Deriving the Y Combinator

Deriving the Y Combinator

Yuta Okazaki!Recurse Center Spring 2 2015

Recursive Function

factorial = function(n){ if(n == 0){ return 1 } else{ n * factorial(n - 1) } }

Pseudo Code

Recursive Functionlet’s simplify this to an ‘infinite’ function.

fact-forever = function(n){ n * fact-forever(n - 1) }

Recursive FunctionWhat if we can’t use a variable?

fact-forever = function(n){ n * fact-forever(n - 1) }

Recursive Function

What is this inside ‘fact-forever’ gonna be?

fact-forever = function(n){ n * fact-forever(n - 1) }

function(n){!

n * (n - 1)!

}

Recursive FunctionI’m imagining if we have a ‘magical function’, that prepares itself again when we need it…

Thought Experimentwith Scheme(Lisp)

Thought Experiment #1Is it possible to think of a lambda, which

returns a lambda, that takes ‘itself’ as an argument?

((lambda(me)(me me)) (lambda(x)(quote hi)))

Thought Experiment #1Is it possible to think of a lambda, which

returns a lambda, that takes ‘itself’ as an argument?

((lambda(me)(me me)) (lambda(x)(quote hi)))

=> ‘hi

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #2Since ‘me’ function takes ‘me’ as an argument, isn’t it fun if we can somehow use it inside it?

quote hi

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #2Since ‘me’ function takes ‘me’ as an argument, isn’t it fun if we can somehow use it inside it?

x

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #2Since ‘me’ function takes ‘me’ as an argument, isn’t it fun if we can somehow use it inside it?

=> expects 1 argument, but found none.

x

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #3Yeah my bad, x is a lambda expecting

one argument. so… pass in x itself maybe?

x

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #3Yeah my bad, x is a lambda expecting

one argument. so… pass in x itself maybe?

xx

=>

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #3Yeah my bad, x is a lambda expecting

one argument. so… pass in x itself maybe?

xx

((lambda(x)(x x)) (lambda(x)(x x)))=>

((lambda(me)(me me)) (lambda(x)( )))

Thought Experiment #3Yeah my bad, x is a lambda expecting

one argument. so… pass in x itself maybe?

=> infinite loop!

xx

((lambda(x)(x x)) (lambda(x)(x x)))=>

Thought Experiment #3

This is the Javascript version of same thing.

function(x){ return x(x)}(function(x){return x(x)})

((lambda(me)(me me)) (lambda(x)( )))

Resume the ProblemSo, how can we use this piece

with the original problem?

xx

function(n) n * (n - 1)

((lambda(x) x )(quote hi))

Functional RefactoringWrap Function

((lambda(y)((lambda(x)x)y))(quote hi))

if you have a lambda that takes one argument, you can wrap it with another lambda which also takes one argument, and call it with that argument, you still get the same result.

((lambda(me)( me me )) (lambda(x)

Combine themSo, here I use a twisted version of Wrap Function,

in order to accumulate the calculation.

(lambda(number)(number * (

( x x )))

(number - 1))))

((lambda(me)( me me )) (lambda(x)

Combine themSo, here I use a twisted version of Wrap Function,

in order to accumulate the calculation.

(lambda(number)(number * (

( x x )

)) (number - 1))))

((lambda(me)( me me )) (lambda(x)

Combine themSo, here I use a twisted version of Wrap Function,

in order to accumulate the calculation.

(lambda(number)(number * (( x x )

)) (number - 1))))

((lambda(me)( me me )) (lambda(x)

Combine themSo, here I use a twisted version of Wrap Function,

in order to accumulate the calculation.

(lambda(number)(number * (( x x )

)) (number - 1))))

Ω Combinator

((lambda(me)( me me )) (lambda(x)

The code below is an application of Ω Combinator.

(lambda(number)(number * (( x x )

)) (number - 1))))

Demo Available

Where’s Y Combinator?

Y Combinator is a nicer form of Omega Combinator. If you want to derive one, please talk to me!

(lambda(le) ((lambda(f)(f f)) (lambda(f) (le (lambda(x) ((f f) x))))))

Twitter @kenzan100