/**
* Returns {@code n!}, that is, the product of the first {@code n} positive integers, or {@code 1}
* if {@code n == 0}.
*
* <p><b>Warning:</b> the result takes <i>O(n log n)</i> space, so use cautiously.
*
* <p>This uses an efficient binary recursive algorithm to compute the factorial with balanced
* multiplies. It also removes all the 2s from the intermediate products (shifting them back in at
* the end).
*
* @throws IllegalArgumentException if {@code n < 0}
*/
public static BigInteger factorial(int n) {
checkNonNegative("n", n);
// If the factorial is small enough, just use LongMath to do it.
if (n < LongMath.factorials.length) {
return BigInteger.valueOf(LongMath.factorials[n]);
}
// Pre-allocate space for our list of intermediate BigIntegers.
int approxSize = IntMath.divide(n * IntMath.log2(n, CEILING), Long.SIZE, CEILING);
ArrayList<BigInteger> bignums = new ArrayList<BigInteger>(approxSize);
// Start from the pre-computed maximum long factorial.
int startingNumber = LongMath.factorials.length;
long product = LongMath.factorials[startingNumber - 1];
// Strip off 2s from this value.
int shift = Long.numberOfTrailingZeros(product);
product >>= shift;
// Use floor(log2(num)) + 1 to prevent overflow of multiplication.
int productBits = LongMath.log2(product, FLOOR) + 1;
int bits = LongMath.log2(startingNumber, FLOOR) + 1;
// Check for the next power of two boundary, to save us a CLZ operation.
int nextPowerOfTwo = 1 << (bits - 1);
// Iteratively multiply the longs as big as they can go.
for (long num = startingNumber; num <= n; num++) {
// Check to see if the floor(log2(num)) + 1 has changed.
if ((num & nextPowerOfTwo) != 0) {
nextPowerOfTwo <<= 1;
bits++;
}
// Get rid of the 2s in num.
int tz = Long.numberOfTrailingZeros(num);
long normalizedNum = num >> tz;
shift += tz;
// Adjust floor(log2(num)) + 1.
int normalizedBits = bits - tz;
// If it won't fit in a long, then we store off the intermediate product.
if (normalizedBits + productBits >= Long.SIZE) {
bignums.add(BigInteger.valueOf(product));
product = 1;
productBits = 0;
}
product *= normalizedNum;
productBits = LongMath.log2(product, FLOOR) + 1;
}
// Check for leftovers.
if (product > 1) {
bignums.add(BigInteger.valueOf(product));
}
// Efficiently multiply all the intermediate products together.
return listProduct(bignums).shiftLeft(shift);
}
/**
* Returns {@code n} choose {@code k}, also known as the binomial coefficient of {@code n} and
* {@code k}, that is, {@code n! / (k! (n - k)!)}.
*
* <p><b>Warning:</b> the result can take as much as <i>O(k log n)</i> space.
*
* @throws IllegalArgumentException if {@code n < 0}, {@code k < 0}, or {@code k > n}
*/
public static BigInteger binomial(int n, int k) {
checkNonNegative("n", n);
checkNonNegative("k", k);
checkArgument(k <= n, "k (%s) > n (%s)", k, n);
if (k > (n >> 1)) {
k = n - k;
}
if (k < LongMath.biggestBinomials.length && n <= LongMath.biggestBinomials[k]) {
return BigInteger.valueOf(LongMath.binomial(n, k));
}
BigInteger accum = BigInteger.ONE;
long numeratorAccum = n;
long denominatorAccum = 1;
int bits = LongMath.log2(n, RoundingMode.CEILING);
int numeratorBits = bits;
for (int i = 1; i < k; i++) {
int p = n - i;
int q = i + 1;
// log2(p) >= bits - 1, because p >= n/2
if (numeratorBits + bits >= Long.SIZE - 1) {
// The numerator is as big as it can get without risking overflow.
// Multiply numeratorAccum / denominatorAccum into accum.
accum =
accum
.multiply(BigInteger.valueOf(numeratorAccum))
.divide(BigInteger.valueOf(denominatorAccum));
numeratorAccum = p;
denominatorAccum = q;
numeratorBits = bits;
} else {
// We can definitely multiply into the long accumulators without overflowing them.
numeratorAccum *= p;
denominatorAccum *= q;
numeratorBits += bits;
}
}
return accum
.multiply(BigInteger.valueOf(numeratorAccum))
.divide(BigInteger.valueOf(denominatorAccum));
}
