Added Chinese remainder theorem

Number_Theory_Functions/Chinese_remainder_theorem/Chinese_remainder_theorem(README).md

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+
+# **Chinese Remainder Theorem**
+
+The Chinese Remainder Theorem (CRT) was discovered by the Chinese mathematician Sun Zi.
+
+## Formulation
+
+Let \( m = m₁ × m₂ × ⋯ × mₖ \), where \( mᵢ \) are pairwise coprime. In addition to \( mᵢ \), we are also given a system of congruences:
+
+a ≡ a₁ (mod m₁) a ≡ a₂ (mod m₂) ⋮ a ≡ aₖ (mod mₖ)
+
+where \( aᵢ \) are given constants. The original form of CRT states that the given system of congruences always has **one and exactly one solution modulo \( m \)**.
+
+
+## Solution for General Case
+
+### Direct Construction
+A direct construction similar to Lagrange interpolation is possible.
+
+Let:
+
+`Mᵢ := Π_{j ≠ i} mⱼ`
+
+(the product of all moduli except `mᵢ`) and:
+
+`Nᵢ := Mᵢ⁻¹ (mod mᵢ)`
+
+Then, a solution to the system of congruences is:
+
+`a ≡ Σ_{i=1}^{k} aᵢMᵢNᵢ (mod m₁m₂...mₖ)`
+
+---
+
+### Verification
+We can check this is indeed a solution by computing `a (mod mᵢ)` for all `i`. Because `Mⱼ` is a multiple of `mᵢ` for `i ≠ j`, we have:
+
+`a ≡ Σ_{j=1}^{k} aⱼMⱼNⱼ (mod mᵢ)`
+
+`≡ aᵢMᵢNᵢ (mod mᵢ)`
+
+`≡ aᵢMᵢMᵢ⁻¹ (mod mᵢ)`
+
+`≡ aᵢ (mod mᵢ)`
+
+
+
+### Generalized Chinese Remainder Theorem (CRT) for Non-Coprime Moduli
+
+When the moduli m₁, m₂, ..., mₖ are **not pairwise coprime**, the standard Chinese Remainder Theorem cannot be directly applied. However, such systems can still be solved under certain conditions.
+
+---
+
+To solve a system of congruences with non-coprime moduli:
+
+1. **Define the System**:
+   x ≡ aᵢ (mod mᵢ), for i = 1, 2, ..., k
+
+2. **Check Pairwise Consistency**:
+   - For each pair of congruences:
+     x ≡ aᵢ (mod mᵢ)
+     x ≡ aⱼ (mod mⱼ)
+
+     - Compute gcd(mᵢ, mⱼ) = g.
+     - Verify aᵢ ≡ aⱼ (mod g). If this condition is violated, the system is inconsistent.
+
+3. **Combine Consistent Congruences**:
+   - Combine congruences iteratively using the least common multiple (lcm) of moduli:
+     x ≡ a (mod lcm(mᵢ, mⱼ))
+
+4. **Solve the Final Reduced System**:
+   - After reducing all congruences, solve the resulting single congruence.
+
+---
+
+## Example: Solving a System with Non-Coprime Moduli
+
+Solve the system:
+x ≡ 2 (mod 6)
+x ≡ 3 (mod 9)
+
+### Step 1: Check Consistency
+- Compute gcd(6, 9) = 3.
+- Verify 2 ≡ 3 (mod 3). This condition is satisfied because both leave a remainder of 2 modulo 3.
+
+### Step 2: Combine Congruences
+- Combine the two congruences into one:
+  x ≡ 2 (mod lcm(6, 9)) = x ≡ 2 (mod 18)
+
+### Final Solution:
+x ≡ 2 (mod 18)
+
+---
+
+## Notes:
+- If the system is inconsistent, no solution exists.
+- If all moduli are coprime, the standard Chinese Remainder Theorem applies.
+
+## USE OF CHINESE REMAINDER THEOREM IN SOLVING PROBLEMS
+Problem: Multiple Congruences with Mixed Moduli and Constraints
+
+Sure! Let's tackle a more complex problem that involves the Chinese Remainder Theorem (CRT). Here's a more challenging problem that would require using CRT, along with some advanced concepts like handling large inputs, dealing with constraints, and considering edge cases.
+
+Problem: Multiple Congruences with Mixed Moduli and Constraints
+You are given several systems of congruences. Each system can have different moduli and constraints. Your goal is to use the Chinese Remainder Theorem to solve this.
+
+### Problem Statement:
+
+Given a system of `n` congruences of the form:
+x ≡ a₁ (mod m₁) x ≡ a₂ (mod m₂) ... x ≡ aₙ (mod mₙ)
+
+Where `mᵢ` and `mⱼ` (for all `i ≠ j`) are **not necessarily coprime** (i.e., they may share common factors). The task is to find the smallest `x` that satisfies all these congruences, or determine that there is **no solution**.
+
+### Input Format:
+- The first line contains a single integer `n` — the number of congruences.
+- The second line contains `n` space-separated integers, each representing a modulus: `m₁, m₂, ..., mₙ`.
+- The third line contains `n` space-separated integers, each representing the corresponding remainder: `a₁, a₂, ..., aₙ`.
+
+
+
+### Output Format:
+- If a solution exists, output the smallest `x` that satisfies all the congruences.
+- If no solution exists, output `No solution`.
+
+Here,to solve this problem, we have to use the extended form of chinese remainder theorem as the given moduli are not given to be coprime i.e.
+gcd(m<sub>i</sub>,m<sub>j</sub>)|abs(a<sub>i</sub>-a<sub>j</sub>)
+
+
+The following code shows the implementation of above logic
+
+
+```cpp
+#include <iostream>
+#include <vector>
+#include <numeric> // For gcd
+using namespace std;
+
+// Function to find the modular inverse of a under modulo m (when gcd(a, m) == 1)
+int modInverse(int a, int m) {
+    int m0 = m, t, q;
+    int x0 = 0, x1 = 1;
+
+    while (a > 1) {
+        q = a / m;
+        t = m;
+        m = a % m;
+        a = t;
+        t = x0;
+        x0 = x1 - q * x0;
+        x1 = t;
+    }
+
+    if (x1 < 0)
+        x1 += m0;
+
+    return x1;
+}
+
+// Function to solve the system of congruences using the Chinese Remainder Theorem
+int chineseRemainder(vector<int> num, vector<int> rem) {
+    int prod = 1;
+
+    // Compute the product of all numbers in num (product of all moduli)
+    for (int n : num)
+        prod *= n;
+
+    int result = 0;
+
+    // Loop through each congruence
+    for (size_t i = 0; i < num.size(); i++) {
+        int pp = prod / num[i];  // Partial product excluding num[i]
+        result += rem[i] * modInverse(pp, num[i]) * pp;
+    }
+
+    // Return the result modulo the product of all moduli
+    return result % prod;
+}
+
+// Function to check if the system of congruences is solvable
+bool isSolvable(const vector<int>& num, const vector<int>& rem) {
+    int n = num.size();
+    for (int i = 0; i < n; i++) {
+        for (int j = i + 1; j < n; j++) {
+            int g = gcd(num[i], num[j]);
+            if (g > 1 && abs(rem[i] - rem[j]) % g != 0) {
+                return false;  // If gcd doesn't divide the difference of remainders, no solution
+            }
+        }
+    }
+    return true;
+}
+
+// Function to solve the system of congruences even when moduli are not coprime
+bool extendedChineseRemainder(vector<int>& num, vector<int>& rem) {
+    int n = num.size();
+
+    // Loop through the moduli and solve pairwise
+    for (int i = 0; i < n; i++) {
+        for (int j = i + 1; j < n; j++) {
+            int g = gcd(num[i], num[j]);
+            // Check if the difference of remainders is divisible by the gcd
+            if (abs(rem[i] - rem[j]) % g != 0) {
+                return false; // No solution if this condition fails
+            }
+
+            // Solve pairwise congruences using the Extended Euclidean Algorithm
+            int x1 = rem[i], x2 = rem[j];
+            int m1 = num[i], m2 = num[j];
+
+            // Calculate the combined modulus
+            int newMod = (m1 * m2) / g;
+            int newRem = 0;
+
+            // Solve the pairwise system (x1 mod m1) and (x2 mod m2)
+            int g1, x, y;
+            extendedGCD(m1, m2, g1, x, y);
+            x = ((x * (x2 - x1) / g) % (m2)) + m1;
+            newRem = (x * m1 + x1) % newMod;
+
+            // Update moduli and remainders for the new system
+            num[i] = newMod;
+            rem[i] = newRem;
+        }
+    }
+
+    return true;
+}
+
+// Function to find the Extended GCD of two numbers (also computes the coefficients x and y such that ax + by = gcd(a, b))
+void extendedGCD(int a, int b, int& g, int& x, int& y) {
+    if (a == 0) {
+        g = b;
+        x = 0;
+        y = 1;
+        return;
+    }
+
+    int x1, y1;
+    extendedGCD(b % a, a, g, x1, y1);
+    x = y1 - (b / a) * x1;
+    y = x1;
+}
+
+// Driver code
+int main() {
+    int n;
+    cin >> n;
+
+    vector<int> num(n);
+    vector<int> rem(n);
+
+    // Input the moduli and remainders
+    for (int i = 0; i < n; i++) {
+        cin >> num[i];
+    }
+
+    for (int i = 0; i < n; i++) {
+        cin >> rem[i];
+    }
+
+    // Check if the system is solvable
+    if (!isSolvable(num, rem)) {
+        cout << "No solution" << endl;
+        return 0;
+    }
+
+    // Solve the system of congruences using the Extended Chinese Remainder Theorem
+    if (!extendedChineseRemainder(num, rem)) {
+        cout << "No solution" << endl;
+    } else {
+        int result = chineseRemainder(num, rem);
+        cout << result << endl;
+    }
+
+    return 0;
+}
+
+
+
+
+```

Number_Theory_Functions/Chinese_remainder_theorem/Chinese_remainder_theorem.cpp

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+#include <iostream>
+#include <vector>
+using namespace std;
+
+// Function to find the smallest x such that:
+// x % num[i] = rem[i] for all i in [0, k-1]
+int chineseRemainder(vector<int> num, vector<int> rem) {
+    int prod = 1;
+
+    // Compute the product of all numbers in num (product of all moduli)
+    for (int n : num)
+        prod *= n;
+
+    // Lambda function to compute the modular inverse of a under modulo m
+    auto modInverse = [](int a, int m) {
+        int m0 = m, t, q;
+        int x0 = 0, x1 = 1;
+
+        // Apply the Extended Euclidean Algorithm
+        while (a > 1) {
+            // q is the quotient
+            q = a / m;
+            t = m;
+            // Update m and a (remainder and divisor)
+            m = a % m, a = t;
+            t = x0;
+            // Update coefficients
+            x0 = x1 - q * x0;
+            x1 = t;
+        }
+
+        // Make x1 positive if it is negative
+        if (x1 < 0)
+            x1 += m0;
+
+        return x1;
+    };
+
+    int result = 0;
+
+    // Loop through each congruence
+    for (size_t i = 0; i < num.size(); i++) {
+        int pp = prod / num[i];  // Partial product excluding num[i]
+        // Add the contribution of the current remainder to the result
+        result += rem[i] * modInverse(pp, num[i]) * pp;
+    }
+
+    // Return the result modulo the product of all moduli
+    return result % prod;
+}
+
+int main() {
+    // Example usage
+    vector<int> num = {3, 4, 5};  // Moduli
+    vector<int> rem = {2, 3, 1};  // Remainders
+
+    // Solve the system of congruences using the Chinese Remainder Theorem
+    int result = chineseRemainder(num, rem);
+
+    // Output the result
+    cout << "The solution is " << result << endl;
+
+    return 0;
+}