Saturday, October 29, 2016

Implementing type-classes as OCaml modules

Implementing type-classes as OCaml modules

Modular type classes

We revisit the idea of type-classes first explored in this post. This time though, the implementation technique will be by via OCaml modules inspired by the paper "Modular Type Classes" [2] by Harper et. al.

Starting with the basics, consider the class of types whose values can be compared for equality. Call this type-class Eq. We represent the class as a module signature.

    module type EQ = sig
      type t

      val eq : t * t → bool
Specific instances of Eq are modules that implement this signature. Here are two examples.
    module Eq_bool : EQ with type t = bool = struct
      type t = bool
      let eq (a, b) = a = b
    module Eq_int : EQ with type t = int = struct
      type t = int
      let eq (a, b) = a = b

Given instances of class Eq (X and Y say,) we realize that products of those instances are also in Eq. This idea can be expressed as a functor with the following type.

    module type EQ_PROD =
      functor (X : EQ) (Y : EQ) → EQ with type t = X.t * Y.t
The implementation of this functor is simply stated as the following.
    module Eq_prod : EQ_PROD =
      functor (X : EQ) (Y : EQ) → struct
        type t = X.t * Y.t

        let eq ((x1, y1), (x2, y2)) =  X.eq (x1, x2) && Y.eq(y1, y2)
With this functor we can build concrete instances for products. Here's one example.
    module Eq_bool_int : 
       EQ with type t = (bool * int) = Eq_prod (Eq_bool) (Eq_int)

The class Eq can be used as a building block for the construction of new type classes. For example, we might define a new type-class Ord that admits types that are equality comparable and whose values can be ordered with a "less-than" relation. We introduce a new module type to describe this class.

    module type ORD = sig
      include EQ

      val lt : t * t → bool
Here's an example instance of this class.
    module Ord_int : ORD with type t = int = struct
      include Eq_int
      let lt (x, y) = Pervasives.( < ) x y
As before, given two instances of this class, we observe that products of these instances also reside in the class. Accordingly, we have this functor type
    module type ORD_PROD =
      functor (X : ORD) (Y : ORD) → ORD with type t = X.t * Y.t
with the following implementation.
    module Ord_prod : ORD_PROD =
      functor (X : ORD) (Y : ORD) → struct
        include Eq_prod (X) (Y)
        let lt ((x1, y1), (x2, y2)) =
 (x1, x2) || X.eq (x1, x2) && (y1, y2)
This is the corresponding instance for pairs of intgers.
    module Ord_int_int = Ord_prod (Ord_int) (Ord_int)
Here's a simple usage example.
    let test_ord_int_int = 
      let x = (1, 2) and y = (1, 4) in
      assert ( not (Ord_int_int.eq (x, y)) && (x, y))

Using type-classes to implement parameteric polymorphism

This section begins with the Show type-class.

     module type SHOW = sig
      type t

      val show : t → string
In what follows, it is convenient to make an alias for module values of this type.
   type α show_impl = (module SHOW with type t = α)
Here are two instances of this class...
    module Show_int : SHOW with type t = int = struct
      type t = int
      let show = Pervasives.string_of_int

    module Show_bool : SHOW with type t = bool = struct
      type t = bool

      let show = function | true → "True" | false → "False"
... and here these instances are "packed" as values.
    let show_int : int show_impl = 
      (module Show_int : SHOW with type t = int)

    let show_bool : bool show_impl = 
      (module Show_bool : SHOW with type t = bool)
The existence of the Show class is all that is required to enable the writing of our first parametrically polymorphic function.
    let print : α show_impl → α → unit =
      fun (type a) (show : a show_impl) (x : a) →
      let module Show = (val show : SHOW with type t = a) in
      print_endline@@ x
    let test_print_1 : unit = print show_bool true
    let test_print_2 : unit = print show_int 3
The function print can be used with values of any type α as long as the caller can produce evidence of α's membership in Show (in the form of a compatible instance).

The next example begins with the definition of a type-class Num (the class of additive numbers) together with some example instances.

    module type NUM = sig
      type t
      val from_int : int → t
      val ( + ) : t → t → t
    module Num_int : NUM with type t = int = struct
      type t = int
      let from_int x = x
      let ( + ) = Pervasives.( + )
    let num_int = (module Num_int : NUM with type t = int)
    module Num_bool : NUM with type t = bool = struct
      type t = bool
      let from_int = function | 0 → false | _ → true
      let ( + ) = function | true → fun _ → true | false → fun x → x
    let num_bool = (module Num_bool : NUM with type t = bool)
The existence of Num admits writing a polymorphic function sum that will work for any α list of values if only α can be shown to be in Num.
    let sum : α num_impl → α list → α =
      fun (type a) (num : a num_impl) (ls : a list) →
        let module Num = (val num : NUM with type t = a) in
        List.fold_right Num.( + ) ls (Num.from_int 0)
    let test_sum = sum num_int [1; 2; 3; 4]
This next function requires evidence of membership in two classes.
    let print_incr : (α show_impl * α num_impl) → α → unit =
      fun (type a) ((show : a show_impl), (num : a num_impl)) (x : a) →
        let module Num = (val num : NUM with type t = a) in
        let open Num
        in print show (x + from_int 1)
    (*An instantiation*)
    let print_incr_int (x : int) : unit = print_incr (show_int, num_int) x

If α is in Show then we can easily extend Show to include the type α list. As we saw earlier, this kind of thing can be done with an approriate functor.

    module type LIST_SHOW =
      functor (X : SHOW) → SHOW with type t = X.t list
    module List_show : LIST_SHOW =
      functor (X : SHOW) → struct
        type t = X.t list
        let show =
            fun xs →
              let rec go first = function
                | [] → "]"
                | h :: t →
                  (if (first) then "" else ", ") ^ h ^ go false t 
              in "[" ^ go true xs
There is also another way : one can write a function to dynamically compute an α list show_impl from an α show_impl.
  let show_list : α show_impl → α list show_impl =
    fun (type a) (show : a show_impl) →
      let module Show = (val show : SHOW with type t = a) in
      (module struct
        type t = a list
        let show : t → string =
          fun xs →
            let rec go first = function
              | [] → "]"
              | h :: t →
                (if (first) then "" else ", ") ^ h ^ go false t
            in "[" ^ go true xs
      end : SHOW with type t = a list)
   let testls : string = let module Show =
       (val (show_list show_int) : SHOW with type t = int list) in (1 :: 2 :: 3 :: [])

The type-class Mul is an aggregation of the type-classes Eq and Num together with a function to perform multiplication.

   module type MUL = sig
      include EQ
      include NUM with type t := t
      val mul : t → t → t
   type α mul_impl = (module MUL with type t = α)
   module type MUL_F =
     functor (E : EQ) (N : NUM with type t = E.t) → MUL with type t = E.t
A default instance of Mul can be provided given compatible instances of Eq and Num.
    module Mul_default : MUL_F =
      functor (E : EQ) (N : NUM with type t = E.t)  → struct
        include (E : EQ with type t = E.t)
        include (N : NUM with type t := E.t)
        let mul : t → t → t =
          let rec loop x y = begin match () with
            | () when eq (x, (from_int 0)) → from_int 0
            | () when eq (x, (from_int 1)) → y
            | () → y + loop (x + (from_int (-1))) y
          end in loop
    module Mul_bool : MUL with type t = bool = 
      Mul_default (Eq_bool) (Num_bool)
Specific instances can be constructed as needs demand.
   module Mul_int : MUL with type t = int = struct
     include (Eq_int : EQ with type t = int)
     include (Num_int : NUM with type t := int)
     let mul = Pervasives.( * )
   let dot : α mul_impl → α list → α list → α =
     fun (type a) (mul : a mul_impl) →
       fun xs ys →
         let module M = (val mul : MUL with type t = a) in
         sum (module M : NUM with type t = a)@@ List.map2 M.mul xs ys
   let test_dot =
     dot (module Mul_int : MUL with type t = int) [1; 2; 3] [4; 5; 6]
Note that in this definition of dot, coercision of the provided Mul instance to its base Num instance is performed.

This last section provides an example of polymorphic recursion utilizing the dynamic production of evidence by way of the show_list function presented earlier.

   let rec replicate : int → α → α list =
     fun n x → if n <= 0 then [] else x :: replicate (n - 1) x
   let rec print_nested : α. α show_impl → int → α → unit =
     fun show_mod → function
     | 0 → fun x → print show_mod x
     | n → fun x → print_nested (show_list show_mod) (n - 1) (replicate n x)
   let test_nested =
     let n = read_int () in
     print_nested (module Show_int : SHOW with type t = int) n 5

[1] Implementing, and Understanding Type Classes -- Oleg Kiselyov
[2] Modular Type Classes -- Harper et. al.

Wednesday, October 26, 2016

Haskell type-classes in OCaml and C++

Haskell type-classes in OCaml and C++

This article examines the emulation of Haskell like type-classes in OCaml and C++. It follows [1] closely (recommended for further reading), extending on some of the example code given there to include C++.

First stop, a simplified version of the Show type-class with a couple of simple instances.

    class Show a where
      show :: a -> string

    instance Show Int where
      show x = x -- internal

    instance Show Bool where
      str True = "True"
      str False = "False"
The OCaml equivalent shown here uses the "dictionary passing" technique for implementation. The type-class declaration Show in Haskell translates to a data-type declaration for a polymorphic record α show in OCaml.
    type α show = {
      show : α → string

    let show_bool : bool show = {
      show = function | true → "True" | false → "False"

    let show_int : int show = {
      show = string_of_int
In C++ we can use a template class to represent the type-class and specializations to represent the instances.
      template <class A> struct Show {};

      template <>
      struct Show<int> {
        static std::string (*show)(int);
      std::string(*Show<int>::show)(int) = &std::to_string;

      template <>
      struct Show<bool> {
        static std::string show (bool);
      std::string Show<bool>::show (bool b) { return b ? "true" : "false"; }

Next up print, a parametrically polymorphic function.

      print :: Show a => a -> IO ()
      print x = putStrLn$ show x
According to our dictionary passing scheme in OCaml, this renders as the following.
      let print : α show → α → unit = 
        fun {show} → fun x → print_endline@@ show x
The key point to note here is that in OCaml, evidence of the α value's membership in the Show class must be produced explicitly by the programmer. In C++, like Haskell, no evidence of the argument's membership is required, the compiler keeps track of that implicitly.
    template <class A>
    void print (A const& a) {
      std::cout << Show<A>::show (a) << std::endl;

This next simplified type-class shows a different pattern of overloading : the function fromInt is overloaded on the result type and the (+) function is binary.

    class Num a where
      fromInt :: Int -> a
      (+)     :: a -> a -> a

    sum :: Num a => [a] -> a
    sum ls = foldr (+) (fromInt 0) ls
Translation into OCaml is as in the following.
    type α num = {
      from_int : int → α;
      add      : α → α → α;

    let sum : α num → α list → α = 
      fun {from_int; add= ( + )} → 
        fun ls →
          List.fold_right ( + ) ls (from_int 0)
Translation into C++, reasonably mechanical. One slight disappointment is that it doesn't seem possible to get the operator '+' syntax as observed in both the Haskell and OCaml versions.
    template <class A>
    struct Num {};
    namespace detail {
      template <class F, class A, class ItT>
      A fold_right (F f, A z, ItT begin, ItT end) {
        if (begin == end) return z;
        return f (fold_right (f, z, std::next (begin), end), *begin);
    template <class ItT>
    typename std::iterator_traits<ItT>::value_type 
    sum (ItT begin, ItT end) {
      using A = typename std::iterator_traits<ItT>::value_type;
      auto add = Num<A>::add;
      auto from_int = Num<A>::from_int;
      return detail::fold_right (add, from_int (0), begin, end);
In Haskell, Int is made a member of Num with this declaration.
     instance Num Int where
       fromInt x = x
       (+)       = (Prelude.+)
Returning to OCaml, we can define a couple of instances including the one above like this.
    let int_num  : int num  = { 
      from_int = (fun x → x); 
      add      = Pervasives.( + ); 
    let bool_num : bool num = {
      from_int = (function | 0 → false | _ → true);
      add = function | true → fun _ → true | false → fun x → x
The code defining those above instances in C++ follows.
    template <>
    struct Num<int> {
      static int from_int (int);
      static int add (int, int);
    int Num<int>::from_int (int i) { return i; }
    int Num<int>::add (int x, int y) { return x + y; }
    template <>
    struct Num<bool> {
      static bool from_int (int);
      static bool add (bool, bool);
    bool Num<bool>::from_int (int i) { return i != 0; }
    bool Num<bool>::add (bool x, bool y) { if (x) return true; return y; }
Here now is a function with two type-class constraints.
    print_incr :: (Show a, Num a) => a -> IO ()
    print_incr x = print$ x + fromInt 1
In OCaml this can be written like so.
    let print_incr : (α show * α num) → α → unit = 
      fun (show, {from_int; add= ( + )}) → 
        fun x → print show (x + (from_int 1))
In C++, this is said as you see below.
    template <class A>
    void print_incr (A x) {
      print (Num<A>::add (x, Num<A>::from_int (1)));
Naturally, the above function will only be defined for types A that are members of both the Show and Num classes and will yield compile errors for types that are not.

Moving on, we now look at another common pattern, an instance with a constraint : a Show instance for all list types [a] when the element instance is a member of Show.

    instance Show a => Show [a] where
      show xs = "[" ++ go True xs
          go _ [] = "]"
          go first (h:t) =
            (if first then "" else ", ") ++ show h ++ go False t

In OCaml, this takes the form of a function. The idea is, given evidence of a type α's membership in Show the function produces evidence that the type α list is also in Show.
    let show_list : α show → α list show =
      fun {show} →
        {show = fun xs →
          let rec go first = function
            | [] → "]"
            | h :: t →
              (if (first) then "" else ", ") ^ show h ^ go false t in
          "[" ^ go true xs
It might be possible to do better than the following partial specialization over vector<> in C++ (that is, to write something generic, just once, that works for a wider set ofsequence types) using some advanced meta-programming "hackery", I don't really know. I suspect finding out might be a bit of a rabbit hole best avoided for now.
    template <class A>
    struct Show<std::vector<A>> {
      static std::string show (std::vector<A> const& ls);
    template <class A>
    std::string Show<std::vector<A>>::show (std::vector<A> const& ls) {
      bool first=true;
      typename std::vector<A>::const_iterator begin=ls.begin (), end=ls.end ();
      std::string s="[";
      while (begin != end) {
        if (first) first = false;
        else s += ", ";
        //A compile time error will result here if if there is no
        //evidence that `A` is in `Show`
        s += Show<A>::show (*begin++);
      s += "]";
      return s;

In this next example, we need a type-class describing types that can be compared for equality, Eq. That property and the Num class can be combined to produce a type-class with a super-class and a default.

    class Eq where
      (==) :: a -> a -> bool
      (/=) :: a -> a -> bool
    deriving instance Eq Bool
    deriving instance Eq Int

    class (Eq a, Num a) => Mul a where
      (*) :: a -> a -> a
      x * _ | x == fromInt 0 = fromInt 0
      x * y | x == fromInt 1 = y
      x * y | y + (x + (fromInt (-1))) * y

    dot :: Mul a => [a] -> [a] -> a
    dot xs ys = sum$ zipWith ( * ) xs ys
Modeling the above in OCaml is done with a dictionary for the Mul type-class and a function to generate instances from super-class instances.
    type α mul = {
      mul_super : α eq * α num;
      mul : α → α → α
    let mul_default : α eq * α num → α mul = 
      fun (({eq}, {from_int; add = ( + )}) as super) → 
          mul_super = super;
          mul = let rec loop x y = begin match () with
          | () when eq x (from_int 0) → from_int 0
          | () when eq x (from_int 1) → y
          | () → y + loop (x + (from_int (-1))) y 
          end in loop
    let bool_mul : bool mul = 
      mul_default (bool_eq, bool_num)
    let int_mul : int mul = {
      mul_super = (int_eq, int_num);
      mul = Pervasives.( * )

    let dot : α mul → α list → α list → α = 
      fun {mul_super = (eq, num); mul} →
        fun xs ys → sum num@@ List.map2 mul xs ys
As one would expect, expressing the base class/derived class relationships in C++ is playing to its strengths.
    template <class A> struct Eq {};
    template <>
    struct Eq<bool> {
      static bool eq (bool, bool);
      static bool neq (bool, bool);
    bool Eq<bool>::eq (bool s, bool t) { return s == t; }
    bool Eq<bool>::neq (bool s, bool t) { return s != t; }
    template <>
    struct Eq<int> {
      static int eq (int, int);
      static int neq (int, int);
    int Eq<int>::eq (int s, int t) { return s == t; }
    int Eq<int>::neq (int s, int t) { return s != t; }

    template <class A>
    struct Mul : Eq<A>, Num <A> {
      using Eq<A>::eq;
      using Num<A>::add;
      using Num<A>::from_int;
      static A mul (A x, A y);
    template <class A>
    A Mul<A>::mul (A x, A y) {
      if (eq (x, from_int (0))) return from_int (0);
      if (eq (x, from_int (1))) return y;
      return add (y, mul ((add (x, from_int (-1))), y));
    template struct Mul<bool>;
    template <> int Mul<int>::mul (int x, int y) { return x * y; }
    namespace detail{
      template <class F, class It, class Acc>
      Acc map2 (F f
       , It xs_begin, It xs_end, It ys_begin, It ys_end, Acc acc) {
        if ((xs_begin == xs_end) || (ys_begin == ys_end)) return acc;
        return map2 (f
              , std::next (xs_begin)
              , xs_end
              , std::next (ys_begin)
              , ys_end
              , *acc++ = f (*xs_begin, *ys_begin));
    }//namespace detail
    template <class A>
    A dot (std::vector<A> const& xs, std::vector<A> const& ys) {
      std::vector<A> buf;
      detail::map2 (
       , xs.begin (), xs.end()
       , ys.begin (), ys.end ()
       , std::back_inserter(buf));
      return sum (buf.begin (), buf.end ());

This very last example is in polymorphic recursion. The Haskell reads as follows.

   print_nested :: Show a => Int -> a -> IO ()
   print_nested 0 x = print x
   print_nested n x = print_nested (n - 1) (replicate n x)

   test_nested = do
     n <- getLine
     print_nested (read n) (5::Int)

Those two functions are very interesting! Translating it to OCaml yields the following.
    let rec replicate : int → α → α list = 
      fun n x → if n >= 0 then [] else x :: replicate (n - 1) x
    let rec print_nested : α. α show → int → α → unit =
      fun show_dict → function
      | 0 →
          fun x →
            print show_dict x
      | n → 
          fun x →
            print_nested (show_list show_dict) (n - 1) (replicate n x)

    let test_nested =
      let n = read_int () in
      print_nested show_int n 5
Now if you examine the output of the above if '4' (say) was entered, you'll see something like this:
   [[[[5, 5, 5, 5], [5, 5, 5, 5], [5, 5, 5, 5]], [[5, 5, 5, 5], [5, 5,
   5, 5], [5, 5, 5, 5]]]]
You can see, looking at this, that the type of the printed list is not determinable at compile-time. It is dependent on a runtime parameter! It follows that the evidence that the type is in the Show class can not be produced statically. It has to be computed dynamically which is what you see there in the application of show_list to the current show_dict in the n <> 0 branch of the print_nested function. Note also the requirement for the universal quantifier in the function signature. It's mandatory.

OK, so how about the above code in C++? Well a naive transliteration gives the following.

    namespace detail {
      template<class A, class ItT>
      ItT replicate (int n, A x, ItT dst) {
        if (n <= 0) return dst;
        return replicate ((n - 1), x, *dst++ = x);
    }//namespace detail
    template <class A>
    void print_nested (int n, A const& x) {
      if (n == 0)
        print (x);
      else {
        std::vector<A> buf;
        detail::replicate(n, x, std::back_inserter(buf));
        print_nested (n - 1, buf);
    void test_nested () {
      int n;
      std::cin >> n;
      print_nested (n, 5);
Unfortunately though, this program though exhibits unbounded compile time recursion (compilation doesn't terminate).

[1] Implementing, and Understanding Type Classes -- Oleg Kiselyov

Wednesday, October 12, 2016

Monty Hall

Suppose you're on a game show, and you're given the choice of three doors : Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No.3 which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

What do you think?

This problem is known as the "Monty Hall" problem. It's named after the host of the American television game show "Let's make a deal".

Paul Erdős, one of the most prolific mathematicians in history remained unconvinced (of the correct answer to the above problem) until he was shown a computer simulation confirming the predicted result.

Here's a simulation in OCaml one hopes, may have convinced Paul!

module Monty = struct

  (*[dtr w p n] where [n] is the number of doors, selects which door
    to remain (closed) given the winning door [w] and the player
    chosen door [p]*)
  let rec dtr w p n =
    if p <> w then  w 
    else if p = 0 then n - 1 else 0

  (*[gen_game d] generates a game with [d] doors and returns a game
    with a winning door, a player selected door and a door to keep
    closed before if the player wants to switch*)
  let gen_game (d : int) : (int * int * int) =
    let w = d and p = d in 
    (w, p, dtr w p d)

  let num_wins = ref 0 (*To keep track of scores*)
  type strategy = Hold | Switch (*The type of strategies*)

  (*Play a single game*)
  let play_game (d : int) (s : strategy) : unit =
    let w, p, r = gen_game d in
    match s with
    | Hold → num_wins := !num_wins + if p = w then 1 else 0
    | Switch → num_wins := !num_wins + if r = w then 1 else 0

  (*Play a set of [n] games*)
  let play_games (d : int) (n : int) (s : strategy ) : unit = 
    let rec loop (i : int) : unit = 
      if i = n then ()
      else  begin
        play_game d s;
        loop (i + 1)
    in loop 0


open Monty

(*Initialized from the command line*)
let version       = ref false
let num_doors     = ref 0
let num_sims      = ref 0
let read_args () =
  let specification =
    [("-v", Arg.Set version, "Print the version number");
     ("-d", Arg.Set_int num_doors, "Number of doors (>= 3)" );
     ("-n", Arg.Set_int num_sims, "Number of simulations (>= 1)");
  in Arg.parse specification
  (fun s →
    Printf.printf "Warning : Ignoring unrecognized argument \"%s\"\n" s)
  "Usage : monty -d <number of doors> -n <number of simulations>"

(*[fabs e] computes the absolute value of [e]*)
let fabs (e : float) : float = if e < 0. then ~-.e else e

let () = 
    read_args ();
    if !version then Printf.printf "1.0.0\n"
      let n = !num_sims and d = !num_doors in
      if d < 3 then
        raise (Invalid_argument "Number of doors must be >= than 3");
      if n < 1 then
        raise (Invalid_argument "Number of simulations must be >= 1");
        num_wins := 0;
        play_games d n Hold;
        Printf.printf "Num wins (hold): %d\n" !num_wins;
        let err=fabs (float_of_int (!num_wins) /. 
                    (float_of_int n) -. 1.0 /. (float_of_int d)) in
        Printf.printf "Error %f\n" err;
        num_wins := 0;
        play_games d n Switch;
        Printf.printf "Num wins (switch): %d\n" !num_wins;
        let err=fabs (float_of_int (!num_wins) /. 
                   (float_of_int n) -. (float_of_int (d - 1) /. 
                                                (float_of_int d))) in
        Printf.printf "Error %f\n" err ;

  | Invalid_argument s → Printf.printf "%s\n" s

Wednesday, October 5, 2016

Conversion operations of the lambda-calculus



This note provides a super lightweight explanation of the three conversion operations of the $\lambda$-calculus known (cryptically) as $\alpha$, $\beta$ and $\eta$ conversions respectively (borrowed fairly freely from the delightful reference given at the bottom.)

Syntax vs. semantics

The $\textbf{syntax}$ of the language of $\textit{$\lambda$- expressions}$ is \[ \begin{eqnarray} <exp> & ::= & <constant>\;\; & \text{Constants} \nonumber \\ & \mid & <variable>\;\; & \text{Variables} \nonumber \\ & \mid & <exp> <exp>\;\; & \text{Applications} \nonumber \\ & \mid & \lambda<variable>.<exp>\;\; & \text{Abstractions} \nonumber \end{eqnarray} \] The $\textbf{semantics}$ of the the $\lambda$-calculus is defined by three $\textit{conversion rules}$. To understand them requires the terminology of $\textit{free}$ and $\textit{bound}$ variables. An occurence of a variable in a $\lambda$-expression is bound if there is an enclosing abstraction that binds it, and is free otherwise. For example, in $\lambda x.+\; ((\lambda y. +\;y\; z)\;7)\;x$, $x$ and $y$ appear bound whereas $z$ appears free.


$\beta$-reduction describes how to apply a function to an argument. For example, $\left(\lambda x.+\;x\;1\right)\; 4$ denotes the application of a particular $\lambda$-abstraction to the argument $4$. The result of applying a $\lambda$-abstraction to an argument is an instance of the body of the $\lambda$-abstraction in which (free) occurences of the formal parameter in the body are replaced with (copies of) the argument. Thus, $\left(\lambda x.+\;x\;1\right)\; 4 \rightarrow +\;4\;1 \rightarrow 5$. In the event there are no occurences of the formal parameter in the abstraction body, the argument is discarded unused so, $(\lambda x.\;3)\;4 \rightarrow 3$. Care is needed when formal parameter names are not unique. For example, \[ \begin{eqnarray} & & \left(\lambda x.\;\left(\lambda x.+ \left(-\;x\;1\right)\right)\;x\;3\right)\; 9 \nonumber \\ & \rightarrow & \left(\lambda x.+ \left(-\;x\;1\right)\right)\;9\;3 \nonumber \\ & \rightarrow & +\;\left(-\;9\;1\right)\;3 \nonumber \\ & \rightarrow & +\;8\;3 \nonumber \\ & \rightarrow & 11 \nonumber \end{eqnarray} \] The key point of that example is that we did not substitue for the inner $x$ in the first reduction because it was not free in the body of the outer $\lambda x$ abstraction. Indeed, in the OCaml top-level we observe

       # (fun x -> (fun x -> ( + ) (( - ) x 1)) x 3) 9 ;;
       - : int = 11
or equivalently, in C++,
      auto add = [](int x) { return [=](int y) { return x + y; }; };
      auto sub = [](int x) { return [=](int y) { return x - y; }; };
      [=](int x) { 
        return [=](int x) {
          return add (sub (x) (1)); 
          } (x) (3); 
      } (9) ; //is the value '11'
The $\beta$-rule applied backwards is called $\beta$-abstraction and written with a backwards reduction arrow '$\leftarrow$'. Thus, $+\;4\;1 \leftarrow (\lambda x.\;+\;1\;x)\;4$. $\beta$-conversion means reduction or abstraction and is written with a double-ended arrow augmented with a $\beta$ (in order to distinguish it from other forms of conversion). So, $+\;4\;1 \underset{\beta}{\leftrightarrow} (\lambda x.\;+\;1\;x)\;4$. One way to look at $\beta$ conversion is that it is saying something about $\lambda$-expressions that look different but mean the same thing.


It seems obvious that the two abstractions $\lambda x.+\;x\;1$ and $\lambda y.+\;y\;1$ "ought" to be equivalent. $\alpha$-conversion allows us to change the name of a formal parameter as long as it is done consistently. So we write $\lambda x.+\;x\;1 \underset{\alpha}{\leftrightarrow} \lambda y.+\;y\;1$. Of course, the newly introduced name must not occur free in the body of the original $\lambda$-abstraction.


This last conversion rule exists to to complete our intuition about what $\lambda$-abstractions "ought" to be equivalent. The rule is this : If $f$ denotes a function, $x$ a variable that does not occur free in $f$, then $\lambda x.f\;x \underset{\eta}{\leftrightarrow} f$. For example, in OCaml if we define f by let f x = x + 1 then clearly fun x -> f x produces the same results for all values x in the domain of f.


The first section provides a set of formal rules for constructing expressions (the BNF grammar). Using the notation $E\;\left[M/x\right]$ to mean the expression $E$ with $M$ substituted for free occurrences of $x$ we can succintly state the the rules for converting one expression into an equivalent one as \[ \begin{eqnarray} x\;\left[M/x\right] & = & M \nonumber \\ c\;\left[M/x\right] & = & c\;\;\text{where $c$ is any variable or constant other than $x$} \nonumber \\ \left(E\;F\right)\;\left[M/x\right] & = & E\left[M/x\right]\; F\left[M/x\right]\; \nonumber \\ \left(\lambda x.E\right)\;\left[M/x\right] & = & \lambda x.E \nonumber \\ \left(\lambda y.E\right)\;\left[M/x\right] & & \text{where $y$ is any variable other than $x$} \nonumber \\ & = & \lambda y.E\left[M/x\right]\;\text{if $x$ does not occur free in E or $y$ does not occur free in $M$} \nonumber \\ & = & \lambda z.\left(E\left[z/y\right]\right)\left[M/x\right]\;\text{otherwise}\nonumber \\ \end{eqnarray} \]

[1] The Implementation of Functional Programming Languages by Simon L. Peyton Jones. 1987.

Tuesday, October 4, 2016

Eliminating left-recursion

Eliminating left recursion

Consider a non-terminal $A$ with two productions $A \rightarrow A \alpha \mid \beta$ where $\alpha$, $\beta$ are sequences of terminals and non-terminals that do not start with $A$. It produces strings of the form $\beta\alpha^{*}$. This rule for $A$ exhibits direct left recursion. The left recursion can be turned into right recursion by rewriting in terms of a new non-terminal $R$ as $A \rightarrow \beta R \nonumber$ and $R \rightarrow \alpha R \mid \epsilon$.

Generally, immediate left recursion can be eliminated by the following technique which works for any number of $A$ productions. First group as \[ \begin{equation} A \rightarrow A\alpha_{1} \mid A\alpha_{2} \mid \cdots \mid A\alpha_{m} \mid \beta_{1} \mid \cdots \mid \beta_{n} \end{equation} \] where no $\beta_{i}$ begins with an $A$. Then replace the $A$ productions by \[ \begin{eqnarray} A &\rightarrow& \beta_{1}A^{\prime} \mid \beta_{2}A^{\prime} \mid \cdots \mid \beta_{n}A^{\prime} \nonumber \\ A^{\prime} &\rightarrow& \alpha_{1}A^{\prime} \mid \alpha_{2}A^{\prime} \mid \cdots \mid \alpha_{m}A^{\prime} \mid \epsilon \nonumber \end{eqnarray} \] This procedure eliminates all direct left recursion from the $A$ and $A^{\prime}$ rules (provided no $\alpha_{i}$ is $\epsilon$). For example, the language of arithmetic expressions might be written \[ \begin{eqnarray} & E &\rightarrow E + T \mid E - T \mid T \nonumber \\ & T &\rightarrow T * F \mid T \mathbin{/} F \mid F \nonumber \\ & F &\rightarrow \left( E \right) \mid \mathbf{id} \nonumber \end{eqnarray} \] which, on applying the above procedure yields \[ \begin{eqnarray} & E &\rightarrow T\;E^{\prime} \nonumber \\ & E^{\prime} &\rightarrow +\;T\;E^{\prime} \mid -\;T\;E^{\prime}\nonumber \mid \epsilon \nonumber \\ & T &\rightarrow F\;T^{\prime} \nonumber \\ & T^{\prime} &\rightarrow *\;F\;T^{\prime} \mid \mathbin{/}\;F\;T^{\prime} \mid \epsilon \nonumber \\ & F &\rightarrow \left( E \right) \mid \mathbf{id} \nonumber \end{eqnarray} \] Consider the grammar \[ \begin{eqnarray} S &\rightarrow& A\;a \mid b \nonumber \\ A &\rightarrow& A\;c \mid S\;d \mid \epsilon \nonumber \end{eqnarray} \] The non-terminal $S$ is left recursive because $S \Rightarrow A\;a \Rightarrow S\;d\;a$ but it is not immediately left recursive. The procedure given above does not eliminate left recursion of this kind. It is however amenable to the following approach. First order the non-terminals $S$ then $A$. We'd start by eliminating direct left recursion from the $S$-productions but there is none among the $S$-productions so we move onto the $A$-productions. First we eliminate $S$ from the $A$-productions by substitution to obtain the following $A$-productions \[ \begin{equation} A \rightarrow A\;c \mid A\;a\;d \mid b\;d \mid \epsilon \nonumber \end{equation} \] and now eliminate the direct left recursion in the $A$ to get \[ \begin{eqnarray} & S & \rightarrow A\;a \mid b \nonumber \\ & A & \rightarrow b\;d\;A^{\prime} \mid A^{\prime} \nonumber \\ & A^{\prime} & \rightarrow c\;A^{\prime} \mid a\;d\;A^{\prime} \mid \epsilon \nonumber \end{eqnarray} \] Technically, the above approach is only guaranteed to work when the grammar to which it is applied has no cycles or $\epsilon$-productions. The above example violates this in that the rule for $A$ contained an $\epsilon$-production but it turns out in this case to be harmless. Generalizing, assuming an input grammar $G$ with no cycles or $\epsilon$-productions, an equivalent grammar with no left recursion can be found by, arranging the nonterminals of $G$, $A_{1}, A_{2}, \dots, A_{n}$ say, then visiting each in order, for each $A_{i}$, replace each production of the form $A_{i} \rightarrow A_{j}\gamma$ by the productions $A_{i} \rightarrow \delta_{1}\gamma \mid \delta_{2}\gamma \mid \cdots \mid \delta_{k}\gamma$ where $A_{j} \rightarrow \delta_{1} \mid \delta_{2} \mid \cdots \mid \delta_{k}$ are all the current $A_{j}$ productions, $j < i$ and then elminiate the immediate left recursion among the $A_{i}$ productions.

One of the pre-conditions of the algorithm of the previous section is that the input grammar $G$ contain no $\epsilon$-productions. So, we seek a method for eliminating $\epsilon$-productions where we can. To begin, we define a non-terminal $A$ of a grammar $G$ $\textit{nullable}$ if, $A \overset{*}{\Rightarrow} \epsilon$. A non-terminal is nullable if, in $G$, $A \rightarrow \epsilon$ or if $A \rightarrow A_{1}A_{2} \cdots A_{k}$ and each $A_{i}$ is nullable. To illustrate the procedure, let $G$ be given as: \[ \begin{eqnarray} S &\rightarrow& A\;B \nonumber \\ A &\rightarrow& A\;a\;A \mid \epsilon \nonumber \\ B &\rightarrow& B\;b\;B \mid \epsilon \nonumber \end{eqnarray} \] In this grammar all of $S$, $A$ and $B$ are nullable. The new grammar introduces a new start rule $S^{\prime} \rightarrow S$ and since $S$ is nullable we also add an $\epsilon$ alternative to conclude $S^{\prime} \rightarrow S \mid \epsilon$. Now, for each rule $A \rightarrow X_{1} X_{2} \dots X_{k}$ create rules, $A \rightarrow \alpha_{1}\alpha_{2}\cdots\alpha_{k}$ where \[ \begin{equation} \alpha_{i} = \begin{cases} X_{i} & \text{if $X_{i}$ is a terminal/non-nullable non-terminal} \\ X_{i}\; \text{or}\;\epsilon & \text{if $X_{i}$ is nullable} \end{cases} \end{equation} \] and not all $\alpha_{i}$ are nullable. Applying this procedure then, we get \[ \begin{eqnarray} & S^{\prime} &\rightarrow S \mid \epsilon \nonumber \\ & S &\rightarrow A\;B \mid A \mid B \nonumber \\ & A &\rightarrow A\;a\;A \mid a\;A \mid A\; a \mid a \nonumber \\ & B &\rightarrow B\;b\;B \mid b\;B \mid B\; b \mid b \nonumber \end{eqnarray} \] The net effect is that $\epsilon$-productions have been eliminated but for the $S^{\prime}$ production which does not appear on the right-hand-side of any other rule.

[1] Compilers Principles, Techniques, & Tools by Aho et. al. 2nd Ed. 2007.

Tuesday, September 27, 2016

The fixpoint combinator

Consider the following recursive definition of the factorial function. \[ FAC = \lambda n.\;IF \left(=\;n\;0\right)\;1\;\left(*\;n\;\left(FAC\;\left(-\;n\;1\right)\right)\right) \nonumber \] The definition relies on the ability to name a $\lambda$-abstraction and then to refer to this name inside the $\lambda$-abstraction itself. No such facility is provided by the $\lambda$-calculus. $\beta$-abstraction is applying $\beta$-reduction backwards to introduce new $\lambda$-abstractions, thus $+\;4\;1\leftarrow \left(\lambda x.\;+\;x\;1\right)\; 4$. By $\beta$-abstraction on $FAC$, its definition can be written \[ FAC = \left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right) FAC \nonumber \] This definition has taken the form $FAC = g\;FAC$ where $g = \left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right)$ is without recursion. We see also that $FAC$ is a fixed point ("fixpoint") of $g$. It is clear this fixed point can only depend on $g$ so supposing there were a function $Y$ which takes a function and delivers a fixpoint of the function as the result, we'd have $FAC = Y\;g = g\;(Y\;g)$. Under the assumption such a function exists, in order to build confidence this definition of $FAC$ works, we will try to compute $FAC\;1$. Recall \[ \begin{eqnarray} &FAC& = Y\;g \nonumber \\ &g& = \lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right) \nonumber \end{eqnarray} \] So, \[ \begin{eqnarray} FAC\;1 &\rightarrow& (Y\;g)\; 1 \nonumber \\ &\rightarrow& (g\;(Y\;g))\;1 \nonumber \\ &\rightarrow& (\left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right) (Y\;g))\; 1 \nonumber \\ &\rightarrow& \left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(\left(Y\;g\right)\;\left(-\;n\;1\right)\right)\right)\right)\; 1 \nonumber \\ &\rightarrow& *\;1\;\left(\left(Y\;g\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(g\;\left(Y\;g\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(\left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right)\;\left(Y\;g\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(\left(Y\;g\right)\;\left(-\;n\;1\right)\right)\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;1 \nonumber \\ &=& 1 \nonumber \end{eqnarray} \]

The $Y$ combinator of the $\lambda$-calculus is defined as the $\lambda$-term $Y = \lambda f.\;\left(\lambda x.\;f\;\left(x\;x\right)\right)\left(\lambda x.\;f\;\left(x\;x\right)\right)$. $\beta$ reduction of this term applied to an arbitrary function $g$ proceeds like this: \[ \begin{eqnarray} Y\;g &\rightarrow& \left(\lambda f.\;\left(\lambda x.\;f\;\left(x\;x\right)\right) \left(\lambda x.\;f\;\left(x\;x\right)\right)\right)\;g \nonumber \\ &\rightarrow& \left(\lambda x.\;g\;\left(x\;x\right)\right) \left(\lambda x.\;g\;\left(x\;x\right)\right) \nonumber \\ &\rightarrow& g\;\left(\left(\lambda x.\;g\;\left(x\;x\right)\right)\;\left(\lambda x.\;g\;\left(x\;x\right)\right)\right) \nonumber \\ &=& g\;\left(Y\;g\right) \end{eqnarray} \] The application of this term has produced a fixpoint of $g$. That is, we are satisfied that this term will serve as a definition for $Y$ having the property we need and call it the "fixpoint combinator".

In the untyped $\lambda$-calculus, $Y$ can be defined and that is sufficient for expressing all the functions that can be computed without having to add a special construction to get recursive functions. In typed $\lambda$-calculus, $Y$ cannot be defined as the term $\lambda x.\;f\;(x\;x)$ does not have a finite type. Thus, when implementing recursion in a functional programming language it is usual to implement $Y$ as a built-in function with the reduction rule $Y\;g \rightarrow g\;(Y\;g)$ or, in a strict language, $(Y\; g)\;x \rightarrow (g\;(Y\;g))\;x$ to avoid infinite recursion.

For an OCaml like language, the idea then is to introduce a built-in constant $\mathbf{Y}$ and to denote the function defined by $\mathbf{let\;rec}\;f\;x = e$ as $\mathbf{Y}(\mathbf{fun}\;f\;x \rightarrow e)$. Intuitivly, $\mathbf{Y}$ is a fixpoint operator that associates a functional $F$ of type $\alpha \rightarrow \beta$ with a fixpoint $Y(F)$ of type $\alpha \rightarrow \beta \rightarrow \alpha$, that is, a value having the property $\mathbf{Y}\;F = F\;\left(\mathbf{Y}\;F\right)$. The relevant deduction rules involving this constant are: \[ \begin{equation} \frac{\vdash f\;(Y\;f)\;x \Rightarrow v} {\vdash (Y\;f)\;x \Rightarrow v} \tag{App-rec} \end{equation} \] \[ \begin{equation} \frac{\vdash e_{2}\left[Y(\mathbf{fun}\;f\;x \rightarrow e_{1})/f\right] \Rightarrow v} {\vdash \mathbf{let\;rec}\;f\;x=e_{1}\;\mathbf{in}\;e_{2} \Rightarrow v} \nonumber \tag {Let-rec} \end{equation} \]

[1] The Implementation of Functional Programming Languages,Simon Peyton Jones, 1987.
[2] The Functional Approach to Programming, Guy Cousineau, Michel Mauny, 1998.

Tuesday, September 20, 2016

Custom operators in OCaml

If like me, you've always been a little hazy on the rules for defining OCaml operators then, this little post might help!

It is possible to "inject" user-defined operator syntax into OCaml programs. Here's how it works. First we define a set of characters called "symbol characters".

Symbol character (definition)

A character that is one of

! $ % & * + - . / : < = > ? @ ^ | ~

Prefix operators

The ! ("bang") prefix operator, has a predefined semantic as the operation of "de-referencing" a reference cell. A custom prefix operator can made by from a ! followed by one or more symbol characters.

So, to give some examples, one can define prefix operators like !!, !~ or even something as exotic as !::>. For example, one might write something like

let ( !+ ) x : int ref → unit = incr x
as a syntactic sugar equivalent to fun x → incr x

Additionally, prefix operators can begin with one of ~ and ? and, as in the case of !, must be followed by one or more symbol characters. So, in summary, a prefix operator begins with one of

! ~ ?
and is followed by one or more symbol characters.

For example let ( ~! ) x = incr x defines an alternative syntax equivalent to the !+ operator presented earlier.

Prefix operators have the highest possible precedence.

Infix operators

It is in fact possible to define operators in 5 different categories. What distinguish these categories from each other are their associativity and precedence properties.

Level 0

Level 0 operators are left associative with the same precedence as =. A level 0 operator starts with one of

= < > | & $
and is followed by zero or more symbol chars. For example, >>= is an operator much beloved by monadic programmers and |> (pipe operator) is a builtin equivalent to let ( |> ) x f = f x.

Level 1

Level 1 operators are right associative, have a precedence just above = and start with one of

@ ^
. That is, these operators are consistent with operations involving joining things. @@ (the "command" operator) of course has a predefined semantic as function application, that is, equivalent to the definition let ( @@ ) f x = f x.

Level 2

Level 2 operators are left associative have a precedence level shared with + and - and indeed, are defined with a leading (one of)

+ -
and, as usual, followed by a sequence of symbol characters. These operators are consistent for usage with operations generalizing addition or difference like operations. Some potential operators of this kind are +~, ++ and so on.

Level 3

Level 3 operators are also left associative and have a precedence level shared with * and /. Operators of this kind start with one of

* / %
followed by zero or more symbol characters and are evocative of operations akin to multiplication, division. For example, *~ might make a good companion for +~ of the previous section.

Level 4

Level 4 operators are right associative and have a precedence above *. The level 4 operators begin with

and are followed by zero or more symbol characters. The operation associated with ** is exponentiation (binds tight and associates to the right). The syntax **~ would fit nicely into the +~, *~ set of the earlier sections.

Saturday, August 27, 2016

Balanced binary search trees

The type of "association tables" (binary search trees).

type (α, β) t =
| Empty
| Node of (α , β) t * α * β * (α, β) t * int
There are two cases : a tree that is empty or, a node consisting of a left sub-tree, a key, the value associated with that key, a right sub-tree and, an integer representing the "height" of the tree (the number of nodes to traverse before reaching the most distant leaf).

The binary search tree invariant will be made to apply in that for any non empty tree $n$, every node in the left sub-tree is ordered less than $n$ and every node in the right sub-tree of $n$ is ordered greater than $n$ (in this program, ordering of keys is performed using the function).

This function, height, given a tree, extracts its height.

let height : (α, β) t -> int = function
  | Empty -> 0
  | Node (_, _, _, _, h) -> h

The value empty, is a constant, the empty tree.

let empty : (α, β) t = Empty

create l x d r creates a new non-empty tree with left sub-tree l, right sub-tree r and the binding of key x to the data d. The height of the tree created is computed from the heights of the two sub-trees.

let create (l : (α, β) t) (x : α) (d : β) (r : (α, β) t) : (α, β) t =
  let hl = height l and hr = height r in
  Node (l, x, d, r, (max hl hr) + 1)

This next function, balance is where all the action is at. Like the preceding function create, it is a factory function for interior nodes and so takes the same argument list as create. It has an additional duty though in that the tree that it produces takes balancing into consideration.

let balance (l : (α, β) t) (x : α) (d : β) (r : (α, β) t) : (α, β) t =
  let hl = height l and hr = height r in
  if hl > hr + 1 then
    match l with
In this branch of the program, it has determined that production of a node with the given left and right sub-trees (denoted $l$ and $r$ respectively) would be unbalanced because $h(l) > hr(1) + 1$ (where $h$ denotes the height function).

There are two possible reasons to account for this. They are considered in turn.

    (*Case 1*)
    | Node (ll, lv, ld, lr, _) when height ll >= height lr ->
      create ll lv ld (create lr x d r)
So here, we find that $h(l) > h(r) + 1$, because of the height of the left sub-tree of $l$.
    (*Case 2*)
    | Node (ll, lv, ld, Node (lrl, lrv, lrd, lrr, _), _) ->
      create (create ll lv ld lrl) lrv lrd (create lrr x d r)
In this case, $h(l) > h(r) + 1$ because of the height of the right sub-tree of $l$.
    | _ -> assert false
We assert false for all other patterns as we aim to admit by construction no further possibilities.

We now consider the case $h(r) > h(l) + 1$, that is the right sub-tree being "too long".

  else if hr > hl + 1 then
    match r with

There are two possible reasons.

    (*Case 3*)
    | Node (rl, rv, rd, rr, _) when height rr >= height rl ->
      create (create l x d rl) rv rd rr
Here $h(r) > h(l) + 1$ because of the right sub-tree of $r$.
    (*Case 4*)
    | Node (Node (rll, rlv, rld, rlr, _), rv, rd, rr, _) ->
      create (create l x d rll) rlv rld (create rlr rv rd rr)
Lastly, $h(r) > h(l) + 1$ because of the left sub-tree of $r$.
    | _ -> assert false
Again, all other patterns are (if we write this program correctly according to our intentions,) impossible and so, assert false as there are no further possibilities.

In the last case, neither $h(l) > h(r) + 1$ or $h(r) > h(l) + 1$ so no rotation is required.

    create l x d r

add x data t computes a new tree from t containing a binding of x to data. It resembles standard insertion into a binary search tree except that it propagates rotations through the tree to maintain balance after the insertion.

let rec add (x : α) (data : β) : (α, β) t -> (α, β) t = function
    | Empty -> Node (Empty, x, data, Empty, 1)
    | Node (l, v, d, r, h) ->
      let c = compare x v in
      if c = 0 then
        Node (l, x, data, r, h)
      else if c < 0 then
        balance (add x data l) v d r
        balance l v d (add x data r)

To implement removal of nodes from a tree, we'll find ourselves needing a function to "merge" two binary searchtrees $l$ and $r$ say where we can assume that all the elements of $l$ are ordered before the elements of $r$.

let rec merge (l : (α, β) t) (r : (α, β) t) : (α, β) t = 
  match (l, r) with
  | Empty, t -> t
  | t, Empty -> t
  | Node (l1, v1, d1, r1, h1), Node (l2, v2, d2, r2, h2) ->
    balance l1 v1 d1 (balance (merge r1 l2) v2 d2 r2)
Again, rotations are propagated through the tree to ensure the result of the merge results in a balanced tree.

With merge available, implementing remove becomes tractable.

let remove (id : α) (t : (α, β) t) : (α, β) t = 
  let rec remove_rec = function
    | Empty -> Empty
    | Node (l, k, d, r, _) ->
      let c = compare id k in
      if c = 0 then merge l r else
        if c < 0 then balance (remove_rec l) k d r
        else balance l k d (remove_rec r) in
  remove_rec t

The remaining algorithms below are "stock" algorithms for binary search trees with no particular consideration of balancing necessary and so we won't dwell on them here.

let rec find (x : α) : (α, β) t -> β = function
  | Empty ->  raise Not_found
  | Node (l, v, d, r, _) ->
    let c = compare x v in
    if c = 0 then d
    else find x (if c < 0 then l else r)

let rec mem (x : α) : (α, β) t -> bool = function
  | Empty -> false
  | Node (l, v, d, r, _) ->
    let c = compare x v in
    c = 0 || mem x (if c < 0 then l else r)
let rec iter (f : α -> β -> unit) : (α, β) t -> unit = function
  | Empty -> ()
  | Node (l, v, d, r, _) ->
    iter f l; f v d; iter f r

let rec map (f : α -> β -> γ) : (α, β) t -> (α, γ) t = function
  | Empty -> Empty
  | Node (l, k, d, r, h) -> 
    Node (map f l, k, f k d, map f r, h)

let rec fold (f : α -> β -> γ -> γ) (m : (α, β) t) (acc : γ) : γ =
  match m with
  | Empty -> acc
  | Node (l, k, d, r, _) -> fold f r (f k d (fold f l acc))

open Format

let print 
    (print_key : formatter -> α -> unit)
    (print_data : formatter -> β -> unit)
    (ppf : formatter)
    (tbl : (α, β) t) : unit =
  let print_tbl ppf tbl =
    iter (fun k d -> 
           fprintf ppf "@[<2>%a ->@ %a;@]@ " print_key k print_data d)
      tbl in
  fprintf ppf "@[[[%a]]@]" print_tbl tbl

The source code for this post can be found in the file 'ocaml/misc/' in the OCaml source distribution. More information on balanced binary search trees including similar but different implementation techniques and complexity analyses can be found in this Cornell lecture and this one.

Friday, August 19, 2016

Even Sillier C++

The C++ try...catch construct provides a facility for discrimination of exceptions based on their types. This is a primitive "match" construct. It turns out, this is enough to encode sum types.

The program to follow uses the above idea to implement an interpreter for the language of additive expressions using exception handling for case discrimination.

#include <iostream>
#include <cassert>
#include <exception>
#include <memory>

struct expr {
  virtual ~expr() {}

  virtual void throw_ () const = 0;

using expr_ptr = std::shared_ptr<expr const>;

class expr is an abstract base class, class int_ and class add derived classes corresponding to the two cases of expressions. Sub-expressions are represented as std::shared_ptr<expr> instances.

struct int_ : expr { 
  int val; 
  int_ (int val) : val{val}

  void throw_ () const { throw *this; } 

struct add : expr { 
  expr_ptr left; 
  expr_ptr right; 

  template <class U, class V>
  add (U const& left, V const& right) 
    : left {expr_ptr{new U{left}}}
    , right {expr_ptr{new V{right}}}

  void throw_ () const { throw *this; } 

With the above machinery in place, here then is the "interpreter". It is implemented as a pair of mutually recursive functions.

int eval_rec ();

int eval (expr const& xpr) {
  try {
    xpr.throw_ ();
  catch (...) {
    return eval_rec ();

int eval_rec () {
  assert (std::current_exception());

  try {
  catch (int_ const& i) {
    return i.val;
  catch (add const& op) {
    return eval (*op.left) +  eval (*op.right);

This little program exercises the interpreter on the expression $(1 + 2) + 3$.

int main () {

    // (1 + 2) + 3
    std::cout << eval (add{add{int_{1}, int_{2}}, int_{3}}) << std::endl;
  catch (...){
    std::cerr << "Unhandled exception\n";
  return 0;

Credit to Mathias Gaunard who pointed out using a virtual function for the throwing of an expression, removed the need for explicit dynamic_cast operations in an earlier version of this program.

Saturday, July 23, 2016

Simple category theory constructions

This post is about some very simple category theory constructions and how one can model them in C++.

First a type for binary products.

template <class A, class B>
using product = std::pair<A, B>;

//`fst (x, y)` is the projection `x`
template <class A, class B>
inline auto fst (product<A, B> const& p) {
  return p.first;

//`snd (x, y)` is the projection `y`
template <class A, class B>
inline auto snd (product<A, B> const& p) {
  return p.second;

//`mk_product (a, b) computes the product of `a` and `b`
template <class A, class B>
inline product<A, B> mk_product (A&& a, B&& b) {
  return  std::make_pair (std::forward<A> (a), std::forward<B> (b));

Now $dup : X \rightarrow X \times X$ defined by $x \mapsto (x,\; x)$.

//`dup a` computes the product `(a, a)`
template <class A>
inline product<A, A> dup (A const& x) { return mk_product (x, x); }

Next up, $twist : X \times Y \rightarrow Y \times X$ defined by $(x,\; y) \mapsto (y,\; x)$.

//`twist (x, y)` is the product `(y, x)`
template <class A, class B>
inline product<B, A> twist (product<A, B> const& p) {
  return mk_product (snd (p), fst (p));

If $f : U \rightarrow R$ and $g : V \rightarrow S$ then we have $ravel : U \times V \rightarrow R \times S$ defined by $(x,\; y) \mapsto (f (x),\; g (y))$.

//`ravel f g (x, y)` is the product `(f x, g y)` (credit to Max
//Skaller on the name)
auto ravel = [=](auto f) {
  return [=](auto g) {
    return [=](auto const& x) { 
      return mk_product (f (fst (x)), g (snd (x))); 

If $X \times Y$ is a (binary) product with projections $\pi_{x}$ and $\pi_{y}$, $Z$ an object, morphisms $f : Z \rightarrow X$, $g : Z \rightarrow Y$ then $\left\langle f, g \right\rangle : Z \rightarrow X \times Y$ is the mediating arrow $z \mapsto (f (z),\;g (z))$.

//The product of morphisms <`f`, `g`> (see
auto prod = [=](auto f, auto g) {
  return [=](auto const& z) { return mk_product (f (z), g (z)); };

We can use the Pretty good sum type library to define a type suitable for modeling (binary) coproducts.

template <class A>
struct Left { 
  A a; 
  template <class U> explicit Left (U&& u) : a {std::forward<U> (u)} {}
  A const& value () const { return a; }

template <class B>
struct Right { 
  B b; 
  template <class U> explicit Right (U&& u) : b {std::forward<U> (u)} {}
  B const& value () const { return b; }

template <class A, class B>
using sum = pgs::sum_type<Left<A>, Right<B>>;
template <class> struct sum_fst_type;
template <class A, class B>
  struct sum_fst_type<sum<A, B>> { using type = A; };
template <class S> struct sum_snd_type;
template <class A, class B>
  struct sum_snd_type<sum<A, B>> { using type = B; };

template <class S>
using sum_fst_t = typename sum_fst_type<S>::type;
template <class S>
using sum_snd_t = typename sum_snd_type<S>::type;

If $X + Y$ is a (binary) sum with injections $i_{x}$ and $i_{y}$, $Z$ an object, morphisms $f : X \rightarrow Z$, $g : Y \rightarrow Z$ then $\left[ f, g \right] : X \times Y \rightarrow Z $ is the mediating arrow $\begin{align*} e \mapsto \begin{cases} f (e) & \text{if $e \in X$} \\ g (e) & \text{if $e \in Y$} \end{cases} \end{align*}$.

//The coproduct of morphisms [`f, `g`] (see
template <class S>
auto co_product = [=](auto f) {
  return [=](auto g) {
    return [=](S const& s) {
      using A = sum_fst_t<S>;
      using B = sum_snd_t<S>;
      using lres_t = decltype (f (std::declval<A>()));
      using rres_t = decltype (g (std::declval<B>()));
      static_assert (
        std::is_same<lres_t, rres_t>::value
       , "co_product : result types differ");
      using res_t = lres_t;
      return s.match<lres_t>(
        [=](Left<A> const& l) { return f (l.value ()); },
        [=](Right<B> const& r) { return g (r.value ()); }

That's it. Here's some example usage of all this stuff.

  auto succ=[](int i) { return i + 1; };
  auto pred=[](int i) { return i - 1; };
  auto twice=[](int i) { return 2 * i; };
  auto square=[](int i) { return i * i; };
  auto add=[](product<int, int> const& s) { return (fst (s) + snd (s)); };



  auto p = dup (1);
  std::cout << p << std::endl; //Prints '(1, 1)'

  p = prod (succ, pred) (4);
  std::cout << p << std::endl; //Prints '(5, 3)'

  p = twist (p);
  std::cout << p << std::endl; //Prints '(3, 5)'
  p = ravel (succ) (pred) (p);
  std::cout << p << std::endl; //Prints '(4, 4)'



  sum<int, float> l{pgs::constructor<Left<int>>{}, 1};
  sum<int, float> r{pgs::constructor<Right<float>>{}, 1.0f};
  std::cout << 
    co_product<sum<int, float>> 
      ([=](int i) { return std::to_string(i); })
      ([=](float f) { return std::to_string(f); })
    << std::endl;
  ;//Prints '1'
  std::cout << 
    co_product<sum<int, float>> 
      ([=](int i) { return std::to_string(i); })
      ([=](float f) { return std::to_string(f); })
    << std::endl;
  ;//Prints '1.000000'


Friday, June 17, 2016

Generic mappings over pairs

Browsing around on Oleg Kiselyov's excellent site, I came across a very interesting paper about "Advanced Polymorphism in Simpler-Typed Languages". One of the neat examples I'm about to present is concerned with expressing mappings over pairs that are generic not only in the datatypes involved but also over the number of arguments. The idea is to produce a family of functions $pair\_map_{i}$ such that

pair_map_1 f g (x, y) (x', y') → (f x, g y) 
pair_map_2 f g (x, y) (x', y') → (f x x', g y y') 
pair_map_3 f g (x, y) (x', y') (x'', y'', z'') → (f x x' x'', g y y' y'')
The technique used to achieve this brings a whole bunch of functional programming ideas together : higher order functions, combinators and continuation passing style (and also leads into topics like the "value restriction" typing rules in the Hindley-Milner system).
let ( ** ) app k = fun x y -> k (app x y)
let pc k a b = k (a, b)
let papp (f1, f2) (x1, x2) = (f1 x1, f2 x2)
let pu x = x
With the above definitions, $pair\_map_{i}$ is generated like so.
(*The argument [f] in the below is for the sake of value restriction*)
let pair_map_1 f = pc (papp ** pu) (f : α -> β)
let pair_map_2 f = pc (papp ** papp ** pu) (f : α -> β -> γ)
let pair_map_3 f = pc (papp ** papp ** papp ** pu) (f : α -> β -> γ -> δ)
For example,
# pair_map_2 ( + ) ( - ) (1, 2) (3, 4) ;;
- : int * int = (4, -2)

Reverse engineering how this works requires a bit of algebra.

Let's tackle $pair\_map_{1}$. First

pc (papp ** pu) = (λk f g. k (f, g)) (papp ** pu) = λf g. (papp ** pu) (f, g)
papp ** pu = λx y. pu (papp x y) = λx y. papp x y
λf g. (papp ** pu) (f, g) =
    λf g. (λ(a, b) (x, y). (a x, b y)) (f, g) =
    λf g (x, y). (f x, g y)
that is, pair_map_1 = pc (papp ** pu) = λf g (x, y). (f x, g y) and, we can read the type off from that last equation as (α → β) → (γ → δ) → α * γ → β * δ.

Now for $pair\_map_{2}$. We have

pc (papp ** papp ** pu) =
    (λk f g. k (f, g)) (papp ** papp ** pu) =
    λf g. (papp ** papp ** pu) (f, g)
papp ** papp ** pu = papp ** (papp ** pu) =
    papp ** (λa' b'. pu (papp a' b')) =
    papp ** (λa' b'. papp a' b') = 
    λa b. (λa' b'. papp a' b') (papp a b)
which means,
pc (papp ** papp ** pu) = 
    λf g. (papp ** papp ** pu) (f, g) =
    λf g. (λa b.(λa' b'. papp a' b') (papp a b)) (f, g) =
    λf g. (λb. (λa' b'. papp a' b') (papp (f, g) b)) =
    λf g. λ(x, y). λa' b'. (papp a' b') (papp (f, g) (x, y)) =
    λf g. λ(x, y). λa' b'. (papp a' b') (f x, g y) =
    λf g. λ(x, y). λb'. papp (f x, g y) b' =
    λf g. λ(x, y). λ(x', y'). papp (f x, g y) (x', y') =
    λf g (x, y) (x', y'). (f x x', g y y')
that is, a function in two binary functions and two pairs as we expect. Phew! The type in this instance is (α → β → γ) → (δ → ε → ζ) → α * δ → β * ε → γ * ζ.

To finish off, here's the program transliterated into C++(14).

#include <utility>
#include <iostream>

//let pu x = x
auto pu = [](auto x) { return x; };

//let ( ** ) app k  = fun x y -> k (app x y)
template <class F, class K>
auto operator ^ (F app, K k) {
  return [=](auto x) {
    return [=] (auto y) {
      return k ((app (x)) (y));

//let pc k a b = k (a, b)
auto pc = [](auto k) {
  return [=](auto a) {
    return [=](auto b) { 
      return k (std::make_pair (a, b)); };

//let papp (f, g) (x, y) = (f x, g y)
auto papp = [](auto f) { 
  return [=](auto x) { 
    return std::make_pair (f.first (x.first), f.second (x.second)); };

int main () {

  auto pair = &std::make_pair<int, int>;

  auto succ= [](int x){ return x + 1; };
  auto pred= [](int x){ return x - 1; };
  auto p  = (pc (papp ^ pu)) (succ) (pred) (pair (1, 2));
  std::cout << p.first << ", " << p.second << std::endl;

  auto add = [](int x) { return [=](int y) { return x + y; }; };
  auto sub = [](int x) { return [=](int y) { return x - y; }; };
  auto p = pc (papp ^ papp ^ pu) (add) (sub) (pair(1, 2)) (pair (3, 4));
  std::cout << p.first << ", " << p.second << std::endl;

  return 0;

Thursday, April 21, 2016

Oh! Pascal!

I can't help but want to share my joy at coming across this pearl of a program from the "Pascal User Manual and Report" - Jensen and Wirth (circa 1974). In my edition, it's program 4.7 (graph1.pas).

This is it, rephrased in OCaml.

(* Graph of f x = exp (-x) * sin (2 * pi * x)

  Program 4.7, Pascal User Manual and Report, Jensen & Wirth

let round (x : float) : int =
  let f, i = 
    let t = modf x in 
    (fst t, int_of_float@@ snd t) in
  if f = 0.0 then i
  else if i >= 0 then
    if f >= 0.5 then i + 1 else i
  else if -.f >= 0.5 then i - 1 else i

let graph (oc : out_channel) : unit =
  (*The x-axis runs vertically...*)
  let s = 32. in (*32 char widths for [y, y + 1]*)
  let h = 34 in (*char position of x-axis*)
  let d = 0.0625 in (*1/16, 16 lines for [x, x + 1]*)
  let c = 6.28318 in (* 2pi *)
  let lim = 32 in
  for i = 0 to lim do
    let x = d *. (float_of_int i) in
    let y = exp (-.x) *. sin (c *. x) in
    let n = round (s *. y) + h in
    for _ = n downto 0 do output_char oc ' '; done;
    output_string oc "*\n"

let () = print_newline (); graph stdout; print_newline ()

The output from the above is wonderful :)


Wednesday, April 13, 2016

Dictionaries as functions

This is an "oldie but a goodie". It's super easy.

A dictionary is a data structure that represents a map from keys to values. The question is, can this data structure and its characteristic operations be encoded using only functions?

The answer of course is yes and indeed, here's one such an encoding in OCaml.

(*The type of a dictionary with keys of type [α] and values of type
type (α, β) dict = α -> β option

(*The empty dictionary maps every key to [None]*)
let empty (k : α) : β option = None

(*[add d k v] is the dictionary [d] together with a binding of [k] to
let add (d : (α, β) dict) (k : α) (v : β) : (α, β) dict = 
  fun l -> 
    if l = k then Some v else d l

(*[find d k] retrieves the value bound to [k]*)
let find (d : (α, β) dict) (k : α) : β option = d k
Test it like this.

  Name                            | Age
  "Felonius Gru"                  |  53
  "Dave the Minion"               | 4.54e9
  "Dr. Joseph Albert Nefario"     |  80

let despicable = 
          empty "Felonius Gru" 53
       "Dave the Minion" (int_of_float 4.54e9)
    "Dr. Nefario" 80 

let _ = 
  find despicable "Dave the Minion" |> 
      function | Some x -> x | _ -> failwith "Not found"

Moving on, can we implement this in C++? Sure. Here's one way.

#include <pgs/pgs.hpp>

#include <functional>
#include <iostream>
#include <cstdint>

using namespace pgs;

// -- A rough and ready `'a option` (given the absence of
// `std::experimental::optional`

struct None {};

template <class A>
struct Some { 
  A val;
  template <class Arg>
  explicit Some (Arg&& s) : val { std::forward<Arg> (s) }

template <class B>
using option = sum_type<None, Some<B>>;

template <class B>
std::ostream& operator << (std::ostream& os, option<B> const& o) {
  return o.match<std::ostream&>(
    [&](Some<B> const& a) -> std::ostream& { return os << a.val; },
    [&](None) -> std::ostream& { return os << "<empty>"; }

//-- Encoding of dictionaries as functions

template <class K, class V>
using dict_type = std::function<option<V>(K)>;

//`empty` is a dictionary constant (a function that maps any key to
template <class A, class B>
dict_type<A, B> empty = 
  [](A const&) { 
    return option<B>{ constructor<None>{} }; 

//`add (d, k, v)` extends `d` with a binding of `k` to `v`
template <class A, class B>
dict_type<A, B> add (dict_type<A, B> const& d, A const& k, B const& v) {
  return [=](A const& l) {
    return (k == l) ? option<B>{ constructor<Some<B>>{}, v} : d (l);

//`find (d, k)` searches for a binding in `d` for `k`
template <class A, class B>
option<B> find (dict_type<A, B> const& d, A const& k) {
  return d (k);

//-- Test driver

int main () {

  using dict_t = dict_type<std::string, std::int64_t>;

  auto nil = empty<std::string, std::int64_t>;
  dict_t(*insert)(dict_t const&, std::string const&, std::int64_t const&) = &add;

  dict_t despicable = 
    insert (
      insert (
        insert (nil
           , std::string {"Felonius Gru"}, std::int64_t{53})
           , std::string {"Dave the Minion"}, std::int64_t{4530000000})
          , std::string {"Dr. Joseph Albert Nefario"}, std::int64_t{80})

  std::cout << 
    find (despicable, std::string {"Dave the Minion"}) << std::endl;

  return 0;

Sunday, April 3, 2016

C++ : Streams

In this blog post, types and functions were presented in OCaml for modeling streams. This post takes the action to C++.

First, the type definition for a stream.

struct Nil {};
template <class T> class Cons;

template <class T>
using stream = sum_type <
  , recursive_wrapper<Cons<T>>
The definition is in terms of the sum_type<> type from the "pretty good sum" library talked about here.

The definition of Cons<>, will be in terms of "thunks" (suspensions). They're modeled as procedures that when evaluated, compute streams.

template <class T>
using stream_thunk = std::function<stream<T>()>;
To complete the abstraction, a function that given a suspension, "thaws" it.
template <class T> inline 
stream<T> force (stream_thunk<T> const& s) { 
  return s (); 

The above choices made, here is the definition for Cons<>.

template <class T>
class Cons {
  using value_type = T;
  using reference = value_type&;
  using const_reference = value_type const&;
  using stream_type = stream<value_type>;

  using stream_thunk_type = stream_thunk<value_type>;

  template <class U, class V>
  Cons (U&& h, V&& t) : 
    h {std::forward<U> (h)}, t {std::forward<V> (t)}

  const_reference hd () const { return h; }
  stream_type tl () const { return force (t); }

  value_type h;
  stream_thunk_type t;

Next, utility functions for working with streams.

The function hd () gets the head of a stream and tl () gets the stream that remains when the head is stripped off.

template <class T>
T const hd (stream<T> const& s) {
  return s.template match<T const&> (
      [](Cons<T> const& l) -> T const& { return l.hd (); }
    , [](otherwise) -> T const & { throw std::runtime_error { "hd" }; }

template <class T>
stream<T> tl (stream<T> const& l) {
  return l.template match <stream<T>> (
    [] (Cons<T> const& s) -> stream <T> { return (); }
  , [] (otherwise) -> stream<T> { throw std::runtime_error{"tl"}; }

The function take () returns the the first $n$ values of a stream.

template <class T, class D>
D take (unsigned int n, stream <T> const& s, D dst) {
  return (n == 0) ? dst :
    s.template match<D>(
       [&](Nil const& _) -> D { return  dst; },
       [&](Cons<T> const& l) -> D { 
         return take (n - 1, (), *dst++ = l.hd ()); }

It's time to share a little "hack" I picked up for writing infinite lists.

  • To start, forget about streams;
  • Write your list using regular lists;
  • Ignore the fact that it won't terminate;
  • Rewrite in terms of Cons and convert the tail to a thunk.

For example, in OCaml the (non-terminating!) code

  let naturals = 
    let rec loop x = x :: loop (x + 1) in
  next 0
leads to this definition of the stream of natural numbers.
let naturals =
 let rec loop x = Cons (x, lazy (loop (x + 1))) in
loop 0

Putting the above to work, a generator for the stream of natural numbers can be written like this.

class natural_numbers_gen {
  using int_stream = stream<int>;
  int start;

  int_stream from (int x) const {
    return int_stream{
      constructor<Cons<int>>{}, x, [=]() { return this->from (x + 1); }
  explicit natural_numbers_gen (int start) : start (start) 

  explicit operator int_stream() const { return from (start); }
The first $10$ (say) natural numbers can then be harvested like this.
std::vector<int> s;
take (10, stream<int> (natural_numbers_gen{0}), std::back_inserter (s));

The last example, a generator of the Fibonacci sequence. Applying the hack, start with the following OCaml code.

  let fibonacci_numbers = 
    let rec fib a b = a :: fib b (a + b) in
    fib 0 1
The rewrite of this code into streams then leads to this definition.
let fibonnaci_sequence = 
  let rec fib a b = Cons (a, lazy (fib b (a + b))) in
fib 0 1
Finally, casting the above function into C++ yields the following.
class fibonacci_numbers_gen {
  using int_stream = stream<int>;
  int start;

  int_stream loop (int a, int b) const {
    return int_stream{
      constructor<Cons<int>>{}, a, [=]() {return this->loop (b, a + b); }
  explicit fibonacci_numbers_gen () 

  explicit operator int_stream() const { return loop (0, 1); }

Saturday, April 2, 2016



This post is inspired by one of those classic "99 problems in Prolog".What we are looking for here are two functions that satisfy these signatures.

val rotate_left : int -> α list -> α list
val rotate_right : int -> α list -> α list 
rotate_left n rotates a list $n$ places to the left, rotate_right n rotates a list $n$ places to the right. Examples:
# rotate_left 3 ['a';'b';'c';'d';'e';'f';'g';'h'] ;;
- : char list = ['d'; 'e'; 'f'; 'g'; 'h'; 'a'; 'b'; 'c']

# rotate_left (-2) ['a';'b';'c';'d';'e';'f';'g';'h'] ;;
- : char list = ['g'; 'h'; 'a'; 'b'; 'c'; 'd'; 'e'; 'f']
Of course, rotate_left and rotate_right are inverse functions of each other so we expect, for any int $x$ and list $l$, rotate_right x @@ rotate_left x l $=$ rotate_left x @@ rotate_right x l $=$ l.

Well, there are a variety of solutions to this problem with differing degrees of verbosity, complexity and efficiency. My own attempt at a solution resulted in this.

let rec drop (k : int) (l : α list) : α list =
  match k, l with
  | i, _ when i <= 0 -> l
  | _, [] -> []
  | _, (_ :: xs) -> drop (k - 1) xs

let rec take (k : int) (l : α list) : α list =
  match k, l with
  | i, _ when i <= 0 -> []
  | _, [] -> []
  | _, (x :: xs)  -> x :: take (k - 1) xs

let split_at (n : int) (l : α list) : α list * α list = 
  (take n l), (drop n l)

let rec rotate_left (n : int) (l : α list) : α list =
  match n with
  | _ when n = 0 -> l
  | _ when n < 0 ->  rotate_right (-n) l
  | _ -> 
    let m : int = List.length l in
    let k : int = n mod m in
    let (l : α list), (r : α list) = split_at k l in 
    r @ l

and rotate_right (n : int) (l : α list) : α list =
  match n with
  | _ when n = 0 -> l
  | _ when n < 0 ->  rotate_left (-n) l
  | _ -> 
    let m : int = List.length l in
    let k : int = m - n mod m in
    let (l : α list), (r : α list) = split_at k l in 
    r @ l

So far so good, but then I was shown the following solution in Haskell.

rotateLeft n xs 
  | n >= 0     = take (length xs) $ drop n $ concat $ repeat xs
  | otherwise  = rotateLeft (length xs + n) xs

rotateRight n = rotateLeft (-n)
I found that pretty nifty! See, in the function rotateLeft, repeat xs creates an infinite list of lists, (each a copy of xs), "joins" that infinite list of lists into one infinite list, then the first $n$ elements are dropped from that the list and we take the next length xs which gets us the original list rotated left $n$ places.

I felt compelled to attempt to emulate the program above in OCaml.

The phrasing "works" in Haskell due to the feature of lazy evaluation. OCaml on the other hand is eagerly evaluated. Lazy evaluation is possible in OCaml however, you just need to be explicit about it. Here's a type for "lazy lists" aka "streams".

type α stream =  Nil | Cons of α * α stream Lazy.t
A value of type α Lazy.t is a deferred computation, called a suspension that has the result type α. The syntax lazy$(expr)$ makes a suspension of $expr$, without yet evaluating $expr$. "Forcing" the suspension (using Lazy.force) evaluates $expr$ and returns its result.

Next up, functions to get the head and tail of a stream.

let hd = function | Nil -> failwith "hd" | Cons (h, _) -> h
let tl = function | Nil -> failwith "tl" | Cons (_, t) -> Lazy.force t
Also useful, a function to lift an α list to an α stream.
let from_list (l : α list) : α stream =
  List.fold_right (fun x s -> Cons (x, lazy s)) l Nil

Those are the basic building blocks. Now we turn attention to implementing repeat x to create infinite lists of the repeated value $x$.

let rec repeat (x : α) : α stream = Cons (x, lazy (repeat x))

Now to implement concat (I prefer to call this function by its alternative name flatten).

The characteristic operation of flatten is the joining together of two lists. For eager lists, we can write a function join that appends two lists like this.

let rec join l m =
  match l with
  | [] -> m
  | h :: t -> h :: (join t m)
This generalizes naturally to streams.
let rec join (l : α stream) (m : α stream) =
  match l with
  | Nil -> m
  | Cons (h, t) -> Cons (h, lazy (join (Lazy.force t) m))
For eager lists, we can write flatten in terms of join.
let rec flatten : α list list -> α list = function
   | [] -> []
   | (h :: tl) -> join h (flatten tl)
Emboldened by our earlier success we might try to generalize it to streams like this.
let rec flatten (l : α stream stream) : α stream =
   match l with
   | Nil -> lazy Nil
   | Cons (l, r) ->  join l (flatten (Lazy.force r))
Sadly, no. This definition is going to result in stack overflow. There is an alternative phrasing of flatten we might try.
let rec flatten = function
  | [] -> []
  | [] :: t -> flatten t
  | (x :: xs) :: t -> x :: (flatten (xs :: t))
Happy to say, this one generalizes and gets around the eager evaluation problem that causes the unbounded recursion.
let rec flatten : α stream stream -> α stream = function
  | Nil -> Nil
  | Cons (Nil, t) -> flatten (Lazy.force t)
  | Cons (Cons (x, xs), t) ->
      Cons (x, lazy (flatten (Cons (Lazy.force xs, t))))

take and drop are straight forward generalizations of their eager counterparts.

let rec drop (n : int) (lst : α stream ) : α stream = 
  match (n, lst) with
  | (n, _) when n < 0 -> invalid_arg "negative index in drop"
  | (n, xs) when n = 0 -> xs
  | (_, Nil) -> Nil
  | (n, Cons (_, t)) -> drop (n - 1) (Lazy.force t)

let rec take (n : int) (lst : α stream) : α list = 
  match (n, lst) with
  | (n, _) when n < 0 -> invalid_arg "negative index in take"
  | (n, _) when n = 0 -> []
  | (_, Nil) -> []
  | (n, Cons (h, t)) -> h :: (take (n - 1) (Lazy.force t))

Which brings us to the lazy version of rotate expressed in about the same number of lines of code!

let rec rotate_left (k : int) (l : α list) : α list =
  let n = List.length l in
  if k >= 0 then
    l |> from_list |> repeat |> flatten |> drop k |> take n
  else rotate_left (n + k) l

let rotate_right (n : int) : α list -> α list = rotate_left (-n)