How can I perform multiplication without the '*' operator?

Question

I was just going through some basic stuff as I am learning C. I came upon a question to multiply a number by 7 without using the * operator. Basically it\'s like this

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Answer 1

If you can use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Find the 2^b which is larger than "a" which turns out to be the 
    //ceiling of (Log base 2 of b) == numbers of digits to shift
    double logBase2 = Math.log(absB) / Math.log(2);
    long bits = (long)Math.ceil(logBase2);

    //Get the value of 2^bits
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<<bits;

    //Subtract the "difference" "a" times
    int diffLoop = Math.abs(a);
    while (diffLoop>0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

If you cannot use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Get the number of bits for a 2^(b+1) larger number
    int bits = 0;
    int bitInteger = absB;
    while (bitInteger>0) {
        bitInteger /= 2;
        bits++;
    }

    //Get the value of 2^bit
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<<bits;

    //Subtract the "difference" "a" times
    int diffLoop = absA;
    while (diffLoop>0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

I found this to be more efficient:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    long result = 0L;
    while (absA>0) {
        if ((absA&1)>0) result += absB; //Is odd
        absA >>= 1;
        absB <<= 1;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

and yet another way.

public static final long multiplyUsingLogs(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);
    long result = Math.round(Math.pow(10, (Math.log10(absA)+Math.log10(absB))));
    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

Answer 2

Let N be the number that we want to multiply with 7.

N x 7 = N + N + N + N + N + N + N
N x 7 = N + N + N + N + N + N + N + (N - N)
N x 7 = (N + N + N + N + N + N + N + N) - N
N x 7 = 8xN - N

As we know that, left shifting any number by one bit multiply it by 2. Hence, multiplying any number with 8 is equivalent to right shifting it by 3 bits

N x 7 = (N << 3) - N 

I found this description here http://www.techcrashcourse.com/2016/02/c-program-to-multiply-number-by-7-bitwise-operator.html

Hope it helps.

Answer 3

By making use of recursion, we can multiply two integers with the given constraints.

To multiply x and y, recursively add x y times.

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

Answer 4

Think about the normal multiplication method we use

         1101 x        =>13
         0101          =>5
---------------------
         1101
        0000
       1101
      0000
===================        
      1000001 .        => 65

Writing the same above in the code

#include<stdio.h>

int multiply(int a, int b){
    int res = 0,count =0;
    while(b>0) {
        if(b & 0x1)
            res = res + (a << count);
        b = b>>1;
        count++;
    }
    return res;
}
int main() {
    printf("Sum of x+y = %d", multiply(5,10));
    return 0;
}

Answer 5

Think about how you multiply in decimal using pencil and paper:

  12
x 26
----
  72
 24
----
 312

What does multiplication look like in binary?

   0111
x  0101
-------
   0111
  0000
 0111
-------
 100011

Notice anything? Unlike multiplication in decimal, where you need to memorize the "times table," when multiplying in binary, you are always multiplying one of the terms by either 0 or 1 before writing it down in the list addends. There's no times table needed. If the digit of the second term is 1, you add in the first term. If it's 0, you don't. Also note how the addends are progressively shifted over to the left.

If you're unsure of this, do a few binary multiplications on paper. When you're done, convert the result back to decimal and see if it's correct. After you've done a few, I think you'll get the idea how binary multiplication can be implemented using shifts and adds.

Answer 6

JAVA:
Considering the fact, that every number can be splitted into powers of two:

1 = 2 ^ 0
2 = 2 ^ 1
3 = 2 ^ 1 + 2 ^ 0
...

We want to get x where:
x = n * m

So we can achieve that by doing following steps:

1.   while m is greater or equal to 2^pow:
     1.1  get the biggest number pow, such as 2^pow is lower or equal to m
     1.2  multiply n*2^pow and decrease m to m-2^pow
2.   sum the results

Sample implementation using recursion:

long multiply(int n, int m) {
    int pow = 0;
    while (m >= (1 << ++pow)) ;
    pow--;
    if (m == 1 << pow) return (n << pow);
    return (n << pow) + multiply(n, m - (1 << pow));
}

I got this question in last job interview and this answer was accepted.

EDIT: solution for positive numbers