471 lines
16 KiB
Java
471 lines
16 KiB
Java
/*
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* Copyright (C) 2011 The Guava Authors
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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package com.google.common.math;
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import static com.google.common.base.Preconditions.checkArgument;
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import static com.google.common.base.Preconditions.checkNotNull;
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import static com.google.common.math.MathPreconditions.checkNonNegative;
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import static com.google.common.math.MathPreconditions.checkPositive;
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import static com.google.common.math.MathPreconditions.checkRoundingUnnecessary;
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import static java.math.RoundingMode.CEILING;
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import static java.math.RoundingMode.FLOOR;
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import static java.math.RoundingMode.HALF_EVEN;
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import java.math.BigDecimal;
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import java.math.BigInteger;
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import java.math.RoundingMode;
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import java.util.ArrayList;
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import java.util.List;
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import com.google.common.annotations.GwtCompatible;
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import com.google.common.annotations.GwtIncompatible;
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import com.google.common.annotations.VisibleForTesting;
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/**
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* A class for arithmetic on values of type {@code BigInteger}.
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*
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* <p>
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* The implementations of many methods in this class are based on material from
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* Henry S. Warren, Jr.'s <i>Hacker's Delight</i>, (Addison Wesley, 2002).
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*
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* <p>
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* Similar functionality for {@code int} and for {@code long} can be found in
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* {@link IntMath} and {@link LongMath} respectively.
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*
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* @author Louis Wasserman
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* @since 11.0
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*/
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@GwtCompatible(emulated = true)
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public final class BigIntegerMath {
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/**
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* Returns {@code true} if {@code x} represents a power of two.
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*/
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public static boolean isPowerOfTwo(BigInteger x) {
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checkNotNull(x);
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return x.signum() > 0 && x.getLowestSetBit() == x.bitLength() - 1;
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}
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/**
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* Returns the base-2 logarithm of {@code x}, rounded according to the specified
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* rounding mode.
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*
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* @throws IllegalArgumentException if {@code x <= 0}
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* @throws ArithmeticException if {@code mode} is
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* {@link RoundingMode#UNNECESSARY} and
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* {@code x} is not a power of two
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*/
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@SuppressWarnings("fallthrough")
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// TODO(kevinb): remove after this warning is disabled globally
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public static int log2(BigInteger x, RoundingMode mode) {
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checkPositive("x", checkNotNull(x));
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int logFloor = x.bitLength() - 1;
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switch (mode) {
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case UNNECESSARY:
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checkRoundingUnnecessary(isPowerOfTwo(x)); // fall through
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case DOWN:
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case FLOOR:
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return logFloor;
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case UP:
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case CEILING:
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return isPowerOfTwo(x) ? logFloor : logFloor + 1;
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case HALF_DOWN:
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case HALF_UP:
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case HALF_EVEN:
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if (logFloor < SQRT2_PRECOMPUTE_THRESHOLD) {
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BigInteger halfPower = SQRT2_PRECOMPUTED_BITS.shiftRight(SQRT2_PRECOMPUTE_THRESHOLD - logFloor);
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if (x.compareTo(halfPower) <= 0) {
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return logFloor;
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} else {
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return logFloor + 1;
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}
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}
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/*
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* Since sqrt(2) is irrational, log2(x) - logFloor cannot be exactly 0.5
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*
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* To determine which side of logFloor.5 the logarithm is, we compare x^2 to
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* 2^(2 * logFloor + 1).
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*/
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BigInteger x2 = x.pow(2);
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int logX2Floor = x2.bitLength() - 1;
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return (logX2Floor < 2 * logFloor + 1) ? logFloor : logFloor + 1;
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default:
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throw new AssertionError();
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}
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}
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/*
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* The maximum number of bits in a square root for which we'll precompute an
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* explicit half power of two. This can be any value, but higher values incur
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* more class load time and linearly increasing memory consumption.
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*/
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@VisibleForTesting
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static final int SQRT2_PRECOMPUTE_THRESHOLD = 256;
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@VisibleForTesting
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static final BigInteger SQRT2_PRECOMPUTED_BITS = new BigInteger(
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"16a09e667f3bcc908b2fb1366ea957d3e3adec17512775099da2f590b0667322a", 16);
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/**
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* Returns the base-10 logarithm of {@code x}, rounded according to the
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* specified rounding mode.
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*
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* @throws IllegalArgumentException if {@code x <= 0}
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* @throws ArithmeticException if {@code mode} is
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* {@link RoundingMode#UNNECESSARY} and
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* {@code x} is not a power of ten
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*/
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@GwtIncompatible("TODO")
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@SuppressWarnings("fallthrough")
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public static int log10(BigInteger x, RoundingMode mode) {
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checkPositive("x", x);
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if (fitsInLong(x)) {
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return LongMath.log10(x.longValue(), mode);
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}
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int approxLog10 = (int) (log2(x, FLOOR) * LN_2 / LN_10);
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BigInteger approxPow = BigInteger.TEN.pow(approxLog10);
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int approxCmp = approxPow.compareTo(x);
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/*
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* We adjust approxLog10 and approxPow until they're equal to floor(log10(x))
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* and 10^floor(log10(x)).
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*/
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if (approxCmp > 0) {
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/*
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* The code is written so that even completely incorrect approximations will
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* still yield the correct answer eventually, but in practice this branch should
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* almost never be entered, and even then the loop should not run more than
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* once.
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*/
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do {
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approxLog10--;
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approxPow = approxPow.divide(BigInteger.TEN);
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approxCmp = approxPow.compareTo(x);
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} while (approxCmp > 0);
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} else {
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BigInteger nextPow = BigInteger.TEN.multiply(approxPow);
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int nextCmp = nextPow.compareTo(x);
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while (nextCmp <= 0) {
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approxLog10++;
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approxPow = nextPow;
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approxCmp = nextCmp;
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nextPow = BigInteger.TEN.multiply(approxPow);
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nextCmp = nextPow.compareTo(x);
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}
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}
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int floorLog = approxLog10;
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BigInteger floorPow = approxPow;
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int floorCmp = approxCmp;
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switch (mode) {
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case UNNECESSARY:
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checkRoundingUnnecessary(floorCmp == 0);
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// fall through
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case FLOOR:
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case DOWN:
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return floorLog;
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case CEILING:
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case UP:
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return floorPow.equals(x) ? floorLog : floorLog + 1;
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case HALF_DOWN:
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case HALF_UP:
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case HALF_EVEN:
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// Since sqrt(10) is irrational, log10(x) - floorLog can never be exactly 0.5
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BigInteger x2 = x.pow(2);
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BigInteger halfPowerSquared = floorPow.pow(2).multiply(BigInteger.TEN);
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return (x2.compareTo(halfPowerSquared) <= 0) ? floorLog : floorLog + 1;
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default:
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throw new AssertionError();
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}
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}
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private static final double LN_10 = Math.log(10);
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private static final double LN_2 = Math.log(2);
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/**
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* Returns the square root of {@code x}, rounded with the specified rounding
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* mode.
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*
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* @throws IllegalArgumentException if {@code x < 0}
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* @throws ArithmeticException if {@code mode} is
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* {@link RoundingMode#UNNECESSARY} and
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* {@code sqrt(x)} is not an integer
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*/
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@GwtIncompatible("TODO")
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@SuppressWarnings("fallthrough")
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public static BigInteger sqrt(BigInteger x, RoundingMode mode) {
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checkNonNegative("x", x);
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if (fitsInLong(x)) {
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return BigInteger.valueOf(LongMath.sqrt(x.longValue(), mode));
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}
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BigInteger sqrtFloor = sqrtFloor(x);
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switch (mode) {
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case UNNECESSARY:
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checkRoundingUnnecessary(sqrtFloor.pow(2).equals(x)); // fall through
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case FLOOR:
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case DOWN:
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return sqrtFloor;
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case CEILING:
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case UP:
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int sqrtFloorInt = sqrtFloor.intValue();
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boolean sqrtFloorIsExact = (sqrtFloorInt * sqrtFloorInt == x.intValue()) // fast check mod 2^32
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&& sqrtFloor.pow(2).equals(x); // slow exact check
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return sqrtFloorIsExact ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
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case HALF_DOWN:
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case HALF_UP:
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case HALF_EVEN:
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BigInteger halfSquare = sqrtFloor.pow(2).add(sqrtFloor);
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/*
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* We wish to test whether or not x <= (sqrtFloor + 0.5)^2 = halfSquare + 0.25.
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* Since both x and halfSquare are integers, this is equivalent to testing
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* whether or not x <= halfSquare.
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*/
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return (halfSquare.compareTo(x) >= 0) ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
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default:
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throw new AssertionError();
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}
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}
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@GwtIncompatible("TODO")
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private static BigInteger sqrtFloor(BigInteger x) {
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/*
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* Adapted from Hacker's Delight, Figure 11-1.
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*
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* Using DoubleUtils.bigToDouble, getting a double approximation of x is
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* extremely fast, and then we can get a double approximation of the square
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* root. Then, we iteratively improve this guess with an application of Newton's
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* method, which sets guess := (guess + (x / guess)) / 2. This iteration has the
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* following two properties:
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*
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* a) every iteration (except potentially the first) has guess >=
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* floor(sqrt(x)). This is because guess' is the arithmetic mean of guess and x
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* / guess, sqrt(x) is the geometric mean, and the arithmetic mean is always
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* higher than the geometric mean.
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*
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* b) this iteration converges to floor(sqrt(x)). In fact, the number of correct
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* digits doubles with each iteration, so this algorithm takes O(log(digits))
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* iterations.
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*
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* We start out with a double-precision approximation, which may be higher or
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* lower than the true value. Therefore, we perform at least one Newton
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* iteration to get a guess that's definitely >= floor(sqrt(x)), and then
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* continue the iteration until we reach a fixed point.
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*/
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BigInteger sqrt0;
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int log2 = log2(x, FLOOR);
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if (log2 < Double.MAX_EXPONENT) {
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sqrt0 = sqrtApproxWithDoubles(x);
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} else {
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int shift = (log2 - DoubleUtils.SIGNIFICAND_BITS) & ~1; // even!
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/*
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* We have that x / 2^shift < 2^54. Our initial approximation to sqrtFloor(x)
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* will be 2^(shift/2) * sqrtApproxWithDoubles(x / 2^shift).
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*/
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sqrt0 = sqrtApproxWithDoubles(x.shiftRight(shift)).shiftLeft(shift >> 1);
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}
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BigInteger sqrt1 = sqrt0.add(x.divide(sqrt0)).shiftRight(1);
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if (sqrt0.equals(sqrt1)) {
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return sqrt0;
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}
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do {
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sqrt0 = sqrt1;
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sqrt1 = sqrt0.add(x.divide(sqrt0)).shiftRight(1);
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} while (sqrt1.compareTo(sqrt0) < 0);
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return sqrt0;
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}
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@GwtIncompatible("TODO")
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private static BigInteger sqrtApproxWithDoubles(BigInteger x) {
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return DoubleMath.roundToBigInteger(Math.sqrt(DoubleUtils.bigToDouble(x)), HALF_EVEN);
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}
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/**
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* Returns the result of dividing {@code p} by {@code q}, rounding using the
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* specified {@code RoundingMode}.
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*
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* @throws ArithmeticException if {@code q == 0}, or if
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* {@code mode == UNNECESSARY} and {@code a} is not
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* an integer multiple of {@code b}
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*/
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@GwtIncompatible("TODO")
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public static BigInteger divide(BigInteger p, BigInteger q, RoundingMode mode) {
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BigDecimal pDec = new BigDecimal(p);
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BigDecimal qDec = new BigDecimal(q);
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return pDec.divide(qDec, 0, mode).toBigIntegerExact();
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}
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/**
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* Returns {@code n!}, that is, the product of the first {@code n} positive
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* integers, or {@code 1} if {@code n == 0}.
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*
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* <p>
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* <b>Warning</b>: the result takes <i>O(n log n)</i> space, so use cautiously.
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*
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* <p>
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* This uses an efficient binary recursive algorithm to compute the factorial
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* with balanced multiplies. It also removes all the 2s from the intermediate
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* products (shifting them back in at the end).
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*
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* @throws IllegalArgumentException if {@code n < 0}
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*/
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public static BigInteger factorial(int n) {
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checkNonNegative("n", n);
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// If the factorial is small enough, just use LongMath to do it.
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if (n < LongMath.factorials.length) {
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return BigInteger.valueOf(LongMath.factorials[n]);
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}
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// Pre-allocate space for our list of intermediate BigIntegers.
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int approxSize = IntMath.divide(n * IntMath.log2(n, CEILING), Long.SIZE, CEILING);
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ArrayList<BigInteger> bignums = new ArrayList<BigInteger>(approxSize);
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// Start from the pre-computed maximum long factorial.
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int startingNumber = LongMath.factorials.length;
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long product = LongMath.factorials[startingNumber - 1];
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// Strip off 2s from this value.
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int shift = Long.numberOfTrailingZeros(product);
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product >>= shift;
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// Use floor(log2(num)) + 1 to prevent overflow of multiplication.
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int productBits = LongMath.log2(product, FLOOR) + 1;
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int bits = LongMath.log2(startingNumber, FLOOR) + 1;
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// Check for the next power of two boundary, to save us a CLZ operation.
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int nextPowerOfTwo = 1 << (bits - 1);
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// Iteratively multiply the longs as big as they can go.
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for (long num = startingNumber; num <= n; num++) {
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// Check to see if the floor(log2(num)) + 1 has changed.
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if ((num & nextPowerOfTwo) != 0) {
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nextPowerOfTwo <<= 1;
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bits++;
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}
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// Get rid of the 2s in num.
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int tz = Long.numberOfTrailingZeros(num);
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long normalizedNum = num >> tz;
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shift += tz;
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// Adjust floor(log2(num)) + 1.
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int normalizedBits = bits - tz;
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// If it won't fit in a long, then we store off the intermediate product.
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if (normalizedBits + productBits >= Long.SIZE) {
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bignums.add(BigInteger.valueOf(product));
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product = 1;
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productBits = 0;
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}
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product *= normalizedNum;
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productBits = LongMath.log2(product, FLOOR) + 1;
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}
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// Check for leftovers.
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if (product > 1) {
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bignums.add(BigInteger.valueOf(product));
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}
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// Efficiently multiply all the intermediate products together.
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return listProduct(bignums).shiftLeft(shift);
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}
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static BigInteger listProduct(List<BigInteger> nums) {
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return listProduct(nums, 0, nums.size());
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}
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static BigInteger listProduct(List<BigInteger> nums, int start, int end) {
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switch (end - start) {
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case 0:
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return BigInteger.ONE;
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case 1:
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return nums.get(start);
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case 2:
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return nums.get(start).multiply(nums.get(start + 1));
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case 3:
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return nums.get(start).multiply(nums.get(start + 1)).multiply(nums.get(start + 2));
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default:
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// Otherwise, split the list in half and recursively do this.
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int m = (end + start) >>> 1;
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return listProduct(nums, start, m).multiply(listProduct(nums, m, end));
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}
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}
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/**
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* Returns {@code n} choose {@code k}, also known as the binomial coefficient of
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* {@code n} and {@code k}, that is, {@code n! / (k! (n - k)!)}.
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*
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* <p>
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* <b>Warning</b>: the result can take as much as <i>O(k log n)</i> space.
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*
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* @throws IllegalArgumentException if {@code n < 0}, {@code k < 0}, or
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* {@code k > n}
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*/
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public static BigInteger binomial(int n, int k) {
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checkNonNegative("n", n);
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checkNonNegative("k", k);
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checkArgument(k <= n, "k (%s) > n (%s)", k, n);
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if (k > (n >> 1)) {
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k = n - k;
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}
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if (k < LongMath.biggestBinomials.length && n <= LongMath.biggestBinomials[k]) {
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return BigInteger.valueOf(LongMath.binomial(n, k));
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}
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BigInteger accum = BigInteger.ONE;
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long numeratorAccum = n;
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long denominatorAccum = 1;
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int bits = LongMath.log2(n, RoundingMode.CEILING);
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int numeratorBits = bits;
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for (int i = 1; i < k; i++) {
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int p = n - i;
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int q = i + 1;
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// log2(p) >= bits - 1, because p >= n/2
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if (numeratorBits + bits >= Long.SIZE - 1) {
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// The numerator is as big as it can get without risking overflow.
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// Multiply numeratorAccum / denominatorAccum into accum.
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accum = accum.multiply(BigInteger.valueOf(numeratorAccum)).divide(BigInteger.valueOf(denominatorAccum));
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numeratorAccum = p;
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denominatorAccum = q;
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numeratorBits = bits;
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} else {
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// We can definitely multiply into the long accumulators without overflowing
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// them.
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numeratorAccum *= p;
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denominatorAccum *= q;
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numeratorBits += bits;
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}
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}
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return accum.multiply(BigInteger.valueOf(numeratorAccum)).divide(BigInteger.valueOf(denominatorAccum));
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}
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// Returns true if BigInteger.valueOf(x.longValue()).equals(x).
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@GwtIncompatible("TODO")
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static boolean fitsInLong(BigInteger x) {
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return x.bitLength() <= Long.SIZE - 1;
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}
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private BigIntegerMath() {
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}
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}
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