com.google.common.math.LongMath.java Source code

Java tutorial

Introduction

Here is the source code for com.google.common.math.LongMath.java

Source

/*
 * 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 com.google.common.primitives.UnsignedLongs;

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.
 *
 * <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 1 if {@code x < y} as unsigned longs, and 0 otherwise.  Assumes that x - y fits into a
     * signed long.  The implementation is branch-free, and benchmarks suggest it is measurably
     * faster than the straightforward ternary expression.
     */
    @VisibleForTesting
    static int lessThanBranchFree(long x, long y) {
        // Returns the sign bit of x - y.
        return (int) (~~(x - y) >>> (Long.SIZE - 1));
    }

    /**
     * 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 logFloor + lessThanBranchFree(cmp, x);

        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);
        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 logFloor + lessThanBranchFree(floorPow, x);
        case HALF_DOWN:
        case HALF_UP:
        case HALF_EVEN:
            // sqrt(10) is irrational, so log10(x)-logFloor is never exactly 0.5
            return logFloor + lessThanBranchFree(halfPowersOf10[logFloor], x);
        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)). If x < 10^y, then we want the
         * lower of the two possible values, or y - 1, otherwise, we want y.
         */
        return y - lessThanBranchFree(x, powersOf10[y]);
    }

    // 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);
        }
        /*
         * Let k be the true value of floor(sqrt(x)), so that
         *
         *            k * k <= x          <  (k + 1) * (k + 1)
         * (double) (k * k) <= (double) x <= (double) ((k + 1) * (k + 1))
         *          since casting to double is nondecreasing.
         *          Note that the right-hand inequality is no longer strict.
         * Math.sqrt(k * k) <= Math.sqrt(x) <= Math.sqrt((k + 1) * (k + 1))
         *          since Math.sqrt is monotonic.
         * (long) Math.sqrt(k * k) <= (long) Math.sqrt(x) <= (long) Math.sqrt((k + 1) * (k + 1))
         *          since casting to long is monotonic
         * k <= (long) Math.sqrt(x) <= k + 1
         *          since (long) Math.sqrt(k * k) == k, as checked exhaustively in
         *          {@link LongMathTest#testSqrtOfPerfectSquareAsDoubleIsPerfect}
         */
        long guess = (long) Math.sqrt(x);
        // Note: guess is always <= FLOOR_SQRT_MAX_LONG.
        long guessSquared = guess * guess;
        // Note (2013-2-26): benchmarks indicate that, inscrutably enough, using if statements is
        // faster here than using lessThanBranchFree.
        switch (mode) {
        case UNNECESSARY:
            checkRoundingUnnecessary(guessSquared == x);
            return guess;
        case FLOOR:
        case DOWN:
            if (x < guessSquared) {
                return guess - 1;
            }
            return guess;
        case CEILING:
        case UP:
            if (x > guessSquared) {
                return guess + 1;
            }
            return guess;
        case HALF_DOWN:
        case HALF_UP:
        case HALF_EVEN:
            long sqrtFloor = guess - ((x < guessSquared) ? 1 : 0);
            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.)
             *
             * If we treat halfSquare as an unsigned long, we know that
             *            sqrtFloor^2 <= x < (sqrtFloor + 1)^2
             * halfSquare - sqrtFloor <= x < halfSquare + sqrtFloor + 1
             * so |x - halfSquare| <= sqrtFloor.  Therefore, it's safe to treat x - halfSquare as a
             * signed long, so lessThanBranchFree is safe for use.
             */
            return sqrtFloor + lessThanBranchFree(halfSquare, x);
        default:
            throw new AssertionError();
        }
    }

    /**
     * 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}, a non-negative value less than {@code m}.
     * This differs from {@code x % m}, which might be negative.
     *
     * <p>For example:
     *
     * <pre> {@code
     *
     * mod(7, 4) == 3
     * mod(-7, 4) == 1
     * mod(-1, 4) == 3
     * mod(-8, 4) == 0
     * mod(8, 4) == 0}</pre>
     *
     * @throws ArithmeticException if {@code m <= 0}
     * @see <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.17.3">
     *      Remainder Operator</a>
     */
    @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}, a non-negative value less than {@code m}.
     * This differs from {@code x % m}, which might be negative.
     *
     * <p>For example:
     *
     * <pre> {@code
     *
     * mod(7, 4) == 3
     * mod(-7, 4) == 1
     * mod(-1, 4) == 3
     * mod(-8, 4) == 0
     * mod(8, 4) == 0}</pre>
     *
     * @throws ArithmeticException if {@code m <= 0}
     * @see <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.17.3">
     *      Remainder Operator</a>
     */
    @GwtIncompatible("TODO")
    public static long mod(long x, long m) {
        if (m <= 0) {
            throw new ArithmeticException("Modulus must be positive");
        }
        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(-FLOOR_SQRT_MAX_LONG <= b && b <= FLOOR_SQRT_MAX_LONG);
                    b *= b;
                }
            }
        }
    }

    @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.

    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);
    }

    /*
     * If n <= millerRabinBases[i][0], then testing n against bases millerRabinBases[i][1..]
     * suffices to prove its primality.  Values from miller-rabin.appspot.com.
     *
     * NOTE: We could get slightly better bases that would be treated as unsigned, but benchmarks
     * showed negligible performance improvements.
     */
    private static final long[][] millerRabinBaseSets = { { 291830, 126401071349994536L },
            { 885594168, 725270293939359937L, 3569819667048198375L },
            { 273919523040L, 15, 7363882082L, 992620450144556L },
            { 47636622961200L, 2, 2570940, 211991001, 3749873356L },
            { 7999252175582850L, 2, 4130806001517L, 149795463772692060L, 186635894390467037L,
                    3967304179347715805L },
            { 585226005592931976L, 2, 123635709730000L, 9233062284813009L, 43835965440333360L, 761179012939631437L,
                    1263739024124850375L },
            { Long.MAX_VALUE, 2, 325, 9375, 28178, 450775, 9780504, 1795265022 } };

    private enum MillerRabinTester {
        /**
         * Works for inputs <= FLOOR_SQRT_MAX_LONG.
         */
        SMALL {
            @Override
            long mulMod(long a, long b, long m) {
                /*
                 * NOTE(lowasser, 2015-Feb-12): Benchmarks suggest that changing this to
                 * UnsignedLongs.remainder and increasing the threshold to 2^32 doesn't pay for itself,
                 * and adding another enum constant hurts performance further -- I suspect because
                 * bimorphic implementation is a sweet spot for the JVM.
                 */
                return (a * b) % m;
            }

            @Override
            long squareMod(long a, long m) {
                return (a * a) % m;
            }
        },
        /**
         * Works for all nonnegative signed longs.
         */
        LARGE {
            /**
             * Returns (a + b) mod m.  Precondition: 0 <= a, b < m < 2^63.
             */
            private long plusMod(long a, long b, long m) {
                return (a >= m - b) ? (a + b - m) : (a + b);
            }

            /**
             * Returns (a * 2^32) mod m.  a may be any unsigned long.
             */
            private long times2ToThe32Mod(long a, long m) {
                int remainingPowersOf2 = 32;
                do {
                    int shift = Math.min(remainingPowersOf2, Long.numberOfLeadingZeros(a));
                    // shift is either the number of powers of 2 left to multiply a by, or the biggest shift
                    // possible while keeping a in an unsigned long.
                    a = UnsignedLongs.remainder(a << shift, m);
                    remainingPowersOf2 -= shift;
                } while (remainingPowersOf2 > 0);
                return a;
            }

            @Override
            long mulMod(long a, long b, long m) {
                long aHi = a >>> 32; // < 2^31
                long bHi = b >>> 32; // < 2^31
                long aLo = a & 0xFFFFFFFFL; // < 2^32
                long bLo = b & 0xFFFFFFFFL; // < 2^32

                /*
                 * a * b == aHi * bHi * 2^64 + (aHi * bLo + aLo * bHi) * 2^63 + aLo * bLo.
                 *       == (aHi * bHi * 2^32 + aHi * bLo + aLo * bHi) * 2^32 + aLo * bLo
                 *
                 * We carry out this computation in modular arithmetic.  Since times2ToThe32Mod accepts
                 * any unsigned long, we don't have to do a mod on every operation, only when intermediate
                 * results can exceed 2^63.
                 */
                long result = times2ToThe32Mod(aHi * bHi /* < 2^62 */, m); // < m < 2^63
                result += aHi * bLo; // aHi * bLo < 2^63, result < 2^64
                if (result < 0) {
                    result = UnsignedLongs.remainder(result, m);
                }
                // result < 2^63 again
                result += aLo * bHi; // aLo * bHi < 2^63, result < 2^64
                result = times2ToThe32Mod(result, m); // result < m < 2^63
                return plusMod(result, UnsignedLongs.remainder(aLo * bLo /* < 2^64 */, m), m);
            }

            @Override
            long squareMod(long a, long m) {
                long aHi = a >>> 32; // < 2^31
                long aLo = a & 0xFFFFFFFFL; // < 2^32

                /*
                 * a^2 == aHi^2 * 2^64 + aHi * aLo * 2^33 + aLo^2
                 *     == (aHi^2 * 2^32 + aHi * aLo * 2) * 2^32 + aLo^2
                 * We carry out this computation in modular arithmetic.  Since times2ToThe32Mod accepts
                 * any unsigned long, we don't have to do a mod on every operation, only when intermediate
                 * results can exceed 2^63.
                 */
                long result = times2ToThe32Mod(aHi * aHi /* < 2^62 */, m); // < m < 2^63
                long hiLo = aHi * aLo * 2;
                if (hiLo < 0) {
                    hiLo = UnsignedLongs.remainder(hiLo, m);
                }
                // hiLo < 2^63
                result += hiLo; // result < 2^64
                result = times2ToThe32Mod(result, m); // result < m < 2^63
                return plusMod(result, UnsignedLongs.remainder(aLo * aLo /* < 2^64 */, m), m);
            }
        };

        static boolean test(long base, long n) {
            // Since base will be considered % n, it's okay if base > FLOOR_SQRT_MAX_LONG,
            // so long as n <= FLOOR_SQRT_MAX_LONG.
            return ((n <= FLOOR_SQRT_MAX_LONG) ? SMALL : LARGE).testWitness(base, n);
        }

        /**
         * Returns a * b mod m.
         */
        abstract long mulMod(long a, long b, long m);

        /**
         * Returns a^2 mod m.
         */
        abstract long squareMod(long a, long m);

        /**
         * Returns a^p mod m.
         */
        private long powMod(long a, long p, long m) {
            long res = 1;
            for (; p != 0; p >>= 1) {
                if ((p & 1) != 0) {
                    res = mulMod(res, a, m);
                }
                a = squareMod(a, m);
            }
            return res;
        }

        /**
         * Returns true if n is a strong probable prime relative to the specified base.
         */
        private boolean testWitness(long base, long n) {
            int r = Long.numberOfTrailingZeros(n - 1);
            long d = (n - 1) >> r;
            base %= n;
            if (base == 0) {
                return true;
            }
            // Calculate a := base^d mod n.
            long a = powMod(base, d, n);
            // n passes this test if
            //    base^d = 1 (mod n)
            // or base^(2^j * d) = -1 (mod n) for some 0 <= j < r.
            if (a == 1) {
                return true;
            }
            int j = 0;
            while (a != n - 1) {
                if (++j == r) {
                    return false;
                }
                a = squareMod(a, n);
            }
            return true;
        }
    }

    private LongMath() {
    }
}