example: Bipolar maps with faces of degree p + 2 We have S(x, y) = x¯ y + ¯ xp + ¯ xp−1y + · · · + ¯ xyp−1 + yp. The equation S(x, y) = S(x′, y), seen as an equation in x′, admits p + 1 solutions, denoted x0 = x, x1, . . . xp. Denote xp+1 = ¯ y, and yi := 1/xi for 0 ≤ i ≤ p + 1.

• Proposition. The orbit of (x, y) is the set of all (xi, yj), for

0 ≤ i = j ≤ p + 1. It contains (p + 1)(p + 2) elements.

Step 2: Compute the “orbit” of (x1, . . . , xd)

• Proposition. The orbit of (x, y) is the set of all (xi, yj), for

0 ≤ i = j ≤ p + 1. Thus, in total, (p + 1)(p + 2) elements. Examples: p = 1 (bipolar triangulations) (x0, y1) (x0, y2) (x1, y2) (x2, y1) (x2, y0) (x1, y0)

Step 2: Compute the “orbit” of (x1, . . . , xd)

• Proposition. The orbit of (x, y) is the set of all (xi, yj), for

0 ≤ i = j ≤ p + 1. Thus, in total, (p + 1)(p + 2) elements. Examples: p = 2 (bipolar quadrangulations) (x0, y1) (x0, y2) (x3, y2) (x1, y2) (x2, y1) (x2, y3) (x0, y3) (x3, y0) (x1, y3) (x1, y0) (x3, y1) (x2, y0)

Step 3: Find a linear combination free from sections

We assume that the orbit is finite

Step 3: Find a linear combination free from sections

We assume that the orbit is finite There may be several such linear combinations

• Example. For d = 1 and S = {−1, 2}, we have

K(x, y)xQ(x) = x − tQ(0), K(x, y)x1Q(x1) = x1 − tQ(0), K(x, y)x2Q(x2) = x2 − tQ(0), and any linear combination of K(x, y) (xQ(x) − x1Q(x1)) = x − x1 K(x, y) (xQ(x) − x2Q(x2)) = x − x2 is free from Q(0).

Step 3: Find a linear combination free from sections

We assume that the orbit is finite There may be several such linear combinations No example with no such linear combination

Step 3: Find a linear combination free from sections

We assume that the orbit is finite There may be several such linear combinations No example with no such linear combination Bipolar maps: several section free equations when p ≥ 2

Step 4: Extract the main series Q(x1, . . . , xd; t)

Extraction can be impossible (the equation does not characterize the series)

Step 4: Extract the main series Q(x1, . . . , xd; t)

Extraction can be impossible (the equation does not characterize the series) Example: Gessel’s walks. If S = {→, ր, ←, ւ}, the (unique) linear combination free from sections is xyQ(x, y) − ¯ xQ(¯ x¯ y, y) + xQ(¯ x¯ y, x2y) − xyQ(¯ x, x2y) + ¯ x¯ yQ(¯ x, ¯ y) − xQ(xy, ¯ y) + ¯ xQ(xy, ¯ x2¯ y) − ¯ x¯ yQ(x, ¯ x2¯ y) = 0 and has many solutions, for instance 1, x, xy, y − x2, . . .

Step 4: Extract the main series Q(x1, . . . , xd; t)

Extraction can be impossible (the equation does not characterize the series) Extraction can be tricky

• Example. For d = 1 and S = {−1, 2}, we have
• 1 − t(¯

x + x2)

• (xQ(x) − x1Q(x1)) = x − x1

that is, xQ(x) − x1Q(x1) = x − x1 1 − t(¯ x + x2) (1) with x1 = −x2 +

• x (x3 + 4)

2x

Step 4: Extract the main series Q(x1, . . . , xd; t)

Extraction can be impossible (the equation does not characterize the series) Extraction can be tricky

• Example. For d = 1 and S = {−1, 2}, we have
• 1 − t(¯

x + x2)

• (xQ(x) − x1Q(x1)) = x − x1

that is, xQ(x) − x1Q(x1) = x − x1 1 − t(¯ x + x2) (1) with x1 = −x2 +

• x (x3 + 4)

2x = ¯ x2 − ¯ x5 + 2¯ x8 + O(¯ x11) as a series in ¯ x = 1/x. ⇒ Expand (1) in t and ¯ x, and extract the positive part in x to get xQ(x).

• IV. Some cases that work

Walks on a half-line (d = 1)

Let S ⊂ Z with min S = m.

Proposition [Bostan, mbm, Melczer 16+]

Q(x) = [x≥0] m

j=1(1 − ¯

xxj) 1 − tS(x) , where the xj are the roots of S(xj) = S(x) whose expansion in ¯ x involves no positive power of x.

Walks on a half-line (d = 1)

Let S ⊂ Z with min S = m.

Proposition [Bostan, mbm, Melczer 16+]

Q(x) = [x≥0] m

j=1(1 − ¯

xxj) 1 − tS(x) , where the xj are the roots of S(xj) = S(x) whose expansion in ¯ x involves no positive power of x.

Classical solution [Gessel 80, mbm-Petkovšek 00, Banderier-Flajolet 02...]

Q(x) = m

j=1(1 − ¯

xXj) 1 − tS(x) . where the Xj ≡ Xj(t) are the roots of 1 − tS(x) whose expansion in t involves no negative power of t. The series Q(x) is algebraic. These solutions are (of course) equivalent.

Hadamard walks in 2D

Assume S(x, y) = U(x) + V (x)T(y)

Hadamard walks in 2D

Assume S(x, y) = U(x) + V (x)T(y)

Proposition [Bostan, mbm, Melczer 16+]

The series Q(x, y) is D-finite, and reads Q(x, y) = [x≥y≥] m

i=1(1 − ¯

xxi(y)) m′

j=1(1 − ¯

yyj) 1 − tS(x, y) , where the xi(y) are the roots of S(x, y) = S(x′, y) (solved for x′), whose expansion in ¯ x involves no positive power of x, the yj are the roots of S(x, y) = S(x, y′), or T(y) = T(y′) (solved for y′) whose expansion in ¯ y involve no positive powers of y.

Bipolar maps

Proposition [mbm, Fusy, Raschel 16+]

The generating function of bipolar maps with faces of degree p + 2 is Q(x, y) = [x≥y≥](y − ¯ x1) (1 − ¯ x¯ y) Sx(x, y) 1 − tS(x, y) , where S(x, y) is the step polynomial: S(x, y) = x¯ y + ¯ xp + ¯ xp−1y + · · · + ¯ xyp−1 + yp. and x1 is the only root of S(x, y) = S(x′, y) (solved for x′) whose expansion in ¯ y involves a positive power of y. It is D-finite.

Quadrant walks with steps in {−2, ±1, 0}2

Models with at least one occurrence of −2 quadrant models: 13110 α rational: 29 + 227 |orbit| < ∞: 13 + 227 OS = 0: 4 + 227 D-finite OS = 0: 9 ??? |orbit| = ∞ (?): 16 Not D-finite? α irrational: 12854 Not D-finite

Quadrant walks with steps in {−2, ±1, 0}2

Models with at least one occurrence of −2 quadrant models: 13110 α rational: 29 + 227 |orbit| < ∞: 13 + 227 OS = 0: 4 + 227 D-finite OS = 0: 9 ??? |orbit| = ∞ (?): 16 Not D-finite? α irrational: 12854 Not D-finite Number of excursions of length n: q(0, 0; n) ∼ κµnnα [Denisov-Wachtel 15], [Bostan-Raschel-Salvy 14]

Quadrant walks with steps in {−2, ±1, 0}2

Models with at least one occurrence of −2 quadrant models: 13110 α rational: 29 + 227 |orbit| < ∞: 13 + 227 OS = 0: 4 + 227 D-finite OS = 0: 9 ??? |orbit| = ∞ (?): 16 Not D-finite? α irrational: 12854 Not D-finite Number of excursions of length n: q(0, 0; n) ∼ κµnnα [Denisov-Wachtel 15], [Bostan-Raschel-Salvy 14] 227: The number of Hadamard models OS: The orbit sum (the RHS of the section free equation)

Quadrant walks with steps in {−2, ±1, 0}2

Non-singular models with no occurrence of −2 quadrant models: 74 α rational: 7 + 16 |orbit| < ∞: 7 + 16 OS = 0: 3 + 16 D-finite OS = 0: 4 algebraic |orbit| = ∞: 0 α irrational: 51 Not D-finite Singular: all steps lie in a half-plane Number of excursions of length n: q(0, 0; n) ∼ κµnnα [Denisov-Wachtel 15], [Bostan-Raschel-Salvy 14] 16: The number of Hadamard models OS: The orbit sum (the RHS of the section free equation)

Some interesting models

• Non-Hadamard, solvable via our approach (and D-finite):

Some interesting models

• Non-Hadamard, solvable via our approach (and D-finite):

For the first model, Q(x, y) = [x≥0y≥0]

• x3 − 2 y2 − x

y2 − x x2y2 − y2 − 2 x

• x5y4 (1 − t(y + x¯

y + ¯ x¯ y + ¯ x2y)) . The coefficients are nice: for n = 2i + j + 4m, q(i, j; n) = (i + 1)(j + 1)(i + j + 2)n!(n + 2)! m!(3m + 2i + j + 2)!(2m + i + 1)!(2m + i + j + 2)!.

Some interesting models

• Non-Hadamard, solvable via our approach (and D-finite):
• Non-Hadamard, orbit sum zero:

Conj.: DF DF DF ? ? Conj.: DF Alg Alg ?

Final comments

Still a lot to be done... Is there a unique section free equation when there are no large forward steps? Closer study for tricky examples (the 9 analogues of Kreweras’ and Gessel’s algebraic models) Walks where α is rational but the orbit looks infinite quadrant models: 13110 α rational: 29 + 227 |orbit| < ∞: 13 + 227 OS = 0: 4 + 227 D-finite OS = 0: 9 ??? |orbit| = ∞ (?): 16 Not D-finite? α irrational: 12854 Not D-finite

Final comments

Still a lot to be done... Is there a unique section free equation when there are no large forward steps? Closer study for tricky examples (the 9 analogues of Kreweras’ and Gessel’s algebraic models) Walks where α is rational but the orbit looks infinite