在每次重复的函数调用之后创建允许类型更改的折叠，以便在没有递归的情况下调用函数n次

Question

我正在尝试使用dfold定义的here

dfold 
    :: KnownNat k    
    => Proxy (p :: TyFun Nat * -> *)    
    -> (forall l. SNat l -> a -> (p @@ l) -> p @@ (l + 1))  
    -> (p @@ 0) 
    -> Vec k a  
    -> p @@ k

基本上它是一个折叠，允许您在每个循环后返回一个新类型。

我试图概括在这个项目中定义的bitonicSort：https://github.com/adamwalker/clash-utils/blob/master/src/Clash/Sort.hs

我有两个函数对dfold生成的类型很重要：

bitonicSort
    :: forall n a. (KnownNat n, Ord a)
    => (Vec n a -> Vec n a)             -- ^ The recursive step
    -> (Vec (2 * n) a -> Vec (2 * n) a) -- ^ Merge step
    -> Vec (2 * n) a                    -- ^ Input vector
    -> Vec (2 * n) a                    -- ^ Output vector
bitonicMerge
    :: forall n a. (Ord a , KnownNat n)
    => (Vec n a -> Vec n a) -- ^ The recursive step
    -> Vec (2 * n) a        -- ^ Input vector
    -> Vec (2 * n) a        -- ^ Output vector

上述项目中使用的示例是：

bitonicSorterExample 
    :: forall a. (Ord a) 
    => Vec 16 a -- ^ Input vector
    -> Vec 16 a -- ^ Sorted output vector
bitonicSorterExample = sort16
    where
    sort16 = bitonicSort sort8 merge16
    merge16 = bitonicMerge merge8

    sort8  = bitonicSort  sort4  merge8
    merge8 = bitonicMerge merge4

    sort4  = bitonicSort  sort2 merge4
    merge4 = bitonicMerge merge2

    sort2  = bitonicSort  id merge2
    merge2 = bitonicMerge id 

我继续做了一个更通用的版本。

genBitonic :: (Ord a, KnownNat n) =>
  (Vec n a -> Vec n a, Vec (2 * n) a -> Vec (2 * n) a)
  -> (Vec (2 * n) a -> Vec (2 * n) a, Vec (4 * n) a -> Vec (4 * n) a)
genBitonic (bSort,bMerge) = (bitonicSort bSort bMerge, bitonicMerge bMerge)

bitonicBase :: Ord a =>  (Vec 1 a -> Vec 1 a, Vec 2 a -> Vec 2 a)
bitonicBase = (id, bitonicMerge id)

有了这个版本，我可以快速制作新的Bitonic Sorts，如下所示：

bSort16 :: Ord a => Vec 16 a -> Vec 16 a
bSort16 = fst $ genBitonic $ genBitonic $ genBitonic $ genBitonic bitonicBase

bSort8 :: Ord a => Vec 8 a -> Vec 8 a
bSort8 = fst $ genBitonic $ genBitonic $ genBitonic bitonicBase

bSort4 :: Ord a => Vec 4 a -> Vec 4 a
bSort4 = fst $ genBitonic $ genBitonic bitonicBase

bSort2 :: Ord a => Vec 2 a -> Vec 2 a
bSort2 = fst $ genBitonic bitonicBase

每个Sort都使用指定大小的向量。

testVec16 :: Num a => Vec 16 a
testVec16 =  9 :> 2 :> 8 :> 6 :> 3 :> 7 :> 0 :> 1 :> 4 :> 5 :> 2 :> 8 :> 6 :> 3 :> 7 :> 0 :> Nil

testVec8 :: Num a => Vec 8 a
testVec8 =  9 :> 2 :> 8 :> 6 :> 3 :> 7 :> 0 :> 1 :> Nil

testVec4 :: Num a => Vec 4 a
testVec4 =  9 :> 2 :> 8 :> 6 :> Nil

testVec2 :: Num a => Vec 2 a
testVec2 =  2 :> 9 :>  Nil

快速说明：

• 我正在尝试苹果“gen Bitonic”到“bitonic Base”开球时间。
• 我正在使用CLaSH将其合成为VHDL，因此我不能使用递归来应用t次
• 我们将始终将vec尺寸2 ^ t分类到相同尺寸的vec
• “Vec n a”是大小为n且类型为a的向量

我想创建一个为给定Vec生成函数的函数。我相信使用dfold或dtfold，这是正确的路径。

我想用功能genBitonic做折叠。

然后使用fst来获得我需要的排序功能。

我有两种可能的设计：

一：使用合成折叠来获得一个基础的函数。

bSort8 :: Ord a => Vec 8 a -> Vec 8 a
bSort8 = fst $ genBitonic.genBitonic.genBitonic $ bitonicBase

在基地被回复之前，它会产生类似的结果

**If composition was performed three times**

foo3 ::
  (Ord a, KnownNat n) =>
  (Vec n a -> Vec n a, Vec (2 * n) a -> Vec (2 * n) a)
  -> (Vec (2 * (2 * (2 * n))) a -> Vec (2 * (2 * (2 * n))) a,
      Vec (4 * (2 * (2 * n))) a -> Vec (4 * (2 * (2 * n))) a)

二：第二个想法是使用bitonicBase作为值b来开始积累。在我应用fst之前，这将直接导致我需要它的形式。

编辑vecAcum只是意味着在dfold内部建立的价值。

在dfold示例中，他们使用:>进行折叠，:只是列表运算符

>>> :t (:>)
(:>) :: a -> Vec n a -> Vec (n + 1) a

的向量形式

genBitonic :: (Ord a, KnownNat n) =>
  (Vec n a -> Vec n a, Vec (2 * n) a -> Vec (2 * n) a)
  -> (Vec (2 * n) a -> Vec (2 * n) a, Vec (4 * n) a -> Vec (4 * n) a)

我想要做的是取两个函数的元组，如：

genBitonic . genBitonic

并撰写它们。所以

(Vec n a -> Vec n a, Vec (2 * n) a -> Vec (2 * n) a)
-> (Vec (2 * (2 * n)) a -> Vec (2 * (2 * n)) a, Vec (4 * (2 * n)) a -> Vec (4 * (2 * n)) a)

会有类型：

bitonicBase :: Ord a =>  (Vec 1 a -> Vec 1 a, Vec 2 a -> Vec 2 a)
bitonicBase = (id, bitonicMerge id)
bSort4 :: Ord a => Vec 4 a -> Vec 4 a
bSort4 = fst $ genBitonic $ genBitonic bitonicBase

那么基本函数就是固化类型的东西。例如

data SplitHalf (a :: *) (f :: TyFun Nat *) :: *
type instance Apply (SplitHalf a) l = (Vec (2^l) a -> Vec (2^l) a, Vec (2 ^ (l + 1)) a -> Vec (2 ^ (l + 1)) a)

generateBitonicSortN2 :: forall k a . (Ord a, KnownNat k) => SNat k ->  Vec (2^k) a -> Vec (2^k) a
generateBitonicSortN2 k =  fst $ dfold (Proxy :: Proxy (SplitHalf a)) vecAcum base vecMath
  where
    vecMath = operationList k


vecAcum :: (KnownNat l, KnownNat gl,  Ord a) => SNat l
                                -> (SNat gl -> SplitHalf a @@ gl -> SplitHalf a @@ (gl+1))
                                -> SplitHalf a @@ l
                                -> SplitHalf a @@ (l+1)
vecAcum l0 f acc = undefined --  (f l0) acc

base :: (Ord a) => SplitHalf a @@ 0
base = (id,id)

general :: (KnownNat l,  Ord a)
        => SNat l
        -> SplitHalf a @@ l
        -> SplitHalf a @@ (l+1)
general _ (x,y) = (bitonicSort x y, bitonicMerge y )

operationList :: (KnownNat k, KnownNat l, Ord a)
              => SNat k
              -> Vec k
                   (SNat l
                 -> SplitHalf a @@ l
                 -> SplitHalf a @@ (l+1))
operationList k0 = replicate k0 general

我正在使用dfold为长度为n的向量构建函数，这相当于在长度为n的向量上进行递归。

我试过了：

我尝试按照dfold下列出的示例进行操作

{-# LANGUAGE BangPatterns         #-}
{-# LANGUAGE DataKinds            #-}
{-# LANGUAGE GADTs                #-}
{-# LANGUAGE KindSignatures       #-}
{-# LANGUAGE MagicHash            #-}
{-# LANGUAGE PatternSynonyms      #-}
{-# LANGUAGE Rank2Types           #-}
{-# LANGUAGE ScopedTypeVariables  #-}
{-# LANGUAGE TemplateHaskell      #-}
{-# LANGUAGE TupleSections        #-}
{-# LANGUAGE TypeApplications     #-}
{-# LANGUAGE TypeFamilies         #-}
{-# LANGUAGE TypeOperators        #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE ViewPatterns         #-}

{-# LANGUAGE Trustworthy #-}

我正在使用dfold源代码使用的扩展

   Sort.hs:182:71: error:
    * Could not deduce (KnownNat l) arising from a use of `vecAcum'
      from the context: (Ord a, KnownNat k)
        bound by the type signature for:
                   generateBitonicSortN2 :: (Ord a, KnownNat k) =>
                                            SNat k -> Vec (2 ^ k) a -> Vec (2 ^ k) a
        at Sort.hs:181:1-98
      Possible fix:
        add (KnownNat l) to the context of
          a type expected by the context:
            SNat l
            -> (SNat l0
                -> (Vec (2 ^ l0) a -> Vec (2 ^ l0) a,
                    Vec (2 ^ (l0 + 1)) a -> Vec (2 ^ (l0 + 1)) a)
                -> (Vec (2 ^ (l0 + 1)) a -> Vec (2 ^ (l0 + 1)) a,
                    Vec (2 ^ ((l0 + 1) + 1)) a -> Vec (2 ^ ((l0 + 1) + 1)) a))
            -> SplitHalf a @@ l
            -> SplitHalf a @@ (l + 1)
    * In the second argument of `dfold', namely `vecAcum'
      In the second argument of `($)', namely
        `dfold (Proxy :: Proxy (SplitHalf a)) vecAcum base vecMath'
      In the expression:
        fst $ dfold (Proxy :: Proxy (SplitHalf a)) vecAcum base vecMath

Sort.hs:182:84: error:
    * Could not deduce (KnownNat l0) arising from a use of `vecMath'
      from the context: (Ord a, KnownNat k)
        bound by the type signature for:
                   generateBitonicSortN2 :: (Ord a, KnownNat k) =>
                                            SNat k -> Vec (2 ^ k) a -> Vec (2 ^ k) a
        at Sort.hs:181:1-98
      The type variable `l0' is ambiguous
    * In the fourth argument of `dfold', namely `vecMath'
      In the second argument of `($)', namely
        `dfold (Proxy :: Proxy (SplitHalf a)) vecAcum base vecMath'
      In the expression:
        fst $ dfold (Proxy :: Proxy (SplitHalf a)) vecAcum base vecMath
Failed, modules loaded: none.

错误消息：

base

**编辑**添加了更多细节。

Answer 1

你的

base :: (Ord a) => SplitHalf a @@ 0
base = (id, bitonicMerge id)

案是错的;它应该是

Prelude

总而言之，这是一个完全可用的版本，在GHC 8.0.2上进行了测试（但它应该在基于GHC 8.0.2的CLaSH上完全相同，以operationList导入的方式为模）。事实证明Vec n ()的东西除了它的脊椎之外没有被使用，所以我们可以使用

{-# LANGUAGE DataKinds, GADTs, KindSignatures #-}
{-# LANGUAGE Rank2Types, ScopedTypeVariables  #-}
{-# LANGUAGE TypeFamilies, TypeOperators, UndecidableInstances #-}

{-# OPTIONS_GHC -fplugin GHC.TypeLits.Normalise -fplugin GHC.TypeLits.KnownNat.Solver #-}
{-# OPTIONS_GHC -fno-warn-incomplete-patterns -fno-warn-redundant-constraints #-}

{-# LANGUAGE NoImplicitPrelude #-}
import Prelude (Integer, (+), Num, ($), undefined, id, fst, Int, otherwise)
import CLaSH.Sized.Vector
import CLaSH.Promoted.Nat
import Data.Singletons
import GHC.TypeLits
import Data.Ord

type ExpVec k a = Vec (2 ^ k) a

data SplitHalf (a :: *) (f :: TyFun Nat *) :: *
type instance Apply (SplitHalf a) k = (ExpVec k a -> ExpVec k a, ExpVec (k + 1) a -> ExpVec (k + 1) a)

generateBitonicSortN2 :: forall k a . (Ord a, KnownNat k) => SNat k -> ExpVec k a -> ExpVec k a
generateBitonicSortN2 k = fst $ dfold (Proxy :: Proxy (SplitHalf a)) step base (replicate k ())
  where
    step :: SNat l -> () -> SplitHalf a @@ l -> SplitHalf a @@ (l+1)
    step SNat _ (sort, merge) = (bitonicSort sort merge, bitonicMerge merge)

    base = (id, bitonicMerge id)

代替。

*Main> generateBitonicSortN2  (snatProxy Proxy)  testVec2
<9,2>
*Main> generateBitonicSortN2  (snatProxy Proxy)  testVec4
<9,8,6,2>
*Main> generateBitonicSortN2  (snatProxy Proxy)  testVec8
<9,8,7,6,3,2,1,0>
*Main> generateBitonicSortN2  (snatProxy Proxy)  testVec16
<9,8,8,7,7,6,6,5,4,3,3,2,2,1,0,0>
*Main>

这按预期工作，例如：

{-# LANGUAGE GADTs, DataKinds, TypeFamilies, UndecidableInstances,
      FlexibleInstances, FlexibleContexts, ConstraintKinds,
      UndecidableSuperClasses, TypeOperators #-}

import GHC.TypeLits
import GHC.Exts (Constraint)
import Data.Proxy

data Peano = Z | S Peano

data SPeano n where
  SZ :: SPeano Z
  SS :: SPeano n -> SPeano (S n)

type family PowerOfTwo p where
  PowerOfTwo  Z    = 1
  PowerOfTwo (S p) = 2 * PowerOfTwo p

type family KnownPowersOfTwo p :: Constraint where
  KnownPowersOfTwo  Z    = ()
  KnownPowersOfTwo (S p) = (KnownNat (PowerOfTwo p), KnownPowersOfTwo p)

data Vec (n :: Nat) a -- abstract

type OnVec n a = Vec n a -> Vec n a
type GenBitonic n a = (OnVec n a, OnVec (2 * n) a)

genBitonic :: (Ord a, KnownNat n) => GenBitonic n a -> GenBitonic (2 * n) a
genBitonic = undefined

bitonicBase :: Ord a => GenBitonic 1 a
bitonicBase = undefined

genBitonicN :: (Ord a, KnownPowersOfTwo p) => SPeano p -> GenBitonic (PowerOfTwo p) a
genBitonicN  SZ    = bitonicBase
genBitonicN (SS p) = genBitonic (genBitonicN p)

Answer 2

我正在使用CLaSH将其合成为VHDL，因此我不能使用递归来应用t次

我不理解这句话，但除此之外：

genBitonicN

KnownNat (PowerOfTwo p)是通过对代表幂的peano数的递归来定义的。在每个递归步骤中弹出一个新的KnownPowersOfTwo（通过genBitonicN类型系列）。在价值水平bSortN是微不足道的，只是做你想要的。但是我们需要一些额外的机器来定义一个方便的

type family Lit n where
  Lit 0 = Z
  Lit n = S (Lit (n - 1))

class IPeano n where
  speano :: SPeano n

instance IPeano Z where
  speano = SZ

instance IPeano n => IPeano (S n) where
  speano = SS speano

class (n ~ PowerOfTwo (PowerOf n), KnownPowersOfTwo (PowerOf n)) =>
         IsPowerOfTwo n where
  type PowerOf n :: Peano
  getPower :: SPeano (PowerOf n)

instance IsPowerOfTwo 1 where
  type PowerOf 1 = Lit 0
  getPower = speano

instance IsPowerOfTwo 2 where
  type PowerOf 2 = Lit 1
  getPower = speano

instance IsPowerOfTwo 4 where
  type PowerOf 4 = Lit 2
  getPower = speano

instance IsPowerOfTwo 8 where
  type PowerOf 8 = Lit 3
  getPower = speano

instance IsPowerOfTwo 16 where
  type PowerOf 16 = Lit 4
  getPower = speano

-- more powers go here

bSortN :: (IsPowerOfTwo n, Ord a) => OnVec n a
bSortN = fst $ genBitonicN getPower

：

bSort1 :: Ord a => OnVec 1 a
bSort1 = bSortN

bSort2 :: Ord a => OnVec 2 a
bSort2 = bSortN

bSort4 :: Ord a => OnVec 4 a
bSort4 = bSortN

bSort8 :: Ord a => OnVec 8 a
bSort8 = bSortN

bSort16 :: Ord a => OnVec 16 a
bSort16 = bSortN

这里有些例子：

IsPowerOfTwo

我不知道我们是否可以避免为每个2的幂定义bSortN。

更新：这是

pnatToSPeano :: IPeano (Lit n) => proxy n -> SPeano (Lit n)
pnatToSPeano _ = speano

bSortNP :: (IPeano (Lit p), KnownPowersOfTwo (Lit p), Ord a)
        => proxy p -> OnVec (PowerOfTwo (Lit p)) a
bSortNP = fst . genBitonicN . pnatToSPeano

的另一个变种：

bSort16 :: Ord a => OnVec 16 a
bSort16 = bSortNP (Proxy :: Proxy 4)

一个例子：

qazxswpoi