/**
* 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 the square root of {@code x}, rounded with the specified rounding mode.
*
* @throws IllegalArgumentException if {@code x < 0}
* @throws ArithmeticException if {@code mode} is {@link RoundingMode#UNNECESSARY} and {@code
* sqrt(x)} is not an integer
*/
@GwtIncompatible // TODO
@SuppressWarnings("fallthrough")
public static BigInteger sqrt(BigInteger x, RoundingMode mode) {
checkNonNegative("x", x);
if (fitsInLong(x)) {
return BigInteger.valueOf(LongMath.sqrt(x.longValue(), mode));
}
BigInteger sqrtFloor = sqrtFloor(x);
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(sqrtFloor.pow(2).equals(x)); // fall through
case FLOOR:
case DOWN:
return sqrtFloor;
case CEILING:
case UP:
int sqrtFloorInt = sqrtFloor.intValue();
boolean sqrtFloorIsExact =
(sqrtFloorInt * sqrtFloorInt == x.intValue()) // fast check mod 2^32
&& sqrtFloor.pow(2).equals(x); // slow exact check
return sqrtFloorIsExact ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
BigInteger halfSquare = sqrtFloor.pow(2).add(sqrtFloor);
/*
* We wish to test whether or not x <= (sqrtFloor + 0.5)^2 = halfSquare + 0.25. Since both x
* and halfSquare are integers, this is equivalent to testing whether or not x <=
* halfSquare.
*/
return (halfSquare.compareTo(x) >= 0) ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
default:
throw new AssertionError();
}
}
/**
* 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));
}
/**
* Returns the base-10 logarithm of {@code x}, rounded according to the specified rounding mode.
*
* @throws IllegalArgumentException if {@code x <= 0}
* @throws ArithmeticException if {@code mode} is {@link RoundingMode#UNNECESSARY} and {@code x}
* is not a power of ten
*/
@GwtIncompatible // TODO
@SuppressWarnings("fallthrough")
public static int log10(BigInteger x, RoundingMode mode) {
checkPositive("x", x);
if (fitsInLong(x)) {
return LongMath.log10(x.longValue(), mode);
}
int approxLog10 = (int) (log2(x, FLOOR) * LN_2 / LN_10);
BigInteger approxPow = BigInteger.TEN.pow(approxLog10);
int approxCmp = approxPow.compareTo(x);
/*
* We adjust approxLog10 and approxPow until they're equal to floor(log10(x)) and
* 10^floor(log10(x)).
*/
if (approxCmp > 0) {
/*
* The code is written so that even completely incorrect approximations will still yield the
* correct answer eventually, but in practice this branch should almost never be entered, and
* even then the loop should not run more than once.
*/
do {
approxLog10--;
approxPow = approxPow.divide(BigInteger.TEN);
approxCmp = approxPow.compareTo(x);
} while (approxCmp > 0);
} else {
BigInteger nextPow = BigInteger.TEN.multiply(approxPow);
int nextCmp = nextPow.compareTo(x);
while (nextCmp <= 0) {
approxLog10++;
approxPow = nextPow;
approxCmp = nextCmp;
nextPow = BigInteger.TEN.multiply(approxPow);
nextCmp = nextPow.compareTo(x);
}
}
int floorLog = approxLog10;
BigInteger floorPow = approxPow;
int floorCmp = approxCmp;
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(floorCmp == 0);
// fall through
case FLOOR:
case DOWN:
return floorLog;
case CEILING:
case UP:
return floorPow.equals(x) ? floorLog : floorLog + 1;
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
// Since sqrt(10) is irrational, log10(x) - floorLog can never be exactly 0.5
BigInteger x2 = x.pow(2);
BigInteger halfPowerSquared = floorPow.pow(2).multiply(BigInteger.TEN);
return (x2.compareTo(halfPowerSquared) <= 0) ? floorLog : floorLog + 1;
default:
throw new AssertionError();
}
}
