Integers too big for floats
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
Dividing large integers to floats
factorial :: (Enum a, Num a) => a -> a factorial k = product [1..k]
You will find that
factorial 777 is not representable by
However it is representable by an
You will find that
factorial 777 / factorial 778 is representable as
Double but not as
but the temporary results are representable by
Integers and not by
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 as
and convert that to a floating point number:
fromRational (factorial 777 % factorial 778)
fromRational is clever enough to handle big numerators and denominators.
But there is an efficiency problem:
fromRational can perform the imprecise division,
% operator will cancel the fraction precisely.
You may use the
However that's a hack, since it is not sure that other operations work well on non-cancelled fractions.
You had to import
But since we talk about efficiency let's go on to the next paragraph, where we talk about real performance.
Avoid big integers at all
The example seems to be stupid, because you may think that nobody divides
factorial 777 by
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 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 + 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
For these two terms it is not a problem to convert the
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