high bits of long multiplication in Java?

Question

Is there any way to get the high half of the multiplication of two longs in Java? I.e. the part that vanishes due to overflow. (So the upper 64 bits of the 128-bit result)

I'm used to writing OpenCL code where the command mul_hi does exactly this: http://www.khronos.org/registry/cl/sdk/1.0/docs/man/xhtml/mul_hi.html

Since OpenCL can do it efficiently on my CPU, Java should be able to do so as well, but I can't find how I should do this (or even mimic its behaviour efficiently) in Java. Is this possible in Java, and if so, how?

Answer 1

If either x or y can be negative, you should use Hacker's Delight function (Henry S. Warren, Hacker's Delight, Addison-Wesley, 2nd edition, Fig. 8.2):

long x_high = x >>> 32;
long x_low = x & 0xFFFFFFFFL;
long y_high = y >>> 32;
long y_low = y & 0xFFFFFFFFL;
long z2 = x_low * y_low;
long t = x_high * y_low + (z2 >>> 32);
long z1 = t & 0xFFFFFFFFL;
long z0 = t >>> 32;
z1 += x_low * y_high;
return x_high * y_high + z0 + (z1 >>> 32);

Answer 2

Let's say you have two longs, x and y, and x = x_hi * 2^32 + x_lo, and y = y_hi * 2^32 + y_lo.

Then x * y == (x_hi * y_hi) * 2^64 + (x_hi * y_lo + x_lo * y_hi) * 2^32 + (x_lo * y_lo).

The high 64 bits of that product can, therefore, be computed as follows:

long x_hi = x >>> 32;
long y_hi = y >>> 32;
long x_lo = x & 0xFFFFFFFFL;
long y_lo = y & 0xFFFFFFFFL;
long prod_hi = (x_hi * y_hi) + ((x_ hi * y_lo) >>> 32) + ((x_lo * y_hi) >>> 32);

Answer 3

Here is a code snippet from Java's Math.multiplyHigh(long,long)

    public static long multiplyHigh(long x, long y) {
        if (x < 0 || y < 0) {
            // Use technique from section 8-2 of Henry S. Warren, Jr.,
            // Hacker's Delight (2nd ed.) (Addison Wesley, 2013), 173-174.
            long x1 = x >> 32;
            long x2 = x & 0xFFFFFFFFL;
            long y1 = y >> 32;
            long y2 = y & 0xFFFFFFFFL;
            long z2 = x2 * y2;
            long t = x1 * y2 + (z2 >>> 32);
            long z1 = t & 0xFFFFFFFFL;
            long z0 = t >> 32;
            z1 += x2 * y1;
            return x1 * y1 + z0 + (z1 >> 32);
        } else {
            // Use Karatsuba technique with two base 2^32 digits.
            long x1 = x >>> 32;
            long y1 = y >>> 32;
            long x2 = x & 0xFFFFFFFFL;
            long y2 = y & 0xFFFFFFFFL;
            long A = x1 * y1;
            long B = x2 * y2;
            long C = (x1 + x2) * (y1 + y2);
            long K = C - A - B;
            return (((B >>> 32) + K) >>> 32) + A;
        }
    }

As from Java 9, this is included in java.lang.Math and probably direct call to it should be made. Posting the source just to show what is going on "under the hood".

Answer 4

The accepted solution is wrong most of the time (66%), though the error is bounded (it can be smaller than the exact result by at most 2 and it can never be bigger). This comes from

• ignoring the x_lo * y_lo product
• first shifting and then adding x_hi * y_lo and x_lo * y_hi

My solution seems to always work for non-negative operands.

final long x_hi = x >>> 32;
final long y_hi = y >>> 32;
final long x_lo = x & 0xFFFFFFFFL;
final long y_lo = y & 0xFFFFFFFFL;
long result = x_lo * y_lo;
result >>>= 32;

result += x_hi * y_lo + x_lo * y_hi;
result >>>= 32;
result += x_hi * y_hi;

Tested on a billion random operands. There should be a special test for corner cases and some analysis.

Dealing with negative operands would be more complicated as it'd prohibit using the unsigned shift and force us to handle intermediate result overflow.

In case speed doesn't matter much (and it rarely does), I'd go for

 BigInteger.valueOf(x).multiply(BigInteger.valueOf(y))
     .shiftRight(64).longValue();

Answer 5

You should look at using a BigInteger.

Answer 6

Java 9 has Math.multiplyHigh, which according to the Javadocs "Returns as a long the most significant 64 bits of the 128-bit product of two 64-bit factors."