# Recursion vs. Iteration

We added an accumulator as an extra argument to make the factorial function be tail recursive. That's a trick we've seen before. Let's abstract and see how to do it in general.

Suppose we have a recursive function over integers:

let rec f_r n =
if n = 0 then i else op n (f_r (n - 1))


Here, the r in f_r is meant to suggest that f_r is a recursive function. The i and op are pieces of the function that are meant to be replaced by some concrete value i and operator op. For example, with the factorial function, we have:

f_r = fact
i = 1
op = ( * )


Such a function can be made tail recursive by rewriting it as follows:

let f_i acc n =
if n = 0 then acc
else f_i (op acc n) (n - 1)

let f_tr = f_i i


Here, the i in fi is meant to suggest that fi is an iterative function, and i and op are the same as in the recursive version of the function. For example, with factorial we have:

f_i = fact_i
i = 1
op = ( * )
f_tr = fact_tr


We can prove that f_r and f_tr compute the same function. During the proof, next, we will discover certain conditions that must hold of i and op to make the transformation to tail recursion be correct.

Theorem: f_r = f_tr

Proof:  By extensionality, it suffices to show that forall n, f_r n = f_tr n.

As in the previous proof for fact, we will need a strengthed induction
hypothesis.  So we first prove this lemma, which quantifies over all
accumulators that could be input to f_i, rather than only i:

Lemma: forall n, forall acc, op acc (f_r n) = f_i acc n

Proof of Lemma: by induction on n.
P(n) = forall acc, op acc (f_r n) = f_i acc n

Base: n = 0
Show: forall acc, op acc (f_r 0) = f_i acc 0

op acc (f_r 0)
=   { evaluation }
op acc i
=   { if we assume forall x, op x i = x }
acc

f_i acc 0
=
acc

Inductive case: n = k + 1
Show: forall acc, op acc (f_r (k + 1)) = f_i acc (k + 1)
IH: forall acc, op acc (f_r k) = f_i acc k

op acc (f_r (k + 1))
=   { evaluation }
op acc (op (k + 1) (f_r k))
=   { if we assume forall x y z, op x (op y z) = op (op x y) z }
op (op acc (k + 1)) (f_r k)

f_i acc (k + 1)
=   { evaluation }
f_i (op acc (k + 1)) k
=   { IH, instantiating acc as op acc (k + 1)}
op (op acc (k + 1)) (f_r k)

QED

The proof then follows almost immediately from the lemma:

f_r n
=   { if we assume forall x, op i x = x }
op i (f_r n)
=   { lemma, instantiating acc as i }
f_i i n
=   { evaluation }
f_tr n

QED


1. forall x, op x i = x
2. op x (op y z) = op (op x y) z
3. forall x, op i x = x

The first and third say that i is an identity of op: using it on the left or right side leaves the other argument x unchanged. The second says that op is associative. Both those assumptions held for the values we used in the factorial functions:

• op is multiplication, which is associative.
• i is 1, which is an identity of multiplication: multiplication by 1 leaves the other argument unchanged.

So our transformation from a recursive to a tail-recursive function is valid as long as the operator applied in the recursive call is associative, and the value returned in the base case is an identity of that operator.

Returning to the sumto function, we can apply the theorem we just proved to immediately get a tail-recursive version:

let rec sumto_r n =
if n = 0 then 0 else n + sumto_r (n - 1)


Here, the operator is addition, which is associative; and the base case is zero, which is an identity of addition. Therefore our theorem applies, and we can use it to produce the tail-recursive version without even having to think about it:

let rec sumto_i acc n =
if n = 0 then acc else sumto_i (acc + n) (n - 1)

let sumto_tr = sumto_i 0


We already know that sumto_tr is correct, thanks to our theorem.