广度优先搜索实施不工作

时间：2017-04-28 16:19:18

标签： java

我对广度优先搜索算法的实现有问题，我有一个方法，它以0-8的顺序给出一个0到8的整数数组。我还有一个整数m，告诉我哪个数字是空白的。以下是规则：

我得到了一个数字块，例如：

456           
782        
301

并且假设8是空值，我可以用5,7,2和0交换它，因为它们直接在它旁边。我必须使用广度优先搜索来解决这个难题。这是我到目前为止编写的代码：

package application;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Vector;

public class Solution {

    /******************************************
     * Implementation Here
     ***************************************/

    /*
     * Implementation here: you need to implement the Breadth First Search
     * Method
     */
    /* Please refer the instruction document for this function in details */

    public static LinkedHashSet<int[]> OPEN = new LinkedHashSet<int[]>();
        public static HashSet<int[]> CLOSED = new HashSet<int[]>();
    public static boolean STATE = false;
    public static int empty;

    public static void breadthFirstSearch(int[] num, int m, Vector solution1) {
        int statesVisited = 0;
        for(int i : num) {
            if(num[i] == m) {
                empty = i;
            }
        }

        int[] start = num;
        int[] goal = {0,1,2,3,4,5,6,7,8};
        int[] X;
        int[] temp = {};

        OPEN.add(start);

        while (OPEN.isEmpty() == false && STATE == false) {

            X = OPEN.iterator().next();
            OPEN.remove(X);

            int pos = empty; // get position of ZERO or EMPTY SPACE
            if (compareArray(X,goal)) {
                System.out.println("SUCCESS");

                STATE = true;
            } else {
                // generate child nodes
                CLOSED.add(X);

                temp = up(X, pos);
                if (temp != null)
                    OPEN.add(temp);
                temp = left(X, pos);
                if (temp != null)
                    OPEN.add(temp);
                temp = down(X, pos);
                if (temp != null)
                    OPEN.add(temp);
                temp = right(X, pos);
                if (temp != null)
                    OPEN.add(temp);
                if(OPEN.isEmpty()) 
                    System.out.println("Ending loop");
            }
        }

    }
    public static boolean compareArray(int[] a, int[] b) {
        for(int i: a) 
            if(a[i] != b[i])
                return false;

        return true;

    }

    public static int[] up(int[] s, int p) {
        int[] str = s;
        if (p > 3) {
            int temp = str[p-3];
            str[p-3] = str[p];
            str[p] = temp;


        }
        // Eliminates child of X if its on OPEN or CLOSED
        if (!OPEN.contains(str) && CLOSED.contains(str) == false)
            return str;
        else
            return null;
    }


    public static int[] down(int[] s, int p) {
        int[] str = s;
        if (p < 6) {
            int temp = str[p+3];
            str[p+3] = str[p];
            str[p] = temp;

        }

        // Eliminates child of X if its on OPEN or CLOSED
        if (!OPEN.contains(str) && CLOSED.contains(str) == false)
            return str;
        else
            return null;
    }


    public static int[] left(int[] s, int p) {
        int[] str = s;
        if (p != 0 && p != 3 && p != 6) {
            int temp = str[p-1];
            str[p-1] = str[p];
            str[p] = temp;

        }
        // Eliminates child of X if its on OPEN or CLOSED
        if (!OPEN.contains(str) && CLOSED.contains(str) == false)
            return str;
        else
            return null;
    }


    public static int[] right(int[] s, int p) {
        int[] str = s;
        if (p != 2 && p != 5 && p != 8) {
            int temp = str[p+1];
            str[p+1] = str[p];
            str[p] = temp;
        }
        // Eliminates child of X if its on OPEN or CLOSED
        if (!OPEN.contains(str) && CLOSED.contains(str) == false)
            return str;
        else
            return null;
    }
    public static void print(String s) {
        System.out.println(s.substring(0, 3));
        System.out.println(s.substring(3, 6));
        System.out.println(s.substring(6, 9));
        System.out.println();
    }
}

此代码刚刚结束，从未找到答案。也许我做错了什么？请帮忙。

请注意：这是我关于StackOverFlow的第一个问题，所以如果有人有任何批评，请告诉我，我会马上修复它们。

2 个答案:

答案 0 :(得分：2)

首先，你有一个没有做任何事情的参数，Vector solution in：

public static void breadthFirstSearch(int[] num, int m, Vector solution1)

你也传递了你所代表的零元素的位置为m，然后将一个局部变量赋值给那个位置，对我来说似乎有点无意义如果你要去的话就没有必要传递零位置无论如何要搜索它。

更新了广度优先搜索方法：

public static void breadthFirstSearch(int[] num) {
    for (int i : num) {
        if (num[i] == 0) {
            empty = i;
        }
    }

    int[] start = num;
    int[] goal = {1, 2, 3, 4, 5, 6, 7, 8, 0};
    int[] X;
    int[] temp = {};

    OPEN.add(start);

    while (OPEN.isEmpty() == false && STATE == false) {

        X = OPEN.iterator().next();
        OPEN.remove(X);
        int pos = empty; // get position of ZERO or EMPTY SPACE
        if (Arrays.equals(X, goal)) {
            System.out.println("SUCCESS");
            STATE = true;
        } else {
            // generate child nodes
            CLOSED.add(X);
            temp = up(X, pos);
            if (temp != null) {
                OPEN.add(temp);
            }
            temp = left(X, pos);
            if (temp != null) {
                OPEN.add(temp);
            }
            temp = down(X, pos);
            if (temp != null) {
                OPEN.add(temp);
            }
            temp = right(X, pos);
            if (temp != null) {
                OPEN.add(temp);
            }
            if (OPEN.isEmpty()) {
                System.out.println("Ending loop");
            }
        }
    }
}

您的计划的主要问题在于您的移动方法up()，down()，left()，right()。您没有创建数组的完整副本，因此导致对原始数组进行修改。

因此这项任务： int[] str = s;

必须更改为：

   int[] str = new int[s.length];
   System.arraycopy(s, 0, str, 0, s.length);

以下是已完成方法的示例：

public static int[] up(int[] s, int p) {
    int[] str = new int[s.length];
    System.arraycopy(s, 0, str, 0, s.length);

    if (p > 3) {
        int temp = str[p - 3];
        str[p - 3] = str[p];
        str[p] = temp;
    }
    // Eliminates child of X if its on OPEN or CLOSED
    if (!OPEN.contains(str) && !CLOSED.contains(str)) {
        return str;
    } else {
        return null;
    }
}

SIDE NOTE （不是必需的）：

阵列的某些排列不会导致目标状态。此拼图本身可以有9!个配置，但实际上只有9!/2个可以解决。

我写了一个算法来检查拼图的parity，这可以作为一种预处理来完成，我用它来创建随机实例来测试数据。

public  boolean isSolvable(int[] puzzle) {
    boolean parity = true;
    int gridWidth = (int) Math.sqrt(puzzle.length);
    boolean blankRowEven = true; // the row with the blank tile

    for (int i = 0; i < puzzle.length; i++) {
        if (puzzle[i] == 0) { // the blank tile
            blankRowEven = (i / gridWidth) % 2==0;
            continue;
        }
        for (int j = i + 1; j < puzzle.length; j++) {
            if (puzzle[i] > puzzle[j] && puzzle[j] != 0) {
                parity = !parity;
            }
        }
    }

    // even grid with blank on even row; counting from top
    if (gridWidth % 2 == 0 && blankRowEven) { 
        return !parity;
    }
    return parity;
}

对于矢量

您希望能够打印出达到目标状态的路径，我建议您为State设置一个类：

private State previousState;
private int[] current;

public State(int[] current, State previousState) {
    this.current = current;
    this.previousState = previousState
}

public State getPreviouState(){
    return previousState;
}
public int[] getCurrentState(){
    return currentState;
}

然后当你有目标State时，你可以遍历所有以前的状态，看看它所采用的路径。

State current = GOAL;
while(current != null){
    System.out.println(Arrays.toString(current));
    current = current.getPreviousState();
}

答案 1 :(得分：1)

方法up（...）有一个错误：

你有：

i

我猜应该是：

.iter.1
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