Multiply two integers without using multiplication, division and bitwise operators, and no loops

By making use of recursion, we can multiply two integers with the given constraints. 
To multiply x and y, recursively add x y times. 

Approach:

Since we cannot use any of the given symbols, the only way left is to use recursion, with the fact that x is to be added to x y times.

Base case: When the numbers of times  x has to be added becomes 0. 

Recursive call: If the base case is not met, then add x to the current resultant value and pass it to the next iteration.

C++

// C++ program to Multiply two integers without
// using multiplication, division and bitwise
//  operators, and no loops
#include<iostream>

using namespace std;
class GFG
{

/* function to multiply two numbers x and y*/
public : int multiply(int x, int y)
{
    /* 0 multiplied with anything gives 0 */
    if(y == 0)
    return 0;

    /* Add x one by one */
    if(y > 0 )
    return (x + multiply(x, y-1));

    /* the case where y is negative */
    if(y < 0 )
    return -multiply(x, -y);
}
};

// Driver code
int main()
{
    GFG g;
    cout << endl << g.multiply(5, -11);
    getchar();
    return 0;
}

// This code is contributed by SoM15242

C

#include<stdio.h>
/* function to multiply two numbers x and y*/
int multiply(int x, int y)
{
   /* 0  multiplied with anything gives 0 */
   if(y == 0)
     return 0;

   /* Add x one by one */
   if(y > 0 )
     return (x + multiply(x, y-1));

  /* the case where y is negative */
   if(y < 0 )
     return -multiply(x, -y);
}

int main()
{
  printf("\n %d", multiply(5, -11));
  getchar();
  return 0;
}

Java

class GFG {

    /* function to multiply two numbers x and y*/
    static int multiply(int x, int y) {

        /* 0 multiplied with anything gives 0 */
        if (y == 0)
            return 0;

        /* Add x one by one */
        if (y > 0)
            return (x + multiply(x, y - 1));

        /* the case where y is negative */
        if (y < 0)
            return -multiply(x, -y);

        return -1;
    }

    // Driver code
    public static void main(String[] args) {

        System.out.print("\n" + multiply(5, -11));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Function to multiply two numbers
# x and y
def multiply(x,y):

    # 0 multiplied with anything
    # gives 0
    if(y == 0):
        return 0

    # Add x one by one
    if(y > 0 ):
        return (x + multiply(x, y - 1))

    # The case where y is negative
    if(y < 0 ):
        return -multiply(x, -y)

# Driver code
print(multiply(5, -11))

# This code is contributed by Anant Agarwal.

C#

// Multiply two integers without
// using multiplication, division
// and bitwise operators, and no
// loops
using System;

class GFG {

    // function to multiply two numbers
    // x and y
    static int multiply(int x, int y) {

        // 0 multiplied with anything gives 0
        if (y == 0)
            return 0;

        // Add x one by one
        if (y > 0)
            return (x + multiply(x, y - 1));

        // the case where y is negative
        if (y < 0)
            return -multiply(x, -y);

        return -1;
    }

    // Driver code
    public static void Main() {

        Console.WriteLine(multiply(5, -11));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// function to multiply
// two numbers x and y
function multiply($x, $y)
{
/* 0 multiplied with
anything gives 0 */
if($y == 0)
    return 0;

/* Add x one by one */
if($y > 0 )
    return ($x + multiply($x,
                          $y - 1));

/* the case where
y is negative */
if($y < 0 )
    return -multiply($x, -$y);
}

// Driver Code
echo multiply(5, -11);

// This code is contributed by mits.
?>

Javascript

<script>

// javascript program to Multiply two integers without
// using multiplication, division and bitwise
//  operators, and no loops

/* function to multiply two numbers x and y*/
function multiply( x,  y)
{
    /* 0 multiplied with anything gives 0 */
    if(y == 0)
    return 0;

    /* Add x one by one */
    if(y > 0 )
    return (x + multiply(x, y-1));

    /* the case where y is negative */
    if(y < 0 )
    return -multiply(x, -y);
}


// Driver code

   document.write( multiply(5, -11));

// This code is contributed by todaysgaurav

</script>

Output

-55

Time Complexity: O(y) where y is the second argument to function multiply().

Auxiliary Space: O(y) for the recursion stack

Another approach: The problem can also be solved using basic math property

(a+b)² = a² + b² + 2a*b

⇒  a*b = ((a+b)² – a² – b²) / 2

For computing the square of numbers, we can use the power function in C++ and for dividing by 2 in the above expression we can write a recursive function.

Below is the implementation of the above approach: 

C++

// C++ program to Multiply two integers without
// using multiplication, division and bitwise
// operators, and no loops
#include<bits/stdc++.h>
using namespace std;

// divide a number by 2 recursively
int divideby2(int num)
{
   if(num<2)
    return 0;
   return 1 + divideby2(num-2);
}

int multiply(int a,int b)
{
    int whole_square=pow(a+b,2);
    int a_square=pow(a,2);
    int b_square=pow(b,2);

    int val= whole_square- a_square - b_square;

    int product;

    // for positive value of variable val
    if(val>=0)
    product = divideby2(val);
    // for negative value of variable val
    // we first compute the division by 2 for
    // positive val and by subtracting from
    // 0 we can make it negative
    else
    product = 0 - divideby2(abs(val));

    return product;
}

// Driver code
int main()
{
    int a=5;
    int b=-11;
    cout << multiply(a,b);
    return 0;
}

// This code is contributed by Pushpesh raj.

Output

-55

Russian Peasant (Multiply two numbers using bitwise operators)
Please write comments if you find any of the above code/algorithm incorrect, or find better ways to solve the same problem.