
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.




// version 1.2.10
 
import java.util.Random
 
val rand = Random()
const val RAND_MAX = 32767
 
// cell states
const val FILL  = 1
const val RWALL = 2  // right wall
const val BWALL = 4  // bottom wall
 
val x = 10
val y = 10
var grid = IntArray(x * (y + 2))
var cells = 0
var end = 0
var m = 0
var n = 0
 
fun makeGrid(p: Double) {
    val thresh = (p * RAND_MAX).toInt()
    m = x
    n = y
    grid.fill(0)  // clears grid
    for (i in 0 until m) grid[i] = BWALL or RWALL
    cells = m
    end = m
    for (i in 0 until y) {
        for (j in x - 1 downTo 1) {
            val r1 = rand.nextInt(RAND_MAX + 1)
            val r2 = rand.nextInt(RAND_MAX + 1)
            grid[end++] = (if (r1 < thresh) BWALL else 0) or
                          (if (r2 < thresh) RWALL else 0)
        }
        val r3 = rand.nextInt(RAND_MAX + 1)
        grid[end++] = RWALL or (if (r3 < thresh) BWALL else 0)
    }
}
 
fun showGrid() {
    for (j in 0 until m) print("+--")
    println("+")
 
    for (i in 0..n) {
        print(if (i == n) " " else "|")
        for (j in 0 until m) {
            print(if ((grid[i * m + j + cells] and FILL) != 0) "[]" else "  ")
            print(if ((grid[i * m + j + cells] and RWALL) != 0) "|" else " ")
        }
        println()
        if (i == n) return
        for (j in 0 until m) {
            print(if ((grid[i * m + j + cells] and BWALL) != 0) "+--" else "+  ")
        }
        println("+")
    }
}
 
fun fill(p: Int): Boolean {
    if ((grid[p] and FILL) != 0) return false
    grid[p] = grid[p] or FILL
    if (p >= end) return true  // success: reached bottom row
    return (((grid[p + 0] and BWALL) == 0) && fill(p + m)) ||
           (((grid[p + 0] and RWALL) == 0) && fill(p + 1)) ||
           (((grid[p - 1] and RWALL) == 0) && fill(p - 1)) ||
           (((grid[p - m] and BWALL) == 0) && fill(p - m))
}
 
fun percolate(): Boolean {
    var i = 0
    while (i < m && !fill(cells + i)) i++
    return i < m
}
 
fun main(args: Array<String>) {
    makeGrid(0.5)
    percolate()
    showGrid()
 
    println("\nrunning $x x $y grids 10,000 times for each p:")
    for (p in 1..9) {
        var cnt = 0
        val pp = p / 10.0
        for (i in 0 until 10_000) {
            makeGrid(pp)
            if (percolate()) cnt++
        }
        println("p = %3g: %.4f".format(pp, cnt.toDouble() / 10_000))
    }
}