TimSort.java

/*
 * Copyright (C) 2008 The Android Open Source Project
 * 
 * Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the
 * License. You may obtain a copy of the License at
 * 
 * http://www.apache.org/licenses/LICENSE-2.0
 * 
 * Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS"
 * BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language
 * governing permissions and limitations under the License.
 */

package com.badlogic.gdx.utils;

import java.util.Arrays;
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. */
class TimSort<T> {
	/** 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 T[] a;

	/** The comparator for this sort. */
	private Comparator<? super T> 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. */
	private T[] tmp; // Actual runtime type will be Object[], regardless of T
	private int tmpCount;

	/** 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;

	/** Asserts have been placed in if-statements for performance. To enable them, set this field to true and enable them in VM with
	 * a command line flag. If you modify this class, please do test the asserts! */
	private static final boolean DEBUG = false;

	TimSort () {
		tmp = (T[])new Object[INITIAL_TMP_STORAGE_LENGTH];
		runBase = new int[40];
		runLen = new int[40];
	}

	public void doSort (T[] a, Comparator<T> c, int lo, int hi) {
		stackSize = 0;
		rangeCheck(a.length, lo, hi);
		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;
		}

		this.a = a;
		this.c = c;
		tmpCount = 0;

		/** 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. */
		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
			pushRun(lo, runLen);
			mergeCollapse();

			// Advance to find next run
			lo += runLen;
			nRemaining -= runLen;
		} while (nRemaining != 0);

		// Merge all remaining runs to complete sort
		if (DEBUG) assert lo == hi;
		mergeForceCollapse();
		if (DEBUG) assert stackSize == 1;

		this.a = null;
		this.c = null;
		T[] tmp = this.tmp;
		for (int i = 0, n = tmpCount; i < n; i++)
			tmp[i] = null;
	}

	/** Creates a TimSort instance to maintain the state of an ongoing sort.
	 * 
	 * @param a the array to be sorted
	 * @param c the comparator to determine the order of the sort */
	private TimSort (T[] a, Comparator<? super T> c) {
		this.a = a;
		this.c = c;

		// Allocate temp storage (which may be increased later if necessary)
		int len = a.length;
		T[] newArray = (T[])new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
		tmp = newArray;

		/*
		 * Allocate runs-to-be-merged stack (which cannot be expanded). The stack length requirements are described in listsort.txt.
		 * The C version always uses the same stack length (85), but this was measured to be too expensive when sorting "mid-sized"
		 * arrays (e.g., 100 elements) in Java. Therefore, we use smaller (but sufficiently large) stack lengths for smaller arrays.
		 * The "magic numbers" in the computation below must be changed if MIN_MERGE is decreased. See the MIN_MERGE declaration
		 * above for more information.
		 */
		int stackLen = (len < 120 ? 5 : len < 1542 ? 10 : len < 119151 ? 19 : 40);
		runBase = new int[stackLen];
		runLen = new int[stackLen];
	}

	/*
	 * The next two methods (which are package private and static) constitute the entire API of this class. Each of these methods
	 * obeys the contract of the public method with the same signature in java.util.Arrays.
	 */

	static <T> void sort (T[] a, Comparator<? super T> c) {
		sort(a, 0, a.length, c);
	}

	static <T> void sort (T[] a, int lo, int hi, Comparator<? super T> c) {
		if (c == null) {
			Arrays.sort(a, lo, hi);
			return;
		}

		rangeCheck(a.length, lo, hi);
		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<T> ts = new TimSort<T>(a, c);
		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
		if (DEBUG) assert lo == hi;
		ts.mergeForceCollapse();
		if (DEBUG) 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 <T> void binarySort (T[] a, int lo, int hi, int start, Comparator<? super T> c) {
		if (DEBUG) 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;
			if (DEBUG) 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;
			}
			if (DEBUG) assert left == right;

			/*
			 * The invariants still hold: pivot >= all in [lo, left) and pivot < all in [left, start), so pivot belongs at left. Note
			 * that if there are elements equal to pivot, left points to the first slot after them -- that's why this sort is stable.
			 * Slide elements over to make room for pivot.
			 */
			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 <T> int countRunAndMakeAscending (T[] a, int lo, int hi, Comparator<? super T> c) {
		if (DEBUG) 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) {
		if (DEBUG) 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[n - 2] > runLen[n - 1] + runLen[n] 2. runLen[n - 1] > runLen[n]
	 * 
	 * where n is the index of the last run in runLen.
	 * 
	 * This method has been formally verified to be correct after checking the last 4 runs.
	 * Checking for 3 runs results in an exception for large arrays.
	 * (Source: http://envisage-project.eu/proving-android-java-and-python-sorting-algorithm-is-broken-and-how-to-fix-it/)
	 * 
	 * 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. */
	private void mergeCollapse () {
		while (stackSize > 1) {
			int n = stackSize - 2;
			if ((n >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) || (n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1])) {
				if (runLen[n - 1] < runLen[n + 1]) n--;
			} else if (runLen[n] > runLen[n + 1]) {
				break; // Invariant is established
			}
			mergeAt(n);
		}
	}

	/** Merges all runs on the stack until only one remains. This method is called once, to complete the sort. */
	private 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) {
		if (DEBUG) assert stackSize >= 2;
		if (DEBUG) assert i >= 0;
		if (DEBUG) 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];
		if (DEBUG) assert len1 > 0 && len2 > 0;
		if (DEBUG) assert base1 + len1 == base2;

		/*
		 * Record the length of the combined runs; if i is the 3rd-last run now, also slide over the last run (which isn't involved
		 * in this merge). The current run (i+1) goes away in any case.
		 */
		runLen[i] = len1 + len2;
		if (i == stackSize - 3) {
			runBase[i + 1] = runBase[i + 2];
			runLen[i + 1] = runLen[i + 2];
		}
		stackSize--;

		/*
		 * Find where the first element of run2 goes in run1. Prior elements in run1 can be ignored (because they're already in
		 * place).
		 */
		int k = gallopRight(a[base2], a, base1, len1, 0, c);
		if (DEBUG) assert k >= 0;
		base1 += k;
		len1 -= k;
		if (len1 == 0) return;

		/*
		 * Find where the last element of run1 goes in run2. Subsequent elements in run2 can be ignored (because they're already in
		 * place).
		 */
		len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
		if (DEBUG) assert len2 >= 0;
		if (len2 == 0) return;

		// Merge remaining runs, using tmp array with min(len1, len2) elements
		if (len1 <= len2)
			mergeLo(base1, len1, base2, len2);
		else
			mergeHi(base1, len1, base2, len2);
	}

	/** 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 <T> int gallopLeft (T key, T[] a, int base, int len, int hint, Comparator<? super T> c) {
		if (DEBUG) 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;
		}
		if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

		/*
		 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
		 * Do a binary search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
		 */
		lastOfs++;
		while (lastOfs < ofs) {
			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]
		}
		if (DEBUG) assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
		return ofs;
	}

	/** Like gallopLeft, except that if the range contains an element equal to key, gallopRight returns the index after the
	 * rightmost equal element.
	 * 
	 * @param 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 <T> int gallopRight (T key, T[] a, int base, int len, int hint, Comparator<? super T> c) {
		if (DEBUG) 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;
		}
		if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

		/*
		 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
		 * Do a binary search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
		 */
		lastOfs++;
		while (lastOfs < ofs) {
			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
		}
		if (DEBUG) 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) {
		if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

		// Copy first run into temp array
		T[] a = this.a; // For performance
		T[] tmp = ensureCapacity(len1);
		System.arraycopy(a, base1, tmp, 0, len1);

		int cursor1 = 0; // Indexes into tmp array
		int cursor2 = base2; // Indexes int a
		int dest = base1; // Indexes int a

		// 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<? super T> 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 {
				if (DEBUG) 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 {
				if (DEBUG) 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) {
			if (DEBUG) 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 {
			if (DEBUG) assert len2 == 0;
			if (DEBUG) 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) {
		if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

		// Copy second run into temp array
		T[] a = this.a; // For performance
		T[] tmp = ensureCapacity(len2);
		System.arraycopy(a, base2, tmp, 0, len2);

		int cursor1 = base1 + len1 - 1; // Indexes into a
		int cursor2 = 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, 0, 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<? super T> 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 {
				if (DEBUG) 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 {
				if (DEBUG) 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, 0, 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) {
			if (DEBUG) 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 {
			if (DEBUG) assert len1 == 0;
			if (DEBUG) assert len2 > 0;
			System.arraycopy(tmp, 0, 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) {
		tmpCount = Math.max(tmpCount, minCapacity);
		if (tmp.length < minCapacity) {
			// Compute smallest power of 2 > minCapacity
			int newSize = minCapacity;
			newSize |= newSize >> 1;
			newSize |= newSize >> 2;
			newSize |= newSize >> 4;
			newSize |= newSize >> 8;
			newSize |= newSize >> 16;
			newSize++;

			if (newSize < 0) // Not bloody likely!
				newSize = minCapacity;
			else
				newSize = Math.min(newSize, a.length >>> 1);

			T[] newArray = (T[])new Object[newSize];
			tmp = newArray;
		}
		return tmp;
	}

	/** Checks that fromIndex and toIndex are in range, and throws an appropriate exception if they aren't.
	 * 
	 * @param arrayLen the length of the array
	 * @param fromIndex the index of the first element of the range
	 * @param toIndex the index after the last element of the range
	 * @throws IllegalArgumentException if fromIndex > toIndex
	 * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex > arrayLen */
	private static void rangeCheck (int arrayLen, int fromIndex, int toIndex) {
		if (fromIndex > toIndex) throw new IllegalArgumentException("fromIndex(" + fromIndex + ") > toIndex(" + toIndex + ")");
		if (fromIndex < 0) throw new ArrayIndexOutOfBoundsException(fromIndex);
		if (toIndex > arrayLen) throw new ArrayIndexOutOfBoundsException(toIndex);
	}
}