Question

我正在尝试实施生活游戏，重点关注效率，以及模式匹配的功能。模式是眼罩，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。为了找到我想要计算摩尔邻域的索引的邻居，然后检查它们是否是哈希值，如果它们增加了get_neighbours函数的返回变量。南北似乎有效，但东西没有。 NE，SE，SW，NW都基于先前的逻辑（即向西北方向）。

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，我尝试使用与上面相同的逻辑来构建5x5子阵列。这主要使用“读头”。从提供的位置向东穿过世界，直到它移动了5个空间。然后，它返回到原始位置并向南移动正确的行数，然后再向东移动，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] = '\0';
    return subarray;

}

正如它所做的那样，它构建了来自世界的子数组，然后我可以strcmp（）将其作为模式的字符串。

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

我很好奇是什么逻辑允许我在保留边界条件的同时计算摩尔邻域的指数，这样我就可以正确地计算邻域和子阵列（因为这两者都使用相同的逻辑）。

编辑：如果有任何googlers想要它的子阵列功能。

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;
}

Answer 1

您正在以行方式为数组建立索引：index(i, j) = j * grid_width + i为i=0..grid_width-1, j=0..grid_height-1。我们将loc的结果称为index(i, j)，然后反转index以获取i和j：

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

向东增加i一个，向西减少一个，两者都以宽度为模：

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;

注意：

(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，我们将i替换为grid_width的任何表达式loc = j * grid_width + i，以避免计算i第一名;）

将j增加一个模高度等于添加grid_width并按total包裹，因为total是宽x高。使其更明确，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

1 个答案: