X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=exercises%2F_assignment6.mdwn;h=ebc20e064bc1ad1f7d02756ad8a015913dfaea9d;hp=d05f111327164d977dbe25be823994bca6a040d0;hb=b13412acbd429e1cc83669d6e881e3d9e98fbda3;hpb=937c6218cf7d2b947b3966342a38d883853a6492 diff --git a/exercises/_assignment6.mdwn b/exercises/_assignment6.mdwn index d05f1113..ebc20e06 100644 --- a/exercises/_assignment6.mdwn +++ b/exercises/_assignment6.mdwn @@ -8,39 +8,37 @@ 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 either the OCaml CL evaluator or the Haskell -evaluator developed in the wiki notes, 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.) +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 -Once you grok reduction and evaluation order in Combinatory Logic, +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. 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.) +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_reduction.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. +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 @@ -54,15 +52,16 @@ as this: - : bool = true 2. Once you get the `free_in` function working, you'll need to -complete the `substitute` function. 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. -Sample target: +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: @@ -70,15 +69,112 @@ get it working: # reduce (App (Abstract ("x", Var "x"), Constant (Num 3)));; - : lambdaTerm = Constant (Num 3) -What kind of evaluation strategy does this evaluator use? In +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 +## 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. +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]