Merge branch 'working'

[lambda.git] / exercises / assignment8-9_answers.mdwn

This is the assignment for weeks 8-9, on Reader and State monads.

<!--
1. When discussing safe division, we worked with operators like `map2 (+)` which just did their ordinary thing, only now lifted up into working on "boxed" or monadic values. Also there was a special division operator, which interacted with the new possibilities presented by the Option/Maybe monad. Instead of this special division operator and the Option monad, show us how to write expressions in the Reader monad with special `getx` operators. That is, instead of representing computations like `(2+3)/0/1`, now you will represent computations like `(2+x)+1`. To keep things simple, suppose that your language only allows three variables `x`, `y`, and `z`, and you can represent the `env` as a triple of `int`s. For this problem you don't need to demonstrate how to implement binding expressions like `let x = 3 in ...`. You just need to compute the value of possibly open expressions, relative to a supplied `env` that gives values for `x` and `y` and `z`.

2. Okay, now what changes do you need to make to add in expressions like `let x = 3 in ...`
-->


1. Jacobson's reader monad only allows for establishing a single binding
relationship at a time.  It requires considerable cleverness to deploy
her combinators in a way that establishes multiple binding
relationships, as in 

        John_x thinks Mary_y said he_x likes her_y.

    See her 1999 paper for details. Essentially, she ends up layering several
Reader monads over each other.

    Here is [[code for the arithmetic tree Chris presented in week 8|code/arith1.ml]]. It computes
`\n. (+ 1 (* (/ 6 n) 4))`.  Your task is to modify it to compute
`\n m. (+ 1 (* (/ 6 n) m))`.  You will need to modify five lines.
The first one is the type of a boxed int.  Instead of `type num = int
-> int`, you'll need

        type num = int -> int -> int

    The second and third are the definitions of `mid` and `map2`. The fourth
is the one that encodes the variable `n`, the line that begins `(Leaf
(Num (fun n -> ...`.  The fifth line you need to modify is the one
that replaces "4" with "m".  When you have these lines modified,
you should be able to execute the following expression:

        # match eval t2 with Leaf (Num f) -> f 2 4;;
        - : int = 13

2. Based on [[the evaluator code from assignment 7|/exercises/assignment7/#index3h2]], and what you've learned about the Reader monad,
enhance the arithmetic tree code to handle an arbitrary set of free variables. Don't use Juli8 libraries for this; just do it by hand.
Return to the original code (that is, before the modifications required by the previous problem).

    Start like this:

        type env = string -> int
        type num = env -> int
        let my_env = fun var -> match var with "x" -> 2 | "y" -> 4 | _ -> 0;;

    When you have it working, try

        # match eval t2 with Leaf (Num f) -> f my_env;;
        - : int = 13

    For this problem you don't need to demonstrate how to implement binding expressions like `let x = 3 in ...`. You just need to compute the value of possibly open expressions, relative to the supplied `env`.

3. Okay, now what changes do you need to make to add in expressions like `let x = 3 in ...`

4. Add in the Option/Maybe monad.  Start here:

        type num = env -> int option

    Show that your code handles division by zero gracefully.

5. Consider the following code which uses the Juli8 libraries for OCaml.

        module S = Monad.State(struct type store = int end);;
        let xx = S.(mid 1 >>= fun x -> put 20 >> modify succ >> get >>= fun y -> mid [x;y]) in
        S.run xx 0

    Recall that `xx >> yy` is short for `xx >>= fun _ -> yy`. The equivalent Haskell code is:

        import Control.Monad.State
        let { xx :: State Int [Int];
              xx = return 1 >>= \x -> put 20 >> modify succ >> get >>= \y -> return [x,y] } in
        runState xx 0

    Or:

        import Control.Monad.State
        let { xx :: State Int [Int];
              xx = do { x <- return 1;
                        put 20;
                        modify succ;
                        y <- get;
                        return [x,y] } } in
        runState xx 0

    Don't try running the code yet. Instead, get yourself into a position to predict what it will do, by reading the past few discussions about the State monad. After you've made a prediction, then run the code and see if you got it right.

6. Here's another one:

        (* start with module S = ... as before *)
        let yy = S.(let xx = modify succ >> get in
           xx >>= fun x1 -> xx >>= fun x2 -> xx >>= fun x3 -> mid [x1;x2;x3]) in
        S.run yy 0

    The equivalent Haskell code is:

        import Control.Monad.State
        let { xx :: State Int Int;
              xx = modify succ >> get;
              yy = xx >>= \x1 -> xx >>= \x2 -> xx >>= \x3 -> return [x1,x2,x3] } in
        runState yy 0

    What is your prediction? What did OCaml or Haskell actually evaluate this to?

7. Suppose you're trying to use the State monad to keep a running tally of how often certain arithmetic operations have been used in computing a complex expression. You've come upon the design plan of using the same State monad module `S` from the previous problems, and defining a function like this:

        let counting_plus xx yy = S.(tick >> map2 (+) xx yy)

    How should you define the operation `tick` to make this work? The intended behavior is that after running:

        let zz = counting_plus (S.mid 1) (counting_plus (S.mid 2) (S.mid 3)) in
        S.run zz 0

    you should get a payload of `6` (`1+(2+3)`) and a final `store` of `2` (because `+` was used twice).

8. Instead of the design in the previous problem, suppose you had instead chosen to do things this way:

        let counting_plus xx yy = S.(map2 (+) xx yy >>= tock)

    How should you define the operation `tock` to make this work, with the same behavior as before?

    <!-- How would you expand your strategy, if you also wanted to be safe from division by zero? This is a deep question. How should you combine two monads into a single system? If you don't arrive at working code, you can still discuss the issues and design choices. -->

9. Here is how to create a monadic stack of a Reader monad transformer wrapped around an underlying Option monad:

        module O = Monad.Option (* not really necessary *)
        module R = Monad.Reader(struct type env = (* whatever *) end)
        module RO = R.T(O) (* wrap R's Transformer around O *)

    You can inspect the types that result by saying `#show RO.result` (in OCaml version >= 4.02), or by running:

        let env0 = (* some appropriate env, depending on how you defined R *) in
        let xx = RO.(mid 1) in RO.run xx env0

    and inspecting the type of the result. In Haskell:

        import Control.Monad.Reader
        -- substitute your own choices for the type Env and value env0
        let { xx :: ReaderT Env Maybe Int; xx = return 1 } in runReaderT xx env0

    Okay, here are some questions about various monad transformers. Use OCaml or Haskell to help you answer them. Which combined monad has the type of an optional list (that is, either `None` or `Some [...]`): an Option transformer wrapped around an underlying List monad, or a List transformer wrapped around an underlying Option monad? Which combined monad has the type of a function from `store`s to a pair `('a list, store)`: a List transformer wrapped around an underlying State monad or a State transformer wrapped around an underlying List monad?

The last two problems are non-monadic.

10. This is a question about native mutation mechanisms in languages that have them, like OCaml or Scheme. What an expression like this:

        let cell = ref 0 in
        let incr c = (let old = !cell in let () = cell := old + 1 in ()) in
        (incr cell, !cell, incr cell, incr cell)

    will evaluate to will be `((), n, (), ())` for some number `n` between `0` and `3`. But what number is sensitive to the details of OCaml's evaluation strategy for evaluating tuple expressions. How can you avoid that dependence? That is, how can you rewrite such code to force it that the values in the 4-tuple have been evaluated left-to-right? Show us a strategy that works no matter what the expressions in the tuple are, not just these particular ones. (But you can assume that the expressions all terminate.)

11. In the evaluator code for [[Week 7 homework|/exercises/assignment7]], we left the `LetRec` portions unimplemented. How might we implement these for the second, `env`-using interpreter? One strategy would be to interpret expressions like:

        letrec f = \x. BODY in
        TERM

    as though they really read:

        let f = FIX (\f x. BODY) in
        TERM

    for some fixed-point combinator `FIX`. And that would work, supposing you use some fixed point combinator like the "primed" ones we showed you earlier that work with eager/call-by-value evaluation strategies. But for this problem, we want you to approach the task a different way.

    Begin by deleting all the `module VA = ...` code that implements the substitute-and-repeat interpreter. Next, change the type of `env` to be an `(identifier * bound) list`. Add a line after the definition of that type that says `and bound = Plain of result | Recursive of identifier * identifier * term * env`. The idea here is that some variables will be bound to ordinary `result`s, and others will be bound to special structures we've made to keep track of the recursive definitions. These special structures are akin to the `Closure of identifier * term * env` we already added to the `term` (or really more properly `result`) datatype. For `Closure`s, the single `identifier` is the bound variable, the `term` is the body of the lambda abstract, and the `env` is the environment that is in place when some variable is bound to this lambda abstract. Those same parameters make up the last three arguments of our `Recursive` structure. The first argument in the `Recursive` structure is to hold the variable that our `letrec` construction binds to the lambda abstract. That is, in:

        letrec f = \x. BODY in
        TERM

    both of the variables `f` and `x` need to be interpreted specially when we evaluate `BODY`, and this is how we keep track of which variable is `f`.

    Just making those changes will require you to change some other parts of the interpreter to make it still work. Before trying to do anything further with `letrec`, try finding what parts of the code need to be changed to accommodate these modifications to our types. See if you can get the interpreter working again as well as it was before.

    OK, once you've done that, then add an extra line:

        | LetRec of identifier * term * term

    to the definition of the `term` datatype. (For `letrec IDENT1 = TERM1 in TERM2`. You can assume that `TERM1` is always a `Lambda` term.) Now what will you need to add to the `eval` function to get it to interpret these terms properly? This will take some thought, and a good understanding of how the other clauses in the `eval` function are working.

    Here's a conceptual question: why did we point you in the direction of complicating the type that environments associate variables with, rather than just adding a new clause to the `result` type, as we did with Closures?