# Mutating Recursion

Posted on 2014-05-22 by nbloomf

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

{-# LANGUAGE NoImplicitPrelude #-}
module MutatingRecursion
( mutatingRec, _test_mutatingrec, main_mutatingrec
) where

import Testing
import Booleans
import DisjointUnions
import NaturalNumbers

Note that both simple recursion and bailout recursion produce functions with type $\nats \times A \rightarrow B;$ we can call that $$A$$ argument the parameter. Now simple and bailout recursion use the parameter in different ways. Simple recursion is only allowed to change $$A$$ “outside” the recursive call, while bailout recursion can only change $$A$$ “inside” the recursive call. These restrictions were necessary so that simple and bailout recursion would have tail-recursive implementations. But there are times when we will want a recursive function with signature $$\nats \times A \rightarrow B$$ that can change its $$A$$ parameter both inside and outside the recursion.

For this situation we introduce yet another recursion operator on $$\nats$$, which we’ll call mutating recursion.

Let $$A$$ and $$B$$ be sets. Suppose we have mappings $$\varepsilon : A \rightarrow B$$, $$\beta : \nats \times A \rightarrow B$$, $$\psi : \nats \times A \rightarrow B$$, $$\chi : \nats \times A \times B \rightarrow B$$, and $$\omega : \nats \times A \rightarrow A$$. Then there is a unique map $$Θ : \nats \times A \rightarrow B$$ such that $Θ(\zero,a) = \varepsilon(a)$ and $Θ(\next(n),a) = \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,Θ(n,\omega(n,a)))}.$ We denote this $$Θ$$ by $$\mutrec(\varepsilon)(\beta)(\psi)(\chi)(\omega)$$.

Define $$δ \in B^{A \times \nats}$$ by $δ(a,n) = \varepsilon(a)$ and $$σ : B^{A \times \nats} \rightarrow B^{A \times \nats}$$ by $σ(g)(a,n) = \bif{\beta(\prev(n),a)}{\psi(\prev(n),a)}{\chi(\prev(n),a,g(\omega(\prev(n),a)),\prev(n))}.$ Now $$(B^{A \times \nats},δ,σ)$$ is an iterative set. We now define $Θ(n,a) = \natrec(δ)(σ)(n)(\tup(a)(n)).$

To see that $$Θ$$ has the claimed properties, note that $\begin{eqnarray*} & & Θ(\zero,a) \\ & \let{Θ(n,a) = \natrec(δ)(σ)(n)(\tup(a)(n))} = & \natrec(δ)(σ)(\zero)(\tup(a)(\zero)) \\ & \href{/posts/arithmetic-made-difficult/natural-numbers.html#cor-natrec-zero} = & δ(\tup(a)(\zero)) \\ & \let{δ(\tup(a)(n)) = \varepsilon(a)} = & \varepsilon(a) \end{eqnarray*}$ and $\begin{eqnarray*} & & Θ(\next(n),a) \\ & \let{Θ(n,a) = \natrec(δ)(σ)(n)(\tup(a)(n))} = & \natrec(δ)(σ)(\next(n))(\tup(a)(\next(n))) \\ & \href{/posts/arithmetic-made-difficult/natural-numbers.html#cor-natrec-next} = & σ(\natrec(δ)(σ)(n))(\tup(a)(\next(n))) \\ & \let{σ(g)(\tup(a)(n)) = \bif{\beta(\prev(n),a)}{\psi(\prev(n),a)}{\chi(\prev(n),a,g(\omega(\prev(n),a),\prev(n)))}} = & \bif{\beta(\prev(\next(n)),a)}{\psi(\prev(\next(n)),a)}{\chi(\prev(\next(n)),a,\natrec(δ)(σ)(n)(\omega(\prev(\next(n)),a),\prev(\next(n))))} \\ & \href{/posts/arithmetic-made-difficult/Unary.html#thm-prev-next} = & \bif{\beta(n,a)}{\psi(\prev(\next(n)),a)}{\chi(\prev(\next(n)),a,\natrec(δ)(σ)(n)(\omega(\prev(\next(n)),a),\prev(\next(n))))} \\ & \href{/posts/arithmetic-made-difficult/Unary.html#thm-prev-next} = & \bif{\beta(n,a)}{\psi(\prev(\next(n)),a)}{\chi(n,a,\natrec(δ)(σ)(n)(\omega(\prev(\next(n)),a),\prev(\next(n))))} \\ & \href{/posts/arithmetic-made-difficult/Unary.html#thm-prev-next} = & \bif{\beta(n,a)}{\psi(\prev(\next(n)),a)}{\chi(n,a,\natrec(δ)(σ)(n)(\omega(n,a),\prev(\next(n))))} \\ & \href{/posts/arithmetic-made-difficult/Unary.html#thm-prev-next} = & \bif{\beta(n,a)}{\psi(\prev(\next(n)),a)}{\chi(n,a,\natrec(δ)(σ)(n)(\omega(n,a),n))} \\ & \href{/posts/arithmetic-made-difficult/Unary.html#thm-prev-next} = & \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,\natrec(δ)(σ)(n)(\omega(n,a),n))} \end{eqnarray*}$ as claimed.

Now we show that $$Θ$$ is unique with this property. To that end, suppose $$\Psi$$ does as well; we show that $$Θ(n,a) = \Psi(n,a)$$ for all $$n$$ by induction. For the base case $$n = \zero$$, we have $\begin{eqnarray*} & & Θ(\zero,a) \\ & \hyp{Θ(\zero,a) = \varepsilon(a)} = & \varepsilon(a) \\ & \hyp{\Psi(\zero,a) = \varepsilon(a)} = & \Psi(\zero,a) \end{eqnarray*}$ For the inductive step, suppose the equality holds for all $$a$$ for some $$n$$. Now $\begin{eqnarray*} & & Θ(\next(n),a) \\ & \hyp{Θ(\next(n),a) = \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,Θ(n,\omega(n,a)))}} = & \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,Θ(n,\omega(n,a)))} \\ & \hyp{Θ(n,a) = \Psi(n,a)} = & \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,\Psi(n,\omega(n,a)))} \\ & \hyp{\Psi(\next(n),a) = \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,\Psi(n,\omega(n,a)))}} = & \Psi(\next(n),a) \end{eqnarray*}$ as needed.

## Implementation

As usual we now want to implement $$\mutrec$$ in software, and there are a couple of ways to go about this. First, the signature.

mutatingRec, mutatingRec' :: (Natural n, Boolean bool)
=> (a -> b)
-> (n -> a -> bool)
-> (n -> a -> b)
-> (n -> a -> b -> b)
-> (n -> a -> a)
-> n
-> a
-> b

There’s the naive way:

mutatingRec epsilon beta psi chi omega n a =
case unnext n of
Left () -> epsilon a
Right m -> ifThenElse (beta m a)
(psi m a)
(chi m a (mutatingRec epsilon beta psi chi omega m (omega m a)))

And there’s the definition from the proof:

mutatingRec' epsilon beta psi chi omega n a =
naturalRec delta sigma n (a,n)
where
delta (b,_) = epsilon b

sigma g (b,m) = ifThenElse (beta (prev m) b)
(psi (prev m) b)
(chi (prev m) b (g (omega (prev m) b, (prev m))))

The naive implementation of mutating recursion is not tail recursive, and I think (without proof) that no truly tail recursive implementation exists (that is sort of the reason for this operator).

While we’re at it, we should test that the two are not not equivalent.

_test_mutatingrec_equiv :: (Natural n, Boolean bool, Equal b)
=> n -> a -> b -> bool
-> Test ((a -> b) -> (n -> a -> bool) -> (n -> a -> b) -> (n -> a -> b -> b) -> (n -> a -> a) -> n -> a -> Bool)
_test_mutatingrec_equiv _ _ _ _ =
testName "mutatingRec(delta,beta,psi,chi,omega)(n,a) == mutatingRec'(delta,beta,psi,chi,omega)(n,a)" \$
\delta beta psi chi omega n a -> eq
(mutatingRec delta beta psi chi omega n a)
(mutatingRec' delta beta psi chi omega n a)

## What it does

As with the other recursion operators, the “uniqueness” part of mutating recursion is also handy. To be a little more explicit, it says the following.

Let $$A$$ and $$B$$ be sets, with mappings $\begin{eqnarray*} \varepsilon & : & A \rightarrow B \\ \beta & : & \nats \times A \rightarrow \bool \\ \psi & : & \nats \times A \rightarrow B \\ \chi & : & \nats \times A \times B \rightarrow B \\ \omega & : & \nats \times A \rightarrow A. \end{eqnarray*}$ Then $$\mutrec(\varepsilon)(\beta)(\psi)(\chi)(\omega)$$ is the unique solution $$f : \nats \times A \rightarrow B$$ to the following system of functional equations for all $$k \in \nats$$, $$a \in A$$: $\left\{\begin{array}{l} f(\zero,a) = \varepsilon(a) \\ f(\next(n),a) = \bif{\beta(n,a)}{\psi(n,a)}{\chi(n,a,f(n,\omega(n,a)))} \end{array}\right.$

## Testing

Suite:

_test_mutatingrec
:: (TypeName n, Natural n, Equal n, Show n, Arbitrary n
, Equal b, Arbitrary a, CoArbitrary a, Arbitrary b, CoArbitrary b
, Boolean bool, Arbitrary bool
, Show a, Show b, TypeName a, TypeName b, CoArbitrary n)
=> Int -> Int -> n -> a -> b -> bool -> IO ()
_test_mutatingrec size cases n a b p = do
testLabel3 "mutatingRec" n a b

let args = testArgs size cases

runTest args (_test_mutatingrec_equiv n a b p)

Main:

main_mutatingrec :: IO ()
main_mutatingrec = do
_test_mutatingrec 50 500 (zero :: Unary) (true :: Bool)  (true :: Bool)  (true :: Bool)
_test_mutatingrec 50 500 (zero :: Unary) (zero :: Unary) (true :: Bool)  (true :: Bool)
_test_mutatingrec 50 500 (zero :: Unary) (true :: Bool)  (zero :: Unary) (true :: Bool)
_test_mutatingrec 50 500 (zero :: Unary) (zero :: Unary) (zero :: Unary) (true :: Bool)