如何在 JavaScript 中实现 Karatsuba 乘法？

Question

我尝试使用以下代码实现karatsuba algorithm。当 x 和 y（参数）中的位数与 recursive call 不匹配时，问题就开始了，在这种情况下，下面的逻辑不起作用。截至目前，当 x 和 y 的位数相同时，我得到了正确的输出。

更准确地说，我认为问题始于 z1 和 z3 的计算，因为这就是 x 和 y 的位数经常不匹配的地方。

对于如何定义m 的逻辑，我也有点困惑，这是这里基础的力量。我相信我正在明确我的问题？ （任何关于更多优化的建议都会更有帮助，因为我刚刚开始了我的 java 脚本之旅）。

function karatSuba(x,y)
{
  var x1,x0,y1,y0,base,m;
  var dummy_x = x.toString();
  var dummy_y = y.toString();
  var n = (dummy_x.length>dummy_y.length) ? dummy_y.length : dummy_x.length;
  m = Math.round(n/2);
  base  = 10;

  //base case
  if((x<base)||(y<base))
  return x * y;
  //base case

  var bm = Math.pow(base ,m);

  var dummy_x1 = dummy_x.substring(0,n/2);
  var x1 = parseInt(dummy_x1);
  dummy_x1 = null;

  var dummy_x1 = dummy_x.substring(n/2,n);
  var x0 = parseInt(dummy_x1);
  dummy_x1 = null;

  var dummy_y1 = dummy_y.substring(0,n/2);
  var y1 = parseInt(dummy_y1);
  dummy_y1 = null;

  var dummy_y0 = dummy_y.substring(n/2,n);
  var y0 = parseInt(dummy_y0);
  dummy_y = null;

  var p = x1 + x0;
  var q = y1 + y0;

  var a   =   karatSuba(x1,y1);
  var b   =   karatSuba(x0,y0);
  var z1  =   karatSuba(a,Math.pow(bm,2));
  var z2  =   b;
  //var z3  =   karatSmul(bm ,((karatSmul(p,q) - a - b)));
  var z3  = bm * ((p*q) - (a) - (b)); 
  var z   =   z1 + z2 + z3;
  return z;

 }

console.log(karatSuba(344,100));

Answer 1

您编写算法的方式有几个错误。下面的代码应该可以工作。

function karatSuba(x,y)
{

  var x1,x0,y1,y0,base,m;
  base  = 10;


  if((x<base)||(y<base)){
    console.log( " X - y = " , x,y, x*y)
    return x * y;
  }

  var dummy_x = x.toString();
  var dummy_y = y.toString();

  var n = (dummy_x.length > dummy_y.length) ? dummy_y.length : dummy_x.length;
  m = Math.round(n/2);



  var high1 = parseInt(dummy_x.substring(0,dummy_x.length-m));
  var low1 = parseInt(dummy_x.substring(dummy_x.length-m,dummy_x.length  )) ;

  var high2 = parseInt(dummy_y.substring(0,dummy_y.length-m)); 
  var low2 = parseInt(dummy_y.substring(dummy_y.length-m,dummy_y.length));


  var z0   =   karatSuba( low1, low2);
  var z1   =   karatSuba(low1+high1, low2+high2);
  var z2   =   karatSuba(high1,high2);

  var res  =   (z2 *  Math.pow(10, 2 * m )  ) + ( (z1-z2-z0) * Math.pow(10,  m )) + z0;

  return res;

 }

var a = 12345;
var b = 6789;
console.log(karatSuba(a,b));
console.log(a * b);

Answer 2

function range(start, stop, step) {
    if (typeof stop == 'undefined') {
        // one param defined
        stop = start;
        start = 0;
    }
    if (typeof step == 'undefined') {
        step = 1;
    }
    if ((step > 0 && start >= stop) || (step < 0 && start <= stop)) {
        return [];
    }
    var result = [];
    for (var i = start; step > 0 ? i < stop : i > stop; i += step) {
        result.push(i);
    }
    return result;
};
function zeroPad(numberString, zeros, left = true) {
    //Return the string with zeros added to the left or right.
    for (var i in range(zeros)) {
        if (left)
            numberString = '0' + numberString
        else
            numberString = numberString + '0'
    }

    return numberString
}
function largeMultiplcation(x, y) {
    x = x.toString();
    y = y.toString();

    if (x.length == 1 && y.length == 1)
        return int(x) * int(y)

    if (x.length() < y.length)
        x = zeroPad(x, y.length - x.length);

    else
        y = zeroPad(y, x.length - y.length);

    n = x.length
    j = Math.floor(n/2);

    //for odd digit integers
    if ( n % 2 != 0)
        j += 1    
    var BZeroPadding = n - j
    var AZeroPadding = BZeroPadding * 2

    a = parseInt(x.substring(0,j));
    b = parseInt(x.substring(j));
    c = parseInt(y.substring(0,j));
    d = parseInt(y.substring(j));

    //recursively calculate
    ac = LargeMultiplication(a, c)
    bd = LargeMultiplication(b, d)
    k = LargeMultiplication(a + b, c + d)
    A = int(zeroPad(str(ac), AZeroPadding, false))
    B = int(zeroPad(str(k - ac - bd), BZeroPadding, false))
    return A + B + bd
}

Answer 3

如果你的输入是数字，而不是字符串，你会做这样的事情：

function karatsuba(x, y) {
  if (x<10 && y<10) {
    return x*y;
  }
  let maxLength = Math.max(x.toString().length, y.toString().length);
  let m = Math.round(maxLength/2);

  let xHigh = Math.floor(x/ Math.pow(10,m));
  let yHigh = Math.floor(y/ Math.pow(10,m));
  let xLow = x %  Math.pow(10,m);
  let yLow = y %  Math.pow(10,m);

  let a = karatsuba(xHigh, yHigh);
  let d = karatsuba(xLow, yLow);
  let e = karatsuba(xLow+xHigh, yLow+yHigh)-a-d;

  return a * Math.pow(10, m*2) + e * Math.pow(10,m) + d;
}

当然，如果您使用这种方法，当您将两个 64 位数字等较大的数字相乘时，您会损失一些精度。 (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/MAX_SAFE_INTEGER)

解决方法是使用 BigInt 原语 (https://github.com/tc39/proposal-bigint) 或为 Python 放弃 JS。 (https://www.geeksforgeeks.org/what-is-maximum-possible-value-of-an-integer-in-python/)