
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.




with javascript_semantics
constant w = 10, h = 10
 
sequence wall = join(repeat("+",w+1),"---")&"\n",
         cell = join(repeat("|",w+1),"   ")&"\n",
         grid
 
procedure new_grid(atom p)
    grid = split(join(repeat(wall,h+1),cell),'\n')
    -- now knock down some walls
    for i=1 to length(grid)-1 do
        integer jstart = 5-mod(i,2)*3,
                jlimit = length(grid[i])-3
        -- (ie 2..38 on odd lines, 5..37 on even)
        for j=jstart to jlimit by 4 do
            if rnd()>p then
                grid[i][j..j+2] = "   "
            end if
        end for
    end for
end procedure
 
function percolate(integer x=0, y=0)
    if x=0 then
        for j=3 to length(grid[1])-2 by 4 do
            if grid[1][j]=' ' and percolate(1,j) then
                return true
            end if
        end for
    elsif grid[x][y]=' ' then
        grid[x][y] = '*'
        if (x=length(grid)-1)
        or (         grid[x+1][y]=' ' and percolate(x+1,y))
        or (y>6  and grid[x][y-2]=' ' and percolate(x,y-4))
        or (y<36 and grid[x][y+2]=' ' and percolate(x,y+4))
        or (x>1  and grid[x-1][y]=' ' and percolate(x-1,y)) then
            return true
        end if
    end if
    return false
end function
 
constant LIM=1000
 
for p=0 to 10 do
    integer count = 0
    for t=1 to LIM do
        new_grid(p/10)
        count += percolate()
    end for
    printf(1,"p=%.1f: %5.3f\n",{p/10,count/LIM})
end for
puts(1,"sample grid for p=0.6:\n")
new_grid(0.6)
{} = percolate()
printf(1,"%s\n",{join(grid,'\n')})
