How to define a recursive function in the (pure) calculus of constructions? I do not see any fixpoint combinator there.
People in the CS stack exchange might be able to provide some more insight, but here is an attempt.
Inductive data types are defined in the calculus of constructions with a Church encoding: the data type is the type of its fold function. The most basic example are the natural numbers, which are defined as follows, using a Coq-like notation:
nat := forall (T : Type), T -> (T -> T) -> T
This encoding yields two things: (1) terms zero : nat and succ : nat -> nat for constructing natural numbers, and (2) an operator nat_rec for writing recursive functions.
zero : nat
zero T x f := x
succ : nat -> nat
succ n T x f := f (n T x f)
nat_rec : forall T, T -> (T -> T) -> nat -> T
nat_rec T x f n := n T x f
If we pose F := nat_rec T x f for terms x : T and f : T -> T, we see that the following equations are valid:
F zero = x
F (succ n) = f (F n)
Thus, nat_rec allows us to define recursive functions by specifying a return value x for the base case, and a function f to process the value of the recursive call. Note that this does not allow us to define arbitrary recursive functions on the natural numbers, but only those that perform recursive calls on the predecessor of their argument. Allowing arbitrary recursion would open the door to partial functions, which would compromise the soundness of the calculus.
This example can be generalized to other inductive data types. For instance, we can define the type of lists of natural numbers as the type of their fold right function:
list_nat := forall T, T -> (nat -> T -> T) -> T
Related
How do you make an anonymous recursive function (something simple for example factorial n?) I have heard it is possible but no idea how to make it work in OCaml.
let a =
fun x -> ....
I just don't know how to keep it going...
Here is a definition of factorial using only anonymous functions:
let fact =
(fun f -> (fun x a -> f (x x) a) (fun x a -> f (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1))
It requires the use of the -rectypes flag.
Here's a session showing that it works:
$ rlwrap ocaml -rectypes
OCaml version 4.03.0
let fact =
(fun f -> (fun x a -> f (x x) a) (fun x a -> f (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1));;
val fact : int -> int = <fun>
# fact 8;;
- : int = 40320
I cheated somewhat by looking up the Y Combinator here: Rosetta Code: Y Combinator
Update
Disclaimer: you would do better to read up on lambda calculus, fixed points, and the Y Combinator than to get your info from me. I'm not a theorist, just a humble practitioner.
Following the actual computation is almost impossible (but definitely worth doing I'm sure). But at a high level the ideas are like this.
The first line of the definition is the Y Combinator, which in general calculates the fixed point of a function. It so happens that a recursive function is the fixed point of a function from functions to functions.
So the first goal is to find the function whose fixed point is the factorial function. That's the second line of the definition. If you give it a function of type int -> int, it gives you back another function of type int -> int. And if you give it the factorial function, it gives you back the factorial function. This means that the factorial function is its fixed point.
So then when you apply the Y Combinator to this function, you do indeed get the factorial function.
Let me try to expand a bit on Jeffrey Scofield's answer. A non-anonymous recursive definition of the factorial function could be
let rec fact n =
if n < 2 then 1 else n * fact (n - 1)
The first problem you encounter when you try to define an anonymous recursive function is how to do the actual recursive call (fact (n - 1) in our case). For a call we need a name and we do not have a name for an anonymous function. The solution is to use a temporary name. With the temporary name f, the definition body is just
fun n -> if n < 2 then 1 else n * f (n - 1)
This term does not have a type, because the "temporary name" f is unbound. But we can turn it into a value that does have a type by bounding f as well. Let us call the result g:
let g = fun f n -> if n < 2 then 1 else n * f (n - 1)
g is not yet anonymous at the moment, but only because I want to refer to it again.
Observe that g has type (int -> int) -> (int -> int). What we want (the factorial function) will have type (int -> int). So g takes something of the type we want (a function type in this case) and produces something of the same type. The intuition is that g takes an approximation of the factorial function, namely a function f which works for all n up to some limit N and returns a better approximation, namely a function that works for all n up to N+1.
Finally we need something that turns g into an actual recursive definition.
Doing so is a very generic task. Recall that g improves the approximation quality. The final factorial function fact is one which cannot be further improved. So applying g to fact should be the same as just fact. (Actually that is only true from a value point of view. The actual computation inherent in g fact n for some n is different from that of just fact n. But the returned values are the same.) In other words, fact is a fixed point of g. So what we need is something that computes fixed points.
Luckily, there is a single function that does so: The Y combinator. From a value point of view, the Y combinator (let us use y in OCaml, as uppercase is reserved for constructors) is defined by the fact that y g = g (y g) for all g: given some function g, the combinator returns one of its fixed points.
Consequently,
y : (`a -> `a) -> `a
In our case the type variable is instantiated by (int -> int).
One possible way to define y would be
let y = fun g -> (fun x -> g (x x)) (fun x -> g (x x))
but this works only with lazy evaluation (as, I believe, Haskell has). As OCaml has eager evaluation, it produces a stack overflow when used. The reason is that OCaml tries to turn something like y g 8 into
g (y g) 8
g (g (y g)) 8
g (g (g (y g))) 8
...
without ever getting to call g.
The solution is to use deferred computation inside of y:
let y = fun g -> (fun x a -> g (x x) a) (fun x a -> g (x x) a)
One drawback is that y does not work for arbitrary types any more. It only works for function types.
y : ((`b -> `c) -> (`b -> `c)) -> (`b -> `c)
But you asked for recursive definitions of functions anyway, not for recursive definitions of other values. So, our definition of the factorial function is y g with y and g defined as above. Neither y nor g are anonymous yet, but that can be remedied easily:
(fun g -> (fun x a -> g (x x) a) (fun x a -> g (x x) a))
(fun f n -> if n < 2 then 1 else n * f (n - 1))
UPDATE:
Defining y only works with the -rectypes option. The reason is that we apply x to itself.
There is also an "intuitive" way to accomplish anonymous recursion without resorting to Y combinators.
It makes use of a let binding to store the value of a lambda that accepts itself as an argument, so that it can call itself with itself as the first parameter, like so:
let fact = (let fact0 = (fun self n -> if n < 2 then 1 else n * self self (n - 1)) in (fun n -> fact0 fact0 n));;
It's anonymous only to the extent that it is not defined with let rec.
I have the following recursive function that creates a list of 0s (i.e. [0,...,0]) in VDM. How can this be translated to Isabelle using fun-where?
VDM:
NewList: nat1 * seq of nat -> seq of nat
NewList(n, l) ==
if len l = n then l
else NewList(n, l ^ [0])
-- pre/post-conditions excluded here
My attempts are horribly wrong due to my lack of understanding of Isabelle (but below at least proves that I tried...).
Isabelle:
fun
NewList:: "N ⇒ (VDMNat VDMSeq) ⇒ (VDMNat VDMSeq)"
where
"NewList n [] = NewList n [0]"
| "NewList n [x] = (if len [x] = n then [x] else NewList n (x#[0]))"
| "NewList n (x # xs) = (if len (x # xs) = n then (x # xs) else NewList n ((x # xs) # [(0::VDMNat)]))"
*The data types VDMNat and VDMSeq are defined in some library. Please ignore the VDMNat and VDMSeq for now - any sort of implementation using Isabelle's data types are welcome (at least it would provide a good reference for my implementation). Please refer to the VDM code for the data types intended.
Could you also please explain what x, xs, and (x # xs) are referring to? I've seen this in several recursive function examples (though none helps me).
Thank you for your help!
First of all, x and xs are variables. When definiting recursive functions on lists, these are often used to denote the first element of the list (x) and the remaining list (xs). The expression x # xs means ‘x prepended to the list xs’, and that is the reason why (x # xs) # [0] in your question does not work: x # xs is a list and [0] is also a list. You would have to do x # xs # [0}, where # is the function to concatenate two lists.
Now, to your function: My interpretation of your function definition is that you have a natural number n and a list l and want to pad the list l with zeros at the back up to length n.
However, when the list l is of length > n to begin with, your function does not terminate. You would have to think about what to do in that case.
Here are my suggestions for what you could do:
Possibility 1
Change the = n to a ≥ n. Then you can prove termination of the function by looking at
function new_list :: "nat ⇒ nat list ⇒ nat list" where
"new_list n l = (if length l ≥ n then l else new_list n (l # [0]))"
by pat_completeness auto
termination by (relation "measure (λ(n, l). n - length l)") auto
However, proving theorems about this will probably get ugly. I would therefore urge you to do something like the following two possibilities. Ideally, use functions from Isabelle's standard library, because there is usually good automation setup for them. Alternatively, define your own small building blocks (like take and replicate) for your datatypes and prove reusable facts on them and combine them to do what you want. A ‘monolithic’ function definition like yours is difficult to work with when doing proofs.
Possibility 2
Use the builtin function replicate, which takes a natural number n and an element and returns a list of n times that element:
definition new_list :: "nat ⇒ nat list ⇒ nat list" where
"new_list n l = l # replicate (n - length l) 0"
You can also do the same thing with fun, but definition is the more low-level tool. Note that definition does not add the function definition theorem new_list_def as a simplifier rule; you can do this by writing declare new_list_def [simp].
Possibility 3
You can combine possibility 2 with the builtin function take to ensure that you always get a list of length exactly n, even when the input list is longer (it is then possibly truncated):
definition new_list :: "nat ⇒ nat list ⇒ nat list" where
"new_list n l = take n l # replicate (n - length l) 0"
Summary
In the first two cases, you can prove the theorems
length l ≤ n ⟹ length (new_list n l) = n
take (length l) (new_list n l) = l
(in the first case by induction using new_list.induct; in the second case just by unfolding the definition and simplifying)
In the third case, you can prove
length (new_list n l) = n
take (length l) (new_list n l) = take n l
Obviously, if length l ≤ n, the first two and the last one coincide completely.
The easy solution is: replicate n (0::nat) using the function replicate of Isabelle/HOL's library.
If you want to implement the function yourself via fun then do what you should always do in functional programming ;) try to split your problem into smaller problems that can be solved recursively:
fun newlist :: "nat => nat list"
where
"newlist 0 = []" -- "the only list of length 0*)
| "newlist (Suc n) = ..." -- "use result for 'n' to obtain result for 'n+1'"
I am interested in how would one define f to the n in Coq:
Basically, as an exercise, I would like to write this definition and then confirm that my
algorithm implements this specification. Inductive definition seems appropriate here, but I was not able to make it clean as above. What would be a clean Coq implementation of the above?
With the pow_func function that gallais defined, you can state your specification as lemmas, such as:
Lemma pow_func0: forall (A:Type) (f: A -> A) (x: A), pow_fun f O x = f x.
and
Lemma pow_funcS: forall (n:nat) (A: Type) (f: A->A) (x:A), pow_fun f (S n) x = f (pow_fun f n x).
The proof should be trivial by unfolding the definition
Inductive is used to define types closed under some operations; this is not what you are looking for here. What you want to build is a recursive function iterating over n. This can be done using the Fixpoint keyword:
Fixpoint pow_func {A : Type} (f : A -> A) (n : nat) (a : A) : A :=
match n with
| O => f a
| S n => f (pow_func f n a)
end.
If you want a nicer syntax for this function, you can introduce a Notation:
Notation "f ^ n" := (pow_func f n).
However, note that this is not a well-behaved definition of a notion of power: if you compose f ^ m and f ^ n, you don't get f ^ (m + n) but rather f ^ (1 + m + n). To fix that, you should pick the base case f ^ 0 to be the neutral element for composition id rather than f itself. Which would give you:
Fixpoint pow_func' {A : Type} (f : A -> A) (n : nat) (a : A) : A :=
match n with
| O => a
| S n => f (pow_func' f n a)
end.
From categorical point of view, functor is pair of two maps (one between objects and another between arrows of categories), following some axioms.
I have assumed, what every Functor instance is similar to mathematical definition i.e. can map both objects and functions, but Haskell's Functor class has only function fmap which maps functions.
Why so?
UPD In other words:
Every Monad type M has an function return :: a -> M a.
And Functor type F has no function return :: a -> F a, but only F x constructor.
There are two levels: types and values. As objects of Hask are types, you can map them only with type constructors, which have the * -> * kind:
α -> F α (for Functor F),
β -> M β (for Monad M).
Then for a functor you need a map on morphisms (i.e. functions, which are values):
fmap :: (α -> β) -> (F α -> F β)
The important thing is that return :: α -> M α of Monad is not a mapper of a type α to the M α as you might think. According to the mathematical definition of a monad, return corresponds to a natural transformation from Id functor to the M functor. And the Id functor is kind of implicit in Hask. The standard definition of a monad also requires another natural transformation M ◦ M -> M. So translating it to Haskell would look like
class Functor m => Monad m where
return :: Id α -> m α
join :: m (m α) -> m α
(As a side-note: these two natural transformations are actually the unit and multiplication, which make monad a monoid in the category of endofunctors)
The actual definition differs, but is equivalent. See Haskell/wiki on that.
If you take the composition-like operator derived from the standard bind >>= :: m α -> (α -> m β) -> m β:
(>=>) :: Monad m => (α -> m β) -> (β -> m γ) -> (α -> m γ)
f >=> g = \a => f a >>= g
You can see, that it's all actually about the Kleisli category. See also the article on nLab about monads in computer science.
Objects of a category are not the same as objects in a OO programming language (we prefer to call those values in Haskell; what they mean in category theory was discussed here). Rather, the objects of Hask are types. Haskell Functors are endofunctors in Hask, i.e. associate types to types, by the following means:
Prelude> :k Maybe
Maybe :: * -> *
Prelude> :k Int
Int :: *
Prelude> :k Maybe Int
Maybe Int :: *
OTOH, the arrows of Hask are in fact values, of some function type a -> b. These are associated in the following way:
fmap :: ( Functor (f :: t -> f t {- type-level -} ) )
=> (a->b) -> fmap(a->b) {- value-level -}
≡ (a->b) -> (f a->f b)
Though you were using those fancy categorical terms in your question and should be completely satisfied with the existing answers, here is an attempt for a rather trivial explanation:
Suppose there would be a function return (or pure or unit or ...) in the Functor type class.
Now try to define some common instances of Functor: [] (Lists), Maybe, ((,) a) (Tuples with a left component)
Easy enough, eh?
Here are the ordinary Functor instances:
instance Functor [] where
fmap f (x : xs) = f x : fmap xs
fmap _ [] = []
instance Functor Maybe where
fmap f (Just x) = Just (f x)
fmap _ Nothing = Nothing
instance Functor ((,) a) where
fmap f (x, y) = (x, f y)
What about return for Functor now?
Lists:
instance Functor [] where
return x = [x]
Alright. What about Maybe?
instance Functor Maybe where
return x = Just x
Okay. Now Tuples:
instance Functor ((,) a) where
return x = (??? , x)
You see, it is unknown which value should be filled into the left component of that tuple. The instance declaration says it has a type a but we do not know a value from that type. Maybe the type a is the Unit type with only one value. But if its Bool, should we take True or False? If it is Either Int Bool should we take Left 0 or Right False or Left 1?
So you see, if you had a return on Functors, you could not define a lot of valid functor instances in general (You would need to impose a constraint of something like a FunctorEmpty type class).
If you look at the documentation for Functor and Monad you will see that there are indeed instances for Functor ((,) a) but not for Monad ((,) a). This is because you just can't define return for that thing.
If you have
instance Functor F where
fmap = ...
Then the type constructor F is the action on objects (which are types) taking a type T to the type F T, and fmap is the action on morphisms (which are functions) taking a function f :: T -> U to fmap f :: F T -> F U.
In category theory, a functor maps all the objects from a category to another, but a functor doesn't map the elements in the objects.
I'm trying to write the sieve of Eratosthenes in Coq. I have a function crossout : forall {n:nat}, vector bool n -> nat -> vector bool n. When the sieve finds a number that is prime, it uses crossout to mark all the numbers that are not prime and then recurses on the resulting vector. The sieve obviously can't be structurally recursive on the vector itself, but it is structurally recursive on the length of the vector. What I want is to do something like this:
Fixpoint sieve {n:nat} (v:vector bool n) (acc:nat) {struct n} : list nat :=
match v with
| [] => Datatypes.nil
| false :: v' => sieve v' (S acc)
| true :: v' => Datatypes.cons acc (sieve (crossout v' acc) (S acc))
end.
But if I write it like this, Coq complains that the length of v' is not a subterm of n. I know that it is, but no matter how I structure the function, I can't seem to convince Coq that it is. Does anyone know how I can?
This is one of the most common pitfalls with dependent types in Coq. What is happening intuitively is that as soon as you pattern match on v, Coq "forgets" that the length of that vector is actually n, and loses the connection between the length of v' and the predecessor of n. The solution here is to apply what Adam Chlipala calls the convoy pattern, and make the pattern match return a function. While it is possible to do it by pattern matching on v, I think it is easier to do it by pattern matching on n:
Require Import Vector.
Axiom crossout : forall {n}, t bool n -> nat -> t bool n.
Fixpoint sieve {n:nat} : t bool n -> nat -> list nat :=
match n with
| 0 => fun _ _ => Datatypes.nil
| S n' => fun v acc =>
if hd v then
Datatypes.cons acc (sieve (crossout (tl v) acc) (S acc))
else
sieve (tl v) (S acc)
end.
Notice how the header of sieve has changed a little bit: now the return type is actually a function to help Coq's type inference.
For more information, check out Adam's book: http://adam.chlipala.net/cpdt/html/MoreDep.html.