Counting Walks in the Quarter Plane
By Mireille Bousquet-M'elou. 2017
61 citations recorded by the source corpus.
We study planar walks that start from a given point (i0j0), take their steps in a finite set \(\mathfrak{S}\), and are confined in the first quadrant x ≥ 0, y ≥ 0. Their enumeration can be attacked in a systematic way: the generating function Q(x, y; t) that counts them by their length (variable t) and the coordinates of their endpoint (variables x, y) satisfies a linear functional equation encoding the step-by-step description of walks. For instance, for the square lattice walks starting from the origin this equation reads $$\left( {xy - t(x + y + {x^2}y + x{y^2})} \right) {\text{Q(x,y;t)}} {\text{ = }} {\text{xy - xtQ(x,0;t) - ytQ(0,y;t)}}{\text{.}}$$
Subfields: Discrete Mathematics and Combinatorics, Mathematical Physics
View Paper: https://www.semanticscholar.org/paper/a76af1640027a20637c9a3e9bc02bb8dad5039e2
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In Discrete Mathematics and Combinatorics
In Mathematical Physics
Citations in the corpus
- ON PARTITIONS AVOIDING 3-CROSSINGS (2005, unknown): “First of all, a number of equations with a finite diagram have been solved, and shown to have a D-finite solution [3, 4, 5, 18].”
- On 3-Dimensional Lattice Walks Confined to the Positive Octant (2014, unknown): “=2; 1; 1=2;0;1;2;3;4;5]; 1; [ 1;0;1;2;3;4;5;6;7;8;9]; 1; [1;5=3;2;7=3;8=3;3;10=3;11=3;4;13=3;5]; p 3 9 ; [0;1;3=2;2;5=2;3;3;4;5;6;7]; 3 2 p 2; [0;1=2;1;3=2;2;5=2;3;4;5;6;7]; 55 apparentsingularities; [0;1;2;3;4;5;6;7;8;9;11]: Therefore,Lhas6 finitetruesingularities,andthisimpliesthatdegA 1 isatmost6.”
- Walks with small steps in the quarter plane (2008, background): “Ever since it was imported from probability theory to combinatorics, this group has proved useful (in several disguises, like the obstinate, algebraic, or iterated kernel method) to solve various enumeration problems, dealing with walks [11, 13, 15, 33, 40, 41], but also with other objects, like per”
- Walks in the quarter plane: Kreweras’ algebraic model (2004, background): “Taking the difference (an alternative derivation of Kreweras’ result, obtained by considering the product (R(Y0) − xY0)(R(Y1) − xY1), is presented in [5]) and dividing by Y0 − Y1 gives R(Y0) − R(Y1) Y0 − Y1 − x = tx 2R(x) + 2x̄ − 1/t √ (x) , (8)”