Types in a function in Haskell

Question

I am trying to write a function called check that when applied to n returns a Bool indicating whether the ∑r=0 to n (n choose r) = 2^n

I have made the factorial and choose functions which work but I am having difficulty understanding what the type would be for the check function. I have written what I have done for it and I think logically it might work but I am not sure. The error I get when running says 'couldn't match type a -> Bool to actual type Bool' but I am not entirely sure how to make it an actual Bool type.

factorial :: (Integral a) => a -> a
factorial 0 = 1 
factorial n = n * factorial (n - 1)

choose :: (Integral a) => a -> a -> a
n `choose` r
    | r < 0     = 0
    | r > n     = 0
    | otherwise = factorial n `div` (factorial r * factorial (n-r))

check :: (Integral a) => a -> a -> Bool 
check n = (if (factorial (choose n)) == 2^n then True else False)

1) Fix the type: a->a->Bool takes two arguments, but you only need one. 2) Try a list comprehension e.g. sum [ ... | r<-[0..n] ] == 2^n. 3) Never use if condition then True else False -- it's equivalent to just condition, only more complex to read. — chi
– chi, Commented Oct 17, 2022 at 12:49
Style note: if (factorial (choose n)) == 2^n then True else False is equivalent to just writing: factorial (choose n) == 2^n — Chris
– Chris, Commented Oct 17, 2022 at 18:25
Some observations. choose takes two arguments, but you have given it only one in factorial (choose n); perhaps this is related to the problem you are having. Where is the code reflection in check of r, and its bounds of 0 to n, and the summation of the results? What math is being reflected by the factorial in check? — Daniel Wagner
– Daniel Wagner, Commented Oct 18, 2022 at 1:39

Francis King · Accepted Answer · 2022-10-17 12:58:26Z

2

The correct type declaration for check is a -> Bool

The types are separated by -> and grouped using parentheses.

The last type is the returned type, here Bool.

The previous types are the input types, here a.

Your error message is couldn't match type a -> Bool to actual type Bool. What's happening here is that Haskell matches the a to variable n, but is then left with a -> Bool, when actually the return type is Bool.

answered Oct 17, 2022 at 12:58

Francis King

1,7421 gold badge10 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Types in a function in Haskell

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related