-Assignment 5
+<!-- λ Λ ∀ ≡ α β ω Ω -->
-ω
-Ω
-λ
-Λ
-≡
-α
-β
+This is a long and substantial assignment. On the one hand, it doesn't have any stumpers like we gave you in past weeks, such as defining `pred` or mutual recursion. (Well, we *do* ask you to define `pred` again, but now you understand a basic strategy for doing so, so it's no longer a stumper.) On the other hand, there are a bunch of problems; many of them demand a modest amount of time; and this week you're coming to terms with both System F *and* with OCaml or Haskell. So it's a lot to do.
+
+The upside is that there won't be any new homework assigned this week, and you can take longer to complete this assignment if you need to. As always, don't get stuck on the "More challenging" questions if you find them too hard. Be sure to send us your work even if it's not entirely completed, so we can see where you are. And consult with us (over email or in person on Wednesday) about things you don't understand, especially issues you're having working with OCaml or Haskell, or with translating between System F and them. We will update this homework page with clarifications or explanations that your questions prompt.
+
+We will also be assigning some philosophy of language readings for you to look at before this Thursday's seminar meeting.
+
+Those, and lecture notes from this past week, will be posted shortly.
## Option / Maybe Types ##
-You've already defined and worked with `map` as a function on lists. Now we're going to work instead with the type OCaml defines like this:
+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:
type ('a) option = None | Some of 'a
data Maybe a = Nothing | Just a
-That is, instances of this type are either some `'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 list guaranteed to have a length ≤ 1.
+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.
-In one of the homework sessions, 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?
+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?
-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`. We won't confuse a bin that contains a bin with an empty bin. (Not even if the contained bin is itself empty.)
+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.)
1. Your first problem is to write a `maybe_map` function for these types. Here is the type of the function you should write:
-- Haskell
maybe_map :: (a -> b) -> Maybe a -> Maybe b
- 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`.
+ 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.)
- One way to extract the contents of a option or Maybe value is to pattern match on that value, as you did with lists. In OCaml:
+ 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:
- match x with
+ match m with
| None -> ...
| Some y -> ...
In Haskell:
- case x of {
+ case m of {
Nothing -> ...;
Just y -> ...
}
- 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. You may want to review the [[Rosetta pages here]] and also read some of the tutorials we linked to [[for OCaml]] or [[for Haskell]]. [WHERE]
+ 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`.
+
+ You may want to review [[Rosetta pages|/rosetta1]] and also read some of the tutorials we linked to for [[/learning OCaml]] or [[/learning Haskell]].
2. Next write a `maybe_map2` function. Its type should be:
(* OCaml *)
- maybe_map2 ('a -> 'b -> 'c) -> ('a) option -> ('b) option -> ('c) option
+ maybe_map2 : ('a -> 'b -> 'c) -> ('a) option -> ('b) option -> ('c) option
-- Haskell
maybe_map2 :: (a -> b -> c) -> Maybe a -> Maybe b -> Maybe c
data Color = Red | Green | Blue | ... deriving (Eq, Show)
data Color_tree a = Leaf a | Branch (Color_tree a) Color (Color_tree a) deriving (Show)
-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 lists to you when they are the result of evaluating an expression.
+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.
Here's how you create an instance of such a tree:
Branch _ c _ -> c == Red
}
-These expressions query whether `t` is a branching `color_tree` (not a leaf) whose root is labeled `Red`.
+These expressions query whether `t` is a branching `color_tree` (that is, not a leaf) whose root is labeled `Red`.
+
+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.
Choose one of these languages and write the following functions.
-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 labels as the original, but with all of its leaves now having had `f` applied to their original value. So for example, `map (2*) t1` would return `t1` with all of its leaf values doubled.
+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.
-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.
+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.
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`?
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.)
-7. How would you use the function defined in problem 4 to make a copy of a tree with the same structure and inner node labels, but where the leftmost leaf is now labeled `0`, the second-leftmost leaf is now labeled `1`, and so on.
+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.
-8. (More challenging.) Write a recursive function that makes a copy of a tree with the same structure and inner node labels, 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:
+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:
<pre>
Red
2 2
</pre>
-9. (More challenging.) Assume you have a `color_tree` whose leaves are labeled with `int`s (which might be negative). For this problem, assume also that the the same color never labels multiple inner nodes. 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.
+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 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 Tree ##
+## Search Trees ##
(More challenging.) For the next problem, assume the following type definition:
-- 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 nodes descending to the left from a given node will be less than the `int` of that parent node, and all the `int`s in nodes 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.
+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:
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]`.
+ 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.
+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.
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 that do this. This also corresponds to what you can write as a list comprehension in Haskell like this:
+ 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 ]
+ <!-- or `f <$/fmap> ZipList xs <*/ap> ZipList ys`; or `pure f <*> ...`; or `liftA2 f (ZipList xs) (ZipList 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 an earlier problem?
+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 list
+ 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.
+ 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.
+ <!-- in Haskell, `liftA2 f xs ys` -->
A similar choice between "zipping" and "crossing" could be made when `map2`-ing two trees. For example, the trees:
"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/Maybes, or for trees, the structure of the result exactly matched the structure of the argument.
-
-[LOOKUP in APPLICATIVE]
+* 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/Maybes 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.
+* 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.
+* 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.
+ (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.)
-These three levels of how radical a transformation you are making to a structure, and the parallels between the transformations to lists, to option/Maybes, and to trees, will be ideas we build on in coming weeks.
+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.
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 _____ ;;
+ 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 ________
+ data Lambda_term = Var Identifier | Abstract Identifier _____ | App ________ deriving (Show)
-15. Again, we've left some gaps. Choose one of these languages and fill in the gaps to complete the definition.
+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.)
+
+15. 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
+ 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`.
+ 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 ##
-(These questions are adapted from web materials by Umut Acar. See
-<http://www.mpi-sws.org/~umut/>.)
-
-
-(For the System F questions, you can either download and compile Pierce's evaluator for system F to test your work [WHERE], or work on paper.)
-
+<!-- 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. -->
+(For the System F questions, you can either work on paper, or [download and compile](http://www.cis.upenn.edu/~bcpierce/tapl/) 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.)
Let's think about the encodings of booleans, numerals and lists in System F,
-and get data-structures with the same form working in OCaml or Haskell. (Of course, OCaml and Haskell
-have *native* versions of these datas-structures: OCaml's `true`, `1`, and `[1;2;3]`.
+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 ::= type_constants | α ... | type1 -> type2 | ∀α. type
+ 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
+ Bool ≡ ∀α. α -> α -> α
true ≡ Λα. λy:α. λn:α. y
false ≡ Λα. λy:α. λn:α. n
b [T] res1 res2
-where `b` is a boolean value, and `T` is the shared type of `res1` and `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`.
+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
+ (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:
+The type `Nat` (for "natural number") may be encoded as follows:
- nat ≡ ∀α. (α -> α) -> α -> α
+ 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)
+ 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
+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 you have a function `s : α -> α` and a base element `z : α`.
-
-18. Get booleans and Church numbers working in OCaml or Haskell,
-including versions of `bool`, `true`, `false`, `zero`, `zero?` (though this is not a legal function name in either of those languages, use something else), `succ`, and `pred`.
-It's especially useful to do a version of `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
+any type `α`, as long as your function is of type `α -> α` and you have a base element of type `α`.
- -- 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:
+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.
- 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)
+ 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.
-A more general polymorphic list type is:
+ To get you started, here is how to define `sysf_bool` and `sysf_true` in OCaml:
- list ≡ ∀β. ∀α. (β -> α -> α) -> α -> α
+ type ('a) sysf_bool = 'a -> 'a -> 'a
+ let sysf_true : ('a) sysf_bool = fun y n -> y
-As with nats, the natural recursion is built into our encoding of this datatype. So we can write functions like `map`:
+ And here in Haskell:
- map : (S -> T) -> list_S -> list_T
+ type Sysf_bool a = a -> a -> a -- this is another case where Haskell uses `type` instead of `data`
+ -- To my mind the natural thing to write next 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
-<!--
- = λf:S -> T. λxs:list. xs [S] [list [T]] (λx:S. λys:list [T]. cons [T] (f x) ys) (nil [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.
+ Note that in both OCaml and the Haskell code, the generalization `∀'a` on the free type variable `'a` is implicit. If you really want to, you can supply it explicitly in Haskell by saying:
-20. Also give us the type and definition for a `sysf_head` function. Think about what value to give back if the 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.
-
-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.)
+ :set -XExplicitForAll
+ let { sysf_true :: forall a. Sysf_bool a; ... }
+ -- or
+ let { sysf_true :: forall a. a -> a -> a; ... }
+ You can't however, put a `forall a.` in the `type Sysf_bool ...` declaration. The reasons for this are too complicated to explain here.
-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.
-
+ Note also that `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 [[assignment3 answers]].
+Consider the following list type, specified using OCaml or Haskell datatypes:
-## More on Types ##
+ (* OCaml *)
+ type ('a) my_list = Nil | Cons of 'a * 'a my_list
-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. To get you started, here's one typing for **K**:
+ -- Haskell
+ data My_list a = Nil | Cons a (My_list a) deriving (Show)
- # let k (y:'a) (n:'b) = y;;
- val k : 'a -> 'b -> 'a = [fun]
- # k 1 true;;
- - : int = 1
+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:
- If you can't understand how one term can have several types, recall our discussion in this week's notes [WHERE] of "principal types".
+ 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 on the type is built into our encoding of it.
+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<sub>ω</sub>. System F *can* already define a more polymorphic list type:
+ list ≡ ∀β. ∀α. (β -> α -> α) -> α -> α
-## Evaluation Order ##
+But this is more awkward to work with, because for functions like `map` we want to give them not just the type:
-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.
+ (S -> T) -> list -> list
+but more specifically, the type:
-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?
+ (S -> T) -> list [S] -> list [T]
- let rec f x = f x;;
+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:
- let rec f x = f f;;
+ (S -> T) -> list_S -> list_T
- let rec f x = f x in f f;;
+<!--
+ = λf:S -> T. λxs:list. xs [S] [list [T]] (λx:S. λys:list [T]. cons [T] (f x) ys) (nil [T])
+-->
- let rec f x = f x in f ();;
+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 `('a) sysf_list` or `Sysf_list a`.)
- let rec f () = f f;;
+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.
- let rec f () = f ();;
+21. Modify your implementation of the predecessor function for Church numerals, above, to implement a `sysf_tail` function for your lists.
- let rec f () = f () in f f;;
+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.)
- let rec f () = f () in f ();;
-24. Throughout this problem, assume that we have:
- let rec blackhole x = blackhole x;;
- <!-- Haskell could say: `let blackhole = \x -> fix (\f -> f)` -->
- All of the following are well-typed. Which ones terminate? What are the generalizations?
- blackhole;;
+## More on Types ##
- blackhole ();;
+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**:
- fun () -> blackhole ();;
+ # let k (y:'a) (n:'b) = y ;;
+ val k : 'a -> 'b -> 'a = [fun]
+ # k 1 true ;;
+ - : int = 1
- (fun () -> blackhole ()) ();;
+ If you can't understand how one term can have several types, recall our discussion in this week's notes of "principal types". (WHERE?)
- if true then blackhole else blackhole;;
- if false then blackhole else blackhole;;
- if true then blackhole else blackhole ();;
- if false then blackhole else blackhole ();;
+## Evaluation Order ##
- if true then blackhole () else blackhole;;
+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.
- if false then blackhole () else blackhole;;
- if true then blackhole () else blackhole ();;
+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?
- if false then blackhole () else blackhole ();;
+ 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 ()
- let _ = blackhole in 2;;
+24. Throughout this problem, assume that we have:
- let _ = blackhole () in 2;;
+ let rec blackhole x = blackhole x
-25. This problem is to think about how to control order of evaluation.
-The following expression is an attempt to make explicit the
-behavior of `if ... then ... else ...` explored in the previous question.
+ <!-- Haskell could say: `let blackhole = \x -> fix (\f -> f)` -->
+ 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`!),
This almost works. For instance,
- if true then 1 else 2;;
+ if true then 1 else 2
- evaluates to 1, and
+ evaluates to `1`, and
let b = true in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+ match b with true -> y | false -> n
- also evaluates to 1. Likewise,
+ also evaluates to `1`. Likewise,
- if false then 1 else 2;;
+ 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;;
+ match b with true -> y | false -> n
- both evaluate to 2.
+ both evaluate to `2`.
However,
let rec blackhole x = blackhole x in
- if true then blackhole else blackhole ();;
+ if true then blackhole else blackhole ()
terminates, but
let b = true in
let y = blackhole in
let n = blackhole () in
- match b with true -> y | false -> n;;
+ 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.
+ 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
+ Here's a [[hint|assignment5 hint1]].