How to avoid overflow in modular multiplication?

Consider below simple method to multiply two numbers.

// A Simple solution that causes overflow when  
// value of (a % mod) * (b % mod) becomes more than 
// maximum value of long long int  
#define ll long long 
  
ll multiply(ll a, ll b, ll mod) 
{ 
   return ((a % mod) * (b % mod)) % mod; 
} 

The above function works fine when multiplication doesn’t result in overflow. But if input numbers are such that the result of multiplication is more than maximum limit.

For example, the above method fails when mod = 10¹¹, a = 9223372036854775807 (largest long long int) and b = 9223372036854775807 (largest long long int). Note that there can be smaller values for which it may fail. There can be many more examples of smaller values. In fact any set of values for which multiplication can cause a value greater than maximum limit.

How to avoid overflow?
We can multiply recursively to overcome the difficulty of overflow. To multiply a*b, first calculate a*b/2 then add it twice. For calculating a*b/2 calculate a*b/4 and so on (similar to log n exponentiation algorithm).

// To compute (a * b) % mod
multiply(a,  b, mod)
1)  ll res = 0; // Initialize result
2)  a = a % mod.
3)  While (b > 0)
        a) If b is odd, then add 'a' to result.
               res = (res + a) % mod
        b) Multiply 'a' with 2
           a = (a * 2) % mod
        c) Divide 'b' by 2
           b = b/2  
4)  Return res

Below is the implementation.

C++

// C++ program for modular multiplication without 
// any overflow 
#include<iostream> 
using namespace std; 
  
typedef long long int ll; 
  
// To compute (a * b) % mod 
ll mulmod(ll a, ll b, ll mod) 
{ 
    ll res = 0; // Initialize result 
    a = a % mod; 
    while (b > 0) 
    { 
        // If b is odd, add 'a' to result 
        if (b % 2 == 1) 
            res = (res + a) % mod; 
  
        // Multiply 'a' with 2 
        a = (a * 2) % mod; 
  
        // Divide b by 2 
        b /= 2; 
    } 
  
    // Return result 
    return res % mod; 
} 
  
// Driver program 
int main() 
{ 
   ll a = 9223372036854775807, b = 9223372036854775807; 
   cout << mulmod(a, b, 100000000000); 
   return 0; 
} 

Java

// Java program for modular multiplication   
// without any overflow  
  
class GFG  
{ 
  
    // To compute (a * b) % mod  
    static long mulmod(long a, long b,  
                            long mod)  
    { 
        long res = 0; // Initialize result  
        a = a % mod; 
        while (b > 0) 
        { 
            // If b is odd, add 'a' to result  
            if (b % 2 == 1)  
            { 
                res = (res + a) % mod; 
            } 
  
            // Multiply 'a' with 2  
            a = (a * 2) % mod; 
  
            // Divide b by 2  
            b /= 2; 
        } 
  
        // Return result  
        return res % mod; 
    } 
  
    // Driver code  
    public static void main(String[] args) 
    { 
  
        long a = 9223372036854775807L, b = 9223372036854775807L; 
        System.out.println(mulmod(a, b, 100000000000L)); 
    } 
}  
  
// This code is contributed by Rajput-JI 

Python3

# Python3 program for modular multiplication  
# without any overflow 
  
# To compute (a * b) % mod 
def mulmod(a, b, mod): 
  
    res = 0; # Initialize result 
    a = a % mod; 
    while (b > 0): 
      
        # If b is odd, add 'a' to result 
        if (b % 2 == 1): 
            res = (res + a) % mod; 
  
        # Multiply 'a' with 2 
        a = (a * 2) % mod; 
  
        # Divide b by 2 
        b //= 2; 
  
    # Return result 
    return res % mod; 
  
# Driver Code 
a = 9223372036854775807; 
b = 9223372036854775807; 
print(mulmod(a, b, 100000000000)); 
  
# This code is contributed by mits 

C#

// C# program for modular multiplication  
// without any overflow  
using System; 
  
class GFG  
{  
  
// To compute (a * b) % mod  
static long mulmod(long a, long b, long mod)  
{  
    long res = 0; // Initialize result  
    a = a % mod;  
    while (b > 0)  
    {  
        // If b is odd, add 'a' to result  
        if (b % 2 == 1)  
        {  
            res = (res + a) % mod;  
        }  
  
        // Multiply 'a' with 2  
        a = (a * 2) % mod;  
  
        // Divide b by 2  
        b /= 2;  
    }  
  
    // Return result  
    return res % mod;  
}  
  
// Driver code  
public static void Main(String[] args)  
{  
    long a = 9223372036854775807L,  
         b = 9223372036854775807L;  
    Console.WriteLine(mulmod(a, b, 100000000000L));  
}  
}  
  
// This code is contributed by 29AjayKumar 

Output:

84232501249

Thanks to Utkarsh Trivedi for suggesting above solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

C++

Java

Python3

C#

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