Integers too big for floats

From HaskellWiki

(Difference between revisions)

Current revision

Although floating point types can represent a large range of magnitudes,

you will sometimes have to cope with integers that are larger than what is representable by

Double

1 Dividing large integers to floats

Consider

factorial :: (Enum a, Num a) => a -> a
factorial k = product [1..k]

You will find that

factorial 777

is not representable by

Double

. However it is representable by an

Integer

. You will find that

factorial 777 / factorial 778

is representable as

Double

but not as

Integer

, but the temporary results are representable by

Integer

s and not by

Double

Is there a variant of division which accepts big integers and emits floating point numbers?

Actually you can represent the fraction

factorial 777 / factorial 778

Rational

and convert that to a floating point number:

fromRational (factorial 777 % factorial 778)

Fortunately

fromRational

is clever enough to handle big numerators and denominators.

But there is an efficiency problem:

Before

fromRational

can perform the imprecise division, the

%

operator will cancel the fraction precisely. You may use the

Rational

constructor

:%

instead.

However that's a hack, since it is not sure that other operations work well on non-cancelled fractions.

You had to import

GHC.Real

But since we talk about efficiency let's go on to the next paragraph, where we talk about real performance.

2 Avoid big integers at all

The example seems to be stupid, because you may think that nobody divides

factorial 777

factorial 778

without noticing, that this can be greatly simplified. So let's take the original task which led to the problem above. The problem is to compute the reciprocal of $π$ using Chudnovsky's algorithm:

$\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}\ .$

A possible Haskell implementation is:

-- An exact division
-- Courtesy of Max Rabkin
(/.) :: (Real a, Fractional b) => a -> a -> b
x /. y = fromRational $ toRational x / toRational y
 
-- Compute n!
fac :: Integer -> Integer
fac n = product [1..n]
 
-- Compute n! / m! efficiently
facDiv :: Integer -> Integer -> Integer
facDiv n m 
    | n > m = product [n, n - 1 .. m + 1]
    | n == m = 1
    | otherwise = facDiv m n
 
 
-- Compute pi using the specified number of iterations
pi' :: Integer -> Double
pi' steps = 1.0 / (12.0 * s / f)
    where
      s = sum [chudnovsky n | n <- [0..steps]]
      f = fromIntegral c ** (3.0 / 2.0) -- Common factor in the sum
 
      -- k-th term of the Chudnovsky serie
      chudnovsky :: Integer -> Double
      chudnovsky k 
          | even k = num /. den
          | otherwise = -num /. den
          where
            num = (facDiv (6 * k) (3 * k)) * (a + b * k)
            den = (fac k) ^ 3 * (c ^ (3 * k))
 
      a = 13591409
      b = 545140134
      c = 640320
 
main = print $ pi' 1000

To be honest, this program doesn't really need much more optimization than limiting the number of terms to 2,

since the subsequent terms are much below the precision of

Double

. For these two terms it is not a problem to convert the

Integer

s to

Double

But assume these conversions are a problem. We will show a way to avoid them. The trick is to compute the terms incrementally. We do not need to compute the factorials from scratch for each term, instead we compute each term using the term before.

start :: Floating a => a
start =
   12 / sqrt 640320 ^ 3
 
arithmeticSeq :: Num a => [a]
arithmeticSeq =
   iterate (545140134+) 13591409
 
factors :: Floating a => [a]
factors =
   -- note canceling of product[(6*k+1)..6*(k+1)] / product[(3*k+1)..3*(k+1)]
   map (\k -> -(6*k+1)*(6*k+3)*(6*k+5)/(320160*(k+1))^3) $ iterate (1+) 0
 
summands :: Floating a => [a]
summands =
   zipWith (*) arithmeticSeq $ scanl (*) start factors
 
recipPi :: Floating a => a
recipPi =
   sum $ take 2 summands