Binary Exponentiation for Competitive Programming

In competitive programming, we often need to do a lot of big number calculations fast. Binary exponentiation is like a super shortcut for doing powers and can make programs faster. This article will show you how to use this powerful trick to enhance your coding skills.

What is Binary Exponentiation?

Binary Exponentiation or Exponentiation by squaring is the process of calculating a number raised to the power another number (A^B) in Logarithmic time of the exponent or power, which speeds up the execution time of the program.

Why to Use Binary Exponentiation?

Whenever we need to calculate (A^B), we can simple calculate the result by taking the result as 1 and multiplying A for exactly B times. The time complexity for this approach is O(B) and will fail when values of B in order of 10⁸ or greater. This is when we can use Binary exponentiation because it can calculate the result in O(log(B)) time complexity, so we can easily calculate the results for larger values of B in order of 10¹⁸ or less.

Idea Behind Binary Exponentiation:

When we are calculating (A^B), we can have 3 possible positive values of B:

• Case 1: If B = 0, whatever be the value of A, our result will be 1.
• Case 2: If B is an even number, then instead of calculating (A^B), we can calculate ((A²)^B/2) and the result will be same.
• Case 3: If B is an odd number, then instead of calculating (A^B), we can calculate (A * (A^{(B – 1)/2})²),

Examples:

2 ¹² = (2) ^{2 * 6}
= (4) ⁶
= (4) ^{2 * 3}
= (16) ³
= 16 * (16) ²
= 16 * (256) ¹

Recursive Implementation of Binary Exponentiation:

C++

#include <bits/stdc++.h>
using namespace std;

long long power(long long A, long long B)
{
    if (B == 0)
        return 1;

    long long res = power(A, B / 2);

    if (B % 2)
        return res * res * A;
    else
        return res * res;
}

int main()
{
    cout << power(2, 12) << "\n";
    return 0;
}

Java

import java.io.*;
class GFG {
    static long power(long A, long B) {
        if (B == 0)
            return 1;

        long res = power(A, B / 2);

        if (B % 2 == 1)
            return res * res * A;
        else
            return res * res;
    }

    public static void main(String[] args) {
        System.out.println(power(2, 12));
    }
}
//This code is contributed by Rohit Singh

Python3

def power(A, B):
    if B == 0:
        return 1

    res = power(A, B // 2)

    if B % 2:
        return res * res * A
    else:
        return res * res

if __name__ == "__main__":
    print(power(2, 12))

C#

using System;

class Program
{
    // Recursive function to calculate the power of a number (A^B)
    static long Power(long A, long B)
    {
        // Base case: A^0 is always 1
        if (B == 0)
            return 1;

        // Recursive calculation of A^(B/2)
        long res = Power(A, B / 2);

        // Multiply the result by itself
        res *= res;

        // If B is odd, multiply by A one more time
        if (B % 2 != 0)
            res *= A;

        return res;
    }

    static void Main()
    {
        // Example: Calculate and print the result of 2^12
        Console.WriteLine(Power(2, 12));
    }
}

Javascript

// Function to calculate the power of a number A raised to the power of B
function power(A, B) {
    if (B === 0) {
        return 1;
    }

    let res = power(A, Math.floor(B / 2));

    if (B % 2 === 1) {
        return res * res * A;
    } else {
        return res * res;
    }
}

// Driver code
console.log(power(2, 12));

Output

4096

Iterative Implementation of Binary Exponentiation:

C++

#include <bits/stdc++.h>
using namespace std;

long long power(long long a, long long b) {
    long long result = 1;
    while(b) {
        if (b & 1) 
        result = result * a;
        a = a * a;
        b >>= 1;
    }
    return result;
}

int main() {
    cout<<power(2, 12)<<"\n";
    return 0;
}

Java

public class Main {
    public static long power(long a, long b) {
        long result = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                result *= a;
            }
            a *= a;
            b >>= 1;
        }
        return result;
    }

    public static void main(String[] args) {
        System.out.println(power(2, 12));
    }
}

Python3

def power(a, b):
    result = 1
    while b:
        if b & 1:
            result = result * a
        a = a * a
        b >>= 1
    return result

if __name__ == "__main__":
    print(power(2, 12))

C#

using System;

class Program {
    // Function to calculate power
    static long Power(long a, long b)
    {
        long result = 1;
        while (b > 0) {
            if ((b & 1) == 1)
                result *= a;
            a *= a;
            b >>= 1;
        }
        return result;
    }

    static void Main() { Console.WriteLine(Power(2, 12)); }
}

Javascript

// Function to calculate the power of a number
function power(a, b) {
    let result = 1;
    while (b > 0) {
        // If b is odd, multiply result by a
        if (b & 1) {
            result *= a;
        }
        // Square a
        a *= a;
        // Divide b by 2 using bitwise right shift
        b >>= 1;
    }
    return result;
}

// Main function
function main() {
    // Output the result of power(2, 12)
    console.log(power(2, 12));
}

// Call the main function
main();

Output

4096

Even though, both the iterative and recursive approach have the same time complexity, the iterative approach is still faster because there are no overheads for recursive calls.

Use Cases of Binary Exponentiation in Competitive Programming:

1. Fast Computation of N^th Fibonacci Number:

We can compute N^th Fibonacci Number by simply running a loop till N and in every iteration i, we calculate the i^th Fibonacci number using (i-1)^th and (i-2)^th iteration. But this approach runs in linear time complexity, that is O(N). But, if we are concerned with simply the N^th Fibonacci number and not every number before it, then we can compute it in O(logN) by using matrix exponentiation, where we build a 2⨯2 matrix to transition from (i-2)^th and (i-1)^th Fibonacci number to (i-1)^th and i^th Fibonacci number.

Click here to expand your knowledge about Matrix Exponentiation

2. Compute a large number modulo M:

It is hardly to see any problem which is all about to compute A^B, but Binary Exponentiation comes in handy when our answer becomes too large to be stored as an integer, that is greater than INT_MAX. There are many problems which asks us to count the number of ways to do something and as the number of ways are too large to be stored in an Integer variable, the question asks us to print the answer modulo 1000000007 or modulo 998244353. Since, it is proved here, that

(A * B) mod M = ((A mod M) * (B mod M)) mod M

Therefore, Binary Exponentiation is very useful in computing A^B mod M.

C++

#include <bits/stdc++.h>;
using namespace std;

const int mod = 1e9 + 7;

long long power(long long a, long long b)
{
    long long result = 1;
    while (b) {
        if (b & 1)
            result = (result * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return result;
}

int main()
{
    cout << power(2, 42) << "\n";
    return 0;
}

Java

import java.util.*;

public class Main {
    // Constant representing the modulo value
    static final int mod = 1000000007;

    // Function to calculate the power of a number (a) raised to the power of b modulo mod
    public static long power(long a, long b) {
        long result = 1;
        while (b > 0) {
            // If the current bit of b is set, multiply the result by a
            if ((b & 1) == 1)
                result = (result * a) % mod;
            
            // Square the value of a and reduce it modulo mod
            a = (a * a) % mod;
            
            // Right shift b to move to the next bit
            b >>= 1;
        }
        return result;
    }

    public static void main(String[] args) {
        // Output the result of 2^42 modulo mod
        System.out.println(power(2, 42));
    }
}

Python3

# Python Implementation
mod = 10**9 + 7

def power(a, b):
    result = 1
    while b:
        if b & 1:
            result = (result * a) % mod
        a = (a * a) % mod
        b >>= 1
    return result

print(power(2, 42))

# This code is contributed by Sakshi

C#

using System;

class Program
{
    const int mod = 1000000007;

    static long power(long a, long b)
    {
        long result = 1;
        while (b > 0)
        {
            if ((b & 1) == 1)
                result = (result * a) % mod;
            a = (a * a) % mod;
            b >>= 1;
        }
        return result;
    }

    static void Main()
    {
        Console.WriteLine(power(2, 42));
    }
}
//this code is contributed by Aman.

Javascript

// Function to calculate the power of a number (a) raised to the power of b modulo mod
function power(a, b) {
    const mod = 1000000007;
    let result = 1;
    while (b > 0) {
        // If the current bit of b is set, multiply the result by a and take modulo mod
        if (b & 1)
            result = (result * a) % mod;

        // Square the value of a and reduce it modulo mod
        a = (a * a) % mod;

        // Right shift b to move to the next bit
        b >>= 1;
    }
    return result;
}

// Output the result of 2^42 modulo mod
console.log(power(2, 42));

Output

46480318

3. Apply Permutation of a given Sequence large number of times:

Let’s suppose, we are given a Permutation P and a Sequence S and we need to apply P to S for a large number of times (say K), then we can easily compute the final sequence by using Binary Exponentiation.

Examples:

P = [2, 3, 4, 1], S = [2, 1, 3, 4]
After applying permutation P to Sequence S once,
S’ = [1, 3, 4, 2]
Explanation:
S'[1] = S[P[1]] = S[2] = 1
S'[2] = S[P[2]] = S[3] = 3
S'[3] = S[P[3]] = S[4] = 4
S'[4] = S[P[4]] = S[1] = 2

After applying permutation P to Sequence S twice,
S” = [3, 4, 2, 1]
Explanation:
S”[1] = S'[P[1]] = S'[2] = S[P[2]] = S[3] = 3
S”[2] = S'[P[2]] = S'[3] = S[P[3]] = S[4] = 4
S”[3] = S'[P[3]] = S'[4] = S[P[4]] = S[1] = 2
S”[4] = S'[P[4]] = S'[1] = S[P[1]] = S[2] = 1

Observations: If we observe carefully, in the above example instead of applying permutation P to S twice, if we apply permutation P in itself (P’) and then apply it on S once, we will get the same result.

P = [2, 3, 4, 1], S = [2, 1, 3, 4]
After applying permutation P to itself once,
P’ = [3, 4, 1, 2]
Explanation:
P'[1] = P[P[1]] = P[2] = 3
P'[2] = P[P[2]] = P[3] = 4
P'[3] = P[P[3]] = P[4] = 1
P'[4] = P[P[4]] = P[1] = 2

Now, applying permutation P’ to S,
S”[1] = S[P'[1]] = S[3] = 3
S”[2] = S[P'[2]] = S[4] = 4
S”[3] = S[P'[3]] = S[1] = 2
S”[4] = S[P'[4]] = S[2] = 1

Therefore, it is clear that applying a permutation P to a sequence S for N times is equal to applying permutation P’ to sequence S for N/2 times and we can simply solve this using Binary Exponentiation:

C++

#include <bits/stdc++.h>
using namespace std;

// binary exponentiation
void apply(vector<int>& S, vector<int>& P)
{
    vector<int> temp(S.size());
    for (int i = 1; i < S.size(); i++) {
        temp[i] = S[P[i]];
    }
    for (int i = 1; i < S.size(); i++)
        S[i] = temp[i];
}

// Function to apply Permutation P to Sequence S
// K number of times
void solve(vector<int>& S, vector<int>& P, int K)
{
    while (K) {
        if (K & 1)
            apply(S, P);
        apply(P, P);
        K >>= 1;
    }
}
int main()
{
    vector<int> P{ 0, 2, 3, 4, 1 };
    vector<int> S{ 0, 2, 1, 3, 4 };
    int K = 2;
    solve(S, P, K);
    cout << "New Sequence = ";
    for (int i = 1; i < S.size(); i++)
        cout << S[i] << " ";
    return 0;
}

Java

import java.util.Arrays;

public class BinaryExponentiation {

    // Binary exponentiation
    static void apply(int[] S, int[] P) {
        int[] temp = Arrays.copyOf(S, S.length);
        for (int i = 1; i < S.length; i++) {
            temp[i] = S[P[i]];
        }
        System.arraycopy(temp, 1, S, 1, S.length - 1);
    }

    // Function to apply Permutation P to Sequence S
    // K number of times
    static void solve(int[] S, int[] P, int K) {
        while (K > 0) {
            if ((K & 1) == 1) {
                apply(S, P);
            }
            apply(P, P);
            K >>= 1;
        }
    }

    public static void main(String[] args) {
        int[] P = {0, 2, 3, 4, 1};
        int[] S = {0, 2, 1, 3, 4};
        int K = 2;
        solve(S, P, K);
        System.out.print("New Sequence = ");
        for (int i = 1; i < S.length; i++) {
            System.out.print(S[i] + " ");
        }
    }
}

Python3

# Binary exponentiation
def apply(S, P):
    temp = [0] * len(S)
    for i in range(1, len(S)):
        temp[i] = S[P[i]]
    for i in range(1, len(S)):
        S[i] = temp[i]

# Function to apply Permutation P to Sequence S K number of times
def solve(S, P, K):
    while K:
        if K & 1:
            apply(S, P)
        apply(P, P)
        K >>= 1

if __name__ == "__main__":
    P = [0, 2, 3, 4, 1]
    S = [0, 2, 1, 3, 4]
    K = 2
    solve(S, P, K)
    print("New Sequence =", end=" ")
    for i in range(1, len(S)):
        print(S[i], end=" ")

JavaScript

// Function to apply permutation P to sequence S
// K number of times
function solve(S, P, K) {
    // Helper function to apply permutation P to sequence S
    function apply(S, P) {
        const temp = Array(S.length);
        for (let i = 1; i < S.length; i++) {
            temp[i] = S[P[i]];
        }
        for (let i = 1; i < S.length; i++) {
            S[i] = temp[i];
        }
    }

    // Perform binary exponentiation to optimize
    // the application of permutations
    while (K > 0) {
        if (K & 1) {
            apply(S, P);
        }
        apply(P, P);
        K >>= 1;
    }
}

// Main function
function main() {
    // Initial sequence S and permutation P
    const P = [0, 2, 3, 4, 1];
    const S = [0, 2, 1, 3, 4];
    const K = 2; // Number of times to apply permutation P to sequence S

    // Apply permutation P to sequence S K times
    solve(S, P, K);

    // Output the new sequence
    process.stdout.write("New Sequence = ");
    for (let i = 1; i < S.length; i++) {
        process.stdout.write(S[i] + " ");
    }
}

// Call the main function to execute the program
main();

Output

New Sequence = 3 4 2 1 

Complexity Analysis:

• Brute Force: O(N * K)
• Binary Exponentiation: O(N * logK)

4. Compute Product of 2 very Large Numbers Modulo M:

If we have 2 very large number A and B and we need no compute (A * B) mod M but (A * B) in 64-bit integer, then we can use the concept of binary exponentiation to compute the product of these 2 numbers by adding A to the answer B times.

When we are calculating (A * B), we can have 3 possible positive values of B:

Case 1: If B = 0, whatever be the value of A, our result will be 0.
Case 2: If B is an even number, then instead of calculating (A * B) mod M , we can calculate (((A * 2) mod M) * B/2) mod M and the result will be same.
Case 3: If B is an odd number, then instead of calculating (A * B), we can calculate (A + A * (B-1)) mod M, which is same as case 2.

We can recursively follow the above steps to get our result.

Examples:

(25 * 10) mod 60 = ((25 * 2) mod 60 * 5) mod 60
= ((50 mod 60) * 5) mod 60
= (50 + (50 mod 60) * 4) mod 60
= (50 + (100 mod 60) * 2) mod 60
= (50 + (40 mod 60) * 2) mod 60
= (50 + (80 mod 60)) mod 60
= (50 + 20) mod 60
= 10

Therefore, we can compute (A * B) mod M using Binary Exponentiation:

C++

#include <bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;

// function to multiply to very large numbers
int multiply(int A, int B) {
    int ans = 0;
    while(B) {
        if(B & 1) {
            ans = (ans + A) % mod;
        }
        A = (A + A) % mod;
        B >>= 1;
    }
    return ans;
}

int main() {

    int A = 1e9, B = 1e9;
    cout << multiply(A, B);
    return 0;
}

Java

import java.math.BigInteger;

public class Main {
    // Modulo constant
    static final BigInteger MOD = new BigInteger("1000000007");

    // Function to multiply two very large numbers
    static BigInteger multiply(BigInteger A, BigInteger B) {
        BigInteger ans = BigInteger.ZERO;
        while (!B.equals(BigInteger.ZERO)) {
            if (B.and(BigInteger.ONE).equals(BigInteger.ONE)) {
                ans = (ans.add(A)).mod(MOD);
            }
            A = (A.add(A)).mod(MOD);
            B = B.shiftRight(1);
        }
        return ans;
    }

    public static void main(String[] args) {
        BigInteger A = new BigInteger("1000000000");
        BigInteger B = new BigInteger("1000000000");
        System.out.println(multiply(A, B));
    }
}

Python

# Function to multiply two very large numbers
def multiply(A, B, mod):
    ans = 0
    while B:
        if B & 1:
            ans = (ans + A) % mod
        A = (A + A) % mod
        B >>= 1
    return ans

# Main function
if __name__ == "__main__":
    mod = 10**9 + 7
    A = int(1e9)
    B = int(1e9)
    print(multiply(A, B, mod))

JavaScript

const mod = BigInt(1e9 + 7); // BigInt constant for the modulo operation

// Function to multiply two very large numbers
function multiply(A, B) {
    let ans = BigInt(0); // Initialize ans as BigInt
    while (B) {
        if (B & BigInt(1)) { // Bitwise AND operation, checking if the least significant bit is set
            ans = (ans + A) % mod; // BigInt arithmetic
        }
        A = (A + A) % mod; // BigInt arithmetic
        B >>= BigInt(1); // Bitwise right shift by 1 position
    }
    return ans;
}

// Main function
function main() {
    const A = BigInt(1e9); // BigInt constant for A
    const B = BigInt(1e9); // BigInt constant for B
    console.log(multiply(A, B).toString()); // Output the result as string
}

// Call the main function
main();
//this code is cintriuted by Adarsh.

Output

49

Complexity Analysis:

• Brute Force: O(1), but not possible for large numbers
• Binary Exponentiation: O(log B), as we have distributed the multiplication operation to a series of log(B) addition operations.

Practice Problems on Binary Exponentiation for Competitive Programming:

Easy Level Problems on Binary Exponentiation:

Problem	Problem Link
Padovan Sequence	Practice Now
Count ways to express N as the sum of 1, 3 and 4	Practice Now
Modified Fibonacci	Practice Now
Geometric Progression	Practice Now
Carol Numbers	Practice Now

Medium Level Problems on Binary Exponentiation:

Problem	Problem Link
Matrix Exponentiation	Practice Now
Challenge by Nikitasha	Practice Now
Left Most Divisor	Practice Now
Rahul and the Lift	Practice Now

Hard Level Problems on Binary Exponentiation:

Problem	Problem Link
nCr mod M	Practice Now
Generalized Fibonacci Numbers	Practice Now
Find the pattern	Practice Now