Difference between revisions of "Integers too big for floats"

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(describe the problem)
 
(implementations and optimizations)
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Before <hask>fromRational</hask> can perform the imprecise division,
 
Before <hask>fromRational</hask> can perform the imprecise division,
 
the <hask>%</hask> operator will cancel the fraction precisely.
 
the <hask>%</hask> operator will cancel the fraction precisely.
You may use the <hask>Rational</hask> constructor <hask:>%</hask> instead.
+
You may use the <hask>Rational</hask> constructor <hask>:%</hask> instead.
 
However that's a hack, since it is not sure that other operations work well on non-cancelled fractions.
 
However that's a hack, since it is not sure that other operations work well on non-cancelled fractions.
 
You had to import <hask>GHC.Real</hask>.
 
You had to import <hask>GHC.Real</hask>.
  +
  +
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 <hask>factorial 777</hask> by <hask>factorial 778</hask>
  +
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 <math>\pi</math> using [http://en.wikipedia.org/wiki/Chudnovsky_algorithm Chudnovsky's] algorithm:
  +
:<math>
  +
\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}\ .
  +
</math>
  +
  +
A possible [http://paste.pocoo.org/show/102801/ Haskell implementation] is:
  +
<haskell>
  +
-- 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
  +
</haskell>
  +
  +
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 <hask>Double</hask>.
  +
For these two terms it is not a problem to convert the <hask>Integer</hask>s to <hask>Double</hask>s.
  +
  +
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 for each term,
  +
instead we compute each term using the term before.
  +
<haskell>
  +
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
  +
</haskell>
   
   

Revision as of 16:17, 6 February 2009

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.

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 Integers and not by Doubles. 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 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.


Avoid big integers at all

The example seems to be stupid, because you may think that nobody divides factorial 777 by 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 \pi 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]
    | 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 Integers to Doubles.

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 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


See also