Reusing a Lambda function in Haskell

Question

I'm supposed to take this code:

f x y z = x^3 - g (x + g (y - g z) + g (z^2))
 where g x = 2*x^2 + 10*x + 1

And rewrite it without where (or let).

They mean to write it with a Lambda function (\x ->...)

I'm trying to reuse a Lambda function on Haskell. Any ideas?

Answer 1

That question seems kinda curious and interesting for me. So, I'm trying to figured out what is lambda calculus is, find an answer and want to show it to OP (all hints have already been showed actually, spoiler alert).

Firstly, lets try to redefine f:

λ> let f = (\g x y z -> x^3 - g(x + g(y - g z) + g(z^2)))
f ::
  (Integer -> Integer) -> Integer -> Integer -> Integer -> Integer

So, we've got function, which get function and 3 numbers and return the answer. Using curring we can add g definition right here, like f_new = f g:

λ> let f = (\g x y z -> x^3 - g(x + g(y - g z) + g(z^2))) (\x -> 2*x^2 + 10*x + 1)
f :: Integer -> Integer -> Integer -> Integer

We're done. Let's check it:

λ> f 0 0 0
-13

The answer is correct.

UPD:

In those examples let is just a way to declare function in the interpreter, so final answer is:

f :: Num a => a -> a -> a -> a
f = (\g x y z -> x^3 - g(x + g(y - g z) + g(z^2))) (\x -> 2*x^2 + 10*x + 1)

Answer 2

Using lambda calculus g is (\x -> 2*x^2 + 10*x + 1)

So you need to substitute g with that in f x y z = x^3 - g (x + g (y - g z) + g (z^2))

$> echo "f x y z = x^3 - g (x + g (y - g z) + g (z^2))" | sed -r -e 's/g/(\\x -> 2*x^2 + 10*x + 1)/g'
f x y z = x^3 - (\x -> 2*x^2 + 10*x + 1) (x + (\x -> 2*x^2 + 10*x + 1) (y - (\x -> 2*x^2 + 10*x + 1) z) + (\x -> 2*x^2 + 10*x + 1) (z^2))

I'm just kidding, sorry.

Answer 3

To expand on hammar's and bravit's hints, your solution is going to require not just one lambda, but two - one of which will look a great deal like g, and the other of which will look a great deal like the second half of f

Answer 4

As bravit hints at, you can rewrite a non-recursive let using a lambda in the following way:

let x = A in B     ==>     (\x -> B) A 

where x is a variable and A and B are expressions.

Answer 5

I think the intention is what bravit hints at.
The smartypants follow-the-letters-of-the-law workaround is binding g with a case ;)

Answer 6

To reuse something you can make it an argument to something.