-# Assignment 6 (week 7)
+There is no separate assignment 6. (There was a single big assignment for weeks 5 and 6, and
+we're going to keep the assignment numbers in synch with the weeks.)
+
+[[!toc]]
## 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
+normal form, that is, 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
+reduce subterms 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
+would be for CL. Using the eager version of the OCaml CL evaluator (`reduce_try2`),
+prove that the evaluator does reduce terms inside of at least
+some "functional" CL terms. 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
+## Evaluation in the untyped lambda calculus: substitute-and-repeat
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
+substitution in the obvious way, and keeps reducing-and-repeating until
+there are no more eligible redexes. 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
+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
+Instead, we're going to give you almost the complete program, with a few gaps
+in it that you have to complete. You have to understand enough to
+add the last pieces to make the program function.
-2. Once you get the `free_in` function working, you'll need to
-complete the `substitute` function. Sample target:
+You can find the skeleton code [[here|/code/untyped_evaluator.ml]].
- # 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))
+The first place you need to add code is in the `free_in` function. You already
+wrote such a function [[in a previous homework|exercises/assignment5-6#occurs_free]], so this
+part should be easy. The intended behavior is for the function to return results
+like this:
-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.
+ # free_in "x" (App (Lambda("x, Var "x"), Var "x"));;
+ - : bool = 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:
+The second place you need to add code is in the `reduce_head_once` function. As we
+explain in the code comments, this is similar to the `reduce_if_redex` function in the
+combinatory logic interpreters. Three of the clauses near the end of this function are incomplete.
- # reduce (App (Abstract ("x", Var "x"), Constant (Num 3)));;
- - : lambdaTerm = Constant (Num 3)
+As we explain in the code comments, this interpreter uses an eager/call-by-value reduction
+strategy. What are its other evaluation order properties? Does it perform beta-reduction
+underneath lambdas? Does it perform eta-reduction?
-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
+## Evaluation in the untyped lambda calculus: environments
-Ok, the previous strategy sucked: tracking free and bound variables,
-computing fresh variables, it's all super complicated.
+The previous interpreter strategy is nice because it corresponds so closely to the
+reduction rules we give when specifying our lambda calculus. (Including
+specifying evaluation order, which redexes it's allowed to reduce, and
+so on.) But keeping track of free and bound variables, computing fresh
+variables when needed, that's all a pain.
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
+inside of the terms, we'll keep a separate 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
+since it amounts to evaluating terms relative to an assignment
+function. The difference between the substitute-and-repeat 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.
+The skeleton code for this is at the [[same link as before|/code/untyped_evaluator.ml]].
+This part of the exercise is the "VB" part of that code.
+
+You'll see that in the `eval` function, a new kind of value `Closure (ident) (term) (env)`
+is used. What's that about?
-6. A snag: what happens when we want to replace a variable with a term
-that itself contains a free variable?
+[[Read about Closures here.|/topics/week7_environments_and_closures]]
- 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?
+So that's what's going on with those `Closure`s. In the simple code we gave you to work with, we just made these another clause in the `term` datatype. That's really not correct. `Closure`s aren't terms. The syntax for our language doesn't have any constituents that get parsed into `Closure`s. `Closure`s are only created *during the course of evaluating terms*: specifically, when a variable gets bound to an abstract, which may itself contain variables that are locally free (not bound by the abstract itself). So really we should have two datatypes, one for terms, and another for the *results* (sometimes called "values") that terms can evaluate to. `Closure`s are results, but they aren't terms. (In later weeks, we will see more examples of results that aren't terms, but can only be generated during the course of a computation.) `App`s are terms, but not results. If we had primitive numbers or other constants in our language, they might be both terms and results. In the fuller code from which your homework is simplified, this is how the types are in fact defined. But it makes things more complicated. So to keep things simple for the homework, we just pretended like `Closure`s were a new, exotic kind of `term`.
-We'll guide you to a solution involving closures. The first step is
-to allow values to carry around a specific environment with them:
+In any case, now you know what's going on with the `Closure`s, and you should be able to complete the missing pieces of the `eval` function in the code skeleton linked above.
- type value = LiteralV of literal | Closure of lambdaTerm * env
+If you've completed all the missing parts correctly (there are six gaps for the previous stage of the homework, and two gaps for this stage), then you should be able to compile the code skeleton, and use it as described in the comments at the start of the code.
-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 *)
+## Fuller interpreter
-What should it be in order to solve the problem?
+We've also prepared [[a fuller version of the interpreter, that has user-friendly input
+and printing of results|/topics/week7_untyped_evaluator]]. It
+will be easiest for you to understand that code if you've
+completed the gaps in the simplified skeleton linked above.
+There's nothing you need to do with this; it's just for you to play with. If you're interested,
+you can compare the code you completed for the previous two segments of the homework
+to the (only moderately more complex) code in the `engine.ml` file of this fuller program.
## Monads
-Mappables (functors), MapNables (applicatives functors), and Monads
+Mappables (functors), MapNables (applicative 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.
+boxed types. Here, a "boxed type" is a type function with one unsaturated
+hole (which may have several occurrences, as in `(α,α) tree`). We can think of the box type
+as a function from a type to a type.
+
+Recall that a monad requires a singleton function <code>⇧ (\* mid \*) : P-> <u>P</u></code>, and a
+composition operator like <code>>=> : (P-><u>Q</u>) -> (Q-><u>R</u>) -> (P-><u>R</u>)</code>.
+
+As we said in the notes, we'll move freely back and forth between using `>=>` and using `<=<` (aka `mcomp`), which
+is just `>=>` with its arguments flipped. `<=<` has the virtue that it corresponds more
+closely to the ordinary mathematical symbol `○`. But `>=>` has the virtue
+that its types flow more naturally from left to right.
+
+Anyway, `mid`/`⇧` and (let's say) `<=<` have to obey the Monad Laws:
+
+ ⇧ <=< k == k
+ k <=< ⇧ == k
+ j <=< (k <=< l) == (j <=< k) <=< l
+
+For example, the Identity monad has the identity function `I` for `⇧`
+and ordinary function composition `○` for `<=<`. It is easy to prove
+that the laws hold for any terms `j`, `k`, and `l` whose types are
+suitable for `⇧` and `<=<`:
+
+ ⇧ <=< k == I ○ k == \p. I (k p) ~~> \p. k p ~~> k
+ k <=< ⇧ == k ○ I == \p. k (I p) ~~> \p. k p ~~> k
+
+ (j <=< k) <=< l == (\p. j (k p)) ○ l == \q. (\p. j (k p)) (l q) ~~> \q. j (k (l q))
+ j <=< (k <=< l) == j ○ (k ○ l) == j ○ (\p. k (l p)) == \q. j ((\p. k (l p)) q) ~~> \q. j (k (l q))
+
+1. On a number of occasions, we've used the Option/Maybe type to make our
+conceptual world neat and tidy (for instance, think of [[our discussion
+of Kaplan's Plexy|topics/week6_plexy]]). As we learned in class, there is a natural monad
+for the Option type. Using the vocabulary 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
+ type 'a option = None | Some 'a
+
+ (If you have trouble keeping straight what is the OCaml terminology for this and what is the Haskell terminology, don't worry, we do too.)
-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.
+ Now the obvious singleton for the Option monad is `\p. Some p`. Give
+(or reconstruct) either of the composition operators `>=>` or `<=<`.
+Show that 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
+2. Do the same with lists. That is, given an arbitrary type
+`'a`, let the boxed type be `['a]` or `'a list`, that is, lists of values of type `'a`. The `⇧`
+is the singleton function `\p. [p]`, and the composition operator is:
- >=> (first:P->[Q]) (second:Q->[R]) :(P->[R]) = List.flatten (List.map f (g a))
+ let (>=>) (j : 'a -> 'b list) (k : 'b -> 'c list) : 'a -> 'c list =
+ fun a -> List.flatten (List.map k (j a))
-For example:
+ For example:
- f p = [p, p+1]
- s q = [q*q, q+q]
- >=> f s 7 = [49, 14, 64, 16]
+ let j a = [a; a+1];;
+ let k b = [b*b; b+b];;
+ (j >=> k) 7
+ (* which OCaml evaluates to:
+ - : int list = [49; 14; 64; 16]
+ *)
+ Show that these obey the Monad Laws.