Percolation/Site percolation

Given an $M\times N$ rectangular array of cells numbered $\mathrm {cell} [0..M-1,0..N-1]$ assume $M$ is horizontal and $N$ is downwards.

Assume that the probability of any cell being filled is a constant $p$ where

0.0\leq p\leq 1.0

The task

Simulate creating the array of cells with probability $p$ and then testing if there is a route through adjacent filled cells from any on row $0$ to any on row $N$ cell, i.e. testing for site percolation.

Given $p$ repeat the percolation $t$ times to estimate the proportion of times that the fluid can percolate to the bottom for any given $p$ .

Show how the probability of percolating through the random grid changes with $p$ going from $0.0$ to $1.0$ in $0.1$ increments and with the number of repetitions to estimate the fraction at any given $p$ as $t=100$ .

Use an $M=15,N=15$ grid of cells for all cases.

Optionally depict a successfull percolation path through a cell grid graphically.

Show all output on this page.

See

Percolation theory

Python

<lang python>from random import random import string from pprint import pprint as pp

M, N, t = 15, 15, 100

cell2char = ' #' + string.ascii_letters NOT_VISITED = 1 # filled cell not walked

class PercolatedException(Exception): pass

def newgrid(p):

   return [[int(random() < p) for m in range(M)] for n in range(N)] # cell

def pgrid(cell, percolated=None):

   for n in range(N):
       print( '%i)  ' % (n % 10) 
              + ' '.join(cell2char[cell[n][m]] for m in range(M)))
   if percolated: 
       where = percolated.args[0][0]
       print('!)  ' + '  ' * where + cell2char[cell[n][where]])

def check_from_top(cell):

   n, walk_index = 0, 1
   try:
       for m in range(M):
           if cell[n][m] == NOT_VISITED:
               walk_index += 1
               walk_maze(m, n, cell, walk_index)
   except PercolatedException as ex:
       return ex
   return None

def walk_maze(m, n, cell, indx):

   # fill cell 
   cell[n][m] = indx
   # down
   if n < N - 1 and cell[n+1][m] == NOT_VISITED:
       walk_maze(m, n+1, cell, indx)
   # THE bottom
   elif n == N - 1:
       raise PercolatedException((m, indx))
   # left
   if m and cell[n][m - 1] == NOT_VISITED:
       walk_maze(m-1, n, cell, indx)
   # right
   if m < M - 1 and cell[n][m + 1] == NOT_VISITED:
       walk_maze(m+1, n, cell, indx)
   # up
   if n and cell[n-1][m] == NOT_VISITED:
       walk_maze(m, n-1, cell, indx)

if __name__ == '__main__':

   sample_printed = False
   pcount = {}
   for p10 in range(11):
       p = p10 / 10.0
       pcount[p] = 0
       for tries in range(t):
           cell = newgrid(p)
           percolated = check_from_top(cell)
           if percolated:
               pcount[p] += 1
               if not sample_printed:
                   print('\nSample percolating %i x %i, p = %5.2f grid\n' % (M, N, p))
                   pgrid(cell, percolated)
                   sample_printed = True
   print('\n p: Fraction of %i tries that percolate through\n' % t )
   
   pp({p:c/float(t) for p, c in pcount.items()})</lang>

Output:

The Ascii art grid of cells has blanks for cells that were not filled. Filled cells start off as the '#', hash character and are changed to a succession of printable characters by successive tries to navigate from the top, (top - left actually), filled cell to the bottom.

The '!)' row shows where the percolation finished and you can follow the letter backwards from that row, (letter 'c' in this case), to get the route. The program stops after finding its first route through.

Sample percolating 15 x 15, p =  0.40 grid

0)    a a a       b   c #        
1)    a a   #         c c   #   #
2)        # #   # #     c # # #  
3)  #   #       # # #   c        
4)    #     #         c c c c c c
5)  # # # # # #         c   c   c
6)        # # #         c   c   c
7)  #   #     # #     #   #   # c
8)  #   # #     #   #       c c c
9)    #       #         #   c    
0)  #       #   # # # #   c c # #
1)      #     #   #     # c      
2)  #     # # # # #   c c c   c  
3)  #   # # #         c   c c c  
4)      #           # c         #
!)                    c

 p: Fraction of 100 tries that percolate through

{0.0: 0.0,
 0.1: 0.0,
 0.2: 0.0,
 0.3: 0.0,
 0.4: 0.01,
 0.5: 0.11,
 0.6: 0.59,
 0.7: 0.94,
 0.8: 1.0,
 0.9: 1.0,
 1.0: 1.0}

Note the abrupt change in percolation at around p = 0.6. These abrupt changes are expected.