EmptyStackException - 2D数组上的Java深度优先搜索算法

问题描述 投票:-3回答:3

我已经看过这个问题here我从那里尝试了大部分代码示例但是当我在我的代码中使用它时它只是跳过算法。

我有这个使用堆栈的DFS算法,我得到了一个EmptyStackException,我调试了算法,在第一次递归搜索后,堆栈是空的,第一次搜索工作但是然后堆栈的大小设置为0,我在这里缺少什么? github

第一次搜索后,如何确保堆栈不为空?

我得到例外的那一行是while(true){AddBridges state = gameTree.peek(); ...

我正在使用2d数组从0到4随机生成节点0 = null 1-4 = island每次用户启动游戏时,数组都会生成随机整数,这会导致算法制动,

经过一个周末的调试后,我发现算法有时会在4-6次搜索后刹车,有时在第一次搜索后会中断。

public int[][] debug_board_state_easy = new int[4][4];

//This Generates random 2d array
private void InitializeEasy() {
   Random rand = new Random();

   setCurrentState(new State(WIDTH_EASY));
   for (int row = 0; row < debug_board_state_easy.length; row++) {
      for (int column = 0; column < debug_board_state_easy[row].length; column++) {
          debug_board_state_easy[row][column] = Integer.valueOf(rand.nextInt(5));
      }
  }

  for (int row = 0; row < debug_board_state_easy.length; row++) {
     for (int column = 0; column < debug_board_state_easy[row].length; column++) {
        System.out.print(debug_board_state_easy[row][column] + " ");
     }
     System.out.println(debug_board_state_easy);
  }
//I am applying the search algorithm here...
  this.search();

  for (int row = 0; row < WIDTH_EASY; ++row) {
     for (int column = 0; column < WIDTH_EASY; ++column) {
         getCurrentState().board_elements[row][column] = new BoardElement();
         getCurrentState().board_elements[row][column].max_connecting_bridges = Integer.valueOf(debug_board_state_easy[row][column]);
              getCurrentState().board_elements[row][column].row = row;
              getCurrentState().board_elements[row][column].col = column;

         if (getCurrentState().board_elements[row][column].max_connecting_bridges > 0) {
            getCurrentState().board_elements[row][column].is_island = true;
         }
      }
   }
}

void search() {
   Map<Point, List<Direction>> remainingOptions = new HashMap<>();
   Stack<Land> gameTree = new Stack<>();
   gameTree.push(new AddBridges(debug_board_state_easy));

   while(true) {
      AddBridges state = gameTree.peek();
      int[] p = state.lowestTodo();
      if (p == null)
         System.out.println("solution found");
      // move to next game state
         int row = p[0];
         int column = p[1];
         System.out.println("expanding game state for node at (" + row + ", " + column + ")");

         List<Direction> ds = null;
         if(remainingOptions.containsKey(new Point(row,column)))
            ds = remainingOptions.get(new Point(row,column));
         else{
            ds = new ArrayList<>();
            for(Direction dir : Direction.values()) {
               int[] tmp = state.nextIsland(row, column, dir);
               if(tmp == null)
                  continue;
               if(state.canBuildBridge(row,column,tmp[0], tmp[1]))
                  ds.add(dir);
            }
            remainingOptions.put(new Point(row,column), ds);
         }
      // if the node can no longer be expanded, and backtracking is not possible we quit
         if(ds.isEmpty() && gameTree.isEmpty()){
            System.out.println("no valid configuration found");
            return;
         }
      // if the node can no longer be expanded, we need to backtrack
         if(ds.isEmpty()){
            gameTree.pop();
            remainingOptions.remove(new Point(row,column));
            System.out.println("going back to previous decision");
            continue;
         }

         Direction dir = ds.remove(0);
         System.out.println("connecting " + dir.name());
         remainingOptions.put(new Point(row,column), ds);

         AddBridgesnextState = new AddBridges(state);
         int[] tmp = state.nextIsland(row,column,dir);
         nextState.connect(row,column, tmp[0], tmp[1]);
         gameTree.push(nextState);
      }
   }
}

添加桥类

public class AddBridges {

    private int[][] BRIDGES_TO_BUILD;

    private boolean[][] IS_ISLAND;
    private Direction[][] BRIDGES_ALREADY_BUILT;

    public Land(int[][] bridgesToDo){
        BRIDGES_TO_BUILD = copy(bridgesToDo);

        int numberRows = bridgesToDo.length;
        int numberColumns = bridgesToDo[0].length;
        BRIDGES_ALREADY_BUILT = new Direction[numberRows][numberColumns];
        IS_ISLAND = new boolean[numberRows][numberColumns];
        for(int i=0;i<numberRows;i++) {
            for (int j = 0; j < numberColumns; j++) {
                BRIDGES_ALREADY_BUILT[i][j] = null;
                IS_ISLAND[i][j] = bridgesToDo[i][j] > 0;
            }
        }
    }

    public AddBridges (AddBridges other){
        BRIDGES_TO_BUILD = copy(other.BRIDGES_TO_BUILD);
        int numberRows = BRIDGES_TO_BUILD.length;
        int numberColumns =  BRIDGES_TO_BUILD[0].length;
        BRIDGES_ALREADY_BUILT = new Direction[numberRows][numberColumns];
        IS_ISLAND = new boolean[numberRows][numberColumns];
        for(int i=0;i<numberRows;i++) {
            for (int j = 0; j < numberColumns; j++) {
                BRIDGES_ALREADY_BUILT[i][j] = other.BRIDGES_ALREADY_BUILT[i][j];
                IS_ISLAND[i][j] = other.IS_ISLAND[i][j];
            }
        }
    }

    public int[] next(int r, int c, Direction dir){
        int numberRows = BRIDGES_TO_BUILD.length;
        int numberColumns = BRIDGES_TO_BUILD[0].length;

        // out of bounds
        if(r < 0 || r >=numberRows || c < 0 || c >= numberColumns)
            return null;


        // motion vectors
        int[][] motionVector = {{-1, 0},{0,1},{1,0},{0,-1}};
        int i = Arrays.asList(Direction.values()).indexOf(dir);

        // calculate next
        int[] out = new int[]{r + motionVector[i][0], c + motionVector[i][1]};

        r = out[0];
        c = out[1];

        // out of bounds
        if(r < 0 || r >=numberRows || c < 0 || c >= numberColumns)
            return null;

        // return
        return out;
    }

    public int[] nextIsland(int row, int column, Direction dir){
        int[] tmp = next(row,column,dir);
        if(tmp == null)
            return null;
        while(!IS_ISLAND[tmp[0]][tmp[1]]){
            tmp = next(tmp[0], tmp[1], dir);
            if(tmp == null)
                return null;
        }
        return tmp;
    }

    public boolean canBuildBridge(int row0, int column0, int row1, int column1){
        if(row0 == row1 && column0 > column1){
            return canBuildBridge(row0, column1, row1, column0);
        }
        if(column0 == column1 && row0 > row1){
            return canBuildBridge(row1, column0, row0, column1);
        }
        if(row0 == row1){
            int[] tmp = nextIsland(row0, column0, Direction.EAST);
            if(tmp == null)
                return false;
            if(tmp[0] != row1 || tmp[1] != column1)
                return false;
            if(BRIDGES_TO_BUILD[row0][column0] == 0)
                return false;
            if(BRIDGES_TO_BUILD[row1][column1] == 0)
                return false;
            for (int i = column0; i <= column1 ; i++) {
                if(IS_ISLAND[row0][i])
                    continue;
                if(BRIDGES_ALREADY_BUILT[row0][i] == Direction.NORTH)
                    return false;
            }
        }
        if(column0 == column1){
            int[] tmp = nextIsland(row0, column0, Direction.SOUTH);
            if(tmp == null)
                return false;
            if(tmp[0] != row1 || tmp[1] != column1)
                return false;
            if(BRIDGES_TO_BUILD[row0][column0] == 0 || BRIDGES_TO_BUILD[row1][column1] == 0)
                return false;
            for (int i = row0; i <= row1 ; i++) {
                if(IS_ISLAND[i][column0])
                    continue;
                if(BRIDGES_ALREADY_BUILT[i][column0] == Direction.EAST)
                    return false;
            }
        }
        // default
        return true;
    }

    public int[] lowestTodo(){
        int R = BRIDGES_TO_BUILD.length;
        int C = BRIDGES_TO_BUILD[0].length;

        int[] out = {0, 0};
        for (int i=0;i<R;i++) {
            for (int j = 0; j < C; j++) {
                if(BRIDGES_TO_BUILD[i][j] == 0)
                    continue;
                if (BRIDGES_TO_BUILD[out[0]][out[1]] == 0)
                    out = new int[]{i, j};
                if (BRIDGES_TO_BUILD[i][j] < BRIDGES_TO_BUILD[out[0]][out[1]])
                    out = new int[]{i, j};
            }
        }
        if (BRIDGES_TO_BUILD[out[0]][out[1]] == 0) {
            return null;
        }
        return out;
    }

    @TargetApi(Build.VERSION_CODES.GINGERBREAD)
    private int[][] copy(int[][] other){
        int[][] out = new int[other.length][other.length == 0 ? 0 : other[0].length];
        for(int r=0;r<other.length;r++)
            out[r] = Arrays.copyOf(other[r], other[r].length);
        return out;
    }

    public void connect(int r0, int c0, int r1, int c1){
        if(r0 == r1 && c0 > c1){
            connect(r0, c1, r1, c0);
            return;
        }
        if(c0 == c1 && r0 > r1){
            connect(r1, c0, r0, c1);
            return;
        }
        if(!canBuildBridge(r0, c0, r1, c1))
            return;

        BRIDGES_TO_BUILD[r0][c0]--;
        BRIDGES_TO_BUILD[r1][c1]--;

        if(r0 == r1){
            for (int i = c0; i <= c1 ; i++) {
                if(IS_ISLAND[r0][i])
                    continue;
                BRIDGES_ALREADY_BUILT[r0][i] = Direction.EAST;
            }
        }
        if(c0 == c1){
            for (int i = r0; i <= r1 ; i++) {
                if(IS_ISLAND[i][c0])
                    continue;
                BRIDGES_ALREADY_BUILT[i][c0] = Direction.NORTH;
            }
        }
    }
}
java algorithm exception stack depth-first-search
3个回答
4
投票

问题的一部分对我来说是问题的根源:“我在这里缺少什么”?单元测试,除非我在你的项目中没有看到它们。

诸如“每次用户启动游戏时阵列生成随机整数,这会导致算法制动吗?”等问题是单元测试存在的原因,以及以下内容:

  1. 在编写不会成为问题的代码段的测试过程中,您将明确证明它们不是问题所在。
  2. 如果您正在使用的代码无法按原样进行测试,或者“太复杂”无法测试,重写它将使您成为更好的设计师,并且经常会导致“我不相信我没有”看到“时刻。
  3. 当您重构此程序(降低复杂性,重命名变量以使其更易于理解等)时,如果出现问题,您将立即得到通知,并且您不必花费另一个周末来解决问题。

作为一个例子,不是在搜索它的方法中随机化板,而是将其随机化到别处,然后将其作为参数(或者在类的构造函数中)放入该方法中。这样,您可以使用自己提供的值初始化自己的测试板,并查看某些电路板是否比其他电路板工作得更好以及原因。将较大的方法拆分为较小的方法,每个方法都有自己的参数和测试。目的是用较小的确认工件制作一个程序,而不是制作一个巨大的东西,然后想知道问题是你认为它是什么或其他什么的小部分。

从长远来看,你将节省大量的时间和挫折,并且你将在那些编码数小时然后调试数小时的人之前结束联赛。


0
投票

代码中有很多内容,但我注意到的第一件事可能会有所帮助:

// if the node can no longer be expanded, we need to backtrack
            if(ds.isEmpty()){
                gameTree.pop();
                remainingOptions.remove(new Point(row,column));
                System.out.println("going back to previous decision");
                continue;
            }

你从堆栈弹出,并继续下一个while(true)迭代,此时,堆栈上可能没有任何东西,因为你没有添加任何其他内容。


0
投票

我同意@Rosa -

删除或查找空Stack时应发生EmptyStackException-

======代码中的迭代/状态======

 **if(ds.isEmpty()){**  //HashMap isEmpty = true and gameTree.size() = 1
            gameTree.pop();    // After removing element gameTree.size() = 0 (no elements in stack to peek or pop)
            remainingOptions.remove(new Point(row,column));
            System.out.println("going back to previous decision");
            continue;  //Skip all the instruction below this, continue to next iteration
        }

========下一次迭代========

  while(true){
            AddBridges state = gameTree.peek(); // gameTree.size() = 0 and a peek 

操作失败,程序将返回EmptyStackException。

要求是空的检查 -

if(gameTree.isEmpty()){ 
            System.out.println("no valid configuration found");
            return;
        }
        AddBridges state = gameTree.peek(); 

因为,函数没有执行任何操作,而是循环。如果要处理其他实例,则需要“中断”。

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