
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.




from collections import namedtuple
from random import random
from pprint import pprint as pp
 
Grid = namedtuple('Grid', 'cell, hwall, vwall')
 
M, N, t = 10, 10, 100
 
class PercolatedException(Exception): pass
 
HVF = [(' .', ' _'), (':', '|'), (' ', '#')]  # Horiz, vert, fill chars
 
def newgrid(p):
    hwall = [[int(random() < p) for m in range(M)] 
             for n in range(N+1)]
    vwall = [[(1 if m in (0, M) else int(random() < p)) for m in range(M+1)] 
             for n in range(N)]
    cell = [[0 for m in range(M)] 
             for n in range(N)]
    return Grid(cell, hwall, vwall)
 
def pgrid(grid, percolated=None):
    cell, hwall, vwall = grid
    h, v, f = HVF
    for n in range(N):
        print('    ' + ''.join(h[hwall[n][m]] for m in range(M)))
        print('%i)  ' % (n % 10) + ''.join(v[vwall[n][m]] + f[cell[n][m] if m < M else 0]
                                          for m in range(M+1))[:-1])
    n = N
    print('    ' + ''.join(h[hwall[n][m]] for m in range(M)))
    if percolated: 
        where = percolated.args[0][0]
        print('!)  ' + '  ' * where + ' ' + f[1])
 
def pour_on_top(grid):
    cell, hwall, vwall = grid
    n = 0
    try:
        for m in range(M):
            if not hwall[n][m]:
                flood_fill(m, n, cell, hwall, vwall)
    except PercolatedException as ex:
        return ex
    return None
 
 
def flood_fill(m, n, cell, hwall, vwall):
    # fill cell 
    cell[n][m] = 1
    # bottom
    if n < N - 1 and not hwall[n + 1][m] and not cell[n+1][m]:
        flood_fill(m, n+1, cell, hwall, vwall)
    # THE bottom
    elif n == N - 1 and not hwall[n + 1][m]:
        raise PercolatedException((m, n+1))
    # left
    if m and not vwall[n][m] and not cell[n][m - 1]:
        flood_fill(m-1, n, cell, hwall, vwall)
    # right
    if m < M - 1 and not vwall[n][m + 1] and not cell[n][m + 1]:
        flood_fill(m+1, n, cell, hwall, vwall)
    # top
    if n and not hwall[n][m] and not cell[n-1][m]:
        flood_fill(m, n-1, cell, hwall, vwall)
 
if __name__ == '__main__':
    sample_printed = False
    pcount = {}
    for p10 in range(11):
        p = (10 - p10) / 10.0    # count down so sample print is interesting
        pcount[p] = 0
        for tries in range(t):
            grid = newgrid(p)
            percolated = pour_on_top(grid)
            if percolated:
                pcount[p] += 1
                if not sample_printed:
                    print('\nSample percolating %i x %i grid' % (M, N))
                    pgrid(grid, percolated)
                    sample_printed = True
    print('\n p: Fraction of %i tries that percolate through' % t )
 
    pp({p:c/float(t) for p, c in pcount.items()})