This is my solution to the problem on the percolation assignment on the course Algorithms I on Coursera by Princeton University.
You can view the fully project specifications more in depth here: https://coursera.cs.princeton.edu/algs4/assignments/percolation/specification.php
Percolation. Given a composite systems comprised of randomly distributed insulating and metallic materials: what fraction of the materials need to be metallic so that the composite system is an electrical conductor? Given a porous landscape with water on the surface (or oil below), under what conditions will the water be able to drain through to the bottom (or the oil to gush through to the surface)? Scientists have defined an abstract process known as percolation to model such situations.
The model. We model a percolation system using an n-by-n grid of sites. Each site is either open or blocked. A full site is an open site that can be connected to an open site in the top row via a chain of neighboring (left, right, up, down) open sites. We say the system percolates if there is a full site in the bottom row. In other words, a system percolates if we fill all open sites connected to the top row and that process fills some open site on the bottom row. (For the insulating/metallic materials example, the open sites correspond to metallic materials, so that a system that percolates has a metallic path from top to bottom, with full sites conducting. For the porous substance example, the open sites correspond to empty space through which water might flow, so that a system that percolates lets water fill open sites, flowing from top to bottom.)
percolates does not percolate The problem. In a famous scientific problem, researchers are interested in the following question: if sites are independently set to be open with probability p (and therefore blocked with probability 1 − p), what is the probability that the system percolates? When p equals 0, the system does not percolate; when p equals 1, the system percolates. The plots below show the site vacancy probability p versus the percolation probability for 20-by-20 random grid (left) and 100-by-100 random grid (right).
Percolation threshold for 20-by-20 grid Percolation threshold for 100-by-100 grid
When n is sufficiently large, there is a threshold value p* such that when p < p* a random n-by-n grid almost never percolates, and when p > p*, a random n-by-n grid almost always percolates. No mathematical solution for determining the percolation threshold p* has yet been derived. Your task is to write a computer program to estimate p*.
Percolation data type. To model a percolation system, create a data type Percolation with the following API:
public class Percolation {
// creates n-by-n grid, with all sites initially blocked
public Percolation(int n)
// opens the site (row, col) if it is not open already
public void open(int row, int col)
// is the site (row, col) open?
public boolean isOpen(int row, int col)
// is the site (row, col) full?
public boolean isFull(int row, int col)
// returns the number of open sites
public int numberOfOpenSites()
// does the system percolate?
public boolean percolates()
// test client (optional)
public static void main(String[] args)
} Corner cases. By convention, the row and column indices are integers between 1 and n, where (1, 1) is the upper-left site: Throw an IllegalArgumentException if any argument to open(), isOpen(), or isFull() is outside its prescribed range. Throw an IllegalArgumentException in the constructor if n ≤ 0.
Performance requirements. The constructor must take time proportional to n2; all instance methods must take constant time plus a constant number of calls to union() and find().
Monte Carlo simulation. To estimate the percolation threshold, consider the following computational experiment:
Initialize all sites to be blocked. Repeat the following until the system percolates: Choose a site uniformly at random among all blocked sites. Open the site. The fraction of sites that are opened when the system percolates provides an estimate of the percolation threshold. For example, if sites are opened in a 20-by-20 lattice according to the snapshots below, then our estimate of the percolation threshold is 204/400 = 0.51 because the system percolates when the 204th site is opened.
Percolation 50 sites
50 open sites Percolation 100 sites 100 open sites Percolation 150 sites 150 open sites Percolation 204 sites 204 open sites
By repeating this computation experiment T times and averaging the results, we obtain a more accurate estimate of the percolation threshold. Let xt be the fraction of open sites in computational experiment t. The sample mean x¯¯¯ provides an estimate of the percolation threshold; the sample standard deviation s; measures the sharpness of the threshold.
x¯¯¯=x1+x2+⋯+xTT,s2=(x1−x¯¯¯)2+(x2−x¯¯¯)2+⋯+(xT−x¯¯¯)2T−1 Assuming T is sufficiently large (say, at least 30), the following provides a 95% confidence interval for the percolation threshold: [x¯¯¯−1.96sT−−√,x¯¯¯+1.96sT−−√] To perform a series of computational experiments, create a data type PercolationStats with the following API.
public class PercolationStats {
// perform independent trials on an n-by-n grid
public PercolationStats(int n, int trials)
// sample mean of percolation threshold
public double mean()
// sample standard deviation of percolation threshold
public double stddev()
// low endpoint of 95% confidence interval
public double confidenceLo()
// high endpoint of 95% confidence interval
public double confidenceHi()
// test client (see below) public static void main(String[] args)
} Throw an IllegalArgumentException in the constructor if either n ≤ 0 or trials ≤ 0. Also, include a main() method that takes two command-line arguments n and T, performs T independent computational experiments (discussed above) on an n-by-n grid, and prints the sample mean, sample standard deviation, and the 95% confidence interval for the percolation threshold. Use StdRandom to generate random numbers; use StdStats to compute the sample mean and sample standard deviation.
~/Desktop/percolation> java-algs4 PercolationStats 200 100 mean = 0.5929934999999997 stddev = 0.00876990421552567 95% confidence interval = [0.5912745987737567, 0.5947124012262428]
~/Desktop/percolation> java-algs4 PercolationStats 200 100 mean = 0.592877 stddev = 0.009990523717073799 95% confidence interval = [0.5909188573514536, 0.5948351426485464]
~/Desktop/percolation> java-algs4 PercolationStats 2 10000 mean = 0.666925 stddev = 0.11776536521033558 95% confidence interval = [0.6646167988418774, 0.6692332011581226]
~/Desktop/percolation> java-algs4 PercolationStats 2 100000 mean = 0.6669475 stddev = 0.11775205263262094 95% confidence interval = [0.666217665216461, 0.6676773347835391]