Lazily Tying the Knot for 1 Dimensional Dynamic Programming

Question

Several years ago I took an algorithms course where we were giving the following problem (or one like it):

There is a building of n floor

Answer 1

The problem is that min needs to fully evaluate both calls to f, so if one of them loops infinitly min will never return. So you have to create a new type, encoding that the number returned by f is Zero or a Successor of Zero.

data Natural = Next Natural 
             | Zero

toNum :: Num n => Natural -> n
toNum Zero     = 0
toNum (Next n) = 1 + (toNum n)

minimal :: Natural -> Natural -> Natural
minimal Zero _            = Zero
minimal _ Zero            = Zero
minimal (Next a) (Next b) = Next $ minimal a b

f i j | i == j = Zero
      | otherwise = Next $ minimal (f l j) (f r j)
      where l = i + 2
            r = i - 3

This code actually works.