# 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]