## Fixed points ## 1. Recall that `ω ≡ \x. x x`, and `Ω ≡ ω ω`. Is `Ω` a fixed point for `ω`? Find a fixed point for `ω`, and prove that it is a fixed point. 2. Consider `Ω ξ` for an arbitrary term `ξ`. `Ω` is so busy reducing itself (the eternal narcissist) that it never gets around to noticing whether it has an argument, let alone doing anything with that argument. If so, how could `Ω` have a fixed point? That is, how could there be an `ξ` such that `Ω ξ <~~> ξ`? To answer this question, begin by constructing `Y Ω`. Prove that `Y Ω` is a fixed point for `Ω`. 3. Find two different terms that have the same fixed point. That is, find terms `F`, `G`, and `ξ` such that `F ξ <~~> ξ` and `G ξ <~~> ξ`. (If you need a hint, reread the notes on fixed points.) ## Writing recursive functions ## 4. 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 ~~> 6`, and `fact 4 ~~> 24`. 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 succ = \n s z. s (n s z) 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 let fact = ... in 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_]]. ; all functions from the previous question, plus let num_equal? = ??? 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 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 ... The functions `take_while` and `drop_while` work as described in [[Week 1's homework|assignment1]]. 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. 6. Questions about trees. ## Arithmetic infinity? ## The next few questions involve reasoning about Church arithmetic and infinity. Let's choose some arithmetic functions: succ = \n s z. s (n s z) add = \l r. r succ l in mult = \l r. r (add l) 0 in exp = \base r. r (mult base) 1 in There is a pleasing pattern here: addition is defined in terms of the successor function, multiplication is defined in terms of addition, and exponentiation is defined in terms of multiplication. 1. Find a fixed point `ξ` for the successor function. Prove it's a fixed point, i.e., demonstrate that `succ ξ <~~> ξ`. 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`: n + inf == inf n * inf == inf 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`.) 2. Prove that `add 1 ξ <~~> ξ`, where `ξ` is the fixed point you found in (1). What about `add 2 ξ <~~> ξ`? 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 ξ` 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`. ## Mutually-recursive functions ## 10. (Challenging.) One way to define the function `even?` is to have it hand off part of the work to another function `odd?`: let even? = \x. (zero? x) ; if x == 0 then result is true ; else result turns on whether x-1 is odd (odd? (pred x)) At the same tme, though, it's natural to define `odd?` in such a way that it hands off part of the work to `even?`: let odd? = \x. (zero? x) ; if x == 0 then result is false ; else result turns on whether x-1 is even (even? (pred x)) Such a definition of `even?` and `odd?` is called **mutually recursive**. If you trace through the evaluation of some sample numerical arguments, you can see that eventually we'll always reach a base step. So the recursion should be perfectly well-grounded: even? 3 ~~> (zero? 3) true (odd? (pred 3)) ~~> odd? 2 ~~> (zero? 2) false (even? (pred 2)) ~~> even? 1 ~~> (zero? 1) true (odd? (pred 1)) ~~> odd? 0 ~~> (zero? 0) false (even? (pred 0)) ~~> false But we don't yet know how to implement this kind of recursion in the Lambda Calculus. The fixed point operators we've been working with so far worked like this: let ξ = Y h in ξ <~~> h ξ Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on a *pair* of functions `h` and `g`, as follows: let ξ1 = Y1 h g in let ξ2 = Y2 h g in ξ1 <~~> h ξ1 ξ2 and ξ2 <~~> g ξ1 ξ2 If we gave you such a `Y1` and `Y2`, how would you implement the above definitions of `even?` and `odd?`? 11. (More challenging.) Using our derivation of `Y` from [[this week's notes|topics/week4_fixed_point_combinators_]] 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`.