Knight movement.... " how to output all possible moves. "
- by josh kant
The following is the code i wrote.. I have to write it for nXn but for easyness i tried to test it for 5X5. It does not display my output... could anybody tell me whats wrong with the following code:
{
#include <iostream>
#include <iomanip>
using namespace std;
void knight ( int startx, int starty, int n, int p[][5], int used [][5], int &count);
int main ( )
{
const int n = 5; // no. of cloumns and rows
int startx = 0;
int starty = 0;
int p[5][5];
int used[5][5];
int count = 1;
int i= 0;
int j = 0;
//initializing the array
for ( i = 0; i < 5; i++)
{
for ( j = 0; j < 5; j++)
{
p[i][j] = 0;
used [i][j] = 0;
}
}
//outputting the initialized array..
i=0;
while ( i< 5)
{
for ( j = 0; j < 5; j++)
{
cout << setw(3) << p[i][j];
}
i++;
cout << endl;
}
knight (startx,starty,n,p,used,count);
return 0;
}
void knight ( int x, int y, int n, int p[][5], int used [][5], int &count)
{
int i = 0;
//knight (x,y,n,p,used,count)
for ( i = 0; i < n*n; i++)
{
if ( used [x][y] == 0 )
{
used[x][y] = 1; // mark it used;
p[x][y] += count; //inserting step no. into the solution
//go for the next possible steps;
//move 1
//2 squares up and 1 to the left
if (x-1 < 0 && y+2 < n && p[x-1][y+2] == 0)
{
used[x-1][y+2] = 1;
p[x-1][y+2] += count;
knight (x-1,y+2,n,p,used,count);
used[x-1][y+2] = 0;
}
//move 2
//2 squares up and 1 to the right
if ( x+1 < n && y+2 < n && p[x+1][y+2] == 0 )
{
used[x+1][y+2] = 1;
p[x+1][y+2] += count;
knight (x+1,y+2,n,p,used,count);
used[x+1][y+2] = 0;
}
//move 3
//1 square up and 2 to the right
if ( x+2 < n && y+1 < n && p[x+2][y+1] == 0 )
{
used[x+2][y+1] = 1;
p[x+2][y+1] += count;
knight (x+2,y+1,n,p,used,count);
used[x+2][y+1] = 0;
}
//move 4
//1 square down and 2 to the right
if ( x+2 < n && y-1 < n && p[x+2][y-1] == 0 )
{
used[x+2][y-1] = 1;
p[x+2][y-1] += count;
knight (x+2,y-1,n,p,used,count);
used[x+2][y-1] = 0;
}
//move 5
//2 squares down and 1 to the right
if ( x+1 < n && y-2 < n && p[x+1][y-2] == 0 )
{
used[x+1][y-2] = 1;
p[x+1][y-2] += count;
knight (x+1,y-2,n,p,used,count);
used[x+1][y-2] = 0;
}
//move 6
//2 squares down and 1 to the left
if ( x-1 < n && y-2 < n && p[x-1][y-2] == 0 )
{
used[x-1][y-2] = 1;
p[x-1][y-2] += count;
knight (x-1,y-2,n,p,used,count);
used[x-1][y-2] = 0;
}
//move 7
//1 square down and 2 to the left
if ( x-2 < n && y-1 < n && p[x-2][y-1] == 0 )
{
used[x-2][y-1] = 1;
p[x-2][y-1] += count;
knight (x-2,y-1,n,p,used,count);
used[x-2][y-1] = 0;
}
//move 8
//one square up and 2 to the left
if ( x-2 < n && y+1< n && p[x-2][y+1] == 0 )
{
used[x-2][y+1] = 1;
p[x-2][y+1] += count;
knight (x-2,y+1,n,p,used,count);
used[x-2][y+1] = 0;
}
}
}
if ( x == n-1 && y == n-1)
{
while ( i != n)
{
for ( int j = 0; j < n; j++)
cout << setw(3) << p[i][j];
i++;
}
}
}
Thank you!!
