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• Mathematical Algorithms

# 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.#includeusing 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 Codeint 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 valueN = 1000000007;     # Function codedef 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 codebas = 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.

## PHP

 

## 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.#includeusing 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 codeint 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 valueN = 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 Codebas = 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 valuestatic 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 valuelet 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 Codelet 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 multiplydef 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 Codebase = 5exp = 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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