In the 8X8 chess board, taking only knight into consideration, if we start the knight from any square of the chessboard, the aim is to cover max number of square , without repeating any square . so far I have found the most efficient solution with my code below is:
60 29 34 49 0 15 46 0
35 50 1 16 45 48 11 0
30 59 28 33 2 9 14 47
51 36 31 44 17 12 3 10
58 43 52 27 32 25 8 13
37 40 55 18 23 6 21 4
42 57 38 53 26 19 24 7
39 54 41 56 0 22 5 20
Where the number starting with 1, shows the path followed by knight. My question is can this code be corrected for a perfect answer which is 64 (mine reaches only 60)?
package game;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Knight {
static int board[][]=new int[8][8];
static int value=1;
public static void zero()
{
for(int i=0;i<7;i++)
for(int j=0;j<7;j++)
board[i][j]=0;
}
public static void knightpos(int x,int y)throws IOException
{
if(value==61)
{ System.out.println();
for(int i=0;i<8;i++)
{
System.out.println();
System.out.println();
for(int j=0;j<8;j++)
System.out.print(" "+board[i][j]);
}
System.exit(0);
}
if(x+1<=7&&y+2<=7)
{
if(board[x+1][y+2]==0)
{ board[x+1][y+2]=value++;
knightpos(x+1,y+2);
}
}
if(x+2<=7&&y+1<=7)
{
if(board[x+2][y+1]==0)
{
board[x+2][y+1]=value++;
knightpos(x+2,y+1);
}
}
if(x-2>=0&&y-1>=0)
{
if(board[x-2][y-1]==0)
{board[x-2][y-1]=value++;
knightpos(x-2,y-1);
}
}
if(x+2<=7&&y-1>=0)
{
if(board[x+2][y-1]==0)
{board[x+2][y-1]=value++;
knightpos(x+2,y-1);
}
}
if(x+1<=7&&y-2>=0)
{
if(board[x+1][y-2]==0)
{board[x+1][y-2]=value++;
knightpos(x+1,y-2);}
}
if(x-1>=0&&y-2>=0)
{
if(board[x-1][y-2]==0)
{board[x-1][y-2]=value++;
knightpos(x-1,y-2);}
}
if(x-2>=0&&y+1<=7)
{
if(board[x-2][y+1]==0)
{board[x-2][y+1]=value++;
knightpos(x-2,y+1);}
}
if(x-1>=0&&y+2<=7)
{
if(board[x-1][y+2]==0)
{board[x-1][y+2]=value++;
knightpos(x-1,y+2);}
}
board[x][y]=0;
value--;
return;
}
public static boolean chk() {
for(int i=0;i<7;i++)
for(int j=0;j<7;j++)
if(board[i][j]==0)
return false;
return true;
}
public static void main(String args[])throws IOException
{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
System.out.println("Knight chess game input x,y position ");
int x=Integer.parseInt(br.readLine());
int y=Integer.parseInt(br.readLine());
{
if(!chk())
{
zero();
value=1;
knightpos(x,y);
}
}
}
}
If you look at your solution, the first step, where you could branch to a zero is step 3. As additional information, we see, that you could have jumped from 60 to 1 again, building a cycle (but you couldn’t, because you aren’t allowed to visit a place twice.
But if you start at 4, you can move to 60, from there to 1, 2, 3 and now you can visit one of the zero fields.
However, that’s only an improvement for 1 field. Since the other unvisited fields aren’t visitable in sequence, this can’t be improved further in the same way.