跟着学习Algorithm的课程也有好几周了，感觉作业难度还算适中，但要想得满分实在需要一些姿势水平。对算法时间和内存的高阶要求对我这个外行简直吃力，从第三周开始就已经不能100/100了，又苦于不知道向谁求救，很难受。加上自己记性一直不好，一些之前理解很到位的算法隔一段时间不用就只能记个模棱两可，所以决定把作业发到网上（前三周的有时间再补），一方面希望和大家分享交流，听听大家的建议（如果有人看的话[手动捂脸.jpg]），一方面也给自己留个备案，糊涂时就翻出来看看。（希望没有违反honor code...）
1.Specification
8 Puzzle
Write a program to solve the 8-puzzle problem (and its natural generalizations) using the A* search algorithm.
The problem. The 8-puzzle problem is a puzzle invented and popularized by Noyes Palmer Chapman in the 1870s. It is played on a 3-by-3 grid with 8 square blocks labeled 1 through 8 and a blank square. Your goal is to rearrange the blocks so that they are in order, using as few moves as possible. You are permitted to slide blocks horizontally or vertically into the blank square. The following shows a sequence of legal moves from an initial board (left) to the goal board (right).
1 3 1 3 1 2 3 1 2 3 1 2 3 4 2 5 => 4 2 5 => 4 5 => 4 5 => 4 5 6 7 8 6 7 8 6 7 8 6 7 8 6 7 8 initial 1 left 2 up 5 left goal
Best-first search. Now, we describe a solution to the problem that illustrates a general artificial intelligence methodology known as the A* search algorithm. We define a search node of the game to be a board, the number of moves made to reach the board, and the previous search node. First, insert the initial search node (the initial board, 0 moves, and a null previous search node) into a priority queue. Then, delete from the priority queue the search node with the minimum priority, and insert onto the priority queue all neighboring search nodes (those that can be reached in one move from the dequeued search node). Repeat this procedure until the search node dequeued corresponds to a goal board. The success of this approach hinges on the choice of priority function for a search node. We consider two priority functions:
- Hamming priority function. The number of blocks in the wrong position, plus the number of moves made so far to get to the search node. Intuitively, a search node with a small number of blocks in the wrong position is close to the goal, and we prefer a search node that have been reached using a small number of moves.
- Manhattan priority function. The sum of the Manhattan distances (sum of the vertical and horizontal distance) from the blocks to their goal positions, plus the number of moves made so far to get to the search node.
8 1 3 1 2 3 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 4 2 4 5 6 ---------------------- ---------------------- 7 6 5 7 8 1 1 0 0 1 1 0 1 1 2 0 0 2 2 0 3 initial goal Hamming = 5 + 0 Manhattan = 10 + 0
We make a key observation: To solve the puzzle from a given search node on the priority queue, the total number of moves we need to make (including those already made) is at least its priority, using either the Hamming or Manhattan priority function. (For Hamming priority, this is true because each block that is out of place must move at least once to reach its goal position. For Manhattan priority, this is true because each block must move its Manhattan distance from its goal position. Note that we do not count the blank square when computing the Hamming or Manhattan priorities.) Consequently, when the goal board is dequeued, we have discovered not only a sequence of moves from the initial board to the goal board, but one that makes the fewest number of moves. (Challenge for the mathematically inclined: prove this fact.)
A critical optimization. Best-first search has one annoying feature: search nodes corresponding to the same board are enqueued on the priority queue many times. To reduce unnecessary exploration of useless search nodes, when considering the neighbors of a search node, don't enqueue a neighbor if its board is the same as the board of the previous search node.
8 1 3 8 1 3 8 1 8 1 3 8 1 3 4 2 4 2 4 2 3 4 2 4 2 5 7 6 5 7 6 5 7 6 5 7 6 5 7 6 previous search node neighbor neighbor neighbor (disallow)
Game tree. One way to view the computation is as a game tree, where each search node is a node in the game tree and the children of a node correspond to its neighboring search nodes. The root of the game tree is the initial search node; the internal nodes have already been processed; the leaf nodes are maintained in a priority queue; at each step, the A* algorithm removes the node with the smallest priority from the priority queue and processes it (by adding its children to both the game tree and the priority queue).
Detecting unsolvable puzzles. Not all initial boards can lead to the goal board by a sequence of legal moves, including the two below:
1 2 3 1 2 3 4 4 5 6 5 6 7 8 8 7 9 10 11 12 13 15 14 unsolvable unsolvableTo detect such situations, use the fact that boards are divided into two equivalence classes with respect to reachability: (i) those that lead to the goal board and (ii) those that lead to the goal board if we modify the initial board by swapping any pair of blocks (the blank square is not a block). (Difficult challenge for the mathematically inclined: prove this fact.) To apply the fact, run the A* algorithm on two puzzle instances—one with the initial board and one with the initial board modified by swapping a pair of blocks—in lockstep (alternating back and forth between exploring search nodes in each of the two game trees). Exactly one of the two will lead to the goal board.
Board and Solver data types. Organize your program by creating an immutable data type Board with the following API:
public class Board {
public Board(int[][] blocks) // construct a board from an n-by-n array of blocks
// (where blocks[i][j] = block in row i, column j)
public int dimension() // board dimension n
public int hamming() // number of blocks out of place
public int manhattan() // sum of Manhattan distances between blocks and goal
public boolean isGoal() // is this board the goal board?
public Board twin() // a board that is obtained by exchanging any pair of blocks
public boolean equals(Object y) // does this board equal y?
public Iterable<Board> neighbors() // all neighboring boards
public String toString() // string representation of this board (in the output format specified below)
public static void main(String[] args) // unit tests (not graded)
}
Corner cases. You may assume that the constructor receives an n-by-n array containing the n2 integers between 0 and n2 − 1, where 0 represents the blank square.
Performance requirements. Your implementation should support all Board methods in time proportional to n2 (or better) in the worst case.
Also, create an immutable data type Solver with the following API:
public class Solver {
public Solver(Board initial) // find a solution to the initial board (using the A* algorithm)
public boolean isSolvable() // is the initial board solvable?
public int moves() // min number of moves to solve initial board; -1 if unsolvable
public Iterable<Board> solution() // sequence of boards in a shortest solution; null if unsolvable
public static void main(String[] args) // solve a slider puzzle (given below)
}
To implement the A* algorithm, you must use MinPQ from algs4.jar for the priority queue(s).
Corner cases. The constructor should throw a java.lang.NullPointerException if passed a null argument.
Solver test client. Use the following test client to read a puzzle from a file (specified as a command-line argument) and print the solution to standard output.
public static void main(String[] args) {
// create initial board from file
In in = new In(args[0]);
int n = in.readInt();
int[][] blocks = new int[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
blocks[i][j] = in.readInt();
Board initial = new Board(blocks);
// solve the puzzle
Solver solver = new Solver(initial);
// print solution to standard output
if (!solver.isSolvable())
StdOut.println("No solution possible");
else {
StdOut.println("Minimum number of moves = " + solver.moves());
for (Board board : solver.solution())
StdOut.println(board);
}
}
Input and output formats. The input and output format for a board is the board dimension n followed by the n-by-n initial board, using 0 to represent the blank square. As an example,
% more puzzle04.txt 3 0 1 3 4 2 5 7 8 6 % java Solver puzzle04.txt Minimum number of moves = 4 3 0 1 3 4 2 5 7 8 6 3 1 0 3 4 2 5 7 8 6 3 1 2 3 4 0 5 7 8 6 3 1 2 3 4 5 0 7 8 6 3 1 2 3 4 5 6 7 8 0
% more puzzle3x3-unsolvable.txt 3 1 2 3 4 5 6 8 7 0 % java Solver puzzle3x3-unsolvable.txt No solution possibleYour program should work correctly for arbitrary n-by-n boards (for any 2 ≤ n < 128), even if it is too slow to solve some of them in a reasonable amount of time.
Deliverables. Submit only the files Board.java and Solver.java (with the Manhattan priority). We will supply algs4.jar. You may not call any library functions other those in java.lang, java.util, andalgs4.jar. You must use MinPQ for the priority queue(s).
2.思路
基本思路specification里已经给的比较详细了，主要就是利用A*算法进行最优路径的选取，中间利用Priority Queue进行高效的排序筛选，数据结构层次为：
- Game Tree
- search node
- Board
Solver.java重点在构造函数，要求实现A*算法。这里需要提一点的是，在使用MinPQ时，需要自己添加Comparator作为Board排优先级的依据。
debug后提交，correctness是没什么问题，memory和timing就很惨淡了，还好老师给了checklist，那就照着改呗。
3. Tips
checklist中给了很多建议，我这里亲测有效的有下面几个：
- 构造Board时用char[] 代替int[][]，道理checklist里面讲的很明白，实现起来思路也不麻烦，就是代码会多一些强制类型转换和/，%。
- toString 方法用StringBuilder，不要用+，看来从python里带出的习惯要改一改了
- Solver.java 中使用一个MinPQ来存储origin Board和twin Board。原理checklist里面没讲，个人感觉可能是两个twins board中有解的那个的priority至少不会比无解的大，这样循环几轮后应该就只会pop有解的board了。没有严格的数学证明做支撑，纯是个人感觉。至于怎么只用一个PQ一开始也没想通，后来搜索到了一位同学的帖子http://blog.csdn.net/liuweiran900217/article/details/19818289#t4，参考着完成了。
4.代码
/**
* Created by zhqch on 2017/2/25.
* Board class for 8puzzle problem in assignment 4: Priority Queue
*/
public class Board {
private char[] blocks;
private int dim;
public Board(int[][] blocks) {
// construct a board from an n-by-n array of blocks
// (where blocks[i][j] = block in row i, column j)s
// save as 1d array to speed up the algorithm
// use char[] instead of int[] to save memory
this.dim = blocks.length;
int n = this.dim * this.dim;
this.blocks = new char[n];
for (int i = 0; i < n; i++) {
this.blocks[i] = (char) blocks[i / dim][i % dim];
}
}
public int dimension() { // board dimension n
return this.dim;
}
public int hamming() { // number of blocks out of place
int hamming = 0;
for (int i = 0; i < this.dim * this.dim; i++) { // don't count blank block
if (blocks[i] != (char) (i + 1) && blocks[i] != (char) 0) {
hamming++;
}
}
return hamming;
}
public int manhattan() { // sum of Manhattan distances between blocks and goal
int manhattan = 0;
for (int i = 0; i < this.dim * this.dim; i++) {
if (blocks[i] != (char) (i + 1) && blocks[i] != (char) 0) { // don't count blank block
int block_val = (int) blocks[i] - 1;
manhattan += Math.abs(block_val / this.dim - i / this.dim) + Math.abs(block_val % this.dim - i % this.dim);
}
}
return manhattan;
}
public boolean isGoal() { // is this board the goal board?
return this.hamming() == 0;
}
public Board twin() { // a board that is obtained by exchanging any pair of blocks
if (blocks[0] == (char) 0 || blocks[1] == (char) 0)
return swap(2, 3);
else
return swap(0, 1);
}
private Board swap(int i, int j) { // swap two blocks of index i and j
int[][] newBlocks = new int[this.dim][this.dim];
for (int k = 0; k < this.dim * this.dim; k++) {
newBlocks[k / this.dim][k % this.dim] = (int) this.blocks[k];
}
int temp = newBlocks[i / this.dim][i % this.dim];
newBlocks[i / this.dim][i % this.dim] = newBlocks[j / this.dim][j % this.dim];
newBlocks[j / this.dim][j % this.dim] = temp;
return new Board(newBlocks);
}
public boolean equals(Object y) { // does this board equal y?
if (y == this) return true;
if (y == null) return false;
if (y.getClass() != this.getClass()) return false;
Board that = (Board) y;
if (that.dimension() != this.dimension()) return false;
for (int i = 0; i < this.dim * this.dim; i++) {
if (this.blocks[i] != that.blocks[i])
return false;
}
return true;
}
public Iterable<Board> neighbors() { // all neighboring boards
Queue<Board> neighborQueue = new Queue<Board>();
int blankPos = 0;
// find the position of blank block
for (int i = 0; i < this.dim * this.dim; i++) {
if (this.blocks[i] == (char) 0) {
blankPos = i;
break;
}
}
if (blankPos / this.dim != 0)
neighborQueue.enqueue(swap(blankPos, blankPos - this.dim));
if (blankPos / this.dim != this.dim - 1)
neighborQueue.enqueue(swap(blankPos, blankPos + this.dim));
if (blankPos % this.dim != 0)
neighborQueue.enqueue(swap(blankPos, blankPos - 1));
if (blankPos % this.dim != this.dim - 1)
neighborQueue.enqueue(swap(blankPos, blankPos + 1));
return neighborQueue;
}
public String toString() { // string representation of this board (in the output format specified below)
StringBuilder output = new StringBuilder(""); //use StringBuilder to save time
output.append(this.dim);
output.append("\n");
for (int i = 0; i < this.dim * this.dim; i++) {
output.append ((int) blocks[i]);
output.append(" ");
if (i % this.dim == this.dim - 1)
output.append("\n");
}
return output.toString();
}
public static void main(String[] args) { // unit tests (not graded)
// int[][] test = {{0, 1, 3},{4, 2, 5},{7 ,8 ,6}};
// Board testBoard2 = new Board(test);
// // read in the board specified in the filename
// In in = new In(args[0]);
// int n = in.readInt();
// int[][] tiles = new int[n][n];
// for (int i = 0; i < n; i++) {
// for (int j = 0; j < n; j++) {
// tiles[i][j] = in.readInt();
// }
// }
// Board testBoard = new Board(tiles);
// StdOut.println(testBoard);
// StdOut.println(testBoard.swap(1, 2));
// StdOut.println(testBoard.twin());
// StdOut.println(testBoard.hamming());
// StdOut.println(testBoard.manhattan());
// StdOut.println(testBoard.isGoal());
// StdOut.println(testBoard.neighbors());
// StdOut.println(testBoard.equals(testBoard2));
}
}import edu.princeton.cs.algs4.*;
import edu.princeton.cs.algs4.Stack;
import java.util.*;
/**
* Created by Cifer Zhang on 2017/2/27.
*/
public class Solver {
private class SearchNode{ // SearchNode for A* algorithm tree
public Board board;
public int moves;
public SearchNode parent;
public boolean isTwin;
public SearchNode(Board board, int moves, SearchNode parent, boolean isTwin) {
this.board = board;
this.moves = moves;
this.parent = parent;
this.isTwin = isTwin;
}
}
private SearchNode sN;
private int moves = 0;
private boolean solvable = false;
// use Comparator out of class, use Comparable in class
private MinPQ<SearchNode> minPQ = new MinPQ<>(new Comparator<SearchNode>(){
public int compare(SearchNode s1, SearchNode s2){
return s1.board.manhattan() + s1.moves - s2.board.manhattan() - s2.moves;
}
});
private Stack<Board> solutionStack = new Stack<>();
public Solver(Board initial) { // find a solution to the initial board (using the A* algorithm)
minPQ.insert(new SearchNode(initial, 0, null, false));
minPQ.insert(new SearchNode(initial.twin(), 0, null, true));
while(true) {
this.sN = minPQ.delMin();
// terminate search when a goal searchNode dequeue
if (sN.board.isGoal()){
if (sN.isTwin)
this.moves = -1;
else {
this.solvable = true;
this.moves = sN.moves;
// trace back its parent boards as solution
solutionStack.push(sN.board);
while (sN.parent != null) {
sN = sN.parent;
solutionStack.push(sN.board);
}
}
break;
}
for (Board neighbor: sN.board.neighbors()){
if (sN.parent == null || !neighbor.equals(sN.parent.board)) // don't enqueue a visited board
minPQ.insert(new SearchNode(neighbor, sN.moves + 1, sN, sN.isTwin));
}
}
}
public boolean isSolvable() { // is the initial board solvable?
return this.solvable;
}
public int moves() { // min number of moves to solve initial board; -1 if unsolvable
return this.moves;
}
public Iterable<Board> solution() { // sequence of boards in a shortest solution; null if unsolvable
if (this.solvable)
return solutionStack;
else
return null;
}
public static void main(String[] args) {// solve a slider puzzle (given below)
// create initial board from file
In in = new In(args[0]);
int n = in.readInt();
int[][] blocks = new int[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
blocks[i][j] = in.readInt();
Board initial = new Board(blocks);
// solve the puzzle
Solver solver = new Solver(initial);
// print solution to standard output
if (!solver.isSolvable())
StdOut.println("No solution possible");
else {
StdOut.println("Minimum number of moves = " + solver.moves());
for (Board board : solver.solution())
StdOut.println(board);
}
}
}
5.结果
最后得分是98/100，有三个memory测试通过无望，也就没再细改了。如果大家有什么好的改进想法还希望多多提出来，先谢过了。
Computing memory of Solver
*-----------------------------------------------------------
Running 3 total tests.
Test 1: memory with puzzle20.txt (must be <= 2.0x reference solution)
- memory of student Solver = 108568 bytes
- memory of reference Solver = 4896 bytes
- student / reference = 22.17
==> FAILED
Test 2: memory with puzzle25.txt (must be <= 2.0x reference solution)
- memory of student Solver = 1387704 bytes
- memory of reference Solver = 6056 bytes
- student / reference = 229.15
==> FAILED
Test 3: memory with puzzle30.txt (must be <= 2.0x reference solution)
- memory of student Solver = 5719512 bytes
- memory of reference Solver = 7216 bytes
- student / reference = 792.62
==> FAILED
Total: 0/3 tests passed!