I recently came across this question:
其中主要询问如何实现此函数来计算f(n)的极限:
我如何在haskell中实现它?我正在努力学习函数式编程,这对我来说似乎是一个很好的挑战
答案 0 :(得分:10)
有很多方法!
这里使用递归辅助函数:
f :: (Eq a, Floating a) => a -> a
f n = f' n n
where f' 1 x = x
f' n x = let n' = n-1 in f' n' (n' / (1 + x))
手工解决:
f 1 = f' 1 1
= 1
f 2 = f' 2 2
= f' 1 (1 / (1 + 2))
= 1/(1+2)
f 3 = f' 3 3
= f' 2 (2 / (1 + 3))
= f' 1 (1 / (1 + (2 / (1 + 3))))
= 1 / (1 + (2 / (1 + 3)))
这是使用递归辅助函数执行此操作的另一种方法:
f :: (Eq a, Floating a) => a -> a
f n = f' 1 n
where f' a n | a == n = a
| otherwise = a / (1 + f' (a+1) n)
手工解决:
f 1 = f' 1 1
= 1
f 2 = f' 1 2
= 1 / (1 + f' 2 2)
= 1 / (1 + 2)
f 3 = f' 1 3
= 1 / (1 + f' 2 3)
= 1 / (1 + (2 / (1 + f' 3 3)))
= 1 / (1 + (2 / (1 + 3)))
第一种方法是尾递归,而第二种方法只是递归。
或者,如链接所示,通过折叠
f :: (Eq a, Floating a) => a -> a
f n = foldr1 (\n x -> n / (1 + x)) [1..n]
再次,手工解决:
f 5 = foldr1 (\n x -> n / (1 + x)) [1,2,3,4,5]
= g 1 (g 2 (g 3 (g 4 5)))
= g 1 (g 2 (g 3 (4 / (1 + 5))))
= g 1 (g 2 (3 / (1 + (4 / (1 + 5)))))
= g 1 (2 / ( 1 + (3 / (1 + (4 / (1 + 5))))))
= 1 / (1 + (2 / ( 1 + (3 / (1 + (4 / (1 + 5)))))))
where g = \n x -> n / (1 + x)