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/*
 * 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 int}. 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 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 */ @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. * *

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") // TODO(kevinb): remove after this warning is disabled globally 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; /** * 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("need BigIntegerMath to adequately test") @SuppressWarnings("fallthrough") public static int log10(int x, RoundingMode mode) { checkPositive("x", x); int logFloor = log10Floor(x); int 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(); } } 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 = maxLog10ForLeadingZeros[Integer.numberOfLeadingZeros(x)]; // y is the higher of the two possible values of floor(log10(x)) int sgn = (x - powersOf10[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; } // maxLog10ForLeadingZeros[i] == floor(log10(2^(Long.SIZE - i))) @VisibleForTesting static final byte[] maxLog10ForLeadingZeros = {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[] powersOf10 = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000}; // halfPowersOf10[i] = largest int less than 10^(i + 0.5) @VisibleForTesting static final int[] halfPowersOf10 = {3, 31, 316, 3162, 31622, 316227, 3162277, 31622776, 316227766, Integer.MAX_VALUE}; /** * 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).intValue()}. This implementation runs in {@code O(log k)} * time. * *

Compare {@link #checkedPow}, which throws an {@link ArithmeticException} upon overflow. * * @throws IllegalArgumentException if {@code k < 0} */ @GwtIncompatible("failing tests") public static int pow(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: return (k < Integer.SIZE) ? (1 << k) : 0; case (-2): if (k < Integer.SIZE) { return ((k & 1) == 0) ? (1 << k) : -(1 << k); } else { return 0; } default: // continue below to handle the general case } for (int accum = 1;; k >>= 1) { switch (k) { case 0: return accum; case 1: return b * accum; 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("need BigIntegerMath to adequately test") @SuppressWarnings("fallthrough") public static int sqrt(int x, RoundingMode mode) { checkNonNegative("x", x); int 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: int 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 (x <= halfSquare | halfSquare < 0) ? sqrtFloor : sqrtFloor + 1; default: throw new AssertionError(); } } 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. * *

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} */ 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. * *

{@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; default: // continue below to handle the general case } 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; } private 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}; /** * Returns {@code n} choose {@code k}, also known as the binomial coefficient of {@code n} and * {@code k}, or {@link Integer#MAX_VALUE} if the result does not fit in an {@code int}. * * @throws IllegalArgumentException if {@code n < 0}, {@code k < 0} or {@code k > n} */ @GwtIncompatible("need BigIntegerMath to adequately test") public static int 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; } if (k >= biggestBinomials.length || n > biggestBinomials[k]) { return Integer.MAX_VALUE; } switch (k) { case 0: return 1; case 1: return n; default: long result = 1; for (int i = 0; i < k; i++) { result *= n - i; result /= i + 1; } return (int) result; } } // binomial(biggestBinomials[k], k) fits in an int, but not binomial(biggestBinomials[k]+1,k). @VisibleForTesting static int[] biggestBinomials = { 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() {} }





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