• Key Insight
• Code Implementations
• Complexity Analysis
• Conclusion

Codeforces Problem 1904A - Forked!

Both Logical Approach and Optimized Approach: All Confusions Cleared

problem link: Codeforces Problem 1904A - Forked! from Codeforces.

Welcome to R.A.S Coders! In today’s post, we tackle an intriguing problem involving an infinite chessboard and knights with custom movement patterns. We’ll explore how to break the problem into a bounded solution, visualize the knight’s moves in the context of real chess, and discuss approaches to arrive at the final solution with code in multiple languages.

Part 1: Converting the Infinite Problem to a Bounded Problem

At first glance, the problem seems daunting because the chessboard is infinite, and the knight could theoretically be placed anywhere. However, by observing key constraints, we can transform this infinite problem into a manageable one.

The knight’s ability to attack the king and queen is directly dependent on their positions:

• If the king and queen are positioned on the board, the potential positions from which a knight can attack them are naturally bounded by the knight’s movement pattern.

Bounding the Problem

Let’s consider a custom knight with moves “(a, b)” and their permutations. These moves define the possible cells the knight can reach in a single step:

• Positive and negative combinations of “(a, b):”
• (a, b), (a, -b), (-a, b), (-a, -b)
• (b, a), (b, -a), (-b, a), (-b, -a)

Given the positions of the king “(xK, yK)” and queen “(xQ, yQ):”

1. Compute all cells from which a knight could attack the king by applying the moves above relative to “(xK, yK)”.
2. Similarly, compute all cells from which a knight could attack the queen by applying the moves above relative to “(xQ, yQ)”.

This creates two finite sets of possible positions for the knight, dramatically reducing the search space. By focusing only on these bounded sets, we avoid unnecessary computations across the infinite chessboard.

Part 2: Visualizing with a Real Chess Example

To make the knight’s movements more intuitive, let’s use a standard chess example. In classical chess, a knight moves in an “L” shape, jumping 2 squares in one direction and 1 square in a perpendicular direction. This results in up to 8 possible positions from any given cell.

Example Scenario

Suppose the king is placed at (4, 4), and the queen is at (6, 6). To attack both pieces simultaneously, the knight must occupy a cell that satisfies the attack criteria for both. Let’s break this down:

1. From the king’s position (4, 4), a standard chess knight can move to the following positions:
• (6, 5), (6, 3), (5, 6), (5, 2), (2, 3), (2, 5), (3, 2), (3, 6)
2. From the queen’s position (6, 6), the knight’s potential positions are:
• (8, 7), (8, 5), (7, 8), (7, 4), (4, 5), (4, 7), (5, 4), (5, 8)

Finding Overlaps

By comparing these sets of positions, we identify common cells where the knight can simultaneously attack both the king and queen. This overlap determines the solution to our problem.

By visualizing the knight’s moves relative to the king and queen, we gain a clearer understanding of how the solution emerges.

Part 3: Deriving the Final Approach

Now that we understand the problem and its visualization, let’s formalize the solution. Here’s how we can compute the knight’s valid positions step by step:

Step 1: Calculate Knight’s Attack Positions for the King

Using the custom moves (a, b) and their permutations, calculate all positions from which the knight can attack the king’s position (xK, yK). Store these positions in a set called kingAttacks.

Step 2: Calculate Knight’s Attack Positions for the Queen

Similarly, calculate all positions from which the knight can attack the queen’s position (xQ, yQ). Store these positions in another set called queenAttacks.

Step 3: Find Common Positions

To find where the knight can attack both the king and queen, compute the intersection of kingAttacks and queenAttacks. The size of this intersection gives the number of valid positions.

Code Implementations

Below are the implementations in Java and C++. Note: To understand optimised code please refer the description given in part 4 below this code

First Approach (Set-based)

import java.util.*;

public class KnightAttack {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // Number of test cases

        while (t-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int xK = sc.nextInt();
            int yK = sc.nextInt();
            int xQ = sc.nextInt();
            int yQ = sc.nextInt();

            System.out.println(solve(a, b, xK, yK, xQ, yQ));
        }

        sc.close();
    }

    // Function to solve a single test case
    private static int solve(int a, int b, int xK, int yK, int xQ, int yQ) {
        int[][] moves = {
            {a, b}, {a, -b}, {-a, b}, {-a, -b},
            {b, a}, {b, -a}, {-b, a}, {-b, -a}
        };

        Set kingAttacks = new HashSet<>();
        Set queenAttacks = new HashSet<>();

        for (int[] move : moves) {
            kingAttacks.add((xK + move[0]) + "," + (yK + move[1]));
            queenAttacks.add((xQ + move[0]) + "," + (yQ + move[1]));
        }

        kingAttacks.retainAll(queenAttacks);
        return kingAttacks.size();
    }
}
        

First Approach (Set-based)

     
#include <iostream>
#include <unordered_set>
#include <string>
#include <vector>

using namespace std;

// Function to solve a single test case
int solve(int a, int b, int xK, int yK, int xQ, int yQ) {
    // All possible knight moves
    vector<pair<int, int>> moves = {
        {a, b}, {a, -b}, {-a, b}, {-a, -b},
        {b, a}, {b, -a}, {-b, a}, {-b, -a}
    };

    unordered_set<string> kingAttacks, queenAttacks;

    // Calculate all attack positions for the king and queen
    for (auto move : moves) {
        kingAttacks.insert(to_string(xK + move.first) + "," + to_string(yK + move.second));
        queenAttacks.insert(to_string(xQ + move.first) + "," + to_string(yQ + move.second));
    }

    // Find the intersection of attack positions
    int count = 0;
    for (const string& pos : kingAttacks) {
        if (queenAttacks.find(pos) != queenAttacks.end()) {
            count++;
        }
    }

    return count;
}

int main() {
    int t; // Number of test cases
    cin >> t;

    while (t-- > 0) {
        int a, b, xK, yK, xQ, yQ;
        cin >> a >> b >> xK >> yK >> xQ >> yQ;

        cout << solve(a, b, xK, yK, xQ, yQ) << endl;
    }

    return 0;
}

        

Optimized Approach (Java)

import java.util.*;
 
public class KnightAttack {
 
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // Number of test cases
 
        while (t-- > 0) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int xK = sc.nextInt();
            int yK = sc.nextInt();
            int xQ = sc.nextInt();
            int yQ = sc.nextInt();
 
            System.out.println(solve(a, b, xK, yK, xQ, yQ));
        }
 
        sc.close();
    }
 
    // Optimized function to solve a single test case
    private static int solve(int a, int b, int xK, int yK, int xQ, int yQ) {
        int[][] moves = {
            {a, b}, {a, -b}, {-a, b}, {-a, -b},
            {b, a}, {b, -a}, {-b, a}, {-b, -a}
        };
 
        int count = 0;
 
        // Check each possible knight move
        for (int[] move : moves) {
            int knightX = xK + move[0];
            int knightY = yK + move[1];
 
            // Check if the knight at (knightX, knightY) can also attack the queen
            if (Math.abs(knightX - xQ) == a && Math.abs(knightY - yQ) == b ||
                Math.abs(knightX - xQ) == b && Math.abs(knightY - yQ) == a) {
                count++;
            }
        }
        if(a==b) count/=2;
        return count;
    }
}
        

Optimized Approach (C++)

#include <iostream>
#include <unordered_set>
#include <string>

using namespace std;

// Function to solve a single test case using the optimized approach
int solve(int a, int b, int xK, int yK, int xQ, int yQ) {
    int moves[8][2] = {
        {a, b}, {a, -b}, {-a, b}, {-a, -b},
        {b, a}, {b, -a}, {-b, a}, {-b, -a}
    };

    int count = 0;

    // Check each possible knight move
    for (int i = 0; i < 8; i++) {
        int knightX = xK + moves[i][0];
        int knightY = yK + moves[i][1];

        // Check if the knight at (knightX, knightY) can also attack the queen
        if ((abs(knightX - xQ) == a && abs(knightY - yQ) == b) ||
            (abs(knightX - xQ) == b && abs(knightY - yQ) == a)) {
            count++;
        }
    }

    // Handle symmetry if a == b
    if (a == b) count /= 2;

    return count;
}

int main() {
    int t; // Number of test cases
    cin >> t;

    while (t-- > 0) {
        int a, b, xK, yK, xQ, yQ;
        cin >> a >> b >> xK >> yK >> xQ >> yQ;

        cout << solve(a, b, xK, yK, xQ, yQ) << endl;
    }

    return 0;
}

        

Diving Deeper: Optimized Approach for Knight's Moves

In the previous implementation, we used a set-based approach to compute and compare attack positions for the king and queen. While effective, we can further optimize the process by leveraging a direct computation approach that focuses only on valid overlapping positions.

This optimized method reduces unnecessary storage and operations, ensuring that we only compute what is strictly required.

Breaking Down the Optimized Solution

1. Defining Knight’s Moves:

   The knight’s movement is characterized by all permutations of its custom moves (a, b) and (b, a) with both positive and negative signs. These moves are represented in an array for easy iteration:

   
   int[][] moves = {
       {a, b}, {a, -b}, {-a, b}, {-a, -b},
       {b, a}, {b, -a}, {-b, a}, {-b, -a}
   };
           

2. Counting Valid Positions:

   Instead of storing all possible attack positions for the king and queen, the code directly checks each potential knight position derived from the king’s location (xK, yK). For each computed position (knightX, knightY), it verifies if the knight can also attack the queen’s position (xQ, yQ):

   
   if (Math.abs(knightX - xQ) == a && Math.abs(knightY - yQ) == b ||
       Math.abs(knightX - xQ) == b && Math.abs(knightY - yQ) == a) {
       count++;
   }
           

3. Handling Symmetry:

   If the knight’s moves are symmetric (i.e., a == b), certain positions may be counted twice. The code accounts for this by halving the count of valid positions:

   
   if (a == b) count /= 2;
           

4. Returning the Result:

   The total count of positions where the knight can simultaneously attack both the king and queen is returned for each test case.

Why This Approach Works

This method eliminates the need for additional storage (e.g., sets for attack positions) by directly checking the overlapping conditions for each possible knight move. By doing so, it not only simplifies the logic but also optimizes performance, especially for scenarios with a large number of test cases or varied input sizes.

Benefits of the Optimized Solution

• Efficiency: Reduces memory usage and unnecessary computations by avoiding intermediate data structures.
• Clarity: Focuses on the core logic—validating overlapping attack positions—making it straightforward to follow.

Time Complexity and Space Complexity Explanation

We have two different approaches to solve this problem. Let's break down the time and space complexities of both:

Optimized Approach

In the optimized approach, we make use of efficient arithmetic and logic to calculate the possible knight attack positions. The main operations involve constant-time calculations for determining the attackable positions relative to the king and queen's positions.

Time Complexity: The time complexity for the optimized approach is O(1). This is because we are only performing a fixed number of arithmetic operations regardless of the input values. No iteration over the board or any other complex operations are required.

Space Complexity: The space complexity is O(1), as we only need a few variables to store the positions of the king, queen, and the movement parameters of the knight. No additional memory is required for storing large data structures.

Set-based Approach

The second approach involves using a set to track the positions where the knight can attack the king and queen. This method requires us to check all potential attack positions using the set data structure.

Time Complexity: The time complexity for this approach is O(1), as the set operations (insertion and lookup) are typically constant-time operations. The logic for checking the knight's attack positions remains the same.

Space Complexity: The space complexity is O(n), where 'n' is the number of knight positions stored in the set. Although the set only stores unique positions, the size of the set might grow depending on the number of valid attack positions. However, in this case, 'n' remains constant due to the limited number of knight moves. Since the number of possible knight moves is fixed and does not change with the input size, 'n' is effectively constant, making the space complexity O(1).

Why the Optimized Approach is Better

Although both approaches have the same time complexity of O(1), the optimized approach is generally faster because it avoids the overhead of using additional data structures like sets. The optimized method directly calculates the attack positions using arithmetic, making it more efficient in practice. The space complexity is also reduced to O(1), as no extra storage is needed.

Conclusion

This problem demonstrates how infinite scenarios can often be reduced to finite, manageable solutions through critical observations. By focusing on the king and queen’s positions, we bounded the problem space. A real chess analogy helped us visualize the knight’s movements, and we solved it using both set-based and optimized approaches

We hope you found this problem as engaging as we did! Stay tuned for more coding insights and solutions on R.A.S Coders.