
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 



c


{\displaystyle c}

 be a 2D boolean square matrix of 



n
×
n


{\displaystyle n\times n}

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



p


{\displaystyle p}

, (and of 0 is therefore 



1
−
p


{\displaystyle 1-p}

).
We define a cluster of 1's as being a group of 1's connected vertically or
horizontally (i.e., using the Von Neumann neighborhood rule) and bounded by either 



0


{\displaystyle 0}

 or by the limits of the matrix.
Let the number of such clusters in such a randomly constructed matrix be 




C

n




{\displaystyle C_{n}}

.

Percolation theory states that 



K
(
p
)


{\displaystyle K(p)}

 (the mean cluster density) will satisfy 



K
(
p
)
=

C

n



/


n

2




{\displaystyle K(p)=C_{n}/n^{2}}

 as 



n


{\displaystyle n}

 tends to infinity. For 



p
=
0.5


{\displaystyle p=0.5}

, 



K
(
p
)


{\displaystyle K(p)}

 is found numerically to approximate 



0.065770


{\displaystyle 0.065770}

...

Task
Show the effect of varying 



n


{\displaystyle n}

 on the accuracy of simulated 



K
(
p
)


{\displaystyle K(p)}

 for 



p
=
0.5


{\displaystyle p=0.5}

 and
for values of 



n


{\displaystyle n}

 up to at least 



1000


{\displaystyle 1000}

.
Any calculation of 




C

n




{\displaystyle C_{n}}

 for finite 



n


{\displaystyle n}

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



t


{\displaystyle t}

 runs, where 



t


{\displaystyle t}

 ≥ 



5


{\displaystyle 5}

.

For extra credit, graphically show clusters in a 



15
×
15


{\displaystyle 15\times 15}

, 



p
=
0.5


{\displaystyle p=0.5}

 grid.

Show your output here.

See also
 s-Cluster on Wolfram mathworld.
UInt32 seed = 0
F nonrandom()
   :seed = 1664525 * :seed + 1013904223
   R (:seed >> 16) / Float(FF'FF)
 
V nn = 15
V tt = 5
V pp = 0.5
V NotClustered = 1
V Cell2Char = ‘ #abcdefghijklmnopqrstuvwxyz’
V NRange = [4, 64, 256, 1024, 4096]
 
F newGrid(n, p)
   R (0 .< n).map(i -> (0 .< @n).map(i -> Int(nonrandom() < @@p)))
 
F walkMaze(&grid, m, n, idx) -> N
   grid[n][m] = idx
   I n < grid.len - 1 & grid[n + 1][m] == NotClustered
      walkMaze(&grid, m, n + 1, idx)
   I m < grid[0].len - 1 & grid[n][m + 1] == NotClustered
      walkMaze(&grid, m + 1, n, idx)
   I m > 0 & grid[n][m - 1] == NotClustered
      walkMaze(&grid, m - 1, n, idx)
   I n > 0 & grid[n - 1][m] == NotClustered
      walkMaze(&grid, m, n - 1, idx)
 
F clusterCount(&grid)
   V walkIndex = 1
   L(n) 0 .< grid.len
      L(m) 0 .< grid[0].len
         I grid[n][m] == NotClustered
            walkIndex++
            walkMaze(&grid, m, n, walkIndex)
   R walkIndex - 1
 
F clusterDensity(n, p)
   V grid = newGrid(n, p)
   R clusterCount(&grid) / Float(n * n)
 
F print_grid(grid)
   L(row) grid
      print(L.index % 10, end' ‘) ’)
      L(cell) row
         print(‘ ’Cell2Char[cell], end' ‘’)
      print()
 
V grid = newGrid(nn, 0.5)
print(‘Found ’clusterCount(&grid)‘ clusters in this ’nn‘ by ’nn" grid\n")
print_grid(grid)
print()
 
L(n) NRange
   V sum = 0.0
   L 0 .< tt
      sum += clusterDensity(n, pp)
   V sim = sum / tt
   print(‘t = #.  p = #.2  n = #4  sim = #.5’.format(tt, pp, n, sim))