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,它是一个文件夹,允许您在每个循环后返回一个新类型。
我正在尝试推广这个项目中定义的bitonicSort:https://github.com/adamwalker/clash-utils/blob/master/src/Clash/Sort.hs
我有两个函数,这两个函数对于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
上面提到的项目中使用的示例如下:
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)
使用这个版本,我可以快速创建新的Bitonic排序,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
每个排序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:
我想做一个函数,为给定的Vec生成函数。我相信在这里使用dfold或dtfold是正确的方法。
我想用像genBitonic
这样的函数来做折叠。
然后使用fst
获取排序所需的函数。
我有两种可能的设计:
One:使用组合进行折叠,得到一个带基的函数。
bSort8 :: Ord a => Vec 8 a -> Vec 8 a
bSort8 = fst $ genBitonic.genBitonic.genBitonic $ bitonicBase
在基础回复之前,它可能会导致类似这样的结果
**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)
Two:第二个想法是使用bitonicBase作为开始累加的值b。这将直接导致在应用fst
之前我需要它的表单。
编辑 vecAcum
只是为了在dfold
中建立价值。
在dfold示例中,它们使用列表运算符:
的向量形式的:>
进行折叠
>>> :t (:>)
(:>) :: a -> Vec n a -> Vec (n + 1) a
我想要做的是获取两个函数的元组,如下所示:
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)
然后把它们组合起来。因此,genBitonic . genBitonic
的类型为:
(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)
因此,基函数将是巩固类型的基础。例如:
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
我使用dfold来构建长度为n的向量的函数,该函数等同于对长度为n的向量进行递归。
我尝试过的:
我尝试遵循dfold下面列出的示例。
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
我使用的是dfold源代码使用的扩展。
{-# 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 #-}
错误消息:
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.
**编辑**添加了更多细节。
发布于 2017-03-07 16:17:30
您的base
案例是错误的;它应该是错误的
base :: (Ord a) => SplitHalf a @@ 0
base = (id, bitonicMerge id)
综上所述,这是一个完整的工作版本,在GHC8.0.2上进行了测试(但它应该同样可以在基于GHC8.0.2的CLaSH上工作,对Prelude
导入内容进行模块化)。事实证明,除了脊椎之外,不使用operationList
对象,因此我们可以使用Vec n ()
。
{-# 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)
这就像预期的那样工作,例如:
*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>
发布于 2017-03-08 23:27:55
我用CLaSH把它综合成
,所以我不能用递归来应用t次
我不明白这句话,但除此之外:
{-# 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)
genBitonicN
是由代表幂的peano数的递归定义的。在每个递归步骤中,都会弹出一个新的KnownNat (PowerOfTwo p)
(通过KnownPowersOfTwo
类型族)。在值级别,genBitonicN
是微不足道的,它只是做您想做的事情。然而,我们需要一些额外的机制来定义一个方便的bSortN
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
下面是一些示例:
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
我不知道我们是否可以避免为2的每一个幂定义IsPowerOfTwo
。
更新:这是bSortN
的另一个变体
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
举个例子:
bSort16 :: Ord a => OnVec 16 a
bSort16 = bSortNP (Proxy :: Proxy 4)
https://stackoverflow.com/questions/42191163
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