I have code which is calculating catalan numbers with method of Binominal Coefficients.
def BinominalCoefficient(n,k):
res = 1;
if (k > n - k):
k = n - k
for i in range(k):
res *= (n - i)
res /= (i + 1)
return res
def CatalanNumbers(n):
c = BinominalCoefficient(2*n, n)
return (c//(n+1))
print (CatalanNumbers(510))
I have a "nan" result when i try to calculate Catalan number which n is more than 510. Why this is happening? And how can i solve it?
Using Binomial Coefficient:Catalan numbers can also be represented as Catalan(n)=2nCn/(n+1). This reduces the time complexity to:O(n).
This can be proved by using the asymptotic growth of the central binomial coefficients, by Stirling's approximation for. , or via generating functions. The only Catalan numbers Cn that are odd are those for which n = 2k − 1; all others are even. The only prime Catalan numbers are C2 = 2 and C3 = 5.
I assume you're using Python 3.
Your res /= (i + 1)
should be res //= (i + 1)
to force integer arithmetic:
def BinominalCoefficient(n,k):
res = 1
if (k > n - k):
k = n - k
for i in range(k):
res *= (n - i)
res //= (i + 1)
return res
def CatalanNumbers(n):
c = BinominalCoefficient(2*n, n)
return (c//(n+1))
print (CatalanNumbers(511))
returns
2190251491739477424254235019785597839694676372955883183976582551028726151813997871354391075304454574949251922785248583970189394756782256529178824038918189668852236486561863197470752363343641524451529091938039960955474280081989297135147411990495428867310575974835605457151854594468879961981363032236839645
You get nan
because the divison /= in Python 3 returns a float which overflows to inf
.
In addition to xnx's answer, note that starting Python 3.8
, with the addition of math.comb
(binomial coefficient) in the standard library, we can also calculate Catalan numbers as such:
import math
def catalan(n):
return math.comb(2*n, n) / (n+1)
catalan(511) # 2.1902514917394773e+303
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