How to perform multiplication, using bitwise operators?

Question

I am working through a problem which i was able to solve, all but for the last piece - i am not sure how can one do multiplication using bitwise operators:

0*8 = 0

1*8 = 8

2*8 = 16 

3*8 = 24 

4*8 = 32

Can you please recommend an approach to solve this?

Answer 1

To multiply by any value of 2 to the power of N (i.e. 2^N) shift the bits N times to the left.

0000 0001 = 1 

times 4 = (2^2 => N = 2) = 2 bit shift : 0000 0100 = 4

times 8 = (2^3 -> N = 3) = 3 bit shift : 0010 0000 = 32

etc..

To divide shift the bits to the right.

The bits are whole 1 or 0 - you can't shift by a part of a bit thus if the number you're multiplying by is does not factor a whole value of N ie.

since: 17 = 16  + 1 
thus:  17 = 2^4 + 1

therefore: x * 17 = (x * 16) + x in other words 17 x's  

thus to multiply by 17 you have to do a 4 bit shift to the left, and then add the original number again:

==> x * 17 = (x * 16) + x 
==> x * 17 = (x * 2^4) + x 
==> x * 17 = (x shifted to left by 4 bits) + x 

so let x = 3 = 0000 0011 

times 16 = (2^4 => N = 4) = 4 bit shift : 0011 0000 = 48

plus the x (0000 0011)

ie.

    0011 0000  (48)  
+   0000 0011   (3)
=============
    0011 0011  (51)

Edit: Update to the original answer. Charles Petzold has written a fantastic book 'Code' that will explain all of this and more in the easiest of ways. I thoroughly recommend this.

Answer 2

To multiply two binary encoded numbers without a multiply instruction. It would be simple to iteratively add to reach the product.

unsigned int mult(x, y)
unsigned int x, y;
{
    unsigned int reg = 0;

    while(y--)
        reg += x;
    return reg;
}

Using bit operations, the characteristic of the data encoding can be exploited. As explained previously, a bit shift is the same as multiply by two. Using this an adder can be used on the powers of two.

// multiply two numbers with bit operations

unsigned int mult(x, y)
unsigned int x, y;
{
    unsigned int reg = 0;

    while (y != 0)
    {
        if (y & 1)
        {
            reg += x;
        }
        x <<= 1;
        y >>= 1;
    }
    return reg;
}

Answer 3

You'd factor the multiplicand into powers of 2.
3*17 = 3*(16+1) = 3*16 + 3*1 ... = 0011b << 4 + 0011b

Answer 4

public static int multi(int x, int y){
        boolean neg = false;
        if(x < 0 && y >= 0){
            x = -x;
            neg = true;
        }
        else if(y < 0 && x >= 0){
            y = -y;
            neg = true;
        }else if( x < 0 && y < 0){
            x = -x;
            y = -y;
        }

        int res = 0;
        while(y!=0){
            if((y & 1) == 1) res += x;
            x <<= 1;
            y >>= 1;
        }
        return neg ? (-res) : res;
    }

Answer 5

I believe this should be a left shift. 8 is 2^3, so left shift 3 bits:

2 << 3 = 8

Answer 6

-(int)multiplyNumber:(int)num1 withNumber:(int)num2
{
    int mulResult =0;
    int ithBit;

    BOOL isNegativeSign = (num1<0 && num2>0) || (num1>0 && num2<0)   ;
    num1 = abs(num1);
    num2 = abs(num2);


    for(int i=0;i<sizeof(num2)*8;i++)
    {
        ithBit =  num2 & (1<<i);
        if(ithBit>0){
            mulResult +=(num1<<i);
        }

    }

    if (isNegativeSign) {
        mulResult =  ((~mulResult)+1 );
    }

    return mulResult;
}

Answer 7

I have just realized that this is the same answer as the previous one. LOL sorry.

public static uint Multiply(uint a, uint b)
{
   uint c = 0;
   while(b > 0)
   {
      c += ((b & 1) > 0) ? a : 0;
      a <<= 1;
      b >>= 1;
   }
   return c;
}

Answer 8

I was working on a recursive multiplication problem without the * operator and came up with a solution that was informed by the top answer here.

I thought it was worth posting because I really like the explanation in the top answer here, but wanted to expand on it in a way that:

1. Had a function representation.
2. Handled cases where your "remainder" was arbitrary.

This only handles positive integers, but you could wrap it in a check for negatives like some of the other answers.

def rec_mult_bitwise(a,b):
    # Base cases for recursion
    if b == 0:
        return 0
    if b == 1:
        return a

    # Get the most significant bit and the power of two it represents
    msb = 1
    pwr_of_2 = 0
    while True:
        next_msb = msb << 1
        if next_msb > b:
            break
        pwr_of_2 += 1
        msb = next_msb
        if next_msb == b:
            break

    # To understand the return value, remember:
    # 1: Left shifting by the power of two is the same as multiplying by the number itself (ie x*16=x<<4)
    # 2: Once we've done that, we still need to multiply by the remainder, hence b - msb
    return (a << pwr_of_2) + rec_mult_bitwise(a, b - msb)

Answer 9

Using Bitwise operator reduces the time complexity.

In cpp:

#include<iostream>
using name space std;

int main(){
   int a, b, res = 0;           // read the elements
   cin>>a>>b;

   // find the small number to reduce the iterations

   small = (a<b)?a:b;           // small number using terinary operator
   big = (small^a)?a:b;         // big number using bitwise XOR operator

   while(small > 0)             
   {
      if(small & 1)             
      {
         res += big;
      }
      big = big << 1;           // it increases the number << is big * (2 power of big)
      small = small >> 1;       // it decreases the number >> is small / (2 power of small)
   }
   cout<<res;
}

In Python:

a = int(input())
b = int(input())
res = 0    

small = a if(a < b) else b
big  = a if(small ^ a) else b

def multiplication(small, big):
    res = 0
    while small > 0:
        if small & 1:
            res += big
        big = big << 1
        small = small >> 1

        return res

 answer = multiplication(small, big)
 print(answer)