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Problem 8: All-Ys Has Been (2pts)

Problem

Given mystery function Y, complete fib_maker and number_of_six_maker so that the given doctests work correctly.

When Y is called on fib_maker, it should return a function which takes a positive integer n and returns the nth Fibonacci number.

Similarly, when Y is called on number_of_six_maker it should return a function that takes a positive integer x and returns the number of times the digit 6 appears in x.

给定神秘函数 Y，完成 fib_maker 和 number_of_six_maker，使给定的 doctest 能够正常工作。

当 Y 在 fib_maker 上被调用时，它应该返回一个函数，该函数接受一个正整数 n 并返回第 n 个斐波那契数。

类似地，当 Y 在 number_of_six_maker 上被调用时，它应该返回一个函数，该函数接受一个正整数 x 并返回数字 6 在 x 中出现的次数。

def Y(f): return (lambda x: x(x))(lambda x: f(lambda z: x(x)(z)))
def fib_maker(f): return lambda r: 'YOUR_EXPRESSION_HERE'
def number_of_six_maker(f): return lambda r: 'YOUR_EXPRESSION_HERE'


my_fib = Y(fib_maker)
my_number_of_six = Y(number_of_six_maker)

# This code sets up doctests for my_fib and my_number_of_six.

my_fib.__name__ = 'my_fib'
my_fib.__doc__ = """Given n, returns the nth Fibonacci nuimber.

>>> my_fib(0)
0
>>> my_fib(1)
1
>>> my_fib(2)
1
>>> my_fib(3)
2
>>> my_fib(4)
3
>>> my_fib(5)
5
"""

my_number_of_six.__name__ = 'my_number_of_six'
my_number_of_six.__doc__ = """Return the number of 6 in each digit of a positive integer n.

>>> my_number_of_six(666)
3
>>> my_number_of_six(123456)
1
"""

Hints

Hint: You may use the ternary operator <a> if <bool-exp> else <b>, which evaluates to <a> if <bool-exp> is truthy and evaluates to <b> if <bool-exp> is false-y.

  • 你可以使用三元运算符 <a> if <bool-exp> else <b>，它在 <bool-exp> 为真值时求值为 <a>，在 <bool-exp> 为假值时求值为 <b>。

  • 不要把题目想的太复杂。

  • 这个“神秘函数”叫“Y 组合子”。有兴趣可以查查资料。