closing to keep a single issue

Fixed

There is not enough constraint here
"EqlS :: Equal n m -> Equal (S n) (S m)"
and
|=| is not covering all cases

With the operation |=|, any 2 natural numbers are equal and I can prove the following:

p :: Equal (S Z) (S (S Z))
p = NumS NumZ |=| NumS (NumS NumZ)

I am new to Haskell, but I do not understand how |=| compiles

I mean you need up to sqrt(n)+1 no n/2+1

No, try the previous algorithm. It will time out.

The logic behind this is to greatly reduce numbers looped, sometimes up to one fourth, especially on large numbers. When I get a divisor I get its complement. When I reach a number that is already in the list, I know that I have already reached the second side of divisors, thus I have got everything. I can get out of the loop after that. For example, let's see how 12 is processed:

divisors(12)
# iteration 1
[1,12] # identifies 1 as divisor and then gets the complement of it to 12, which is 12. The largest divisor is half of that one, so we can break out of the loop halfway through anyway
# iteration 2
[1,12,2,6]
# iteration 3
[1,12,2,6,3,4]
#iteration 4
break # because I have ran into a number already in the divisors list, that means I have got all the divisors and all the complements.

Actually I can optimize it further by using fromkeys

You are looping over much more integers than needed

Please check the fork; the fork can view over less than one half.

But your algorithm is still great!

Both algorithms are linear on the input; but yes, this one is faster

I think that no number has a divisor larger than it's half other than itself too .

This program works faster because it doesn't go through the entire range, you only need to view half of it

The sample test for laziness is calling foldr instead of fixFoldr in Haskell

I am new in Haskell and I do not know how to use the type

type Lens s t a b = forall f . Functor f => (a -> f b) -> (s -> f t)

Any pointers or link to a tutorial on how to instanciate certain functor f?

When the test asks for the page number of the item element at -1, it expect Just 0; in Haskell