-- Opgaver før lecture
sum' x | x > 0 = x + sum' (x-1)
       | otherwise = 0

flip' = map (\(x, y) -> (y, x))
-- Here i would guess that the type is [(a, b)] -> [(b, a)]
-- When quering i get flip' :: [(b, a)] -> [(a, b)]
-- Yeah they are the same


-- I would say the type of fib is (Eq a, Num a, Num b) => a -> b
-- Okay det vil nok være bedre at bruge Integral, eftersom fib tal kun er integer

fib 0 = 1
fib 1 = 1
fib n = fib (n-1) + fib (n-2)


-- I would say that the complexity is O(2^n), because we apply
-- fib two times for every invocation of fib.
-- Which does kind of suck

reverse' [] = []
reverse' (x : xs) = reverse' xs ++ [x]


-- Okay so it's clear that we accept lists.
-- However it's a bit unclear which types we accept.
-- Fortunately it does not really matter as do not do anything with x itself.
-- I would therefore way that the type of reverse' :: [a] -> [a].

-- Running `:t` on reverse' reveals that we where correct.

-- Hmm i would say that a ispalindrome function would have type (Eq c) => [c] -> Bool

ispalindrome x = x == reverse' x


-- Okay assuming that a are all integers, i
-- would say that the fype of cfrac :: (Real a, Integral b) -> a -> b -> [b]

cfrac _ 0 = []
cfrac x n = let intPart = truncate x in
              intPart : cfrac (1 / (x - fromIntegral intPart)) (n-1)
-- REMEMBER fromInteger for ***'s sake.
-- Without it, x will be tagged with (RealFrac a, Integral a) => a,
-- which destroys everything.