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Assignment 5
-Types and OCAML
+Types and OCaml
---------------
-0. Recall that the S combinator is given by \x y z. x z (y z).
- Give two different typings for this function in OCAML.
- To get you started, here's one typing for K:
+0. Recall that the S combinator is given by \x y z. x z (y z).
+ Give two different typings for this function in OCaml.
+ 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
+ # let k (y:'a) (n:'b) = y;;
+ val k : 'a -> 'b -> 'a = [fun]
+ # k 1 true;;
+ - : int = 1
-1. 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?
+1. 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 x;;
- let rec f x = f f;;
+ let rec f x = f f;;
- let rec f x = f x in f f;;
+ let rec f x = f x in f f;;
- let rec f x = f x in f ();;
+ let rec f x = f x in f ();;
- let rec f () = f f;;
+ let rec f () = f f;;
- let rec f () = f ();;
+ let rec f () = f ();;
- let rec f () = f () in f f;;
+ let rec f () = f () in f f;;
- let rec f () = f () in f ();;
+ let rec f () = f () in f ();;
-2. Throughout this problem, assume that we have
+2. Throughout this problem, assume that we have
- let rec omega x = omega x;;
+ let rec blackhole x = blackhole x;;
- All of the following are well-typed.
- Which ones terminate? What are the generalizations?
+ All of the following are well-typed.
+ Which ones terminate? What are the generalizations?
- omega;;
+ blackhole;;
- omega ();;
+ blackhole ();;
- fun () -> omega ();;
+ fun () -> blackhole ();;
- (fun () -> omega ()) ();;
+ (fun () -> blackhole ()) ();;
- if true then omega else omega;;
+ if true then blackhole else blackhole;;
- if false then omega else omega;;
+ if false then blackhole else blackhole;;
- if true then omega else omega ();;
+ if true then blackhole else blackhole ();;
- if false then omega else omega ();;
+ if false then blackhole else blackhole ();;
- if true then omega () else omega;;
+ if true then blackhole () else blackhole;;
- if false then omega () else omega;;
+ if false then blackhole () else blackhole;;
- if true then omega () else omega ();;
+ if true then blackhole () else blackhole ();;
- if false then omega () else omega ();;
+ if false then blackhole () else blackhole ();;
- let _ = omega in 2;;
+ let _ = blackhole in 2;;
- let _ = omega () in 2;;
+ let _ = blackhole () in 2;;
-3. This problem is to begin thinking about controlling order of evaluation.
+3. This problem is to begin thinking about controlling order of evaluation.
The following expression 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 OCAML expression,
-and "no" is any other OCAML expression (of the same type as "yes"!),
+The idea is to define an `if`-`then`-`else` expression using
+other expression types. So assume that "yes" is any OCaml expression,
+and "no" is any other OCaml expression (of the same type as "yes"!),
and that "bool" is any boolean. 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
+ let b = bool in
+ let y = yes in
+ let n = no in
+ match b with true -> y | false -> n
-This almost works. For instance,
+ 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;;
+ let b = true in let y = 1 in let n = 2 in
+ 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
+ and
- let b = false in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+ let b = false in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
-both evaluate to 2.
+ both evaluate to 2.
-However,
+ However,
- let rec omega x = omega x in
- if true then omega else omega ();;
+ let rec blackhole x = blackhole x in
+ if true then blackhole else blackhole ();;
-terminates, but
+ terminates, but
- let rec omega x = omega x in
- let b = true in
- let y = omega in
- let n = omega () in
- match b with true -> y | false -> n;;
+ 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, `match bool with true -> yes |
-false -> no;;` works as desired, but your assignment is to solve it
-without using the magical evaluation order properties of either `if`
-or of `match`. That is, you must keep the `let` statements, though
-you're allowed to adjust what `b`, `y`, and `n` get assigned to.
+ does not terminate. Incidentally, `match bool with true -> yes |
+ false -> no;;` works as desired, but your assignment is to solve it
+ without using the magical evaluation order properties of either `if`
+ or of `match`. That is, you must keep the `let` statements, though
+ you're allowed to adjust what `b`, `y`, and `n` get assigned to.
-[[Hint assignment 5 problem 3]]
+ [[hints/assignment 5 hint 1]]
-Baby monads
------------
+Booleans, Church numerals, and v3 lists in OCaml
+------------------------------------------------
+
+(These questions adapted from web materials by Umut Acar. See
+.)
+
+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. (Of course, OCaml
+has *native* versions of these datas-structures: its `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 Î»-calculus.
+
+ types Ï ::= c | 'a | Ï1 â Ï2 | â'a. Ï
+ expressions e ::= x | Î»x:Ï. e | e1 e2 | Î'a. e | e [Ï]
+
+The boolean type, and its two values, may be encoded as follows:
+
+ bool := â'a. 'a â 'a â 'a
+ true := Î'a. Î»t:'a. Î»f :'a. t
+ false := Î'a. Î»t:'a. Î»f :'a. f
- Read the lecture notes for week 6, then write a
- function `lift` that generalized the correspondence between + and
- `add`: that is, `lift` takes any two-place operation on integers
- and returns a version that takes arguments of type `int option`
- instead, returning a result of `int option`. In other words,
- `lift` will have type
+It's used like this:
- (int -> int -> int) -> (int option) -> (int option) -> (int option)
+ b [Ï] e1 e2
- so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
- Don't worry about why you need to put `+` inside of parentheses.
- You should make use of `bind` in your definition of `lift`:
+where b is a boolean value, and Ï is the shared type of e1 and e2.
- let bind (x: int option) (f: int -> (int option)) =
- match x with None -> None | Some n -> f n;;
+**Exercise**. 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.
-Booleans, Church numbers, and Church lists in OCAML
----------------------------------------------------
-These questions adapted from web materials written by some smart dude named Acar.
-The idea is to get booleans, Church numbers, "Church" lists, and
-binary trees working in OCAML.
+The type nat (for "natural number") may be encoded as follows:
- Recall from class System F, or the polymorphic Î»-calculus.
+ nat := â'a. 'a â ('a â 'a) â 'a
+ zero := Î'a. Î»z:'a. Î»s:'a â 'a. z
+ succ := Î»n:nat. Î'a. Î»z:'a. Î»s:'a â 'a. s (n ['a] z s)
- ÏÂ ::=Â Î±Â |Â Ï1Â âÂ Ï2Â |Â âÎ±.Â Ï
- eÂ ::=Â xÂ |Â Î»x:Ï.Â eÂ |Â e1Â e2Â |Â ÎÎ±.Â eÂ |Â eÂ [ÏÂ ]
+A nat n is deï¬ned 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 'a, as long as you have a base element z : 'a and a function s : 'a â 'a.
- RecallÂ thatÂ boolÂ mayÂ beÂ encodedÂ asÂ follows:
+**Exercise**: get booleans and Church numbers working in OCaml,
+including OCaml versions of bool, true, false, zero, iszero, 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. The point of the excercise
+is to do these things on your own, so avoid using the built-in
+OCaml booleans and integers.
- boolÂ :=Â âÎ±.Â Î±Â âÂ Î±Â âÂ Î±
- trueÂ :=Â ÎÎ±.Â Î»t:Î±.Â Î»fÂ :Î±.Â t
- falseÂ :=Â ÎÎ±.Â Î»t:Î±.Â Î»fÂ :Î±.Â f
+Consider the following list type:
- (whereÂ ÏÂ indicatesÂ theÂ typeÂ ofÂ e1Â andÂ e2)
+ type 'a list = Nil | Cons of 'a * 'a list
- NoteÂ thatÂ eachÂ ofÂ the followingÂ terms,Â whenÂ appliedÂ toÂ the
- appropriateÂ arguments,Â returnÂ aÂ resultÂ ofÂ typeÂ bool.
+We can encode Ï lists, lists of elements of type Ï as follows:
- (a)Â theÂ termÂ notÂ thatÂ takesÂ anÂ argumentÂ ofÂ typeÂ boolÂ andÂ computesÂ itsÂ negation;
- (b)Â theÂ termÂ andÂ thatÂ takesÂ twoÂ argumentsÂ ofÂ typeÂ boolÂ andÂ computesÂ theirÂ conjunction;
- (c)Â theÂ termÂ orÂ thatÂ takesÂ twoÂ argumentsÂ ofÂ typeÂ boolÂ andÂ computesÂ theirÂ disjunction.
+ Ï list := â'a. 'a â (Ï â 'a â 'a) â 'a
+ nil Ï := Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. n
+ make_list Ï := Î»h:Ï. Î»t:Ï list. Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. c h (t ['a] n c)
- TheÂ typeÂ nat (for "natural number") mayÂ beÂ encodedÂ asÂ follows:
+More generally, the polymorphic list type is:
- natÂ :=Â âÎ±.Â Î±Â âÂ (Î±Â âÂ Î±)Â âÂ Î±
- zeroÂ :=Â ÎÎ±.Â Î»z:Î±.Â Î»s:Î±Â âÂ Î±.Â z
- succÂ :=Â Î»n:nat.Â ÎÎ±.Â Î»z:Î±.Â Î»s:Î±Â âÂ Î±.Â sÂ (nÂ [Î±]Â zÂ s)
+ list := â'b. â'a. 'a â ('b â 'a â 'a) â 'a
- AÂ natÂ nÂ isÂ deï¬nedÂ byÂ whatÂ itÂ canÂ do,Â whichÂ isÂ toÂ computeÂ aÂ functionÂ iteratedÂ nÂ times.Â InÂ theÂ polymorphic
- encodingÂ above,Â theÂ resultÂ ofÂ thatÂ iterationÂ canÂ beÂ anyÂ typeÂ Î±,Â asÂ longÂ asÂ youÂ haveÂ aÂ baseÂ elementÂ zÂ :Â Î±Â and
- aÂ functionÂ sÂ :Â Î±Â âÂ Î±.
+As with nats, recursion is built into the datatype.
- **Excercise**: get booleans and Church numbers working in OCAML,
- including OCAML versions of bool, true, false, zero, succ, add.
+We can write functions like map:
- ConsiderÂ theÂ followingÂ listÂ type:
+ map : (Ï â Ï ) â Ï list â Ï list
- datatypeÂ âaÂ listÂ = Nil |Â ConsÂ ofÂ âaÂ *Â âaÂ list
+
- WeÂ canÂ encodeÂ ÏÂ lists,Â listsÂ ofÂ elementsÂ ofÂ typeÂ ÏÂ asÂ follows:
+**Excercise** convert this function to OCaml. We've given you the type; you
+only need to give the term.
- ÏÂ listÂ :=Â âÎ±.Â Î±Â âÂ (ÏÂ âÂ Î±Â âÂ Î±)Â âÂ Î±
- nilÏÂ :=Â ÎÎ±.Â Î»n:Î±.Â Î»c:ÏÂ âÂ Î±Â âÂ Î±.Â n
- makeListÏÂ :=Â Î»h:Ï.Â Î»t:ÏÂ list.Â ÎÎ±.Â Î»n:Î±.Â Î»c:ÏÂ âÂ Î±Â âÂ Î±.Â cÂ hÂ (tÂ [Î±]Â nÂ c)
+Also give us the type and definition for a `head` function. Think about what
+value to give back if the argument is the empty list. Ultimately, we might
+want to make use of our `'a option` technique, but for this assignment, just
+pick a strategy, no matter how clunky.
- AsÂ withÂ nats,Â recursion is built into the datatype.
+Be sure to test your proposals with simple lists. (You'll have to `make_list`
+the lists yourself; don't expect OCaml to magically translate between its
+native lists and the ones you buil.d)
- WeÂ canÂ writeÂ functions likeÂ map:
- mapÂ :Â (ÏÂ âÂ ÏÂ )Â âÂ ÏÂ listÂ âÂ ÏÂ list
- :=Â Î»fÂ :ÏÂ âÂ Ï.Â Î»l:ÏÂ list.Â lÂ [ÏÂ list]Â nilÏÂ (Î»x:Ï.Â Î»y:ÏÂ list.Â consÏÂ (fÂ x)Â y
+
+
+
+Baby monads
+-----------
+
+Read the material on dividing by zero/towards monads from ~~the end of lecture
+notes for week 6~~ the start of lecture notes for week 7, then write a function `lift'` that generalized the
+correspondence between + and `add'`: that is, `lift'` takes any two-place
+operation on integers and returns a version that takes arguments of type `int
+option` instead, returning a result of `int option`. In other words, `lift'`
+will have type:
- **Excercise** convert this function to OCAML. Also write an `append` function.
- Test with simple lists.
+ (int -> int -> int) -> (int option) -> (int option) -> (int option)
- ConsiderÂ theÂ followingÂ simpleÂ binaryÂ treeÂ type:
+so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
+Don't worry about why you need to put `+` inside of parentheses.
+You should make use of `bind'` in your definition of `lift'`:
- typeÂ âaÂ treeÂ = Leaf |Â NodeÂ ofÂ âaÂ treeÂ *Â âaÂ *Â âaÂ tree
+ let bind' (u: int option) (f: int -> (int option)) =
+ match u with None -> None | Some x -> f x;;
- **Excercise**
- Write a function `sumLeaves` that computes the sum of all the
- leaves in an int tree.
- WriteÂ aÂ functionÂ `inOrder`Â :Â ÏÂ treeÂ âÂ ÏÂ listÂ thatÂ computesÂ theÂ in-orderÂ traversalÂ ofÂ aÂ binaryÂ tree.Â You
- mayÂ assumeÂ theÂ aboveÂ encodingÂ ofÂ lists;Â deï¬neÂ anyÂ auxiliaryÂ functionsÂ youÂ need.