Solving CSPs with Backtracking

1. What is a Constraint Satisfaction Problem?

At its core, a CSP consists of three components. To understand them, let's start with a high-level overview.

As the article explains, a CSP is defined by:

• Variables: The decision points or objects you need to assign values to.
• Domains: The set of possible values each variable can take.
• Constraints: The rules that a valid solution must obey.

Our goal is to find an assignment of values to variables from their domains that satisfies all constraints. Backtracking is a natural fit for solving CSPs because it systematically explores assignments and backtracks as soon as a constraint is violated.

2. The N-Queens Puzzle as a CSP

The N-Queens puzzle is a classic CSP. The goal is to place N chess queens on an N×N chessboard so that no two queens threaten each other.

To understand the rules and how they translate into constraints, watch the beginning of this video from NeetCode.

Let's formally define the N-Queens puzzle as a CSP:

• Variables: The video gives us a crucial hint. Since each queen must be in a different row, we can simplify the problem. Let's say our variables are q_0, q_1, ..., q_{N-1}, where q_r represents the queen in row r.
• Domain: The value we need to assign to each variable q_r is its column index. So, for an N×N board, the domain for each variable is {0, 1, ..., N-1}.
• Constraints: Given any two queens, q_r1 at (r1, c1) and q_r2 at (r2, c2):
  1. Same Row: r1 ≠ r2. This is automatically satisfied by our choice of variables!
  2. Same Column: c1 ≠ c2.
  3. Same Diagonal: They cannot be on the same diagonal.

The main challenge is how to efficiently check the column and diagonal constraints.

3. The Backtracking Algorithm and Efficient Constraint Checking

Our backtracking strategy will be to place one queen per row, from row 0 to row N-1. For each row, we will iterate through all possible columns and check if placing a queen there is "safe".

• Choose: Pick a safe column c for the current row r.
• Explore: Recursively call the function for the next row, r + 1.
• Un-choose: If the recursive call returns (meaning it has explored all paths from that choice), remove the queen from (r, c) to explore other columns for row r.

But how do we check if (r, c) is safe in constant time? Checking the column is easy—we can use a set to store occupied columns. The diagonals are trickier, but there's a clever mathematical property we can use.

To summarize the key insight:

• For all squares (r, c) on a positive diagonal (top-left to bottom-right), the value r + c is constant.
• For all squares (r, c) on a negative diagonal (top-right to bottom-left), the value r - c is constant.

This means we can use three sets to keep track of our constraints in O(1) time:

1. A set for occupied columns.
2. A set for occupied positive diagonals (identified by r + c).
3. A set for occupied negative diagonals (identified by r - c).

When we want to place a queen at (r, c), we just check if c, r+c, and r-c are already in our sets.

4. Code Implementation

Now, let's see how this all comes together in code. The following video provides a complete walkthrough of a Python implementation.

Here's a Python implementation similar to the one in the video.

def solveNQueens(n):
    # Sets to store occupied columns and diagonals
    col_set = set()
    pos_diag_set = set() # (r + c)
    neg_diag_set = set() # (r - c)
    
    result = []
    # Initialize an empty board
    board = [["."] * n for _ in range(n)]

    def backtrack(r):
        # Base Case: If we've placed queens in all rows
        if r == n:
            # Format the board and add it to the result
            copy = ["".join(row) for row in board]
            result.append(copy)
            return

        # Try placing a queen in each column of the current row 'r'
        for c in range(n):
            # --- Constraint Check ---
            if c in col_set or (r + c) in pos_diag_set or (r - c) in neg_diag_set:
                continue # This position is not safe, try next column

            # --- Choose ---
            col_set.add(c)
            pos_diag_set.add(r + c)
            neg_diag_set.add(r - c)
            board[r][c] = "Q"

            # --- Explore ---
            backtrack(r + 1)

            # --- Un-choose (Backtrack) ---
            col_set.remove(c)
            pos_diag_set.remove(r + c)
            neg_diag_set.remove(r - c)
            board[r][c] = "."
            
    backtrack(0) # Start the process from row 0
    return result

# Example for n=4
solutions = solveNQueens(4)
for sol in solutions:
    for row in sol:
        print(row)
    print("-" * 10)

# Expected Output (order might differ):
# .Q..
# ...Q
# Q...
# ..Q.
# ----------
# ..Q.
# Q...
# ...Q
# .Q..
# ----------

5. Walkthrough and Complexity

To solidify your understanding, let's trace the algorithm for a smaller n.

Complexity Analysis

For competitive programming, understanding the performance of your algorithm is critical.

• Time Complexity: O(N!). This is a loose upper bound, but it captures the factorial nature of the problem. For the first row, we have N choices, for the second at most N-1, and so on. The actual number of operations is much smaller due to early pruning by the diagonal constraints, but O(N!) is the standard accepted complexity. (Note: A factor of N^2 is sometimes included for building the string representation of each solution found, making it O(N^2 * N!)).
• Space Complexity: O(N^2) to store the board and the results. The recursion depth is N, and the sets also store up to O(N) elements, but the O(N^2) space for the board dominates.

Conclusion

Congratulations on completing the backtracking module! You have progressed from simple generation problems to a complex constraint satisfaction problem. This is a significant step in your competitive programming journey.

Key Takeaways:

• The N-Queens puzzle is a classic Constraint Satisfaction Problem (CSP) that can be modeled with variables (queen in each row), domains (columns), and constraints (no attacks).
• Backtracking provides a natural template to solve CSPs by exploring choices and pruning the search space as soon as a constraint is violated.
• For N-Queens, the key to an efficient solution is using sets and the properties of diagonals (r+c and r-c) to perform O(1) constraint checks.
• The "Choose, Explore, Un-choose" pattern remains central, but the "Choose" step now includes a critical validity check.

Up Next:
We will now move to a completely new topic: Heaps and Priority Queues. We'll leave the world of recursive search and explore a tree-based data structure that is incredibly efficient for problems involving finding the minimum or maximum element, such as "Top K" problems and implementing Dijkstra's algorithm.