Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Functional programming, design and analysis of algorithms / 1 / f / Solution

Assign $y:u$ and $x:v$. The application s y x forces $s:u\to v\to w$ for a fresh type $w$; applying f to that result forces $f:w\to r$. No further equations identify these four type variables. The principal type is therefore

k :: (u -> v -> w) -> v -> (w -> r) -> u -> r
k s x f y = f (s y x)

Every compatible typing is an instance of this one, because type inference imposed precisely the two application constraints and no class constraints.