Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 30 / 1 / c / Solution

Count partially directed self-avoiding walks using north, east and west steps. Within one horizontal level, a self-avoiding walk must move consistently east or consistently west. It can change horizontal direction only after a north step. Conversely, these rules guarantee self-avoidance: every visited horizontal level is new, and its horizontal run is monotone.

Let $a_n$ count these walks, including $a_0=1$. A horizontal run has generating functionEvery walk decomposes uniquely into an initial horizontal run followed by zero or more pairs consisting of a north step and a horizontal run. Therefore the generating function isComparing coefficients yields $a_0=1$, $a_1=3$, and $a_n=2a_{n-1}+a_{n-2}$ for $n\ge 2$. Solving this linear recurrence relation givesThese are a subset of all self-avoiding walks, so $c_n\ge a_n$ andAllowing either horizontal direction at successive heights gives the strict gain over the north-east-only family.