Counting Pairs of Lattice Paths by Intersections

Abstract

We count the pairs of walks between diagonally opposite corners of a given lattice rectangle by the number of points in which they intersect. We note that the number of such pairs with one intersection is twice the number with no intersection and we give a bijective proof of that fact. Some probabilistic variants of the problem are also investigated.

On Pairs of Lattice Paths with a Given Number of Intersections

“…Moreover, we may drop the condition that endpoints do not coincide (as is the case in Theorem 1): Remember that we are dealing with ordered pairs of paths, i.e., every pair ( p, q) of paths p, q, where both paths start at the origin and end in the same point P (and where p{q), is considered different from the``reversed'' pair (q, p). Here, the common endpoint counts as point of intersection (contrary to the definition in [1,2]; pointed out by Sulanke [6]): Corollary 1. Let 0 t v and u w 0; let P=(t, u), Q=(v, w) be two points in the integer lattice Z_Z (P=Q is possible).…”

“…5]. The definition for M n, k r, s in [1,2] differs slightly from the one given above insofar as intersections at the endpoint (if r=s) are also not counted: This difference is irrelevant for the following theorem, but will simplify notation of the subsequent corollary.…”

“…We now consider the number of ordered pairs of lattice paths (as defined in Corollary 1) for the special case P=Q=(x, y): Adapting the notation from [1,2] (remember that the intersection at P=Q is not counted there), we denote this number by N x+ y, x k&1 . First, it is clear that if k is maximal, i.e., k=x+ y, then the number of pairs of lattice paths is simply ( x+ y x ) (since the paths must be equal in this case).…”

“…A first formula for M n, k r, s was given in [1,Thm. 2]; the formula we want to prove here was given in [2,Thm.…”

Neighborhood monotonicity, the extended Zermelo model, and symmetric knockout tournaments

“…By Theorem 4 of [5], the number of ordered pairs of lattice paths that begin at the origin and proceed with a total of k + 1 north or east steps and do not intersect except at the origin is 2k + 2 k + 1 . Numbering the steps from k down to 0 (with 0 being the last step) and encoding north steps with a ''1'' and east steps with a ''0'', we see that this collection is equipollent with…”