Rajat Mondal - 3 years ago 159

C++ Question

I am using this crude monte-carlo integration technique to find out the value of $\pi$ and noticed that the integration value steadily deviates from the actual value as I increase the number of sample points. The code is:

`#include<iostream>`

#include<cmath>

#include<cstdlib>

using namespace std;

float f(float x)//definition of the integrand

{

return sqrt(1-x*x);

}

float rand1()//random number generator between 0 and 1

{

float s=rand();

return s/(RAND_MAX+1.0);

}

float calcint(float xi,float xf,float yi,float yf,float N)//integrator

{

float n=0;

for(int i=0;i<N;i++)

{

float x=(xf-xi)*rand1();float y=(yf-yi)*rand1();

if (y<f(x))

{

n=n+1;

}

}

return n/N*(xf-xi)*(yf-yi);

}

int main()

{

float N=100000000;

for (int i=1; i<N; i=i+N/10)//lists integration value for different sampling

{

cout<<i<<"\t"<<4*calcint(0,1,0,1,i)<<endl;

}

return 0;

}

The output is,

10000000 3.14188

20000000 3.14059

30000000 2.23696

40000000 1.67772

50000000 1.34218

60000000 1.11848

70000000 0.958698

80000000 0.838861

90000000 0.745654

Why does it happen? Are Monte-carlo integration techniques guaranteed to converge with larger number of sample points?

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Answer Source

The problem is limited precision of `float`

type. It has 24 significant precision bits, and the maximal possible integer number that can be represented by `float`

type is 16777216 while 16777217 cannot be represented because it requires 25 significant bits (1 0000 0000 0000 0000 0000 0001 in binary). See more details here: Which is the first integer that an IEEE 754 float is incapable of representing exactly?

It means that when you add 1.0f to 16777216.0f, the result will be 16777216.0f instead of 16777217.0f. Therefore, you should use integer type for `n`

instead of floating point type to count the number of events.

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