猜测同时运行的线程的下一个位置

Question

I am making a maze and two threads are moving on it simultaneously.The problem they mustn't share the same position at same time. I don't know how I can do it. Is it possible to know where they will move next and prevent the other thread from moving to that position? Please give me an idea. Thanks.

Here the code :

        public int[][] visitedCell=new int[15][15]; // holds how many times visited for each cell
    public boolean cntrl=true,repeat=true,end;
    public int r=0;
    public int cnt;
    public boolean find;
    public void  decision(int posX,int posY){
       int distanceToExit;
       cnt=0; // holds how many cell is free around the cell which thread is on
       r=0;
       end=false; // checks robot found exit
       find=false; // checks suitable cell found to move in next step
       posX=posX/40; // all cells are 40*40 dimension.since it is divided 40 to find poisition
       posY=posY/40; // found y position
       int[][] neighbours={{posX,posY+1},{posX+1,posY+1},{posX+1,posY},{posX+1,posY-1},{posX,posY-1},{posX-1,posY-1},{posX-1,posY},{posX-1,posY+1}}; // all 8 neighbours of a cell
       int[][] freeCellChoises = new int[8][2]; // holds free cells to move
       int[][] distanceCell=new int[8][2];
       for(int i=0;i<8;i++){ // checks which neighbour cells are free
             if((neighbours[i][0] >0 && neighbours[i][0] <14) && (neighbours[i][1] >0 && neighbours[i][1] < 14) || (neighbours[i][0]==1 && neighbours[i][1]==14) || (neighbours[i][0]==14 && neighbours[i][1]==1) ) // [1,14] = enter position and [14,1]= exit position
             {
               if(Draw.paintArray[neighbours[i][0]][neighbours[i][1]]==0){// cell is free.it is eligible
                   freeCellChoises[cnt][0]=neighbours[i][0]; // load eligible cells this array
                   freeCellChoises[cnt][1]=neighbours[i][1];
                   distanceToExit=(int) Math.sqrt((Math.pow(neighbours[i][0]-560, 2)+Math.pow(neighbours[i][1]-40,2)));
                   distanceCell[cnt][0]=cnt;
                   distanceCell[cnt][1]=distanceToExit;
                   cnt++;}
             }
       } // eligible cells are ready anymore
       if(Frame.radButSel==1){ // random movement
           int no=rndm.nextInt(cnt); // choose one of the eligible cell randomly
           x=freeCellChoises[no][0] * 40;
           y=freeCellChoises[no][1] * 40;
       }
       if(Frame.radButSel==2){ // closed way movement ( find the most clodes cell to the exit ) .Exit is [14,1].So x must be max, y must be min to a cell has priority
           int maxX=freeCellChoises[0][0];
           int minY=freeCellChoises[0][1];
           int selection1=0,selection2=0;
           for(int i=0;i<cnt;i++){ // x i byk y si kck  sec
               if(freeCellChoises[i][0]> maxX){
                   maxX=freeCellChoises[i][0];
                   selection1=i;}
               if(freeCellChoises[i][1]<minY){
                   minY=freeCellChoises[i][1];
                   selection2=i;
               }


           }
         if(cnt!=0) // checks there is a priority cell
             r=rndm.nextInt(2)+1; // selects one of the priority cell

         if(r==1 && visitedCell[freeCellChoises[selection1][0]][freeCellChoises[selection1][1]] <2){ //selection1.same cell musnt be visited more than 2 times
               x=freeCellChoises[selection1][0] * 40;
               y=freeCellChoises[selection1][1] * 40;}

         else if(r==2 && visitedCell[freeCellChoises[selection2][0]][freeCellChoises[selection2][1]] <2){//selection2
              x=freeCellChoises[selection2][0] * 40;
              y=freeCellChoises[selection2][1] * 40;}

          else{ // applies when there is not any priority cell
               System.out.println("oncelik yok");
               int repeat =0;
               while(repeat<cnt){
                  r=rndm.nextInt(cnt); // choose one of the eligible cell
                  x=freeCellChoises[r][0] * 40;
                  y=freeCellChoises[r][1] * 40;
                  if(visitedCell[freeCellChoises[r][0]][freeCellChoises[r][1]] <2){
                      repeat=10;
                  }
                  else
                      repeat++;

               }System.out.println("x="+x+"y="+y);

          }
      }
      if(Frame.radButSel==3){

      }

      if(x==560 && y==40){ // checks decided cell is exit point
           Action.pool.shutdownNow();// thread finished
           end=true;
           Main.butAct++; // when butAct=2 , "RESULT" button will be active
           timer.stopTime();} // stops time for the thread

      distance=(int) Math.sqrt(Math.pow(x-560,2) + Math.pow(y-40, 2));// calculates distance between thread - exit

}

public Action() throws InterruptedException{
    pool=Executors.newFixedThreadPool(2); // two thread in the pool
    robot1=new Robot(40,560); // starts enter position
    robot2=new Robot(40,560); // starts enter position
    pool.submit(robot1); // loads robot1 to pool
    pool.submit(robot2);// loadss robot2 to pool
}
public void run() {
    while(true){ // run threads always
    try {
        Frame.worker.pauseIfNeeded();} // checks whether pause button is pressed
    catch (InterruptedException ex) {
        Logger.getLogger(Robot.class.getName()).log(Level.SEVERE, null, ex);}
    if(end==false){// not reach exit
       try{
            System.out.println(Thread.currentThread().getName());// displays current thread name
            System.out.println("pozisyon x="+x+"y="+y);
            decision(x,y); // thread makes decision to move
            visitedCell[x/40][y/40]+=1; // increade number of visitide cell count for that cell in the array
            visCell++; //increase visited cell count for the thread
            Thread.sleep(300);} // thread sleeps for a while to observe movement changing
       catch(Exception ex){
       }
    }
    else{// found exit
            Thread.currentThread().interrupt(); // Thread killed
            if(Main.butAct==2)// after a thread found exit, checks if result button can be active anymore
               Frame.button4.setEnabled(true); // activates result button
    }
  }//end while

}

Answer 1

The most simple solution would be to "divide" the work in such a way that the two work regions (solutions/paths to be tried out) are mutually exclusive. A workaround would be to have a concurrent set of positions which would be checked by each thread before making a move.

Answer 2

You will have to use the concept of so called mutual exclusion. In the Java programming language, you will have to use the synchronized keyword to do the job for you. A simple example maybe seen at [1]:

public class SynchronizedCounter {
  private int i = 0;

  public synchronized void increment() {
    i++;
  }

  public synchronized void decrement() {
    i--;
  }

  public synchronized int value() {
    return i;
  }
}

Here you see a code that makes sure that only one thread is able to modify the value of the shared variable i . Note that the code uses this as the so-called "lock object". You may rewrite the code in the following way:

public class SynchronizedCounter {
  private int i = 0;

  public void increment() {
    synchronized(this) {
      i++;
    }
  }

  public void decrement() {
    synchronized(this) {
      i--;
    }
  }

  public int value() {
    synchronized(this) {
      return i;
    }
  }
}

Or you may want to create your own lock object and use it:

public class SynchronizedCounter {
  private int i = 0;
  private Object lock = new Object();

  public void increment() {
    synchronized(lock) {
      i++;
    }
  }

  public void decrement() {
    synchronized(lock) {
      i--;
    }
  }

  public int value() {
    synchronized(lock) {
      return i;
    }
  }
}

Now any number of threads may call the methods of this object randomly, but only one thread at one time will be able to pass the lock and do the actual modification.

[1] http://docs.oracle.com/javase/tutorial/essential/concurrency/syncmeth.html

Answer 3

Or you could do it something like this. The manager class is aware of the position of both threads, and the moveTo method checks that they don't coincide at the same location.

class MazeManager {
  int x1, x2, y1, y2;

  public synchronized boolean moveTo(int threadId, int x, int y) {
    ..
  }
}

Answer 4

Have the class that manages the maze force the threads to synchronize through itself, and make requests to change their position. The maze manager can then allow or deny requests to move based on the position of other threads.

Answer 5

Like Sanjay says, prevention is better than cure. You can partition the cells traversed by the two robots such that they never share the same cell. If that is not possible, the cure is to use locks for each of the cells. Robots obtain the corresponding lock before moving into the cell and release the lock when done. This will cause a robot to wait if it is trying to enter the same cell as another robot. This is a more decentralized approach than having the maze manager control the moves and will result in less contention.