java - 编程简介 : Conways Game of Life

Question

我已经研究这段代码很长时间了，但我似乎无法弄清楚。我的代码的顶部部分用于预填充网格。但由于某种原因，我无法让我的形象按照预期移动和成长。请帮助我!!

import java.util.Random;
public class Life {
    public static void main(String []args){

        //Declaring and initializing grid variables
        int gridSize = 200;
        int cellSize = 3;
        Grid grid = new Grid(gridSize, cellSize, "The Game of Life");
        grid.setDelay(10);

        Random r = new Random();
        int aliveColor = 1;
        int deadColor= 0;
        int aliveCells;
        int row = 0; 
        int column =  0;
        int val = grid.getPos(row,column);
        int generation;
        int aliveNeighbors;
        int cell;

        //loop statement pre-populating the grid
        for (row = 0; row <= gridSize-1; row++){
            for(column = 0; column <= gridSize-1; column++){
                if (r.nextInt(100) > 49){
                    grid.setPos(row,  column, aliveColor);
                }
                else
                    grid.setPos(row, column, deadColor);
            }
        }
        grid.initialize();


        //Loop executing the rules of the Game of Life
        do
        {
            row = 0;
            column = 0;
            generation = 0;
            while (row <= gridSize-1){
                while (column <= gridSize-1){
                    cell = grid.getPos(row, column);
                    aliveNeighbors = grid.matchingNeighbors(row, column,aliveColor);
                    if (cell == aliveColor)
                    {
                        if (aliveNeighbors == 2 || aliveNeighbors == 3){
                            val =1;
                            grid.setPos(row,  column,val);
                        }
                    }
                    else if (cell == deadColor)
                    {   
                        if (aliveNeighbors == 3){
                            val =1;
                            grid.setPos(row, column,val);
                        }
                    }
                    else{
                        val = 0;
                        grid.setPos(row,column,val);

                    }
                    column++;
                }   
                row++;
                grid.update();
            } 
            grid.update();
            generation++;
        } while (generation >= 0);
    }
}

编辑

//Loop executing the rules of the Game of Life

    do
    {   
        row = 0;
        column = 0;
        generation = 0;
        while (row <= gridSize-1){
            while (column <= gridSize-1){
                cell = grid.getPos(row, column);
                aliveNeighbors = grid.matchingNeighbors(row, column,aliveColor);
                if (cell == aliveColor)
                {
                    if (aliveNeighbors == 2 || aliveNeighbors == 3){
                        val =1;
                        grid.setPos(row,  column,val);
                    }
                    if (aliveNeighbors ==1 || aliveNeighbors == 4){
                        val = 0;
                        grid.setPos(row,column,val);
                    }
                }
                else
                {
                    if (cell == deadColor)
                    {   
                        if (aliveNeighbors == 3){
                            val =1;
                            grid.setPos(row, column,val);
                        }
                        if (aliveNeighbors == 1 || aliveNeighbors == 2 || aliveNeighbors == 4){
                            val = 0;
                            grid.setPos(row, column, val);
                        }
                    }
                }   
                column++;
            }
            row++;
        } 
        grid.update();
        generation++;
    } while (generation >= 0);
}








    EDIT****
      do
    {   
        row = 0;
        column = 0;
        generation = 0;
        while (row <= gridSize-1){
            while (column <= gridSize-1){
                cell = grid.getPos(row, column);
                aliveNeighbors = grid.matchingNeighbors(row, column,aliveColor);
                if (cell == aliveColor)
                {
                    if (aliveNeighbors == 2 || aliveNeighbors == 3){
                        grid.setPos(row, column, aliveColor);
                    } else {
                        grid.setPos(row,column, deadColor); 
                    }
                }
                else
                {
                    if (cell == deadColor){ 
                        if (aliveNeighbors == 3){
                            grid.setPos(row, column,aliveColor);
                        } else {
                            grid.setPos(row,column,deadColor);
                        }
                    }
                }   
                column++;
            }
            row++;
        } 
        grid.update();
        generation++;
    } while (generation >= 0);
}

Answer 1

我注意到的两个问题:

1. 您应该仅在完成所有单元格的更新后更新网格。

2. 您的缩进已关闭，并且 } 放错了位置。重新格式化后，可以清楚地看到您没有覆盖所有情况，因此有些单元格没有获得新值。例如，如果一个单元格是活的，则只有当它有 2 或 3 个邻居时才处理这种情况，但是该 if 没有 else，因此，一个活的单元格具有 1 个或 4 个或更多邻居的单元格永远不会更新。

   您可能会认为在这种情况下代码不知何故会落入一般的“else”，但事实上，它永远不会到达那里。所有单元格值要么是“活”，要么是“死”。因此，构造如下:

   if ( cell == aliveColor ) {
       // Handle alive cell
   } else if ( cell == deadColor ) {
       // Handle dead cell
   } else {
       // Handle all other cases - but there are none!
   }
   

   相当于:

   if ( cell == aliveColor ) {
      // Handle alive cell
   } else {
      // Handle dead cell
   }
   

   最初的“else”永远不会到达，因为除了“aliveColor”和“deadColor”之外没有其他情况。

<小时/>

对您的编辑的回复:

            if (cell == aliveColor)
            {
                if (aliveNeighbors == 2 || aliveNeighbors == 3){
                    val =1;
                    grid.setPos(row,  column,val);
                }
                if (aliveNeighbors ==1 || aliveNeighbors == 4){
                    val = 0;
                    grid.setPos(row,column,val);
                }
            }

出于某种原因，您似乎认为每个单元格最多可以有 4 个邻居。再次查看实验室文档中的示例单元格:

⬜︎⬜︎⬛︎⬛︎⬜︎⬛︎⬛︎⬛︎⬜︎

This cell has 5 dead neighbors, and 3 live ones. In fact, each cell can have up to 8 neighbors - the diagonal neighbors also count!

But your rules were:

1. If x is alive and has exactly 2 or 3 live neighbors, then x survives to the next generation. Otherwise it dies.
2. If x is dead and has exactly 3 live neighbors, then x is alive in the next generation. Otherwise it remains dead

This otherwise translates directly to else in programming. So instead of having a second if with the complementary condition (if there are 1, 4,5,6,7 or 8 live neighbors...), just give the first if an else:

                if (aliveNeighbors == 2 || aliveNeighbors == 3) {
                    grid.setPos(row, column, aliveColor);
                } else {
                    grid.setPos(row, column, deadColor);
                }

当然，同样的事情也适用于死亡细胞的规则。

(请注意，我还删除了 val=1 并直接使用该值，val 是多余的。此外，我还使用了解释该值的名称比 1 和 0 (如果你想将游戏更改为使用绿色和红色怎么办？)。

<小时/>

我之前没有注意到的一件重要的事情:

do
{   
    row = 0;
    column = 0;
    generation = 0;
    while (row <= gridSize-1){
        while (column <= gridSize-1){

在行循环之前将行和列初始化为零。

这意味着在第一行末尾，column 保持 gridSize-1 并且不会再次设置为零。因此，实际上您只更新一行，然后只更新其他每一行的最后一列。

应该是:

  do
{   
    row = 0;
    generation = 0;
    while (row <= gridSize-1){
        column = 0;
        while (column <= gridSize-1){

或者你可以使用 for 循环:

for ( row = 0; row < gridSize; row++ ) {
    for ( column = 0; column < gridSize; column++ ) {
        ...

如果您选择这样做，请不要忘记从循环体内删除 row++ 和 column++，这样它们就不会递增两次。