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.)

Evaluation order in Combinatory Logic

  1. Give a term that the lazy evaluators (either the Haskell evaluator, or the lazy version of the OCaml evaluator) do not evaluate all the way to a 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 strategies is whether they 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 (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: 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, 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 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 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.

You can find the skeleton code here.

The first place you need to add code is in the free_in function. You already wrote such a function in a previous homework, so this part should be easy. The intended behavior is for the function to return results like this:

# free_in "x" (App (Lambda("x, Var "x"), Var "x"));;
- : bool = true

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.

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?

Evaluation in the untyped lambda calculus: environments

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 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 terms relative to an assignment function. The difference between the substitute-and-repeat approach above, and this approach, is one huge step towards monads.

The skeleton code for this is at the same link as before. 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?

Read about Closures here.

So that's what's going on with those Closures. 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. Closures aren't terms. The syntax for our language doesn't have any constituents that get parsed into Closures. Closures 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. Closures 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.) Apps 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 Closures were a new, exotic kind of term.

In any case, now you know what's going on with the Closures, and you should be able to complete the missing pieces of the eval function in the code skeleton linked above.

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.

Fuller interpreter

We've also prepared a fuller version of the interpreter, that has user-friendly input and printing of results. 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 file of this fuller program.


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 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 ⇧ (* mid *) : P-> P, and a composition operator like >=> : (P->Q) -> (Q->R) -> (P->R).

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). 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,

      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.)

    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] or 'a list, that is, lists of values of type 'a. The is the singleton function \p. [p], and the composition operator is:

      let (>=>) (j : 'a -> 'b list) (k : 'b -> 'c list) : 'a -> 'c list =
        fun a -> List.flatten ( k (j a))

    For example:

      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.