org.testifyproject.google.common.math.Quantiles Maven / Gradle / Ivy
/*
* Copyright (C) 2014 The Guava Authors
*
* Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except
* in org.testifyproject.testifyprojectpliance with the License. You may obtain a copy of the License at
*
* http://www.apache.org.testifyproject/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 org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.math;
import static org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.base.Preconditions.checkArgument;
import static java.lang.Double.NEGATIVE_INFINITY;
import static java.lang.Double.NaN;
import static java.lang.Double.POSITIVE_INFINITY;
import static java.util.Arrays.sort;
import static java.util.Collections.unmodifiableMap;
import org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.annotations.Beta;
import org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.annotations.GwtIncompatible;
import org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.primitives.Doubles;
import org.testifyproject.testifyproject.google.org.testifyproject.testifyprojectmon.primitives.Ints;
import java.math.RoundingMode;
import java.util.Collection;
import java.util.HashMap;
import java.util.Map;
/**
* Provides a fluent API for calculating
* quantiles.
*
* Examples
*
* To org.testifyproject.testifyprojectpute the median:
*
{@code
*
* double myMedian = median().org.testifyproject.testifyprojectpute(myDataset);}
*
* where {@link #median()} has been statically imported.
*
* To org.testifyproject.testifyprojectpute the 99th percentile:
*
{@code
*
* double myPercentile99 = percentiles().index(99).org.testifyproject.testifyprojectpute(myDataset);}
*
* where {@link #percentiles()} has been statically imported.
*
* To org.testifyproject.testifyprojectpute median and the 90th and 99th percentiles:
*
{@code
*
* Map myPercentiles =
* percentiles().indexes(50, 90, 99).org.testifyproject.testifyprojectpute(myDataset);}
*
* where {@link #percentiles()} has been statically imported: {@code myPercentiles} maps the keys
* 50, 90, and 99, to their corresponding quantile values.
*
* To org.testifyproject.testifyprojectpute quartiles, use {@link #quartiles()} instead of {@link #percentiles()}. To org.testifyproject.testifyprojectpute
* arbitrary q-quantiles, use {@link #scale scale(q)}.
*
*
These examples all take a copy of your dataset. If you have a double array, you are okay with
* it being arbitrarily reordered, and you want to avoid that copy, you can use
* {@code org.testifyproject.testifyprojectputeInPlace} instead of {@code org.testifyproject.testifyprojectpute}.
*
*
Definition and notes on interpolation
*
* The definition of the kth q-quantile of N values is as follows: define x = k * (N - 1) / q; if
* x is an integer, the result is the value which would appear at index x in the sorted dataset
* (unless there are {@link Double#NaN NaN} values, see below); otherwise, the result is the average
* of the values which would appear at the indexes floor(x) and ceil(x) weighted by (1-frac(x)) and
* frac(x) respectively. This is the same definition as used by Excel and by S, it is the Type 7
* definition in
* R, and it is
* described by
*
* wikipedia as providing "Linear interpolation of the modes for the order statistics for the
* uniform distribution on [0,1]."
*
*
Handling of non-finite values
*
* If any values in the input are {@link Double#NaN NaN} then all values returned are
* {@link Double#NaN NaN}. (This is the one occasion when the behaviour is not the same as you'd get
* from sorting with {@link java.util.Arrays#sort(double[]) Arrays.sort(double[])} or
* {@link java.util.Collections#sort(java.util.List) Collections.sort(List<Double>)} and
* selecting the required value(s). Those methods would sort {@link Double#NaN NaN} as if it is
* greater than any other value and place them at the end of the dataset, even after
* {@link Double#POSITIVE_INFINITY POSITIVE_INFINITY}.)
*
*
Otherwise, {@link Double#NEGATIVE_INFINITY NEGATIVE_INFINITY} and
* {@link Double#POSITIVE_INFINITY POSITIVE_INFINITY} sort to the beginning and the end of the
* dataset, as you would expect.
*
*
If required to do a weighted average between an infinity and a finite value, or between an
* infinite value and itself, the infinite value is returned. If required to do a weighted average
* between {@link Double#NEGATIVE_INFINITY NEGATIVE_INFINITY} and {@link Double#POSITIVE_INFINITY
* POSITIVE_INFINITY}, {@link Double#NaN NaN} is returned (note that this will only happen if the
* dataset contains no finite values).
*
*
Performance
*
* The average time org.testifyproject.testifyprojectplexity of the org.testifyproject.testifyprojectputation is O(N) in the size of the dataset. There is a
* worst case time org.testifyproject.testifyprojectplexity of O(N^2). You are extremely unlikely to hit this quadratic case on
* randomly ordered data (the probability decreases faster than exponentially in N), but if you are
* passing in unsanitized user data then a malicious user could force it. A light shuffle of the
* data using an unpredictable seed should normally be enough to thwart this attack.
*
*
The time taken to org.testifyproject.testifyprojectpute multiple quantiles on the same dataset using {@link Scale#indexes
* indexes} is generally less than the total time taken to org.testifyproject.testifyprojectpute each of them separately, and
* sometimes much less. For example, on a large enough dataset, org.testifyproject.testifyprojectputing the 90th and 99th
* percentiles together takes about 55% as long as org.testifyproject.testifyprojectputing them separately.
*
*
When calling {@link ScaleAndIndex#org.testifyproject.testifyprojectpute} (in {@linkplain ScaleAndIndexes#org.testifyproject.testifyprojectpute either
* form}), the memory requirement is 8*N bytes for the copy of the dataset plus an overhead which is
* independent of N (but depends on the quantiles being org.testifyproject.testifyprojectputed). When calling
* {@link ScaleAndIndex#org.testifyproject.testifyprojectputeInPlace org.testifyproject.testifyprojectputeInPlace} (in
* {@linkplain ScaleAndIndexes#org.testifyproject.testifyprojectputeInPlace either form}), only the overhead is required. The
* number of object allocations is independent of N in both cases.
*
* @author Pete Gillin
* @since 20.0
*/
@Beta
@GwtIncompatible
public final class Quantiles {
/**
* Specifies the org.testifyproject.testifyprojectputation of a median (i.e. the 1st 2-quantile).
*/
public static ScaleAndIndex median() {
return scale(2).index(1);
}
/**
* Specifies the org.testifyproject.testifyprojectputation of quartiles (i.e. 4-quantiles).
*/
public static Scale quartiles() {
return scale(4);
}
/**
* Specifies the org.testifyproject.testifyprojectputation of percentiles (i.e. 100-quantiles).
*/
public static Scale percentiles() {
return scale(100);
}
/**
* Specifies the org.testifyproject.testifyprojectputation of q-quantiles.
*
* @param scale the scale for the quantiles to be calculated, i.e. the q of the q-quantiles, which
* must be positive
*/
public static Scale scale(int scale) {
return new Scale(scale);
}
/**
* Describes the point in a fluent API chain where only the scale (i.e. the q in q-quantiles) has
* been specified.
*/
public static final class Scale {
private final int scale;
private Scale(int scale) {
checkArgument(scale > 0, "Quantile scale must be positive");
this.scale = scale;
}
/**
* Specifies a single quantile index to be calculated, i.e. the k in the kth q-quantile.
*
* @param index the quantile index, which must be in the inclusive range [0, q] for q-quantiles
*/
public ScaleAndIndex index(int index) {
return new ScaleAndIndex(scale, index);
}
/**
* Specifies multiple quantile indexes to be calculated, each index being the k in the kth
* q-quantile.
*
* @param indexes the quantile indexes, each of which must be in the inclusive range [0, q] for
* q-quantiles; the order of the indexes is unimportant, duplicates will be ignored, and the
* set will be snapshotted when this method is called
*/
public ScaleAndIndexes indexes(int... indexes) {
return new ScaleAndIndexes(scale, indexes.clone());
}
/**
* Specifies multiple quantile indexes to be calculated, each index being the k in the kth
* q-quantile.
*
* @param indexes the quantile indexes, each of which must be in the inclusive range [0, q] for
* q-quantiles; the order of the indexes is unimportant, duplicates will be ignored, and the
* set will be snapshotted when this method is called
*/
public ScaleAndIndexes indexes(Collection indexes) {
return new ScaleAndIndexes(scale, Ints.toArray(indexes));
}
}
/**
* Describes the point in a fluent API chain where the scale and a single quantile index (i.e. the
* q and the k in the kth q-quantile) have been specified.
*/
public static final class ScaleAndIndex {
private final int scale;
private final int index;
private ScaleAndIndex(int scale, int index) {
checkIndex(index, scale);
this.scale = scale;
this.index = index;
}
/**
* Computes the quantile value of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles (with any associated lost of precision), and which will not be mutated by
* this call (it is copied instead)
* @return the quantile value
*/
public double org.testifyproject.testifyprojectpute(Collection dataset) {
return org.testifyproject.testifyprojectputeInPlace(Doubles.toArray(dataset));
}
/**
* Computes the quantile value of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will not
* be mutated by this call (it is copied instead)
* @return the quantile value
*/
public double org.testifyproject.testifyprojectpute(double... dataset) {
return org.testifyproject.testifyprojectputeInPlace(dataset.clone());
}
/**
* Computes the quantile value of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles (with any associated lost of precision), and which will not be mutated by
* this call (it is copied instead)
* @return the quantile value
*/
public double org.testifyproject.testifyprojectpute(long... dataset) {
return org.testifyproject.testifyprojectputeInPlace(longsToDoubles(dataset));
}
/**
* Computes the quantile value of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles, and which will not be mutated by this call (it is copied instead)
* @return the quantile value
*/
public double org.testifyproject.testifyprojectpute(int... dataset) {
return org.testifyproject.testifyprojectputeInPlace(intsToDoubles(dataset));
}
/**
* Computes the quantile value of the given dataset, performing the org.testifyproject.testifyprojectputation in-place.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, and which will
* be arbitrarily reordered by this method call
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public double org.testifyproject.testifyprojectputeInPlace(double... dataset) {
checkArgument(dataset.length > 0, "Cannot calculate quantiles of an empty dataset");
if (containsNaN(dataset)) {
return NaN;
}
// Calculate the quotient and remainder in the integer division x = k * (N-1) / q, i.e.
// index * (dataset.length - 1) / scale. If there is no remainder, we can just find the value
// whose index in the sorted dataset equals the quotient; if there is a remainder, we
// interpolate between that and the next value.
// Since index and (dataset.length - 1) are non-negative ints, their product can be expressed
// as a long, without risk of overflow:
long numerator = (long) index * (dataset.length - 1);
// Since scale is a positive int, index is in [0, scale], and (dataset.length - 1) is a
// non-negative int, we can do long-arithmetic on index * (dataset.length - 1) / scale to get
// a rounded ratio and a remainder which can be expressed as ints, without risk of overflow:
int quotient = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
int remainder = (int) (numerator - (long) quotient * scale);
selectInPlace(quotient, dataset, 0, dataset.length - 1);
if (remainder == 0) {
return dataset[quotient];
} else {
selectInPlace(quotient + 1, dataset, quotient + 1, dataset.length - 1);
return interpolate(dataset[quotient], dataset[quotient + 1], remainder, scale);
}
}
}
/**
* Describes the point in a fluent API chain where the scale and a multiple quantile indexes (i.e.
* the q and a set of values for the k in the kth q-quantile) have been specified.
*/
public static final class ScaleAndIndexes {
private final int scale;
private final int[] indexes;
private ScaleAndIndexes(int scale, int[] indexes) {
for (int index : indexes) {
checkIndex(index, scale);
}
this.scale = scale;
this.indexes = indexes;
}
/**
* Computes the quantile values of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles (with any associated lost of precision), and which will not be mutated by
* this call (it is copied instead)
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public Map org.testifyproject.testifyprojectpute(Collection dataset) {
return org.testifyproject.testifyprojectputeInPlace(Doubles.toArray(dataset));
}
/**
* Computes the quantile values of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will not
* be mutated by this call (it is copied instead)
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public Map org.testifyproject.testifyprojectpute(double... dataset) {
return org.testifyproject.testifyprojectputeInPlace(dataset.clone());
}
/**
* Computes the quantile values of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles (with any associated lost of precision), and which will not be mutated by
* this call (it is copied instead)
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public Map org.testifyproject.testifyprojectpute(long... dataset) {
return org.testifyproject.testifyprojectputeInPlace(longsToDoubles(dataset));
}
/**
* Computes the quantile values of the given dataset.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, which will be
* cast to doubles, and which will not be mutated by this call (it is copied instead)
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public Map org.testifyproject.testifyprojectpute(int... dataset) {
return org.testifyproject.testifyprojectputeInPlace(intsToDoubles(dataset));
}
/**
* Computes the quantile values of the given dataset, performing the org.testifyproject.testifyprojectputation in-place.
*
* @param dataset the dataset to do the calculation on, which must be non-empty, and which will
* be arbitrarily reordered by this method call
* @return an unmodifiable map of results: the keys will be the specified quantile indexes, and
* the values the corresponding quantile values
*/
public Map org.testifyproject.testifyprojectputeInPlace(double... dataset) {
checkArgument(dataset.length > 0, "Cannot calculate quantiles of an empty dataset");
if (containsNaN(dataset)) {
Map nanMap = new HashMap();
for (int index : indexes) {
nanMap.put(index, NaN);
}
return unmodifiableMap(nanMap);
}
// Calculate the quotients and remainders in the integer division x = k * (N - 1) / q, i.e.
// index * (dataset.length - 1) / scale for each index in indexes. For each, if there is no
// remainder, we can just select the value whose index in the sorted dataset equals the
// quotient; if there is a remainder, we interpolate between that and the next value.
int[] quotients = new int[indexes.length];
int[] remainders = new int[indexes.length];
// The indexes to select. In the worst case, we'll need one each side of each quantile.
int[] requiredSelections = new int[indexes.length * 2];
int requiredSelectionsCount = 0;
for (int i = 0; i < indexes.length; i++) {
// Since index and (dataset.length - 1) are non-negative ints, their product can be
// expressed as a long, without risk of overflow:
long numerator = (long) indexes[i] * (dataset.length - 1);
// Since scale is a positive int, index is in [0, scale], and (dataset.length - 1) is a
// non-negative int, we can do long-arithmetic on index * (dataset.length - 1) / scale to
// get a rounded ratio and a remainder which can be expressed as ints, without risk of
// overflow:
int quotient = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
int remainder = (int) (numerator - (long) quotient * scale);
quotients[i] = quotient;
remainders[i] = remainder;
requiredSelections[requiredSelectionsCount] = quotient;
requiredSelectionsCount++;
if (remainder != 0) {
requiredSelections[requiredSelectionsCount] = quotient + 1;
requiredSelectionsCount++;
}
}
sort(requiredSelections, 0, requiredSelectionsCount);
selectAllInPlace(
requiredSelections, 0, requiredSelectionsCount - 1, dataset, 0, dataset.length - 1);
Map ret = new HashMap();
for (int i = 0; i < indexes.length; i++) {
int quotient = quotients[i];
int remainder = remainders[i];
if (remainder == 0) {
ret.put(indexes[i], dataset[quotient]);
} else {
ret.put(
indexes[i], interpolate(dataset[quotient], dataset[quotient + 1], remainder, scale));
}
}
return unmodifiableMap(ret);
}
}
/**
* Returns whether any of the values in {@code dataset} are {@code NaN}.
*/
private static boolean containsNaN(double... dataset) {
for (double value : dataset) {
if (Double.isNaN(value)) {
return true;
}
}
return false;
}
/**
* Returns a value a fraction {@code (remainder / scale)} of the way between {@code lower} and
* {@code upper}. Assumes that {@code lower <= upper}. Correctly handles infinities (but not
* {@code NaN}).
*/
private static double interpolate(double lower, double upper, double remainder, double scale) {
if (lower == NEGATIVE_INFINITY) {
if (upper == POSITIVE_INFINITY) {
// Return NaN when lower == NEGATIVE_INFINITY and upper == POSITIVE_INFINITY:
return NaN;
}
// Return NEGATIVE_INFINITY when NEGATIVE_INFINITY == lower <= upper < POSITIVE_INFINITY:
return NEGATIVE_INFINITY;
}
if (upper == POSITIVE_INFINITY) {
// Return POSITIVE_INFINITY when NEGATIVE_INFINITY < lower <= upper == POSITIVE_INFINITY:
return POSITIVE_INFINITY;
}
return lower + (upper - lower) * remainder / scale;
}
private static void checkIndex(int index, int scale) {
if (index < 0 || index > scale) {
throw new IllegalArgumentException(
"Quantile indexes must be between 0 and the scale, which is " + scale);
}
}
private static double[] longsToDoubles(long[] longs) {
int len = longs.length;
double[] doubles = new double[len];
for (int i = 0; i < len; i++) {
doubles[i] = longs[i];
}
return doubles;
}
private static double[] intsToDoubles(int[] ints) {
int len = ints.length;
double[] doubles = new double[len];
for (int i = 0; i < len; i++) {
doubles[i] = ints[i];
}
return doubles;
}
/**
* Performs an in-place selection to find the element which would appear at a given index in a
* dataset if it were sorted. The following preconditions should hold:
*
* - {@code required}, {@code from}, and {@code to} should all be indexes into {@code array};
*
- {@code required} should be in the range [{@code from}, {@code to}];
*
- all the values with indexes in the range [0, {@code from}) should be less than or equal to
* all the values with indexes in the range [{@code from}, {@code to}];
*
- all the values with indexes in the range ({@code to}, {@code array.length - 1}] should be
* greater than or equal to all the values with indexes in the range [{@code from}, {@code to}].
*
* This method will reorder the values with indexes in the range [{@code from}, {@code to}] such
* that all the values with indexes in the range [{@code from}, {@code required}) are less than or
* equal to the value with index {@code required}, and all the values with indexes in the range
* ({@code required}, {@code to}] are greater than or equal to that value. Therefore, the value at
* {@code required} is the value which would appear at that index in the sorted dataset.
*/
private static void selectInPlace(int required, double[] array, int from, int to) {
// If we are looking for the least element in the range, we can just do a linear search for it.
// (We will hit this whenever we are doing quantile interpolation: our first selection finds
// the lower value, our second one finds the upper value by looking for the next least element.)
if (required == from) {
int min = from;
for (int index = from + 1; index <= to; index++) {
if (array[min] > array[index]) {
min = index;
}
}
if (min != from) {
swap(array, min, from);
}
return;
}
// Let's play quickselect! We'll repeatedly partition the range [from, to] containing the
// required element, as long as it has more than one element.
while (to > from) {
int partitionPoint = partition(array, from, to);
if (partitionPoint >= required) {
to = partitionPoint - 1;
}
if (partitionPoint <= required) {
from = partitionPoint + 1;
}
}
}
/**
* Performs a partition operation on the slice of {@code array} with elements in the range
* [{@code from}, {@code to}]. Uses the median of {@code from}, {@code to}, and the midpoint
* between them as a pivot. Returns the index which the slice is partitioned around, i.e. if it
* returns {@code ret} then we know that the values with indexes in [{@code from}, {@code ret})
* are less than or equal to the value at {@code ret} and the values with indexes in ({@code ret},
* {@code to}] are greater than or equal to that.
*/
private static int partition(double[] array, int from, int to) {
// Select a pivot, and move it to the start of the slice i.e. to index from.
movePivotToStartOfSlice(array, from, to);
double pivot = array[from];
// Move all elements with indexes in (from, to] which are greater than the pivot to the end of
// the array. Keep track of where those elements begin.
int partitionPoint = to;
for (int i = to; i > from; i--) {
if (array[i] > pivot) {
swap(array, partitionPoint, i);
partitionPoint--;
}
}
// We now know that all elements with indexes in (from, partitionPoint] are less than or equal
// to the pivot at from, and all elements with indexes in (partitionPoint, to] are greater than
// it. We swap the pivot into partitionPoint and we know the array is partitioned around that.
swap(array, from, partitionPoint);
return partitionPoint;
}
/**
* Selects the pivot to use, namely the median of the values at {@code from}, {@code to}, and
* halfway between the two (rounded down), from {@code array}, and ensure (by swapping elements if
* necessary) that that pivot value appears at the start of the slice i.e. at {@code from}.
* Expects that {@code from} is strictly less than {@code to}.
*/
private static void movePivotToStartOfSlice(double[] array, int from, int to) {
int mid = (from + to) >>> 1;
// We want to make a swap such that either array[to] <= array[from] <= array[mid], or
// array[mid] <= array[from] <= array[to]. We know that from < to, so we know mid < to
// (although it's possible that mid == from, if to == from + 1). Note that the postcondition
// would be impossible to fulfil if mid == to unless we also have array[from] == array[to].
boolean toLessThanMid = (array[to] < array[mid]);
boolean midLessThanFrom = (array[mid] < array[from]);
boolean toLessThanFrom = (array[to] < array[from]);
if (toLessThanMid == midLessThanFrom) {
// Either array[to] < array[mid] < array[from] or array[from] <= array[mid] <= array[to].
swap(array, mid, from);
} else if (toLessThanMid != toLessThanFrom) {
// Either array[from] <= array[to] < array[mid] or array[mid] <= array[to] < array[from].
swap(array, from, to);
}
// The postcondition now holds. So the median, our chosen pivot, is at from.
}
/**
* Performs an in-place selection, like {@link #selectInPlace}, to select all the indexes
* {@code allRequired[i]} for {@code i} in the range [{@code requiredFrom}, {@code requiredTo}].
* These indexes must be sorted in the array and must all be in the range [{@code from},
* {@code to}].
*/
private static void selectAllInPlace(
int[] allRequired, int requiredFrom, int requiredTo, double[] array, int from, int to) {
// Choose the first selection to do...
int requiredChosen = chooseNextSelection(allRequired, requiredFrom, requiredTo, from, to);
int required = allRequired[requiredChosen];
// ...do the first selection...
selectInPlace(required, array, from, to);
// ...then recursively perform the selections in the range below...
int requiredBelow = requiredChosen - 1;
while (requiredBelow >= requiredFrom && allRequired[requiredBelow] == required) {
requiredBelow--; // skip duplicates of required in the range below
}
if (requiredBelow >= requiredFrom) {
selectAllInPlace(allRequired, requiredFrom, requiredBelow, array, from, required - 1);
}
// ...and then recursively perform the selections in the range above.
int requiredAbove = requiredChosen + 1;
while (requiredAbove <= requiredTo && allRequired[requiredAbove] == required) {
requiredAbove++; // skip duplicates of required in the range above
}
if (requiredAbove <= requiredTo) {
selectAllInPlace(allRequired, requiredAbove, requiredTo, array, required + 1, to);
}
}
/**
* Chooses the next selection to do from the required selections. It is required that the array
* {@code allRequired} is sorted and that {@code allRequired[i]} are in the range [{@code from},
* {@code to}] for all {@code i} in the range [{@code requiredFrom}, {@requiredTo}]. The value
* returned by this method is the {@code i} in that range such that {@code allRequired[i]} is as
* close as possible to the center of the range [{@code from}, {@code to}]. Choosing the value
* closest to the center of the range first is the most efficient strategy because it minimizes
* the size of the subranges from which the remaining selections must be done.
*/
private static int chooseNextSelection(
int[] allRequired, int requiredFrom, int requiredTo, int from, int to) {
if (requiredFrom == requiredTo) {
return requiredFrom; // only one thing to choose, so choose it
}
// Find the center and round down. The true center is either centerFloor or halfway between
// centerFloor and centerFloor + 1.
int centerFloor = (from + to) >>> 1;
// Do a binary search until we're down to the range of two which encloses centerFloor (unless
// all values are lower or higher than centerFloor, in which case we find the two highest or
// lowest respectively). If centerFloor is in allRequired, we will definitely find it. If not,
// but centerFloor + 1 is, we'll definitely find that. The closest value to the true (unrounded)
// center will be at either low or high.
int low = requiredFrom;
int high = requiredTo;
while (high > low + 1) {
int mid = (low + high) >>> 1;
if (allRequired[mid] > centerFloor) {
high = mid;
} else if (allRequired[mid] < centerFloor) {
low = mid;
} else {
return mid; // allRequired[mid] = centerFloor, so we can't get closer than that
}
}
// Now pick the closest of the two candidates. Note that there is no rounding here.
if (from + to - allRequired[low] - allRequired[high] > 0) {
return high;
} else {
return low;
}
}
/**
* Swaps the values at {@code i} and {@code j} in {@code array}.
*/
private static void swap(double[] array, int i, int j) {
double temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}