Description says: the numbers are in the range (1  ≤  a, b, c  ≤  10).

Is this correct? (0, 1, 1) returns 1 instead of 2.

The description now says:

we want to find all integers x, y (x >= 0, y >= 0) solutions...

It's been 9 months, and the description still disagrees with the sample cases.

Oh yeah, I did figured out I commented the function signature.

Wow this is easier than I expected as a 5kyu problem.
The problem asks us to optimize, but the test cases seems too easy. :p

I thought you were on holidays!-)

Can't reproduce. Did you relax the types of your solution?

What? Getting error from the haskell test.

/tmp/haskell11563-18-fmwhrh/Codewars/Kata/Fib/Test.hs:17:43:
No instance for (Arbitrary t0) arising from a use of property' The type variable t0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Arbitrary a => Arbitrary (Blind a)
-- Defined in Test.QuickCheck.Modifiers' instance Arbitrary a => Arbitrary (Fixed a) -- Defined in Test.QuickCheck.Modifiers'
instance (Integral a, Bounded a) => Arbitrary (Large a)
-- Defined in Test.QuickCheck.Modifiers' ...plus 37 others In the expression: property In the second argument of ($)', namely
property $ \ (Positive n) -> productFib n shouldBesolution n' In a stmt of a 'do' block: it "should work for random numbers" $ property $ \ (Positive n) -> productFib nshouldBe` solution n

/tmp/haskell11563-18-fmwhrh/Codewars/Kata/Fib/Test.hs:18:20:
No instance for (Eq t0) arising from a use of shouldBe' The type variable t0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Eq a => Eq (GHC.Real.Ratio a) -- Defined in GHC.Real' instance Eq time-1.4.0.1:Data.Time.LocalTime.LocalTime.LocalTime -- Defined in time-1.4.0.1:Data.Time.LocalTime.LocalTime'
instance (Eq e, Data.Functor.Classes.Eq1 m, Eq a) =>
Eq (Control.Monad.Trans.Error.ErrorT e m a)
-- Defined in Control.Monad.Trans.Error' ...plus 210 others In the expression: productFib n shouldBesolution n In the second argument of($)', namely
\ (Positive n) -> productFib n shouldBesolution n' In the second argument of($)', namely
property $ \ (Positive n) -> productFib n shouldBe` solution n'

Same problem with me 25 Passed 1 Failed

:/ Thought I have an equation, but figured out Haskell case requires us to work with Infinite list.

When will this issue be fixed for Haskell :( :(

I can't understand this. Can anyone provide an example for 3 factorial?

This kata needs to give more description about what is expected. I am totally confused. Like what would be the Codewars.Group is?

Impossible to correct now, I can't edit the kata anymore:-( Thanks for the feedback!