Most efficient way of checking for shared row, column, diagonal? - c++

Question

In C++ if I have a square array int board[8][8] that's filled like this:
0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0
0 0 0 1 0 0 0 0
1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0
0 1 0 0 0 0 0 0
What's the shortest way to check if any of the 1's share a row, column, or diagonal with another 1?
edit: I said most efficient when I really meant shortest

Answer 1

8 x 8 board? This must be related to chess or something.
Here's a clever way to test if any piece is on hit by the queen (i.e. almost identical to whether a 1 shares a row, column or diagonal with another 1).
bool CG_queen::move(File f_to, Rank r_to, File f_from, Rank r_from)
{
bool canMakeMove = false;
//Check to see if Queen is moving only by File or only by Rank.
//aka, only vertically or horizontally.
if ( f_from == f_to || r_from == r_to )
{
canMakeMove = true;
}
//Check to see if Queen only moves diagonally.
if ( abs(f_from - f_to) == abs(r_to - r_from) )
{
canMakeMove = true;
}
return canMakeMove;
}

Answer 2

You can use a bitmask for the rows, columns and diagonals to indicate if there is a 1 on any of them:
int rowMask = 0;
int ColumnMask = 0;
int diagonalMask0 = 0;
int diagonalMask1 = 0;
for(int i = 0; i < 8; i++)
{
for(int j = 0; j < 8; j++)
{
if(board[i][j])
{
// test row:
if(rowMask & (1 << i))
return true;
rowMask |= 1 << i; // mark row set
// test column:
if(columnMask & (1 << j))
return true;
columnMask |= 1 << j; // mark column set
// test first diagonal:
if(diagonalMask0 & (1 << (i + j)))
return true;
diagonalMask0 |= 1 << (i + j); // mark diagonal set
// test first diagonal:
if(diagonalMask1 & (1 << (8 + i - j)))
return true;
diagonalMask1 |= 1 << (8 + i - j); // mark diagonal set
}
}
}
return false;
If there is an element set in a particular row, the bit for that row is tested in rowMask. If it is already set then return true, otherwise set it using a bitwise OR so other elements can be tested against it. Do likewise for columns and the diagonals.