I have always wanted to find an algorithm that did this. I do not care how slow it is, just as long as it can return the nth digit of Pi:
ex:
size_t piAt(long long int n)
{
}
Preferably, not using an infinite series.
If anyone has a function or class that does 开发者_C百科this, in C or C++ I'd really be interested in seeing it.
Thanks
Here is a solution of Simon Plouffe coded by Fabrice Bellard:
/*
* 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).
*/
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
/* uncomment the following line to use 'long long' integers */
/* #define HAS_LONG_LONG */
#ifdef HAS_LONG_LONG
#define mul_mod(a,b,m) (( (long long) (a) * (long long) (b) ) % (m))
#else
#define mul_mod(a,b,m) fmod( (double) a * (double) b, m)
#endif
/* return the inverse of x mod y */
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 */
int pow_mod(int a, int b, int m)
{
int r, aa;
r = 1;
aa = a;
while (1) {
if (b & 1)
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 */
int is_prime(int n)
{
int r, i;
if ((n % 2) == 0)
return 0;
r = (int) (sqrt(n));
for (i = 3; i <= r; i += 2)
if ((n % i) == 0)
return 0;
return 1;
}
/* return the prime number immediatly after n */
int next_prime(int n)
{
do {
n++;
} while (!is_prime(n));
return n;
}
int main(int argc, char *argv[])
{
int av, a, vmax, N, n, num, den, k, kq, kq2, t, v, s, i;
double sum;
if (argc < 2 || (n = atoi(argv[1])) <= 0) {
printf("This program computes the n'th decimal digit of \\pi\n"
"usage: pi n , where n is the digit you want\n");
exit(1);
}
N = (int) ((n + 20) * log(10) / log(2));
sum = 0;
for (a = 3; a <= (2 * N); a = next_prime(a)) {
vmax = (int) (log(2 * N) / log(a));
av = 1;
for (i = 0; i < vmax; i++)
av = av * a;
s = 0;
num = 1;
den = 1;
v = 0;
kq = 1;
kq2 = 1;
for (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 (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 = fmod(sum + (double) s / (double) av, 1.0);
}
printf("Decimal digits of pi at position %d: %09d\n", n,
(int) (sum * 1e9));
return 0;
}
And it works:
C:\tcc>tcc -I libtcc libtcc/libtcc.def -run examples/pi.c 1
Decimal digits of pi at position 1: 141592653
C:\tcc>tcc -I libtcc libtcc/libtcc.def -run examples/pi.c 1000
Decimal digits of pi at position 1000: 938095257
http://bellard.org/pi/pi_n2/pi_n2.html
This remarkable solution shows how to compute the Nth digit of π in O(N) time and O(log·N) space, and to do so without having to compute all the digits leading up to it.
Oh, and it’s in hex.
If you don’t want to do that, you can do this from the shell easily enough:
% perl -Mbignum=bpi -wle 'print bpi(20)'
3.1415926535897932385
% perl -Mbignum=bpi -wle 'print bpi(50)'
3.1415926535897932384626433832795028841971693993751
% perl -Mbignum=bpi -wle 'print bpi(200)'
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303820
% perl -Mbignum=bpi -wle 'print bpi(1000)'
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420199
Several of the solutions including this one actually print several digits starting with the nth digit.
There is a fine ( and strikingly faster than previous answers) solution at http://numbers.computation.free.fr/Constants/Algorithms/pidec.cpp
The code only needs modification of "typedef int64 ModInt;" line to "typedef int64_t ModInt;"
gcc pidec.cpp && time echo 20000 | ./a.out
Pidec, direct computation of decimal digits of pi at a given position n.
(http://numbers.computation.free.fr/Constants/constants.html for more details)
Enter n : Parameters : M=122, N=7094, M*N+M=872562
Series time : 0.17
Digits of pi after n-th decimal digit : 203856539
Total time: 0.68
real 0m0.691s
user 0m0.678s
sys 0m0.006s
Compilation finished at Wed Feb 3 13:34:48
This on a "early 2015" macbook pro.
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