Red / \ Blue \ / \ Green a b / \ c Red / \ d e(for any leaf values `a` through `e`), it should return:

Red / \ Blue \ / \ Green 1 1 / \ 1 Red / \ 2 29. (More challenging.) Assume you have a `color_tree` whose leaves are labeled with `int`s (which may be negative). For this problem, assume also that the the same color never labels multiple inner branches. Write a recursive function that reports which color has the greatest "score" when you sum up all the values of its descendent leaves. Since some leaves may have negative values, the answer won't always be the color at the tree root. In the case of ties, you can return whichever of the highest scoring colors you like. ## Search Trees ## (More challenging.) For the next problem, assume the following type definition: (* OCaml *) type search_tree = Nil | Inner of search_tree * int * search_tree -- Haskell data Search_tree = Nil | Inner Search_tree Int Search_tree deriving (Show) That is, its leaves have no labels and its inner nodes are labeled with `int`s. Additionally, assume that all the `int`s in branches descending to the left from a given node will be less than the `int` of that parent node, and all the `int`s in branches descending to the right will be greater. We can't straightforwardly specify this constraint in OCaml's or Haskell's type definitions. We just have to be sure to maintain it by hand. 10. Write a function `search_for` with the following type, as displayed by OCaml: type direction = Left | Right search_for : int -> search_tree -> direction list option Haskell would say instead: data Direction = Left | Right deriving (Eq, Show) search_for :: Int -> Search_tree -> Maybe [Direction] Your function should search through the tree for the specified `int`. If it's never found, it should return the value OCaml calls `None` and Haskell calls `Nothing`. If it finds the `int` right at the root of the `search_tree`, it should return the value OCaml calls `Some []` and Haskell calls `Just []`. If it finds the `int` by first going down the left branch from the tree root, and then going right twice, it should return `Some [Left; Right; Right]` or `Just [Left, Right, Right]`. ## More Map2s ## Above, you defined `maybe_map2` [WHERE]. Before we encountered `map2` for lists. There are in fact several different approaches to mapping two lists together. 11. One approach is to apply the supplied function to the first element of each list, and then to the second element of each list, and so on, until the lists are exhausted. If the lists are of different lengths, you might stop with the shortest, or you might raise an error. Different implementations make different choices about that. Let's call this function: (* OCaml *) map2_zip : ('a -> 'b -> 'c) -> ('a) list -> ('b) list -> ('c) list Write a recursive function that implements this, in Haskell or OCaml. Let's say you can stop when the shorter list runs out, if they're of different lengths. (OCaml and Haskell each already have functions in their standard libraries --- `map2` or `zipWith` -- that do this. And it also corresponds to a list comprehension you can write in Haskell like this: :set -XParallelListComp [ f x y | x <- xs | y <- ys ] But we want you to write this function from scratch.) 12. What is the relation between the function you just wrote, and the `maybe_map2` function you wrote for problem 2, above? 13. Another strategy is to take the *cross product* of the two lists. If the function: (* OCaml *) map2_cross : ('a -> 'b -> 'c) -> ('a) list -> ('b) list -> ('c) list is applied to the arguments `f`, `[x0, x1, x2]`, and `[y0, y1]`, then the result should be: `[f x0 y0, f x0 y1, f x1 y0, f x1 y1, f x2 y0, f x2 y1]`. Write this function. A similar choice between "zipping" and "crossing" could be made when `map2`-ing two trees. For example, the trees:

0 5 / \ / \ 1 2 6 7 / \ / \ 3 4 8 9could be "zipped" like this (ignoring any parts of branches on the one tree that extend farther than the corresponding branch on the other):

f 0 5 / \ f 1 6 f 2 714. You can try defining that if you like, for extra credit. "Crossing" the trees would instead add copies of the second tree as subtrees replacing each leaf of the original tree, with the leaves of that larger tree labeled with `f` applied to `3` and `6`, then `f` applied to `3` and `8`, and so on across the fringe of the second tree; then beginning again (in the subtree that replaces the `4` leaf) with `f` applied to `4` and `6`, and so on. * In all the plain `map` functions, whether for lists, or for `option`/`Maybe`s, or for trees, the structure of the result exactly matched the structure of the argument. * In the `map2` functions, whether for lists or for `option`/`Maybe`s or for trees, and whether done in the "zipping" style or in the "crossing" style, the structure of the result may be a bit different from the structure of the arguments. But the *structure* of the arguments is enough to determine the structure of the result; you don't have to look at the specific list elements or labels on a tree's leaves or nodes to know what the *structure* of the result will be. * We can imagine more radical transformations, where the structure of the result *does* depend on what specific elements the original structure(s) had. For example, what if we had to transform a tree by turning every leaf into a subtree that contained all of those leaf's prime factors? Or consider our problem from last week [WHERE] where you converted `[3, 2, 0, 1]` not into `[[3,3,3], [2,2], [], [1]]` --- which still has the same structure, that is length, as the original --- but rather into `[3, 3, 3, 2, 2, 1]` --- which doesn't. (Some of you had the idea last week to define this last transformation in Haskell as `[x | x <- [3,2,0,1], y <- [0..(x-1)]]`, which just looks like a cross product, that we counted under the *previous* bullet point. However, in that expression, the second list's structure depends upon the specific values of the elements in the first list. So it's still true, as I said, that you can't specify the structure of the output list without looking at those elements.) These three levels of how radical a transformation you are making to a structure, and the parallels between the transformations to lists, to `option`/`Maybe`s, and to trees, will be ideas we build on in coming weeks. ## Untyped Lambda Terms ## In OCaml, you can define some datatypes that represent terms in the untyped Lambda Calculus like this: type identifier = string type lambda_term = Var of identifier | Abstract of identifier * _____ | App of _____ We've left some gaps. In Haskell, you'd define it instead like this: type Identifier = String data Lambda_term = Var Identifier | Abstract Identifier _____ | App ________ 15. Again, we've left some gaps. Choose one of these languages and fill in the gaps to complete the definition. 16. Write a function `occurs_free` that has the following type: occurs_free : identifier -> lambda_term -> bool That's how OCaml would show it. Haskell would use double colons `::` instead, and would also capitalize all the type names. Your function should tell us whether the supplied identifier ever occurs free in the supplied `lambda_term`. ## Encoding Booleans, Church numerals, and Right-Fold Lists in System F ## (For the System F questions, you can either work on paper, or download and compile Pierce's evaluator for system F to test your work [WHERE].) Let's think about the encodings of booleans, numerals and lists in System F, and get datatypes with the same form working in OCaml or Haskell. (Of course, OCaml and Haskell have *native* versions of these types: OCaml's `true`, `1`, and `[1;2;3]`. But the point of our exercise requires that we ignore those.) Recall from class System F, or the polymorphic λ-calculus, with this grammar: types ::= constants | α ... | type1 -> type2 | ∀α. type expressions ::= x ... | λx:type. expr | expr1 expr2 | Λα. expr | expr [type] The boolean type, and its two values, may be encoded as follows: Bool ≡ ∀α. α -> α -> α true ≡ Λα. λy:α. λn:α. y false ≡ Λα. λy:α. λn:α. n It's used like this: b [T] res1 res2 where `b` is a `Bool` value, and `T` is the shared type of `res1` and `res2`. 17. How should we implement the following terms? Note that the result of applying them to the appropriate arguments should also give us a term of type `Bool`. (a) the term `not` that takes an argument of type `Bool` and computes its negation (b) the term `and` that takes two arguments of type `Bool` and computes their conjunction (c) the term `or` that takes two arguments of type `Bool` and computes their disjunction The type `Nat` (for "natural number") may be encoded as follows: Nat ≡ ∀α. (α -> α) -> α -> α zero ≡ Λα. λs:α -> α. λz:α. z succ ≡ λn:Nat. Λα. λs:α -> α. λz:α. s (n [α] s z) A number `n` is deﬁned by what it can do, which is to compute a function iterated `n` times. In the polymorphic encoding above, the result of that iteration can be any type `α`, as long as your function is of type `α -> α` and you have a base element of type `α`. 18. Translate these encodings of booleans and Church numbers into OCaml or Haskell, implementing versions of `sysf_bool`, `sysf_true`, `sysf_false`, `sysf_nat`, `sysf_zero`, `sysf_iszero` (this is what we'd earlier write as `zero?`, but you can't use `?`s in function names in OCaml or Haskell), `sysf_succ`, and `sysf_pred`. We include the `sysf_` prefixes so as not to collide with any similarly-named native functions or values in these languages. Keep in mind the capitalization rules. In OCaml, types are written `sysf_bool`, and in Haskell, they are capitalized `Sysf_bool`. In both languages, variant/constructor tags (like `None` or `Some`) are capitalized, and function names start lowercase. But for this problem, you shouldn't need to use any variant/constructor tags. To get you started, here is how to define `sysf_bool` and `sysf_true` in OCaml: type ('a) sysf_bool = 'a -> 'a -> 'a let sysf_true : ('a) sysf_bool = fun y n -> y And here in Haskell: type Sysf_bool a = a -> a -> a -- this is a case where Haskell does use `type` instead of `data` -- Now, to my mind the natural thing to write here would be: let sysf_true :: Sysf_bool a = \y n -> y -- But for complicated reasons, that won't work, and you need to do this instead: let { sysf_true :: Sysf_bool a; sysf_true = \y n -> y } -- Or this: let sysf_true = (\y n -> y) :: Sysf_bool a Note that in both OCaml and the Haskell code, `sysf_true` can be applied to further arguments directly: sysf_true 10 20 You don't do anything like System F's `true [int] 10 20`. The OCaml and Haskell interpreters figure out what type `sysf_true` needs to be specialized to (in this case, to `int`), and do that automatically. It's especially useful for you to implement a version of a System F encoding `pred`, starting with one of the (untyped) versions available in the lambda library accessible from the main wiki page. [WHERE] The point of the exercise is to do these things on your own, so avoid using the built-in OCaml or Haskell booleans and integers. Consider the following list type, specified using OCaml or Haskell datatypes: (* OCaml *) type ('a) my_list = Nil | Cons of 'a * 'a my_list -- Haskell data My_list a = Nil | Cons a (My_list a) We can encode that type into System F as a right-fold, just as we did in the untyped Lambda Calculus, like this: list_T ≡ ∀α. (T -> α -> α) -> α -> α nil_T ≡ Λα. λc:T -> α -> α. λn:α. n cons_T ≡ λx:T. λxs:list_T. Λα. λc:T -> α -> α. λn:α. c x (xs [α] c n) As with `Nat`s, the natural recursion is built into our encoding of the list datatype. There is some awkwardness here, because System F doesn't have any parameterized types like OCaml's `('a) list` or Haskell's `[a]`. For those, we need to use a more complex system called System F_ω. System F *can* already define a more general polymorphic list type: list ≡ ∀β. ∀α. (β -> α -> α) -> α -> α But this is more awkward to work with, because for functions like `map` we want to give them not just the type: (S -> T) -> list -> list but more specifically, the type: (S -> T) -> list [S] -> list [T] Yet we haven't given ourselves the capacity to talk about `list [S]` and so on as a type. Hence, I'll just use the more clumsy, ad hoc specification of `map`'s type as: FIXME qua (S -> T) -> list_S -> list_T 19. Convert this list encoding and the `map` function to OCaml or Haskell. Call it `sysf_list`, `sysf_nil` and so on, to avoid collision with the names for native lists in these languages. (In OCaml and Haskell you *can* say `('a) sysf_list` or `Sysf_list a`.) 20. Also give us the type and definition for a `sysf_head` function. Think about what value to give back if its argument is the empty list. Ultimately, we might want to make use of the `option`/`Maybe` technique explored in questions 1--2, but for this assignment, just pick a strategy, no matter how clunky. 21. Modify the implementation of the predecessor function [[given in the class notes|topics/week5_system_f]] [WHERE] to implement a `sysf_tail` function for your lists. Be sure to test your proposals with simple lists. (You'll have to `sysf_cons` up a few sample lists yourself; don't expect OCaml or Haskell to magically translate between their native lists and the ones you've just defined.) ## More on Types ## 22. Recall that the **S** combinator is given by `\f g x. f x (g x)`. Give two different typings for this term in OCaml or Haskell. To get you started, here's one typing for **K**: # let k (y:'a) (n:'b) = y ;; val k : 'a -> 'b -> 'a = [fun] # k 1 true ;; - : int = 1 If you can't understand how one term can have several types, recall our discussion in this week's notes [WHERE] of "principal types". ## Evaluation Order ## Do these last three problems specifically with OCaml in mind, not Haskell. Analogues of the questions exist in Haskell, but because the default evaluation rules for these languages are different, it's too complicated to look at how these questions should be translated into the Haskell setting. 23. Which of the following expressions is well-typed in OCaml? For those that are, give the type of the expression as a whole. For those that are not, why not? let rec f x = f x let rec f x = f f let rec f x = f x in f f let rec f x = f x in f () let rec f () = f f let rec f () = f () let rec f () = f () in f f let rec f () = f () in f () 24. Throughout this problem, assume that we have: let rec blackhole x = blackhole x All of the following are well-typed. Which ones terminate? What generalizations can you make? blackhole blackhole () fun () -> blackhole () (fun () -> blackhole ()) () if true then blackhole else blackhole if false then blackhole else blackhole if true then blackhole else blackhole () if false then blackhole else blackhole () if true then blackhole () else blackhole if false then blackhole () else blackhole if true then blackhole () else blackhole () if false then blackhole () else blackhole () let _ = blackhole in 2 let _ = blackhole () in 2 25. This problem aims to get you thinking about how to control order of evaluation. Here is an attempt to make explicit the behavior of `if ... then ... else ...` explored in the previous question. The idea is to define an `if ... then ... else ...` expression using other expression types. So assume that `yes` is any (possibly complex) OCaml expression, and `no` is any other OCaml expression (of the same type as `yes`!), and that `bool` is any boolean expression. Then we can try the following: `if bool then yes else no` should be equivalent to let b = bool in let y = yes in let n = no in match b with true -> y | false -> n This almost works. For instance, if true then 1 else 2 evaluates to 1, and let b = true in let y = 1 in let n = 2 in match b with true -> y | false -> n also evaluates to 1. Likewise, if false then 1 else 2 and let b = false in let y = 1 in let n = 2 in match b with true -> y | false -> n both evaluate to 2. However, let rec blackhole x = blackhole x in if true then blackhole else blackhole () terminates, but let rec blackhole x = blackhole x in let b = true in let y = blackhole in let n = blackhole () in match b with true -> y | false -> n does not terminate. Incidentally, using the shorter `match bool with true -> yes | false -> no` rather than the longer `let b = bool ... in match b with ...` *would* work as we desire. But your assignment is to control the evaluation order *without* using the special evaluation order properties of OCaml's native `if` or of its `match`. That is, you must keep the `let b = ... in match b with ...` structure in your answer, though you are allowed to adjust what `b`, `y`, and `n` get assigned to. [[hints/assignment 5 hint 1]] WHERE