+++ /dev/null
-# Assignment 6 (week 7)
-
-## Evaluation order in Combinatory Logic
-
-1. Give a term that the lazy evaluators (either [[the Haskell
-evaluator|code/ski_evaluator.hs]], or the lazy version of [[the OCaml
-evaluator|code/ski_evaluator.ml]]) do not evaluate all the way to a
-normal form, i.e., that contains a redex somewhere inside of it after
-it has been reduced.
-
-2. One of the [[criteria we established for classifying reduction
-strategies|topics/week3_evaluation_order]] strategies is whether they
-reduce subexpressions hidden under lambdas. That is, for a term like
-`(\x y. x z) (\x. x)`, do we reduce to `\y.(\x.x) z` and stop, or do
-we reduce further to `\y.z`? Explain what the corresponding question
-would be for CL. Using the eager version of the OCaml CL evaluator,
-prove that the evaluator does reduce expressions inside of at least
-some "functional" CL expressions. Then provide a modified evaluator
-that does not perform reductions in those positions. (Just give the
-modified version of your recursive reduction function.)
-
-
-## Evaluation in the untyped lambda calculus: substitution
-
-Having sketched the issues with a discussion of Combinatory Logic,
-we're going to begin to construct an evaluator for a simple language
-that includes lambda abstraction. In this problem set, we're going to
-work through the issues twice: once with a function that does
-substitution in the obvious way. You'll see it's somewhat
-complicated. The complications come from the need to worry about
-variable capture. (Seeing these complications should give you an
-inkling of why we presented the evaluation order discussion using
-Combinatory Logic, since we don't need to worry about variables in
-CL.)
-
-We're not going to ask you to write the entire program yourself.
-Instead, we're going to give you [[the complete program, minus a few
-little bits of glue|code/reduction_with_substitution.ml]]. What you
-need to do is understand how it all fits together. When you do,
-you'll understand how to add the last little bits to make functioning
-program.
-
-1. In the previous homework, you built a function that took an
-identifier and a lambda term and returned a boolean representing
-whether that identifier occured free inside of the term. Your first
-task is to complete the `free_in` function, which has been crippled in
-the code base (look for lines that say `COMPLETE THIS LINE`). Once
-you have your function working, you should be able to run queries such
-as this:
-
- # free_in "x" (App (Abstract ("x", Var "x"), Var "x"));;
- - : bool = true
-
-2. Once you get the `free_in` function working, you'll need to
-complete the `substitute` function. Sample target:
-
- # substitute (App (Abstract ("x", ((App (Abstract ("x", Var "x"), Var "y")))), Constant (Num 3))) "y" (Constant (Num 4));;
- - : lambdaTerm = App (Abstract ("x", App (Abstract ("x", Var "x"), Constant (Num 4))), Constant (Num 3))
-
-By the way, you'll see a new wrinkle on OCaml's pattern-matching
-construction: `| PATTERN when x = 2 -> RESULT`. This means that a
-match with PATTERN is only triggered if the boolean condition in the
-`when` clause evaluates to true.
-
-3. Once you have completed the previous two problems, you'll have a
-complete evaluation program. Here's a simple sanity check for when you
-get it working:
-
- # reduce (App (Abstract ("x", Var "x"), Constant (Num 3)));;
- - : lambdaTerm = Constant (Num 3)
-
-4. What kind of evaluation strategy does this evaluator use? In
-particular, what are the answers to the three questions about
-evaluation strategy as given in the discussion of [[evaluation
-strategies|topics/week3_evaluation_order]] as Q1, Q2, and Q3?
-
-## Evaluation in the untyped calculus: environments and closures
-
-Ok, the previous strategy sucked: tracking free and bound variables,
-computing fresh variables, it's all super complicated.
-
-Here's a better strategy. Instead of keeping all of the information
-about which variables have been bound or are still free implicitly
-inside of the terms, we'll keep score. This will require us to carry
-around a scorecard, which we will call an "environment". This is a
-familiar strategy for philosophers of language and for linguists,
-since it amounts to evaluating expressions relative to an assignment
-function. The difference between the assignment function approach
-above, and this approach, is one huge step towards monads.
-
-5. First, you need to get [[the evaluation
-code|code/reduction_with_environments.ml]] working. Look in the
-code for places where you see "not yet implemented", and get enough of
-those places working that you can use the code to evaluate terms.
-
-6. A snag: what happens when we want to replace a variable with a term
-that itself contains a free variable?
-
- term environment
- ------------- -------------
- (\w.(\y.y)w)2 []
- (\y.y)w [w->2]
- y [w->2, y->w]
-
-In the first step, we bind `w` to the argument `2`. In the second
-step, we bind `y` to the argument `w`. In the third step, we would
-like to replace `y` with whatever its current value is according to
-our scorecard. On the simple-minded view, we would replace it with
-`w`. But that's not the right result, because `w` itself has been
-mapped onto 2. What does your evaluator code do?
-
-We'll guide you to a solution involving closures. The first step is
-to allow values to carry around a specific environment with them:
-
- type value = LiteralV of literal | Closure of lambdaTerm * env
-
-This will provide the extra information we need to evaluate an
-identifier all the way down to the correct final result. Here is a
-[[modified version of the evaluator that provides all the scaffoling for
-passing around closures|exercises/reduction_with_closures]].
-The problem is with the following line:
-
- | Closure (Abstract(bound_ident, body), saved_r) -> eval body (push bound_ident arg saved_r) (* FIX ME *)
-
-What should it be in order to solve the problem?
-
-
-## Monads
-
-Mappables (functors), MapNables (applicatives functors), and Monads
-(composables) are ways of lifting computations from unboxed types into
-boxed types. Here, a "boxed type" is a type function with one missing
-piece, which we can think of as a function from a type to a type.
-Call this type function M, and let P, Q, R, and S be variables over types.
-
-Recall that a monad requires a singleton function 1:P-> MP, and a
-composition operator >=>: (P->MQ) -> (Q->MR) -> (P->MR) [the type for
-the composition operator given here corrects a "type"-o from the class handout]
-that obey the following laws:
-
- 1 >=> k = k
- k >=> 1 = k
- j >=> (k >=> l) = (j >=> k) >=> l
-
-For instance, the identity monad has the identity function I for 1
-and ordinary function composition (o) for >=>. It is easy to prove
-that the laws hold for any expressions j, k, and l whose types are
-suitable for 1 and >=>:
-
- 1 >=> k == I o k == \p. I (kp) ~~> \p.kp ~~> k
- k >=> 1 == k o I == \p. k (Ip) ~~> \p.kp ~~> k
-
- (j >=> k) >=> l == (\p.j(kp)) o l == \q.(\p.j(kp))(lq) ~~> \q.j(k(lq))
- j >=> (k >=> l) == j o (k o l) == j o \p.k(lp) == \q.j(\p.k(lp)q) ~~> \q.j(k(lq))
-
-1. On a number of occasions, we've used the Option type to make our
-conceptual world neat and tidy (for instance, think of the discussion
-of Kaplan's Plexy). As we learned in class, there is a natural monad
-for the Option type. Borrowing the notation of OCaml, let's say that
-"`'a option`" is the type of a boxed `'a`, whatever type `'a` is.
-More specifically,
-
- 'a option = Nothing | Just 'a
-
-Then the obvious singleton for the Option monad is \p.Just p. Give
-(or reconstruct) the composition operator >=> we discussed in class.
-Show your composition operator obeys the monad laws.
-
-2. Do the same with lists. That is, given an arbitrary type
-'a, let the boxed type be ['a], i.e., a list of objects of type 'a. The singleton
-is `\p.[p]`, and the composition operator is
-
- >=> (first:P->[Q]) (second:Q->[R]) :(P->[R]) = List.flatten (List.map f (g a))
-
-For example:
-
- f p = [p, p+1]
- s q = [q*q, q+q]
- >=> f s 7 = [49, 14, 64, 16]
-