[Haskell-cafe] Fast Integer Input
Bertram Felgenhauer
bertram.felgenhauer at googlemail.com
Mon Aug 23 11:46:29 EDT 2010
Serguey Zefirov wrote:
> 2010/8/23 <200901104 at daiict.ac.in>:
> > This function takes 1.8 seconds to
> > convert 2000 integers of length 10^13000. I need it to be smaller that
> > 0.5 sec. Is it possible?
>
> 2000 integers of magnitude 10^13000 equals to about 26 MBytes of data
> (2000 numbers each 13000 digits long). Rounding 1.8 seconds to two
> seconds we conclude that you proceed with speed about 13MBytes per
> second. Assuming you have CPU clock frequency near 2.6GHz, you
> performing about 200 clock cycles per input digit.
>
> 10^13000 roughly equal to 2^39000. Or (2^32)^1219 - 1219 32-bit words
> of representation. So you're doing some last nextN =
> (n*10)+currentDigit conversion operations in less that one clock cycle
> per word.
You can do better than calculating (n*10 + d) repeatedly, using a
divide and conquer scheme, which is in fact implemented in ByteString's
readInteger:
((a*10 + b) * 100 + (c*10 + d)) * 10000 + ...
This helps because now we're multiplying numbers of roughly equal
size, which using FFT and related methods, as gmp uses, can be sped
up immensely, beating the quadratic complexity that you get with
the naive approach.
The timings seem about right.
HTH,
Bertram
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