
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

Given an 



M
×
N


{\displaystyle M\times N}

 rectangular array of cells numbered 




c
e
l
l

[
0..
M
−
1
,
0..
N
−
1
]


{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}

, assume 



M


{\displaystyle M}

 is horizontal and 



N


{\displaystyle N}

 is downwards. Each 




c
e
l
l

[
m
,
n
]


{\displaystyle \mathrm {cell} [m,n]}

 is bounded by (horizontal) walls 




h
w
a
l
l

[
m
,
n
]


{\displaystyle \mathrm {hwall} [m,n]}

 and 




h
w
a
l
l

[
m
+
1
,
n
]


{\displaystyle \mathrm {hwall} [m+1,n]}

; (vertical) walls 




v
w
a
l
l

[
m
,
n
]


{\displaystyle \mathrm {vwall} [m,n]}

 and 




v
w
a
l
l

[
m
,
n
+
1
]


{\displaystyle \mathrm {vwall} [m,n+1]}



Assume that the probability of any wall being present is a constant 



p


{\displaystyle p}

 where

 



0.0
≤
p
≤
1.0


{\displaystyle 0.0\leq p\leq 1.0}


Except for the outer horizontal walls at 



m
=
0


{\displaystyle m=0}

 and 



m
=
M


{\displaystyle m=M}

 which are always present.

The task

Simulate pouring a fluid onto the top surface (



n
=
0


{\displaystyle n=0}

) where the fluid will enter any empty cell it is adjacent to if there is no wall between where it currently is and the cell on the other side of the (missing) wall.

The fluid does not move beyond the horizontal constraints of the grid.

The fluid may move “up” within the confines of the grid of cells. If the fluid reaches a bottom cell that has a missing bottom wall then the fluid can be said to 'drip' out the bottom at that point.

Given 



p


{\displaystyle p}

 repeat the percolation 



t


{\displaystyle t}

 times to estimate the proportion of times that the fluid can percolate to the bottom for any given 



p


{\displaystyle p}

.

Show how the probability of percolating through the random grid changes with 



p


{\displaystyle p}

 going from 



0.0


{\displaystyle 0.0}

 to 



1.0


{\displaystyle 1.0}

 in 



0.1


{\displaystyle 0.1}

 increments and with the number of repetitions to estimate the fraction at any given 



p


{\displaystyle p}

 as 



t
=
100


{\displaystyle t=100}

.

Use an 



M
=
10
,
N
=
10


{\displaystyle M=10,N=10}

 grid of cells for all cases.

Optionally depict fluid successfully percolating through a grid graphically.

Show all output on this page.




#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
 
using namespace std;
 
class Grid {
public:
    Grid(const double p, const int x, const int y) : m(x), n(y) {
        const int thresh = static_cast<int>(RAND_MAX * p);
 
        // Allocate two addition rows to avoid checking bounds.
        // Bottom row is also required by drippage
        start = new cell[m * (n + 2)];
        cells = start + m;
        for (auto i = 0; i < m; i++) start[i] = RBWALL;
        end = cells;
        for (auto i = 0; i < y; i++) {
            for (auto j = x; --j;)
                *end++ = (rand() < thresh ? BWALL : 0) | (rand() < thresh ? RWALL : 0);
            *end++ = RWALL | (rand() < thresh ? BWALL : 0);
        }
        memset(end, 0u, sizeof(cell) * m);
    }
 
    ~Grid() {
        delete[] start;
        cells = 0;
        start = 0;
        end = 0;
    }
 
    int percolate() const {
        auto i = 0;
        for (; i < m && !fill(cells + i); i++);
        return i < m;
    }
 
    void show() const {
        for (auto j = 0; j < m; j++)
            cout << ("+-");
        cout << '+' << endl;
 
        for (auto i = 0; i <= n; i++) {
            cout << (i == n ? ' ' : '|');
            for (auto j = 0; j < m; j++) {
                cout << ((cells[i * m + j] & FILL) ? "#" : " ");
                cout << ((cells[i * m + j] & RWALL) ? '|' : ' ');
            }
            cout << endl;
 
            if (i == n) return;
 
            for (auto j = 0; j < m; j++)
                cout << ((cells[i * m + j] & BWALL) ? "+-" : "+ ");
            cout << '+' << endl;
        }
    }
 
private:
    enum cell_state {
        FILL   = 1 << 0,
        RWALL  = 1 << 1,       // right wall
        BWALL  = 1 << 2,       // bottom wall
        RBWALL = RWALL | BWALL // right/bottom wall
    };
 
    typedef unsigned int cell;
 
    bool fill(cell* p) const {
        if ((*p & FILL)) return false;
        *p |= FILL;
        if (p >= end) return true; // success: reached bottom row
 
        return (!(p[0] & BWALL) && fill(p + m)) || (!(p[0] & RWALL) && fill(p + 1))
                ||(!(p[-1] & RWALL) && fill(p - 1)) || (!(p[-m] & BWALL) && fill(p - m));
    }
 
    cell* cells;
    cell* start;
    cell* end;
    const int m;
    const int n;
};
 
int main() {
    const auto M = 10, N = 10;
    const Grid grid(.5, M, N);
    grid.percolate();
    grid.show();
 
    const auto C = 10000;
    cout << endl << "running " << M << "x" << N << " grids " << C << " times for each p:" << endl;
    for (auto p = 1; p < M; p++) {
        auto cnt = 0, i = 0;
        for (; i < C; i++)
            cnt += Grid(p / static_cast<double>(M), M, N).percolate();
        cout << "p = " << p / static_cast<double>(M) << ": " << static_cast<double>(cnt) / i << endl;
    }
 
    return EXIT_SUCCESS;
}