[Haskell-beginners] Why is it so slow to solve "10^(10^10)"?

[Bob Ippolito]

Why wouldn't it be slow? Integers in Haskell are implemented with GMP, and
the representation is simply a sign, a magnitude (number of bits) and then
all of the bits. http://gmplib.org/manual/Integer-Internals.html

10^(10^10) is a very large number. With the representation that GMP uses,
it will take at least 3.86 GB of RAM to represent the bits required by the
result! Of course it takes ages to work with numbers of that magnitude.

    h> ((10^10) * logBase 256 10) * (2**(-30))
    3.8672332825224878

It is possible to encode values like this, but there's nothing that ships
with Haskell Platform that'll do it efficiently. You might look at the
numbers package though.

    h> import Data.Number.BigFloat
    h> ((10 :: BigFloat Eps1)^10)^10
    1.e100

Note that the above example is trivial, only one digit of precision, but
that's all that's needed to represent this result since BigFloat is base 10.

-bob

On Sat, Sep 21, 2013 at 3:32 PM, yi lu <zhiwudazhanjiangshi at gmail.com>wrote:

> Why is it so slow to solve "10^(10^10)" in Haskell?
>
> Yi
>
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