com.google.common.math.LongMath Maven / Gradle / Ivy
/*
* Copyright (C) 2011 The Guava Authors
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.google.common.math;
import static com.google.common.base.Preconditions.checkArgument;
import static com.google.common.base.Preconditions.checkNotNull;
import static com.google.common.math.MathPreconditions.checkNoOverflow;
import static com.google.common.math.MathPreconditions.checkNonNegative;
import static com.google.common.math.MathPreconditions.checkPositive;
import static com.google.common.math.MathPreconditions.checkRoundingUnnecessary;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static java.math.RoundingMode.HALF_EVEN;
import static java.math.RoundingMode.HALF_UP;
import com.google.common.annotations.GwtCompatible;
import com.google.common.annotations.GwtIncompatible;
import com.google.common.annotations.VisibleForTesting;
import java.math.BigInteger;
import java.math.RoundingMode;
/**
* A class for arithmetic on values of type {@code long}. Where possible, methods are defined and
* named analogously to their {@code BigInteger} counterparts.
*
* The implementations of many methods in this class are based on material from Henry S. Warren,
* Jr.'s Hacker's Delight, (Addison Wesley, 2002).
*
*
Similar functionality for {@code int} and for {@link BigInteger} can be found in
* {@link IntMath} and {@link BigIntegerMath} respectively. For other common operations on
* {@code long} values, see {@link com.google.common.primitives.Longs}.
*
* @author Louis Wasserman
* @since 11.0
*/
@GwtCompatible(emulated = true)
public final class LongMath {
// NOTE: Whenever both tests are cheap and functional, it's faster to use &, | instead of &&, ||
/**
* Returns {@code true} if {@code x} represents a power of two.
*
*
This differs from {@code Long.bitCount(x) == 1}, because
* {@code Long.bitCount(Long.MIN_VALUE) == 1}, but {@link Long#MIN_VALUE} is not a power of two.
*/
public static boolean isPowerOfTwo(long x) {
return x > 0 & (x & (x - 1)) == 0;
}
/**
* Returns the base-2 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 two
*/
@SuppressWarnings("fallthrough")
// TODO(kevinb): remove after this warning is disabled globally
public static int log2(long x, RoundingMode mode) {
checkPositive("x", x);
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(isPowerOfTwo(x));
// fall through
case DOWN:
case FLOOR:
return (Long.SIZE - 1) - Long.numberOfLeadingZeros(x);
case UP:
case CEILING:
return Long.SIZE - Long.numberOfLeadingZeros(x - 1);
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
// Since sqrt(2) is irrational, log2(x) - logFloor cannot be exactly 0.5
int leadingZeros = Long.numberOfLeadingZeros(x);
long cmp = MAX_POWER_OF_SQRT2_UNSIGNED >>> leadingZeros;
// floor(2^(logFloor + 0.5))
int logFloor = (Long.SIZE - 1) - leadingZeros;
return (x <= cmp) ? logFloor : logFloor + 1;
default:
throw new AssertionError("impossible");
}
}
/** The biggest half power of two that fits into an unsigned long */
@VisibleForTesting static final long MAX_POWER_OF_SQRT2_UNSIGNED = 0xB504F333F9DE6484L;
/**
* 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")
// TODO(kevinb): remove after this warning is disabled globally
public static int log10(long x, RoundingMode mode) {
checkPositive("x", x);
if (fitsInInt(x)) {
return IntMath.log10((int) x, mode);
}
int logFloor = log10Floor(x);
long floorPow = powersOf10[logFloor];
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(x == floorPow);
// fall through
case FLOOR:
case DOWN:
return logFloor;
case CEILING:
case UP:
return (x == floorPow) ? logFloor : logFloor + 1;
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
// sqrt(10) is irrational, so log10(x)-logFloor is never exactly 0.5
return (x <= halfPowersOf10[logFloor]) ? logFloor : logFloor + 1;
default:
throw new AssertionError();
}
}
@GwtIncompatible("TODO")
static int log10Floor(long x) {
/*
* Based on Hacker's Delight Fig. 11-5, the two-table-lookup, branch-free implementation.
*
* The key idea is that based on the number of leading zeros (equivalently, floor(log2(x))),
* we can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x))
* is 6, then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
*/
int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
// y is the higher of the two possible values of floor(log10(x))
long sgn = (x - powersOf10[y]) >>> (Long.SIZE - 1);
/*
* sgn is the sign bit of x - 10^y; it is 1 if x < 10^y, and 0 otherwise. If x < 10^y, then we
* want the lower of the two possible values, or y - 1, otherwise, we want y.
*/
return y - (int) sgn;
}
// maxLog10ForLeadingZeros[i] == floor(log10(2^(Long.SIZE - i)))
@VisibleForTesting static final byte[] maxLog10ForLeadingZeros = {
19, 18, 18, 18, 18, 17, 17, 17, 16, 16, 16, 15, 15, 15, 15, 14, 14, 14, 13, 13, 13, 12, 12,
12, 12, 11, 11, 11, 10, 10, 10, 9, 9, 9, 9, 8, 8, 8, 7, 7, 7, 6, 6, 6, 6, 5, 5, 5, 4, 4, 4,
3, 3, 3, 3, 2, 2, 2, 1, 1, 1, 0, 0, 0 };
@GwtIncompatible("TODO")
@VisibleForTesting
static final long[] powersOf10 = {
1L,
10L,
100L,
1000L,
10000L,
100000L,
1000000L,
10000000L,
100000000L,
1000000000L,
10000000000L,
100000000000L,
1000000000000L,
10000000000000L,
100000000000000L,
1000000000000000L,
10000000000000000L,
100000000000000000L,
1000000000000000000L
};
// halfPowersOf10[i] = largest long less than 10^(i + 0.5)
@GwtIncompatible("TODO")
@VisibleForTesting
static final long[] halfPowersOf10 = {
3L,
31L,
316L,
3162L,
31622L,
316227L,
3162277L,
31622776L,
316227766L,
3162277660L,
31622776601L,
316227766016L,
3162277660168L,
31622776601683L,
316227766016837L,
3162277660168379L,
31622776601683793L,
316227766016837933L,
3162277660168379331L
};
/**
* Returns {@code b} to the {@code k}th power. Even if the result overflows, it will be equal to
* {@code BigInteger.valueOf(b).pow(k).longValue()}. This implementation runs in {@code O(log k)}
* time.
*
* @throws IllegalArgumentException if {@code k < 0}
*/
@GwtIncompatible("TODO")
public static long pow(long b, int k) {
checkNonNegative("exponent", k);
if (-2 <= b && b <= 2) {
switch ((int) b) {
case 0:
return (k == 0) ? 1 : 0;
case 1:
return 1;
case (-1):
return ((k & 1) == 0) ? 1 : -1;
case 2:
return (k < Long.SIZE) ? 1L << k : 0;
case (-2):
if (k < Long.SIZE) {
return ((k & 1) == 0) ? 1L << k : -(1L << k);
} else {
return 0;
}
default:
throw new AssertionError();
}
}
for (long accum = 1;; k >>= 1) {
switch (k) {
case 0:
return accum;
case 1:
return accum * b;
default:
accum *= ((k & 1) == 0) ? 1 : b;
b *= b;
}
}
}
/**
* 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 long sqrt(long x, RoundingMode mode) {
checkNonNegative("x", x);
if (fitsInInt(x)) {
return IntMath.sqrt((int) x, mode);
}
long sqrtFloor = sqrtFloor(x);
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(sqrtFloor * sqrtFloor == x); // fall through
case FLOOR:
case DOWN:
return sqrtFloor;
case CEILING:
case UP:
return (sqrtFloor * sqrtFloor == x) ? sqrtFloor : sqrtFloor + 1;
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
long halfSquare = sqrtFloor * sqrtFloor + 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. (We have to deal with overflow, though.)
*/
return (halfSquare >= x | halfSquare < 0) ? sqrtFloor : sqrtFloor + 1;
default:
throw new AssertionError();
}
}
@GwtIncompatible("TODO")
private static long sqrtFloor(long x) {
long guess = (long) Math.sqrt(x);
/*
* Lemma: For all a, b, if |a - b| <= 1, then |floor(a) - floor(b)| <= 1.
*
* Proof:
* -1 <= a - b <= 1
* b - 1 <= a <= b + 1
* floor(b - 1) <= floor(a) <= floor(b + 1)
* floor(b) - 1 <= floor(a) <= floor(b) + 1
* -1 <= floor(a) - floor(b) <= 1
*
* Theorem: |floor(sqrt(x)) - guess| <= 1.
*
* Proof: By the lemma, it suffices to show that |sqrt(x) - Math.sqrt(x)| <= 1.
* We consider two cases: x <= 2^53 and x > 2^53.
*
* If x <= 2^53, then x is exactly representable as a double, so the only error is in rounding
* sqrt(x) to a double, which introduces at most 2^-52 relative error. Since sqrt(x) < 2^27,
* the absolute error is at most 2^(27-52) = 2^-25 < 1.
*
* Otherwise, x > 2^53. The rounding error introduced by casting x to a double is at most
* 2^63 * 2^-52 = 2^11. Noting that sqrt(x) > 2^26,
*
* sqrt(x) - 0.5 = sqrt((sqrt(x) - 0.5)^2)
* = sqrt(x - sqrt(x) + 0.25)
* < sqrt(x - 2^26 + 0.25)
* < sqrt(x - 2^11)
* <= sqrt((double) x)
* sqrt(x) + 0.5 = sqrt((sqrt(x) + 0.5)^2)
* = sqrt(x + sqrt(x) + 0.25)
* > sqrt(x + 2^26 + 0.25)
* > sqrt(x + 2^11)
* >= sqrt((double) x)
* sqrt(x) - 0.5 < sqrt((double) x) < sqrt(x) + 0.5
*
* Math.sqrt((double) x) is obtained by rounding sqrt((double) x) to a double, increasing the
* error by at most 2^-52 * sqrt(x) <= 2^(32 - 52) <= 2^-20, so clearly
*
* sqrt(x) - 0.5 - 2^-20 <= Math.sqrt((double) x) <= sqrt(x) + 0.5 + 2^-20
*
* Therefore, |sqrt(x) - Math.sqrt((double) x)| <= 1, so
* |floor(sqrt(x)) - (long) Math.sqrt((double) x)| <= 1
* as desired.
*/
long guessSquared = guess * guess;
if (x - guessSquared >= guess + guess + 1) {
// The condition is equivalent to x >= (guess + 1) * (guess + 1), but doesn't overflow.
guess++;
} else if (x < guessSquared) {
guess--;
}
return guess;
}
/**
* Returns the result of dividing {@code p} by {@code q}, rounding using the specified
* {@code RoundingMode}.
*
* @throws ArithmeticException if {@code q == 0}, or if {@code mode == UNNECESSARY} and {@code a}
* is not an integer multiple of {@code b}
*/
@GwtIncompatible("TODO")
@SuppressWarnings("fallthrough")
public static long divide(long p, long q, RoundingMode mode) {
checkNotNull(mode);
long div = p / q; // throws if q == 0
long rem = p - q * div; // equals p % q
if (rem == 0) {
return div;
}
/*
* Normal Java division rounds towards 0, consistently with RoundingMode.DOWN. We just have to
* deal with the cases where rounding towards 0 is wrong, which typically depends on the sign of
* p / q.
*
* signum is 1 if p and q are both nonnegative or both negative, and -1 otherwise.
*/
int signum = 1 | (int) ((p ^ q) >> (Long.SIZE - 1));
boolean increment;
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(rem == 0);
// fall through
case DOWN:
increment = false;
break;
case UP:
increment = true;
break;
case CEILING:
increment = signum > 0;
break;
case FLOOR:
increment = signum < 0;
break;
case HALF_EVEN:
case HALF_DOWN:
case HALF_UP:
long absRem = abs(rem);
long cmpRemToHalfDivisor = absRem - (abs(q) - absRem);
// subtracting two nonnegative longs can't overflow
// cmpRemToHalfDivisor has the same sign as compare(abs(rem), abs(q) / 2).
if (cmpRemToHalfDivisor == 0) { // exactly on the half mark
increment = (mode == HALF_UP | (mode == HALF_EVEN & (div & 1) != 0));
} else {
increment = cmpRemToHalfDivisor > 0; // closer to the UP value
}
break;
default:
throw new AssertionError();
}
return increment ? div + signum : div;
}
/**
* Returns {@code x mod m}. This differs from {@code x % m} in that it always returns a
* non-negative result.
*
*
For example:
*
*
{@code
*
* mod(7, 4) == 3
* mod(-7, 4) == 1
* mod(-1, 4) == 3
* mod(-8, 4) == 0
* mod(8, 4) == 0}
*
* @throws ArithmeticException if {@code m <= 0}
*/
@GwtIncompatible("TODO")
public static int mod(long x, int m) {
// Cast is safe because the result is guaranteed in the range [0, m)
return (int) mod(x, (long) m);
}
/**
* Returns {@code x mod m}. This differs from {@code x % m} in that it always returns a
* non-negative result.
*
* For example:
*
*
{@code
*
* mod(7, 4) == 3
* mod(-7, 4) == 1
* mod(-1, 4) == 3
* mod(-8, 4) == 0
* mod(8, 4) == 0}
*
* @throws ArithmeticException if {@code m <= 0}
*/
@GwtIncompatible("TODO")
public static long mod(long x, long m) {
if (m <= 0) {
throw new ArithmeticException("Modulus " + m + " must be > 0");
}
long result = x % m;
return (result >= 0) ? result : result + m;
}
/**
* Returns the greatest common divisor of {@code a, b}. Returns {@code 0} if
* {@code a == 0 && b == 0}.
*
* @throws IllegalArgumentException if {@code a < 0} or {@code b < 0}
*/
public static long gcd(long a, long b) {
/*
* The reason we require both arguments to be >= 0 is because otherwise, what do you return on
* gcd(0, Long.MIN_VALUE)? BigInteger.gcd would return positive 2^63, but positive 2^63 isn't
* an int.
*/
checkNonNegative("a", a);
checkNonNegative("b", b);
if (a == 0) {
// 0 % b == 0, so b divides a, but the converse doesn't hold.
// BigInteger.gcd is consistent with this decision.
return b;
} else if (b == 0) {
return a; // similar logic
}
/*
* Uses the binary GCD algorithm; see http://en.wikipedia.org/wiki/Binary_GCD_algorithm.
* This is >60% faster than the Euclidean algorithm in benchmarks.
*/
int aTwos = Long.numberOfTrailingZeros(a);
a >>= aTwos; // divide out all 2s
int bTwos = Long.numberOfTrailingZeros(b);
b >>= bTwos; // divide out all 2s
while (a != b) { // both a, b are odd
// The key to the binary GCD algorithm is as follows:
// Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
// But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.
// We bend over backwards to avoid branching, adapting a technique from
// http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax
long delta = a - b; // can't overflow, since a and b are nonnegative
long minDeltaOrZero = delta & (delta >> (Long.SIZE - 1));
// equivalent to Math.min(delta, 0)
a = delta - minDeltaOrZero - minDeltaOrZero; // sets a to Math.abs(a - b)
// a is now nonnegative and even
b += minDeltaOrZero; // sets b to min(old a, b)
a >>= Long.numberOfTrailingZeros(a); // divide out all 2s, since 2 doesn't divide b
}
return a << min(aTwos, bTwos);
}
/**
* Returns the sum of {@code a} and {@code b}, provided it does not overflow.
*
* @throws ArithmeticException if {@code a + b} overflows in signed {@code long} arithmetic
*/
@GwtIncompatible("TODO")
public static long checkedAdd(long a, long b) {
long result = a + b;
checkNoOverflow((a ^ b) < 0 | (a ^ result) >= 0);
return result;
}
/**
* Returns the difference of {@code a} and {@code b}, provided it does not overflow.
*
* @throws ArithmeticException if {@code a - b} overflows in signed {@code long} arithmetic
*/
@GwtIncompatible("TODO")
public static long checkedSubtract(long a, long b) {
long result = a - b;
checkNoOverflow((a ^ b) >= 0 | (a ^ result) >= 0);
return result;
}
/**
* Returns the product of {@code a} and {@code b}, provided it does not overflow.
*
* @throws ArithmeticException if {@code a * b} overflows in signed {@code long} arithmetic
*/
@GwtIncompatible("TODO")
public static long checkedMultiply(long a, long b) {
// Hacker's Delight, Section 2-12
int leadingZeros = Long.numberOfLeadingZeros(a) + Long.numberOfLeadingZeros(~a)
+ Long.numberOfLeadingZeros(b) + Long.numberOfLeadingZeros(~b);
/*
* If leadingZeros > Long.SIZE + 1 it's definitely fine, if it's < Long.SIZE it's definitely
* bad. We do the leadingZeros check to avoid the division below if at all possible.
*
* Otherwise, if b == Long.MIN_VALUE, then the only allowed values of a are 0 and 1. We take
* care of all a < 0 with their own check, because in particular, the case a == -1 will
* incorrectly pass the division check below.
*
* In all other cases, we check that either a is 0 or the result is consistent with division.
*/
if (leadingZeros > Long.SIZE + 1) {
return a * b;
}
checkNoOverflow(leadingZeros >= Long.SIZE);
checkNoOverflow(a >= 0 | b != Long.MIN_VALUE);
long result = a * b;
checkNoOverflow(a == 0 || result / a == b);
return result;
}
/**
* Returns the {@code b} to the {@code k}th power, provided it does not overflow.
*
* @throws ArithmeticException if {@code b} to the {@code k}th power overflows in signed
* {@code long} arithmetic
*/
@GwtIncompatible("TODO")
public static long checkedPow(long b, int k) {
checkNonNegative("exponent", k);
if (b >= -2 & b <= 2) {
switch ((int) b) {
case 0:
return (k == 0) ? 1 : 0;
case 1:
return 1;
case (-1):
return ((k & 1) == 0) ? 1 : -1;
case 2:
checkNoOverflow(k < Long.SIZE - 1);
return 1L << k;
case (-2):
checkNoOverflow(k < Long.SIZE);
return ((k & 1) == 0) ? (1L << k) : (-1L << k);
default:
throw new AssertionError();
}
}
long accum = 1;
while (true) {
switch (k) {
case 0:
return accum;
case 1:
return checkedMultiply(accum, b);
default:
if ((k & 1) != 0) {
accum = checkedMultiply(accum, b);
}
k >>= 1;
if (k > 0) {
checkNoOverflow(b <= FLOOR_SQRT_MAX_LONG);
b *= b;
}
}
}
}
@GwtIncompatible("TODO")
@VisibleForTesting static final long FLOOR_SQRT_MAX_LONG = 3037000499L;
/**
* Returns {@code n!}, that is, the product of the first {@code n} positive
* integers, {@code 1} if {@code n == 0}, or {@link Long#MAX_VALUE} if the
* result does not fit in a {@code long}.
*
* @throws IllegalArgumentException if {@code n < 0}
*/
@GwtIncompatible("TODO")
public static long factorial(int n) {
checkNonNegative("n", n);
return (n < factorials.length) ? factorials[n] : Long.MAX_VALUE;
}
static final long[] factorials = {
1L,
1L,
1L * 2,
1L * 2 * 3,
1L * 2 * 3 * 4,
1L * 2 * 3 * 4 * 5,
1L * 2 * 3 * 4 * 5 * 6,
1L * 2 * 3 * 4 * 5 * 6 * 7,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19,
1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19 * 20
};
/**
* Returns {@code n} choose {@code k}, also known as the binomial coefficient of {@code n} and
* {@code k}, or {@link Long#MAX_VALUE} if the result does not fit in a {@code long}.
*
* @throws IllegalArgumentException if {@code n < 0}, {@code k < 0}, or {@code k > n}
*/
public static long 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;
}
switch (k) {
case 0:
return 1;
case 1:
return n;
default:
if (n < factorials.length) {
return factorials[n] / (factorials[k] * factorials[n - k]);
} else if (k >= biggestBinomials.length || n > biggestBinomials[k]) {
return Long.MAX_VALUE;
} else if (k < biggestSimpleBinomials.length && n <= biggestSimpleBinomials[k]) {
// guaranteed not to overflow
long result = n--;
for (int i = 2; i <= k; n--, i++) {
result *= n;
result /= i;
}
return result;
} else {
int nBits = LongMath.log2(n, RoundingMode.CEILING);
long result = 1;
long numerator = n--;
long denominator = 1;
int numeratorBits = nBits;
// This is an upper bound on log2(numerator, ceiling).
/*
* We want to do this in long math for speed, but want to avoid overflow. We adapt the
* technique previously used by BigIntegerMath: maintain separate numerator and
* denominator accumulators, multiplying the fraction into result when near overflow.
*/
for (int i = 2; i <= k; i++, n--) {
if (numeratorBits + nBits < Long.SIZE - 1) {
// It's definitely safe to multiply into numerator and denominator.
numerator *= n;
denominator *= i;
numeratorBits += nBits;
} else {
// It might not be safe to multiply into numerator and denominator,
// so multiply (numerator / denominator) into result.
result = multiplyFraction(result, numerator, denominator);
numerator = n;
denominator = i;
numeratorBits = nBits;
}
}
return multiplyFraction(result, numerator, denominator);
}
}
}
/**
* Returns (x * numerator / denominator), which is assumed to come out to an integral value.
*/
static long multiplyFraction(long x, long numerator, long denominator) {
if (x == 1) {
return numerator / denominator;
}
long commonDivisor = gcd(x, denominator);
x /= commonDivisor;
denominator /= commonDivisor;
// We know gcd(x, denominator) = 1, and x * numerator / denominator is exact,
// so denominator must be a divisor of numerator.
return x * (numerator / denominator);
}
/*
* binomial(biggestBinomials[k], k) fits in a long, but not
* binomial(biggestBinomials[k] + 1, k).
*/
static final int[] biggestBinomials =
{Integer.MAX_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE, 3810779, 121977, 16175, 4337, 1733,
887, 534, 361, 265, 206, 169, 143, 125, 111, 101, 94, 88, 83, 79, 76, 74, 72, 70, 69, 68,
67, 67, 66, 66, 66, 66};
/*
* binomial(biggestSimpleBinomials[k], k) doesn't need to use the slower GCD-based impl,
* but binomial(biggestSimpleBinomials[k] + 1, k) does.
*/
@VisibleForTesting static final int[] biggestSimpleBinomials =
{Integer.MAX_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE, 2642246, 86251, 11724, 3218, 1313,
684, 419, 287, 214, 169, 139, 119, 105, 95, 87, 81, 76, 73, 70, 68, 66, 64, 63, 62, 62,
61, 61, 61};
// These values were generated by using checkedMultiply to see when the simple multiply/divide
// algorithm would lead to an overflow.
@GwtIncompatible("TODO")
static boolean fitsInInt(long x) {
return (int) x == x;
}
/**
* Returns the arithmetic mean of {@code x} and {@code y}, rounded toward
* negative infinity. This method is resilient to overflow.
*
* @since 14.0
*/
public static long mean(long x, long y) {
// Efficient method for computing the arithmetic mean.
// The alternative (x + y) / 2 fails for large values.
// The alternative (x + y) >>> 1 fails for negative values.
return (x & y) + ((x ^ y) >> 1);
}
private LongMath() {}
}