Reputation: 6745
I am trying to write a higher-order Racket function that takes a first-order function of one variable and returns its inverse. I know that it has to start off something like this:
(let [(inverse (lambda (f) (lambda (y) ... )))])
I figured this because inverse
must take a function which returns a function which takes a y
and returns x
such that (= (f x) y)
. In other words, the contract for inverse is something like:
; inverse : (number? -> number?) -> (number? -> number?)
I'm just stumped trying to figure out what goes where the elipses are?
EDIT:
In response to people saying this is impossible, I am willing to accept an inverse function that when given y
returns a possible x
. In response to comments about the function not having an inverse, please note the contract that I have for f
. It is a (number? -> number?)
mapping, and therefore has an inverse.
Upvotes: 3
Views: 1732
Reputation: 6745
I knew that I had seen this before, but I couldn't remember how it worked. Now I remember, but I realized that I had been misleading in my question because the version I had seen assumed that we already had a function called root
which would return one of the zeros of a provided function. Given that function, it is pretty easy:
(define (inverse f) (lambda (y) (root (lambda (x) (- (f x) y)))))
It's pretty easy to see how this works. The inverse of a function is the x
such that f(x) = y
. Obviously, the root of the function f(x) - y = 0
is that x
.
The place where I had gone wrong is that the best we can do for root
is Newton's method or some other approximation.
Upvotes: 2
Reputation: 235984
For the general case, given an arbitrary function f
you can't tell what's its inverse function. Even worse, a given function might not have an inverse at all - for example: the input function could perform an MD5 hash, which has no inverse. Sorry, your question has no answer.
Upvotes: 6
Reputation: 31147
Consider f(x)=x^2. This is a very simple function without an inverse. (because f(1)=f(-1) there are no unique inverse to y=1).
Since a very simple function might have no inverse, you cant't expect a general Scheme function to have an inverse.
Upvotes: 2