C algorithm to try out all the possible combinations of 12 knights in a chess board

Question

I have been writing this program that tries to find how to put 12 knights on a chess board so all squares are either taken up by a knight or one of 12 knights can reach them in one move. So far I have created 2 functions: one takes a board with 12 knights and fills all the squares they can reach with 'T' and another that takes a filled chess board and checks if there are left any squares not taken up by a knight nor dominated by it (meaning no knight can reach it in one move).

Now I am dealing with the problem of how can I try out all the possible combinations of knights. My idea is that after every single combination of 12 knights on a board, I will send the board with 12 knights to be filled with 'T's for all the squares they can reach, and then send it to another function to check if all squares are dominated.

Here is my code:

#include <stdio.h>
#include <stdlib.h>
int filling(char array[8][8])
{
    int i, j;
    for (i = 0;i < 8;i++)
    {
        for (j = 0; j < 8;j++)
        {
            if(array[i][j] == 'H')
            {
                if(i-1>-1 && j+2<8 && array[i-1][j+2]!='H') array[i-1][j+2]='T';
                if(i+1<8 && j+2<8 && array[i+1][j+2]!='H') array[i+1][j+2]='T';
                if(i-2>-1 && j+1<8 && array[i-2][j+1]!='H') array[i-2][j+1]='T';
                if(i+2<8 && j+1<8 && array[i+2][j+1]!='H') array[i+2][j+1]='T';
                if(i-2>-1 && j-1>-1 && array[i-2][j-1]!='H') array[i-2][j-1]='T';
                if(i+2<8 && j-1>-1 && array[i+2][j-1]!='H') array[i+2][j-1]='T';
                if(i-1>-1 && j-2>-1 && array[i-1][j-2]!='H') array[i-1][j-2]='T';
                if(i+1<8 && j-2>-1 && array[i+1][j-2]!='H') array[i+1][j-2]='T';

            }
        }
    }
}
int checking(char array[8][8])
{
    int i, j;
    for(i = 0;i < 8;i++)
    {
        for(j = 0;j < 8;j++)
        {
            if(array[i][j] != 'H' && array[i][j] != 'T')
                return 0;
        }
    }
    return 1;
}
int main()
{
    int i, j;
    char board[8][8];
    for(i = 0; i < 8; i++)
    {

        for(j = 0; j < 8; j++)
            board[i][j] = '0';
    }
    // The following lines are here to test if the checking and filling work
    /*board[2][1] = 'H';
    board[2][2] = 'H';
    board[3][2] = 'H';
    board[5][2] = 'H';
    board[6][2] = 'H';
    board[5][3] = 'H';
    board[2][4] = 'H';
    board[1][5] = 'H';
    board[2][5] = 'H';
    board[4][5] = 'H';
    board[5][5] = 'H';
    board[5][6] = 'H'; */
    filling(board);
    if(checking(board) == 1) printf (" \n Works");
    else printf ("\n Doesnt work");
    for(i = 0; i < 8; i++)
    {
        printf("\n");
        for(j = 0; j < 8; j++)
            printf("%c ", board[i][j]);
    }
    return 0;
}

What kind of algorithm I could use to try out every combo? Thank you for your answers.

Probably you don't actually want to try every combination. The first knight has 64 possible placements, the second has 63, etc. So you have `64*63*62*...*53` permutations, which is 1573144097507348889600 (a.k.a. "too many"). It would take a while (: So you need to find ways of narrowing that search space, like using symmetry or early-out pruning. — jwd, Apr 19 '16 at 23:46
Very similar question: http://stackoverflow.com/questions/36527317/12-dominating-knights-puzzle-backtracking — jwd, Apr 19 '16 at 23:50

score 1 · Answer 1 · answered Apr 19 '16 at 23:43

You need these things:

A sorting rule that puts all possible combinations in a well-defined order.
An initialization rule that defines the board's first such combination.
An increment rule that transitions a board from its current combination to the next combination in the well-defined order.
A rule that detects when the board is in the last combination.

Then you just use algorithm 2 to put the board in the first state. Then check with algorithm 4 to see if you're in the last state. If not, use algorithm 3 to go to the next state.

Algorithm 1 is probably the hard one. One simple rule is just to convert the board's position to a binary number with a zero for an empty square and a one for a full square in a well-defined order, say starting from the top left and going across, then moving to the next row.

score 0 · Answer 2 · answered Apr 20 '16 at 00:09

If you're on decent, or even semi-decent hardware, you've got a 64-bit unsigned integral type available to you. Let's call that uint64_t.

You can store a chessboard with knight positions as a single uint64_t. You can also store a "dominated" mask in the same type. You simply establish a correlation between the 64 bits in the type and the 64 spaces on the chessboard.

Since there are 64 possible locations, you can pre-compute the possible threat masks for each position, storing that as an array of 64 uint64_t values.

You need to set the position of each knight. But you can safely do them in a certain order, so that knight #1 is always the "highest" knight - his bit position is always highest, or lowest, or whatever. So you could write code like this:

for (k0 = 64; k0 > 11; --k0) 
    for (k1 = k0 - 1; k1 > 10; --k1)
        for (k2 = k1 - 1; k2 > 9; --k2) 
            ...
            for (k11 = k10 - 1; k11 >= 0; --k11)
                /* do stuff */

But that's horrible, because there are a squillion possibilities (squillion = 1573144097507348889600, thanks @jwd!).

That said, the threat mask for each knight can be computed "incrementally" from the masks of the outer knights, so it might be faster to perform all the computations than to try to store/mark/compute the rotations and flips of the board.

Something like this:

for (kXX = kYY - 1; kXX > (11-XX); --kXX) {
    threat_XX = threat_YY | pre_computed_threat[kXX];
    for (kZZ = KXX - 1; kZZ > (11-ZZ); --kZZ) {
        threat_ZZ = threat_XX | pre_computed_threat[kZZ];

The nice thing about this approach is that your objective is total coverage by threats - all 1 bits in the threat_11 map, in other words. You can test just by inverting the bits and comparing with zero.

score 0 · Answer 3 · answered Apr 20 '16 at 03:11

Well thanks to my questions of today I got myself blocked from asking more until my rep gets better, but for what its worth I managed to solve the problem myself. I had to take 3 positions in each quarter of the board that can only be reached by seperate horses, that way making my 12 for cycles shorter, as they only had to go through specific locations, instead of trying every single of of them. The execution times is just above one second and it finds two different layouts for 12 knights, marking them 'H' and the positions they can reach in one turn 'T'. Here is the code if anyone ever comes to a problem like this:

#include <stdio.h>
#include <stdlib.h>
struct horses
{
    int x;
    int y;
};
void fill(struct horses arkliai[12], char array[8][8])
{
    int i,j;
    for(i = 0; i < 8; i++)
    {

        for(j = 0; j < 8; j++)
            array[i][j] = '0';
    }
    for(i = 0; i < 12; i++)
    {
        array[arkliai[i].x][arkliai[i].y] = 'H';
    }
    pildymas(array);
}
void startingpositions(struct horses arkliai[12])
{
    int i = 0;
    int j = 0;
    int a = 0;
    for(a = 0; a < 12; a++)
    {
        if(i > 7)
        {
            i = 0;
            j++;
        }
        arkliai[a].x = i;
        arkliai[a].y = j;
        i++;
    }
}
void combinacijos(struct horses h[12], char array[8][8])
{
    int a,b,c,d,e,f,g,hi,ii,ji,k,l;
    for(a = 0; a < 2; a++)
    {
        if(a == 0)
        {
            h[0].x = 1;
            h[0].y = 2;
        }
        if(a == 1)
        {
            h[0].x = 2;
            h[0].y = 1;
        }
        for(b = 0; b < 2; b++)
        {
            if(b == 0)
            {
                h[1].x = 5;
                h[1].y = 1;
            }
            if(b == 1)
            {
                h[1].x = 6;
                h[1].y = 2;
            }
            for(c = 0; c < 2; c++)
            {
                if(c == 0)
                {
                    h[2].x = 1;
                    h[2].y = 5;
                }
                if(c == 1)
                {
                    h[2].x = 2;
                    h[2].y = 6;
                }
                for(d = 0; d <2;d++)
                {
                    if(d == 0)
                    {
                        h[3].x = 5;
                        h[3].y = 6;
                    }
                    if(d == 1)
                    {
                        h[3].x = 6;
                        h[3].y = 5;
                    }
                    for(e = 0; e < 3; e++)
                    {
                        if(e == 0)
                        {
                            h[4].x = 2;
                            h[4].y = 0;
                        }
                        if(e == 1)
                        {
                            h[4].x = 2;
                            h[4].y = 2;
                        }
                        if(e == 2)
                        {
                            h[4].x = 1;
                            h[4].y = 3;
                        }
                        for (f = 0; f < 3; f++)
                            {
                               if(f == 0)
                                {
                                    h[5].x = 1;
                                    h[5].y = 4;
                                }
                                if(f == 1)
                                {
                                    h[5].x = 2;
                                    h[5].y = 5;
                                }
                                if(f == 2)
                                {
                                    h[5].x = 2;
                                    h[5].y = 7;
                                }
                                for (g = 0; g < 3; g++)
                                {
                                    if(g == 0)
                                    {
                                        h[6].x = 5;
                                        h[6].y = 7;
                                    }
                                    if(g == 1)
                                    {
                                        h[6].x = 5;
                                        h[6].y = 5;
                                    }
                                    if(g == 2)
                                    {
                                        h[6].x = 6;
                                        h[6].y = 4;
                                    }
                                for(hi = 0; hi < 3; hi++)
                                    {
                                        if(hi == 0)
                                        {
                                            h[7].x = 5;
                                            h[7].y = 0;
                                        }
                                        if(hi == 1)
                                        {
                                            h[7].x = 5;
                                            h[7].y = 2;
                                        }
                                        if(hi == 2)
                                        {
                                            h[7].x = 6;
                                            h[7].y = 3;
                                        }
                                        for(ii = 0; ii < 4; ii++)
                                        {
                                            if (ii == 0)
                                            {
                                                h[8].x = 3;
                                                h[8].y = 0;
                                            }
                                            if (ii == 1)
                                            {
                                                h[8].x = 3;
                                                h[8].y = 2;
                                            }
                                            if (ii == 2)
                                            {
                                                h[8].x = 0;
                                                h[8].y = 3;
                                            }
                                            if (ii == 3)
                                            {
                                                h[8].x = 2;
                                                h[8].y = 3;
                                            }
                                            for(ji = 0; ji < 4; ji++)
                                            {
                                                if (ji == 0)
                                                {
                                                    h[9].x = 3;
                                                    h[9].y = 7;
                                                }
                                                if (ji == 1)
                                                {
                                                    h[9].x = 3;
                                                    h[9].y = 5;
                                                }
                                                if (ji == 2)
                                                {
                                                    h[9].x = 2;
                                                    h[9].y = 4;
                                                }
                                                if (ji == 3)
                                                {
                                                    h[9].x = 0;
                                                    h[9].y = 4;
                                                }
                                                for(k = 0; k < 4; k++)
                                                {
                                                    if (k == 0)
                                                    {
                                                        h[10].x = 4;
                                                        h[10].y = 7;
                                                    }
                                                    if (k == 1)
                                                    {
                                                        h[10].x = 4;
                                                        h[10].y = 5;
                                                    }
                                                    if (k == 2)
                                                    {
                                                        h[10].x = 5;
                                                        h[10].y = 4;
                                                    }
                                                    if (k == 3)
                                                    {
                                                        h[10].x = 7;
                                                        h[10].y = 4;
                                                    }
                                                    for(l = 0;l < 64;l++)
                                                    {
                                                        if(h[11].x == 7)
                                                        {
                                                            if(h[11].y == 7)
                                                            {
                                                                h[11].x = 0;
                                                                h[11].y = 0;
                                                                break;
                                                            }
                                                            h[11].x = 0;
                                                            h[11].y = h[11].y +1;
                                                        }
                                                       else { h[11].x= h[11].x+1; }
                                                        fill(h, array);
                                                    }
                                                }
                                            }
                                        }

                                    }
                                }
                            }
                    }
                }
            }
        }
    }
}
int pildymas(char array[8][8])
{
    int i, j;
    for (i = 0;i < 8;i++)
    {
        for (j = 0; j < 8;j++)
        {
            if(array[i][j] == 'H')
            {
                if(i-1>-1 && j+2<8 && array[i-1][j+2]!='H') array[i-1][j+2]='T';
                if(i+1<8 && j+2<8 && array[i+1][j+2]!='H') array[i+1][j+2]='T';
                if(i-2>-1 && j+1<8 && array[i-2][j+1]!='H') array[i-2][j+1]='T';
                if(i+2<8 && j+1<8 && array[i+2][j+1]!='H') array[i+2][j+1]='T';
                if(i-2>-1 && j-1>-1 && array[i-2][j-1]!='H') array[i-2][j-1]='T';
                if(i+2<8 && j-1>-1 && array[i+2][j-1]!='H') array[i+2][j-1]='T';
                if(i-1>-1 && j-2>-1 && array[i-1][j-2]!='H') array[i-1][j-2]='T';
                if(i+1<8 && j-2>-1 && array[i+1][j-2]!='H') array[i+1][j-2]='T';

            }
        }
    }
    tikrinimas(array);
}
int tikrinimas(char array[8][8])
{
    int i, j;
    for(i = 0;i < 8;i++)
    {
        for(j = 0;j < 8;j++)
        {
            if(array[i][j] != 'H' && array[i][j] != 'T')
                return 0;
        }
    }
    printas(array);
}
int printas(char array[8][8])
{
    int i,j;
    for(j = 0; j <8; j++)
    {
        printf("\n");
        for(i = 0; i <8 ; i++)
            printf("%c ", array[i][7-j]);
    }
    printf("\n");
}
int main()
{
    struct horses hr[12];
    char board[8][8];
    startingpositions(hr);
    combinacijos(hr, board);
    return 0;
}

C algorithm to try out all the possible combinations of 12 knights in a chess board

3 Answers3