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06_n_queen_using_bitset.cpp
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/*
Agenda - Clear some of the backtracking concepts
Questions
---------
N-Queens Problem:
- print/count 1 config
- print/count all config
Approaches
---------
1. Naive Backtracking Approach
2. Backtracing + Bitset (efficient)
3. Backtracking + Bitmasks (more efficient) [Advance Backtracking]
(i.e how to use bitmask during backtracking)
*/
#include <iostream>
#include <bitset>
using namespace std;
bitset<30> col, d1, d2;
/*
Note: - bitset is a class in c++
- bitset is memory efficient
Eg: In the above example bitset<30> col, will take memory of 30 bits
But, if we have to created an array of boolean. Then 1 bool = 8 bits
So, Then array of size 30 will take 30*8bits memory.
That's why we will be using bitset
*/
// [Using Bitset Approach] function to place N-queens on NxN board
void solveNQueens_efficient(char board[][20], int r, int n, int &count) // r:row | n: total Queens
{
// Base case
if(r == n)
{
// print the board
/* for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(board[i][j] == 1)
{
cout << "Q ";
}
else
{
cout << "_ ";
}
}
cout << endl;
}
cout << endl; */
count++;
return;
}
// Recursive case
for(int c=0; c<n; c++)
{
// if no queen is placed on the column 'col' & both diagnols 'd1' & 'd2'
if(!col[c] && !d1[r-c+n-1] && !d2[r+c])
{
// placing the queen (i.e 1 means booked)
col[c] = d1[r-c+n-1] = d2[r+c] = board[r][c] = 1;
// recursive call on next row
solveNQueens_efficient(board, r+1, n, count);
// backtracking: unplacing the queen (i.e 0 means available)
col[c] = d1[r-c+n-1] = d2[r+c] = board[r][c] = 0;
}
}
/*
When n=5
__0__ __1__ __2__ __3__ __4__
Board[5][5] = 0|_0,0_|_0,1_|_0,2_|_0,3_|_0,4_|
1|_1,0_|_1,1_|_1,2_|_1,3_|_1,4_|
2|_2,0_|_2,1_|_2,2_|_2,3_|_2,4_|
3|_3,0_|_3,1_|_3,2_|_3,3_|_3,4_|
4|_4,0_|_4,1_|_4,2_|_4,3_|_4,4_|
For Colums(i,j)
- For every ROW in a for-loop, we can refer bitset for mapping Columns
_col0 _col1 _col2 _col3 _col4
row0 |__0__|__1__|__2__|__3__|__4__|
row1 |__0__|__1__|__2__|__3__|__4__|
row2 |__0__|__1__|__2__|__3__|__4__|
row3 |__0__|__1__|__2__|__3__|__4__|
row4 |__0__|__1__|__2__|__3__|__4__|
__0__ __1__ __2__ __3__ __4__
Bitsets (for column) = 0|__1__|__0__|__0__|__0__|__1__|
cols
check if (0,1) is available i.e col[0+1] = col[1]. [Yes, its available]
check if (3,1) is available i.e col[3+1] = col[4]. [No, its not available]
For Diagnals(i,j) from Left to Right:
- i+j can be 0,1,2,3,4,5,6,7,8
__0__ __1__ __2__ __3__ __4__ Eg: 0 = (0,0)
|_0,0_|_0,1_|_0,2_|_0,3_|_0,4_|4 1 = (1,0),(0,1)
|_1,0_|_1,1_|_1,2_|_1,3_|_1,4_|5 2 = (2,0),(1,1),(0,2)
|_2,0_|_2,1_|_2,2_|_2,3_|_2,4_|6 3 = (3,0),(2,1),(1,2),(0,3)
|_3,0_|_3,1_|_3,2_|_3,3_|_3,4_|7 4 = (4,0),(3,1),(2,2),(1,3),(0,4)
|_4,0_|_4,1_|_4,2_|_4,3_|_4,4_|8 5 = (4,1),(3,2),(2,3),(4,1)
6 = (4,2),(3,3)(2,4)
/ / / / Right 7 = (4,3),(3,4)
/ / / / 8 = (4,4)
/ / / /
Left / / / /
- So, each diagnols can be mapped with unique intergers.
__0__ __1__ __2__ __3__ __4__ __5__ __6__ __7__ __8__
Bitsets (for diagnols) = 0|__0__|__0__|__0__|__0__|__0__|__1__|__0__|__1__|__0__|
d1 (diagnols)
check if (2,1) is available i.e d1[2+1] = d1[3] [Yes, its available]
check if (4,3) is available i.e d1[4+3] = d1[7] [No, its not available]
For Diagnals(i,j) Right to Left:
- i-j can be 0,1,2,3,4,5,6,7,8
Mapping with 0 to 8
__0__ _-1__ _-2__ _-3__ _-4__ Eg:-4 = (0,4) | 0 => (0-4)+4
0|_0,0_|_0,1_|_0,2_|_0,3_|_0,4_| -3 = (0,3),(1,4) | 1 => (0-3)+4, (1-4)+4
1|_1,0_|_1,1_|_1,2_|_1,3_|_1,4_| -2 = (0,2),(1,3),(2,4) | 2 => .....
2|_2,0_|_2,1_|_2,2_|_2,3_|_2,4_| -1 = (0,1),(1,2),(2,3),(3,4) | 3 => .....
3|_3,0_|_3,1_|_3,2_|_3,3_|_3,4_| 0 = (0,0),(1,1),(2,2),(3,3),(4,4) | 4 => .....
4|_4,0_|_4,1_|_4,2_|_4,3_|_4,4_| 1 = (1,0),(2,1),(3,2),(4,3) | 5 => .....
2 = (2,0),(3,1)(4,2) | 6 => .....
Right \ \ \ \ 3 = (3,0),(4,1) | 7 => .....
\ \ \ \ 4 = (4,0) | 8 => (4-0)+4
\ \ \ \
\ \ \ \ Left
- Now, the range is -4 to 4. Since, we cann't keep a array with -ve index.
So, we keep the range from 0 to 8 and
we will map -4 to 0 and 4 to 8
Thus to achive this, we will do:
= r - c + 4
= r - c + (n-1)
__0__ __1__ __2__ __3__ __4__ __5__ __6__ __7__ __8__
Bitsets (for diagnols) = 0|__0__|__0__|__0__|__0__|__0__|__1__|__0__|__1__|__0__|
d2 (diagnols)
check if (1,3) is available i.e d2[1-3+4] = d2[2] [Yes, its available]
check if (4,1) is available i.e d2[4-1+4] = d2[7] [No, its not available]
*/
}
// function to check for safe cell when placing queens on board
bool isSafe(char board[][20], int row, int col, int n)
{
//check for column
for(int i=0; i<row; i++)
{
if(board[i][col] == 1)
{
return false;
}
}
// check for left diagonal
int x = row;
int y = col;
while(x>=0 && y>=0)
{
if(board[x][y] == 1)
{
return false;
}
x--;
y--;
}
// check for right diagonal
x = row;
y = col;
while(x>=0 && y<n)
{
if(board[x][y] == 1)
{
return false;
}
x--;
y++;
}
// The position is now safe, after checking with the columns & both diagnols
return true;
}
// [Naive Backtracking Approach] function to place N-queens on NxN board
int solveNQueens(char board[][20], int row, int n) // n: total queens
{
// Base Case
if(row == n)
{
// you have successfully placed queens in n rows (0...n-1);
// print the board
/* for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(board[i][j] == 1)
{
cout << "Q ";
}
else
{
cout << "_ ";
}
}
cout << endl;
}
cout << endl; */
return 1;
}
// Recursive Case
int count = 0;
// try to place the queen on current row and call on the remaining part
for(int col=0; col<n; col++)
{
// check if (row,col)th position is safe to place the queen or not
if(isSafe(board, row, col, n))
{
// place the queen - assuming row,col is the right position
board[row][col] = 1;
count += solveNQueens(board, row+1, n);
// when (row,col) is not the right position i.e above assumption is wrong
board[row][col] = 0; // backtrack
}
}
return count;
}
// function to drive code
int main()
{
int n;
cout << "Enter Number of Queens: ";
cin >> n;
char board[20][20] = {0};
int count = 0;
// count = solveNQueens(board, 0, n); // using Naive Backtracking Approach
solveNQueens_efficient(board, 0, n, count); // using Bitset Approach
cout << "Total number of ways to place N-Queens: " << count;
cout << endl;
return 0;
}
/*
OUTPUT:
Case 1:
Enter Number of Queens: 4
_ Q _ _
_ _ _ Q
Q _ _ _
_ _ Q _
_ _ Q _
Q _ _ _
_ _ _ Q
_ Q _ _
Total number of ways to place N-Queens: 2
Case 2:
Enter Number of Queens: 3
Total number of ways to place N-Queens: 0
Case 3:
Enter Number of Queens: 1
Q
Total number of ways to place N-Queens: 1
*/