Handshakes That Don't Cross

class HandshakeCombinations {
    const int MOD = 1'000'000'007;
    int modularMultiply(int x, int y) { return (long long)x * y % MOD; }

public:
    int calcWays(int people) {
        int pairs = people / 2;
        vector<int> inverse(pairs + 2);
        inverse[1] = 1;
        for (int i = 2; i <= pairs + 1; i++) {
            int div = MOD / i, rem = MOD % i;
            inverse[i] = MOD - modularMultiply(div, inverse[rem]);
        }
        int result = 1;
        for (int i = 0; i < pairs; i++) {
            result = modularMultiply(modularMultiply(2 * (2 * i + 1), inverse[i + 2]), result);
        }
        return result;
    }
};

The given C++ code provides a solution for calculating the number of ways people can perform handshakes without crossing. The solution tackles the problem by considering even numbers of people (pairwise handshakes) and utilizes modular arithmetic to handle large numbers, ensuring results fit within defined limits.

• You calculate the number of non-crossing handshake ways based on the even number of people divided by two (those that can be paired). This is done in the calcWays method of the HandshakeCombinations class.
• To handle modular arithmetic efficiently, there is a modularMultiply method that multiplies two numbers under a modulo defined as 1'000'000'007.
• To invert modular values, an inverse vector is prepared. This vector uses properties of modular arithmetic to compute each number's modular inverse up to the count of pairs plus one. This step is crucial for further calculations.
• Final result computation involves iterative modular multiplications using values from the inverse vector and intermediary results from previous iterations.

Edit the source code to experiment with different values by inputting different numbers of people in the calcWays method. Ensure to maintain integral multiples for the correct calculations of pair handshakes.

class Solution {
    private static int MOD = 1000000007;

    private int multiply(int x, int y) {
        return (int) ((long) x * y % MOD);
    }

    public int countWays(int totalPeople) {
        int half = totalPeople / 2;
        int[] inversion = new int[totalPeople / 2 + 2];
        inversion[1] = 1;
        for (int i = 2; i <= half + 1; i++) {
            int div = MOD / i, rem = MOD % i;
            inversion[i] = MOD - multiply(div, inversion[rem]);
        }
        int combinations = 1;
        for (int i = 0; i < half; i++) {
            combinations = multiply(multiply(2 * (2 * i + 1), inversion[i + 2]), combinations);
        }
        return combinations;
    }
}

The Java program provided calculates the number of ways to make handshakes such that no two handshakes cross each other, given an even number of total participants. The approach follows a mathematical strategy based on the combinatorial number theory related to Catalan numbers. Here’s a brief description of how the solution works:

• Initialize a constant MOD to prevent overflow by taking results modulo (10^9 + 7).
• Implement a helper function multiply(x, y) to compute the product of two numbers under modulo operations safely.
• countWays(int totalPeople) is the main function which takes totalPeople (even) as input and determines the number of non-crossing handshake configurations:
  • Compute half as half the number of totalPeople because each unique non-crossing handshake pattern can be represented as a problem of pairing half pairs of participants.
  • Use an array inversion to store modular multiplicative inverses computed using a formula derived from properties of modular arithmetic.
  • Initialize the combinations to calculate the number of unique ways.
  • Loop through potential handshake positions and use the precomputed inverses, applying the multiplication helper method to consider all pairs and update combinations accordingly.
  • Finally, the function returns the total number of non-crossing handshake combinations computed.

This solution efficiently calculates the result by leveraging the properties of Catalan numbers and modular arithmetic, making it suitable for problems involving a large number of participants due to its optimized approach with precomputation and modular operations.

class Solution:
    def countHandshakes(self, people: int) -> int:
        mod = 1000000007
        pairs = people // 2
        inverse = [None] * (pairs+2)
        inverse[1] = 1
        for j in range(2, pairs+2):
            quotient = mod // j
            remainder = mod % j
            inverse[j] = mod - quotient * inverse[remainder] % mod
        catalan = 1
        for k in range(pairs):
            catalan = 2 * (2 * k + 1) * inverse[k + 2] * catalan % mod
        return catalan

This Python solution implements an algorithm to solve the "Handshakes That Don't Cross" problem. The problem is essentially about counting the number of ways to pair people such that no two handshakes cross each other, under the condition that people is even.

• The algorithm leverages modular arithmetic with a modulus of 1000000007 to ensure the result fits within typical integer limits and avoid overflow.
• The key mathematical concept used here is the Catalan number, which is pivotal in combinatorial problems involving properly nested and non-crossing structures.
• An array called inverse is set up to store modular inverses using the extended Euclidean algorithm. The inverse is calculated through a dynamic programming approach to prevent recalculations and improve efficiency.
• The solution iteratively computes the required Catalan number using a special product formula that involves the values from the inverse array.
• Finally, it returns the computed Catalan number, which is the answer to the problem, representing the number of valid non-crossing handshake pairings.

Assure the people input is even, and if not, the function will not function correctly as it strictly depends on pair-based calculations. Adjust the code to handle odd inputs by returning 0 or a similar error handling, as non-paired people cannot form a valid set of non-crossing handshakes.