diff --git a/README.md b/README.md index 588edc0..6806364 100644 --- a/README.md +++ b/README.md @@ -34,10 +34,16 @@ This can also be done graphically in IntelliJ: ![intellij-gradle.png](intellij-gradle.png) -## Test Cases +### Test Cases Run the task "test": ```shell ./gradlew test ``` + +## TimSort + +Imported from +- src/java.base/share/classes/java/util/TimSort.java +- src/java.base/share/classes/java/util/ComparableTimSort.java diff --git a/app/src/main/java/de/uni_marburg/powersort/ComparableTimSort.java b/app/src/main/java/de/uni_marburg/powersort/ComparableTimSort.java new file mode 100644 index 0000000..cd5f3f7 --- /dev/null +++ b/app/src/main/java/de/uni_marburg/powersort/ComparableTimSort.java @@ -0,0 +1,907 @@ +/* + * Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved. + * Copyright 2009 Google Inc. All Rights Reserved. + * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. + * + * This code is free software; you can redistribute it and/or modify it + * under the terms of the GNU General Public License version 2 only, as + * published by the Free Software Foundation. Oracle designates this + * particular file as subject to the "Classpath" exception as provided + * by Oracle in the LICENSE file that accompanied this code. + * + * This code is distributed in the hope that it will be useful, but WITHOUT + * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or + * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License + * version 2 for more details (a copy is included in the LICENSE file that + * accompanied this code). + * + * You should have received a copy of the GNU General Public License version + * 2 along with this work; if not, write to the Free Software Foundation, + * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. + * + * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA + * or visit www.oracle.com if you need additional information or have any + * questions. + */ + +package de.uni_marburg.powersort; + +/** + * This is a near duplicate of {@link TimSort}, modified for use with + * arrays of objects that implement {@link Comparable}, instead of using + * explicit comparators. + * + *

If you are using an optimizing VM, you may find that ComparableTimSort + * offers no performance benefit over TimSort in conjunction with a + * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}. + * If this is the case, you are better off deleting ComparableTimSort to + * eliminate the code duplication. (See Arrays.java for details.) + * + * @author Josh Bloch + */ +class ComparableTimSort { + /** + * 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 + * {@link #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 static final int MIN_MERGE = 32; + + /** + * The array being sorted. + */ + private final Object[] a; + + /** + * When we get into galloping mode, we stay there until both runs win less + * often than MIN_GALLOP consecutive times. + */ + private static final int MIN_GALLOP = 7; + + /** + * 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 int minGallop = MIN_GALLOP; + + /** + * 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 static final int INITIAL_TMP_STORAGE_LENGTH = 256; + + /** + * Temp storage for merges. A workspace array may optionally be + * provided in constructor, and if so will be used as long as it + * is big enough. + */ + private Object[] tmp; + private int tmpBase; // base of tmp array slice + private int tmpLen; // length of tmp array slice + + /** + * 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 int stackSize = 0; // Number of pending runs on stack + private final int[] runBase; + private final int[] runLen; + + /** + * Creates a TimSort instance to maintain the state of an ongoing sort. + * + * @param a the array to be sorted + * @param work a workspace array (slice) + * @param workBase origin of usable space in work array + * @param workLen usable size of work array + */ + private ComparableTimSort(Object[] a, Object[] work, int workBase, int workLen) { + this.a = a; + + // Allocate temp storage (which may be increased later if necessary) + int len = a.length; + int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? + len >>> 1 : INITIAL_TMP_STORAGE_LENGTH; + if (work == null || workLen < tlen || workBase + tlen > work.length) { + tmp = new Object[tlen]; + tmpBase = 0; + tmpLen = tlen; + } + else { + tmp = work; + tmpBase = workBase; + tmpLen = workLen; + } + + /* + * 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. + * The maximum value of 49 allows for an array up to length + * Integer.MAX_VALUE-4, if array is filled by the worst case stack size + * increasing scenario. More explanations are given in section 4 of: + * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf + */ + int stackLen = (len < 120 ? 5 : + len < 1542 ? 10 : + len < 119151 ? 24 : 49); + runBase = new int[stackLen]; + runLen = new int[stackLen]; + } + + /* + * The next method (package private and static) constitutes the + * entire API of this class. + */ + + /** + * Sorts the given range, using the given workspace array slice + * for temp storage when possible. This method is designed to be + * invoked from public methods (in class Arrays) after performing + * any necessary array bounds checks and expanding parameters into + * the required forms. + * + * @param a the array to be sorted + * @param lo the index of the first element, inclusive, to be sorted + * @param hi the index of the last element, exclusive, to be sorted + * @param work a workspace array (slice) + * @param workBase origin of usable space in work array + * @param workLen usable size of work array + * @since 1.8 + */ + static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) { + assert a != null && lo >= 0 && lo <= hi && hi <= a.length; + + int 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) { + int initRunLen = countRunAndMakeAscending(a, lo, hi); + binarySort(a, lo, hi, lo + initRunLen); + 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. + */ + ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen); + int minRun = minRunLength(nRemaining); + do { + // Identify next run + int runLen = countRunAndMakeAscending(a, lo, hi); + + // If run is short, extend to min(minRun, nRemaining) + if (runLen < minRun) { + int force = nRemaining <= minRun ? nRemaining : minRun; + binarySort(a, lo, lo + force, lo + runLen); + 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 + assert lo == hi; + ts.mergeForceCollapse(); + assert ts.stackSize == 1; + } + + /** + * 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 {@code lo}, inclusive, to {@code 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}) + */ + @SuppressWarnings({"fallthrough", "rawtypes", "unchecked"}) + private static void binarySort(Object[] a, int lo, int hi, int start) { + assert lo <= start && start <= hi; + if (start == lo) + start++; + for ( ; start < hi; start++) { + Comparable pivot = (Comparable) a[start]; + + // Set left (and right) to the index where a[start] (pivot) belongs + int left = lo; + int right = start; + assert left <= right; + /* + * Invariants: + * pivot >= all in [lo, left). + * pivot < all in [right, start). + */ + while (left < right) { + int mid = (left + right) >>> 1; + if (pivot.compareTo(a[mid]) < 0) + right = mid; + else + left = mid + 1; + } + 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. + */ + int n = start - left; // The number of elements to move + // Switch is just an optimization for arraycopy in default case + switch (n) { + case 2: a[left + 2] = a[left + 1]; + case 1: a[left + 1] = a[left]; + break; + default: System.arraycopy(a, left, a, left + 1, n); + } + a[left] = pivot; + } + } + + /** + * 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}. + * @return the length of the run beginning at the specified position in + * the specified array + */ + @SuppressWarnings({"unchecked", "rawtypes"}) + private static int countRunAndMakeAscending(Object[] a, int lo, int hi) { + assert lo < hi; + int runHi = lo + 1; + if (runHi == hi) + return 1; + + // Find end of run, and reverse range if descending + if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending + while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0) + runHi++; + reverseRange(a, lo, runHi); + } else { // Ascending + while (runHi < hi && ((Comparable) a[runHi]).compareTo(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 static void reverseRange(Object[] a, int lo, int hi) { + hi--; + while (lo < hi) { + Object t = a[lo]; + a[lo++] = a[hi]; + a[hi--] = t; + } + } + + /** + * Returns the minimum acceptable run length for an array of the specified + * length. Natural runs shorter than this will be extended with + * {@link #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 static int minRunLength(int n) { + assert n >= 0; + int r = 0; // Becomes 1 if any 1 bits are shifted off + while (n >= MIN_MERGE) { + r |= (n & 1); + n >>= 1; + } + return n + r; + } + + /** + * 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 void pushRun(int runBase, int runLen) { + 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. + * + * Thanks to Stijn de Gouw, Jurriaan Rot, Frank S. de Boer, + * Richard Bubel and Reiner Hahnle, this is fixed with respect to + * the analysis in "On the Worst-Case Complexity of TimSort" by + * Nicolas Auger, Vincent Jug, Cyril Nicaud, and Carine Pivoteau. + */ + private void mergeCollapse() { + while (stackSize > 1) { + int n = stackSize - 2; + if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1] || + n > 1 && runLen[n-2] <= runLen[n] + runLen[n-1]) { + if (runLen[n - 1] < runLen[n + 1]) + n--; + } else if (n < 0 || 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 void mergeForceCollapse() { + while (stackSize > 1) { + int 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 + */ + @SuppressWarnings("unchecked") + private void mergeAt(int i) { + assert stackSize >= 2; + assert i >= 0; + assert i == stackSize - 2 || i == stackSize - 3; + + int base1 = runBase[i]; + int len1 = runLen[i]; + int base2 = runBase[i + 1]; + int len2 = runLen[i + 1]; + assert len1 > 0 && len2 > 0; + 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). + */ + int k = gallopRight((Comparable) a[base2], a, base1, len1, 0); + 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((Comparable) a[base1 + len1 - 1], a, + base2, len2, len2 - 1); + 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); + } + + /** + * 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 key 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. + * @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 static int gallopLeft(Comparable key, Object[] a, + int base, int len, int hint) { + assert len > 0 && hint >= 0 && hint < len; + + int lastOfs = 0; + int ofs = 1; + if (key.compareTo(a[base + hint]) > 0) { + // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] + int maxOfs = len - hint; + while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 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] + final int maxOfs = hint + 1; + while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to base + int tmp = lastOfs; + lastOfs = hint - ofs; + ofs = hint - tmp; + } + 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) { + int m = lastOfs + ((ofs - lastOfs) >>> 1); + + if (key.compareTo(a[base + m]) > 0) + lastOfs = m + 1; // a[base + m] < key + else + ofs = m; // key <= a[base + m] + } + 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 key 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. + * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] + */ + private static int gallopRight(Comparable key, Object[] a, + int base, int len, int hint) { + assert len > 0 && hint >= 0 && hint < len; + + int ofs = 1; + int lastOfs = 0; + if (key.compareTo(a[base + hint]) < 0) { + // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] + int maxOfs = hint + 1; + while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to b + int 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] + int maxOfs = len - hint; + while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to b + lastOfs += hint; + ofs += hint; + } + 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) { + int m = lastOfs + ((ofs - lastOfs) >>> 1); + + if (key.compareTo(a[base + m]) < 0) + ofs = m; // key < a[b + m] + else + lastOfs = m + 1; // a[b + m] <= key + } + assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] + return ofs; + } + + /** + * 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) + */ + @SuppressWarnings({"unchecked", "rawtypes"}) + private void mergeLo(int base1, int len1, int base2, int len2) { + assert len1 > 0 && len2 > 0 && base1 + len1 == base2; + + // Copy first run into temp array + Object[] a = this.a; // For performance + Object[] tmp = ensureCapacity(len1); + + int cursor1 = tmpBase; // Indexes into tmp array + int cursor2 = base2; // Indexes int a + int dest = base1; // Indexes int a + System.arraycopy(a, base1, tmp, cursor1, len1); + + // Move first element of second run and deal with degenerate cases + a[dest++] = a[cursor2++]; + if (--len2 == 0) { + System.arraycopy(tmp, cursor1, a, dest, len1); + return; + } + if (len1 == 1) { + System.arraycopy(a, cursor2, a, dest, len2); + a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge + return; + } + + int minGallop = this.minGallop; // Use local variable for performance + outer: + while (true) { + int count1 = 0; // Number of times in a row that first run won + int 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 { + assert len1 > 1 && len2 > 0; + if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) { + a[dest++] = a[cursor2++]; + count2++; + count1 = 0; + if (--len2 == 0) + break outer; + } else { + a[dest++] = tmp[cursor1++]; + count1++; + count2 = 0; + if (--len1 == 1) + break outer; + } + } while ((count1 | 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 { + assert len1 > 1 && len2 > 0; + count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0); + if (count1 != 0) { + System.arraycopy(tmp, cursor1, a, dest, count1); + dest += count1; + cursor1 += count1; + len1 -= count1; + if (len1 <= 1) // len1 == 1 || len1 == 0 + break outer; + } + a[dest++] = a[cursor2++]; + if (--len2 == 0) + break outer; + + count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0); + if (count2 != 0) { + System.arraycopy(a, cursor2, a, dest, count2); + dest += count2; + cursor2 += count2; + len2 -= count2; + if (len2 == 0) + break outer; + } + a[dest++] = tmp[cursor1++]; + if (--len1 == 1) + break outer; + minGallop--; + } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); + if (minGallop < 0) + minGallop = 0; + minGallop += 2; // Penalize for leaving gallop mode + } // End of "outer" loop + this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field + + if (len1 == 1) { + assert len2 > 0; + System.arraycopy(a, cursor2, a, dest, len2); + a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge + } else if (len1 == 0) { + throw new IllegalArgumentException( + "Comparison method violates its general contract!"); + } else { + assert len2 == 0; + assert len1 > 1; + System.arraycopy(tmp, cursor1, a, dest, len1); + } + } + + /** + * 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) + */ + @SuppressWarnings({"unchecked", "rawtypes"}) + private void mergeHi(int base1, int len1, int base2, int len2) { + assert len1 > 0 && len2 > 0 && base1 + len1 == base2; + + // Copy second run into temp array + Object[] a = this.a; // For performance + Object[] tmp = ensureCapacity(len2); + int tmpBase = this.tmpBase; + System.arraycopy(a, base2, tmp, tmpBase, len2); + + int cursor1 = base1 + len1 - 1; // Indexes into a + int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array + int dest = base2 + len2 - 1; // Indexes into a + + // Move last element of first run and deal with degenerate cases + a[dest--] = a[cursor1--]; + if (--len1 == 0) { + System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); + return; + } + if (len2 == 1) { + dest -= len1; + cursor1 -= len1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); + a[dest] = tmp[cursor2]; + return; + } + + int minGallop = this.minGallop; // Use local variable for performance + outer: + while (true) { + int count1 = 0; // Number of times in a row that first run won + int 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 { + assert len1 > 0 && len2 > 1; + if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) { + a[dest--] = a[cursor1--]; + count1++; + count2 = 0; + if (--len1 == 0) + break outer; + } else { + a[dest--] = tmp[cursor2--]; + count2++; + count1 = 0; + if (--len2 == 1) + break outer; + } + } while ((count1 | 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 { + assert len1 > 0 && len2 > 1; + count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1); + if (count1 != 0) { + dest -= count1; + cursor1 -= count1; + len1 -= count1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); + if (len1 == 0) + break outer; + } + a[dest--] = tmp[cursor2--]; + if (--len2 == 1) + break outer; + + count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, tmpBase, len2, len2 - 1); + if (count2 != 0) { + dest -= count2; + cursor2 -= count2; + len2 -= count2; + System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); + if (len2 <= 1) + break outer; // len2 == 1 || len2 == 0 + } + a[dest--] = a[cursor1--]; + if (--len1 == 0) + break outer; + minGallop--; + } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); + if (minGallop < 0) + minGallop = 0; + minGallop += 2; // Penalize for leaving gallop mode + } // End of "outer" loop + this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field + + if (len2 == 1) { + assert len1 > 0; + dest -= len1; + cursor1 -= len1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); + a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge + } else if (len2 == 0) { + throw new IllegalArgumentException( + "Comparison method violates its general contract!"); + } else { + assert len1 == 0; + assert len2 > 0; + System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); + } + } + + /** + * 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 Object[] ensureCapacity(int minCapacity) { + if (tmpLen < minCapacity) { + // Compute smallest power of 2 > minCapacity + int newSize = -1 >>> Integer.numberOfLeadingZeros(minCapacity); + newSize++; + + if (newSize < 0) // Not bloody likely! + newSize = minCapacity; + else + newSize = Math.min(newSize, a.length >>> 1); + + @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) + Object[] newArray = new Object[newSize]; + tmp = newArray; + tmpLen = newSize; + tmpBase = 0; + } + return tmp; + } + +} \ No newline at end of file diff --git a/app/src/main/java/de/uni_marburg/powersort/TimSort.java b/app/src/main/java/de/uni_marburg/powersort/TimSort.java new file mode 100644 index 0000000..7d79a65 --- /dev/null +++ b/app/src/main/java/de/uni_marburg/powersort/TimSort.java @@ -0,0 +1,942 @@ +/* + * Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved. + * Copyright 2009 Google Inc. All Rights Reserved. + * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. + * + * This code is free software; you can redistribute it and/or modify it + * under the terms of the GNU General Public License version 2 only, as + * published by the Free Software Foundation. Oracle designates this + * particular file as subject to the "Classpath" exception as provided + * by Oracle in the LICENSE file that accompanied this code. + * + * This code is distributed in the hope that it will be useful, but WITHOUT + * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or + * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License + * version 2 for more details (a copy is included in the LICENSE file that + * accompanied this code). + * + * You should have received a copy of the GNU General Public License version + * 2 along with this work; if not, write to the Free Software Foundation, + * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. + * + * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA + * or visit www.oracle.com if you need additional information or have any + * questions. + */ + +package de.uni_marburg.powersort; + +import java.util.Comparator; + +/** + * 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. + * + * @author Josh Bloch + */ +class TimSort { + /** + * 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 + * {@link #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 static final int MIN_MERGE = 32; + + /** + * The array being sorted. + */ + private final T[] a; + + /** + * The comparator for this sort. + */ + private final Comparator c; + + /** + * When we get into galloping mode, we stay there until both runs win less + * often than MIN_GALLOP consecutive times. + */ + private static final int MIN_GALLOP = 7; + + /** + * 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 int minGallop = MIN_GALLOP; + + /** + * 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 static final int INITIAL_TMP_STORAGE_LENGTH = 256; + + /** + * Temp storage for merges. A workspace array may optionally be + * provided in constructor, and if so will be used as long as it + * is big enough. + */ + private T[] tmp; + private int tmpBase; // base of tmp array slice + private int tmpLen; // length of tmp array slice + + /** + * 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 int stackSize = 0; // Number of pending runs on stack + private final int[] runBase; + private final int[] runLen; + + /** + * 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 + * @param work a workspace array (slice) + * @param workBase origin of usable space in work array + * @param workLen usable size of work array + */ + private TimSort(T[] a, Comparator c, T[] work, int workBase, int workLen) { + this.a = a; + this.c = c; + + // Allocate temp storage (which may be increased later if necessary) + int len = a.length; + int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? + len >>> 1 : INITIAL_TMP_STORAGE_LENGTH; + if (work == null || workLen < tlen || workBase + tlen > work.length) { + @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) + T[] newArray = (T[])java.lang.reflect.Array.newInstance + (a.getClass().getComponentType(), tlen); + tmp = newArray; + tmpBase = 0; + tmpLen = tlen; + } + else { + tmp = work; + tmpBase = workBase; + tmpLen = workLen; + } + + /* + * 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. + * The maximum value of 49 allows for an array up to length + * Integer.MAX_VALUE-4, if array is filled by the worst case stack size + * increasing scenario. More explanations are given in section 4 of: + * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf + */ + int stackLen = (len < 120 ? 5 : + len < 1542 ? 10 : + len < 119151 ? 24 : 49); + runBase = new int[stackLen]; + runLen = new int[stackLen]; + } + + /* + * The next method (package private and static) constitutes the + * entire API of this class. + */ + + /** + * Sorts the given range, using the given workspace array slice + * for temp storage when possible. This method is designed to be + * invoked from public methods (in class Arrays) after performing + * any necessary array bounds checks and expanding parameters into + * the required forms. + * + * @param a the array to be sorted + * @param lo the index of the first element, inclusive, to be sorted + * @param hi the index of the last element, exclusive, to be sorted + * @param c the comparator to use + * @param work a workspace array (slice) + * @param workBase origin of usable space in work array + * @param workLen usable size of work array + * @since 1.8 + */ + static void sort(T[] a, int lo, int hi, Comparator c, + T[] work, int workBase, int workLen) { + assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; + + int 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) { + int initRunLen = countRunAndMakeAscending(a, lo, hi, c); + binarySort(a, lo, hi, lo + initRunLen, 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. + */ + TimSort ts = new TimSort<>(a, c, work, workBase, workLen); + int minRun = minRunLength(nRemaining); + do { + // Identify next run + int runLen = countRunAndMakeAscending(a, lo, hi, c); + + // If run is short, extend to min(minRun, nRemaining) + if (runLen < minRun) { + int force = nRemaining <= minRun ? nRemaining : minRun; + binarySort(a, lo, lo + force, lo + runLen, 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 + assert lo == hi; + ts.mergeForceCollapse(); + assert ts.stackSize == 1; + } + + /** + * 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 {@code lo}, inclusive, to {@code 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 + */ + @SuppressWarnings("fallthrough") + private static void binarySort(T[] a, int lo, int hi, int start, + Comparator c) { + assert lo <= start && start <= hi; + if (start == lo) + start++; + for ( ; start < hi; start++) { + T pivot = a[start]; + + // Set left (and right) to the index where a[start] (pivot) belongs + int left = lo; + int right = start; + assert left <= right; + /* + * Invariants: + * pivot >= all in [lo, left). + * pivot < all in [right, start). + */ + while (left < right) { + int mid = (left + right) >>> 1; + if (c.compare(pivot, a[mid]) < 0) + right = mid; + else + left = mid + 1; + } + 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. + */ + int n = start - left; // The number of elements to move + // Switch is just an optimization for arraycopy in default case + switch (n) { + case 2: a[left + 2] = a[left + 1]; + case 1: a[left + 1] = a[left]; + break; + default: System.arraycopy(a, left, a, left + 1, n); + } + a[left] = pivot; + } + } + + /** + * 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 static int countRunAndMakeAscending(T[] a, int lo, int hi, + Comparator c) { + assert lo < hi; + int runHi = lo + 1; + if (runHi == hi) + return 1; + + // Find end of run, and reverse range if descending + if (c.compare(a[runHi++], a[lo]) < 0) { // Descending + while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) + runHi++; + reverseRange(a, lo, runHi); + } else { // Ascending + while (runHi < hi && c.compare(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 static void reverseRange(Object[] a, int lo, int hi) { + hi--; + while (lo < hi) { + Object t = a[lo]; + a[lo++] = a[hi]; + a[hi--] = t; + } + } + + /** + * Returns the minimum acceptable run length for an array of the specified + * length. Natural runs shorter than this will be extended with + * {@link #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 static int minRunLength(int n) { + assert n >= 0; + int r = 0; // Becomes 1 if any 1 bits are shifted off + while (n >= MIN_MERGE) { + r |= (n & 1); + n >>= 1; + } + return n + r; + } + + /** + * 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 void pushRun(int runBase, int runLen) { + 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. + * + * Thanks to Stijn de Gouw, Jurriaan Rot, Frank S. de Boer, + * Richard Bubel and Reiner Hahnle, this is fixed with respect to + * the analysis in "On the Worst-Case Complexity of TimSort" by + * Nicolas Auger, Vincent Jug, Cyril Nicaud, and Carine Pivoteau. + */ + private void mergeCollapse() { + while (stackSize > 1) { + int n = stackSize - 2; + if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1] || + n > 1 && runLen[n-2] <= runLen[n] + runLen[n-1]) { + if (runLen[n - 1] < runLen[n + 1]) + n--; + } else if (n < 0 || 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 void mergeForceCollapse() { + while (stackSize > 1) { + int 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 void mergeAt(int i) { + assert stackSize >= 2; + assert i >= 0; + assert i == stackSize - 2 || i == stackSize - 3; + + int base1 = runBase[i]; + int len1 = runLen[i]; + int base2 = runBase[i + 1]; + int len2 = runLen[i + 1]; + assert len1 > 0 && len2 > 0; + 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). + */ + int k = gallopRight(a[base2], a, base1, len1, 0, c); + 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, c); + 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); + } + + /** + * 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 key 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 static int gallopLeft(T key, T[] a, int base, int len, int hint, + Comparator c) { + assert len > 0 && hint >= 0 && hint < len; + int lastOfs = 0; + int ofs = 1; + if (c.compare(key, a[base + hint]) > 0) { + // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] + int maxOfs = len - hint; + while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 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] + final int maxOfs = hint + 1; + while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to base + int tmp = lastOfs; + lastOfs = hint - ofs; + ofs = hint - tmp; + } + 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) { + int m = lastOfs + ((ofs - lastOfs) >>> 1); + + if (c.compare(key, a[base + m]) > 0) + lastOfs = m + 1; // a[base + m] < key + else + ofs = m; // key <= a[base + m] + } + 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 key 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 static int gallopRight(T key, T[] a, int base, int len, + int hint, Comparator c) { + assert len > 0 && hint >= 0 && hint < len; + + int ofs = 1; + int lastOfs = 0; + if (c.compare(key, a[base + hint]) < 0) { + // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] + int maxOfs = hint + 1; + while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to b + int 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] + int maxOfs = len - hint; + while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { + lastOfs = ofs; + ofs = (ofs << 1) + 1; + if (ofs <= 0) // int overflow + ofs = maxOfs; + } + if (ofs > maxOfs) + ofs = maxOfs; + + // Make offsets relative to b + lastOfs += hint; + ofs += hint; + } + 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) { + int m = lastOfs + ((ofs - lastOfs) >>> 1); + + if (c.compare(key, a[base + m]) < 0) + ofs = m; // key < a[b + m] + else + lastOfs = m + 1; // a[b + m] <= key + } + assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] + return ofs; + } + + /** + * 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 void mergeLo(int base1, int len1, int base2, int len2) { + assert len1 > 0 && len2 > 0 && base1 + len1 == base2; + + // Copy first run into temp array + T[] a = this.a; // For performance + T[] tmp = ensureCapacity(len1); + int cursor1 = tmpBase; // Indexes into tmp array + int cursor2 = base2; // Indexes int a + int dest = base1; // Indexes int a + System.arraycopy(a, base1, tmp, cursor1, len1); + + // Move first element of second run and deal with degenerate cases + a[dest++] = a[cursor2++]; + if (--len2 == 0) { + System.arraycopy(tmp, cursor1, a, dest, len1); + return; + } + if (len1 == 1) { + System.arraycopy(a, cursor2, a, dest, len2); + a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge + return; + } + + Comparator c = this.c; // Use local variable for performance + int minGallop = this.minGallop; // " " " " " + outer: + while (true) { + int count1 = 0; // Number of times in a row that first run won + int 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 { + assert len1 > 1 && len2 > 0; + if (c.compare(a[cursor2], tmp[cursor1]) < 0) { + a[dest++] = a[cursor2++]; + count2++; + count1 = 0; + if (--len2 == 0) + break outer; + } else { + a[dest++] = tmp[cursor1++]; + count1++; + count2 = 0; + if (--len1 == 1) + break outer; + } + } while ((count1 | 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 { + assert len1 > 1 && len2 > 0; + count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); + if (count1 != 0) { + System.arraycopy(tmp, cursor1, a, dest, count1); + dest += count1; + cursor1 += count1; + len1 -= count1; + if (len1 <= 1) // len1 == 1 || len1 == 0 + break outer; + } + a[dest++] = a[cursor2++]; + if (--len2 == 0) + break outer; + + count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); + if (count2 != 0) { + System.arraycopy(a, cursor2, a, dest, count2); + dest += count2; + cursor2 += count2; + len2 -= count2; + if (len2 == 0) + break outer; + } + a[dest++] = tmp[cursor1++]; + if (--len1 == 1) + break outer; + minGallop--; + } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); + if (minGallop < 0) + minGallop = 0; + minGallop += 2; // Penalize for leaving gallop mode + } // End of "outer" loop + this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field + + if (len1 == 1) { + assert len2 > 0; + System.arraycopy(a, cursor2, a, dest, len2); + a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge + } else if (len1 == 0) { + throw new IllegalArgumentException( + "Comparison method violates its general contract!"); + } else { + assert len2 == 0; + assert len1 > 1; + System.arraycopy(tmp, cursor1, a, dest, len1); + } + } + + /** + * 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 void mergeHi(int base1, int len1, int base2, int len2) { + assert len1 > 0 && len2 > 0 && base1 + len1 == base2; + + // Copy second run into temp array + T[] a = this.a; // For performance + T[] tmp = ensureCapacity(len2); + int tmpBase = this.tmpBase; + System.arraycopy(a, base2, tmp, tmpBase, len2); + + int cursor1 = base1 + len1 - 1; // Indexes into a + int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array + int dest = base2 + len2 - 1; // Indexes into a + + // Move last element of first run and deal with degenerate cases + a[dest--] = a[cursor1--]; + if (--len1 == 0) { + System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); + return; + } + if (len2 == 1) { + dest -= len1; + cursor1 -= len1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); + a[dest] = tmp[cursor2]; + return; + } + + Comparator c = this.c; // Use local variable for performance + int minGallop = this.minGallop; // " " " " " + outer: + while (true) { + int count1 = 0; // Number of times in a row that first run won + int 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 { + assert len1 > 0 && len2 > 1; + if (c.compare(tmp[cursor2], a[cursor1]) < 0) { + a[dest--] = a[cursor1--]; + count1++; + count2 = 0; + if (--len1 == 0) + break outer; + } else { + a[dest--] = tmp[cursor2--]; + count2++; + count1 = 0; + if (--len2 == 1) + break outer; + } + } while ((count1 | 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 { + assert len1 > 0 && len2 > 1; + count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); + if (count1 != 0) { + dest -= count1; + cursor1 -= count1; + len1 -= count1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); + if (len1 == 0) + break outer; + } + a[dest--] = tmp[cursor2--]; + if (--len2 == 1) + break outer; + + count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c); + if (count2 != 0) { + dest -= count2; + cursor2 -= count2; + len2 -= count2; + System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); + if (len2 <= 1) // len2 == 1 || len2 == 0 + break outer; + } + a[dest--] = a[cursor1--]; + if (--len1 == 0) + break outer; + minGallop--; + } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); + if (minGallop < 0) + minGallop = 0; + minGallop += 2; // Penalize for leaving gallop mode + } // End of "outer" loop + this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field + + if (len2 == 1) { + assert len1 > 0; + dest -= len1; + cursor1 -= len1; + System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); + a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge + } else if (len2 == 0) { + throw new IllegalArgumentException( + "Comparison method violates its general contract!"); + } else { + assert len1 == 0; + assert len2 > 0; + System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); + } + } + + /** + * 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 T[] ensureCapacity(int minCapacity) { + if (tmpLen < minCapacity) { + // Compute smallest power of 2 > minCapacity + int newSize = -1 >>> Integer.numberOfLeadingZeros(minCapacity); + newSize++; + + if (newSize < 0) // Not bloody likely! + newSize = minCapacity; + else + newSize = Math.min(newSize, a.length >>> 1); + + @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) + T[] newArray = (T[])java.lang.reflect.Array.newInstance + (a.getClass().getComponentType(), newSize); + tmp = newArray; + tmpLen = newSize; + tmpBase = 0; + } + return tmp; + } +} \ No newline at end of file