XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=exercises%2F_assignment4.mdwn;h=f998e7ac73a33c66d353463a62b91aa7b403d710;hp=4dfd5dbf1c1995c50dccb8fd7653cf98fe656c9d;hb=ea4c96609b8d02f5e83a8027cf567bab5562cb5b;hpb=4398d51300777f367c01ef6b88a664b8dfe733b2
diff git a/exercises/_assignment4.mdwn b/exercises/_assignment4.mdwn
index 4dfd5dbf..f998e7ac 100644
 a/exercises/_assignment4.mdwn
+++ b/exercises/_assignment4.mdwn
@@ 16,10 +16,14 @@ find terms `F`, `G`, and `Î¾` such that `F Î¾ <~~> Î¾` and `G Î¾
<~~> Î¾`. (If you need a hint, reread the notes on fixed
points.)
+4. Assume that `Î¨` is some fixed point combinator; we're not telling you which one. (You can just write `Psi` in your homework if you don't know how to generate the symbol `Î¨`.) Prove that `Î¨ Î¨` is a fixed point of itself, that is, that `Î¨ Î¨ <~~> Î¨ Î¨ (Î¨ Î¨)`.
+
+
+
## Writing recursive functions ##
4. Helping yourself to the functions given below,
+5. Helping yourself to the functions given below,
write a recursive function called `fact` that computes the factorial.
The factorial `n! = n * (n  1) * (n  2) * ... * 3 * 2 * 1`.
For instance, `fact 0 ~~> 1`, `fact 1 ~~> 1`, `fact 2 ~~> 2`, `fact 3 ~~>
@@ 27,9 +31,13 @@ For instance, `fact 0 ~~> 1`, `fact 1 ~~> 1`, `fact 2 ~~> 2`, `fact 3 ~~>
let true = \y n. y in
let false = \y n. n in
 let zero? = \n. n (\p. false) true in
 let pred = \n f z. n (\u v. v (u f)) (K z) I in
+ let pair = \a b. \v. v a b in
+ let fst = \a b. a in ; aka true
+ let snd = \a b. b in ; aka false
+ let zero = \s z. z in
let succ = \n s z. s (n s z) in
+ let zero? = \n. n (\p. false) true in
+ let pred = \n. n (\p. p (\a b. pair (succ a) a)) (pair zero zero) snd in
let add = \l r. r succ l in
let mult = \l r. r (add l) 0 in
let Y = \h. (\u. h (u u)) (\u. h (u u)) in
@@ 38,18 +46,24 @@ For instance, `fact 0 ~~> 1`, `fact 1 ~~> 1`, `fact 2 ~~> 2`, `fact 3 ~~>
fac 4
5. For this question, we want to implement **sets** of numbers in terms of lists of numbers, where we make sure as we construct those lists that they never contain a single number more than once. (It would be even more efficient if we made sure that the lists were always sorted, but we won't try to implement that refinement here.) To enforce the idea of modularity, let's suppose you don't know the details of how the lists are implemented. You just are given the functions defined below for them (but pretend you don't see the actual definitions). These define lists in terms of [[one of the new encodings discussed last week/topics/week3_more_lists_]].

+6. For this question, we want to implement **sets** of numbers in terms of lists of numbers, where we make sure as we construct those lists that they never contain a single number more than once. (It would be even more efficient if we made sure that the lists were always sorted, but we won't try to implement that refinement here.) To enforce the idea of modularity, let's suppose you don't know the details of how the lists are implemented. You just are given the functions defined below for them (but pretend you don't see the actual definitions). These define lists in terms of [[one of the new encodings discussed last week/topics/week3_lists#v5lists]].
; all functions from the previous question, plus
 let num_equal? = ??? in
+ ; `num_cmp x y lt eq gt` returns lt when xy
+ let num_cmp = (\base build consume. \l r. r consume (l build base) fst)
+ ; where base is
+ (pair (\a b c. b) (K (\a b c. a)))
+ ; and build is
+ (\p. pair (\a b c. c) p)
+ ; and consume is
+ (\p. p fst p (p snd) (p snd)) in
+ let num_equal? = \x y. num_cmp x y false true false in
let neg = \b y n. b n y in
let empty = \f n. n in
let cons = \x xs. \f n. f x xs in
+ let empty? = \xs. xs (\y ys. false) true in
+ let tail = \xs. xs (\y ys. ys) empty in
+ let append = Y (\append. \xs zs. xs (\y ys. (cons y (append ys zs))) zs) in
let take_while = Y (\take_while. \p xs. xs (\y ys. (p y) (cons y (take_while p ys)) empty) empty) in
let drop_while = Y (\drop_while. \p xs. xs (\y ys. (p y) (drop_while p ys) xs) empty) in
...
@@ 58,10 +72,10 @@ For instance, `fact 0 ~~> 1`, `fact 1 ~~> 1`, `fact 2 ~~> 2`, `fact 3 ~~>
Using those resources, define a `set_cons` and a `set_equal?` function. The first should take a number argument `x` and a set argument `xs` (implemented as a list of numbers assumed to have no repeating elements), and return a (possibly new) set argument which contains `x`. (But make sure `x` doesn't appear in the result twice!) The `set_equal?` function should take two set arguments `xs` and `ys` and say whether they represent the same set. (Be careful, the lists `[1, 2]` and `[2, 1]` are different lists but do represent the same set. Hence, you can't just use the `list_equal?` function you defined in last week's homework.)
 Here are some tips for getting started. Use `drop_while` and `num_equal?` to define a `mem?` function that returns `true` if number `x` is a member of a list of numbers `xs`, else returns `false`. Also use `take_while` and `drop_while` to define a `without` function that returns a copy of a list of numbers `xs` that omits the first occurrence of a number `x`, if there be such. You may find these functions `mem?` and `without` useful in defining `set_cons` and `set_equal?`. Also, for `set_equal?`, you are probably going to want to define the function recursively... as now you know how to do.
+ Here are some tips for getting started. Use `drop_while`, `num_equal?`, and `empty?` to define a `mem?` function that returns `true` if number `x` is a member of a list of numbers `xs`, else returns `false`. Also use `take_while`, `drop_while`, `num_equal?`, `tail` and `append` to define a `without` function that returns a copy of a list of numbers `xs` that omits the first occurrence of a number `x`, if there be such. You may find these functions `mem?` and `without` useful in defining `set_cons` and `set_equal?`. Also, for `set_equal?`, you are probably going to want to define the function recursively... as now you know how to do.
6. Questions about trees.
+7. Questions about trees.
## Arithmetic infinity? ##
@@ 79,11 +93,11 @@ the successor function, multiplication is defined in terms of
addition, and exponentiation is defined in terms of multiplication.
1. Find a fixed point `Î¾` for the successor function. Prove it's a fixed
+8. Find a fixed point `Î¾` for the successor function. Prove it's a fixed
point, i.e., demonstrate that `succ Î¾ <~~> Î¾`.
 We've had surprising success embedding normal arithmetic in the lambda
calculus, modeling the natural numbers, addition, multiplication, and
+ We've had surprising success embedding normal arithmetic in the Lambda
+Calculus, modeling the natural numbers, addition, multiplication, and
so on. But one thing that some versions of arithmetic supply is a
notion of infinity, which we'll write as `inf`. This object usually
satisfies the following constraints, for any finite natural number `n`:
@@ 93,18 +107,18 @@ satisfies the following constraints, for any finite natural number `n`:
n ^ inf == inf
leq n inf == true
 (Note, though, that with some notions of infinite numbers, operations like `+` and `*` are defined in such a way that `inf + n` is different from `n + inf`, and does exceed `inf`.)
+ (Note, though, that with *some* notions of infinite numbers, like [[!wikipedia ordinal numbers]], operations like `+` and `*` are defined in such a way that `inf + n` is different from `n + inf`, and does exceed `inf`.)
2. Prove that `add 1 Î¾ <~~> Î¾`, where `Î¾` is the fixed
point you found in (1). What about `add 2 Î¾ <~~> Î¾`?
+9. Prove that `add Î¾ 1 <~~> Î¾`, where `Î¾` is the fixed
+point you found in (1). What about `add Î¾ 2 <~~> Î¾`?
Comment: a fixed point for the successor function is an object such that it
is unchanged after adding 1 to it. It makes a certain amount of sense
to use this object to model arithmetic infinity. For instance,
depending on implementation details, it might happen that `leq n Î¾` is
true for all (finite) natural numbers `n`. However, the fixed point
you found for `succ` may not be a fixed point for `mult n` or for
`exp n`.
+you found for `succ` and `(+n)` (recall this is shorthand for `\x. add x n`) may not be a fixed point for `(*n)` or for
+`(^n)`.
## Mutuallyrecursive functions ##
@@ 162,6 +176,6 @@ a *pair* of functions `h` and `g`, as follows:
definitions of `even?` and `odd?`?
11. (More challenging.) Using our derivation of `Y` from [[this week's notestopics/week4_fixed_point_combinators_]] as a model, construct a pair `Y1` and `Y2` that behave in the way described above.
+11. (More challenging.) Using our derivation of `Y` from [[this week's notestopics/week4_fixed_point_combinators#derivingy]] as a model, construct a pair `Y1` and `Y2` that behave in the way described above.
Here is one hint to get you started: remember that in the notes, we constructed a fixed point for `h` by evolving it into `H` and using `H H` as `h`'s fixed point. We suggested the thought exercise, how might you instead evolve `h` into some `T` and then use `T T T` as `h`'s fixed point. Try solving this problem first. It may help give you the insights you need to define a `Y1` and `Y2`.