I defined n-dimensional vectors in Idris as follows:
import Data.Vect
NDVect : (Num t) => (rank : Nat) -> (shape : Vect rank Nat) -> (t : Type) -> Type
NDVect Z [] t = t
NDVect (S n) (x::xs) t = Vect x (NDVect n xs t)
Then I defined the following function which maps a function f to every entry in the tensor.
iterateT : (f : t -> t') -> (v : NDVect r s t) -> NDVect r s t'
iterateT {r = Z} {s = []} f v = f v
iterateT {r = S n} {s = x::xs} f v = map (iterateT f) v
But when I try to call iteratorT in the following function:
scale : Num t => (c : t) -> (v : NDVect rank shape t) -> NDVect rank shape t
scale c v = iterateT (*c) v
I get the following error message saying there is a type mismatched, which seems pretty fine to me
When checking right hand side of scale with expected type
NDVect rank shape t
When checking argument v to function Main.iterateT:
Type mismatch between
NDVect rank shape t (Type of v)
and
NDVect r s t (Expected type)
Specifically:
Type mismatch between
NDVect rank shape t
and
NDVect r s t
Specifically:
Type mismatch between
NDVect rank shape t
and
NDVect r s t
I have also been wondering how to express n-dimensional vectors (i.e. tensors) in Idris. I had a play with the type definition in the question, but encountered various issues, so I expressed the NDVect function as a data type:
data NDVect : (rank : Nat) -> (shape : Vect rank Nat) -> Type -> Type where
NDVZ : (value : t) -> NDVect Z [] t
NDV : (values : Vect n (NDVect r s t)) -> NDVect (S r) (n::s) t
And implemented map as follows:
nmap : (t -> u) -> (NDVect r s t) -> NDVect r s u
nmap f (NDVZ value) = NDVZ (f value)
nmap f (NDV values) = NDV (map (nmap f) values)
The following now works:
*Main> NDVZ 5
NDVZ 5 : NDVect 0 [] Integer
*Main> nmap (+4) (NDVZ 5)
NDVZ 9 : NDVect 0 [] Integer
*Main> NDV [NDVZ 1, NDVZ 2, NDVZ 3]
NDV [NDVZ 1, NDVZ 2, NDVZ 3] : NDVect 1 [3] Integer
*Main> nmap (+4) (NDV [NDVZ 1, NDVZ 2, NDVZ 3])
NDV [NDVZ 5, NDVZ 6, NDVZ 7] : NDVect 1 [3] Integer
Unfortunately, having all the type constructors makes things a bit ugly. I'd love to know if there's a cleaner way to solve this.
Edit:
Here's a slightly shorter type signature that doesn't explicitly encode the tensor rank in the type:
data NDVect : (shape : List Nat) -> Type -> Type where
NDVZ : (value : t) -> NDVect [] t
NDV : (values : Vect n (NDVect s t)) -> NDVect (n::s) t
nmap : (t -> u) -> (NDVect s t) -> NDVect s u
nmap f (NDVZ value) = NDVZ (f value)
nmap f (NDV values) = NDV (map (nmap f) values)
Related
In Idris I am trying to write a type-safe(r) version of a splitEvery function for Vect, that on Lists looks like this:
splitEvery : Nat -> List a -> List (List a)
splitEvery _ [] = []
splitEvery n xs = (take n xs) ++ splitEvery n (drop n xs)
So far I've got:
splitEvery : {k : Nat} -> (n : Nat) -> Vect (k * n) a -> Vect k (Vect n a)
splitEvery {k = Z} _ [] = []
splitEvery {k = (S j)} n xs = (take n xs) :: splitEvery {k = j} n (drop n xs)
which will type-check and load in the REPL alright, but when I try to invoke it, it will fail:
*Main> splitEvery 2 (fromList [1..10])
(input):1:1-31:When checking an application of function Main.splitEvery:
Type mismatch between
Vect (length (enumFromTo 1 10))
Integer (Type of fromList (enumFromTo 1 10))
and
Vect (k * 2) Integer (Expected type)
Specifically:
Type mismatch between
length (enumFromTo 1 10)
and
mult k 2
So obviously Idris isn't seeing that a valid choice of k here would be 5. A way to make it work would be to explicitly give it the implicit parameter k, but that's ugly:
*Main> splitEvery {k = 5} 2 (fromList [1..10])
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]] : Vect 5 (Vect 2 Integer)
So my question is: is there a way to make this work that isn't ugly, or is more idiomatic than what I've produced so far?
Idris is only failing to infer the value of k in the REPL for some reason. If you amend your code with
result: Vect 5 (Vect 2 Nat)
result = splitEvery 2 (fromList [1..10])
everything typechecks - as the type is already known and k can be infered. You can achieve the same on the REPL by providing a type hint using the:
the (Vect 5 (Vect 2 Nat)) (splitEvery 2 (fromList [1..10])) works as well.
An example value: [[1, 2], [1]]
Here, I would know that there are 2 lists, and that the first list has length 2, while the second one has length 1. Ideally, this function would compute those types:
func : (n ** Vect n Nat) -> Type
But I don't know how to write it. I'm pretty sure it's something to do with dependent pairs, but I'm not sure how to write it.
To clarify, I know it'd be possible to simply use (n ** Vect n (p ** Vect p Double)) as the type of the example value. However, n only constrains the number of lists, not the number of their elements, because inside the list, p could be anything. I would most likely need something where the first element of the dependent pair is a vector of lengths, not just the number of lists. So something like (Vect n Nat ** Vect n (Vect m Double))--where each m is the corresponding element of the first vector.
You could define a new vector type which contains possibly differently indexed elements of an indexed type at each position:
import Prelude
import Data.Vect
-- heterogeneously indexed vector
data IVect : Vect n ix -> (ix -> Type) -> Type where
Nil : IVect Nil b
(::) : b i -> IVect is b -> IVect (i :: is) b
-- of which a special case is a vector of vectors
VVect : Vect n Nat -> Type -> Type
VVect is a = IVect is (flip Vect a)
test1 : VVect [2, 2, 2] Nat
test1 = [[1, 2], [3, 4], [5, 6]]
test2 : VVect [0, 1, 2] Bool
test2 = [[], [True], [False, True]]
Alternatively, you can define VVect using dependent pairs and map, but this is more cumbersome to use:
VVect' : Vect n Nat -> Type -> Type
VVect' {n = n} is a = (xs : Vect n (m ** Vect m a) ** (map fst xs = is))
test3 : VVect' [0, 1, 2] Bool
test3 = ([(_ ** []), (_ ** [True]), (_ ** [False, False])] ** Refl)
You have some choice though whether to use lists or vectors. With lists as the inner container, values look more compact:
VVect'' : Vect n Nat -> Type -> Type
VVect'' {n = n} is a = (xs : Vect n (List a) ** (map length xs = is))
test4 : VVect'' [0, 1, 2] Bool
test4 = ([[], [True], [False, True]] ** Refl)
I have the following working function:
unMaybe : (t : Type) -> {auto p : t = Maybe x} -> Type
unMaybe {x} _ = x
This function works fine:
> unMaybe (Maybe Int)
Int
I also have another similar function:
unMaybesA : (ts : Vect n Type) -> {xs : Vect n Type} -> {auto p : map Maybe xs = ts} -> Vect n Type
unMaybesA {xs} _ = xs
Unfortunately the following fails:
> unMaybesA [Maybe Int, Maybe String]
(input):1:1-35:When checking argument p to function Main.unMaybesA:
Can't find a value of type
Data.Vect.Vect n implementation of Prelude.Functor.Functor, method map Maybe
xs =
[Maybe Int, Maybe String]
But the following works:
> unMaybesA {xs=[_,_]} [Maybe Int, Maybe String]
[Int, String]
Is the a way to get Idris to automatically do {xs=[_,_]} with however many _ the vector has?
unMaybesB : (ts : Vect n Type) -> {auto p : (xs : Vect n Type ** map Maybe xs = ts)} -> Vect n Type
unMaybesB {p} _ = fst p
Possibly by using an elaborator script to automatically fill p in the function above?
I have the outline of an elab script below. I just need to figure out how to generate n, ts, and xs from the goal.
helper1 : Vect n Type -> Vect n Type -> Type
helper1 ts xs = (map Maybe xs) = ts
unMaybesC : (ts : Vect n Type) -> {auto p : DPair (Vect n Type) (helper1 ts)} -> Vect n Type
unMaybesC {p} _ = fst p
helper2 : (n : Nat) -> (ts : Vect n Type) -> (xs : Vect n Type) -> helper1 ts xs -> DPair (Vect n Type) (helper1 ts)
helper2 _ _ xs p = MkDPair xs p
q : Elab ()
q = do
let n = the Raw `(2 : Nat)
let ts = the Raw `(with Vect [Maybe String, Maybe Int])
let xs = the Raw `(with Vect [String, Int])
fill `(helper2 ~n ~ts ~xs Refl)
solve
qC : Vect 2 Type
qC = unMaybesC {p=%runElab q} [Maybe String, Maybe Int]
map Maybe xs = ts seems idiomatic, but is quite difficult. If you want to auto search for a non-simple proof, write an explicit proof type. Then the proof search will try the constructors and is guided in the right direction.
data IsMaybes : Vect n Type -> Vect n Type -> Type where
None : IsMaybes [] []
Then : IsMaybes xs ms -> IsMaybes (t :: xs) (Maybe t :: ms)
unMaybes : (ts : Vect n Type) -> {xs : Vect n Type} -> {auto p : IsMaybes xs ts} -> Vect n Type
unMaybes ts {xs} = xs
And with this:
> unMaybes [Maybe Nat, Maybe Int, Maybe (Maybe String)]
[Nat, Int, Maybe String] : Vect 3 Type
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)
Given:
*section3> :module Data.Vect
*section3> :let e = the (Vect 0 Int) []
*section3> :let xs = the (Vect _ _) [1,2]
*section3> decEq xs e
(input):1:7:When checking argument x2 to function Decidable.Equality.decEq:
Type mismatch between
Vect 0 Int (Type of e)
and
Vect 2 Integer (Expected type)
Specifically:
Type mismatch between
0
and
2
Why must the Nat arguments equal each other for DecEq?
Note - posted in https://groups.google.com/forum/#!topic/idris-lang/qgtImCLka3I originally
decEq is for homogenous propositional equality:
||| Decision procedures for propositional equality
interface DecEq t where
||| Decide whether two elements of `t` are propositionally equal
total decEq : (x1 : t) -> (x2 : t) -> Dec (x1 = x2)
As you can see, x1 and x2 are both of type t. In your case, you have x1 : Vect 2 Integer and x2 : Vect 0 Int. These are two different types.
You can write your own heterogenous equality decider for Vectors of the same element type by first checking their lengths, then delegating to the homogenous version:
import Data.Vect
vectLength : {xs : Vect n a} -> {ys : Vect m a} -> xs = ys -> n = m
vectLength {n = n} {m = n} Refl = Refl
decEqVect : (DecEq a) => (xs : Vect n a) -> (ys : Vect m a) -> Dec (xs = ys)
decEqVect {n = n} {m = m} xs ys with (decEq n m)
decEqVect xs ys | Yes Refl = decEq xs ys
decEqVect xs ys | No notEq = No (notEq . vectLength)