Poly.v

(** * Poly: Polymorphism and Higher-Order Functions *)

(* Final reminder: Please do not put solutions to the exercises in
   publicly accessible places.  Thank you!! *)

(* Suppress some annoying warnings from Coq: *)
Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From LF Require Export Lists.

(* ################################################################# *)
(** * Polymorphism *)

(** In this chapter we continue our development of basic
    concepts of functional programming.  The critical new ideas are
    _polymorphism_ (abstracting functions over the types of the data
    they manipulate) and _higher-order functions_ (treating functions
    as data).  We begin with polymorphism. *)

(* ================================================================= *)
(** ** Polymorphic Lists *)

(** For the last chapter, we've been working with lists
    containing just numbers.  Obviously, interesting programs also
    need to be able to manipulate lists with elements from other
    types -- lists of booleans, lists of lists, etc.  We _could_ just
    define a new inductive datatype for each of these, for
    example... *)

Inductive boollist : Type :=
  | bool_nil
  | bool_cons (b : bool) (l : boollist).

(** ... but this would quickly become tedious, partly because we
    have to make up different constructor names for each datatype, but
    mostly because we would also need to define new versions of all
    our list manipulating functions ([length], [rev], etc.) and all
    their properties ([rev_length], [app_assoc], etc.) for each
    new datatype definition. *)

(** To avoid all this repetition, Coq supports _polymorphic_
    inductive type definitions.  For example, here is a _polymorphic
    list_ datatype. *)

Inductive list (X:Type) : Type :=
  | nil
  | cons (x : X) (l : list X).

(** This is exactly like the definition of [natlist] from the
    previous chapter, except that the [nat] argument to the [cons]
    constructor has been replaced by an arbitrary type [X], a binding
    for [X] has been added to the function header on the first line,
    and the occurrences of [natlist] in the types of the constructors
    have been replaced by [list X].

    What sort of thing is [list] itself?  A good way to think about it
    is that the definition of [list] is a _function_ from [Type]s to
    [Inductive] definitions; or, to put it more concisely, [list] is a
    function from [Type]s to [Type]s.  For any particular type [X],
    the type [list X] is the [Inductive]ly defined set of lists whose
    elements are of type [X]. *)

Check list : Type -> Type.

(** The [X] in the definition of [list] automatically becomes a
    parameter to the constructors [nil] and [cons] -- that is, [nil]
    and [cons] are now polymorphic constructors; when we use them, we
    must now provide a first argument that is the type of the list
    they are building. For example, [nil nat] constructs the empty
    list of type [nat]. *)

Check (nil nat) : list nat.

(** Similarly, [cons nat] adds an element of type [nat] to a list of
    type [list nat]. Here is an example of forming a list containing
    just the natural number 3. *)

Check (cons nat 3 (nil nat)) : list nat.

(** What might the type of [nil] be? We can read off the type
    [list X] from the definition, but this omits the binding for [X]
    which is the parameter to [list]. [Type -> list X] does not
    explain the meaning of [X]. [(X : Type) -> list X] comes
    closer. Coq's notation for this situation is [forall X : Type,
    list X]. *)

Check nil : forall X : Type, list X.

(** Similarly, the type of [cons] from the definition looks like
    [X -> list X -> list X], but using this convention to explain the
    meaning of [X] results in the type [forall X, X -> list X -> list
    X]. *)

Check cons : forall X : Type, X -> list X -> list X.

(** (A side note on notations: In .v files, the "forall"
    quantifier is spelled out in letters.  In the corresponding HTML
    files (and in the way some IDEs show .v files, depending on the
    settings of their display controls), [forall] is usually typeset
    as the standard mathematical "upside down A," though you'll still
    see the spelled-out "forall" in a few places.  This is just a
    quirk of typesetting -- there is no difference in meaning.) *)

(** Having to supply a type argument for every single use of a
    list constructor would be rather burdensome; we will soon see ways
    of reducing this annotation burden. *)

Check (cons nat 2 (cons nat 1 (nil nat)))
      : list nat.

(** We can now go back and make polymorphic versions of all the
    list-processing functions that we wrote before.  Here is [repeat],
    for example: *)

Fixpoint repeat (X : Type) (x : X) (count : nat) : list X :=
  match count with
  | 0 => nil X
  | S count' => cons X x (repeat X x count')
  end.

(** As with [nil] and [cons], we can use [repeat] by applying it
    first to a type and then to an element of this type (and a number): *)

Example test_repeat1 :
  repeat nat 4 2 = cons nat 4 (cons nat 4 (nil nat)).
Proof. reflexivity. Qed.

(** To use [repeat] to build other kinds of lists, we simply
    instantiate it with an appropriate type parameter: *)

Example test_repeat2 :
  repeat bool false 1 = cons bool false (nil bool).
Proof. reflexivity. Qed.

(** **** Exercise: 2 stars, standard, optional (mumble_grumble)

    Consider the following two inductively defined types. *)

Module MumbleGrumble.

Inductive mumble : Type :=
  | a
  | b (x : mumble) (y : nat)
  | c.

Inductive grumble (X:Type) : Type :=
  | d (m : mumble)
  | e (x : X).

(** Which of the following are well-typed elements of [grumble X] for
    some type [X]?  (Add YES or NO to each line.)
      - [d (b a 5)]           : no
      - [d mumble (b a 5)]    : yes 
      - [d bool (b a 5)]      : yes
      - [e bool true]         : yes
      - [e mumble (b c 0)]    : yes
      - [e bool (b c 0)]      : no
      - [c]                   : no *)
(* FILL IN HERE *)
End MumbleGrumble.
(** [] *)

(* ----------------------------------------------------------------- *)
(** *** Type Annotation Inference *)

(** Let's write the definition of [repeat] again, but this time we
    won't specify the types of any of the arguments.  Will Coq still
    accept it? *)

Fixpoint repeat' X x count : list X :=
  match count with
  | 0        => nil X
  | S count' => cons X x (repeat' X x count')
  end.

(** Indeed it will.  Let's see what type Coq has assigned to [repeat']... *)

Check repeat'
  : forall X : Type, X -> nat -> list X.
Check repeat
  : forall X : Type, X -> nat -> list X.

(** It has exactly the same type as [repeat].  Coq was able to
    use _type inference_ to deduce what the types of [X], [x], and
    [count] must be, based on how they are used.  For example, since
    [X] is used as an argument to [cons], it must be a [Type], since
    [cons] expects a [Type] as its first argument; matching [count]
    with [0] and [S] means it must be a [nat]; and so on.

    This powerful facility means we don't always have to write
    explicit type annotations everywhere, although explicit type
    annotations can still be quite useful as documentation and sanity
    checks, so we will continue to use them much of the time. *)

(* ----------------------------------------------------------------- *)
(** *** Type Argument Synthesis *)

(** To use a polymorphic function, we need to pass it one or
    more types in addition to its other arguments.  For example, the
    recursive call in the body of the [repeat] function above must
    pass along the type [X].  But since the second argument to
    [repeat] is an element of [X], it seems entirely obvious that the
    first argument can only be [X] -- why should we have to write it
    explicitly?

    Fortunately, Coq permits us to avoid this kind of redundancy.  In
    place of any type argument we can write a "hole" [_], which can be
    read as "Please try to figure out for yourself what belongs here."
    More precisely, when Coq encounters a [_], it will attempt to
    _unify_ all locally available information -- the type of the
    function being applied, the types of the other arguments, and the
    type expected by the context in which the application appears --
    to determine what concrete type should replace the [_].

    This may sound similar to type annotation inference -- and, indeed,
    the two procedures rely on the same underlying mechanisms.  Instead
    of simply omitting the types of some arguments to a function, like

      repeat' X x count : list X :=

    we can also replace the types with holes

      repeat' (X : _) (x : _) (count : _) : list X :=

    to tell Coq to attempt to infer the missing information.

    Using holes, the [repeat] function can be written like this: *)

Fixpoint repeat'' X x count : list X :=
  match count with
  | 0        => nil _
  | S count' => cons _ x (repeat'' _ x count')
  end.

(** In this instance, we don't save much by writing [_] instead of
    [X].  But in many cases the difference in both keystrokes and
    readability is nontrivial.  For example, suppose we want to write
    down a list containing the numbers [1], [2], and [3].  Instead of
    this... *)

Definition list123 :=
  cons nat 1 (cons nat 2 (cons nat 3 (nil nat))).

(** ...we can use holes to write this: *)

Definition list123' :=
  cons _ 1 (cons _ 2 (cons _ 3 (nil _))).

(* ----------------------------------------------------------------- *)
(** *** Implicit Arguments *)

(** In fact, we can go further and even avoid writing [_]'s in most
    cases by telling Coq _always_ to infer the type argument(s) of a
    given function.

    The [Arguments] directive specifies the name of the function (or
    constructor) and then lists the (leading) argument names to be
    treated as implicit, each surrounded by curly braces. *)

Arguments nil {X}.
Arguments cons {X}.
Arguments repeat {X}.

(** Now we don't have to supply any type arguments at all in the example: *)

Definition list123'' := cons 1 (cons 2 (cons 3 nil)).

(** Alternatively, we can declare an argument to be implicit
    when defining the function itself, by surrounding it in curly
    braces instead of parens.  For example: *)

Fixpoint repeat''' {X : Type} (x : X) (count : nat) : list X :=
  match count with
  | 0        => nil
  | S count' => cons x (repeat''' x count')
  end.

(** (Note that we didn't even have to provide a type argument to the
    recursive call to [repeat'''].  Indeed, it would be invalid to
    provide one, because Coq is not expecting it.)

    We will use the latter style whenever possible, but we will
    continue to use explicit [Argument] declarations for [Inductive]
    constructors.  The reason for this is that marking the parameter
    of an inductive type as implicit causes it to become implicit for
    the type itself, not just for its constructors.  For instance,
    consider the following alternative definition of the [list]
    type: *)

Inductive list' {X:Type} : Type :=
  | nil'
  | cons' (x : X) (l : list').

(** Because [X] is declared as implicit for the _entire_ inductive
    definition including [list'] itself, we now have to write just
    [list'] whether we are talking about lists of numbers or booleans
    or anything else, rather than [list' nat] or [list' bool] or
    whatever; this is a step too far. *)

(** Let's finish by re-implementing a few other standard list
    functions on our new polymorphic lists... *)

Fixpoint app {X : Type} (l1 l2 : list X) : list X :=
  match l1 with
  | nil      => l2
  | cons h t => cons h (app t l2)
  end.

Fixpoint rev {X:Type} (l:list X) : list X :=
  match l with
  | nil      => nil
  | cons h t => app (rev t) (cons h nil)
  end.

Fixpoint length {X : Type} (l : list X) : nat :=
  match l with
  | nil => 0
  | cons _ l' => S (length l')
  end.

Example test_rev1 :
  rev (cons 1 (cons 2 nil)) = (cons 2 (cons 1 nil)).
Proof. reflexivity. Qed.

Example test_rev2:
  rev (cons true nil) = cons true nil.
Proof. reflexivity. Qed.

Example test_length1: length (cons 1 (cons 2 (cons 3 nil))) = 3.
Proof. reflexivity. Qed.

(* ----------------------------------------------------------------- *)
(** *** Supplying Type Arguments Explicitly *)

(** One small problem with declaring arguments to be implicit is
    that, once in a while, Coq does not have enough local information
    to determine a type argument; in such cases, we need to tell Coq
    that we want to give the argument explicitly just this time.  For
    example, suppose we write this: *)

Fail Definition mynil := nil.

(** (The [Fail] qualifier that appears before [Definition] can be
    used with _any_ command, and is used to ensure that that command
    indeed fails when executed. If the command does fail, Coq prints
    the corresponding error message, but continues processing the rest
    of the file.)

    Here, Coq gives us an error because it doesn't know what type
    argument to supply to [nil].  We can help it by providing an
    explicit type declaration (so that Coq has more information
    available when it gets to the "application" of [nil]): *)

Definition mynil : list nat := nil.

(** Alternatively, we can force the implicit arguments to be explicit by
    prefixing the function name with [@]. *)

Check @nil : forall X : Type, list X.

Definition mynil' := @nil nat.

(** Using argument synthesis and implicit arguments, we can
    define convenient notation for lists, as before.  Since we have
    made the constructor type arguments implicit, Coq will know to
    automatically infer these when we use the notations. *)

Notation "x :: y" := (cons x y)
                     (at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y []) ..).
Notation "x ++ y" := (app x y)
                     (at level 60, right associativity).

(** Now lists can be written just the way we'd hope: *)

Definition list123''' := [1; 2; 3].

(* ----------------------------------------------------------------- *)
(** *** Exercises *)

(** **** Exercise: 2 stars, standard (poly_exercises)

    Here are a few simple exercises, just like ones in the [Lists]
    chapter, for practice with polymorphism.  Complete the proofs
    below. *)

Theorem app_nil_r : forall (X:Type), forall l:list X,
  l ++ [] = l.
Proof.
  intros.
  induction l as [| h t IHl].
  - simpl. reflexivity.
  - simpl. rewrite IHl. reflexivity.  
Qed.

Theorem app_assoc : forall A (l m n:list A),
  l ++ m ++ n = (l ++ m) ++ n.
Proof.
  intros.
  induction l as [| h t IHl].
  - simpl. reflexivity.
  - simpl. rewrite <- IHl. reflexivity.
Qed. 

Lemma app_length : forall (X:Type) (l1 l2 : list X),
  length (l1 ++ l2) = length l1 + length l2.
Proof.
  intros.
  induction l1 as [| h t IHl1].
  - simpl. reflexivity.
  - simpl. rewrite IHl1. reflexivity.
Qed. 
(** [] *)

(** **** Exercise: 2 stars, standard (more_poly_exercises)

    Here are some slightly more interesting ones... *)

Theorem rev_app_distr: forall X (l1 l2 : list X),
  rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof.
  intros.
  induction l1 as [| h t IHl1].
  - simpl. rewrite app_nil_r. reflexivity.
  - simpl. rewrite IHl1. rewrite app_assoc. reflexivity.
Qed.  

Theorem rev_involutive : forall X : Type, forall l : list X,
  rev (rev l) = l.
Proof.
  intros.
  induction l as [| h t IHl].
  - simpl. reflexivity.
  - simpl. rewrite rev_app_distr. rewrite IHl. simpl. reflexivity.
Qed.
(** [] *)

(* ================================================================= *)
(** ** Polymorphic Pairs *)

(** Following the same pattern, the definition for pairs of
    numbers that we gave in the last chapter can be generalized to
    _polymorphic pairs_, often called _products_: *)

Inductive prod (X Y : Type) : Type :=
| pair (x : X) (y : Y).

Arguments pair {X} {Y}.

(** As with lists, we make the type arguments implicit and define the
    familiar concrete notation. *)

Notation "( x , y )" := (pair x y).

(** We can also use the [Notation] mechanism to define the standard
    notation for _product types_ (i.e., the types of pairs): *)

Notation "X * Y" := (prod X Y) : type_scope.

(** (The annotation [: type_scope] tells Coq that this abbreviation
    should only be used when parsing types, not when parsing
    expressions.  This avoids a clash with the multiplication
    symbol.) *)

(** It is easy at first to get [(x,y)] and [X*Y] confused.
    Remember that [(x,y)] is a _value_ built from two other values,
    while [X*Y] is a _type_ built from two other types.  If [x] has
    type [X] and [y] has type [Y], then [(x,y)] has type [X*Y]. *)

(** The first and second projection functions now look pretty
    much as they would in any functional programming language. *)

Definition fst {X Y : Type} (p : X * Y) : X :=
  match p with
  | (x, y) => x
  end.

Definition snd {X Y : Type} (p : X * Y) : Y :=
  match p with
  | (x, y) => y
  end.

(** The following function takes two lists and combines them
    into a list of pairs.  In other functional languages, it is often
    called [zip]; we call it [combine] for consistency with Coq's
    standard library. *)

Fixpoint combine {X Y : Type} (lx : list X) (ly : list Y)
           : list (X*Y) :=
  match lx, ly with
  | [], _ => []
  | _, [] => []
  | x :: tx, y :: ty => (x, y) :: (combine tx ty)
  end.

(** **** Exercise: 1 star, standard, optional (combine_checks)

    Try answering the following questions on paper and
    checking your answers in Coq:
    - What is the type of [combine] (i.e., what does [Check
      @combine] print?)
    - What does

        Compute (combine [1;2] [false;false;true;true]).

      print?

    [] *)

    Compute (combine [1;2] [false;false;true;true]).
    Search (prod).

(** **** Exercise: 2 stars, standard, especially useful (split)

    The function [split] is the right inverse of [combine]: it takes a
    list of pairs and returns a pair of lists.  In many functional
    languages, it is called [unzip].

    Fill in the definition of [split] below.  Make sure it passes the
    given unit test. *)

Fixpoint split {X Y : Type} (l : list (X*Y)) : (list X) * (list Y) :=
  match l with
  | nil => ([], [])
  | h :: t => pair (fst h :: fst (split t)) (snd h :: snd (split t))
  end. 

  (* ([x] , [y]) *)
Example test_split:
  split [(1,false);(2,false)] = ([1;2],[false;false]).
Proof.
  intros. simpl. reflexivity.
Qed.
(** [] *)

(* ================================================================= *)
(** ** Polymorphic Options *)

(** Our last polymorphic type for now is _polymorphic options_,
    which generalize [natoption] from the previous chapter.  (We put
    the definition inside a module because the standard library
    already defines [option] and it's this one that we want to use
    below.) *)

Module OptionPlayground.

Inductive option (X:Type) : Type :=
  | Some (x : X)
  | None.

Arguments Some {X}.
Arguments None {X}.

End OptionPlayground.

(** We can now rewrite the [nth_error] function so that it works
    with any type of lists. *)

Fixpoint nth_error {X : Type} (l : list X) (n : nat)
                   : option X :=
  match l with
  | nil => None
  | a :: l' => match n with
               | O => Some a
               | S n' => nth_error l' n'
               end
  end.

Example test_nth_error1 : nth_error [4;5;6;7] 0 = Some 4.
Proof. reflexivity. Qed.
Example test_nth_error2 : nth_error [[1];[2]] 1 = Some [2].
Proof. reflexivity. Qed.
Example test_nth_error3 : nth_error [true] 2 = None.
Proof. reflexivity. Qed.

(** **** Exercise: 1 star, standard, optional (hd_error_poly)

    Complete the definition of a polymorphic version of the
    [hd_error] function from the last chapter. Be sure that it
    passes the unit tests below. *)

Definition hd_error {X : Type} (l : list X) : option X :=
  match l with
  | [] => None
  | h :: t => Some h
  end. 

(** Once again, to force the implicit arguments to be explicit,
    we can use [@] before the name of the function. *)

Check @hd_error : forall X : Type, list X -> option X.

Example test_hd_error1 : hd_error [1;2] = Some 1.
Proof. simpl. reflexivity. Qed. 

Example test_hd_error2 : hd_error  [[1];[2]]  = Some [1].
Proof. simpl. reflexivity. Qed. 
(** [] *)

(* ################################################################# *)
(** * Functions as Data *)

(** Like most modern programming languages -- especially other
    "functional" languages, including OCaml, Haskell, Racket, Scala,
    Clojure, etc. -- Coq treats functions as first-class citizens,
    allowing them to be passed as arguments to other functions,
    returned as results, stored in data structures, etc. *)

(* ================================================================= *)
(** ** Higher-Order Functions *)

(** Functions that manipulate other functions are often called
    _higher-order_ functions.  Here's a simple one: *)

Definition doit3times {X : Type} (f : X->X) (n : X) : X :=
  f (f (f n)).

(** The argument [f] here is itself a function (from [X] to
    [X]); the body of [doit3times] applies [f] three times to some
    value [n]. *)

Check @doit3times : forall X : Type, (X -> X) -> X -> X.

Example test_doit3times: doit3times minustwo 9 = 3.
Proof. reflexivity. Qed.

Example test_doit3times': doit3times negb true = false.
Proof. reflexivity. Qed.

(* ================================================================= *)
(** ** Filter *)

(** Here is a more useful higher-order function, taking a list
    of [X]s and a _predicate_ on [X] (a function from [X] to [bool])
    and "filtering" the list, returning a new list containing just
    those elements for which the predicate returns [true]. *)

Fixpoint filter {X:Type} (test: X->bool) (l:list X) : list X :=
  match l with
  | [] => []
  | h :: t =>
    if test h then h :: (filter test t)
    else filter test t
  end.

(** For example, if we apply [filter] to the predicate [even]
    and a list of numbers [l], it returns a list containing just the
    even members of [l]. *)

Example test_filter1: filter even [1;2;3;4] = [2;4].
Proof. reflexivity. Qed.

Definition length_is_1 {X : Type} (l : list X) : bool :=
  (length l) =? 1.

Example test_filter2:
    filter length_is_1
           [ [1; 2]; [3]; [4]; [5;6;7]; []; [8] ]
  = [ [3]; [4]; [8] ].
Proof. reflexivity. Qed.

(** We can use [filter] to give a concise version of the
    [countoddmembers] function from the [Lists] chapter. *)

Definition countoddmembers' (l:list nat) : nat :=
  length (filter odd l).

Example test_countoddmembers'1:   countoddmembers' [1;0;3;1;4;5] = 4.
Proof. reflexivity. Qed.
Example test_countoddmembers'2:   countoddmembers' [0;2;4] = 0.
Proof. reflexivity. Qed.
Example test_countoddmembers'3:   countoddmembers' nil = 0.
Proof. reflexivity. Qed.

(* ================================================================= *)
(** ** Anonymous Functions *)

(** It is arguably a little sad, in the example just above, to
    be forced to define the function [length_is_1] and give it a name
    just to be able to pass it as an argument to [filter], since we
    will probably never use it again.  Moreover, this is not an
    isolated example: when using higher-order functions, we often want
    to pass as arguments "one-off" functions that we will never use
    again; having to give each of these functions a name would be
    tedious.

    Fortunately, there is a better way.  We can construct a function
    "on the fly" without declaring it at the top level or giving it a
    name. *)

Example test_anon_fun':
  doit3times (fun n => n * n) 2 = 256.
Proof. reflexivity. Qed.

(** The expression [(fun n => n * n)] can be read as "the function
    that, given a number [n], yields [n * n]." *)

(** Here is the [filter] example, rewritten to use an anonymous
    function. *)

Example test_filter2':
    filter (fun l => (length l) =? 1)
           [ [1; 2]; [3]; [4]; [5;6;7]; []; [8] ]
  = [ [3]; [4]; [8] ].
Proof. reflexivity. Qed.

(** **** Exercise: 2 stars, standard (filter_even_gt7)

    Use [filter] (instead of [Fixpoint]) to write a Coq function
    [filter_even_gt7] that takes a list of natural numbers as input
    and returns a list of just those that are even and greater than
    7. *)

Definition filter_even_gt7 (l : list nat) : list nat :=
  filter ( fun n => andb (even n) (7 <? n) ) l.

Example test_filter_even_gt7_1 :
  filter_even_gt7 [1;2;6;9;10;3;12;8] = [10;12;8].
Proof. reflexivity. Qed.  

Example test_filter_even_gt7_2 :
  filter_even_gt7 [5;2;6;19;129] = [].
Proof. reflexivity. Qed.
  (** [] *)

(** **** Exercise: 3 stars, standard (partition)

    Use [filter] to write a Coq function [partition]:

      partition : forall X : Type,
                  (X -> bool) -> list X -> list X * list X

   Given a set [X], a predicate of type [X -> bool] and a [list X],
   [partition] should return a pair of lists.  The first member of the
   pair is the sublist of the original list containing the elements
   that satisfy the test, and the second is the sublist containing
   those that fail the test.  The order of elements in the two
   sublists should be the same as their order in the original list. *)

Definition partition {X : Type}
                     (test : X -> bool)
                     (l : list X)
                   : list X * list X
:= pair (filter test l) 
        (filter (fun l' =>
                  match test l' with
                  | b => negb b
                  end) l).

Example test_partition1: partition odd [1;2;3;4;5] = ([1;3;5], [2;4]).
Proof. compute. reflexivity. Qed. 
Example test_partition2: partition (fun x => false) [5;9;0] = ([], [5;9;0]).
Proof. compute. reflexivity. Qed.
(** [] *)

(* ================================================================= *)
(** ** Map *)

(** Another handy higher-order function is called [map]. *)

Fixpoint map {X Y : Type} (f : X->Y) (l : list X) : list Y :=
  match l with
  | []     => []
  | h :: t => (f h) :: (map f t)
  end.

(** It takes a function [f] and a list [ l = [n1, n2, n3, ...] ]
    and returns the list [ [f n1, f n2, f n3,...] ], where [f] has
    been applied to each element of [l] in turn.  For example: *)

Example test_map1: map (fun x => plus 3 x) [2;0;2] = [5;3;5].
Proof. compute. reflexivity. Qed.

(** The element types of the input and output lists need not be
    the same, since [map] takes _two_ type arguments, [X] and [Y]; it
    can thus be applied to a list of numbers and a function from
    numbers to booleans to yield a list of booleans: *)

Example test_map2:
  map odd [2;1;2;5] = [false;true;false;true].
Proof. compute. reflexivity. Qed.

(** It can even be applied to a list of numbers and
    a function from numbers to _lists_ of booleans to
    yield a _list of lists_ of booleans: *)

Example test_map3:
    map (fun n => [even n;odd n]) [2;1;2;5]
  = [[true;false];[false;true];[true;false];[false;true]].
Proof. compute. reflexivity. Qed.

(* ----------------------------------------------------------------- *)
(** *** Exercises *)

(** **** Exercise: 3 stars, standard (map_rev)

    Show that [map] and [rev] commute.  You may need to define an
    auxiliary lemma. *)

Lemma map_distr_over_app : forall (X Y : Type) (f: X -> Y) (l1 l2 : list X), map f (l1 ++ l2) = map f l1 ++ map f l2.
Proof. intros.
induction l1 as [| l1h l1t IHl1].
- simpl. reflexivity.
- simpl. rewrite IHl1. reflexivity.
Qed.
    
Theorem map_rev : forall (X Y : Type) (f : X -> Y) (l : list X),
  map f (rev l) = rev (map f l).
Proof.
  intros. 
  induction l as [| h t IHl].
  - simpl. reflexivity.
  - simpl. rewrite <- IHl. rewrite map_distr_over_app. simpl. reflexivity.
Qed. 

(** [] *)

(** **** Exercise: 2 stars, standard, especially useful (flat_map)

    The function [map] maps a [list X] to a [list Y] using a function
    of type [X -> Y].  We can define a similar function, [flat_map],
    which maps a [list X] to a [list Y] using a function [f] of type
    [X -> list Y].  Your definition should work by 'flattening' the
    results of [f], like so:

        flat_map (fun n => [n;n+1;n+2]) [1;5;10]
      = [1; 2; 3; 5; 6; 7; 10; 11; 12].
*)

Fixpoint flat_map {X Y: Type} (f: X -> list Y) (l: list X)
                   : list Y
:= match l with
   | [] => []
   | h :: t => app (f h) (flat_map f t)  
   end. 

Example test_flat_map1:
  flat_map (fun n => [n;n;n]) [1;5;4]
  = [1; 1; 1; 5; 5; 5; 4; 4; 4].
Proof. simpl. reflexivity. Qed.
(** [] *)

(** Lists are not the only inductive type for which [map] makes sense.
    Here is a [map] for the [option] type: *)

Definition option_map {X Y : Type} (f : X -> Y) (xo : option X)
                      : option Y :=
  match xo with
  | None => None
  | Some x => Some (f x)
  end.

(** **** Exercise: 2 stars, standard, optional (implicit_args)

    The definitions and uses of [filter] and [map] use implicit
    arguments in many places.  Replace the curly braces around the
    implicit arguments with parentheses, and then fill in explicit
    type parameters where necessary and use Coq to check that you've
    done so correctly.  (This exercise is not to be turned in; it is
    probably easiest to do it on a _copy_ of this file that you can
    throw away afterwards.)
*)
(** [] *)

(* ================================================================= *)
(** ** Fold *)

(** An even more powerful higher-order function is called
    [fold].  This function is the inspiration for the "[reduce]"
    operation that lies at the heart of Google's map/reduce
    distributed programming framework. *)

Fixpoint fold {X Y: Type} (f : X->Y->Y) (l : list X) (b : Y)
                         : Y :=
  match l with
  | nil => b
  | h :: t => f h (fold f t b)
  end.

(** Intuitively, the behavior of the [fold] operation is to
    insert a given binary operator [f] between every pair of elements
    in a given list.  For example, [ fold plus [1;2;3;4] ] intuitively
    means [1+2+3+4].  To make this precise, we also need a "starting
    element" that serves as the initial second input to [f].  So, for
    example,

       fold plus [1;2;3;4] 0

    yields

       1 + (2 + (3 + (4 + 0))).

    Some more examples: *)

Check (fold andb) : list bool -> bool -> bool.

Example fold_example1 :
  fold mult [1;2;3;4] 1 = 24.
Proof. reflexivity. Qed.

Example fold_example2 :
  fold andb [true;true;false;true] true = false.
Proof. reflexivity. Qed.

Example fold_example3 :
  fold app  [[1];[];[2;3];[4]] [] = [1;2;3;4].
Proof. reflexivity. Qed.

(** **** Exercise: 1 star, standard, optional (fold_types_different)

    Observe that the type of [fold] is parameterized by _two_ type
    variables, [X] and [Y], and the parameter [f] is a binary operator
    that takes an [X] and a [Y] and returns a [Y].  Can you think of a
    situation where it would be useful for [X] and [Y] to be
    different? *)

(* FILL IN HERE

    [] *)

(* ================================================================= *)
(** ** Functions That Construct Functions *)

(** Most of the higher-order functions we have talked about so
    far take functions as arguments.  Let's look at some examples that
    involve _returning_ functions as the results of other functions.
    To begin, here is a function that takes a value [x] (drawn from
    some type [X]) and returns a function from [nat] to [X] that
    yields [x] whenever it is called, ignoring its [nat] argument. *)

Definition constfun {X: Type} (x: X) : nat -> X :=
  fun (k:nat) => x.

Definition ftrue := constfun true.

Example constfun_example1 : ftrue 0 = true.
Proof. reflexivity. Qed.

Example constfun_example2 : (constfun 5) 99 = 5.
Proof. reflexivity. Qed.

(** In fact, the multiple-argument functions we have already
    seen are also examples of passing functions as data.  To see why,
    recall the type of [plus]. *)

Check plus : nat -> nat -> nat.

(** Each [->] in this expression is actually a _binary_ operator
    on types.  This operator is _right-associative_, so the type of
    [plus] is really a shorthand for [nat -> (nat -> nat)] -- i.e., it
    can be read as saying that "[plus] is a one-argument function that
    takes a [nat] and returns a one-argument function that takes
    another [nat] and returns a [nat]."  In the examples above, we
    have always applied [plus] to both of its arguments at once, but
    if we like we can supply just the first.  This is called _partial
    application_. *)

Definition plus3 := plus 3.
Check plus3 : nat -> nat.

Example test_plus3 :    plus3 4 = 7.
Proof. reflexivity. Qed.
Example test_plus3' :   doit3times plus3 0 = 9.
Proof. reflexivity. Qed.
Example test_plus3'' :  doit3times (plus 3) 0 = 9.
Proof. reflexivity. Qed.

(* ################################################################# *)
(** * Additional Exercises *)

Module Exercises.

(** **** Exercise: 2 stars, standard (fold_length)

    Many common functions on lists can be implemented in terms of
    [fold].  For example, here is an alternative definition of [length]: *)

Definition fold_length {X : Type} (l : list X) : nat :=
  fold (fun _ n => S n) l 0.

Example test_fold_length1 : fold_length [4;7;0] = 3.
Proof. reflexivity. Qed.

(** Prove the correctness of [fold_length].  (Hint: It may help to
    know that [reflexivity] simplifies expressions a bit more
    aggressively than [simpl] does -- i.e., you may find yourself in a
    situation where [simpl] does nothing but [reflexivity] solves the
    goal.) *)

Theorem fold_length_correct : forall X (l : list X),
  fold_length l = length l.
Proof.
  intros.
  induction l as [| lh lt IHl].
  - simpl. reflexivity.
  - simpl. rewrite <- IHl. reflexivity.
Qed. 
(** [] *)

(** **** Exercise: 3 stars, standard (fold_map)

    We can also define [map] in terms of [fold].  Finish [fold_map]
    below. *)

Definition fold_map {X Y: Type} (f: X -> Y) (l: list X) : list Y
:= fold ( fun lx recursiveFoldFunction => (f lx)::recursiveFoldFunction ) (l) (nil). 

(** Write down a theorem [fold_map_correct] stating that [fold_map] is
    correct, and prove it in Coq.  (Hint: again, remember that
    [reflexivity] simplifies expressions a bit more aggressively than
    [simpl].) *)

Theorem fold_map_correct: forall (X Y : Type) (f : X -> Y) (lx: list X), fold_map f lx = map f lx.
Proof. 
  intros.
  induction lx as [| lxh lxt IHlx].
  - simpl. reflexivity.
  - simpl. rewrite <- IHlx. reflexivity.
Qed.


(* Do not modify the following line: *)
Definition manual_grade_for_fold_map : option (nat*string) := None.
(** [] *)

(** **** Exercise: 2 stars, advanced (currying)

    The type [X -> Y -> Z] can be read as describing functions that
    take two arguments, one of type [X] and another of type [Y], and
    return an output of type [Z]. Strictly speaking, this type is
    written [X -> (Y -> Z)] when fully parenthesized.  That is, if we
    have [f : X -> Y -> Z], and we give [f] an input of type [X], it
    will give us as output a function of type [Y -> Z].  If we then
    give that function an input of type [Y], it will return an output
    of type [Z]. That is, every function in Coq takes only one input,
    but some functions return a function as output. This is precisely
    what enables partial application, as we saw above with [plus3].

    By contrast, functions of type [X * Y -> Z] -- which when fully
    parenthesized is written [(X * Y) -> Z] -- require their single
    input to be a pair.  Both arguments must be given at once; there
    is no possibility of partial application.

    It is possible to convert a function between these two types.
    Converting from [X * Y -> Z] to [X -> Y -> Z] is called
    _currying_, in honor of the logician Haskell Curry.  Converting
    from [X -> Y -> Z] to [X * Y -> Z] is called _uncurrying_.  *)

(** We can define currying as follows: *)

Definition prod_curry {X Y Z : Type}
  (f : X * Y -> Z) (x : X) (y : Y) : Z := f (x, y).

(** As an exercise, define its inverse, [prod_uncurry].  Then prove
    the theorems below to show that the two are inverses. *)

Definition prod_uncurry {X Y Z : Type}
  (f : X -> Y -> Z) (p : X * Y) : Z
:= f (fst p) (snd p).

(** As a (trivial) example of the usefulness of currying, we can use it
    to shorten one of the examples that we saw above: *)

Example test_map1': map (plus 3) [2;0;2] = [5;3;5].
Proof. simpl. reflexivity. Qed.

(** Thought exercise: before running the following commands, can you
    calculate the types of [prod_curry] and [prod_uncurry]? *)

Check @prod_curry.
Check @prod_uncurry.

Theorem uncurry_curry : forall (X Y Z : Type)
                        (f : X -> Y -> Z)
                        x y,
  prod_curry (prod_uncurry f) x y = f x y.
Proof.
  intros. reflexivity. Qed.

Theorem curry_uncurry : forall (X Y Z : Type)
                        (f : (X * Y) -> Z) (p : X * Y),
  prod_uncurry (prod_curry f) p = f p.
Proof.
  intros.
  destruct p. reflexivity. Qed.
(** [] *)

(** **** Exercise: 2 stars, advanced (nth_error_informal)

    Recall the definition of the [nth_error] function:

   Fixpoint nth_error {X : Type} (l : list X) (n : nat) : option X :=
     match l with
     | [] => None
     | a :: l' => if n =? O then Some a else nth_error l' (pred n)
     end.

   Write a careful informal proof of the following theorem:

   forall X l n, length l = n -> @nth_error X l n = None

   Make sure to state the induction hypothesis _explicitly_.
*)
(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_informal_proof : option (nat*string) := None.
(** [] *)

(* ================================================================= *)
(** ** Church Numerals (Advanced) *)

(** The following exercises explore an alternative way of defining
    natural numbers using the _Church numerals_, which are named after
    their inventor, the mathematician Alonzo Church.  We can represent
    a natural number [n] as a function that takes a function [f] as a
    parameter and returns [f] iterated [n] times. *)

Module Church.
Definition cnat := forall X : Type, (X -> X) -> X -> X.

(** Let's see how to write some numbers with this notation. Iterating
    a function once should be the same as just applying it.  Thus: *)

Definition one : cnat :=
  fun (X : Type) (f : X -> X) (x : X) => f x.

(** Similarly, [two] should apply [f] twice to its argument: *)

Definition two : cnat :=
  fun (X : Type) (f : X -> X) (x : X) => f (f x).

(** Defining [zero] is somewhat trickier: how can we "apply a function
    zero times"?  The answer is actually simple: just return the
    argument untouched. *)

Definition zero : cnat :=
  fun (X : Type) (f : X -> X) (x : X) => x.

(** More generally, a number [n] can be written as [fun X f x => f (f
    ... (f x) ...)], with [n] occurrences of [f].  Let's informally
    notate that as [fun X f x => f^n x], with the convention that [f^0 x]
    is just [x]. Note how the [doit3times] function we've defined
    previously is actually just the Church representation of [3]. *)

Definition three : cnat := @doit3times.

(** So [n X f x] represents "do it [n] times", where [n] is a Church
    numerals and "it" means applying [f] starting with [x].

    Another way to think about the Church representation is that
    function [f] represents the successor operation on [X], and value
    [x] represents the zero element of [X].  We could even rewrite
    with those names to make it clearer: *)

Definition zero' : cnat :=
  fun (X : Type) (succ : X -> X) (zero : X) => zero.
Definition one' : cnat :=
  fun (X : Type) (succ : X -> X) (zero : X) => succ zero.
Definition two' : cnat :=
  fun (X : Type) (succ : X -> X) (zero : X) => succ (succ zero).

(** If we passed in [S] as [succ] and [O] as [zero], we'd even get the Peano
    naturals as a result: *)

Example zero_church_peano : zero nat S O = 0.
Proof. reflexivity. Qed.

Example one_church_peano : one nat S O = 1.
Proof. reflexivity. Qed.

Example two_church_peano : two nat S O = 2.
Proof. reflexivity. Qed.

(** But the intellectually exciting implication of the Church numerals
    is that we don't strictly need the natural numbers to be built-in
    to a functional programming language, or even to be definable with
    an inductive data type. It's possible to represent them purely (if
    not efficiently) with functions.

    Of course, it's not enough to represent numerals; we need to be
    able to do arithmetic with them. Show that we can by completing
    the definitions of the following functions. Make sure that the
    corresponding unit tests pass by proving them with
    [reflexivity]. *)

(** **** Exercise: 2 stars, advanced (church_scc) *)

(** Define a function that computes the successor of a Church numeral.
    Given a Church numeral [n], its successor [scc n] should iterate
    its function argument once more than [n]. That is, given [fun X f x
    => f^n x] as input, [scc] should produce [fun X f x => f^(n+1) x] as
    output. In other words, do it [n] times, then do it once more. *)

Definition scc (n : cnat) : cnat
:= fun (X : Type) ( succ : X -> X ) (zero : X) => succ (n X succ zero).

Example scc_1 : scc zero = one.
Proof. compute. reflexivity. Qed.

Example scc_2 : scc one = two.
Proof. compute. reflexivity. Qed.

Example scc_3 : scc two = three.
Proof. reflexivity. Qed.

(** [] *)

(** **** Exercise: 3 stars, advanced (church_plus) *)

(** Define a function that computes the addition of two Church
    numerals.  Given [fun X f x => f^n x] and [fun X f x => f^m x] as
    input, [plus] should produce [fun X f x => f^(n + m) x] as output.
    In other words, do it [n] times, then do it [m] more times.

    Hint: the "zero" argument to a Church numeral need not be just
    [x]. *)

Definition plus (n m : cnat) : cnat
:= fun (X : Type) ( succ : X -> X ) (zero : X) => n X succ (m X succ zero).

Example plus_1 : plus zero one = one.
Proof. compute. reflexivity. Qed.

Example plus_2 : plus two three = plus three two.
Proof. compute. reflexivity. Qed.

Example plus_3 :
  plus (plus two two) three = plus one (plus three three).
Proof. compute. reflexivity. Qed.

(** [] *)

(** **** Exercise: 3 stars, advanced (church_mult) *)

(** Define a function that computes the multiplication of two Church
    numerals.

    Hint: the "successor" argument to a Church numeral need not be
    just [f].

    Warning: Coq will not let you pass [cnat] itself as the type [X]
    argument to a Church numeral; you will get a "Universe
    inconsistency" error. That is Coq's way of preventing a paradox in
    which a type contains itself. So leave the type argument
    unchanged. *)

Definition mult (n m : cnat) : cnat
:= fun (X : Type) ( succ : X -> X ) (zero : X) => n X (m X succ) zero.

Example mult_1 : mult one one = one.
Proof. compute. reflexivity. Qed.

Example mult_2 : mult zero (plus three three) = zero.
Proof. compute. reflexivity. Qed.

Example mult_3 : mult two three = plus three three.
Proof. compute. reflexivity. Qed.

(** [] *)

(** **** Exercise: 3 stars, advanced (church_exp) *)

(** Exponentiation: *)

(** Define a function that computes the exponentiation of two Church
    numerals.

    Hint: the type argument to a Church numeral need not just be [X].
    But again, you cannot pass [cnat] itself as the type argument.
    Finding the right type can be tricky. *)

Definition exp (n m : cnat) : cnat
:= fun (X : Type) => m (X -> X) (n X).

Example exp_11 : exp two two = plus two two.
Proof. compute. reflexivity. Qed.

Example exp_2 : exp three zero = one.
Proof. compute. reflexivity. Qed.

Example exp_3 : exp three two = plus (mult two (mult two two)) one.
Proof. compute. reflexivity. Qed.

(** [] *)

End Church.
End Exercises.

(* 2023-10-03 16:40 *)

(* DONE :3 *)