用不动点组合子解递归(python实现)

不动点组合子
　　Y = λf. (λx. f (x x)) (λx. f (x x))
　　θ = (λx. λy. (y(x x y))) (λx.λy.(y(x x y)))
　　Y f = f (Y f)
　　θ f = f (θ f)
设
　　h1 =λx.f(x x)
　　h2 =λx.λy.(y(x x y)) 
简化为:
　　Y = λf. h1(h1)
　　θ = h2(h2)

递归求解
ƒ(n) = n*ƒ(n-1)
ƒ(0) = 1
简记为: ƒ = λn.n׃(n-1)                        #ƒ对应代码中的 factorial
用归纳法可证ƒ唯一，
设 g = λƒ.λn.n׃(n-1)                          #g对应代码中的_factorial
则g ƒ = λn.n׃(n-1) = ƒ
因为g(Y g)=(Y g)
所以Y g为ƒ的唯一解
即ƒ(n)=(Y g)(n)=(Θ g)(n)

def _factorial(f):
    def factorial(n):
        if not n:
            return 1
        else:
            return n * f(n - 1)
    return factorial

def Y(f):
    def h(x):
        return f(lambda v: x(x)(v))
    return h(h)

def O(f):
    def h(x):
        return (lambda y: y(lambda v: x(x)(y)(v)))
    return h(h)(f)

#Y(_factorial)(5)
#O(_factorial)(5)

#Y = lambda f: (lambda x:f(lambda v:x(x)(v)))(lambda x:f(lambda v:x(x)(v)))
#YY = lambda f: ((lambda x: x(x))(lambda y: f(lambda v: y(y)(v))))

参考:

http://www.cnblogs.com/ldp615/archive/2013/04/09/recursive-lambda-expressions-1.html

http://www.cnblogs.com/ldp615/archive/2013/04/09/recursive-lambda-expressions-2.html

http://www.cnblogs.com/ldp615/archive/2013/04/10/recursive-lambda-expressions-3.html

http://www.cnblogs.com/ldp615/archive/2013/04/10/recursive-lambda-expressions-4.html

http://www.cnblogs.com/ldp615/archive/2013/04/11/recursive-lambda-expressions-5.html

http://www.pythontip.com/blog/post/12174/