
Percolation Simulation

This is a simulation of aspects of mathematical percolation theory.

For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density 
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density

Let 



v


{\displaystyle v}

 be a vector of 



n


{\displaystyle n}

 values of either 1 or 0 where the probability of any
value being 1 is 



p


{\displaystyle p}

; the probability of a value being 0 is therefore 



1
−
p


{\displaystyle 1-p}

.
Define a run of 1s as being a group of consecutive 1s in the vector bounded
either by the limits of the vector or by a 0. Let the number of such runs in a given
vector of length 



n


{\displaystyle n}

 be 




R

n




{\displaystyle R_{n}}

.

For example, the following vector has 




R

10


=
3


{\displaystyle R_{10}=3}



[1 1 0 0 0 1 0 1 1 1]
 ^^^       ^   ^^^^^

Percolation theory states that





K
(
p
)
=

lim

n
→
∞



R

n



/

n
=
p
(
1
−
p
)


{\displaystyle K(p)=\lim _{n\to \infty }R_{n}/n=p(1-p)}


Task
Any calculation of 




R

n



/

n


{\displaystyle R_{n}/n}

 for finite 



n


{\displaystyle n}

 is subject to randomness so should be
computed as the average of 



t


{\displaystyle t}

 runs, where 



t
≥
100


{\displaystyle t\geq 100}

.

For values of 



p


{\displaystyle p}

 of 0.1, 0.3, 0.5, 0.7, and 0.9, show the effect of varying 



n


{\displaystyle n}


on the accuracy of simulated 



K
(
p
)


{\displaystyle K(p)}

.

Show your output here.

See also
 s-Run on Wolfram mathworld.
#include <stdio.h>
#include <stdlib.h>
 
// just generate 0s and 1s without storing them
double run_test(double p, int len, int runs)
{
	int r, x, y, i, cnt = 0, thresh = p * RAND_MAX;
 
	for (r = 0; r < runs; r++)
		for (x = 0, i = len; i--; x = y)
			cnt += x < (y = rand() < thresh);
 
	return (double)cnt / runs / len;
}
 
int main(void)
{
	double p, p1p, K;
	int ip, n;
 
	puts(	"running 1000 tests each:\n"
		" p\t   n\tK\tp(1-p)\t     diff\n"
		"-----------------------------------------------");
	for (ip = 1; ip < 10; ip += 2) {
		p = ip / 10., p1p = p * (1 - p);
 
		for (n = 100; n <= 100000; n *= 10) {
			K = run_test(p, n, 1000);
			printf("%.1f\t%6d\t%.4f\t%.4f\t%+.4f (%+.2f%%)\n",
				p, n, K, p1p, K - p1p, (K - p1p) / p1p * 100);
		}
		putchar('\n');
	}
 
	return 0;
}