Thanks, chaps, for the replies to my over-hasty post. I should have
checked with a power of 10 value for scale which would have pointed to
types to logBase as the error.
Andrew's comment about floating point not having unlimited precision
clears up a glitch that using fromInteger introduces. If you use it to
typecast b, it appears to set a limit to the number of digits it
generates. It only gives you 308 or so even if you set scale equal to
1000, say.
If you test b against 10^scale, you get as many digits as you wish.
Logesh Pillay
> Message: 10
> Date: Sun, 24 Aug 2008 21:07:53 -0400
> From: ajb(a)spamcop.net
> Subject: Re: [Haskell-beginners] Puzzling type error
> To: beginners(a)haskell.org
> Message-ID: <20080824210753.42c5u3ihw444gc8o-nwo(a)webmail.spamcop.net>
> Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes";
> format="flowed"
>
> G'day all.
>
> Quoting Daniel Fischer <daniel.is.fischer(a)web.de>:
>
>
>> > 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
>
