优化康威的生命游戏

Question

I'm working on speeding up Conway's Game of Life. Right now, the code looks at a cell and then adds up the 3x3 area immediately surrounding the point, then subtracts the value at the point we're looking at. Here's the function that is doing that:

static int neighbors2 (board b, int i, int j)
{
    int n = 0;
    int i_left = max(0,i-1);
    int i_right  = min(HEIGHT, i+2);

    int j_left = max(0,j-1);
    int j_right  = min(WIDTH, j+2);
    int ii, jj;

    for (jj = j_left; jj < j_right; ++jj) {
        for (ii = i_left; ii < i_right; ii++) {
            n += b[ii][jj];
        }
    }

    return n - b[i][j];
}

And here is the code I've been trying to use to iterate through pieces at a time:

//Iterates through the first row of the 3x3 area
static int first_row(board b, int i, int j) {
    int f = 0;
    int i_left = max(0,i-1);

    int j_left = max(0,j-1);
    int j_right  = min(WIDTH, j+2);
    int jj;

    for (jj = j_left; jj < j_right; ++jj) {
        f += b[i_left][jj];
    }

    return f;
}

//Iterates and adds up the second row of the 3x3 area
static int second_row(board b, int i, int j) {
    int g = 0;
    int i_right  = min(HEIGHT, i+2);

    int j_left = max(0,j-1);
    int j_right  = min(WIDTH, j+2);
    int jj;

    if (i_right != i) {
        for (jj = j_left; jj < j_right; ++jj) {
            g += b[i][jj];
        }
    }

    return g;
}

//iterates and adds up the third row of the 3x3 area.
static int third_row(board b, int i, int j) {
    int h = 0;
    int i_right  = min(HEIGHT, i+2);

    int j_left = max(0,j-1);
    int j_right  = min(WIDTH, j+2);
    int jj;

    for (jj = j_left; jj < j_right; ++jj) {
        h += b[i_right][jj];
    }

    return h;
}

//adds up the surrounding spots
//subtracts the spot we're looking at.
static int addUp(board b, int i, int j) {
    int n = first_row(b, i, j) + second_row(b, i, j) + third_row(b, i, j);
    return n - b[i][j];
}

But, for some reason it isn't working. I have no idea why.

Things to note:

1. sometimes i == i_right , so we do not want to add up a row twice.
   
2. The three functions are supposed to do the exact same thing as neighbors2 in separate pieces.
   
3. min and max are functions that were premade for me.
   
4. sometimes sometimes j == j_right , so we do not want to add up something twice. I'm pretty confident the loop takes care of this however.
   
5. Tips and things to consider are appreciated.
   

Thanks all. I've been working on this for a couple hours now and have no idea what is going wrong. It seems like it should work but I keep getting incorrect solutions at random spots among the board.

Answer 1

In neighbors2 , you set i_left and i_right so that the're limited to the rows of the grid. If the current cell is in the top or bottom row, you only loop through two rows instead of 3.

In first_row() and last_row() you also limit it to the rows of the grid. But the result is that these functions will add the cells on the same row as the current cell, which is what second_row does. So you end up adding those rows twice.

You shouldn't call first_row() when i = 0 , and you shouldn't call third_row() when i == HEIGHT .

static int addUp(board b, int i, int j) {
    int n = (i == 0 ? 0 : first_row(b, i, j)) + 
            second_row(b, i, j) + 
            (i == HEIGHT ? 0 : third_row(b, i, j));
    return n - b[i][j];
}

Another option would be to do the check in the functions themselves:

function first_row((board b, int i, int j) {
    if (i == 0) {
        return 0;
    }
    int f = 0;
    int j_left = max(0,j-1);
    int j_right  = min(WIDTH, j+2);
    int jj;

    for (jj = j_left; jj < j_right; ++jj) {
        f += b[i][jj];
    }

    return f;
}    

and similarly for third_row() . But doing it in the caller saves the overhead of the function calls.

BTW, your variable names are very confusing. All the i variables are for rows, which go from top to bottom, not left to right.

Answer 2

#include <stdio.h>
#include <stdlib.h>

#define ROWSDISP 50
#define COLSDISP 100

int rows=ROWSDISP+2, cols=COLSDISP+2;

This is to avoid illegal indexes when stepping over the neighbours.

struct onecell {char alive;
                char neibs;} **cells;

This is the foundation of a (dynamic) 2D-array, of a small struct.

To create space for each row plus the space to hold an array of row pointers:

void init_cells()
{
    int i;
    cells = calloc(rows, sizeof(*cells));
    for(i=0; i<=rows-1; i++)
        cells[i] = calloc(cols, sizeof(**cells));
}

I skip the rand_fill() and glider() funcs. A cell can be set by cells[y][x].alive=1 .

int main(void) {

    struct onecell *c, *n1, *rlow;
    int i, j, loops=0;
    char nbs;

    init_cells();
    rand_fill();
    glider();

    while (loops++ < 1000) {
        printf("\n%d\n", loops);
        for (i = 1; i <= rows-2; i++) {

            for (j = 1; j <= cols-2; j++) {

                c  =  &cells[ i ][ j ];  
                n1 =  &cells[ i ][j+1];
                rlow = cells[i+1];

                nbs = c->neibs + n1->alive + rlow[ j ].alive
                                           + rlow[j+1].alive
                                           + rlow[j-1].alive;
                if(c->alive) {
                    printf("@");

                    n1->neibs++; 
                    rlow[ j ].neibs++;
                    rlow[j+1].neibs++;
                    rlow[j-1].neibs++;

                    if(nbs < 2 || nbs > 3)
                        c->alive = 0;
                } else {
                    printf(" ");
                    if(nbs == 3)
                        c->alive = 1;
                }

            c->neibs = 0;        // reset for next cycle
            }
        printf("\n");
        }
    }
    return(0);
}

There is no iterating a 3x3 square here. Of the 8 neighbours, only the 4 downstream ones are checked; but at the same time their counters are raised.

A benchmark with 100x100 grid:

# time ./a.out >/dev/null
real    0m0.084s
user    0m0.084s
sys     0m0.000s
# bc <<<100*100*1000/.084
119047619

And each of these 100M cells needs to check 8 neighbours, so this is close to the CPU frequency (1 neighbour check per cycle).

It seems twice as fast as the rosetta code solution.

There also is no need to switch the boards. Thanks to the investment in the second field of a cell.