Double Fold

Posted on 2018-01-02 by nbloomf

Tags: arithmetic-made-difficult, literate-haskell

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This page is part of a series on Arithmetic Made Difficult.

This post is literate Haskell; you can load the source into GHCi and play along.

{-# LANGUAGE NoImplicitPrelude #-}
module DoubleFold (
  dfoldr
) where

import Tuples
import DisjointUnions
import Lists

Every \(A\)-inductive homomorphism \(\lists{A} \rightarrow B\) can be defined in terms of \(\foldr(\ast)(\ast)\). But, as with \(\nats\), some specializations of \(\foldr(\ast)(\ast)\) show up often enough to warrant their own name. Today we’ll define one of these, analogous to \(\dnatrec\).

Let \(A\), \(B\), and \(C\) be sets, and suppose we have mappings \[\begin{array}{l} \delta : \lists{B} \rightarrow C \\ \psi : A \times C \rightarrow C \\ \chi : A \times B \times \lists{B} \times C \times C \rightarrow C. \end{array}\] There is a unique solution \(\Omega : \lists{A} \times \lists{B} \rightarrow C\) to the following system of functional equations for all \(a \in A\), \(b \in B\), \(x \in \lists{A}\), and \(y \in \lists{B}\). \[\left\{\begin{array}{l} \Omega(\nil,y) = \delta(y) \\ \Omega(\cons(a,x),\nil) = \psi(a,\Omega(x,\nil)) \\ \Omega(\cons(a,x),\cons(b,y)) = \chi(a,b,y,\Omega(x,y),\Omega(x,\cons(b,y))) \end{array}\right.\] We denote this \(\Omega\) by \(\dfoldr(\delta)(\psi)(\chi)\).

In Haskell:

dfoldr :: (List t)
  => (t b -> c)
  -> (a -> c -> c)
  -> (a -> b -> t b -> c -> c -> c)
  -> t a -> t b -> c
dfoldr delta psi chi = foldr delta phi
  where
    phi a f w = case uncons w of
      Left ()          -> psi a (f nil)
      Right (Pair b y) -> chi a b y (f y) (f (cons b y))

Define a map \(\varphi : A \times C^{\lists{B}} \rightarrow C^{\lists{B}}\) casewise by \[\left\{\begin{array}{l} \varphi(a,g)(\nil) = \psi(a,g(\nil)) \\ \varphi(a,g)(\cons(b,v)) = \chi(a,b,v,g(v),g(\cons(b,v))). \end{array}\right.\] We claim the unique solution \(\Omega : \lists{A} \times \lists{B} \rightarrow C\) is given by \[\Omega(x,y) = \foldr(\delta)(\varphi)(x)(y).\] First we show that \(\Omega\) is actually a solution. To this end let \(a \in A\), \(b \in B\), \(x \in \lists{A}\), and \(y \in \lists{B}\), and note that \[\begin{eqnarray*} & & \Omega(\nil,y) \\ & = & \foldr(\delta)(\varphi)(\nil)(y) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-nil} = & \delta(y) \end{eqnarray*}\] and \[\begin{eqnarray*} & & \Omega(\cons(a,x),\nil) \\ & = & \foldr(\delta)(\varphi)(\cons(a,x))(\nil) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \varphi(a,\foldr(\delta)(\varphi)(x))(\nil) \\ & = & \psi(a,\foldr(\delta)(\varphi)(x)(\nil)) \\ & = & \psi(a,\Omega(x,\nil)), \end{eqnarray*}\] and \[\begin{eqnarray*} & & \Omega(\cons(a,x),\cons(b,y)) \\ & = & \foldr(\delta)(\varphi)(\cons(a,x))(\cons(b,y)) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \varphi(a,\foldr(\delta)(\varphi)(x))(\cons(b,y)) \\ & = & \chi(a,b,y,\foldr(\delta)(\varphi)(x)(y),\foldr(\delta)(\varphi)(x)(\cons(b,y))) \\ & = & \chi(a,b,y,\Omega(x,y),\Omega(x,\cons(b,y))) \end{eqnarray*}\] as needed. To see uniqueness, suppose \(f\) is a solution to the system; we claim that \(f(x,y) = \foldr(\delta)(\varphi)(x)(y)\) for all \(x,y \in \lists{A}\), and show this by list induction on \(x\). For the base case \(x = \nil\), for all \(y \in \lists{B}\) we have \[\begin{eqnarray*} & & f(\nil,y) \\ & = & \delta(y) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-nil} = & \foldr(\delta)(\varphi)(\nil)(y) \end{eqnarray*}\] as claimed. For the inductive step, suppose the equality holds for all \(w \in \lists{B}\) for some \(x \in \lists{A}\), and let \(a \in A\). We consider two possibilities for \(w\). If \(w = \nil\), we have \[\begin{eqnarray*} & & f(\cons(a,x),\nil) \\ & = & \psi(a,f(x,\nil)) \\ & = & \psi(a,\foldr(\delta)(\varphi)(x)(\nil)) \\ & = & \varphi(a,\foldr(\delta)(\varphi)(x))(\nil) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \foldr(\delta)(\varphi)(\cons(a,x))(\nil) \end{eqnarray*}\] as claimed. If \(w = \cons(b,y)\), we instead have \[\begin{eqnarray*} & & f(\cons(a,x),\cons(b,y)) \\ & = & \chi(a,b,y,f(x,y),f(x,\cons(b,y))) \\ & = & \chi(a,b,y,\foldr(\delta)(\varphi)(x)(y),\foldr(\delta)(\varphi)(x)(\cons(b,y))) \\ & = & \varphi(a,\foldr(\delta)(\varphi)(x))(\cons(b,y)) \\ & \href{/posts/arithmetic-made-difficult/Lists.html#def-foldr-cons} = & \foldr(\delta)(\varphi)(\cons(a,x))(\cons(b,y)) \end{eqnarray*}\] as claimed. By induction, we thus have \[f(x,y) = \foldr(\delta)(\varphi)(x)(y) = \Omega(x,y).\]