/*
* 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.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.Beta;
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 int}. 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 long} and for {@link BigInteger} can be found in
* {@link LongMath} and {@link BigIntegerMath} respectively. For other common operations on
* {@code int} values, see {@link com.google.common.primitives.Ints}.
*
* @author Louis Wasserman
* @since 11.0
*/
@Beta
@GwtCompatible(emulated = true)
public final class IntMath {
// 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 Integer.bitCount(x) == 1}, because
* {@code Integer.bitCount(Integer.MIN_VALUE) == 1}, but {@link Integer#MIN_VALUE} is not a power
* of two.
*/
public static boolean isPowerOfTwo(int 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(int x, RoundingMode mode) {
checkPositive("x", x);
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(isPowerOfTwo(x));
// fall through
case DOWN:
case FLOOR:
return (Integer.SIZE - 1) - Integer.numberOfLeadingZeros(x);
case UP:
case CEILING:
return Integer.SIZE - Integer.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 = Integer.numberOfLeadingZeros(x);
int cmp = MAX_POWER_OF_SQRT2_UNSIGNED >>> leadingZeros;
// floor(2^(logFloor + 0.5))
int logFloor = (Integer.SIZE - 1) - leadingZeros;
return (x <= cmp) ? logFloor : logFloor + 1;
default:
throw new AssertionError();
}
}
/** The biggest half power of two that can fit in an unsigned int. */
@VisibleForTesting static final int MAX_POWER_OF_SQRT2_UNSIGNED = 0xB504F333;
private static int log10Floor(int 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 = MAX_LOG10_FOR_LEADING_ZEROS[Integer.numberOfLeadingZeros(x)];
// y is the higher of the two possible values of floor(log10(x))
int sgn = (x - POWERS_OF_10[y]) >>> (Integer.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 - sgn;
}
// MAX_LOG10_FOR_LEADING_ZEROS[i] == floor(log10(2^(Long.SIZE - i)))
@VisibleForTesting static final byte[] MAX_LOG10_FOR_LEADING_ZEROS = {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, 0};
@VisibleForTesting static final int[] POWERS_OF_10 = {1, 10, 100, 1000, 10000,
100000, 1000000, 10000000, 100000000, 1000000000};
// HALF_POWERS_OF_10[i] = largest int less than 10^(i + 0.5)
@VisibleForTesting static final int[] HALF_POWERS_OF_10 =
{3, 31, 316, 3162, 31622, 316227, 3162277, 31622776, 316227766, Integer.MAX_VALUE};
private static int sqrtFloor(int x) {
// There is no loss of precision in converting an int to a double, according to
// http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#5.1.2
return (int) Math.sqrt(x);
}
/**
* 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}
*/
@SuppressWarnings("fallthrough")
public static int divide(int p, int q, RoundingMode mode) {
checkNotNull(mode);
if (q == 0) {
throw new ArithmeticException("/ by zero"); // for GWT
}
int div = p / q;
int rem = p - q * div; // equal to 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 | ((p ^ q) >> (Integer.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:
int absRem = abs(rem);
int cmpRemToHalfDivisor = absRem - (abs(q) - absRem);
// subtracting two nonnegative ints 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.
*
* <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}
*/
public static int mod(int x, int m) {
if (m <= 0) {
throw new ArithmeticException("Modulus " + m + " must be > 0");
}
int 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 int gcd(int a, int b) {
/*
* The reason we require both arguments to be >= 0 is because otherwise, what do you return on
* gcd(0, Integer.MIN_VALUE)? BigInteger.gcd would return positive 2^31, but positive 2^31
* 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 >40% faster than the Euclidean algorithm in benchmarks.
*/
int aTwos = Integer.numberOfTrailingZeros(a);
a >>= aTwos; // divide out all 2s
int bTwos = Integer.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
int delta = a - b; // can't overflow, since a and b are nonnegative
int minDeltaOrZero = delta & (delta >> (Integer.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 >>= Integer.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 int} arithmetic
*/
public static int checkedAdd(int a, int b) {
long result = (long) a + b;
checkNoOverflow(result == (int) result);
return (int) 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 int} arithmetic
*/
public static int checkedSubtract(int a, int b) {
long result = (long) a - b;
checkNoOverflow(result == (int) result);
return (int) 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 int} arithmetic
*/
public static int checkedMultiply(int a, int b) {
long result = (long) a * b;
checkNoOverflow(result == (int) result);
return (int) result;
}
/**
* Returns the {@code b} to the {@code k}th power, provided it does not overflow.
*
* <p>{@link #pow} may be faster, but does not check for overflow.
*
* @throws ArithmeticException if {@code b} to the {@code k}th power overflows in signed
* {@code int} arithmetic
*/
public static int checkedPow(int b, int k) {
checkNonNegative("exponent", k);
switch (b) {
case 0:
return (k == 0) ? 1 : 0;
case 1:
return 1;
case (-1):
return ((k & 1) == 0) ? 1 : -1;
case 2:
checkNoOverflow(k < Integer.SIZE - 1);
return 1 << k;
case (-2):
checkNoOverflow(k < Integer.SIZE);
return ((k & 1) == 0) ? 1 << k : -1 << k;
}
int 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_INT <= b & b <= FLOOR_SQRT_MAX_INT);
b *= b;
}
}
}
}
@VisibleForTesting static final int FLOOR_SQRT_MAX_INT = 46340;
/**
* Returns {@code n!}, that is, the product of the first {@code n} positive
* integers, {@code 1} if {@code n == 0}, or {@link Integer#MAX_VALUE} if the
* result does not fit in a {@code int}.
*
* @throws IllegalArgumentException if {@code n < 0}
*/
public static int factorial(int n) {
checkNonNegative("n", n);
return (n < FACTORIALS.length) ? FACTORIALS[n] : Integer.MAX_VALUE;
}
static final int[] FACTORIALS = {
1,
1,
1 * 2,
1 * 2 * 3,
1 * 2 * 3 * 4,
1 * 2 * 3 * 4 * 5,
1 * 2 * 3 * 4 * 5 * 6,
1 * 2 * 3 * 4 * 5 * 6 * 7,
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8,
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9,
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10,
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11,
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12};
// binomial(BIGGEST_BINOMIALS[k], k) fits in an int, but not binomial(BIGGEST_BINOMIALS[k]+1,k).
@VisibleForTesting static int[] BIGGEST_BINOMIALS = {
Integer.MAX_VALUE,
Integer.MAX_VALUE,
65536,
2345,
477,
193,
110,
75,
58,
49,
43,
39,
37,
35,
34,
34,
33
};
/**
* Returns the arithmetic mean of {@code x} and {@code y}, rounded towards
* negative infinity. This method is overflow resilient.
*
* @since 14.0
*/
public static int mean(int x, int 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 IntMath() {}
}