normal form, i.e., that contains a redex somewhere inside of it after
it has been reduced.
-<!-- reduce3 (FA (K, FA (I, I))) -->
-
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.)
-<!-- just add early no-op cases for Ka and Sab -->
## 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_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.
+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
- : 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:
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.
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, 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.
+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
6. A snag: what happens when we want to replace a variable with a term
that itself contains a free variable?
-<pre>
-term environment
-------------- -------------
-(\w.(\y.y)w)2 []
-(\y.y)w [w->2]
-y [w->2, y->w]
-</pre>
+ 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.
+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) -> (R->MS) that obey the
+following laws:
+
+ 1 >=> k = k
+ k >=> 1 = k
+ j >=> (k >=> l) = (j >=> k) >=> l
-## Baby monads
+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 >=>:
-(Depends on lecture notes for safe division by zero.)
+ 1 >=> k == I o k == \p. I (kp) ~~> \p.kp ~~> k
+ k >=> 1 == k o I == \p. k (Ip) ~~> \p.kp ~~> k
-Write a function `lift'` that generalized the correspondence between +
-and `add'`: that is, `lift'` takes any two-place operation on integers
-and returns a version that takes arguments of type `int option`
-instead, returning a result of `int option`. In other words, `lift'`
-will have type:
+ (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))
- (int -> int -> int) -> (int option) -> (int option) -> (int option)
+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,
-so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
-Don't worry about why you need to put `+` inside of parentheses.
-You should make use of `bind'` in your definition of `lift'`:
+ 'a option = Nothing | Just 'a
- let bind' (u: int option) (f: int -> (int option)) =
- match u with None -> None | Some x -> f x;;
+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.