我正在努力寻找随机生成＆＃34; 15拼图的解决方案步骤＆＃34;。那么告诉我哪个是最好的算法来快速解决它。提供我这样做的方法。
我正在制作一个包含4 * 4数组的节点树，并遍历尚未处理的所有节点，当我得到解决方案时，我停止迭代。
在viewcontroller中我有一些代码
- (IBAction)getSolution:(id)sender {
while (!appDelegate.isResultFound) {
TreeNode *node=[self nodeWithLowestCostAndUnproceessedInRootNode];
[node expandNodeToChilds];
//break;
}
NSLog(@"Result Found");
if([appDelegate.result isEqualToString:@""])
NSLog(@"No move required");
else
NSLog(@"%@",appDelegate.result);
}
-(TreeNode*)nodeWithLowestCostAndUnproceessedInRootNode{
TreeNode *node1;
int lowestCost=200;
for (TreeNode *node in appDelegate.treeNodes) {
if([node myHeuristicsFunction]<lowestCost&&node.isProcessed==NO){
node1=node;
lowestCost=[node.cost intValue];
}
}
return node1;}
并且在节点类中我将节点扩展为（除了父节点使用的移动）
-(void)expandNodeToChilds{
[self checkMovesForEmptyPlace];
if(top.x>=0){
[self addPuzzleBoxToTreeBySwapingPoint:top withMove:@"Bottom"];
}
if(right.y<=3){
[self addPuzzleBoxToTreeBySwapingPoint:right withMove:@"Left"];
}
if(bottom.x<=3){
[self addPuzzleBoxToTreeBySwapingPoint:bottom withMove:@"Top"];
}
if(left.y>=0){
[self addPuzzleBoxToTreeBySwapingPoint:left withMove:@"Right"];
}
self.isProcessed=true;}
目前我正在使用曼哈顿距离A *，但是没有在很长时间内获得结果，应用程序内存增加到1GB并且应用程序崩溃。
答案 0 :(得分：4)
我假设您正在寻找达到this拼图目标的最短路径。您可以将A* algorithm与当前电路板和目标板之间的manhattan distance一起用作成本函数。
following code in Java实现了算法。函数Solver将NxN板的大小作为输入N，然后将相应的N * N个数作为[0，N ^ 2]，给出2d网格中数字的位置。它输出所需的最小移动次数和实际移动次数。 0表示拼图中的空位。
import java.io.InputStreamReader;
import java.util.*;
class Solver{
private int N ;
private int minMoves ;
public static int[] correctRow;
public static int[] correctCol;
private class Node implements Comparable<Node>{
private Board board ;
private int moves ;
private Node prevNode ;
public Node(Board board,int moves,Node prev){
this.board = board ;
this.moves = moves ;
this.prevNode = prev ;
}
public int compareTo(Node that){
int thisPriority = this.moves+this.board.manhattan() ;
int thatPriority = that.moves+that.board.manhattan() ;
if(thisPriority<thatPriority){
return -1 ;
}else if(thisPriority>thatPriority){
return 1 ;
}else{
return 0 ;
}
}
}
private Node lastNode ;
private boolean solvable ;
public Solver(Board initial){
N = initial.dimension() ;
PriorityQueue<Node> pq = new PriorityQueue<Node>() ;
PriorityQueue<Node> pq2 = new PriorityQueue<Node>() ;
pq.add(new Node(initial,0,null)) ;
pq2.add(new Node(initial.twin(),0,null)) ;
while(true){
Node removed = pq.poll();
Node removed2 = pq2.poll();
if(removed.board.isGoal()){
minMoves = removed.moves ;
lastNode = removed ;
solvable = true ;
break ;
}
if(removed2.board.isGoal()){
minMoves = -1 ;
solvable = false ;
break ;
}
Iterable<Board> neighbors = removed.board.neighbors() ;
Iterable<Board> neighbors2 = removed2.board.neighbors() ;
for(Board board : neighbors){
if(removed.prevNode != null && removed.prevNode.board.equals(board) ){
continue ;
}
pq.add(new Node(board,removed.moves+1,removed)) ;
}
for(Board board : neighbors2){
if(removed2.prevNode != null && removed2.prevNode.board.equals(board) ){
continue ;
}
pq2.add(new Node(board,removed2.moves+1,removed2)) ;
}
}
}
public boolean isSolvable(){
return solvable ;
}
public int moves(){
return minMoves ;
}
public Iterable<Board> solution(){
if(!isSolvable()){
return null ;
}
Stack<Board> stack = new Stack<Board>() ;
Node node = lastNode ;
while(true){
if(node == null) break ;
Board board = node.board ;
node = node.prevNode ;
stack.push(board) ;
}
return stack ;
}
static void initCorrectRowsCols(int N){
correctRow = new int[N*N] ;
int z = 0 ;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ){
correctRow[z++] = i ;
}
}
z = 0 ;
correctCol = new int[N*N] ;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ){
correctCol[z++] = j ;
}
}
}
public static void main(String[] args) {
long start = System.currentTimeMillis();
// create initial board from file
Scanner in = new Scanner(new InputStreamReader(System.in));
int N = in.nextInt();
initCorrectRowsCols(N);
int[][] blocks = new int[N][N];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
blocks[i][j] = in.nextInt();
Board initial = new Board(blocks);
// solve the puzzle
Solver solver = new Solver(initial);
long end = System.currentTimeMillis();
System.out.println("time taken " + (end-start) + " milli seconds");
// print solution to standard output
if (!solver.isSolvable())
System.out.println("No solution possible");
else {
System.out.println("Minimum number of moves = " + solver.moves());
Stack<Board> stack = new Stack<Board>();
for (Board board : solver.solution())
stack.push(board);
while(!stack.isEmpty()){
System.out.println(stack.pop());
}
}
}
}
class Board{
private int[][] array ;
private int N ;
int emptyRow;
int emptyCol;
boolean reached;
int manhattan = 0;
public Board(int[][] blocks){
N = blocks.length ;
array = new int[N][N] ;
reached = true;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ) {
array[i][j] = blocks[i][j] ;
if(array[i][j] == 0){
emptyRow = i;
emptyCol = j;
}
if(array[i][j] != N*i + j+1){
if(!(i==N-1 && j==N-1)){
reached = false;
}
}
int num = array[i][j] ;
if(num==0){
continue ;
}
int indManhattan = Math.abs(Solver.correctRow[num-1] - i)
+ Math.abs(Solver.correctCol[num-1]-j) ;
manhattan += indManhattan ;
}
}
}
public int dimension(){
return N ;
}
public int hamming(){
int outOfPlace = 0 ;
for(int i = 0 ; i < N ; i++ ) {
for(int j = 0 ; j < N ; j++ ){
if(i==N-1 && j==N-1) {
break ;
}
if(array[i][j] != i*N+j+1){
outOfPlace++ ;
}
}
}
return outOfPlace ;
}
public int manhattan(){
return manhattan ;
}
public boolean isGoal(){
return reached ;
}
public Board twin(){
int[][] newArray = new int[N][N] ;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ){
newArray[i][j] = array[i][j] ;
}
}
for(int i = 0 ; i < 2 ; i++ ) {
if(newArray[i][0]==0 || newArray[i][5]==0){
continue ;
}
int temp = newArray[i][0] ;
newArray[i][0] = newArray[i][6] ;
newArray[i][7] = temp ;
break ;
}
return new Board(newArray) ;
}
public boolean equals(Object y){
if(y==this){
return true ;
}
if(y == null){
return false ;
}
if(y.getClass() != this.getClass()){
return false ;
}
Board that = (Board)y ;
if(that.array.length != this.array.length){
return false ;
}
for(int i = 0 ; i < N ; i++ ) {
for(int j = 0 ; j < N ; j++ ) {
if(that.array[i][j] != this.array[i][j] ){
return false ;
}
}
}
return true ;
}
public Iterable<Board> neighbors(){
Queue<Board> q = new ArrayDeque<Board>() ;
int firstIndex0 = 0 ;
int secondIndex0 = 0 ;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ) {
if(array[i][j] == 0){
firstIndex0 = i ;
secondIndex0 = j ;
break ;
}
}
}
if(secondIndex0-1>-1){
int[][] newArr = getCopy() ;
exch(newArr,firstIndex0,secondIndex0,firstIndex0,secondIndex0-1) ;
q.add(new Board(newArr)) ;
}
if(secondIndex0+1<N){
int[][] newArr = getCopy() ;
exch(newArr,firstIndex0,secondIndex0,firstIndex0,secondIndex0+1) ;
q.add(new Board(newArr)) ;
}
if(firstIndex0-1>-1){
int[][] newArr = getCopy() ;
exch(newArr,firstIndex0,secondIndex0,firstIndex0-1,secondIndex0) ;
q.add(new Board(newArr)) ;
}
if(firstIndex0+1<N){
int[][] newArr = getCopy() ;
exch(newArr,firstIndex0,secondIndex0,firstIndex0+1,secondIndex0) ;
q.add(new Board(newArr)) ;
}
return q ;
}
private int[][] getCopy(){
int[][] copy = new int[N][N] ;
for(int i = 0 ; i < N ; i++ ) {
for(int j = 0 ; j < N ; j++ ){
copy[i][j] = array[i][j] ;
}
}
return copy ;
}
private void exch(int[][] arr, int firstIndex,int secIndex,int firstIndex2,int secIndex2){
int temp = arr[firstIndex][secIndex] ;
arr[firstIndex][secIndex] = arr[firstIndex2][secIndex2] ;
arr[firstIndex2][secIndex2] = temp ;
}
public String toString(){
StringBuilder s = new StringBuilder() ;
s.append(N + "\n") ;
for(int i = 0 ; i < N ; i++ ){
for(int j = 0 ; j < N ; j++ ) {
s.append(String.format("%4d",array[i][j])) ;
}
s.append("\n") ;
}
return s.toString() ;
}
}
所以输入
3
7 8 5
4 0 2
3 6 1
算法生成输出
Minimum number of moves = 28
3
7 8 5
4 0 2
3 6 1
3
7 0 5
4 8 2
3 6 1
3
7 5 0
4 8 2
3 6 1
3
7 5 2
4 8 0
3 6 1
3
7 5 2
4 0 8
3 6 1
3
7 5 2
4 6 8
3 0 1
3
7 5 2
4 6 8
3 1 0
3
7 5 2
4 6 0
3 1 8
3
7 5 2
4 0 6
3 1 8
3
7 5 2
0 4 6
3 1 8
3
0 5 2
7 4 6
3 1 8
3
5 0 2
7 4 6
3 1 8
3
5 4 2
7 0 6
3 1 8
3
5 4 2
7 1 6
3 0 8
3
5 4 2
7 1 6
0 3 8
3
5 4 2
0 1 6
7 3 8
3
5 4 2
1 0 6
7 3 8
3
5 0 2
1 4 6
7 3 8
3
0 5 2
1 4 6
7 3 8
3
1 5 2
0 4 6
7 3 8
3
1 5 2
4 0 6
7 3 8
3
1 5 2
4 3 6
7 0 8
3
1 5 2
4 3 6
7 8 0
3
1 5 2
4 3 0
7 8 6
3
1 5 2
4 0 3
7 8 6
3
1 0 2
4 5 3
7 8 6
3
1 2 0
4 5 3
7 8 6
3
1 2 3
4 5 0
7 8 6
3
1 2 3
4 5 6
7 8 0
我还想提一下
因此，虽然我给出的解决方案在大多数输入上运行速度很快，但在其他困难的输入上可能会失败。
另请注意，并非所有谜题都可以解决。对于无法解决的谜题，算法会打印拼图无法解决的问题。
PS。上面实现的算法遵循this programming assignment的指导原则。