有没有办法让这个Haskell程序更快？

Question

-- Lambda Calculus ADT
data Term = Fun (Term -> Term) | Num !Double
instance Show Term where
    show (Num x) = "(Num "++(if fromIntegral (floor x) == x then show (floor x) else show x)++")"
    show (Fun _) = "(<function>)"

-- Lambda Calculus term application
(#) :: Term -> Term -> Term
(Fun f) # x = f x
infixl 0 # 

-- We have floats as primitives for performance
float_toChurch :: Term
float_toChurch = Fun (\ (Num n) -> Fun (\f -> Fun (\x -> 
    if n <= 0 
        then x 
        else (f # (float_toChurch # (Num (n - 1)) # f # x)))))

float_add :: Term
float_add = Fun (\ (Num x) -> Fun (\ (Num y) -> Num (x + y)))

-- Function compiled from the Lambda Calculus.
-- It sums all nats from 0 til a number.
sum_til :: Term
sum_til = (Fun(\v0->((((((float_toChurch # v0) # (Fun(\v1->(Fun(\v2->(Fun(\v3->(Fun(\v4->((v3 # v2) # (((v1 # ((float_add # v2) # (Num 1))) # v3) # v4))))))))))) # (Fun(\v1->(Fun(\v2->(Fun(\v3->v3))))))) # (Num 0)) # (Fun(\v1->(Fun(\v2->((float_add # v2) # v1)))))) # (Num 0))))

-- Testing it
main = do
    let n = 512*512*8
    print $ (sum_til # (Num n))

由于没有快速的lambda计算器，我使用上面的策略将非类型化lambda演算的术语编译为Haskell，以便快速计算它们。我对它的性能印象深刻：该程序创建了一个从0到2097152的数字列表，并在我的计算机上在不到一秒钟的时间内将它们相加。这比我预期的要快得多--只比Haskell直接等价物慢4倍--并且足以对我的目标有用。但是，请注意，为了满足类型系统的需要，我必须将函数和术语包装在fun/num构造函数下面。那个拳击很可能不理想。所以，我的问题是：是否有可能运行同样的λ演算程序，甚至更快地得到同样的结果？也就是说，有什么方法可以移除拳击？（另外，它不需要使用Haskell。）

Answer 1

newtype Term = Term (Term -> Term)