Haskell function to test if Int is perfect square using infinite list

Question

Purely for pleasure, and practice, I am trying to write a simple Haskell function to determine if an integer is a perfect square. Now I know that there are other solutions out there but I am wondering if there is a way to do it with an infinite list. I've started with this, but for clear reasons, it is not working (it never stops!)

    isSquare :: Integer -> Bool
    isSquare n = sum[ 1 | x <- [1..], x*x == n] /= 0

Also, if I might add, can someone point out how to search an infinite list for the first instance of something and then STOP! ?

Answer 1

About the infinite search function:

There is already a function that searches a list for a value - elem.

If you assume the infinite list is sorted, we could write a version of elem that works on such a list. This can easily be accomplished by firstly rejecting any elements less than the search value. The first value not rejected then must be either equal to or greater than the search element. If equal - return true, else return false

infiniteElem1 :: (Ord a) => a -> [a] -> Bool
infiniteElem1 x list = (== x) $ head $ dropWhile (< x) list

Example usage:

> infiniteElem1 10 [1..]
True
> infiniteElem1 10 [1,3..]
False

There is one problem though with infiniteElem1 though: If used on a finite list, it may throw an exception if the element isn't found:

> infiniteElem1 100 [1,2,3]
*** Exception: Prelude.head: empty list

This is a reason the function head is best avoided. A better solution is this:

infiniteElem :: (Ord a) => a -> [a] -> Bool
infiniteElem x list = case dropWhile (< x) list of
  [] -> False
  (v:_) -> v == x

Now it works with a finite sorted list as well:

> infiniteElem 100 [1,2,3]
False

With this your problem becomes trivial:

let isSquare n = n `infiniteElem` [ x * x | x <- [1..]]