Understanding Applicative Evaluation Order with the Z-Combinator

Question

I am trying to understand how the Z-combinator (Y-combinator for applicative order languages) definition came about. As Python is applicative I am using Python for this.

So I know Python's evaluation order is applicative. But I seem to be overlooking something in how applicative order works. To compensate for applicative evaluation order I reasoned to build a Y-combinator that does not dive off into infinite recursion it would be sufficient to write it like this:

Y = lambda f : (lambda x : f( lambda z: x(x) (z) )) (lambda x : f(x(x)))

I arrived at this conclusion by first manually deriving Y g like so

Y g =  (lambda f : (lambda x : f(x x)) (lambda x : f(x x))) g      # Definition of Y
    -> (lambda x : g(x x)) (lambda x : g(x x))                   # beta reduction
    -> g((lambda x : g(x x)) (lambda x : g(x x)))                # beta reduction
    -> g((lambda f : (lambda x : f(x x)) (lambda x : f(x x))) g) # lambda abstraction
    =  g(Y g)                                                    # put in Y

And then working my way backwards like this, adding in a lambda abstraction hat would delay the recursion until a value is passed in:

    = g(lambda z : Y g z)                                                   
    =  g(lambda z : (lambda f (lambda x : f(x x)) (lambda x : f(x x))) g z) 
    -> g(lambda z : (lambda x : g(x x)) (lambda x : g(x x)) z)         
    -> (lambda x : g(lambda z : x x z)) (lambda x : g(x x))            
Y g =  (lambda f : (lambda x : f( lambda z : x x z)) (lambda x : f(x x))) g

When I manually evaluate factorial 3 where

factorial_ = lambda f : lambda n : 1 if n == 0 else n * f(n-1)

factorial = Y(factorial_)

in applicative order I get

 factorial 3 =  (lambda n : 1 if n == 0 else n * (lambda z : Y factorial_ z)(n-1)) 3       
            -> 3 * (lambda z : Y factorial_ z)(3-1)                                  
            -> 3 * (Y factorial_ 2)                                             
            =  3 * ((lambda n : 1 if n == 0 else n * (lambda z : Y factorial_ z)(n-1)) 2)
            -> 3 * 2 * ((lambda z : Y factorial_ z)(2-1))
            -> 3 * 2 * (Y factorial_ 1)
            =  3 * 2 * ((lambda n : 1 if n == 0 else n * (lambda z : Y factorial_ z)(n-1)) 1)
            -> 3 * 2 * 1 * ((lambda z : Y factorial_ z)(1-1))
            -> 3 * 2 * 1 * (Y factorial_ 0)
            =  3 * 2 * 1 * ((lambda n : 1 if n == 0 else n * (lambda z : Y factorial_ z)(n-1)) 0)
            -> 3 * 2 * 1 * 1
            -> 6

But when I run

Y = lambda f : (lambda x : f( lambda z: x(x) (z) )) (lambda x : f(x(x)))

factorial_ = lambda f : lambda n : 1 if n == 0 else n * f(n-1)

factorial = Y(factorial_)

print(factorial(3))

I still get the infinite recursion problem:

Y = lambda f : (lambda x : f( lambda z: x(x) (z) )) (lambda x : f(x(x))) [Previous line repeated 994 more times] RecursionError: maximum recursion depth exceeded

So I must not actually have performed correct applicative order on my manual derivation, otherwise I would have gotten infinite recursion like Python gets.

What am I missing here about how applicative order works?

EDIT:

To reiterate:

Let's say I name that version of Y Z':

Let $Z' = \lambda f. \lambda x( f ( \lambda z. x x z)) (\lambda x. f( xx))$

Let $F' = \lambda f. \lambda n. 1 \text{ if } n = 0 \text{ else } n*f(n-1)$ $$ \text{Let } F = Z' F' = (\lambda f . (\lambda x . f( \lambda z . x x z) (\lambda x . f(x x)) F'$$ $$\longrightarrow (\lambda x . F'(\lambda z . x x z)) (\lambda x . F'(x x))$$ $$\longrightarrow F'(\lambda z . (\lambda x . F'(x x)) (\lambda x . F'(x x)) z)$$ Lambda-Abstraction for $F'$
$$= F'(\lambda z . (\lambda f (\lambda x . f(x x)) (\lambda x . f(x x))) F' z)$$ Per definition of $Z'$ $$= F'(\lambda z . Z' F' z)$$

Now applying $F$ to some number:

$$F~ 3 = (\lambda n . 1 \text{ if } n = 0 \text{ else } n * (\lambda z . Z' F' z)(n-1)) 3$$

$$\longrightarrow 3 * (\lambda z . Z' F' z)(3-1) $$

$$\longrightarrow 3 * (Z' F' 2) $$ $$= 3 * ((\lambda n . 1 \text{ if } n = 0 \text{ else } n * (\lambda z . Z' F' z)(n-1)) 2) $$ $$\longrightarrow 3 * 2 * ((\lambda z . Z' F' z)(2-1))$$ $$\longrightarrow 3 * 2 * (Z' F' 1)$$ $$= 3 * 2 * ((\lambda n . 1 \text{ if } n = 0 \text{ else } n * (\lambda z . Z' F' z)(n-1)) 1)$$ $$\longrightarrow 3 * 2 * 1 * ((\lambda z . Z' F' z)(1-1))$$ $$\longrightarrow 3 * 2 * 1 * (Z' F' 0)$$ $$= 3 * 2 * 1 * ((\lambda n . 1 \text{ if } n = 0 \text{ else } n * (\lambda z . Z' F' z)(n-1)) 0)$$ $$\longrightarrow 3 * 2 * 1 * 1$$ $$\longrightarrow 6$$

So, why did this work? It was supposed to go into infinite recursion. What is my mistake?

chi · Accepted Answer · 2018-03-05T22:54:24.490

Let's call your proposal X, instead:

X = lambda f : (lambda x : f( lambda z: x(x) (z) )) (lambda x : f(x(x)))

For convenience, we can rewrite it as

M = (lambda x : f(x(x)))    # depends on f
X = lambda f : (lambda x : f( lambda z: x(x) (z) )) M

Now, when we invoke X(f), we get

X(f) =
(lambda x : f( lambda z: x(x) (z) )) M =
f( lambda z: M(M) (z) )

Assume f "recurses", invoking its argument with some value, e.g. 5. Its code might look as

c = M(M)(5)
do something with c
return something

However, M(M) above is exactly the non terminating Y(f). The additional lambda z: disappears after the first recursive call.

In order to prevent that, we need to add lambda z: inside M as well. Hence we get the actual Z combinator:

Z = lambda f : (lambda x : f( lambda z: x(x) (z) )) (lambda x : f(lambda z: x(x) (z)))

Understanding Applicative Evaluation Order with the Z-Combinator

EDIT:

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