/*
* 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.math.MathPreconditions.checkNonNegative;
import static com.google.common.math.MathPreconditions.checkPositive;
import static com.google.common.math.MathPreconditions.checkRoundingUnnecessary;
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.VisibleForTesting;
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.
*
* <p>The implementations of many methods in this class are based on material from Henry S. Warren,
* Jr.'s <i>Hacker's Delight</i>, (Addison Wesley, 2002).
*
* <p>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.
*
* <p>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")
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;
// 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 };
// halfPowersOf10[i] = largest long less than 10^(i + 0.5)
/**
* 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);
}
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.
/**
* 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() {}
}