Suppose I define my own list type.
data MyVec : Nat -> Type -> Type where
MyNil : MyVec Z a
(::) : a -> MyVec k a -> MyVec (S k) a
And a myMap function serving as fmap for MyVec:
myMap : (a -> b) -> MyVec k a -> MyVec k b
myMap f MyNil = ?rhs_nil
myMap f (x :: xs) = ?rhs_cons
Attempting to solve the ?rhs_nil hole in my mind:
:t ?rhs_nil is MyVec 0 b
fair enough - I need a constructor that returns MyVec parameterized by b and I need k to be 0 (or Z) and I can see that MyNil is indexed by Z and parameterized by whatever, so I can use b easily, therefore ?rhs_nil = MyNil and it typechecks, dandy
:t ?rhs_cons is MyVec (S k)
I need a constructor that returns MyVec parameterized by b and I need k to be (S k)
I do see that the constructor (::) constructs a list indexed by (S k) and I try to use it. The first argument needs to be of type b considering I am building a MyVec <> b and the only way to get it is to apply x to f.
myMap f (x :: xs) = f x :: <>
Now I get confused. The RHS of (::) is supposed to be MyVec k b, why can I not simply use the MyNil constructor, with unification / substitution k == Z (that MyNil) gives me, getting:
myMap f (x :: xs) = f x :: MyNil
I do understand that I need to recurse and have = f x :: myMap f xs, but how does the compiler know the number of times the (::) constructor needs to be applied? How does it infer the correct k for this case, preventing me from using Z there.
The k is already implied by xs : MyVec k a. So you cannot unify k with Z if xs contains some elements.
Related
I am just starting to learn Idris coming from Haskell, and I'm trying to write some simple linear algebra code.
I want to write a Num interface instance for Vect, but specifically for Vect n a with the constraint that a has a Num instance.
In Haskell I would write a typeclass instance like this:
instance Num a => Num (Vect n a) where
(+) a b = (+) <$> a <*> b
(*) a b = (*) <$> a <*> b
fromInteger a = a : Nil
But reading the Idris interface docs does not seem to mention constraints on interface instances.
The best I can do is the following, which predictably causes the compiler to lament about a not being a numeric type:
Num (Vect n a) where
(+) Nil Nil = Nil
(+) (x :: xs) (y :: ys) = x + y :: xs + ys
(*) Nil Nil = Nil
(*) (x :: xs) (y :: ys) = x * y :: xs * ys
fromInteger i = Vect 1 (fromInteger i)
I can work around this by creating my own vector type with a Num constraint (which isn't portable) or by overloading (+) in a namespace (which feels a little clunky):
namespace Vect
(+) : Num a => Vect n a -> Vect n a -> Vect n a
(+) xs ys = (+) <$> xs <*> ys
Is there a way to constrain interface implementations, or is there a better way to accomplish this, eg using dependent types?
In Idris, you'd do (almost) the same as haskell
Num a => Num (Vect n a) where
Like a number of things, this is in the book but not, apparently, in the docs.
Let's say we have a function merge that, well, just merges two lists:
Order : Type -> Type
Order a = a -> a -> Bool
merge : (f : Order a) -> (xs : List a) -> (ys : List a) -> List a
merge f xs [] = xs
merge f [] ys = ys
merge f (x :: xs) (y :: ys) = case x `f` y of
True => x :: merge f xs (y :: ys)
False => y :: merge f (x :: xs) ys
and we'd like to prove something clever about it, for instance, that merging two non-empty lists produces a non-empty list:
mergePreservesNonEmpty : (f : Order a) ->
(xs : List a) -> (ys : List a) ->
{auto xsok : NonEmpty xs} -> {auto ysok : NonEmpty ys} ->
NonEmpty (merge f xs ys)
mergePreservesNonEmpty f (x :: xs) (y :: ys) = ?wut
Inspecting the type of the hole wut gives us
wut : NonEmpty (case f x y of True => x :: merge f xs (y :: ys) False => y :: merge f (x :: xs) ys)
Makes sense so far! So let's proceed and case-split as this type suggests:
mergePreservesNonEmpty f (x :: xs) (y :: ys) = case x `f` y of
True => ?wut_1
False => ?wut_2
It seems reasonable to hope that the types of wut_1 and wut_2 would match the corresponding branches of merge's case expression (so wut_1 would be something like NonEmpty (x :: merge f xs (y :: ys)), which can be instantly satisfied), but our hopes fail: the types are the same as for the original wut.
Indeed, the only way seems to be to use a with-clause:
mergePreservesNonEmpty f (x :: xs) (y :: ys) with (x `f` y)
mergePreservesNonEmpty f (x :: xs) (y :: ys) | True = ?wut_1
mergePreservesNonEmpty f (x :: xs) (y :: ys) | False = ?wut_2
In this case the types would be as expected, but this leads to repeating the function arguments for every with branch (and things get worse once with gets nested), plus with doesn't seem to play nice with implicit arguments (but that's probably worth a question on its own).
So, why doesn't case help here, are there any reasons besides purely implementation-wise behind not matching its behaviour with that of with, and are there any other ways to write this proof?
The stuff to the left of the | is only necessary if the new information somehow propagates backwards to the arguments.
mergePreservesNonEmpty : (f : Order a) ->
(xs : List a) -> (ys : List a) ->
{auto xsok : NonEmpty xs} -> {auto ysok : NonEmpty ys} ->
NonEmpty (merge f xs ys)
mergePreservesNonEmpty f (x :: xs) (y :: ys) with (x `f` y)
| True = IsNonEmpty
| False = IsNonEmpty
-- for contrast
sym' : (() -> x = y) -> y = x
sym' {x} {y} prf with (prf ())
-- matching against Refl needs x and y to be the same
-- now we need to write out the full form
sym' {x} {y=x} prf | Refl = Refl
As for why this is the case, I do believe it's just the implementation, but someone who knows better may dispute that.
There's an issue about proving things with case: https://github.com/idris-lang/Idris-dev/issues/4001
Because of this, in idris-bi we ultimately had to remove all cases in such functions and define separate top-level helpers that match on the case condition, e.g., like here.
I have the following inductive type MyVec:
import Data.Vect
data MyVec: {k: Nat} -> Vect k Nat -> Type where
Nil: MyVec []
(::): {k, n: Nat} -> {v: Vect k Nat} -> Vect n Nat -> MyVec v -> MyVec (n :: v)
-- example:
val: MyVec [3,2,3]
val = [[2,1,2], [0,2], [1,1,0]]
That is, the type specifies the lengths of all vectors inside a MyVec.
The problem is, val will have k = 3 (k is the number of vectors inside a MyVec), but the ctor :: does not know this fact. It will first build a MyVec with k = 1, then with 2, and finally with 3. This makes it impossible to define constraints based on the final shape of the value.
For example, I cannot constrain the values to be strictly less than k. Accepting Vects of Fin (S k) instead of Vects of Nat would rule out some valid values, because the last vectors (the first inserted by the ctor) would "know" a smaller value of k, and thus a stricter contraint.
Or, another example, I cannot enforce the following constraint: the vector at position i cannot contain the number i. Because the final position of a vector in the container is not known to the ctor (it would be automatically known if the final value of k was known).
So the question is, how can I enforce such global properties?
There are (at least) two ways to do it, both of which may require tracking additional information in order to check the property.
Enforcing properties in the data definition
Enforcing all elements < k
I cannot constrain the values to be strictly less than k. Accepting Vects of Fin (S k) instead of Vects of Nat would rule out some valid values...
You are right that simply changing the definition of MyVect to have Vect n (Fin (S k)) in it would not be correct.
However, it is not too hard to fix this by generalizing MyVect to be polymorphic, as follows.
data MyVec: (A : Type) -> {k: Nat} -> Vect k Nat -> Type where
Nil: {A : Type} -> MyVec A []
(::): {A : Type} -> {k, n: Nat} -> {v: Vect k Nat} -> Vect n A -> MyVec A v -> MyVec A (n :: v)
val : MyVec (Fin 3) [3,2,3]
val = [[2,1,2], [0,2], [1,1,0]]
The key to this solution is separating the type of the vector from k in the definition of MyVec, and then, at top level, using the "global value of k to constrain the type of vector elements.
Enforcing vector at position i does not contain i
I cannot enforce that the vector at position i cannot contain the number i because the final position of a vector in the container is not known to the constructor.
Again, the solution is to generalize the data definition to keep track of the necessary information. In this case, we'd like to keep track of what the current position in the final value will be. I call this index. I then generalize A to be passed the current index. Finally, at top level, I pass in a predicate enforcing that the value does not equal the index.
data MyVec': (A : Nat -> Type) -> (index : Nat) -> {k: Nat} -> Vect k Nat -> Type where
Nil: {A : Nat -> Type} -> {index : Nat} -> MyVec' A index []
(::): {A : Nat -> Type} -> {k, n, index: Nat} -> {v: Vect k Nat} ->
Vect n (A index) -> MyVec' A (S index) v -> MyVec' A index (n :: v)
val : MyVec' (\n => (m : Nat ** (n == m = False))) 0 [3,2,3]
val = [[(2 ** Refl),(1 ** Refl),(2 ** Refl)], [(0 ** Refl),(2 ** Refl)], [(1 ** Refl),(1 ** Refl),(0 ** Refl)]]
Enforcing properties after the fact
Another, sometimes simpler way to do it, is to not enforce the property immediately in the data definition, but to write a predicate after the fact.
Enforcing all elements < k
For example, we can write a predicate that checks whether all elements of all vectors are < k, and then assert that our value has this property.
wf : (final_length : Nat) -> {k : Nat} -> {v : Vect k Nat} -> MyVec v -> Bool
wf final_length [] = True
wf final_length (v :: mv) = isNothing (find (\x => x >= final_length) v) && wf final_length mv
val : (mv : MyVec [3,2,3] ** wf 3 mv = True)
val = ([[2,1,2], [0,2], [1,1,0]] ** Refl)
Enforcing vector at position i does not contain i
Again, we can express the property by checking it, and then asserting that the value has the property.
wf : (index : Nat) -> {k : Nat} -> {v : Vect k Nat} -> MyVec v -> Bool
wf index [] = True
wf index (v :: mv) = isNothing (find (\x => x == index) v) && wf (S index) mv
val : (mv : MyVec [3,2,3] ** wf 0 mv = True)
val = ([[2,1,2], [0,2], [1,1,0]] ** Refl)
Following my other question, I tried to implement the actual exercise in Type-Driven Development with Idris for same_cons to prove that, given two equal lists, prepending the same element to each list results in two equal lists.
Example:
prove that 1 :: [1,2,3] == 1 :: [1,2,3]
So I came up with the following code that compiles:
sameS : {xs : List a} -> {ys : List a} -> (x: a) -> xs = ys -> x :: xs = x :: ys
sameS {xs} {ys} x prf = cong prf
same_cons : {xs : List a} -> {ys : List a} -> xs = ys -> x :: xs = x :: ys
same_cons prf = sameS _ prf
I can call it via:
> same_cons {x=5} {xs = [1,2,3]} {ys = [1,2,3]} Refl
Refl : [5, 1, 2, 3] = [5, 1, 2, 3]
Regarding the cong function, my understanding is that it takes a proof, i.e. a = b, but I don't understand its second argument: f a.
> :t cong
cong : (a = b) -> f a = f b
Please explain.
If you have two values u : c and v : c, and a function f : c -> d, then if you know that u = v, it has to follow that f u = f v, following simply from referential transparency.
cong is the proof of the above statement.
In this particular use case, you are setting (via unification) c and d to List a, u to xs, v to ys, and f to (:) x, since you want to prove that xs = ys -> (:) x xs = (:) x ys.
I'm trying to write in Idris a function that create a Vect by passing the size of the Vect and a function taking the index in parameter.
So far, I've this :
import Data.Fin
import Data.Vect
generate: (n:Nat) -> (Nat -> a) ->Vect n a
generate n f = generate' 0 n f where
generate': (idx:Nat) -> (n:Nat) -> (Nat -> a) -> Vect n a
generate' idx Z f = []
generate' idx (S k) f = (f idx) :: generate' (idx + 1) k f
But I would like to ensure that the function passed in parameter is only taking index lesser than the size of the Vect.
I tried that :
generate: (n:Nat) -> (Fin n -> a) ->Vect n a
generate n f = generate' 0 n f where
generate': (idx:Fin n) -> (n:Nat) -> (Fin n -> a) -> Vect n a
generate' idx Z f = []
generate' idx (S k) f = (f idx) :: generate' (idx + 1) k f
But it doesn't compile with the error
Can't convert
Fin n
with
Fin (S k)
My question is : is what I want to do possible and how ?
The key idea is that the first element of the vector is f 0, and for the tail, if you have k : Fin n, then FS k : Fin (S n) is a "shift" of the finite number that increments its value and its type at the same time.
Using this observation, we can rewrite generate as
generate : {n : Nat} -> (f : Fin n -> a) -> Vect n a
generate {n = Z} f = []
generate {n = S _} f = f 0 :: generate (f . FS)
Another possibility is to do what #dfeuer suggested and generate a vector of Fins, then map f over it:
fins : (n : Nat) -> Vect n (Fin n)
fins Z = []
fins (S n) = FZ :: map FS (fins n)
generate' : {n : Nat} -> (f : Fin n -> a) -> Vect n a
generate' f = map f $ fins _
Proving generate f = generate' f is left as en exercise to the reader.
Cactus's answer appears to be about the best way to get what you asked for, but if you want something that can be used at runtime, it will be quite inefficient. The essential reason for this is that to weaken a Fin n to a Fin n+m requires that you completely deconstruct it to change the type of its FZ, and then build it back up again. So there can be no sharing at all between the Fin values produced for each vector element. An alternative is to combine a Nat with a proof that it is below a given bound, which leads to the possibility of erasure:
data NFin : Nat -> Type where
MkNFin : (m : Nat) -> .(LT m n) -> NFin n
lteSuccLeft : LTE (S n) m -> LTE n m
lteSuccLeft {n = Z} prf = LTEZero
lteSuccLeft {n = (S k)} {m = Z} prf = absurd (succNotLTEzero prf)
lteSuccLeft {n = (S k)} {m = (S j)} (LTESucc prf) = LTESucc (lteSuccLeft prf)
countDown' : (m : Nat) -> .(m `LTE` n) -> Vect m (NFin n)
countDown' Z mLTEn = []
countDown' (S k) mLTEn = MkNFin k mLTEn :: countDown' k (lteSuccLeft mLTEn)
countDown : (n : Nat) -> Vect n (NFin n)
countDown n = countDown' n lteRefl
countUp : (n : Nat) -> Vect n (NFin n)
countUp n = reverse $ countDown n
generate : (n : Nat) -> (NFin n -> a) -> Vect n a
generate n f = map f (countUp n)
As in the Fin approach, the function you pass to generate does not need to work on all naturals; it only needs to handle ones less than n.
I used the NFin type to explicitly indicate that I want the LT m n proof to be erased in all cases. If I didn't want/need that, I could just use (m ** LT m n) instead.