[Haskell-cafe] Re: fast integer base-2 log function?

[Simon Marlow]

Stefan O'Rear wrote:
> On Sun, Feb 10, 2008 at 10:15:58PM -0800, Uwe Hollerbach wrote:
>> Hello, haskellers,
>>
>> Is there a fast integer base-2 log function anywhere in the standard
>> libraries? I wandered through the index, but didn't find anything that
>> looked right. I need something that's more robust than logBase, it
>> needs to handle numbers with a few to many thousands of digits. I
>> found a thread from a couple of years ago that suggested there was no
>> such routine, and that simply doing "length (show n)" might be the
>> best. That seems kind of... less than elegant. I've come up with a
>> routine, shown below, that seems reasonably fast (a few seconds of CPU
>> time for a million-bit number, likely adequate for my purposes), but
>> it seems that something with privileged access to the innards of an
>> Integer ought to be even much faster -- it's just a simple walk along
>> a list (array?) after all. Any pointers? Thanks!
> 
> Even easier.
> 
> 
> {-# LANGUAGE MagicHash #-}
> import GHC.Exts
> import Data.Bits
> 
> -- experiment with using a LUT here (hint: FFI + static arrays in C)
> ilog2i0, ilog2i1, ilog2i2, ilog2i3, ilog2i4 :: Int -> Int -> Int
> ilog2i0 k x | x .&. 0xFFFF0000 /= 0 = ilog2i1 (k + 16) (x `shiftR` 16)
>             | otherwise             = ilog2i1 k x
> ilog2i1 k x | x .&. 0xFF00 /= 0     = ilog2i2 (k + 8)  (x `shiftR` 8)
>             | otherwise             = ilog2i2 k x
> ilog2i2 k x | x .&. 0xF0 /= 0       = ilog2i3 (k + 4)  (x `shiftR` 4)
>             | otherwise             = ilog2i3 k x
> ilog2i3 k x | x .&. 0xC /= 0        = ilog2i4 (k + 2)  (x `shiftR` 2)
>             | otherwise             = ilog2i4 k x
> ilog2i4 k x | x .&. 0x2 /= 0        = k + 1 + (x `shiftR` 1)
>             | otherwise             = k + x
> 
> log2i :: Integer -> Int  -- actually returns bit length, and returns garbage on negatives, but do you care?
> log2i (J# len adr) = ilog2i0 0 (I# (indexIntArray# adr (len -# 1#))) + I# (32# *# (len -# 1#))
> log2i (S# sml) = ilog2i0 0 (I# sml)

I don't know if this would be any faster in practice, but it avoids those 
comparisons:

http://aggregate.org/MAGIC/#Log2%20of%20an%20Integer

Cheers,
	Simon