# Integers too big for floats

### From HaskellWiki

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## 1 Dividing large integers to floats

Consider

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

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

Actually you can represent the fractionand convert that to a floating point number:

fromRational (factorial 777 % factorial 778)

But there is an efficiency problem:

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

You had to importBut 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 divideswithout 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:

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] | 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 ofBut 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

## 3 See also

- Haskell Cafe on "about integer and float operations"
- Generic number type