How to get the product of two integers without using *

user_8351535 0
Viewed 131

I am trying to figure out ways to get the products of two integers without using * in my solution. The closest I got was is 

/*Example: 2,5 output ====> 10 */
/*Example: 10,5 output ====> 50 */

const productOfTwoInt = (int,int2) => {
  var data1 = 0;
  var data2 = 0;

  var result;
  var result2;  

  for(var x = 0; x <= int; x++) {
      result = x += data1
      console.log(result)
  }

  for(var j = 0; j <= int2; j++) {
      result2 = j += data2
  }

  return result + result2    
}

console.log(productOfTwoInt(3,5))
math integer javascript
6 Answers

You could take some bit shifting, known as ancient egyptian multiplication or russian multiplication.

  a     b     p   comment
----  ----  ----  ----------------------------------
  6     8     0   skip, because a is even
  3    16    16   add 16 to p, because a is odd
  1    32    48   add 32 to p, because a is odd
  0    64    48   stop iteration, because a is zero


function product(a, b) {
    var p = 0;
    while (a) {
        p += (a & 1) && b;
        a >>= 1;
        b <<= 1;
    }
    return p;
}

console.log(product(6, 8)); // 48

user_14476750

You can do it with inverse like this:

function multiply(a, b) {
  return a / (1 / b)
}

multiply(2, 5)  // 10
multiply(10, 5)  // 50
user_51562800

Your solution looks quite good already. That can be a bit simplified to:

function multiply(a,b){
  //just one minus, lets swap
  if(a<0 && b>0) [a,b] = [b,a];
  //two minuses: let them be positive
  if(a<0 && b<0) ( a = Math.abs(a), b = Math.abs(b) );
  var result = 0;
  while(a--){
    result += b;
  }
  return result;
}

console.log(
  multiply(1,2),
  multiply(3,4),
  multiply(6,7),
  multiply(-1,-2),
  multiply(1,-2)
);

user_52600240

You can do 

const productOfTwoInt = (x, y) => {
    let z = Math.log(x) + Math.log(y);
    return Math.round(Math.exp(z));
}

console.log(productOfTwoInt(3,5))

user_49892840

As suggested by Nina Scholz's answer, the "Russian Multiplication":

const productOfTwoInt = (a, b) => {

    var a0 = a;
    var result = 0;

    while (1) {
        if (a0 % 2 !== 0)
            result += b;
            
        if (a0 == 1 || a0 == -1) break;
        
        a0 = Math.floor(a0 / 2);
        b *= 2;
    }
  return a > 0 ? result : -result;
}

console.log(
    productOfTwoInt(4, 5),
    productOfTwoInt(4, -5),
    productOfTwoInt(-40, 5),
    productOfTwoInt(-4, -50)
);

Traditional solution:

 const productOfTwoInt = (int, int2) => {
        
        let result = 0;
        let isPositive = int > 0;
        let i = 0;
    
    
        while (i != int) {
            result += int2;
            if (isPositive)
                i++;
            else
                i--;
        }
    
 return isPositive ? result : -result;
    
    }
    
    console.log(
      productOfTwoInt(4, 5),
      productOfTwoInt(4, -5),
      productOfTwoInt(-4, 5),
      productOfTwoInt(-4, -5)
    );

user_69042300

You can use do it in this way

Option 1

function mul(m,n){
   // make it more efficient
    if(m<n)
     {
      var temp =m;
      m=n;
      n=temp;
     }   
 var v=0;
 if(n==1) 
      return m;
 v = mul(m,n>>1);
 v = v+v;
 if(n&1) 
      v+= m;
 return v;
}

console.log(mul(-2,5));

Working Example

user_44725400
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