Haskell類型類的重新設計

Question

在獲得一些幫助之後，理解我試圖編譯代碼的問題，在這個問題中（Trouble understanding GHC complaint about ambiguity）Ness建議我重新設計我的類型類以避免我不完全滿意的解決方案。

有問題的類型類是這些：

class (Eq a, Show a) => Genome a where 
    crossover  :: (Fractional b) => b -> a -> a -> IO (a, a) 
    mutate   :: (Fractional b) => b -> a -> IO a 
    develop   :: (Phenotype b) => a -> b 

class (Eq a, Show a) => Phenotype a where 
    --In case of Coevolution where each phenotype needs to be compared to every other in the population 
    fitness   :: [a] -> a -> Int 
    genome   :: (Genome b) => a -> b 

我試圖創建在Haskell一種可伸縮進化算法，應支持不同Genomes和Phenotypes。例如，一個Genome可能是一個位陣列，另一個可能是一個整數列表，而且Phenotypes也可能不同，只是代表部隊運動的雙打列表http://en.wikipedia.org/wiki/Colonel_Blotto，或者它可以代表一個ANN。

由於Phenotype從Genome使用的必須是相當互換的方法開發的，一個Genome類應該能夠通過提供不同的開發方法來支持多個Phenotypes（這可以在代碼中靜態完成，並且沒有在運行時動態完成）。

使用這些類型類的代碼大部分應該是幸福地不知道使用的具體類型，這是什麼導致我問上述問題。

一些，我要適應這些類型類的代碼是：

-- |Full generational replacement selection protocol 
fullGenerational :: (Phenotype b) => 
    (Int -> [b] -> IO [(b, b)]) -> --Selection mechanism 
    Int -> --Elitism 
    Int -> --The number of children to create 
    Double -> --Crossover rate 
    Double -> --Mutation rate 
    [b] -> --Population to select from 
    IO [b] --The new population created 
fullGenerational selection e amount cross mute pop = do 
    parents <- selection (amount - e) pop 
    next <- breed parents cross mute 
    return $ next ++ take e reverseSorted 
      where reverseSorted = reverse $ sortBy (fit pop) pop 

breed :: (Phenotype b, Genome a) => [(b, b)] -> Double -> Double -> IO [b] 
breed parents cross mute = do 
    children <- mapM (\ (dad, mom) -> crossover cross (genome dad) (genome mom)) parents 
    let ch1 = map fst children ++ map snd children 
    mutated <- mapM (mutate mute) ch1 
    return $ map develop mutated 

據我所知，這個代碼將不得不改變，新的限制將被添加，但我想表現我想到的一些代碼使用類型類。例如，上面的全代更換並不需要知道任何關於底層Genome的功能;它只需要知道Phenotypes可以生產Genome，它可以將它們一起繁殖並創建新的孩子。 fullGenerational的代碼應儘可能一般，以便一旦設計了新的Phenotype或創建了更好的Genome，則不需要更改它。

上述類型類如何改變以避免我在類型類模糊性方面遇到的問題，同時保留了我在普通EA代碼中的屬性（應該是可重用的）？

Answer 1

「它只需要知道的表型可以產生產生它的基因組」

這意味着表型確實出現兩種類型的關係，另一個是基因組類型用於產生一個給定表型：

{-# LANGUAGE MultiParamTypeClasses #-} 
{-# LANGUAGE FunctionalDependencies #-} 

import Data.List (sortBy) 

class (Eq a, Show a) => Genome a where 
    crossover  :: (Fractional b) => b -> a -> a -> IO (a, a) 
    mutate   :: (Fractional b) => b -> a -> IO a 
    develop   :: (Phenotype b a) => a -> b 

class (Eq a, Show a, Genome b) => Phenotype a b | a -> b where 
    -- In case of Coevolution where each phenotype needs to be compared to 
    -- every other in the population 
    fitness   :: [a] -> a -> Int 
    genome   :: a -> b 

breed :: (Phenotype b a, Genome a) => [(b, b)] -> Double -> Double -> IO [b] 
breed parents cross mute = do 
    children <- mapM (\(dad, mom)-> crossover cross (genome dad) (genome mom)) 
        parents 
    let ch1 = map fst children ++ map snd children 
    mutated <- mapM (mutate mute) ch1 
    return $ map develop mutated 

-- |Full generational replacement selection protocol 
fullGenerational :: (Phenotype b a, Genome a) => 
    (Int -> [b] -> IO [(b, b)]) -> --Selection mechanism 
    Int -> --Elitism 
    Int -> --The number of children to create 
    Double -> --Crossover rate 
    Double -> --Mutation rate 
    [b] -> --Population to select from 
    IO [b] --The new population created 
fullGenerational selection e amount cross mute pop = do 
    parents <- selection (amount - e) pop 
    next <- breed parents cross mute 
    return $ next ++ take e reverseSorted 
      where reverseSorted = reverse $ sortBy (fit pop) pop 

fit pop a b = LT -- dummy function 

這個編譯。每種表型will have to provide exactly one implementation的genome。