From 05f44e646391e83c89ed8bae87a7c2e48f90531b Mon Sep 17 00:00:00 2001
From: Tatheer Fathima <148070120+T-Fathima@users.noreply.github.com>
Date: Sun, 10 Nov 2024 03:40:55 +0000
Subject: [PATCH] Added modular exponentiation in Number Theory

---
 docs/Number theory/modular-exponentiation.md | 128 +++++++++++++++++++
 1 file changed, 128 insertions(+)
 create mode 100644 docs/Number theory/modular-exponentiation.md

diff --git a/docs/Number theory/modular-exponentiation.md b/docs/Number theory/modular-exponentiation.md
new file mode 100644
index 000000000..65282016c
--- /dev/null
+++ b/docs/Number theory/modular-exponentiation.md	
@@ -0,0 +1,128 @@
+---
+id: modular-exponentiation
+title: "Modular Exponentiation"
+sidebar_label: "Efficient Power Calculation"
+sidebar_position: 15
+description: "A comprehensive guide on Modular Exponentiation for fast computation of large powers in modular arithmetic."
+tags: [Modular Exponentiation, number theory, cryptography, efficient algorithms]
+---
+
+# Modular Exponentiation
+
+Modular exponentiation is an efficient method for computing large powers modulo some integer, commonly used in cryptography and other fields involving large number computations. This technique significantly reduces the time complexity from \(O(b)\) to \(O(\log b)\), making it feasible to compute large powers quickly.
+
+## Basics of Modular Exponentiation
+
+The goal is to compute \(a^b \mod m\) efficiently. Instead of multiplying \(a\) by itself \(b\) times, we use an algorithm that divides the exponentiation process in half each time, working in logarithmic time.
+
+### Example
+
+To compute \(3^{13} \mod 7\):
+1. Break down the exponent 13 into powers of 2 (binary representation).
+2. Compute each power modulo 7 and combine using properties of modular arithmetic.
+
+## Algorithm: Fast Exponentiation
+
+### Recursive Method
+A recursive approach to modular exponentiation calculates each intermediate step modulo \(m\).
+# Python Implementation
+```python
+def mod_exp(base, exponent, modulus):
+    if exponent == 0:
+        return 1
+    half_exp = mod_exp(base, exponent // 2, modulus)
+    half_exp = (half_exp * half_exp) % modulus
+    if exponent % 2 != 0:
+        half_exp = (half_exp * base) % modulus
+    return half_exp
+```
+# C++ Implementation
+```cpp
+#include <iostream>
+using namespace std;
+
+long long mod_exp(long long base, long long exponent, long long modulus) {
+    if (exponent == 0) {
+        return 1;
+    }
+    
+    long long half_exp = mod_exp(base, exponent / 2, modulus);
+    half_exp = (half_exp * half_exp) % modulus;
+    
+    if (exponent % 2 != 0) {
+        half_exp = (half_exp * base) % modulus;
+    }
+    
+    return half_exp;
+}
+
+int main() {
+    long long base, exponent, modulus;
+    cout << "Enter base, exponent, and modulus: ";
+    cin >> base >> exponent >> modulus;
+    
+    cout << "Result: " << mod_exp(base, exponent, modulus) << endl;
+    
+    return 0;
+}
+```
+
+# Iterative Method
+An iterative version avoids recursion by using bitwise operations to halve the exponent at each step.
+
+# Python Implementation
+```python
+def mod_exp(base, exponent, modulus):
+    result = 1
+    base = base % modulus
+    while exponent > 0:
+        if (exponent % 2) == 1:
+            result = (result * base) % modulus
+        exponent = exponent >> 1
+        base = (base * base) % modulus
+    return result
+```
+
+# C++ Implementation
+```cpp
+#include <iostream>
+using namespace std;
+
+long long mod_exp(long long base, long long exponent, long long modulus) {
+    long long result = 1;
+    base = base % modulus;  // Ensure base is within the modulus
+    
+    while (exponent > 0) {
+        // If exponent is odd, multiply result by the base
+        if (exponent % 2 == 1) {
+            result = (result * base) % modulus;
+        }
+        // Halve the exponent and square the base
+        exponent = exponent >> 1;
+        base = (base * base) % modulus;
+    }
+    
+    return result;
+}
+
+int main() {
+    long long base, exponent, modulus;
+    cout << "Enter base, exponent, and modulus: ";
+    cin >> base >> exponent >> modulus;
+    
+    cout << "Result: " << mod_exp(base, exponent, modulus) << endl;
+    
+    return 0;
+}
+```
+
+# Applications of Modular Exponentiation
+Modular exponentiation has applications in areas where fast computation of powers modulo a number is essential:
+
+1. Cryptography: Used in RSA and Diffie-Hellman protocols to secure data by performing computations on large numbers efficiently.
+2. Digital Signatures: Ensures message integrity and authenticity by leveraging fast modular power computations.
+3. Random Number Generation: Modular exponentiation helps generate random sequences within a specific range.
+
+# Time Complexity
+The time complexity of modular exponentiation, both in the recursive and iterative versions, is 
+O(logb), where b is the exponent.
\ No newline at end of file