* 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 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. * *
* 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}
* @see
* Remainder Operator
*/
@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.
*
* * 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}
* @see
* Remainder Operator
*/
@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(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);
}
private LongMath() {
}
}