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我正在尝试以效率和模式匹配功能为重点来实现生活游戏。图案是闪光灯，滑翔机，十字架等。

我有一个世界的一维数组，以及一个宽度和高度。为了找到邻居，我想计算摩尔社区的索引，然后检查这些是否是哈希，如果它们是，这会增加 get_neighbours 函数的返回变量。北方和南方似乎有效，但东方和西方却没有。NE, SE, SW, NW 都是基于前面的逻辑（即向北向西走）。

int get_neighbours(int loc) {
    int neighbours = 0;

    int n = mod(loc - grid_width, total);
    int e = mod(loc + 1, grid_width) + grid_width;
    int s = mod(loc + grid_width, total);
    int w = mod(loc - 1, grid_width) + grid_width;
    int nw = mod(w - grid_width, total);
    int ne = mod(e - grid_width, total);
    int se = mod(e + grid_width, total);
    int sw = mod(w + grid_width, total);

    //Northwest
    if (grid[nw] == '#') {
        neighbours++;
    }
    //North
    if (grid[n] == '#') {
        neighbours++;
    }
    //Northeast
    if (grid[ne] == '#') {
        neighbours++;
    }
    //East
    if (grid[e] == '#') {
        neighbours++;
    }
    //Southeast
    if (grid[se] == '#') {
        neighbours++;
    }
    //South
    if (grid[s] == '#') {
        neighbours++;
    }
    //Southwest
    if (grid[sw] == '#') {
        neighbours++;
    }
    //West
    if (grid[w] == '#') {
        neighbours++;
    }
    return neighbours;
}

int mod(int a, int b) {
    int ret = a % b;
    if (b < 0) {
        return mod(-a, -b);
    }
    else if (ret < 0) {
        ret += b;
    }
    return ret;
}

对于模式匹配，我尝试使用与上述相同的逻辑来构建一个 5x5 子数组。这实质上使用了“读取头”。从提供的位置向东穿过世界，直到它移动了 5 个空间。然后，它返回到原始位置，向南移动正确的行数，然后再次向东移动，直到我们收集到 25 个索引。

char *get_subarray(int loc) {
    char *subarray;
    subarray = malloc(sizeof(char) * 25);

    int i = 0;
    int ptr = loc;

    while (i < 25) {
        subarray[i] = grid[ptr];
        if ((i + 1) % 5 == 0) {
            //return the read head to the original location, then travel south through the grid once for each of the times we have traversed a row
            ptr = loc;
            for (int k = 0; k <= (i / 5); k++) {
                ptr = mod(ptr + grid_width, total);
            }
        } else {
            ptr = mod(ptr + 1, grid_width) + grid_width;
        }
        i++;
    }
    subarray[i] = '\0';
    return subarray;

}

当它这样做时，它从世界构建子数组，然后我可以 strcmp() 将它与字符串作为模式。

int cpu_get_crosses() {
    int crosses = 0;

    for (int i = 0; i < total; i++) {
        if (strcmp(get_subarray(i), "       #   # #   #       ") == 0) {
            crosses++;
        }
    }
    return crosses;
}

作为参考，带有索引的 7x5 网格（带边界）：

34|28 29 30 31 32 33 34|28
--|--------------------|--
6 |0  1  2  3  4  5  6 |0
13|7  8  9  10 11 12 13|7 
20|14 15 16 17 18 19 20|14
27|21 22 23 24 25 26 27|21
34|28 29 30 31 32 33 34|28
--|--------------------|--
6 |0  1  2  3  4  5  6 |0

我很好奇什么逻辑可以让我在保留边界条件的同时计算摩尔邻域的索引，以便我可以正确计算邻居和子数组（因为它们都将使用相同的逻辑）。

编辑：如果有任何 googlers 想要的子数组函数。

char *get_subarray(int loc) {
    char *subarray;
    subarray = malloc(sizeof(char) * 25); //5x5 (=25) subarray

    int i = 0;
    int row = loc / grid_width;
    int ptr = loc;

    while (i < 25) {
        subarray[i] = grid[ptr];
        if ((i + 1) % 5 == 0) {
            //return the read head to the original location, then travel south through the grid once for each of the times we have traversed a row
            ptr = loc;
            for (int k = 0; k <= (i / 5); k++) {
                ptr = mod(ptr + grid_width, total);
            }
            row = ptr / grid_width;
        } else {
            ptr = mod(ptr + 1, grid_width) + row * grid_width;
        }
        i++;
    }
    subarray[i] = '\0';
    return subarray;
}
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1 回答 1

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您正在以逐行方式索引您的数组：index(i, j) = j * grid_width + ifor i=0..grid_width-1, j=0..grid_height-1. 让我们调用and reverseloc的结果来得到and ：index(i, j)indexij

int i = loc % grid_width;
int j = loc / grid_width;

向东加i一，向西减一，均以宽度为模：

int e = j * grid_width + (i + 1) % grid_width
      = j * grid_width + ((j * grid_width + i) + 1) % grid_width
      = j * grid_width + (loc + 1) % grid_width;
int w = j * grid_width + (i + grid_width - 1) % grid_width
      = j * grid_width + ((j * grid_width + i) + grid_width - 1) % grid_width
      = j * grid_width + (loc + grid_width - 1) % grid_width;

笔记：

  1. (i + grid_width - 1) % grid_width等于mod(i - 1, grid_width)
  2. x % M = (k * M + x) % M对于任何积分k，这让我们i在任何表达式中用模grid_width替换，loc = j * grid_width + i以避免i首先计算；）

将高度增加j一个模等于添加grid_width和换行，total因为total是宽度 x 高度。更明确地说，这是推导：

int j1 = (j + 1) % grid_height;
int s = j1 * grid_width + i
      = ((j + 1) % grid_height) * grid_width + i
      = ((j + 1) * grid_width) % (grid_height * grid_width) + i
      = ((j + 1) * grid_width) % total + i
      = (j * grid_width + grid_width + i) % total
      = ((j * grid_width + i) + grid_width) % total
      = (loc + grid_width) % total;
// analogue for j0 = (j + grid_height - 1) % grid_height;
int n = (loc + total - grid_width) % total;
于 2016-05-16T17:50:51.727 回答