8难题解决方案无限执行
督瑞
问题内容：
问题答案：
我正在寻找使用的8难题问题的解决方案A* Algorithm。我在互联网上找到了 这个 项目。请查看文件-
proj1和EightPuzzle。proj1包含程序（main()函数）的入口点，EightPuzzle描述拼图的特定状态。每个状态都是8拼图的对象。
我觉得逻辑上没有错。但是对于我尝试的这两个输入，它永远循环：{8,2,7,5,1,6,3,0,4}和{3,1,6,8,4,5,7,2,0}。它们都是有效的输入状态。代码有什么问题？
注意
- 为了更好地查看，请在Notepad ++或其他文本编辑器（具有识别Java源文件的功能）中复制代码，因为代码中包含很多注释。
- 由于A *需要启发式，因此他们提供了使用曼哈顿距离的选项和计算错位瓷砖数量的启发式方法。为了确保首先执行最佳启发式算法，他们实施了
PriorityQueue。该compareTo()函数在EightPuzzle该类中实现。 - 输入到程序可以通过改变的值来改变
p1d在main()的函数proj1的类。 - 我之所以说存在上述两个输入的解决方案，是因为此处的applet 解决了它们。请确保您从小程序的选项中选择8拼图。
EDIT1*
我给了这个输入{0,5,7,6,8,1,2,4,3}。它进行了 26次动作10 seconds并给出了结果。但小程序给出了与结果中有。
EDIT2 在调试我注意到，随着节点的扩展，新的节点，一段时间后，都有一个启发- 作为或。他们似乎从未减少。因此，一段时间后， *24 moves``0.0001 seconds``A*
f_n``11``12``PriorityQueue(openset)具有11或12的试探法。因此，要扩展到哪个节点没有太多选择。最少的是11，最高的是12。这正常吗？
EDIT3
这是无限循环发生的代码片段（在 proj1-astar（）中 ）。 openset 是包含未扩展节点的PriorityQueue，
closeset 是包含扩展节点的LinkedList。
while（openset.size（）> 0）{
EightPuzzle x = openset.peek();
if(x.mapEquals(goal))
{
Stack<EightPuzzle> toDisplay = reconstruct(x);
System.out.println("Printing solution... ");
System.out.println(start.toString());
print(toDisplay);
return;
}
closedset.add(openset.poll());
LinkedList <EightPuzzle> neighbor = x.getChildren();
while(neighbor.size() > 0)
{
EightPuzzle y = neighbor.removeFirst();
if(closedset.contains(y)){
continue;
}
if(!closedset.contains(y)){
openset.add(y);
}
}
}
EDIT4
我已经知道了这个无限循环的原因 。看我的答案。但是执行大约需要 25-30秒 ，这是相当长的时间。A 应该比这快得多。小程序在
0.003秒内 完成此操作。 我将奖励改善性能的赏金。*
为了 快速参考，我在没有注释的情况下粘贴了两个类：
八拼图
import java.util.*;
public class EightPuzzle implements Comparable <Object> {
int[] puzzle = new int[9];
int h_n= 0;
int hueristic_type = 0;
int g_n = 0;
int f_n = 0;
EightPuzzle parent = null;
public EightPuzzle(int[] p, int h_type, int cost)
{
this.puzzle = p;
this.hueristic_type = h_type;
this.h_n = (h_type == 1) ? h1(p) : h2(p);
this.g_n = cost;
this.f_n = h_n + g_n;
}
public int getF_n()
{
return f_n;
}
public void setParent(EightPuzzle input)
{
this.parent = input;
}
public EightPuzzle getParent()
{
return this.parent;
}
public int inversions()
{
/*
* Definition: For any other configuration besides the goal,
* whenever a tile with a greater number on it precedes a
* tile with a smaller number, the two tiles are said to be inverted
*/
int inversion = 0;
for(int i = 0; i < this.puzzle.length; i++ )
{
for(int j = 0; j < i; j++)
{
if(this.puzzle[i] != 0 && this.puzzle[j] != 0)
{
if(this.puzzle[i] < this.puzzle[j])
inversion++;
}
}
}
return inversion;
}
public int h1(int[] list)
// h1 = the number of misplaced tiles
{
int gn = 0;
for(int i = 0; i < list.length; i++)
{
if(list[i] != i && list[i] != 0)
gn++;
}
return gn;
}
public LinkedList<EightPuzzle> getChildren()
{
LinkedList<EightPuzzle> children = new LinkedList<EightPuzzle>();
int loc = 0;
int temparray[] = new int[this.puzzle.length];
EightPuzzle rightP, upP, downP, leftP;
while(this.puzzle[loc] != 0)
{
loc++;
}
if(loc % 3 == 0){
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc + 1];
temparray[loc + 1] = 0;
rightP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
rightP.setParent(this);
children.add(rightP);
}else if(loc % 3 == 1){
//add one child swaps with right
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc + 1];
temparray[loc + 1] = 0;
rightP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
rightP.setParent(this);
children.add(rightP);
//add one child swaps with left
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc - 1];
temparray[loc - 1] = 0;
leftP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
leftP.setParent(this);
children.add(leftP);
}else if(loc % 3 == 2){
// add one child swaps with left
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc - 1];
temparray[loc - 1] = 0;
leftP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
leftP.setParent(this);
children.add(leftP);
}
if(loc / 3 == 0){
//add one child swaps with lower
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc + 3];
temparray[loc + 3] = 0;
downP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
downP.setParent(this);
children.add(downP);
}else if(loc / 3 == 1 ){
//add one child, swap with upper
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc - 3];
temparray[loc - 3] = 0;
upP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
upP.setParent(this);
children.add(upP);
//add one child, swap with lower
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc + 3];
temparray[loc + 3] = 0;
downP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
downP.setParent(this);
children.add(downP);
}else if (loc / 3 == 2 ){
//add one child, swap with upper
temparray = this.puzzle.clone();
temparray[loc] = temparray[loc - 3];
temparray[loc - 3] = 0;
upP = new EightPuzzle(temparray, this.hueristic_type, this.g_n + 1);
upP.setParent(this);
children.add(upP);
}
return children;
}
public int h2(int[] list)
// h2 = the sum of the distances of the tiles from their goal positions
// for each item find its goal position
// calculate how many positions it needs to move to get into that position
{
int gn = 0;
int row = 0;
int col = 0;
for(int i = 0; i < list.length; i++)
{
if(list[i] != 0)
{
row = list[i] / 3;
col = list[i] % 3;
row = Math.abs(row - (i / 3));
col = Math.abs(col - (i % 3));
gn += row;
gn += col;
}
}
return gn;
}
public String toString()
{
String x = "";
for(int i = 0; i < this.puzzle.length; i++){
x += puzzle[i] + " ";
if((i + 1) % 3 == 0)
x += "\n";
}
return x;
}
public int compareTo(Object input) {
if (this.f_n < ((EightPuzzle) input).getF_n())
return -1;
else if (this.f_n > ((EightPuzzle) input).getF_n())
return 1;
return 0;
}
public boolean equals(EightPuzzle test){
if(this.f_n != test.getF_n())
return false;
for(int i = 0 ; i < this.puzzle.length; i++)
{
if(this.puzzle[i] != test.puzzle[i])
return false;
}
return true;
}
public boolean mapEquals(EightPuzzle test){
for(int i = 0 ; i < this.puzzle.length; i++)
{
if(this.puzzle[i] != test.puzzle[i])
return false;
}
return true;
}
}
proj1
import java.util.*;
public class proj1 {
/**
* @param args
*/
public static void main(String[] args) {
int[] p1d = {1, 4, 2, 3, 0, 5, 6, 7, 8};
int hueristic = 2;
EightPuzzle start = new EightPuzzle(p1d, hueristic, 0);
int[] win = { 0, 1, 2,
3, 4, 5,
6, 7, 8};
EightPuzzle goal = new EightPuzzle(win, hueristic, 0);
astar(start, goal);
}
public static void astar(EightPuzzle start, EightPuzzle goal)
{
if(start.inversions() % 2 == 1)
{
System.out.println("Unsolvable");
return;
}
// function A*(start,goal)
// closedset := the empty set // The set of nodes already evaluated.
LinkedList<EightPuzzle> closedset = new LinkedList<EightPuzzle>();
// openset := set containing the initial node // The set of tentative nodes to be evaluated. priority queue
PriorityQueue<EightPuzzle> openset = new PriorityQueue<EightPuzzle>();
openset.add(start);
while(openset.size() > 0){
// x := the node in openset having the lowest f_score[] value
EightPuzzle x = openset.peek();
// if x = goal
if(x.mapEquals(goal))
{
// return reconstruct_path(came_from, came_from[goal])
Stack<EightPuzzle> toDisplay = reconstruct(x);
System.out.println("Printing solution... ");
System.out.println(start.toString());
print(toDisplay);
return;
}
// remove x from openset
// add x to closedset
closedset.add(openset.poll());
LinkedList <EightPuzzle> neighbor = x.getChildren();
// foreach y in neighbor_nodes(x)
while(neighbor.size() > 0)
{
EightPuzzle y = neighbor.removeFirst();
// if y in closedset
if(closedset.contains(y)){
// continue
continue;
}
// tentative_g_score := g_score[x] + dist_between(x,y)
//
// if y not in openset
if(!closedset.contains(y)){
// add y to openset
openset.add(y);
//
}
//
}
//
}
}
public static void print(Stack<EightPuzzle> x)
{
while(!x.isEmpty())
{
EightPuzzle temp = x.pop();
System.out.println(temp.toString());
}
}
public static Stack<EightPuzzle> reconstruct(EightPuzzle winner)
{
Stack<EightPuzzle> correctOutput = new Stack<EightPuzzle>();
while(winner.getParent() != null)
{
correctOutput.add(winner);
winner = winner.getParent();
}
return correctOutput;
}
}
问题答案：
这是一个建议。我的计时器报告您的示例为0毫秒。在此处给出的难度较大的难题中，需要完成31个动作才能完成，耗时96毫秒。
一个HashSet做了闭集比你的链表更有意义。它具有O（1）时间插入和成员资格测试，其中链接列表所需要的时间与列表的长度成正比，并且该长度在不断增长。
您正在使用额外的数据结构和代码，这些数据和代码会使您的程序变得比所需的复杂和缓慢。多思考，少编写代码，研究其他人的好代码以克服这一问题。我的不是完美的（从来没有代码是完美的），但这是一个起点。
我将每个瓷砖的当前位置到目标的最大曼哈顿距离用作启发式方法。启发式的选择不会影响解决方案中的步骤数，但是 会 极大地影响运行时间。例如，h =
0将产生蛮力广度优先搜索。
请注意，为使A *提供最佳解决方案，启发式方法永远不能 高估目标
的实际最小步数。如果这样做，发现的解决方案可能不是最短的。我不太肯定“反转”启发式方法具有此属性。
package eightpuzzle;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.PriorityQueue;
public class EightPuzzle {
// Tiles for successfully completed puzzle.
static final byte [] goalTiles = { 0, 1, 2, 3, 4, 5, 6, 7, 8 };
// A* priority queue.
final PriorityQueue <State> queue = new PriorityQueue<State>(100, new Comparator<State>() {
@Override
public int compare(State a, State b) {
return a.priority() - b.priority();
}
});
// The closed state set.
final HashSet <State> closed = new HashSet <State>();
// State of the puzzle including its priority and chain to start state.
class State {
final byte [] tiles; // Tiles left to right, top to bottom.
final int spaceIndex; // Index of space (zero) in tiles
final int g; // Number of moves from start.
final int h; // Heuristic value (difference from goal)
final State prev; // Previous state in solution chain.
// A* priority function (often called F in books).
int priority() {
return g + h;
}
// Build a start state.
State(byte [] initial) {
tiles = initial;
spaceIndex = index(tiles, 0);
g = 0;
h = heuristic(tiles);
prev = null;
}
// Build a successor to prev by sliding tile from given index.
State(State prev, int slideFromIndex) {
tiles = Arrays.copyOf(prev.tiles, prev.tiles.length);
tiles[prev.spaceIndex] = tiles[slideFromIndex];
tiles[slideFromIndex] = 0;
spaceIndex = slideFromIndex;
g = prev.g + 1;
h = heuristic(tiles);
this.prev = prev;
}
// Return true iif this is the goal state.
boolean isGoal() {
return Arrays.equals(tiles, goalTiles);
}
// Successor states due to south, north, west, and east moves.
State moveS() { return spaceIndex > 2 ? new State(this, spaceIndex - 3) : null; }
State moveN() { return spaceIndex < 6 ? new State(this, spaceIndex + 3) : null; }
State moveE() { return spaceIndex % 3 > 0 ? new State(this, spaceIndex - 1) : null; }
State moveW() { return spaceIndex % 3 < 2 ? new State(this, spaceIndex + 1) : null; }
// Print this state.
void print() {
System.out.println("p = " + priority() + " = g+h = " + g + "+" + h);
for (int i = 0; i < 9; i += 3)
System.out.println(tiles[i] + " " + tiles[i+1] + " " + tiles[i+2]);
}
// Print the solution chain with start state first.
void printAll() {
if (prev != null) prev.printAll();
System.out.println();
print();
}
@Override
public boolean equals(Object obj) {
if (obj instanceof State) {
State other = (State)obj;
return Arrays.equals(tiles, other.tiles);
}
return false;
}
@Override
public int hashCode() {
return Arrays.hashCode(tiles);
}
}
// Add a valid (non-null and not closed) successor to the A* queue.
void addSuccessor(State successor) {
if (successor != null && !closed.contains(successor))
queue.add(successor);
}
// Run the solver.
void solve(byte [] initial) {
queue.clear();
closed.clear();
// Click the stopwatch.
long start = System.currentTimeMillis();
// Add initial state to queue.
queue.add(new State(initial));
while (!queue.isEmpty()) {
// Get the lowest priority state.
State state = queue.poll();
// If it's the goal, we're done.
if (state.isGoal()) {
long elapsed = System.currentTimeMillis() - start;
state.printAll();
System.out.println("elapsed (ms) = " + elapsed);
return;
}
// Make sure we don't revisit this state.
closed.add(state);
// Add successors to the queue.
addSuccessor(state.moveS());
addSuccessor(state.moveN());
addSuccessor(state.moveW());
addSuccessor(state.moveE());
}
}
// Return the index of val in given byte array or -1 if none found.
static int index(byte [] a, int val) {
for (int i = 0; i < a.length; i++)
if (a[i] == val) return i;
return -1;
}
// Return the Manhatten distance between tiles with indices a and b.
static int manhattanDistance(int a, int b) {
return Math.abs(a / 3 - b / 3) + Math.abs(a % 3 - b % 3);
}
// For our A* heuristic, we just use max of Manhatten distances of all tiles.
static int heuristic(byte [] tiles) {
int h = 0;
for (int i = 0; i < tiles.length; i++)
if (tiles[i] != 0)
h = Math.max(h, manhattanDistance(i, tiles[i]));
return h;
}
public static void main(String[] args) {
// This is a harder puzzle than the SO example
byte [] initial = { 8, 0, 6, 5, 4, 7, 2, 3, 1 };
// This is taken from the SO example.
//byte [] initial = { 1, 4, 2, 3, 0, 5, 6, 7, 8 };
new EightPuzzle().solve(initial);
}
}
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