使用边界条件导航映射到 1D 的 2D 数组

Question

我正在尝试着重于效率以及模式匹配功能来实现生活游戏。图案是方向灯、滑翔机、十字架等。

我有一个世界的一维数组，以及宽度和高度。为了找到邻居，我想计算 Moore 邻域的索引，然后检查这些是否是散列，如果是，这会增加 get_neighbours 函数的 return 变量。北方和南方似乎有效，但东方和西方却不行。 NE、SE、SW、NW 都是基于之前的逻辑（即西向北）。

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。这实际上使用了一个 "read head." 从提供的位置向东穿过世界，直到它移动了 5 个空格。然后，它 returns 到原始位置并向南移动正确的行数，然后再次向东移动，直到我们收集了 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] = '[=11=]';
    return subarray;

}

当它这样做时，它从世界构建子数组，然后我可以针对模式的字符串进行 strcmp()。

int cpu_get_crosses() {
    int crosses = 0;

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

供参考，带有索引（带边界）的 7x5 网格：

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

我很好奇什么逻辑允许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（因为它们都使用相同的逻辑） .

编辑： 子数组函数（如果有任何 google 员工需要的话）。

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] = '[=14=]';
    return subarray;
}

Answer 1

您正在以行方式为数组编制索引：index(i, j) = j * grid_width + i for i=0..grid_width-1, j=0..grid_height-1。让我们调用 loc index(i, j) 的结果并反转 index 得到 i 和 j:

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

向东增加 i 一，向西减少一，两者均以宽度为模：

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;

注：

(i + grid_width - 1) % grid_width 等于 mod(i - 1, grid_width)
x % M = (k * M + x) % M 对于任何积分 k，这让我们用 loc = j * grid_width + i 替换任何表达式中的 i 模 grid_width 以避免计算 i 首先 ;)

将 j 增加一个模数的高度等于添加 grid_width 并换行 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;

使用边界条件导航映射到 1D 的 2D 数组

Navigating a 2D array mapped to 1D with boundary conditions

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conways-game-of-life