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recursionfactorial

How to calculate double factorial (multiply odds) with recursion?[homework]

How should I alter the factorial recursion, to calculate only the odd or only the double elements of the factorial?For example if: 

 multiplyOdds(4)

the result should return 1*3*5*7 =105

I know how recursion works, I just need a bit of help which approach I should use.

Solution

  • Your function multiplyOdds(n) needs to multiply the first n odd numbers? Given that the nth odd number is equal to 2 * n - 1, you can easily write a recursive solution like the one below in Haskell:

    multiplyOdds :: Int -> Int
multiplyOdds n = multiplyOddsTail n 1

multiplyOddsTail :: Int -> Int -> Int
multiplyOddsTail n acc = case n of
  1 -> acc
  n -> multiplyOddsTail (n - 1) (acc * (n * 2 - 1))