(a * b) / c MulDiv and dealing with overflow from intermediate multiplication
Solution 1:
I've been tinkering with an approach that (1) multiplies a and b with the school algorithm on 21-bit limbs (2) proceeds to do long division by c, with an unusual representation of the residual a*b - c*q that uses a double to store the high-order bits and a long to store the low-order bits. I don't know if it can be made to be competitive with standard long division, but for your enjoyment,
public class MulDiv {
public static void main(String[] args) {
java.util.Random r = new java.util.Random();
for (long i = 0; true; i++) {
if (i % 1000000 == 0) {
System.err.println(i);
}
long a = r.nextLong() >> (r.nextInt(8) * 8);
long b = r.nextLong() >> (r.nextInt(8) * 8);
long c = r.nextLong() >> (r.nextInt(8) * 8);
if (c == 0) {
continue;
}
long x = mulDiv(a, b, c);
java.math.BigInteger aa = java.math.BigInteger.valueOf(a);
java.math.BigInteger bb = java.math.BigInteger.valueOf(b);
java.math.BigInteger cc = java.math.BigInteger.valueOf(c);
java.math.BigInteger xx = aa.multiply(bb).divide(cc);
if (java.math.BigInteger.valueOf(xx.longValue()).equals(xx) && x != xx.longValue()) {
System.out.printf("a=%d b=%d c=%d: %d != %s\n", a, b, c, x, xx);
}
}
}
// Returns truncate(a b/c), subject to the precondition that the result is
// defined and can be represented as a long.
private static long mulDiv(long a, long b, long c) {
// Decompose a.
long a2 = a >> 42;
long a10 = a - (a2 << 42);
long a1 = a10 >> 21;
long a0 = a10 - (a1 << 21);
assert a == (((a2 << 21) + a1) << 21) + a0;
// Decompose b.
long b2 = b >> 42;
long b10 = b - (b2 << 42);
long b1 = b10 >> 21;
long b0 = b10 - (b1 << 21);
assert b == (((b2 << 21) + b1) << 21) + b0;
// Compute a b.
long ab4 = a2 * b2;
long ab3 = a2 * b1 + a1 * b2;
long ab2 = a2 * b0 + a1 * b1 + a0 * b2;
long ab1 = a1 * b0 + a0 * b1;
long ab0 = a0 * b0;
// Compute a b/c.
DivBy d = new DivBy(c);
d.shift21Add(ab4);
d.shift21Add(ab3);
d.shift21Add(ab2);
d.shift21Add(ab1);
d.shift21Add(ab0);
return d.getQuotient();
}
}
public strictfp class DivBy {
// Initializes n <- 0.
public DivBy(long d) {
di = d;
df = (double) d;
oneOverD = 1.0 / df;
}
// Updates n <- 2^21 n + i. Assumes |i| <= 3 (2^42).
public void shift21Add(long i) {
// Update the quotient and remainder.
q <<= 21;
ri = (ri << 21) + i;
rf = rf * (double) (1 << 21) + (double) i;
reduce();
}
// Returns truncate(n/d).
public long getQuotient() {
while (rf != (double) ri) {
reduce();
}
// Round toward zero.
if (q > 0) {
if ((di > 0 && ri < 0) || (di < 0 && ri > 0)) {
return q - 1;
}
} else if (q < 0) {
if ((di > 0 && ri > 0) || (di < 0 && ri < 0)) {
return q + 1;
}
}
return q;
}
private void reduce() {
// x is approximately r/d.
long x = Math.round(rf * oneOverD);
q += x;
ri -= di * x;
rf = repairLowOrderBits(rf - df * (double) x, ri);
}
private static double repairLowOrderBits(double f, long i) {
int e = Math.getExponent(f);
if (e < 64) {
return (double) i;
}
long rawBits = Double.doubleToRawLongBits(f);
long lowOrderBits = (rawBits >> 63) ^ (rawBits << (e - 52));
return f + (double) (i - lowOrderBits);
}
private final long di;
private final double df;
private final double oneOverD;
private long q = 0;
private long ri = 0;
private double rf = 0;
}
Solution 2:
You can use the greatest common divisor (gcd) to help.
a * b / c = (a / gcd(a,c)) * (b / (c / gcd(a,c)))
Edit: The OP asked me to explain the above equation. Basically, we have:
a = (a / gcd(a,c)) * gcd(a,c)
c = (c / gcd(a,c)) * gcd(a,c)
Let's say x=gcd(a,c) for brevity, and rewrite this.
a*b/c = (a/x) * x * b
--------------
(c/x) * x
Next, we cancel
a*b/c = (a/x) * b
----------
(c/x)
You can take this a step further. Let y = gcd(b, c/x)
a*b/c = (a/x) * (b/y) * y
------------------
((c/x)/y) * y
a*b/c = (a/x) * (b/y)
------------
(c/(xy))
Here's code to get the gcd.
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
Solution 3:
David Eisenstat got me thinking some more.
I want simple cases to be fast: let double take care of that.
Newton-Raphson may be a better choice for the rest.
/** Multiplies both <code>factor</code>s
* and divides by <code>divisor</code>.
* @return <code>Long.MIN_VALUE</code> if result out of range,<br/>
* else <code>factorA * factor1 / divisor</code> */
public static long
mulDiv(long factorA, long factor1, long divisor) {
final double dd = divisor,
product = (double)factorA * factor1,
a1_d = product / dd;
if (a1_d < -TOO_LARGE || TOO_LARGE < a1_d)
return tooLarge();
if (-ONE_ < a1_d && a1_d < ONE_)
return 0;
if (-EXACT < product && product < EXACT)
return (long) a1_d;
long pLo = factorA * factor1, //diff,
pHi = high64(factorA, factor1);
if (a1_d < -LONG_MAX_ || LONG_MAX_ < a1_d) {
long maxdHi = divisor >> 1;
if (maxdHi < pHi
|| maxdHi == pHi
&& Long.compareUnsigned((divisor << Long.SIZE-1),
pLo) <= 0)
return tooLarge();
}
final double high_dd = TWO_POWER64/dd;
long quotient = (long) a1_d,
loPP = quotient * divisor,
hiPP = high64(quotient, divisor);
long remHi = pHi - hiPP, // xxx overflow/carry
remLo = pLo - loPP;
if (Long.compareUnsigned(pLo, remLo) < 0)
remHi -= 1;
double fudge = remHi * high_dd;
if (remLo < 0)
fudge += high_dd;
fudge += remLo/dd;
long //fHi = (long)fudge/TWO_POWER64,
fLo = (long) Math.floor(fudge); //*round
quotient += fLo;
loPP = quotient * divisor;
hiPP = high64(quotient, divisor);
remHi = pHi - hiPP; // should be 0?!
remLo = pLo - loPP;
if (Long.compareUnsigned(pLo, remLo) < 0)
remHi -= 1;
if (0 == remHi && 0 <= remLo && remLo < divisor)
return quotient;
fudge = remHi * high_dd;
if (remLo < 0)
fudge += high_dd;
fudge += remLo/dd;
fLo = (long) Math.floor(fudge);
return quotient + fLo;
}
/** max <code>double</code> trusted to represent
* a value in the range of <code>long</code> */
static final double
LONG_MAX_ = Double.valueOf(Long.MAX_VALUE - 0xFFF);
/** max <code>double</code> trusted to represent a value below 1 */
static final double
ONE_ = Double.longBitsToDouble(
Double.doubleToRawLongBits(1) - 4);
/** max <code>double</code> trusted to represent a value exactly */
static final double
EXACT = Long.MAX_VALUE >> 12;
static final double
TWO_POWER64 = Double.valueOf(1L<<32)*Double.valueOf(1L<<32);
static long tooLarge() {
// throw new RuntimeException("result too large for long");
return Long.MIN_VALUE;
}
static final long ONES_32 = ~(~0L << 32);
static long high64(long factorA, long factor1) {
long loA = factorA & ONES_32,
hiA = factorA >>> 32,
lo1 = factor1 & ONES_32,
hi1 = factor1 >>> 32;
return ((loA * lo1 >>> 32)
+loA * hi1 + hiA * lo1 >>> 32)
+ hiA * hi1;
}
(I rearranged this code some out of the IDE to have mulDiv() on top.
Being lazy, I have a wrapper for sign handling - might try and do it properly before hell freezes over.
For timing, a model of input is in dire need:
How about such that each result possible is equally likely?)