com.google.common.math.BigIntegerMath Maven / Gradle / Ivy
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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.checkNonNegative;
import static com.google.common.math.MathPreconditions.checkPositive;
import static com.google.common.math.MathPreconditions.checkRoundingUnnecessary;
import static java.math.RoundingMode.CEILING;
import static java.math.RoundingMode.FLOOR;
import static java.math.RoundingMode.HALF_EVEN;
import com.google.common.annotations.Beta;
import com.google.common.annotations.GwtCompatible;
import com.google.common.annotations.GwtIncompatible;
import com.google.common.annotations.VisibleForTesting;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.util.ArrayList;
import java.util.List;
/**
* A class for arithmetic on values of type {@code BigInteger}.
*
* 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 {@code long} can be found in {@link IntMath} and
* {@link LongMath} respectively.
*
* @author Louis Wasserman
* @since 11.0
*/
@GwtCompatible(emulated = true)
public final class BigIntegerMath {
/**
* Returns the smallest power of two greater than or equal to {@code x}. This is equivalent to
* {@code BigInteger.valueOf(2).pow(log2(x, CEILING))}.
*
* @throws IllegalArgumentException if {@code x <= 0}
* @since 20.0
*/
@Beta
public static BigInteger ceilingPowerOfTwo(BigInteger x) {
return BigInteger.ZERO.setBit(log2(x, RoundingMode.CEILING));
}
/**
* Returns the largest power of two less than or equal to {@code x}. This is equivalent to {@code
* BigInteger.valueOf(2).pow(log2(x, FLOOR))}.
*
* @throws IllegalArgumentException if {@code x <= 0}
* @since 20.0
*/
@Beta
public static BigInteger floorPowerOfTwo(BigInteger x) {
return BigInteger.ZERO.setBit(log2(x, RoundingMode.FLOOR));
}
/** Returns {@code true} if {@code x} represents a power of two. */
public static boolean isPowerOfTwo(BigInteger x) {
checkNotNull(x);
return x.signum() > 0 && x.getLowestSetBit() == x.bitLength() - 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(BigInteger x, RoundingMode mode) {
checkPositive("x", checkNotNull(x));
int logFloor = x.bitLength() - 1;
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(isPowerOfTwo(x)); // fall through
case DOWN:
case FLOOR:
return logFloor;
case UP:
case CEILING:
return isPowerOfTwo(x) ? logFloor : logFloor + 1;
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
if (logFloor < SQRT2_PRECOMPUTE_THRESHOLD) {
BigInteger halfPower =
SQRT2_PRECOMPUTED_BITS.shiftRight(SQRT2_PRECOMPUTE_THRESHOLD - logFloor);
if (x.compareTo(halfPower) <= 0) {
return logFloor;
} else {
return logFloor + 1;
}
}
// Since sqrt(2) is irrational, log2(x) - logFloor cannot be exactly 0.5
//
// To determine which side of logFloor.5 the logarithm is,
// we compare x^2 to 2^(2 * logFloor + 1).
BigInteger x2 = x.pow(2);
int logX2Floor = x2.bitLength() - 1;
return (logX2Floor < 2 * logFloor + 1) ? logFloor : logFloor + 1;
default:
throw new AssertionError();
}
}
/*
* The maximum number of bits in a square root for which we'll precompute an explicit half power
* of two. This can be any value, but higher values incur more class load time and linearly
* increasing memory consumption.
*/
@VisibleForTesting static final int SQRT2_PRECOMPUTE_THRESHOLD = 256;
@VisibleForTesting
static final BigInteger SQRT2_PRECOMPUTED_BITS =
new BigInteger("16a09e667f3bcc908b2fb1366ea957d3e3adec17512775099da2f590b0667322a", 16);
/**
* 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")
public static int log10(BigInteger x, RoundingMode mode) {
checkPositive("x", x);
if (fitsInLong(x)) {
return LongMath.log10(x.longValue(), mode);
}
int approxLog10 = (int) (log2(x, FLOOR) * LN_2 / LN_10);
BigInteger approxPow = BigInteger.TEN.pow(approxLog10);
int approxCmp = approxPow.compareTo(x);
/*
* We adjust approxLog10 and approxPow until they're equal to floor(log10(x)) and
* 10^floor(log10(x)).
*/
if (approxCmp > 0) {
/*
* The code is written so that even completely incorrect approximations will still yield the
* correct answer eventually, but in practice this branch should almost never be entered, and
* even then the loop should not run more than once.
*/
do {
approxLog10--;
approxPow = approxPow.divide(BigInteger.TEN);
approxCmp = approxPow.compareTo(x);
} while (approxCmp > 0);
} else {
BigInteger nextPow = BigInteger.TEN.multiply(approxPow);
int nextCmp = nextPow.compareTo(x);
while (nextCmp <= 0) {
approxLog10++;
approxPow = nextPow;
approxCmp = nextCmp;
nextPow = BigInteger.TEN.multiply(approxPow);
nextCmp = nextPow.compareTo(x);
}
}
int floorLog = approxLog10;
BigInteger floorPow = approxPow;
int floorCmp = approxCmp;
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(floorCmp == 0);
// fall through
case FLOOR:
case DOWN:
return floorLog;
case CEILING:
case UP:
return floorPow.equals(x) ? floorLog : floorLog + 1;
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
// Since sqrt(10) is irrational, log10(x) - floorLog can never be exactly 0.5
BigInteger x2 = x.pow(2);
BigInteger halfPowerSquared = floorPow.pow(2).multiply(BigInteger.TEN);
return (x2.compareTo(halfPowerSquared) <= 0) ? floorLog : floorLog + 1;
default:
throw new AssertionError();
}
}
private static final double LN_10 = Math.log(10);
private static final double LN_2 = Math.log(2);
/**
* 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 BigInteger sqrt(BigInteger x, RoundingMode mode) {
checkNonNegative("x", x);
if (fitsInLong(x)) {
return BigInteger.valueOf(LongMath.sqrt(x.longValue(), mode));
}
BigInteger sqrtFloor = sqrtFloor(x);
switch (mode) {
case UNNECESSARY:
checkRoundingUnnecessary(sqrtFloor.pow(2).equals(x)); // fall through
case FLOOR:
case DOWN:
return sqrtFloor;
case CEILING:
case UP:
int sqrtFloorInt = sqrtFloor.intValue();
boolean sqrtFloorIsExact =
(sqrtFloorInt * sqrtFloorInt == x.intValue()) // fast check mod 2^32
&& sqrtFloor.pow(2).equals(x); // slow exact check
return sqrtFloorIsExact ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
case HALF_DOWN:
case HALF_UP:
case HALF_EVEN:
BigInteger halfSquare = sqrtFloor.pow(2).add(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.
*/
return (halfSquare.compareTo(x) >= 0) ? sqrtFloor : sqrtFloor.add(BigInteger.ONE);
default:
throw new AssertionError();
}
}
@GwtIncompatible // TODO
private static BigInteger sqrtFloor(BigInteger x) {
/*
* Adapted from Hacker's Delight, Figure 11-1.
*
* Using DoubleUtils.bigToDouble, getting a double approximation of x is extremely fast, and
* then we can get a double approximation of the square root. Then, we iteratively improve this
* guess with an application of Newton's method, which sets guess := (guess + (x / guess)) / 2.
* This iteration has the following two properties:
*
* a) every iteration (except potentially the first) has guess >= floor(sqrt(x)). This is
* because guess' is the arithmetic mean of guess and x / guess, sqrt(x) is the geometric mean,
* and the arithmetic mean is always higher than the geometric mean.
*
* b) this iteration converges to floor(sqrt(x)). In fact, the number of correct digits doubles
* with each iteration, so this algorithm takes O(log(digits)) iterations.
*
* We start out with a double-precision approximation, which may be higher or lower than the
* true value. Therefore, we perform at least one Newton iteration to get a guess that's
* definitely >= floor(sqrt(x)), and then continue the iteration until we reach a fixed point.
*/
BigInteger sqrt0;
int log2 = log2(x, FLOOR);
if (log2 < Double.MAX_EXPONENT) {
sqrt0 = sqrtApproxWithDoubles(x);
} else {
int shift = (log2 - DoubleUtils.SIGNIFICAND_BITS) & ~1; // even!
/*
* We have that x / 2^shift < 2^54. Our initial approximation to sqrtFloor(x) will be
* 2^(shift/2) * sqrtApproxWithDoubles(x / 2^shift).
*/
sqrt0 = sqrtApproxWithDoubles(x.shiftRight(shift)).shiftLeft(shift >> 1);
}
BigInteger sqrt1 = sqrt0.add(x.divide(sqrt0)).shiftRight(1);
if (sqrt0.equals(sqrt1)) {
return sqrt0;
}
do {
sqrt0 = sqrt1;
sqrt1 = sqrt0.add(x.divide(sqrt0)).shiftRight(1);
} while (sqrt1.compareTo(sqrt0) < 0);
return sqrt0;
}
@GwtIncompatible // TODO
private static BigInteger sqrtApproxWithDoubles(BigInteger x) {
return DoubleMath.roundToBigInteger(Math.sqrt(DoubleUtils.bigToDouble(x)), HALF_EVEN);
}
/**
* 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
public static BigInteger divide(BigInteger p, BigInteger q, RoundingMode mode) {
BigDecimal pDec = new BigDecimal(p);
BigDecimal qDec = new BigDecimal(q);
return pDec.divide(qDec, 0, mode).toBigIntegerExact();
}
/**
* Returns {@code n!}, that is, the product of the first {@code n} positive integers, or {@code 1}
* if {@code n == 0}.
*
*
Warning: the result takes O(n log n) space, so use cautiously.
*
*
This uses an efficient binary recursive algorithm to compute the factorial with balanced
* multiplies. It also removes all the 2s from the intermediate products (shifting them back in at
* the end).
*
* @throws IllegalArgumentException if {@code n < 0}
*/
public static BigInteger factorial(int n) {
checkNonNegative("n", n);
// If the factorial is small enough, just use LongMath to do it.
if (n < LongMath.factorials.length) {
return BigInteger.valueOf(LongMath.factorials[n]);
}
// Pre-allocate space for our list of intermediate BigIntegers.
int approxSize = IntMath.divide(n * IntMath.log2(n, CEILING), Long.SIZE, CEILING);
ArrayList bignums = new ArrayList<>(approxSize);
// Start from the pre-computed maximum long factorial.
int startingNumber = LongMath.factorials.length;
long product = LongMath.factorials[startingNumber - 1];
// Strip off 2s from this value.
int shift = Long.numberOfTrailingZeros(product);
product >>= shift;
// Use floor(log2(num)) + 1 to prevent overflow of multiplication.
int productBits = LongMath.log2(product, FLOOR) + 1;
int bits = LongMath.log2(startingNumber, FLOOR) + 1;
// Check for the next power of two boundary, to save us a CLZ operation.
int nextPowerOfTwo = 1 << (bits - 1);
// Iteratively multiply the longs as big as they can go.
for (long num = startingNumber; num <= n; num++) {
// Check to see if the floor(log2(num)) + 1 has changed.
if ((num & nextPowerOfTwo) != 0) {
nextPowerOfTwo <<= 1;
bits++;
}
// Get rid of the 2s in num.
int tz = Long.numberOfTrailingZeros(num);
long normalizedNum = num >> tz;
shift += tz;
// Adjust floor(log2(num)) + 1.
int normalizedBits = bits - tz;
// If it won't fit in a long, then we store off the intermediate product.
if (normalizedBits + productBits >= Long.SIZE) {
bignums.add(BigInteger.valueOf(product));
product = 1;
productBits = 0;
}
product *= normalizedNum;
productBits = LongMath.log2(product, FLOOR) + 1;
}
// Check for leftovers.
if (product > 1) {
bignums.add(BigInteger.valueOf(product));
}
// Efficiently multiply all the intermediate products together.
return listProduct(bignums).shiftLeft(shift);
}
static BigInteger listProduct(List nums) {
return listProduct(nums, 0, nums.size());
}
static BigInteger listProduct(List nums, int start, int end) {
switch (end - start) {
case 0:
return BigInteger.ONE;
case 1:
return nums.get(start);
case 2:
return nums.get(start).multiply(nums.get(start + 1));
case 3:
return nums.get(start).multiply(nums.get(start + 1)).multiply(nums.get(start + 2));
default:
// Otherwise, split the list in half and recursively do this.
int m = (end + start) >>> 1;
return listProduct(nums, start, m).multiply(listProduct(nums, m, end));
}
}
/**
* Returns {@code n} choose {@code k}, also known as the binomial coefficient of {@code n} and
* {@code k}, that is, {@code n! / (k! (n - k)!)}.
*
* Warning: the result can take as much as O(k log n) space.
*
* @throws IllegalArgumentException if {@code n < 0}, {@code k < 0}, or {@code k > n}
*/
public static BigInteger 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 < LongMath.biggestBinomials.length && n <= LongMath.biggestBinomials[k]) {
return BigInteger.valueOf(LongMath.binomial(n, k));
}
BigInteger accum = BigInteger.ONE;
long numeratorAccum = n;
long denominatorAccum = 1;
int bits = LongMath.log2(n, RoundingMode.CEILING);
int numeratorBits = bits;
for (int i = 1; i < k; i++) {
int p = n - i;
int q = i + 1;
// log2(p) >= bits - 1, because p >= n/2
if (numeratorBits + bits >= Long.SIZE - 1) {
// The numerator is as big as it can get without risking overflow.
// Multiply numeratorAccum / denominatorAccum into accum.
accum =
accum
.multiply(BigInteger.valueOf(numeratorAccum))
.divide(BigInteger.valueOf(denominatorAccum));
numeratorAccum = p;
denominatorAccum = q;
numeratorBits = bits;
} else {
// We can definitely multiply into the long accumulators without overflowing them.
numeratorAccum *= p;
denominatorAccum *= q;
numeratorBits += bits;
}
}
return accum
.multiply(BigInteger.valueOf(numeratorAccum))
.divide(BigInteger.valueOf(denominatorAccum));
}
// Returns true if BigInteger.valueOf(x.longValue()).equals(x).
@GwtIncompatible // TODO
static boolean fitsInLong(BigInteger x) {
return x.bitLength() <= Long.SIZE - 1;
}
private BigIntegerMath() {}
}