Lambda calculus-A brief description of the role of Y-composite

Y Combo: \f. (\x.f (XX)) (\x.f (XX)), accepts a function, returns a higher order function

The y combination is used to generate anonymous recursive functions.

What is anonymous recursive function, consider the following C language recursive function

int sum (int n) {    return n = = 0? 0:n + sum (n-1);}

This function calls itself recursively, calling itself the name of the function body, which is called Sum,sum internal with the name sum, recursive calls to their own

In lambda calculus, you can write a similar expression, sum = \x. x = = 0? 0:sum x

But for a lambda expression, he is anonymous, and Lambda refers to itself in the process of defining it, and even if it is C + +, the lambda expression is not true.

Auto sum = [] (int n) {    return n = = 0? 0:n + sum (n-1);};

The lambda expression itself is anonymous, and we need to bypass this restriction.

One possible solution is to use a higher-order function to wrap the sum above with another function: It takes a function f and returns a function that accepts X, determines the recursive end point, or calls F to continue the recursion:

G = \f. \x. x = = 0? 0:f (x-1)

Written in C + + is like this

Auto G = [] (function<int (int) > F) {    return [&] (int x) {        return x = = 0?: 0:x + f (x-1);    };};

Now we find that when there is a function f that makes g (f) = [] (int x) {return x = = 0? 0:x + f (x-1);} = f, this f is exactly what we need the anonymous recursive function sum

G (f) = f, familiar, remember the concept of fixed point, we need the anonymous recursive function sum is the fixed point of function g

To solve this fixed point sum, we can get an anonymous recursive function, how to solve the attached

The final result: sum = yg,y and g are known before, so that sum is a function with the signature int (int), which is an anonymous recursive function

Y-sub-group is also called fixed point combination, which can be used to solve all anonymous recursive functions.

Attached: sum = YG makes the proof of g (sum) = sum:

Proof: For any g,g (YG) = YG

Make w = \x. G (xx), x = WW//This order is really too TM, anyway, I didn't think

There is x = WW = (\x. G (xx)) W = g (WW) = g (x)

Also because YG = (\x g (XX)) (\x. G (xx)) = WW = X

So g (x) = X is g (YG) = YG

Certificate of Completion

Lambda calculus-A brief description of the role of Y-composite