复杂计算中的浮点精度

Question

For calculating Catalan Numbers, I wrote two codes. One (def "Catalan") works recursively and returns the right Catalan Numbers.

dicatalan = {} 
def catalan(n):
if n == 0:
    return 1
else: 
    res = 0
    if n not in dicatalan:
        for i in range(n):
            res += catalan(i) * catalan(n - i - 1)
        dicatalan[n] = res
return dicatalan[n]

the other (def "catalanFormula") applies the implicit formula, but doesn't calculate accurately starting from n=30. the problem derives from floating points - for k=9 the program returns "6835971.999999999" instead of "6835972" and from this moment on accumulates mistakes till the final wrong answer.

(print line is for checking)

def catalanFormula(n):
result = 1
for k in range(2, n + 1):
    result *= ((n + k) / k)
    print (result)
return int(result)

I tried rounding and failed, tried Decimal import and still got nothing right.

I need the "catalanFormula" work perfectly as "catalan"; Any Ideas?

Thanks!

Answer 1

See the bigfloat package.

from bigfloat import *

setcontext(quadruple_precision)
def catalanFormula(n):
    result = BigFloat(1)
    for k in range(2, n + 1):
        result *= ((BigFloat(n) + BigFloat(k)) / BigFloat(k))
    return result

catalanFormula(30)

Output:

BigFloat.exact('3814986502092304.00000000000000000043', precision=113)

Answer 2

Try calculating the numerator and denominator separately and dividing them at the end. If you do this, you should be able to make it a little bit farther with floating-point.

I'm sure Python has a package for rational numbers. Using rationals is an even better idea.