Henning Thielemann wrote:
factors :: Integer -> Integer -> Bool
factors m n | m == n = False
| m < n = divides m n || factors (m+1) n
Btw. I find the recursion harder to understand than the explicit
definition:
factors n = any (divides n) [2..(n-1)]
For what it's worth, I also found this function un-intuitive. What I'd
expect a function called "factors" to do is exactly what yours does, and
not what the one in the example does.
Cheers,
Daniel.
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