> Here is another solution, due to Martin Bunder and F. Urbanek:
- > let pred = \n. \s z. n (\u v. v (u s)) (K z) I in
+ > let box = \a. \v. v a in
+ > let pred = \n. \s z. n (\b. box (b s)) (K z) I in
> ...
- > That can be hard to wrap your head around. If you trace it through, you'll see that:
+ > That can be hard to wrap your head around. If you trace this expression through, you'll see that:
- > * when `n == 0`, it reduces to `\s z. (\v. z) I`, or `\s z. z`
- > * when `n == 1`, it reduces to `\s z. (\v. v z) I`, or `\s z. z`
- > * when `n == 2`, it reduces to `\s z. (\v. v (s z)) I`, or `\s z. s z`
- > * when `n == 3`, it reduces to `\s z. (\v. v (s (s z))) I`, or `\s z. s (s z)`
+ > * when `n == 0`, it reduces to `\s z. (K z) I`, or `\s z. z`
+ > * when `n == 1`, it reduces to `\s z. (box z) I`, or `\s z. z`
+ > * when `n == 2`, it reduces to `\s z. (box (s z)) I`, or `\s z. s z`
+ > * when `n == 3`, it reduces to `\s z. (box (s (s z))) I`, or `\s z. s (s z)`
- > The technique used here is akin to that used in [[the hint for last week's `reverse`|assignment2_answers#cps-reverse]].
+ > `box a` is `\v. v a`; this stands to `pair a b` as one stands to two. Since boxes (like pairs) are really just functions, the technique used in this definition of `pred` is akin to that used in [[the hint for last week's `reverse`|assignment2_answers#cps-reverse]].
(Want a further challenge? Define `map2` in the Lambda Calculus, using our right-fold encoding for lists, where `map2 g [a, b, c] [d, e, f]` should evaluate to `[g a d, g b e, g c f]`. Doing this will require drawing on a number of different tools we've developed, including that same strategy for defining `tail`. Purely extra credit.)