用函数式编程来测试驱动出Prime Factor题目

近年来,函数式编程Funtional Programming大行其道,成为业界的热门话题。2016年端午节期间,我给中国软件匠艺小组的小伙伴们出了一道题目:用TDD的方式驱动出函数式编程版本的”Prime Factor”,不料各路神仙各显神通,争奇斗艳。

“Prime Factor”题目

这是一道经典的TDD题目:对给定的正整数N,求出其所有质因数,经典算法可以通过TDD驱动出二重循环,如下:

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def calc(i):
result = []
for candidate in xrange(2,i):
while i % candidate == 0 and i > candidate:
result.append(candidate)
i /= candidate
result.append(i)
return result

assert [2,2,2,3,3] == calc(72)

下面按照交作业的顺序展示一下大家的作品。然而TDD是一个推导的过程,仅仅看最终结果是不够的,能看到中间的状态就更好了。

Ronald麦大仙的Python版本

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def factors_from(i, xs):
first = [x for x in xs if i % x == 0][0]
return [first] + factors(i / first)

def factors(i):
return [] if i < 2 else factors_from(i, xrange(2, i+1))

assert factors(3*7*11*13) == [3,7,11,13]

梁辰的Ruby版本

TDD过程存储于http://cyber-dojo.org/kata/edit/9B0ED5F482?avatar=lion

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def first_divisible_factor start_factor, number
(start_factor..number).select{|factor| number%factor == 0}[0]
end

def factors_of number, start_factor=2
return [] if number == 1
factor = first_divisible_factor start_factor, number
[factor] + factors_of(number/factor, factor)
end

武可的Haskell版本

TDD过程存储于http://cyber-dojo.org/kata/edit/2EF003C866?avatar=zebra

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factorOf _ 1 = []
factorOf p n
| exact_div = p : factorOf p (n `div` p)
| no_more = [n]
| otherwise = factorOf next_p n
where
exact_div = n `mod` p == 0
no_more = p * p > n
next_p = p + 1

factor = factorOf 2

Jacky申导的Python版本

TDD过程存储于http://cyber-dojo.org/review/show/2BE7D52F15

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def f(n, p=2):
if n < 2: return []
if n % p == 0: return [p] + f(n/p, p)
return f(n, p+1)

###############################################

from hiker import f
import unittest

class TestHiker(unittest.TestCase):
def test_prime_factor(self):
self.assertEqual([], f(1))
self.assertEqual([2], f(2))
self.assertEqual([3], f(3))
self.assertEqual([2,2], f(4))
self.assertEqual([2,3], f(6))
self.assertEqual([2,2,2], f(8))
self.assertEqual([3,3], f(9))
self.assertEqual([2,2,3,3,3,5,5,19], f(2*2*3*3*3*5*5*19))

if __name__ == '__main__':
unittest.main()

更多函数式编程内容见:


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