# Calculate PI using C#

Posted by OmegaMan at August 14, 2007

Category: How To

Tags:

(Update 4/4/2011)

This article was written back in 2007 and somehow lost(?) between my transferring of this blog from different hosts. It was reported missing and I have found the code and reposted it here. Note this code is not an original creation by myself. I simply brought it into C# from C++. Call the static method Process() with the total # of digits to calculate. HTH

public class CalculatePI
{
/*
* Computation of the n'th decimal digit of \pi with very little memory.
* Written by Fabrice Bellard on January 8, 1997.
*
* We use a slightly modified version of the method described by Simon
* Plouffe in "On the Computation of the n'th decimal digit of various
* transcendental numbers" (November 1996). We have modified the algorithm
* to get a running time of O(n^2) instead of O(n^3log(n)^3).
*
* This program uses mostly integer arithmetic. It may be slow on some
* hardwares where integer multiplications and divisons must be done
* by software. We have supposed that 'int' has a size of 32 bits. If
* your compiler supports 'long long' integers of 64 bits, you may use
* the integer version of 'mul_mod' (see HAS_LONG_LONG).
*/

// Call this static to use.
public static string Process(int digits)
{
StringBuilder result = new StringBuilder();

result.Append("3.");

DateTime StartTime = DateTime.Now;

if (digits > 0)
{

for (int i = 0; i < digits; i += 9)
{
String ds = CalculatePiDigits(i + 1);
int digitCount = Math.Min(digits - i, 9);

if (ds.Length < 9)
ds = string.Format("{0:D9}", int.Parse(ds));

result.Append(ds.Substring(0, digitCount));
}
}

return result.ToString();
}

private static int mul_mod(int a, int b, int m)
{
return (int)(((long)a * (long)b) % m);
}

/* return the inverse of x mod y */
private static int inv_mod(int x, int y)
{
int q, u, v, a, c, t;

u = x;
v = y;
c = 1;
a = 0;

do
{
q = v / u;

t = c;
c = a - q * c;
a = t;

t = u;
u = v - q * u;
v = t;
} while (u != 0);

a = a % y;

if (a < 0)
{
a = y + a;
}

return a;
}

/* return (a^b) mod m */
private static int pow_mod(int a, int b, int m)
{
int r, aa;

r = 1;
aa = a;

while (true)
{
if ((b & 1) != 0)
{
r = mul_mod(r, aa, m);
}

b = b >> 1;

if (b == 0)
{
break;
}

aa = mul_mod(aa, aa, m);
}

return r;
}

/* return true if n is prime */
private static bool is_prime(int n)
{
if ((n % 2) == 0)
{
return false;
}

int r = (int)Math.Sqrt(n);

for (int i = 3; i <= r; i += 2)
{
if ((n % i) == 0)
{
return false;
}
}

return true;
}

/* return the prime number immediatly after n */
private static int next_prime(int n)
{
do
{
n++;
} while (!is_prime(n));

return n;
}

private static String CalculatePiDigits(int n)
{
int av, vmax, num, den, s, t;

int N = (int)((n + 20) * Math.Log(10) / Math.Log(2));

double sum = 0;

for (int a = 3; a <= (2 * N); a = next_prime(a))
{
vmax = (int)(Math.Log(2 * N) / Math.Log(a));

av = 1;

for (int i = 0; i < vmax; i++)
{
av = av * a;
}

s = 0;
num = 1;
den = 1;
int v = 0;
int kq = 1;
int kq2 = 1;

for (int k = 1; k <= N; k++)
{

t = k;

if (kq >= a)
{
do
{
t = t / a;
v--;
} while ((t % a) == 0);

kq = 0;
}
kq++;
num = mul_mod(num, t, av);

t = 2 * k - 1;

if (kq2 >= a)
{
if (kq2 == a)
{
do
{
t = t / a;
v++;
} while ((t % a) == 0);
}
kq2 -= a;
}
den = mul_mod(den, t, av);
kq2 += 2;

if (v > 0)
{
t = inv_mod(den, av);
t = mul_mod(t, num, av);
t = mul_mod(t, k, av);

for (int i = v; i < vmax; i++)
{
t = mul_mod(t, a, av);
}

s += t;

if (s >= av)
{
s -= av;
}
}

}

t = pow_mod(10, n - 1, av);
s = mul_mod(s, t, av);
sum = (sum + (double)s / (double)av) % 1.0;
}

int Result = (int)(sum * 1e9);

String StringResult = String.Format("{0:D9}", Result);

return StringResult;
}

// Put a space between every group of 10 digits.

private static String breakDigitsIntoGroupsOf10(String digits)
{
String result = "";

while (digits.Length > 10)
{
result += digits.Substring(0, 10) + " ";
digits = digits.Substring(10, digits.Length - 10);
}

result += digits;

return result;
}

}


1. Richard says

I’m a peruvian student of computer science, and was looking for code like this, Thanks whoever you are………….

2. Dean Chalk says
3. larry h says

Hey there love this page thank you so much very easy to understand but u keep getting an error with this step here is what im inputing:
” {
int av, vmax, num, den, s, t;

int N = (int)(n + 20) * MathLog(10) / MathLog(2);

double sum = 0;

for (int a = 3; a <= (2 * N); a = next_prime(a)
{
vmax = (int)(MathLog(2 * N) / MathLog(a);

av = 1;

for (int i = 0; i < vmax; i++)
{
av = av * a;
}"
let me know please thanks so much

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