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# Exponential Squaring (Fast Modulo Multiplication)

Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7

Examples:

Input : base = 2, exp = 2
Output : 4

Input  : base = 5, exp = 100000
Output : 754573817

In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the values of is being calculated it can get very large so instead we have to calculate (%modulus value.) We can use the modulus in our naive way by using modulus on all the intermediate steps and take modulus at the end, but in competitions it will definitely show TLE. So, what we can do. The answer is we can try exponentiation by squaring which is a fast method for calculating exponentiation of a number. Here we will be discussing two most common/important methods:

1. Basic Method(Binary Exponentiation)
2. -ary method.

Binary Exponentiation

As described in this article we will be using following formula to recursively calculate (%modulus value):

## C++

 // C++ program to compute exponential // value under modulo using binary // exponentiation. #include using namespace std;   #define N 1000000007 // prime modulo value   long int exponentiation(long int base,                         long int exp) {     if (exp == 0)         return 1;       if (exp == 1)         return base % N;       long int t = exponentiation(base, exp / 2);     t = (t * t) % N;       // if exponent is even value     if (exp % 2 == 0)         return t;       // if exponent is odd value     else         return ((base % N) * t) % N; }   // Driver Code int main() {     long int base = 5;     long int exp = 100000;       long int modulo = exponentiation(base, exp);     cout << modulo << endl;     return 0; }   // This Code is contributed by mits

## Java

 // Java program to compute exponential value under modulo // using binary exponentiation. import java.util.*; import java.lang.*; import java.io.*;   class exp_sq {     static long N = 1000000007L; // prime modulo value     public static void main(String[] args)     {         long base = 5;         long exp = 100000;           long modulo = exponentiation(base, exp);         System.out.println(modulo);     }       static long exponentiation(long base, long exp)     {         if (exp == 0)             return 1;           if (exp == 1)             return base % N;           long t = exponentiation(base, exp / 2);         t = (t * t) % N;           // if exponent is even value         if (exp % 2 == 0)             return t;           // if exponent is odd value         else             return ((base % N) * t) % N;     } }

## Python3

 # Python3 program to compute # exponential value under # modulo using binary # exponentiation.   # prime modulo value N = 1000000007;       # Function code def exponentiation(bas, exp):     if (exp == 0):         return 1;     if (exp == 1):         return bas % N;           t = exponentiation(bas, int(exp / 2));     t = (t * t) % N;           # if exponent is     # even value     if (exp % 2 == 0):         return t;               # if exponent is     # odd value     else:         return ((bas % N) * t) % N;   # Driver code bas = 5; exp = 100000;   modulo = exponentiation(bas, exp); print(modulo);   # This code is contributed # by mits

## C#

 // C# program to compute exponential // value under modulo using binary // exponentiation. using System;   class GFG {           // prime modulo value     static long N = 1000000007L;           // Driver code     public static void Main()     {         long bas = 5;         long exp = 100000;           long modulo = exponentiation(bas, exp);         Console.Write(modulo);     }       static long exponentiation(long bas, long exp)     {         if (exp == 0)             return 1;           if (exp == 1)             return bas % N;           long t = exponentiation(bas, exp / 2);         t = (t * t) % N;           // if exponent is even value         if (exp % 2 == 0)             return t;           // if exponent is odd value         else             return ((bas % N) * t) % N;     } }   // This code is contributed by nitin mittal.



## Javascript



Output :

754573817

Time Complexity: O(log exp) since the binary exponentiation algorithm divides the exponent by 2 at each recursive call, resulting in a logarithmic number of recursive calls.

Space Complexity: O(log exp)

-ary method:

In this algorithm we will be expanding the exponent in base (k>=1), which is somehow similar to above method except we are not using recursion this method uses comparatively less memory and time.

## C++

 // C++ program to compute exponential value using (2^k) // -ary method. #include using namespace std;   #define N 1000000007L; // prime modulo value   long exponentiation(long base, long exp) {     long t = 1L;     while (exp > 0)     {           // for cases where exponent         // is not an even value         if (exp % 2 != 0)             t = (t * base) % N;           base = (base * base) % N;         exp /= 2;     }     return t % N; }   // Driver code int main() {     long base = 5;     long exp = 100000;       long modulo = exponentiation(base, exp);     cout << (modulo);     return 0; }   // This code is contributed by Rajput-Ji

## Java

 // Java program to compute exponential value using (2^k) // -ary method. import java.util.*; import java.lang.*; import java.io.*;   class exp_sq {     static long N = 1000000007L; // prime modulo value     public static void main(String[] args)     {         long base = 5;         long exp = 100000;           long modulo = exponentiation(base, exp);         System.out.println(modulo);     }       static long exponentiation(long base, long exp)     {         long t = 1L;         while (exp > 0) {               // for cases where exponent             // is not an even value             if (exp % 2 != 0)                 t = (t * base) % N;               base = (base * base) % N;             exp /= 2;         }         return t % N;     } }

## Python3

 # Python3 program to compute # exponential value # using (2^k) -ary method.   # prime modulo value N = 1000000007;   def exponentiation(bas, exp):     t = 1;     while(exp > 0):           # for cases where exponent         # is not an even value         if (exp % 2 != 0):             t = (t * bas) % N;           bas = (bas * bas) % N;         exp = int(exp / 2);     return t % N;   # Driver Code bas = 5; exp = 100000;   modulo = exponentiation(bas,exp); print(modulo);   # This code is contributed # by mits

## C#

 // C# program to compute // exponential value // using (2^k) -ary method. using System;   class GFG { // prime modulo value static long N = 1000000007L;   static long exponentiation(long bas,                            long exp) {     long t = 1L;     while (exp > 0)     {           // for cases where exponent         // is not an even value         if (exp % 2 != 0)             t = (t * bas) % N;           bas = (bas * bas) % N;         exp /= 2;     }     return t % N; }   // Driver Code    public static void Main () {     long bas = 5;     long exp = 100000;       long modulo = exponentiation(bas,                                  exp);     Console.WriteLine(modulo); } }   //This code is contributed by ajit

## PHP

 0)     {           // for cases where exponent         // is not an even value         if ($exp % 2 != 0)$t = ($t *$bas) % $N;$bas = ($bas *$bas) % $N;$exp = (int)$exp / 2; } return$t % $N; } // Driver Code$bas = 5; $exp = 100000;$modulo = exponentiation($bas,$exp); echo (\$modulo);   // This code is contributed // by ajit ?>

## Javascript

 // JavaScript program to compute // exponential value // using (2^k) -ary method.   // prime modulo value let N = 1000000007n;   function exponentiation(bas, exp) {     let t = 1n;     while(exp > 0n)     {         // for cases where exponent         // is not an even value         if (exp % 2n != 0n)             t = (t * bas) % N;                             bas = (bas * bas) % N;         exp >>= 1n;     }     return t % N; }     // Driver Code let bas = 5n; let exp = 100000n;   let modulo = exponentiation(bas,exp); console.log(Number(modulo));     // This code is contributed // by phasing17

Output :

754573817

Time Complexity: O(log exp)

Space Complexity: O(1)

### Bit-Manipulation Method

The basic idea behind the algorithm is to use the binary representation of the exponent to compute the power in a faster way.

Specifically, if we can represent the exponent as a sum of powers of 2, then we can use the fact that x^(a+b) = x^a * x^b to compute the power.

Approach :

The steps of the algorithm are as follows :

1. Initialize a result variable to 1, and a base variable to the given base value.
2. Convert the exponent to binary format.
3. Iterate over the bits of the binary representation of the exponent, from right to left.
4. For each bit, square the current value of the base.

5. If the current bit is 1, multiply the result variable by the current value of the base.

6. Divide the exponent by 2, discarding the remainder.

7. Continue the iteration until all bits of the exponent have been processed.

8. Return the result variable modulo the given modulus value.

## C++

 #include using namespace std;   const long long mod = 1000000007LL; // prime modulo value   long long squareMultiply(long long base, long long exp) {     long long b = 1LL;     long long A = base % mod;     if (exp & 1LL) {         b = base % mod;     }     exp >>= 1LL;     while (exp > 0) {         A = (A * A) % mod;         if (exp & 1LL) {             b = (A * b) % mod;         }         exp >>= 1LL;     }     return b % mod; }   int main() {     long long base = 5LL;     long long exp = 100000LL;     long long modulo = squareMultiply(base, exp);     cout << modulo << endl;     return 0; }

## Java

 // Java program to compute exponential value under modulo // using Bit Manipulation. import java.util.*; import java.lang.*; import java.io.*;   class exp_sq {     static long mod = 1000000007L; // prime modulo value     public static void main(String[] args)     {         long base = 5;         long exp = 100000;           long modulo = squareMultiply(base, exp);         System.out.println(modulo);     }       static long squareMultiply(long base, long exp)     {         long b = 1;         long A = base;         if((exp & 1) == 1){             b = base % mod;         }         exp  = (exp >> 1);         while(exp > 0){             A = (A * A) % mod;             if((exp & 1) == 1){                 b = (A * b) % mod;             }             exp = (exp >> 1);         }           return b % mod;     } }

## C#

 using System;   public class Program {     const long mod = 1000000007L; // prime modulo value       static long squareMultiply(long baseNum, long exponent)     {         long b = 1L;         long A = baseNum % mod;         if ((exponent & 1L) == 1L) {             b = baseNum % mod;         }         exponent >>= 1;         while (exponent > 0) {             A = (A * A) % mod;             if ((exponent & 1L) == 1L) {                 b = (A * b) % mod;             }             exponent >>= 1;         }         return b % mod;     }       static public void Main()     {         long baseNum = 5L;         long exponent = 100000L;         long modulo = squareMultiply(baseNum, exponent);         Console.WriteLine(modulo);     } }

## Python3

 mod = 1000000007   # Function to square the multiply def square_multiply(base, exp):     b = 1     A = base % mod     if exp & 1:         b = base % mod     exp >>= 1     while exp > 0:         A = (A * A) % mod         if exp & 1:             b = (A * b) % mod         exp >>= 1     return b % mod     # Driver Code base = 5 exp = 100000   modulo = square_multiply(base, exp) print(modulo)

Output

754573817

Time Complexity — O( log(exp) )

Space Complexity — O(1)

Applications: Besides fast calculation of this method have several other advantages, like it is used in cryptography, in calculating Matrix Exponentiation et cetera.

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