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.0.8.1.source-code.LongPairedTimSort Maven / Gradle / Ivy

/* Copyright (c) 2019 LibJ
 *
 * Permission is hereby granted, free of charge, to any person obtaining a copy
 * of this software and associated documentation files (the "Software"), to deal
 * in the Software without restriction, including without limitation the rights
 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
 * copies of the Software, and to permit persons to whom the Software is
 * furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included in
 * all copies or substantial portions of the Software.
 *
 * You should have received a copy of The MIT License (MIT) along with this
 * program. If not, see .
 */

package org.libj.util.primitive;

import javax.annotation.Generated;

import org.libj.util.primitive.LongComparator;
import org.libj.util.primitive.LongPairedTimSort;

/**
 * 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
 * IntTrimSort instance holds the state of an ongoing sort, assuming the input
 * array is large enough to warrant the full-blown IntTrimSort. Small arrays are
 * sorted in place, using a binary insertion sort.
 *
 * @author Josh Bloch
 * @see java.util.TimSort
 */
@Generated(value="Autogenerated by OpenJAX CodeGen Template (0.2.5)", date="2020-05-23")
class LongPairedTimSort extends PrimitiveTimSort {
  /**
   * the array of sortable values being sorted.
   */
  private final long[] a;

  /**
   * the array of paired values being sorted.
   */
  private final Object[] v;

  /**
   * The comparator for this sort.
   */
  private final LongComparator c;

  /**
   * 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;

  /**
   * 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 long[] tmp;

  /**
   * FIXME: Write this up!
   */
  private Object[] tmpV;

  /**
   * Creates a IntTrimSort instance to maintain the state of an ongoing sort.
   *
   * @param a the array of sortable values
   * @param v the array of paired values
   * @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 LongPairedTimSort(final long[] a, final Object[] v, final LongComparator c, final long[] work, final int workBase, final int workLen) {
    this.a = a;
    this.c = c;
    this.v = v;

    // Allocate temp storage (which may be increased later if necessary)
    final int len = a.length;
    final int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
    if (work == null || workLen < tlen || workBase + tlen > work.length) {
      final long[] newArray = new long[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
     */
    final 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 of sortable values
   * @param v the array of paired values
   * @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(final long[] a, final Object[] v, int lo, final int hi, final LongComparator c, final long[] work, final int workBase, final 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-IntTrimSort" with no merges
    if (nRemaining < MIN_MERGE) {
      int initRunLen = countRunAndMakeAscending(a, v, lo, hi, c);
      binarySort(a, v, 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.
     */
    final LongPairedTimSort ts = new LongPairedTimSort(a, v, c, work, workBase, workLen);
    final int minRun = minRunLength(nRemaining);
    do {
      // Identify next run
      int runLen = countRunAndMakeAscending(a, v, lo, hi, c);

      // If run is short, extend to min(minRun, nRemaining)
      if (runLen < minRun) {
        final int force = nRemaining <= minRun ? nRemaining : minRun;
        binarySort(a, v, 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 of sortable values in which a range is to be sorted
   * @param v the array of paired values 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 static void binarySort(final long[] a, final Object[] v, final int lo, final int hi, int start, final LongComparator c) {
    assert lo <= start && start <= hi;
    if (start == lo)
      ++start;

    for (; start < hi; ++start) {
      final long pivot = a[start];
      final Object v0 = v[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) {
        final 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.
       */
      final 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];
          v[left + 2] = v[left + 1];
        case 1:
          a[left + 1] = a[left];
          v[left + 1] = v[left];
          break;
        default:
          System.arraycopy(a, left, a, left + 1, n);
          System.arraycopy(v, left, v, left + 1, n);
      }
      a[left] = pivot;
      v[left] = v0;
    }
  }

  /**
   * 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 of sortable values in which a run is to be counted and
   *          possibly reversed
   * @param v the array of paired values 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(final long[] a, final Object[] v, final int lo, final int hi, final LongComparator 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, v, 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 of sortable values in which a range is to be reversed
   * @param v the array of paired values 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(final long[] a, final Object[] v, int lo, int hi) {
    --hi;
    while (lo < hi) {
      final long t = a[lo];
      final Object v0 = v[lo];
      a[lo] = a[hi];
      v[lo] = v[hi];
      a[hi] = t;
      v[hi] = v0;
    }
  }

  /**
   * 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
   */
  @Override
  void mergeAt(final int i) {
    assert stackSize >= 2;
    assert i >= 0;
    assert i == stackSize - 2 || i == stackSize - 3;

    int base1 = runBase[i];
    int len1 = runLen[i];
    final 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).
     */
    final 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 of sortable values 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(final long key, final long[] a, final int base, final int len, final int hint, final LongComparator 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]
      final 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
      final 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) {
      final 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 of sortable values 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(final long key, final long[] a, final int base, final int len, final int hint, final LongComparator 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]
      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 b
      final 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]
      final 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) {
      final 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(final int base1, int len1, final int base2, int len2) {
    assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

    // Copy first run into temp array
    final long[] a = this.a; // For performance
    final Object[] v = this.v; // For performance
    final long[] tmp = ensureCapacity(len1);
    final Object[] tmpV = this.tmpV;
    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);
    System.arraycopy(v, base1, tmpV, cursor1, len1);

    // Move first element of second run and deal with degenerate cases
    a[dest] = a[cursor2];
    v[dest++] = v[cursor2++];
    if (--len2 == 0) {
      System.arraycopy(tmp, cursor1, a, dest, len1);
      System.arraycopy(tmpV, cursor1, v, dest, len1);
      return;
    }

    if (len1 == 1) {
      System.arraycopy(a, cursor2, a, dest, len2);
      System.arraycopy(v, cursor2, v, dest, len2);

      a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
      v[dest + len2] = tmpV[cursor1]; // Last elt of run 1 to end of merge
      return;
    }

    final LongComparator 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];
          v[dest++] = v[cursor2++];
          ++count2;
          count1 = 0;
          if (--len2 == 0)
            break outer;
        }
        else {
          a[dest] = tmp[cursor1];
          v[dest] = tmpV[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);
          System.arraycopy(tmpV, cursor1, v, dest, count1);
          dest += count1;
          cursor1 += count1;
          len1 -= count1;
          if (len1 <= 1) // len1 == 1 || len1 == 0
            break outer;
        }

        a[dest] = a[cursor2];
        v[dest++] = v[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);
          System.arraycopy(v, cursor2, v, dest, count2);
          dest += count2;
          cursor2 += count2;
          len2 -= count2;
          if (len2 == 0)
            break outer;
        }

        a[dest] = tmp[cursor1];
        v[dest++] = tmpV[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);
      System.arraycopy(v, cursor2, v, dest, len2);
      a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
      v[dest + len2] = tmpV[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);
      System.arraycopy(tmpV, cursor1, v, 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(final int base1, int len1, final int base2, int len2) {
    assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

    // Copy second run into temp array
    final long[] a = this.a; // For performance
    final Object[] v = this.v; // For performance
    final long[] tmp = ensureCapacity(len2);
    final Object[] tmpV = this.tmpV;
    final int tmpBase = this.tmpBase;
    System.arraycopy(a, base2, tmp, tmpBase, len2);
    System.arraycopy(v, base2, tmpV, 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];
    v[dest--] = v[cursor1--];
    if (--len1 == 0) {
      final int start = dest - (len2 - 1);
      System.arraycopy(tmp, tmpBase, a, start, len2);
      System.arraycopy(tmpV, tmpBase, v, start, len2);
      return;
    }
    if (len2 == 1) {
      dest -= len1;
      cursor1 -= len1;
      System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
      System.arraycopy(v, cursor1 + 1, v, dest + 1, len1);
      a[dest] = tmp[cursor2];
      v[dest] = tmpV[cursor2];
      return;
    }

    final LongComparator 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];
          v[dest--] = v[cursor1--];
          ++count1;
          count2 = 0;
          if (--len1 == 0)
            break outer;
        }
        else {
          a[dest] = tmp[cursor2];
          v[dest--] = tmpV[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);
          System.arraycopy(v, cursor1 + 1, v, dest + 1, count1);
          if (len1 == 0)
            break outer;
        }

        a[dest] = tmp[cursor2];
        v[dest] = tmpV[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);
          System.arraycopy(tmpV, cursor2 + 1, v, dest + 1, count2);
          if (len2 <= 1) // len2 == 1 || len2 == 0
            break outer;
        }

        a[dest] = a[cursor1];
        v[dest] = v[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);
      System.arraycopy(v, cursor1 + 1, v, dest + 1, len1);
      a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
      v[dest] = tmpV[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;
      final int start = dest - (len2 - 1);
      System.arraycopy(tmp, tmpBase, a, start, len2);
      System.arraycopy(tmpV, tmpBase, v, start, 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 long[] ensureCapacity(final 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);

      final long[] newArray = new long[newSize];
      tmp = newArray;
      tmpLen = newSize;
      tmpBase = 0;

      final Object[] newArrayV = new Object[newSize];
      tmpV = newArrayV;
    }

    return tmp;
  }
}