我一直在为学校制作康威的生命游戏版本而且我遇到了一个问题：细胞在错误的地方变得死亡或活着。
我该如何解决？
if (alive == 3 && aryBOARD[x][y] == 0) { //rule 4
aryCHANGE[x][y] = 1;
}
if (alive > 3 && aryBOARD[x][y] == 1) { //rule 3
aryCHANGE[x][y] = 0;
}
if (alive >= 2 && alive <= 3 && aryBOARD[x][y] == 1) { //rule 2
aryCHANGE[x][y] = 1;
}
if (alive < 2 && aryBOARD[x][y] == 1) { //rule 1
aryCHANGE[x][y] = 0;
}
if (dead > 5) { //rule check
aryCHANGE[x][y] = 0;
}
这就是我认为问题所在，但如果整个代码有帮助：
package gameoflife2;
public class Game {
public static void main(String[] args) {
int[][] aryBOARD = new int[5][5];
int x = 0;
int y = 0;
int dead = 0;
int alive = 0;
int i, j;
// Board numbers
// 00011
// 00001
// 01000
// 01100
// 00000
aryBOARD[0][0] = 0;
aryBOARD[0][1] = 0;
aryBOARD[0][2] = 0;
aryBOARD[0][3] = 1;
aryBOARD[0][4] = 1;
aryBOARD[1][0] = 0;
aryBOARD[1][1] = 0;
aryBOARD[1][2] = 0;
aryBOARD[1][3] = 0;
aryBOARD[1][4] = 1;
aryBOARD[2][0] = 0;
aryBOARD[2][1] = 1;
aryBOARD[2][2] = 0;
aryBOARD[2][3] = 0;
aryBOARD[2][4] = 0;
aryBOARD[3][0] = 0;
aryBOARD[3][1] = 1;
aryBOARD[3][2] = 1;
aryBOARD[3][3] = 0;
aryBOARD[3][4] = 0;
aryBOARD[4][0] = 0;
aryBOARD[4][1] = 0;
aryBOARD[4][2] = 0;
aryBOARD[4][3] = 0;
aryBOARD[4][4] = 0;
// end of array
int[][] aryCHANGE = aryBOARD.clone(); // array change is equal to array
// board
// printing array
int rows = 5;
int colums = 5;
for (i = 0; i < rows; i++) {
for (j = 0; j < colums; j++) {
System.out.print(aryBOARD[i][j] + " ");
}
System.out.println("");
}
System.out.println("---------------------------");
// done printing array
// check for dead or alive cells
for (x = 0; x <= 4; x++) {
for (y = 0; y <= 4; y++) {
alive = 0;
dead = 0;
if ((x + 1 < 4) && (x + 1 > 0)) { // right
if (aryBOARD[x + 1][y] == 0) {
dead++;
} else {
alive++;
}
}
if (((y - 1 < 4) && (y - 1 > 0) && (x + 1 < 4) && (x + 1 > 0))) { // bottom
// right
// corner
if (aryBOARD[x + 1][y - 1] == 0) {
dead++;
} else {
alive++;
}
}
if (((y + 1 < 4) && (y + 1 > 0) && (x + 1 < 4) && (x + 1 > 0))) { // top
// right
// corner
if (aryBOARD[x + 1][y + 1] == 0) {
dead++;
} else {
alive++;
}
}
if ((y + 1 < 4) && (y + 1 > 0)) {// top middle
if (aryBOARD[x][y] == 0) {
dead++;
} else {
alive++;
}
}
if (((y + 1 < 4) && (y + 1 > 0) && (x - 1 < 4) && (x - 1 > 0))) {// top
// left
// corner
if (aryBOARD[x - 1][y + 1] == 0) {
dead++;
} else {
alive++;
}
}
if ((x - 1 < 4) && (x - 1 > 0)) {// left
if (aryBOARD[x - 1][y] == 0) {
dead++;
} else {
alive++;
}
}
if (((y - 1 < 4) && (y - 1 > 0) && (x - 1 < 4) && (x - 1 > 0))) {// bottom
// left
// corner
if (aryBOARD[x - 1][y - 1] == 0) {
dead++;
} else {
alive++;
}
}
// x++
if ((y - 1 < 4) && (y - 1 > 0)) {// bottom middle
if (aryBOARD[x][y - 1] == 0) {
dead++;
} else {
alive++;
}
}
// RULES
// 1 Any live cell with fewer than two live neighbors dies, as if caused
// by under-population.
// 2 Any live cell with two or three live neighbors lives on to the next
// generation.
// 3 Any live cell with more than three live neighbors dies, as if by
// overcrowding.
// 4 Any dead cell with exactly three live neighbors becomes a live
// cell, as if by reproduction.
// test alive and dead
if (alive == 3 && aryBOARD[x][y] == 0) {// rule 4
aryCHANGE[x][y] = 1;
}
if (alive > 3 && aryBOARD[x][y] == 1) {// rule 3
aryCHANGE[x][y] = 0;
}
if (alive >= 2 && alive <= 3 && aryBOARD[x][y] == 1) {// rule 2
aryCHANGE[x][y] = 1;
}
if (alive < 2 && aryBOARD[x][y] == 1) {// rule 1
aryCHANGE[x][y] = 0;
}
if (dead > 5) {// rule check
aryCHANGE[x][y] = 0;
}
}
}
for (i = 0; i < rows; i++) {
for (j = 0; j < colums; j++) {
System.out.print(aryCHANGE[i][j] + " ");
}
System.out.println("");
}
System.out.println("---------------------------");
} // end main
} // end class
答案 0 :(得分：1)
您需要更多地考虑构建代码并将其分解为更小的块。这对你有很大帮助，特别是当你将来搬到更大的项目时。
例如，编写一个简单的方法来计算给定单元格周围的活细胞数。
现在你的主要循环变为：
for (int x=0;x<width;x++) {
for (int y=0;y<height;y++) {
switch(countLivingAround(x,y)) {
case 0: // Less than 2 always dies
case 1:
grid(x,y) = 0;
break;
case 2: // Do nothing, keep current state
break;
case 3: // Breed
grid(x,y) = 1;
break;
case 4: // Dies from overcrowding
grid(x,y) = 0;
break;
}
}
}
您的计数功能可以很简单，它只是将[x-1，y]，[x，y + 1]等处的值相加，记得检查板的边缘并正确处理该情况
答案 1 :(得分：0)
除了将逻辑分解为不同的方法之外，您还需要使用调试器来逐步执行代码或使用printlns来显示您评估的x和y值（请参阅下面的示例）您将看到一些你在相邻单元x和y坐标上的计算需要工作。
for (x = 0; x <= 4; x++) {
for (y = 0; y <= 4; y++) {
alive = 0;
dead = 0;
System.out.println("evaluating cell x=" + x + ", y = " + y);
if ((x + 1 < 4) && (x + 1 > 0)) {// right
if (aryBOARD[x + 1][y] == 0) {
dead++;
}
else {
alive++;
}
System.out.println(" check 1 x=" + (x + 1) + ", y = " + y);
}