1 <!-- λ Λ ∀ ≡ α β ω Ω -->
4 ## Option / Maybe Types ##
6 You've already (in [[assignment1]] and [[/topics/week2 encodings]]) defined and worked with `map` as a function on lists. Now we're going to work instead with the type OCaml defines like this:
8 type ('a) option = None | Some of 'a
10 and Haskell defines like this:
12 data Maybe a = Nothing | Just a
14 That is, instances of this type are either an instance of `'a` (this can be any type), wrapped up in a `Some` or `Just` box, or they are a separate value representing a failure. This is sort of like working with a set or a list guaranteed to be either singleton or empty.
16 In one of the homework meetings, Chris posed the challenge: you know those dividers they use in checkout lines to separate your purchases from the next person's? What if you wanted to buy one of those dividers? How could they tell whether it belonged to your purchases or was separating them from others?
18 The OCaml and Haskell solution is to use not supermarket dividers but instead those gray bins from airport security. If you want to buy something, it goes into a bin. (OCaml's `Some`, Haskell's `Just`). If you want to separate your stuff from the next person, you send an empty bin (OCaml's `None`, Haskell's `Nothing`). If you happen to be buying a bin, OK, you put that into a bin. In OCaml it'd be `Some None` (or `Some (Some stuff)` if the bin you're buying itself contains some stuff); in Haskell `Just Nothing`. This way, we can't confuse a bin that contains a bin with an empty bin. (Not even if the contained bin is itself empty.)
20 1. Your first problem is to write a `maybe_map` function for these types. Here is the type of the function you should write:
23 maybe_map : ('a -> 'b) -> ('a) option -> ('b) option
26 maybe_map :: (a -> b) -> Maybe a -> Maybe b
28 If your `maybe_map` function is given a `None` or `Nothing` as its second argument, that should be what it returns. Otherwise, it should apply the function it got as its first argument to the contents of the `Some` or `Just` bin that it got as its second, and return the result, wrapped back up in a `Some` or `Just`. (Yes, we know that the `fmap` function in Haskell already implements this functionality. Your job is to write it yourself.)
30 One way to extract the contents of an `option`/`Maybe` value is to pattern match on that value, as you did with lists. In OCaml:
43 Some other tips: In OCaml you write recursive functions using `let rec`, in Haskell you just use `let` (it's already assumed to be recursive). In OCaml when you finish typing something and want the interpreter to parse it, check and display its type, and evaluate it, type `;;` and then return. In Haskell, if you want to display the type of an expression `expr`, type `:t expr` or `:i expr`.
45 You may want to review [[Rosetta pages|/rosetta1]] and also read some of the tutorials we linked to for [[/learning OCaml]] or [[/learning Haskell]].
47 HERE IS AN OCAML ANSWER:
49 let maybe_map (f: 'a -> 'b) -> (u : 'a option) : 'b option =
51 | Some a -> Some (f a)
55 2. Next write a `maybe_map2` function. Its type should be:
58 maybe_map2 : ('a -> 'b -> 'c) -> ('a) option -> ('b) option -> ('c) option
61 maybe_map2 :: (a -> b -> c) -> Maybe a -> Maybe b -> Maybe c
64 HERE IS AN OCAML ANSWER:
66 let maybe_map2 (f : 'a -> 'b -> 'c) (u : 'a option) (v : 'b option) : 'c option =
68 | Some a, Some b -> Some (f a b)
76 (The questions on Color and Search Trees are adapted from homeworks in Chapters 1 and 2 of Friedman and Wand, *Essentials of Programming Languages*.)
78 Here are type definitions for one kind of binary tree:
81 type color = Red | Green | Blue | ... (* you can add as many as you like *)
82 type ('a) color_tree = Leaf of 'a | Branch of 'a color_tree * color * 'a color_tree
85 data Color = Red | Green | Blue | ... deriving (Eq, Show)
86 data Color_tree a = Leaf a | Branch (Color_tree a) Color (Color_tree a) deriving (Show)
88 These trees always have colors labeling their inner branching nodes, and will have elements of some type `'a` labeling their leaves. `(int) color_tree`s will have `int`s there, `(bool) color_tree`s will have `bool`s there, and so on. The `deriving (Eq, Show)` part at the end of the Haskell declarations is boilerplate to tell Haskell you want to be able to compare the colors for equality, and also that you want the Haskell interpreter to display colors and trees to you when they are the result of evaluating an expression.
90 Here's how you create an instance of such a tree:
93 let t1 = Branch (Leaf 1, Red, Branch (Leaf 2, Green, Leaf 0))
96 let t1 = Branch (Leaf 1) Red (Branch (Leaf 2) Green (Leaf 0))
98 Here's how you pattern match such a tree, binding variables to its components:
103 | Branch (_, col, _) -> col = Red
108 Branch _ col _ -> col == Red
111 These expressions query whether `t` is a branching `color_tree` (that is, not a leaf) whose root is labeled `Red`.
113 Notice that for equality, you should use single `=` in OCaml and double `==` in Haskell. (Double `==` in OCaml will often give the same results, but it is subtly different in ways we're not yet in a position to explain.) *In*equality is expressed as `<>` in OCaml and `/=` in Haskell.
115 Choose one of these languages and write the following functions.
118 3. Define a function `tree_map` whose type is (as shown by OCaml): `('a -> 'b) -> ('a) color_tree -> ('b) color_tree`. It expects a function `f` and an `('a) color_tree`, and returns a new tree with the same structure and inner branch colors as the original, but with all of its leaves now having had `f` applied to their original value. So for example, `map (fun x->2*x) t1` would return `t1` with all of its leaf values doubled.
120 HERE IS A DIRECT OCAML ANSWER:
122 let rec tree_map (f: 'a -> 'b) (t: 'a color_tree) : 'b color_tree =
124 | Leaf a -> Leaf (f a)
125 | Branch (l,col,r) ->
126 let l' = tree_map f l in
127 let r' = tree_map f r in
130 IT MIGHT BE USEFUL TO GENERALIZE THIS PATTERN, LIKE SO:
132 let rec tree_walker (leaf_handler: 'a -> 'h) (joiner : 'h -> color -> 'h -> 'h) (t: 'a color_tree) : 'h =
134 | Leaf a -> leaf_handler a
135 | Branch (l,col,r) ->
136 let lh = tree_walker leaf_handler joiner l in
137 let rh = tree_walker leaf_handler joiner r in
140 <!-- traverse (k: a-> Mb) t = Leaf a -> map Leaf (k a) | Branch(l,col,r) -> map3 Branch l' c' r' -->
142 THEN `tree_map f t` can be defined as `tree_walker (fun a -> Leaf (f a)) (fun l' col r' -> Branch(l',col,r')) t`
145 4. Define a function `tree_foldleft` that accepts an argument `g : 'z -> 'a -> 'z` and a seed value `z : 'z` and a tree `t : ('a) color_tree`, and returns the result of applying `g` first to `z` and `t`'s leftmost leaf, and then applying `g` to *that result* and `t`'s second-leftmost leaf, and so on, all the way across `t`'s fringe. In our examples, only the leaf values affect the result; the inner branch colors are ignored.
147 HERE IS A DIRECT OCAML ANSWER:
149 let rec tree_foldleft (g: 'z -> 'a -> 'z) (z : 'z) (t: 'a color_tree) : 'z =
153 let z' = tree_foldleft g z l in
154 let z'' = tree_foldleft g z' r in
157 HERE IS AN ANSWER THAT RE-USES OUR `tree_walker` FUNCTION FROM THE PREVIOUS ANSWER:
159 let tree_foldleft g z t =
160 let leaf_handler a = fun z -> g z a in
161 let joiner lh _ rh = fun z -> rh (lh z) in
162 let expects_z = tree_walker leaf_handler joiner t in
165 <!-- traverse (k: a-> Mb) t = Leaf a -> map Leaf (k a) | Branch(l,col,r) -> map3 Branch l' c' r' -->
166 <!-- linearize (f: a->Mon) t = Leaf a -> f a | Branch(l,col,r) -> l' && [col' &&] r' -->
168 If you look at the definition of `tree_walker` above, you'll see that its interface doesn't supply the `leaf_handler` function with any input like `z`; the `leaf_handler` only gets the content of each leaf to work on. Thus we're forced to make our `leaf_handler` return a function, that will get its `z` input later. (The strategy used here is like [[the strategy for reversing a list using fold_right in assignment2|assignment2_answers/#cps-reverse]].) Then the `joiner` function chains the results of handling the two branches together, so that when the seed `z` is supplied, we feed it first to `lh` and then the result of that to `rh`. The result of processing any tree then will be a function that expects a `z` argument. Finally, we supply the `z` argument that `tree_foldleft` was invoked with.
171 5. How would you use the function defined in problem 4 (the previous problem) to sum up the values labeling the leaves of an `(int) color_tree`?
173 ANSWER: `tree_foldleft (fun z a -> z + a) 0 your_tree`
175 6. How would you use the function defined in problem 4 to enumerate a tree's fringe? (Don't worry about whether it comes out left-to-right or right-to-left.)
177 ANSWER: `tree_foldleft (fun z a -> a :: z) [] your_tree`
179 7. Write a recursive function to make a copy of a `color_tree` with the same structure and inner branch colors, but where the leftmost leaf is now labeled `0`, the second-leftmost leaf is now labeled `1`, and so on. (Here's a [[hint|assignment5 hint3]], if you need one.)
181 HERE IS A DIRECT OCAML ANSWER, FOLLOWING [[the hint|assignment5_hint3]]:
183 let rec enumerate_from (t:'a color_tree) counter =
185 | Leaf x -> (Leaf counter, counter+1)
186 | Branch (left,col,right) -> let (left',counter') = enumerate_from left counter in
187 let (right',counter'') = enumerate_from right counter' in
188 (Branch (left',col,right'), counter'')
190 (* then this will be the function we were asked for *)
192 let (t', _) = enumerate_from t 0 in
195 IT WOULD ALSO BE POSSIBLE TO WRITE THIS USING OUR `tree_walker` FUNCTION, USING A TECHNIQUE THAT COMBINES THE STRATEGY USED ABOVE WITH THE ONE USED IN `tree_foldleft`:
198 let leaf_handler a = fun counter -> (Leaf counter, counter+1) in
199 let joiner lh col rh =
201 let (left',counter') = lh counter in
202 let (right',counter'') = rh counter' in
203 (Branch (left',col,right'), counter'') in
204 fst (tree_walker leaf_handler joiner t 0)
207 8. (More challenging.) Write a recursive function that makes a copy of a `color_tree` with the same structure and inner branch colors, but replaces each leaf label with the `int` that reports how many of that leaf's ancestors are labeled `Red`. For example, if we give your function a tree:
220 (for any leaf values `a` through `e`), it should return:
233 HERE IS A DIRECT OCAML SOLUTION:
235 let rec tree_red_ancestors_from (cur : int) (t : 'a tree) : int tree =
238 | Branch(l, col, r) ->
239 let cur' = if col = Red then cur + 1 else cur in
240 let l' = tree_red_ancestors_from cur' l in
241 let r' = tree_red_ancestors_from cur' r in
244 (* here is the function we were asked for *)
245 let tree_red_ancestors t = tree_red_ancestors_from 0 t
248 HERE IS HOW TO DO IT USING `tree_walker`:
250 let tree_red_ancestors t =
251 let leaf_handler a = fun cur -> Leaf cur in
252 let joiner lh col rh = fun cur ->
253 let cur' = if col = Red then cur + 1 else cur in
254 Branch(lh cur', col, rh cur') in
255 tree_walker leaf_handler joiner t 0
258 9. (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 no color labels multiple `Branch`s (non-leaf nodes). Write a recursive function that reports which color has the greatest "score" when you sum up all the values of its descendant 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.
260 HERE IS A DIRECT OCAML SOLUTION:
262 type maybe_leader = (color * int) option
264 let rec tree_best_sofar (t : 'a color_tree) (lead : maybe_leader) : maybe_leader * int =
266 | Leaf a -> (None, a)
267 | Branch(l, col, r) ->
268 let (lead',left_score) = tree_best_sofar l lead in
269 let (lead'',right_score) = tree_best_sofar r lead' in
270 let my_score = left_score + right_score in
272 | None -> Some(col,my_score), my_score
273 | Some(_, lead_score) -> (if my_score > lead_score then Some(col,my_score) else lead''), my_score)
275 (* Note that the invocations of that function have to return who-is-the-current-leader?
276 plus their OWN score, even if they are not the current leader. Their parents need the
277 second value to calculate whether they should become the current leader. *)
279 (* here is the function we were asked for *)
281 match tree_best_sofar t None with
282 | Some(leader,leader_score), _ -> leader
283 | None, _ -> failwith "no colors"
286 HERE IS HOW TO DO IT USING `tree_walker`:
288 let tree_best_sofar t =
289 let leaf_handler a = fun leader -> leader, a in
290 let joiner lh col rh = fun leader ->
291 let (leader',left_score) = lh leader in
292 let (leader'',right_score) = rh leader' in
293 let my_score = left_score + right_score in
295 | None -> Some(col,my_score), my_score
296 | Some(_,leader_score) -> (if my_score > leader_score then Some(col,my_score) else leader''), my_score) in
297 tree_walker leaf_handler joiner t
299 Then `tree_best` could be defined as in the direct answer.
304 (More challenging.) For the next problem, assume the following type definition:
307 type search_tree = Nil | Inner of search_tree * int * search_tree
310 data Search_tree = Nil | Inner Search_tree Int Search_tree deriving (Show)
312 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.
314 10. Write a function `search_for` with the following type, as displayed by OCaml:
316 type direction = Left | Right
317 search_for : int -> search_tree -> direction list option
319 Haskell would say instead:
321 data Direction = Left | Right deriving (Eq, Show)
322 search_for :: Int -> Search_tree -> Maybe [Direction]
324 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]`.
326 HERE IS AN OCAML ANSWER:
328 let search_for (sought : int) (t : search_tree) : direction list option =
329 let rec aux (trace : direction list) t =
332 | Inner(l,x,r) when sought < x -> aux (Left::trace) l
333 | Inner(l,x,r) when sought > x -> aux (Right::trace) r
334 | _ -> Some(List.rev trace) in
340 In question 2 above, you defined `maybe_map2`. [[Before|assignment1]] we encountered `map2` for lists. There are in fact several different approaches to mapping two lists together.
342 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:
345 map2_zip : ('a -> 'b -> 'c) -> ('a) list -> ('b) list -> ('c) list
347 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:
349 :set -XParallelListComp
350 [ f x y | x <- xs | y <- ys ]
352 <!-- or `f <$/fmap> ZipList xs <*/ap> ZipList ys`; or `pure f <*> ...`; or `liftA2 f (ZipList xs) (ZipList ys)` -->
353 But we want you to write this function from scratch.)
355 HERE IS AN OCAML ANSWER:
357 let rec map2_zip f xs ys =
361 | x'::xs', y'::ys' -> f x' y' :: map2_zip f xs' ys'
363 12. What is the relation between the function you just wrote, and the `maybe_map2` function you wrote for problem 2, above?
365 ANSWER: option/Maybe types are like lists constrained to be of length 0 or 1.
367 13. Another strategy is to take the *cross product* of the two lists. If the function:
370 map2_cross : ('a -> 'b -> 'c) -> ('a) list -> ('b) list -> ('c) list
372 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.
373 <!-- in Haskell, `liftA2 f xs ys` -->
375 HERE IS AN OCAML ANSWER:
377 let rec map2_cross f xs ys =
380 | x'::xs' -> List.append (List.map (f x') ys) (map2_cross f xs' ys)
382 A similar choice between "zipping" and "crossing" could be made when `map2`-ing two trees. For example, the trees:
392 could be "zipped" like this (ignoring any parts of branches on the one tree that extend farther than the corresponding branch on the other):
400 14. You can try defining that if you like, for extra credit.
402 "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.
404 * 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.
406 * 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.
408 * 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 [[assignment3]] where you converted `[3, 1, 0, 2]` not into `[[3,3,3], [1], [], [2,2]]` --- which still has the same structure, that is length, as the original --- but rather into `[3, 3, 3, 1, 2, 2]` --- which doesn't.
409 (Some of you had the idea last week to define this last transformation in Haskell as `[x | x <- [3,1,0,2], 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.)
411 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.
417 ## Untyped Lambda Terms ##
419 In OCaml, you can define some datatypes that represent terms in the untyped Lambda Calculus like this:
421 type identifier = string
422 type lambda_term = Var of identifier | Abstract of identifier * _____ | App of _____
424 We've left some gaps.
426 In Haskell, you'd define it instead like this:
428 type Identifier = String
429 data Lambda_term = Var Identifier | Abstract Identifier _____ | App ________ deriving (Show)
431 Again, we've left some gaps. (The use of `type` for the first line in Haskell and `data` for the second line is not a typo. The first specifies that `Identifier` will be just a shorthand for an already existing type. The second introduces a new datatype, with its own variant/constructor tags.)
433 15. Choose one of these languages and fill in the gaps to complete the definition.
435 HERE IS AN OCAML DEFINITION:
437 type lambda_term = Var of identifier | Abstract of identifier * lambda_term | App of lambda_term * lambda_term
439 16. Write a function `occurs_free` that has the following type:
441 occurs_free : identifier -> lambda_term -> bool
443 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`.
445 HERE IS AN OCAML ANSWER:
447 let rec occurs_free (ident : identifier) (term : lambda_term) : bool =
449 | Var var_ident -> ident = var_ident (* `x` is free in Var "x" but not in Var "y" *)
450 | Abstract(bound_ident, body) -> ident <> bound_ident && occurs_free ident body (* `x` is free in \y. x but not in \x. blah or \y. y *)
451 | App (head, arg) -> occurs_free ident head || occurs_free ident arg
455 ## Encoding Booleans, Church numerals, and Right-Fold Lists in System F ##
457 <!-- These questions are adapted from web materials by Umut Acar. Were at <http://www.mpi-sws.org/~umut/>. Now he's moved to <http://www.umut-acar.org/> and I can't find the page anymore. -->
459 (For the System F questions, you can either work on paper, or [download and compile](http://www.cis.upenn.edu/~bcpierce/tapl/resources.html#checkers) Pierce's evaluator for system F to test your work. Under the "implementations" link on that page, you want to use Pierce's `fullpoly` or the `fullomega` code. The Chapters of Pierce's book *Types and Programming Languages* most relevant to this week's lectures are 22 and 23; though for context we also recommend at least Chapters 8, 9, 11, 20, and 29. We don't expect most of you to follow these recommendations now, or even to be comfortable enough yet with the material to be *able* to. We're providing the pointers as references that some might conceivably pursue now, and others later.)
462 Let's think about the encodings of booleans, numerals and lists in System F,
463 and get datatypes with the same form working in OCaml or Haskell. (Of course, OCaml and Haskell
464 have *native* versions of these types: OCaml's `true`, `1`, and `[1;2;3]`.
465 But the point of our exercise requires that we ignore those.)
467 Recall from class System F, or the polymorphic λ-calculus, with this grammar:
469 types ::= constants | α ... | type1 -> type2 | ∀α. type
470 expressions ::= x ... | λx:type. expr | expr1 expr2 | Λα. expr | expr [type]
472 The boolean type, and its two values, may be encoded as follows:
474 Bool ≡ ∀α. α -> α -> α
475 true ≡ Λα. λy:α. λn:α. y
476 false ≡ Λα. λy:α. λn:α. n
482 where `b` is a `Bool` value, and `T` is the shared type of `res1` and `res2`.
485 17. How should we implement the following terms? Note that the result
486 of applying them to the appropriate arguments should also give us a term of
489 (a) the term `not` that takes an argument of type `Bool` and computes its negation
490 (b) the term `and` that takes two arguments of type `Bool` and computes their conjunction
491 (c) the term `or` that takes two arguments of type `Bool` and computes their disjunction
495 Bool ≡ ∀α. α -> α -> α
496 true ≡ Λα. λy:α. λn:α. y
497 false ≡ Λα. λy:α. λn:α. n
498 not ≡ λp:Bool. p [Bool] false true
499 and ≡ λp:Bool. λq:Bool. p [Bool] q false
500 or ≡ λp:Bool. λq:Bool. p [Bool] true q
502 When I first wrote up these answers, I had put `(q [Bool])` where I now have just `q` in the body of `and` and `or`. On reflection,
503 this isn't necessary, because `q` is already a `Bool`. But as I learned by checking these answers with Pierce's evaluator, it's
504 also a mistake. What we want is a result whose type _is_ `Bool`, that is, `∀α. α -> α -> α`. `(q [Bool])` doesn't have that type, but
505 rather the type `Bool -> Bool -> Bool`. The first, desired, type has an outermost `∀`. The second, wrong type doesn't; it only has `∀`s
506 inside the antecedents and consequents of the various arrows. The last one of those could be promoted to be an outermost `∀`, since
507 `P -> ∀α. Q ≡ ∀α. P -> Q` when `α` is not free in `P`. But that couldn't be done with the others.
510 The type `Nat` (for "natural number") may be encoded as follows:
512 Nat ≡ ∀α. (α -> α) -> α -> α
513 zero ≡ Λα. λs:α -> α. λz:α. z
514 succ ≡ λn:Nat. Λα. λs:α -> α. λz:α. s (n [α] s z)
516 A number `n` is defined by what it can do, which is to compute a function iterated `n`
517 times. In the polymorphic encoding above, the result of that iteration can be
518 any type `α`, as long as your function is of type `α -> α` and you have a base element of type `α`.
521 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. 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.
523 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.
525 To get you started, here is how to define `sysf_bool` and `sysf_true` in OCaml:
527 type ('a) sysf_bool = 'a -> 'a -> 'a
528 let sysf_true : ('a) sysf_bool = fun y n -> y
532 type Sysf_bool a = a -> a -> a -- this is another case where Haskell uses `type` instead of `data`
533 -- To my mind the natural thing to write next would be:
534 let sysf_true :: Sysf_bool a = \y n -> y
535 -- But for complicated reasons, that won't work, and you need to do this instead:
536 let { sysf_true :: Sysf_bool a; sysf_true = \y n -> y }
538 let sysf_true = (\y n -> y) :: Sysf_bool a
540 Note that in both OCaml and Haskell code, the generalization `∀α` on the free type variable `α` is implicit. If you really want to, you can supply it explicitly in Haskell by saying:
542 :set -XExplicitForAll
543 let { sysf_true :: forall a. Sysf_bool a; ... }
545 let { sysf_true :: forall a. a -> a -> a; ... }
547 You can't however, put a `forall a.` in the `type Sysf_bool ...` declaration. The reasons for this are too complicated to explain here.
549 Note also that `sysf_true` can be applied to further arguments directly:
553 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.
555 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 [[assignment3 answers]].
560 type 'a sysf_bool = 'a -> 'a -> 'a
561 let sysf_true : 'a sysf_bool = fun y n -> y
562 let sysf_false : 'a sysf_bool = fun y n -> n
564 type 'a sysf_nat = ('a -> 'a) -> 'a -> 'a
565 let sysf_zero : 'a sysf_nat = fun s z -> z
567 let sysf_iszero (n : ('a sysf_bool) sysf_nat) : 'a sysf_bool = n (fun _ -> sysf_false) sysf_true
568 (* Annoyingly, though, if you just say sysf_iszero sysf_zero, you'll get an answer that isn't fully polymorphic.
569 This is explained in the comments linked below. The way to get a polymorphic result is to say instead
570 `fun next -> sysf_iszero sysf_zero next`. *)
572 let sysf_succ (n : 'a sysf_nat) : 'a sysf_nat = fun s z -> s (n s z)
573 (* Again, to get a polymorphic result you'll want to call `fun next -> sysf_succ sysf_zero next` *)
575 And here is how to get `sysf_pred`, using a System-F-style encoding of pairs. (For brevity, I'll leave off the `sysf_` prefixes.)
577 type 'a natpair = ('a nat -> 'a nat -> 'a nat) -> 'a nat
578 let natpair (x : 'a nat) (y : 'a nat) : 'a natpair = fun f -> f x y
581 let shift (p : 'a natpair) : 'a natpair = natpair (succ (p fst)) (p fst)
582 let pred (n : ('a natpair) nat) : 'a nat = n shift (natpair zero zero) snd
584 (* As before, to get polymorphic results you need to eta-expand your applications. Witness:
585 # let one = succ zero;;
586 val one : '_a nat = <fun>
587 # let one next = succ zero next;;
588 val one : ('a -> 'a) -> 'a -> 'a = <fun>
589 # let two = succ one;;
590 val two : '_a nat = <fun>
591 # let two next = succ one next;;
592 val two : ('a -> 'a) -> 'a -> 'a = <fun>
595 # fun next -> pred two next;;
596 - : ('a -> 'a) -> 'a -> 'a = <fun>
599 Consider the following list type, specified using OCaml or Haskell datatypes:
602 type ('t) my_list = Nil | Cons of 't * 't my_list
605 data My_list t = Nil | Cons t (My_list t) deriving (Show)
607 We can encode that type into System F in terms of its right-fold, just as we did in the untyped Lambda Calculus, like this:
609 list_T ≡ ∀α. (T -> α -> α) -> α -> α
610 nil_T ≡ Λα. λc:T -> α -> α. λn:α. n
611 cons_T ≡ λx:T. λxs:list_T. Λα. λc:T -> α -> α. λn:α. c x (xs [α] c n)
613 As with `Nat`s, the natural recursion on the type is built into our encoding of it.
615 There is some awkwardness here, because System F doesn't have any parameterized types like OCaml's `('t) my_list` or Haskell's `My_list t`. For those, we need to use a more complex system called System F<sub>ω</sub>. System F *can* already define a more polymorphic list type:
617 list ≡ ∀τ. ∀α. (τ -> α -> α) -> α -> α
619 But this is more awkward to work with, because for functions like `map` we want to give them not just the type:
621 (T -> S) -> list -> list
623 but more specifically, the type:
625 (T -> S) -> list [T] -> list [S]
627 Yet we haven't given ourselves the capacity to talk about `list [S]` and so on as a type in System F. Hence, I'll just use the more clumsy, ad hoc specification of `map`'s type as:
629 (T -> S) -> list_T -> list_S
632 = λf:T -> S. λxs:list. xs [T] [list [S]] (λx:T. λys:list [S]. cons [S] (f x) ys) (nil [S])
635 *Update: Never mind, don't bother with the next three questions. They proved to be more difficult to implement in OCaml than we expected. Here is [[some explanation|assignment5 hint4]].*
637 19. Convert this list encoding and the `map` function to OCaml or Haskell. Again, call the type `sysf_list`, and the functions `sysf_nil`, `sysf_cons`, and `sysf_map`, to avoid collision with the names for native lists and functions in these languages. (In OCaml and Haskell you *can* say `('t) sysf_list` or `Sysf_list t`.)
639 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. It might be cleanest to use the `option`/`Maybe` technique explored in questions 1--2, but for this assignment, just pick a strategy, no matter how clunky.
641 21. Modify your implementation of the predecessor function for Church numerals, above, to implement a `sysf_tail` function for your lists.
643 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.)
651 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**:
653 # let k (y:'a) (n:'b) = y ;;
654 val k : 'a -> 'b -> 'a = [fun]
658 If you can't understand how one term can have several types, recall our discussion in this week's notes of "principal types".
660 ANSWER: OCaml shows the principal typing for S as follows:
662 # let s = fun f g x -> f x (g x);;
663 val s : ('a -> 'b -> 'c) -> ('a -> 'b) -> 'a -> 'c = <fun>
665 You can get a more specific type by unifying any two or more of the above type variables. Thus, here is one other type that S can have:
667 # let s' = (s : ('a -> 'b -> 'a) -> ('a -> 'b) -> 'a -> 'a);;
668 val s' : ('a -> 'b -> 'a) -> ('a -> 'b) -> 'a -> 'a = <fun>
671 ## Evaluation Order ##
673 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.
676 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?
680 c. let rec f x = f x in f f
681 d. let rec f x = f x in f ()
682 e. let rec f () = f f
683 f. let rec f () = f ()
684 g. let rec f () = f () in f f
685 h. let rec f () = f () in f ()
687 ANSWER: All are well-typed except for b, e, and g, which involve self-application. d and h are well-typed but enter an infinite loop. Surprisingly, OCaml's type-checker accepts c, and it too enters an infinite loop.
689 OCaml is not in principle opposed to self-application: `let id x = x in id id` works fine. Neither is it in principle opposed to recursive/infinite types. But it demands that they not be infinite chains of unbroken arrows. There have to be some intervening datatype constructors. Thus this is an acceptable type definition:
691 type 'a ok = Cons of ('a -> 'a ok)
695 type 'a notok = 'a -> 'a notok
697 (In the technical jargon, OCaml has isorecursive not equirecursive types.) In any case, the reason that OCaml rejects b is not the mere fact that it involves self-application, but the fact that typing it would require constructing one of the kinds of infinite chains of unbroken arrows that OCaml forbids. In case c, we can already see that the type of `f` is acceptable (it was ok in case a), and the self-application doesn't impose any new typing constraints because it never returns, so it can have any result type at all.
699 In cases e and g, the typing fails not specifically because of a self-application, but because OCaml has already determined that `f` has to take a `()` argument, and even before settling on `f`'s final type, one thing it knows about `f` is that it isn't `()`. So `let rec f () = f () in f f` fails for the same reason that `let rec f () = f () in f id` would.
702 24. Throughout this problem, assume that we have:
704 let rec blackhole x = blackhole x
706 <!-- Haskell could say: `let blackhole = \x -> fix (\f -> f)` -->
707 All of the following are well-typed. Which ones terminate? What generalizations can you make?
711 c. fun () -> blackhole ()
712 d. (fun () -> blackhole ()) ()
713 e. if true then blackhole else blackhole
714 f. if false then blackhole else blackhole
715 g. if true then blackhole else blackhole ()
716 h. if false then blackhole else blackhole ()
717 i. if true then blackhole () else blackhole
718 j. if false then blackhole () else blackhole
719 k. if true then blackhole () else blackhole ()
720 l. if false then blackhole () else blackhole ()
721 m. let _ = blackhole in 2
722 n. let _ = blackhole () in 2
724 ANSWERS: These terminate: a,c,e,f,g,j,m; these don't: b,d,h,i,k,l,n. Generalization: blackhole is a suspended infinite loop, that is forced by feeding it an argument. The expressions that feed blackhole an argument (in a context that is not itself an unforced suspension) won't terminate. Also, unselected clauses of `if`-terms aren't ever evaluated.
726 25. This problem aims to get you thinking about how to control order of evaluation.
727 Here is an attempt to make explicit the behavior of `if ... then ... else ...` explored in the previous question.
728 The idea is to define an `if ... then ... else ...` expression using
729 other expression types. So assume that `yes` is any (possibly complex) OCaml expression,
730 and `no` is any other OCaml expression (of the same type as `yes`!),
731 and that `bool` is any boolean expression. Then we can try the following:
732 `if bool then yes else no` should be equivalent to
737 match b with true -> y | false -> n
739 This almost works. For instance,
741 if true then 1 else 2
743 evaluates to `1`, and
745 let b = true in let y = 1 in let n = 2 in
746 match b with true -> y | false -> n
748 also evaluates to `1`. Likewise,
750 if false then 1 else 2
754 let b = false in let y = 1 in let n = 2 in
755 match b with true -> y | false -> n
757 both evaluate to `2`.
761 let rec blackhole x = blackhole x in
762 if true then blackhole else blackhole ()
766 let rec blackhole x = blackhole x in
769 let n = blackhole () in
770 match b with true -> y | false -> n
772 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.
774 Here's a [[hint|assignment5 hint1]].
778 let rec blackhole x = blackhole x in
780 let y () = blackhole in
781 let n () = blackhole () in
782 (match b with true -> y | false -> n) ()
784 If you said instead, on the last line:
786 match b with true -> y () | false -> n ()
788 that would arguably still be relying on the special evaluation order properties of OCaml's native `match`. You'd be assuming that `n ()` wouldn't be evaluated in the computation that ends up selecting the other branch. Your assumption would be correct, but to avoid making that assumption, you should instead first select the `y` or `n` result, _and then afterwards_ force the result. That's what we do in the above answer.
790 Question: don't the rhs of all the match clauses in `match b with true -> y | false -> n` have to have the same type? How can they, when one of them is `blackhole` and the other is `blackhole ()`? The answer has two parts. First is that an expression is allowed to have different types when it occurs several times on the same line: consider `let id x = x in (id 5, id true)`, which evaluates just fine. The second is that `blackhole` will get the type `'a -> 'b`, that is, it can have any functional type at all. So the principle types of `y` and `n` end up being `y : unit -> 'a -> 'b` and `n : unit -> 'c`. (The consequent of `n` isn't constrained to use the same type variable as the consequent of `y`.) OCaml can legitimately infer these to be the same type by unifying the types `'c` and `'a -> 'b`; that is, it instantiates `'c` to the functional type had by `y ()`.