G'day all.
Quoting Daniel Fischer
The good fix is to insert calls to the apprpriate conversion function where necessary, for example
sqRoot n scale = sqRoot' (5*n) 5 where sqRoot' a b | floor (logBase 10 $ fromIntegral b) >= scale = div b 10 | a >= b = sqRoot' (a-b) (b+10) | otherwise = sqRoot' (a*100) ((100 * (div b 10)) + (mod b 10))
The problem here is that floating-point numbers do not have arbitrary precision, whereas Integers do. For very large numbers, the conversion might not be possible. Option #1: sqRoot n scale = sqRoot' (5*n) 5 where sqRoot' a b | b >= invScale = div b 10 | a >= b = sqRoot' (a-b) (b+10) | otherwise = sqRoot' (a*100) ((100 * (div b 10)) + (mod b 10)) invScale = 10^scale Option #2: sqRoot n scale = sqRoot' (5*n) 5 where sqRoot' a b | length (show b) >= scale = div b 10 -- May be an off-by-one error here. | a >= b = sqRoot' (a-b) (b+10) | otherwise = sqRoot' (a*100) ((100 * (div b 10)) + (mod b 10)) Sadly, option #2 (and option #3, not shown, which is essentially to implement your own integer-only floor-log-base-10) destroys the pretty "mostly subtraction only" property of the algorithm. Cheers, Andrew Bromage