ByteArrayTimSort.kt
TLDR
This file contains the implementation of a custom sorting algorithm called TimSort. TimSort is a stable, adaptive, iterative mergesort that requires fewer comparisons than traditional mergesort algorithms when working on partially sorted arrays. It offers similar performance to mergesort on random arrays and runs in O(n log n) time complexity.
The provided implementation has been adapted from the TimSort algorithm implemented in Python by Tim Peters. It has been converted to Kotlin and modified to work with byte array chunks. The byte array is virtually divided into entries of a fixed number of bytes, and each entry is compared using a custom comparator.
Methods
The file contains the following methods:
countRunAndMakeAscending
This method counts and makes an ascending run starting at the specified position in the array. It returns the length of the run.
gallopLeft
This method searches for the insert position of a key in a sorted range of the array. It returns the index where the key should be inserted. The search uses a binary search algorithm and compares the key with the elements in the range using the specified comparator.
gallopRight
Similar to gallopLeft, but it returns the index after the rightmost equal element if the range contains an element equal to the key.
binarySort
This method performs a binary insertion sort on a range of the array. It assumes that the range contains an already sorted prefix, and it inserts the remaining elements into the correct position in a way that maintains the sorted order.
pushRun
This method pushes a run (start index and length) onto the pending-run stack.
mergeCollapse
Examines the stack of runs waiting to be merged and merges adjacent runs until the stack invariants are reestablished.
mergeForceCollapse
Merges all runs on the stack until only one run remains. This method is called once, at the end of the sorting process, to complete the sort.
mergeAt
Merges two adjacent runs at the specified indices on the stack.
ensureCapacity
Ensures that the temporary array has at least the specified number of elements, increasing its size if necessary.
Classes
This file contains a single class called ByteArrayTimSort. The class is package-private and has the following constructors and properties:
Constructor
-
ByteArrayTimSort(private val a: ByteArray, private val c: ByteArrayComparator, private val entrySize: Int)
Properties
-
private var minGallop = MIN_GALLOP -
private var tmp: ByteArray? = null -
private var stackSize = 0 -
private val runBase: IntArray -
private val runLen: IntArray
The ByteArrayTimSort class also contains a companion object with the following properties and methods:
-
MIN_MERGE -
MIN_GALLOP -
INITIAL_TMP_STORAGE_LENGTH -
sort(two overloaded versions) -
checkStartAndEnd -
minRunLength
/*
* Copyright (C) 2008 The Android Open Source Project
*
* 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 shark.internal.aosp
import kotlin.math.min
/*
This is TimSort.java from AOSP (Jelly Bean MR2, Apache 2 license), converted to Kotlin and adapted
to work with byte array chunks. The passed in byte array is virtually divided into entries of a
fixed number of bytes N. Each entry is compared by a custom comparator.
Copied from https://android.googlesource.com/platform/libcore/+/jb-mr2-release/luni/src/main/java/java/util/TimSort.java
*/
/**
* A stable, adaptive, iterative mergesort that requires far fewer than
* n lg(n) comparisons when running on partially sorted arrays, while
* offering performance comparable to a traditional mergesort when run
* on random arrays. Like all proper mergesorts, this sort is stable and
* runs O(n log n) time (worst case). In the worst case, this sort requires
* temporary storage space for n/2 object references; in the best case,
* it requires only a small constant amount of space.
*
* This implementation was adapted from Tim Peters's list sort for
* Python, which is described in detail here:
*
* http://svn.python.org/projects/python/trunk/Objects/listsort.txt
*
* Tim's C code may be found here:
*
* http://svn.python.org/projects/python/trunk/Objects/listobject.c
*
* The underlying techniques are described in this paper (and may have
* even earlier origins):
*
* "Optimistic Sorting and Information Theoretic Complexity"
* Peter McIlroy
* SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
* pp 467-474, Austin, Texas, 25-27 January 1993.
*
* While the API to this class consists solely of static methods, it is
* (privately) instantiable; a TimSort instance holds the state of an ongoing
* sort, assuming the input array is large enough to warrant the full-blown
* TimSort. Small arrays are sorted in place, using a binary insertion sort.
*/
@Suppress("detekt.complexity", "detekt.style")
internal class ByteArrayTimSort
/**
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
* @param a the array to be sorted
* @param c the comparator to determine the order of the sort
*/
private constructor(
/**
* The array being sorted.
*/
private val a: ByteArray,
/**
* The comparator for this sort.
*/
private val c: ByteArrayComparator,
private val entrySize: Int
) {
/**
* This controls when we get *into* galloping mode. It is initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* random data, and lower for highly structured data.
*/
private var minGallop = MIN_GALLOP
/**
* Temp storage for merges.
*/
private var tmp: ByteArray? = null // Actual runtime type will be Object[], regardless of T
/**
* A stack of pending runs yet to be merged. Run i starts at
* address `base[i]` and extends for `len[i]` elements. It's always
* true (so long as the indices are in bounds) that:
*
* `runBase[i] + runLen[i] == runBase[i + 1]`
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
private var stackSize = 0 // Number of pending runs on stack
private val runBase: IntArray
private val runLen: IntArray
init {
// Allocate temp storage (which may be increased later if necessary)
val len = a.size / entrySize
val newArray = ByteArray(
entrySize *
if (len < 2 * INITIAL_TMP_STORAGE_LENGTH)
len.ushr(1)
else
INITIAL_TMP_STORAGE_LENGTH
)
tmp = newArray
/*
* Allocate runs-to-be-merged stack (which cannot be expanded). The
* stack length requirements are described in listsort.txt. The C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays (e.g.,
* 100 elements) in Java. Therefore, we use smaller (but sufficiently
* large) stack lengths for smaller arrays. The "magic numbers" in the
* computation below must be changed if MIN_MERGE is decreased. See
* the MIN_MERGE declaration above for more information.
*/
val stackLen = when {
len < 120 -> 5
len < 1542 -> 10
len < 119151 -> 19
else -> 40
}
runBase = IntArray(stackLen)
runLen = IntArray(stackLen)
}
/**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
private fun pushRun(
runBase: Int,
runLen: Int
) {
this.runBase[stackSize] = runBase
this.runLen[stackSize] = runLen
stackSize++
}
/**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
// Fixed with http://www.envisage-project.eu/proving-android-java-and-python-sorting-algorithm-is-broken-and-how-to-fix-it/
private fun mergeCollapse() {
while (stackSize > 1) {
var n = stackSize - 2
if (n >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1] || n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1]) {
if (runLen[n - 1] < runLen[n + 1])
n--
} else if (runLen[n] > runLen[n + 1]) {
break // Invariant is established
}
mergeAt(n)
}
}
/**
* Merges all runs on the stack until only one remains. This method is
* called once, to complete the sort.
*/
private fun mergeForceCollapse() {
while (stackSize > 1) {
var n = stackSize - 2
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--
mergeAt(n)
}
}
/**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
private fun mergeAt(i: Int) {
if (DEBUG) assert(stackSize >= 2)
if (DEBUG) assert(i >= 0)
if (DEBUG) assert(i == stackSize - 2 || i == stackSize - 3)
var base1 = runBase[i]
var len1 = runLen[i]
val base2 = runBase[i + 1]
var len2 = runLen[i + 1]
if (DEBUG) assert(len1 > 0 && len2 > 0)
if (DEBUG) assert(base1 + len1 == base2)
/*
* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2]
runLen[i + 1] = runLen[i + 2]
}
stackSize--
/*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
val k = gallopRight(a, base2, a, base1, len1, 0, entrySize, c)
if (DEBUG) assert(k >= 0)
base1 += k
len1 -= k
if (len1 == 0)
return
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft(a, base1 + len1 - 1, a, base2, len2, len2 - 1, entrySize, c)
if (DEBUG) assert(len2 >= 0)
if (len2 == 0)
return
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2)
else
mergeHi(base1, len1, base2, len2)
}
/**
* Merges two adjacent runs in place, in a stable fashion. The first
* element of the first run must be greater than the first element of the
* second run (a[base1] > a[base2]), and the last element of the first run
* (a[base1 + len1-1]) must be greater than all elements of the second run.
*
* For performance, this method should be called only when len1 <= len2;
* its twin, mergeHi should be called if len1 >= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private fun mergeLo(
base1: Int,
len1: Int,
base2: Int,
len2: Int
) {
var len1 = len1
var len2 = len2
if (DEBUG) assert(len1 > 0 && len2 > 0 && base1 + len1 == base2)
// Copy first run into temp array
val a = this.a // For performance
val entrySize = entrySize
val tmp = ensureCapacity(len1)
System.arraycopy(a, base1 * entrySize, tmp, 0, len1 * entrySize)
var cursor1 = 0 // Indexes into tmp array
var cursor2 = base2 // Indexes int a
var dest = base1 // Indexes int a
// Move first element of second run and deal with degenerate cases
val destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor2Index + i]
}
dest++
cursor2++
if (--len2 == 0) {
System.arraycopy(tmp, cursor1 * entrySize, a, dest * entrySize, len1 * entrySize)
return
}
if (len1 == 1) {
System.arraycopy(a, cursor2 * entrySize, a, dest * entrySize, len2 * entrySize)
val destLen2Index = (dest + len2) * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destLen2Index + i] = tmp[cursor1Index + i] // Last elt of run 1 to end of merge
}
return
}
val c = this.c // Use local variable for performance
var minGallop = this.minGallop // " " " " "
outer@ while (true) {
var count1 = 0 // Number of times in a row that first run won
var count2 = 0 // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
if (DEBUG) assert(len1 > 1 && len2 > 0)
if (c.compare(entrySize, a, cursor2, tmp, cursor1) < 0) {
val destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor2Index + i]
}
dest++
cursor2++
count2++
count1 = 0
if (--len2 == 0)
break@outer
} else {
val destIndex = dest * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor1Index + i]
}
dest++
cursor1++
count1++
count2 = 0
if (--len1 == 1)
break@outer
}
} while (count1 or count2 < minGallop)
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
if (DEBUG) assert(len1 > 1 && len2 > 0)
count1 = gallopRight(a, cursor2, tmp, cursor1, len1, 0, entrySize, c)
if (count1 != 0) {
System.arraycopy(tmp, cursor1 * entrySize, a, dest * entrySize, count1 * entrySize)
dest += count1
cursor1 += count1
len1 -= count1
if (len1 <= 1)
// len1 == 1 || len1 == 0
break@outer
}
var destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor2Index + i]
}
dest++
cursor2++
if (--len2 == 0)
break@outer
count2 = gallopLeft(tmp, cursor1, a, cursor2, len2, 0, entrySize, c)
if (count2 != 0) {
System.arraycopy(a, cursor2 * entrySize, a, dest * entrySize, count2 * entrySize)
dest += count2
cursor2 += count2
len2 -= count2
if (len2 == 0)
break@outer
}
destIndex = dest * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor1Index + i]
}
dest++
cursor1++
if (--len1 == 1)
break@outer
minGallop--
} while ((count1 >= MIN_GALLOP) or (count2 >= MIN_GALLOP))
if (minGallop < 0)
minGallop = 0
minGallop += 2 // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = if (minGallop < 1) 1 else minGallop // Write back to field
when (len1) {
1 -> {
if (DEBUG) assert(len2 > 0)
System.arraycopy(a, cursor2 * entrySize, a, dest * entrySize, len2 * entrySize)
val destLen2Index = (dest + len2) * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destLen2Index + i] = tmp[cursor1Index + i] // Last elt of run 1 to end of merge
}
}
0 -> {
throw IllegalArgumentException(
"Comparison method violates its general contract!"
)
}
else -> {
if (DEBUG) assert(len2 == 0)
if (DEBUG) assert(len1 > 1)
System.arraycopy(tmp, cursor1 * entrySize, a, dest * entrySize, len1 * entrySize)
}
}
}
/**
* Like mergeLo, except that this method should be called only if
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private fun mergeHi(
base1: Int,
len1: Int,
base2: Int,
len2: Int
) {
var len1 = len1
var len2 = len2
if (DEBUG) assert(len1 > 0 && len2 > 0 && base1 + len1 == base2)
// Copy second run into temp array
val a = this.a // For performance
val tmp = ensureCapacity(len2)
val entrySize = entrySize
System.arraycopy(a, base2 * entrySize, tmp, 0, len2 * entrySize)
var cursor1 = base1 + len1 - 1 // Indexes into a
var cursor2 = len2 - 1 // Indexes into tmp array
var dest = base2 + len2 - 1 // Indexes into a
// Move last element of first run and deal with degenerate cases
var destIndex = dest * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor1Index + i]
}
dest--
cursor1--
if (--len1 == 0) {
System.arraycopy(tmp, 0, a, (dest - (len2 - 1)) * entrySize, len2 * entrySize)
return
}
if (len2 == 1) {
dest -= len1
cursor1 -= len1
System.arraycopy(a, (cursor1 + 1) * entrySize, a, (dest + 1) * entrySize, len1 * entrySize)
val destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor2Index + i]
}
return
}
val c = this.c // Use local variable for performance
var minGallop = this.minGallop // " " " " "
outer@ while (true) {
var count1 = 0 // Number of times in a row that first run won
var count2 = 0 // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
if (DEBUG) assert(len1 > 0 && len2 > 1)
if (c.compare(entrySize, tmp, cursor2, a, cursor1) < 0) {
val destIndex = dest * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor1Index + i]
}
dest--
cursor1--
count1++
count2 = 0
if (--len1 == 0)
break@outer
} else {
val destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor2Index + i]
}
dest--
cursor2--
count2++
count1 = 0
if (--len2 == 1)
break@outer
}
} while (count1 or count2 < minGallop)
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
if (DEBUG) assert(len1 > 0 && len2 > 1)
count1 = len1 - gallopRight(tmp, cursor2, a, base1, len1, len1 - 1, entrySize, c)
if (count1 != 0) {
dest -= count1
cursor1 -= count1
len1 -= count1
System.arraycopy(
a, (cursor1 + 1) * entrySize, a, (dest + 1) * entrySize, count1 * entrySize
)
if (len1 == 0)
break@outer
}
destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor2Index + i]
}
dest--
cursor2--
if (--len2 == 1)
break@outer
count2 = len2 - gallopLeft(a, cursor1, tmp, 0, len2, len2 - 1, entrySize, c)
if (count2 != 0) {
dest -= count2
cursor2 -= count2
len2 -= count2
System.arraycopy(
tmp, (cursor2 + 1) * entrySize, a, (dest + 1) * entrySize, count2 * entrySize
)
if (len2 <= 1)
// len2 == 1 || len2 == 0
break@outer
}
val destIndex = dest * entrySize
val cursor1Index = cursor1 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = a[cursor1Index + i]
}
dest--
cursor1--
if (--len1 == 0)
break@outer
minGallop--
} while ((count1 >= MIN_GALLOP) or (count2 >= MIN_GALLOP))
if (minGallop < 0)
minGallop = 0
minGallop += 2 // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = if (minGallop < 1) 1 else minGallop // Write back to field
when (len2) {
1 -> {
if (DEBUG) assert(len1 > 0)
dest -= len1
cursor1 -= len1
System.arraycopy(a, (cursor1 + 1) * entrySize, a, (dest + 1) * entrySize, len1 * entrySize)
val destIndex = dest * entrySize
val cursor2Index = cursor2 * entrySize
for (i in 0 until entrySize) {
a[destIndex + i] = tmp[cursor2Index + i] // Move first elt of run2 to front of merge
}
}
0 -> {
throw IllegalArgumentException(
"Comparison method violates its general contract!"
)
}
else -> {
if (DEBUG) assert(len1 == 0)
if (DEBUG) assert(len2 > 0)
System.arraycopy(tmp, 0, a, (dest - (len2 - 1)) * entrySize, len2 * entrySize)
}
}
}
/**
* Ensures that the external array tmp has at least the specified
* number of elements, increasing its size if necessary. The size
* increases exponentially to ensure amortized linear time complexity.
*
* @param minCapacity the minimum required capacity of the tmp array
* @return tmp, whether or not it grew
*/
private fun ensureCapacity(minCapacity: Int): ByteArray {
if (tmp!!.size < minCapacity * entrySize) {
// Compute smallest power of 2 > minCapacity
var newSize = minCapacity
newSize = newSize or (newSize shr 1)
newSize = newSize or (newSize shr 2)
newSize = newSize or (newSize shr 4)
newSize = newSize or (newSize shr 8)
newSize = newSize or (newSize shr 16)
newSize++
newSize = if (newSize < 0)
// Not bloody likely!
minCapacity
else
min(newSize, (a.size / entrySize).ushr(1))
val newArray = ByteArray(newSize * entrySize)
tmp = newArray
}
return tmp!!
}
companion object {
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* [.minRunLength] computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private const val MIN_MERGE = 32
/**
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
*/
private const val MIN_GALLOP = 7
/**
* Maximum initial size of tmp array, which is used for merging. The array
* can grow to accommodate demand.
*
* Unlike Tim's original C version, we do not allocate this much storage
* when sorting smaller arrays. This change was required for performance.
*/
private const val INITIAL_TMP_STORAGE_LENGTH = 256
/**
* Asserts have been placed in if-statements for performace. To enable them,
* set this field to true and enable them in VM with a command line flag.
* If you modify this class, please do test the asserts!
*/
private const val DEBUG = false
/*
* The next two methods (which are package private and static) constitute
* the entire API of this class. Each of these methods obeys the contract
* of the public method with the same signature in java.util.Arrays.
*/
fun sort(
a: ByteArray,
entrySize: Int,
c: ByteArrayComparator
) {
sort(a, 0, a.size / entrySize, entrySize, c)
}
fun sort(
a: ByteArray,
lo: Int,
hi: Int,
entrySize: Int,
c: ByteArrayComparator
) {
var lo = lo
checkStartAndEnd(a.size / entrySize, lo, hi)
var nRemaining = hi - lo
if (nRemaining < 2)
return // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
val initRunLen = countRunAndMakeAscending(a, lo, hi, entrySize, c)
binarySort(a, lo, hi, lo + initRunLen, entrySize, c)
return
}
/**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
*/
val ts = ByteArrayTimSort(a, c, entrySize)
val minRun = minRunLength(nRemaining)
do {
// Identify next run
var runLen = countRunAndMakeAscending(a, lo, hi, entrySize, c)
// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
val force = if (nRemaining <= minRun) nRemaining else minRun
binarySort(a, lo, lo + force, lo + runLen, entrySize, c)
runLen = force
}
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen)
ts.mergeCollapse()
// Advance to find next run
lo += runLen
nRemaining -= runLen
} while (nRemaining != 0)
// Merge all remaining runs to complete sort
if (DEBUG) assert(lo == hi)
ts.mergeForceCollapse()
if (DEBUG) assert(ts.stackSize == 1)
}
private fun checkStartAndEnd(
len: Int,
start: Int,
end: Int
) {
if (start < 0 || end > len) {
throw ArrayIndexOutOfBoundsException(
"start < 0 || end > len."
+ " start=" + start + ", end=" + end + ", len=" + len
)
}
if (start > end) {
throw IllegalArgumentException("start > end: $start > $end")
}
}
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index `lo`, inclusive, to `start`,
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be sorted
* @param hi the index after the last element in the range to be sorted
* @param start the index of the first element in the range that is
* not already known to be sorted (@code lo <= start <= hi}
* @param c comparator to used for the sort
*/
private fun binarySort(
a: ByteArray,
lo: Int,
hi: Int,
start: Int,
entrySize: Int,
c: ByteArrayComparator
) {
var start = start
if (DEBUG) assert(start in lo..hi)
if (start == lo)
start++
val pivot = ByteArray(entrySize)
while (start < hi) {
val startIndex = start * entrySize
for (i in 0 until entrySize) {
pivot[i] = a[startIndex + i]
}
// Set left (and right) to the index where a[start] (pivot) belongs
var left = lo
var right = start
if (DEBUG) assert(left <= right)
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
val mid = (left + right).ushr(1)
if (c.compare(entrySize, pivot, 0, a, mid) < 0)
right = mid
else
left = mid + 1
}
if (DEBUG) assert(left == right)
/*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
*/
// Switch is just an optimization for arraycopy in default case
when (val n = start - left) { // The number of elements to move
2 -> {
val leftIndex = left * entrySize
val leftPlusOneIndex = (left + 1) * entrySize
val leftPlusTwoIndex = (left + 2) * entrySize
for (i in 0 until entrySize) {
a[leftPlusTwoIndex + i] = a[leftPlusOneIndex + i]
}
for (i in 0 until entrySize) {
a[leftPlusOneIndex + i] = a[leftIndex + i]
}
}
1 -> {
val leftIndex = left * entrySize
val leftPlusOneIndex = (left + 1) * entrySize
for (i in 0 until entrySize) {
a[leftPlusOneIndex + i] = a[leftIndex + i]
}
}
else -> {
System.arraycopy(a, left * entrySize, a, (left + 1) * entrySize, n * entrySize)
}
}
val leftIndex = left * entrySize
for (i in 0 until entrySize) {
a[leftIndex + i] = pivot[i]
}
start++
}
}
/**
* Returns the length of the run beginning at the specified position in
* the specified array and reverses the run if it is descending (ensuring
* that the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
*
* @param a the array in which a run is to be counted and possibly reversed
* @param lo index of the first element in the run
* @param hi index after the last element that may be contained in the run.
* It is required that @code{lo < hi}.
* @param c the comparator to used for the sort
* @return the length of the run beginning at the specified position in
* the specified array
*/
private fun countRunAndMakeAscending(
a: ByteArray,
lo: Int,
hi: Int,
entrySize: Int,
c: ByteArrayComparator
): Int {
if (DEBUG) assert(lo < hi)
var runHi = lo + 1
if (runHi == hi)
return 1
// Find end of run, and reverse range if descending
val comparison = c.compare(entrySize, a, runHi, a, lo)
runHi++
if (comparison < 0) { // Descending
while (runHi < hi && c.compare(entrySize, a, runHi, a, runHi - 1) < 0)
runHi++
reverseRange(a, lo, runHi, entrySize)
} else { // Ascending
while (runHi < hi && c.compare(entrySize, a, runHi, a, runHi - 1) >= 0)
runHi++
}
return runHi - lo
}
/**
* Reverse the specified range of the specified array.
*
* @param a the array in which a range is to be reversed
* @param lo the index of the first element in the range to be reversed
* @param hi the index after the last element in the range to be reversed
*/
private fun reverseRange(
a: ByteArray,
lo: Int,
hi: Int,
entrySize: Int
) {
var lo = lo
var hi = hi
hi--
while (lo < hi) {
val loIndex = lo * entrySize
val hiIndex = hi * entrySize
for (i in 0 until entrySize) {
val t = a[loIndex + i]
a[loIndex + i] = a[hiIndex + i]
a[hiIndex + i] = t
}
lo++
hi--
}
}
/**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* [.binarySort].
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
* is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private fun minRunLength(n: Int): Int {
var n = n
if (DEBUG) assert(n >= 0)
var r = 0 // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r = r or (n and 1)
n = n shr 1
}
return n + r
}
/**
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
*
* @param keyIndex the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
* pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
* In other words, key belongs at index b + k; or in other words,
* the first k elements of a should precede key, and the last n - k
* should follow it.
*/
private fun gallopLeft(
keyArray: ByteArray,
// Index already divided by entrySize
keyIndex: Int,
a: ByteArray,
base: Int,
len: Int,
hint: Int,
entrySize: Int,
c: ByteArrayComparator
): Int {
if (DEBUG) assert(len > 0 && hint >= 0 && hint < len)
var lastOfs = 0
var ofs = 1
if (c.compare(entrySize, keyArray, keyIndex, a, base + hint) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
val maxOfs = len - hint
while (ofs < maxOfs && c.compare(entrySize, keyArray, keyIndex, a, base + hint + ofs) > 0) {
lastOfs = ofs
ofs = ofs * 2 + 1
if (ofs <= 0)
// int overflow
ofs = maxOfs
}
if (ofs > maxOfs)
ofs = maxOfs
// Make offsets relative to base
lastOfs += hint
ofs += hint
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
val maxOfs = hint + 1
while (ofs < maxOfs && c.compare(
entrySize, keyArray, keyIndex, a, base + hint - ofs
) <= 0
) {
lastOfs = ofs
ofs = ofs * 2 + 1
if (ofs <= 0)
// int overflow
ofs = maxOfs
}
if (ofs > maxOfs)
ofs = maxOfs
// Make offsets relative to base
val tmp = lastOfs
lastOfs = hint - ofs
ofs = hint - tmp
}
if (DEBUG) assert(-1 <= lastOfs && lastOfs < ofs && ofs <= len)
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
* to the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
lastOfs++
while (lastOfs < ofs) {
val m = lastOfs + (ofs - lastOfs).ushr(1)
if (c.compare(entrySize, keyArray, keyIndex, a, base + m) > 0)
lastOfs = m + 1 // a[base + m] < key
else
ofs = m // key <= a[base + m]
}
if (DEBUG) assert(lastOfs == ofs) // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs
}
/**
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
*
* @param keyIndex the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private fun gallopRight(
keyArray: ByteArray,
// Index already divided by entrySize
keyIndex: Int,
a: ByteArray,
base: Int,
len: Int,
hint: Int,
entrySize: Int,
c: ByteArrayComparator
): Int {
if (DEBUG) assert(len > 0 && hint >= 0 && hint < len)
var ofs = 1
var lastOfs = 0
if (c.compare(entrySize, keyArray, keyIndex, a, base + hint) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
val maxOfs = hint + 1
while (ofs < maxOfs && c.compare(entrySize, keyArray, keyIndex, a, base + hint - ofs) < 0) {
lastOfs = ofs
ofs = ofs * 2 + 1
if (ofs <= 0)
// int overflow
ofs = maxOfs
}
if (ofs > maxOfs)
ofs = maxOfs
// Make offsets relative to b
val tmp = lastOfs
lastOfs = hint - ofs
ofs = hint - tmp
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
val maxOfs = len - hint
while (ofs < maxOfs && c.compare(
entrySize, keyArray, keyIndex, a, base + hint + ofs
) >= 0
) {
lastOfs = ofs
ofs = ofs * 2 + 1
if (ofs <= 0)
// int overflow
ofs = maxOfs
}
if (ofs > maxOfs)
ofs = maxOfs
// Make offsets relative to b
lastOfs += hint
ofs += hint
}
if (DEBUG) assert(-1 <= lastOfs && lastOfs < ofs && ofs <= len)
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
*/
lastOfs++
while (lastOfs < ofs) {
val m = lastOfs + (ofs - lastOfs).ushr(1)
if (c.compare(entrySize, keyArray, keyIndex, a, base + m) < 0)
ofs = m // key < a[b + m]
else
lastOfs = m + 1 // a[b + m] <= key
}
if (DEBUG) assert(lastOfs == ofs) // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs
}
}
}