132.1.001 Union-Find | 并查集

@(132 - ACM | 算法)

Algorithm | Coursera - by Robert Sedgewick

> Tip: Focus on WHAT is really important!
> Don't just copy it!
> Don't look at the subtitle
> Practice is the key. Just Do it!

Backup

• Coursera - Algorithms
• booksite

P.S. iff == if and only if

0 Introduction

• Dynamic connectivity problem
• union-find data type
• quick find
• quick union
• improvements
  • weighted quick union
  • weighted quick union with path compression
• applications
  • Maze Problem

1 Steps to developing a usable algorithm

• Model the problem
• Find a algorithm
• Fast enough? Fits in memory?
• if not, figure out why
• Find a way to address the problem
• Iterate until satisfied

2 Quick Find | 快速查找

Structure - Linear

Java implementation

public class QuickFindUF
{
	private int[] id;

	//constructor
	public QuickFindUF(int N)
	{
		id = new int[N];//allocate N*int
		for (int i = 0;i<N;i++)
			id[i] = i;
	}

	public boolean connected(int p,int q)
	{
		return id[p] == id[q];
	}

	public void union(int p, int q)
	{
		int pid = id[p];
		int qid = id[q];
		for (int i = 0;i<id.length;i++)
		{
			if(id[i]==pid) id[i] = qid;
		}
	}
}

Quick find is too slow

3 Quick Union

Structure-Tree

• Check if they have the same root
  
  inspire: use the third standard as Reference//第三方标准作为参照物，语言同理
  
  

Java implementation

public class QuickUnionUF
{
	private int[] id;
	//constructor —— set each element to be its own root
	public QuickUnionUF(int N)
	{
		id = new int[N];
		for (int i = 0;i < N;i++) id[i] = i;
	}
	//find the root by chasing parent pointers
	private int root(int i)
	{
		while (i != id[i] i = id[i]);
		return i;
	}

	public boolean connected(int p, int q)
	{
		return root(p) == root(q);
	}
	public void union(int p,int q)
	{
		int i = root(p);
		int j = root(q);
		id[i] = j;
	}
}

Quick Union is also too slow

4 Quik-Union Improvement1 -Weighted quick-union

smaller tree down below - small depends on the bigger( size)

demo

improvement

Java implementation

//Data Structure
//maintain extra array sz[i] to count number of objects in the tree rooted at i
//sz[i] = size of tree rooted i

// Find - identical to quick-union

//Union
//Modify quick-union to:
//1.Link root of smaller tree to root of larger tree.
//2.Update the sz[] array
	int i = root(p);
	int j = root(q);
	if (i == j) return;
	if (sz[i] < sz[j]) { id[i] = j; sz[j] += sz[i]};
	else               { id[j] = i; sz[i] += sz[j]};

Runing time

O(N) = lg N

---

5 Quik-Union Improvement2 -Path compression

Flatten the tree

In practice - Keeps tree almost completely flat.

---

java implementation

• Make every other node in path point to its grandparent

private int root(int i)
{
	while(i != id[i])
	{
		id[i] = id[id[i]];//only one extra line of code!
		i = id[i];
	}
	return i;
}

O(N) = N+M * lgN

6 Summary for solving the dynamic connectivity problem

7 Union-Find Application

Percolation

Monte Carlo simulation //蒙特卡罗模拟

Dynamic connectivity solution to estimate percolation threshold

• Clever Trick
  
  

8 Application - Percolation | 渗滤问题

需要注意的是Timing和Backwash的问题

• Timing：PercolationStats.java里StdRandom.mean()和StdRandom.stddev()都只能调用一次；Percolation.java里实现numberOfOpenSites时切记不能使用循环累加，定义一个私有属性来计数即可；实现open()时相邻的四个sites位置直接加减n或1即可。

• Backwash:实现isFull()时需要另外实例化一个不包含最下端虚拟节点的WeightedQuickUnionUF，可以解决Test 13: check for backwash with predetermined sites，Test 14: check for backwash with predetermined sites that have multiple percolating paths和Test 15: call all methods in random order until all sites are open, allowing isOpen() to be called on a site more than once

三项测试无法通过的问题。Backwash问题是指因为虚拟底部结点的存在，导致底部任一结点渗漏成功的话底部所有结点都会认为渗漏成功。原因是通过底部虚拟结点形成了回流。从而导致isFull()方法出错。

参考链接

• Percolation.java

import edu.princeton.cs.algs4.WeightedQuickUnionUF;

public class Percolation {

	private boolean[] op; // true=open while false=blocked
	private int side; // number of rows or columns
	private int numOp; // number of open sites
	private WeightedQuickUnionUF uf;
	private WeightedQuickUnionUF ufTop;

	public Percolation(int n) {

		if(n <= 0) throw new IllegalArgumentException("Input should be positif!\n");

		this.side = n;
		this.op = new boolean[n*n+2]; // with 2 virtual sites
		this.uf = new WeightedQuickUnionUF(n*n+2);
		this.ufTop = new WeightedQuickUnionUF(n*n+1); // with only the upper virtual site

		for(int i=1; i<n*n+1; i++) op[i] = false;
		op[0] = op[n*n+1] = true;
		this.numOp = 0;

	}

	// both ROW and COL should be integer within 1~n
	private void checkBounds(int row, int col){
		if(row < 1 || row > this.side || col < 1 || col > this.side){
			throw new IllegalArgumentException("Index out of bounds!\n");
		}
	}

	// get position of sites in 3 arrays: op, uf.parent & uf.size
	private int getPosition(int row, int col){
		return (row - 1) * this.side + col;
	}

	private void union(int aPos, int bPos, WeightedQuickUnionUF wq){
		if(!wq.connected(aPos, bPos)){
			wq.union(aPos, bPos);
		}
	}

	private boolean isOpen(int pos){
		return op[pos];
	}

	public void open(int row, int col) {

		checkBounds(row, col);
		if(isOpen(row, col)) return;

		int pos = getPosition(row, col);
		op[pos] = true;
		numOp++;

		// positions of adjacent sites
		int rowPrev = pos - side, rowNext = pos + side,
				colPrev = pos - 1, colNext = pos + 1;

		// try connect the adjacent open sites
		if(row == 1){
			union(0, pos, uf);
			union(0, pos, ufTop);
		}else if(isOpen(rowPrev)){
			union(rowPrev, pos, uf);
			union(rowPrev, pos, ufTop);
		}

		if(row == side){
			union(side * side + 1, pos, uf);
		}else if(isOpen(rowNext)){
			union(rowNext, pos, uf);
			union(rowNext, pos, ufTop);
		}

		if(col != 1 && isOpen(colPrev)) {
			union(colPrev, pos, uf);
			union(colPrev, pos, ufTop);
		}

		if(col != side && isOpen(colNext)) {
			union(colNext, pos, uf);
			union(colNext, pos, ufTop);
		}
	}

	public boolean isOpen(int row, int col) {
		checkBounds(row, col);
		return isOpen(getPosition(row, col));

	}

	/**
	 * check for backwash with predetermined sites that have multiple percolating paths
	 * in this case ufTop should be used instead of uf
	 * @param row
	 * @param col
	 * @return
	 */
	public boolean isFull(int row, int col) {
		checkBounds(row, col);
		//return uf.connected(0, getPosition(row, col)); -> didn't pass the test!
		return ufTop.connected(0, getPosition(row, col));

	}

	// should pass the timing check
	public int numberOfOpenSites(){
		return this.numOp;
	}

	public boolean percolates(){
		return uf.connected(0, side * side + 1);
	}

}

• PercolationStats.java

import edu.princeton.cs.algs4.StdIn;
import edu.princeton.cs.algs4.StdOut;
import edu.princeton.cs.algs4.StdRandom;
import edu.princeton.cs.algs4.StdStats;
import edu.princeton.cs.algs4.Stopwatch;

public class PercolationStats {

	private double[] results; // estimated threshold for each trial
	private double avg;
	private double std;

	public PercolationStats(int n, int trials){

		if(n <= 0 || trials <= 0) throw new IllegalArgumentException();

		results = new double[trials];
		for(int i = 0; i < trials; i++){
			int step = 0;
			Percolation pr = new Percolation(n);
			while(!pr.percolates()){
				int row = StdRandom.uniform(n) + 1;
				int col = StdRandom.uniform(n) + 1;
				if(!pr.isOpen(row, col)){
					pr.open(row, col);
					step++;
				}
			}
			results[i] = (double)step / (n * n);
		}

		this.avg = StdStats.mean(results);
		this.std = StdStats.stddev(results);

	}

	public static void main(String[] args){

		StdOut.printf("%-25s\n", "Please input 2 integers");
		int N = StdIn.readInt();
		int T = StdIn.readInt();

		Stopwatch wt = new Stopwatch();

		PercolationStats ps = new PercolationStats(N, T);

		// elapsed CPU time in seconds
		double elapsed = wt.elapsedTime();

		StdOut.printf("%-25s= %.15f\n", "elapsed CPU time", elapsed);
		StdOut.printf("%-25s= %.7f\n", "mean", ps.mean());
		StdOut.printf("%-25s= %.17f\n", "stddev", ps.stddev());
		StdOut.printf("%-25s= [%.15f, %.15f]\n", "%95 confidence interval",
				ps.confidenceLo(), ps.confidenceHi());
	}

	public double mean(){
		return this.avg;
	}

	public double stddev(){
		return this.std;
	}

	public double confidenceLo(){
		return mean() - 1.96 * stddev() / Math.sqrt(results.length);
	}

	public double confidenceHi(){
		return mean() + 1.96 * stddev() / Math.sqrt(results.length);
	}

}