首页 > 代码库 > Coursera Algorithms Programming Assignment 4: 8 Puzzle (100分)
Coursera Algorithms Programming Assignment 4: 8 Puzzle (100分)
题目原文:http://coursera.cs.princeton.edu/algs4/assignments/8puzzle.html
题目要求:设计一个程序解决8 puzzle问题以及该问题的推广,例如8-puzzle是3*3,程序要能解决n*n的同类问题(2 ≤ n < 128)
典型的8 puzzle如下:
算法设计参照A*搜索算法,即使不了解A*搜索算法,题目也已经将解法解释的很具体了。
Best-first search:设计参照A* 搜索算法。定义一个 search node类,包含board,从初始状态到达当前board状态的移动权重moves,和previous search node。
- 插入初始的search node,其board设为初始board,moves设为0,previous search node设置为0
- 将初始化的search node置于MinPQ类型的优先级队列中
- 删除优先级队列中的min节点,再将该节点的邻居节点放入优先级队列中。
- 重复2和3操作直至从优先级队列中删除的min节点是目标board
A* 搜索算法的优先级判定依据是f(n) = g(n) + h(n),g(n)是从初始节点到达当前节点的代价,h(n)是当前节点到目标节点的预估代价。
在本题中search node中的moves就是g(n),而关于h(n)题目给出了两种候选:
Hamming priority function: 处于错误位置的block的个数(空白处不算block)
Manhatten priority function: 处于错误位置的block距离其各自目标位置的横向和纵向距离之和
h(n)采用这两者均可,根据题目中的图示,显然发现题目推荐采用manhatten方法。
至此优先级队列中的优先级判断依据就是当前search node的moves+manhatten value
A critical optimization: 上述Best-first search中可能会存在刚出队列的节点又被当成其邻居节点的邻居而被放回优先级队列的情况,这种情况会造成很大的性能损失。为了阻止这种情况的发生,可以在Best-first search的第3步“将该节点的邻居节点放入优先级队列”时比较下这个邻居节点的board是否与本节点的board相同。
例如此时{{8,1,3},{4,0,2},{7,6,5}}就不应该放入优先级队列中。
A second optimization: 建议在search node的构造函数中计算其manhattan值,也就是在search node的构造函数中确定其优先级。
Detecting unsolvable puzzles:如果一个board是不可解的,那么随便在该board中选择一对block互换位置,就能将其变为可解的。为此采用同时对board和其互换了一对block的twindboard进行求解,如果board先实现目标解,那么其就是可解的,相反,如果twinboard先实现目标节,那么该board就不可解。
1 import java.util.ArrayList; 2 /** 3 * @author evasean www.cnblogs.com/evasean/ 4 */ 5 public class Board { 6 private static final int BLANK = 0; 7 private final int n; 8 private int[][] blocks; 9 10 public Board(int[][] inBlocks) { 11 // construct a board from an n-by-n array of blocks 12 // (where blocks[i][j] = block in row i, column j) 13 n = inBlocks.length; 14 blocks = new int[n][n]; 15 copy(blocks, inBlocks); 16 } 17 18 private void copy(int[][] toBlocks, int[][] fromBlocks) { 19 for (int row = 0; row < n; row++) 20 for (int col = 0; col < n; col++) 21 toBlocks[row][col] = fromBlocks[row][col]; 22 } 23 24 public int dimension() { 25 // board dimension n 26 return n; 27 } 28 29 private int getRow(int value) { 30 return (value - 1) / n; 31 } 32 33 private int getCol(int value) { 34 return (value - 1) % n; 35 } 36 37 private int getValue(int row, int col) { 38 return row * n + col + 1; 39 } 40 41 public int hamming() { 42 // number of blocks out of place 43 int hamming = 0; 44 for (int row = 0; row < n; row++) 45 for (int col = 0; col < n; col++) 46 if (blocks[row][col] != BLANK && blocks[row][col] != getValue(row, col)) 47 hamming++; 48 return hamming; 49 } 50 51 public int manhattan() { 52 // sum of Manhattan distances between blocks and goal 53 int manhattan = 0; 54 for (int row = 0; row < n; row++) 55 for (int col = 0; col < n; col++) 56 if (blocks[row][col] != BLANK && blocks[row][col] != getValue(row, col)) 57 manhattan += Math.abs(getRow(blocks[row][col]) - row) + Math.abs(getCol(blocks[row][col]) - col); 58 return manhattan; 59 } 60 61 public boolean isGoal() { 62 // is this board the goal board? 63 for (int row = 0; row < n; row++) 64 for (int col = 0; col < n; col++) 65 if (blocks[row][col] != BLANK && blocks[row][col] != getValue(row, col)) 66 return false; 67 return true; 68 } 69 70 public Board twin() { 71 // a board that is obtained by exchanging any pair of blocks 72 Board twinBoard = new Board(blocks); 73 int firRow = 0; 74 int firCol = 0; 75 if (blocks[firRow][firCol] == BLANK) 76 firCol++; 77 for (int row = 0; row < n; row++) { 78 for (int col = 0; col < n; col++) { 79 if (blocks[row][col] != blocks[firRow][firCol] && blocks[row][col] != BLANK) { 80 twinBoard.swap(firRow, firCol, row, col); 81 return twinBoard; 82 } 83 } 84 } 85 return twinBoard; 86 } 87 88 private void swap(int vRow, int vCol, int wRow, int wCol) { 89 int t = blocks[vRow][vCol]; 90 blocks[vRow][vCol] = blocks[wRow][wCol]; 91 blocks[wRow][wCol] = t; 92 } 93 94 public boolean equals(Object y) { 95 // does this board equal y? 96 if (y == null) 97 return false; 98 if (y == this) 99 return true; 100 if (y.getClass().isInstance(this)) { 101 Board yb = (Board) y; 102 if (yb.n != this.n) 103 return false; 104 else { 105 for (int row = 0; row < n; row++) 106 for (int col = 0; col < n; col++) 107 if (yb.blocks[row][col] != blocks[row][col]) 108 return false; 109 return true; 110 } 111 } else 112 return false; 113 } 114 115 public Iterable<Board> neighbors() { 116 // all neighboring boards 117 ArrayList<Board> neighbors = new ArrayList<Board>(); 118 for (int row = 0; row < n; row++) { 119 for (int col = 0; col < n; col++) { 120 if (blocks[row][col] == BLANK) { 121 // 空白的位置分别与上下左右的元素交换一次位置就得到一个邻居board 122 // 与上方元素互换 123 if (row > 0) { 124 Board neighborT = new Board(blocks); 125 neighborT.swap(row, col, row - 1, col); 126 neighbors.add(neighborT); 127 } 128 // 与下方元素互换 129 if (row < n - 1) { 130 Board neighborB = new Board(blocks); 131 neighborB.swap(row, col, row + 1, col); 132 neighbors.add(neighborB); 133 } 134 // 与左边元素互换 135 if (col > 0) { 136 Board neighborL = new Board(blocks); 137 neighborL.swap(row, col, row, col - 1); 138 neighbors.add(neighborL); 139 } 140 // 与右边元素互换 141 if (col < n - 1) { 142 Board neighborR = new Board(blocks); 143 neighborR.swap(row, col, row, col + 1); 144 neighbors.add(neighborR); 145 } 146 } 147 } 148 } 149 return neighbors; 150 } 151 152 public String toString() { 153 // string representation of this board (in the output format specified 154 // below) 155 StringBuilder sb = new StringBuilder(); 156 sb.append(n + "\n"); 157 for (int row = 0; row < n; row++) { 158 for (int col = 0; col < n; col++) { 159 //本来考虑到n<128时元素可能会很大,设置的是%6d,但是提交时不满足校验规则 160 //校验规则要求必须是%2d,很奇怪的校验 161 sb.append(String.format("%2d ", blocks[row][col])); 162 } 163 sb.append("\n"); 164 } 165 return sb.toString(); 166 } 167 168 public static void main(String[] args) { 169 // unit tests (not graded) 170 // int[][] test = { { 0, 1}, {2,3 }}; 171 // Board b = new Board(test); 172 // System.out.println(b); 173 // System.out.println(b.hamming()); 174 // System.out.println(b.manhattan()); 175 } 176 }
1 import edu.princeton.cs.algs4.In; 2 import edu.princeton.cs.algs4.MinPQ; 3 import edu.princeton.cs.algs4.Stack; 4 import edu.princeton.cs.algs4.StdOut; 5 /** 6 * @author evasean www.cnblogs.com/evasean/ 7 */ 8 public class Solver { 9 10 private SearchNode currentNode; 11 private SearchNode twincurrentNode; 12 private Stack<Board> solution; 13 14 private class SearchNode implements Comparable<SearchNode>{ 15 public Board board; 16 public int moves; 17 public SearchNode preSearchNode; 18 19 public final int priority; 20 21 public SearchNode(Board inboard, SearchNode inPreSearchNode){ 22 board = inboard; 23 preSearchNode = inPreSearchNode; 24 if(inPreSearchNode == null) moves = 0; 25 else moves = inPreSearchNode.moves + 1; 26 priority = moves + board.manhattan(); 27 } 28 29 @Override 30 public int compareTo(SearchNode o) { 31 return Integer.compare(this.priority, o.priority); 32 } 33 } 34 35 36 public Solver(Board initial) { 37 // find a solution to the initial board (using the A* algorithm) 38 if(initial == null) 39 throw new IllegalArgumentException("Constructor argument Board is null!"); 40 currentNode = new SearchNode(initial,null); 41 twincurrentNode = new SearchNode(initial.twin(),null); 42 MinPQ<SearchNode> priorityQueue = new MinPQ<SearchNode>(); 43 MinPQ<SearchNode> twinPriorityQueue = new MinPQ<SearchNode>(); 44 priorityQueue.insert(currentNode); 45 twinPriorityQueue.insert(twincurrentNode); 46 while(true){ 47 currentNode = priorityQueue.delMin(); 48 if(currentNode.board.isGoal()) break; 49 putNeighBorsIntoPQ(currentNode,priorityQueue); 50 51 twincurrentNode = twinPriorityQueue.delMin(); 52 if(twincurrentNode.board.isGoal()) break; 53 putNeighBorsIntoPQ(twincurrentNode,twinPriorityQueue); 54 } 55 } 56 57 private void putNeighBorsIntoPQ(SearchNode searchNode, MinPQ<SearchNode> pq){ 58 Iterable<Board> neighbors = searchNode.board.neighbors(); 59 for(Board neighbor : neighbors){ 60 //只有在当前搜索节点的邻居们的borad不与当前节点的preSearchNode的borad相同 61 //才将该邻居放入优先队列 62 63 if(searchNode.preSearchNode==null || !neighbor.equals(searchNode.preSearchNode.board)) 64 pq.insert(new SearchNode(neighbor,searchNode)); 65 } 66 } 67 68 public boolean isSolvable() { 69 // is the initial board solvable? 70 return currentNode.board.isGoal(); 71 } 72 73 public int moves() { 74 // min number of moves to solve initial board; -1 if unsolvable 75 if(currentNode.board.isGoal()) 76 return currentNode.moves; 77 else 78 return -1; 79 } 80 81 public Iterable<Board> solution() { 82 // sequence of boards in a shortest solution; null if unsolvable 83 if(currentNode.board.isGoal()){ 84 solution = new Stack<Board>(); 85 SearchNode node = currentNode; 86 while(node != null){ 87 solution.push(node.board); 88 node = node.preSearchNode; 89 } 90 return solution; 91 }else 92 return null; 93 } 94 95 public static void main(String[] args) { 96 // solve a slider puzzle (given below) 97 // create initial board from file 98 // In in = new In(args[0]); 99 In in = new In("8puzzle/puzzle3x3-unsolvable.txt"); //本地测试之用 100 int n = in.readInt(); 101 int[][] blocks = new int[n][n]; 102 for (int i = 0; i < n; i++) 103 for (int j = 0; j < n; j++) 104 blocks[i][j] = in.readInt(); 105 Board initial = new Board(blocks); 106 107 // solve the puzzle 108 Solver solver = new Solver(initial); 109 110 // print solution to standard output 111 if (!solver.isSolvable()) 112 StdOut.println("No solution possible"); 113 else { 114 StdOut.println("Minimum number of moves = " + solver.moves()); 115 for (Board board : solver.solution()) 116 StdOut.println(board); 117 } 118 } 119 }
Coursera Algorithms Programming Assignment 4: 8 Puzzle (100分)