gods/utils/timsort/timsort.go

// Obtained from: https://github.com/psilva261/timsort

/*
Copyright (c) 2010-2011 Mike Kroutikov

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.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
THE SOFTWARE
*/

// Fast stable sort, uses external 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 derived from Java's TimSort object by Josh Bloch,
// which, in turn, was based on the original code by Tim Peters:
//
// http://svn.python.org/projects/python/trunk/Objects/listsort.txt
//
// Mike K.

package timsort

import (
	"errors"
	"fmt"
)

const (
	/**
	 * 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.
	 */
	_MIN_MERGE = 32
	// mk: tried higher MIN_MERGE and got slower sorting (348->375)
	//	c_MIN_MERGE = 64

	/**
	 * When we get into galloping mode, we stay there until both runs win less
	 * often than c_MIN_GALLOP consecutive times.
	 */
	_MIN_GALLOP = 7

	/**
	 * Maximum initial size of tmp array, which is used for merging.  The array
	 * can grow to accommodate demand.
	 *
	 * Unlike Tim's original C version, we do not allocate this much storage
	 * when sorting smaller arrays.  This change was required for performance.
	 */
	_INITIAL_TMP_STORAGE_LENGTH = 256
)

// Delegate type that sorting uses as a comparator
type LessThan func(a, b interface{}) bool

type timSortHandler struct {

	/**
	 * The array being sorted.
	 */
	a []interface{}

	/**
	 * The comparator for this sort.
	 */
	lt LessThan

	/**
	 * This controls when we get *into* galloping mode.  It is initialized
	 * to c_MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
	 * random data, and lower for highly structured data.
	 */
	minGallop int

	/**
	 * Temp storage for merges.
	 */
	tmp []interface{} // Actual runtime type will be Object[], regardless of T

	/**
	 * A stack of pending runs yet to be merged.  Run i starts at
	 * address base[i] and extends for len[i] elements.  It's always
	 * true (so long as the indices are in bounds) that:
	 *
	 *     runBase[i] + runLen[i] == runBase[i + 1]
	 *
	 * so we could cut the storage for this, but it's a minor amount,
	 * and keeping all the info explicit simplifies the code.
	 */
	stackSize int // Number of pending runs on stack
	runBase   []int
	runLen    []int
}

/**
 * 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
 */
func newTimSort(a []interface{}, lt LessThan) (self *timSortHandler) {
	self = new(timSortHandler)

	self.a = a
	self.lt = lt
	self.minGallop = _MIN_GALLOP
	self.stackSize = 0

	// Allocate temp storage (which may be increased later if necessary)
	len := len(a)

	tmpSize := _INITIAL_TMP_STORAGE_LENGTH
	if len < 2*tmpSize {
		tmpSize = len / 2
	}

	self.tmp = make([]interface{}, tmpSize)

	/*
	 * 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 c_MIN_MERGE is decreased.  See
	 * the c_MIN_MERGE declaration above for more information.
	 */
	// mk: confirmed that for small sorts this optimization gives measurable (albeit small)
	// performance enhancement
	stackLen := 40
	if len < 120 {
		stackLen = 5
	} else if len < 1542 {
		stackLen = 10
	} else if len < 119151 {
		stackLen = 19
	}

	self.runBase = make([]int, stackLen)
	self.runLen = make([]int, stackLen)

	return self
}

// Sorts an array using the provided comparator
func Sort(a []interface{}, lt LessThan) (err error) {
	lo := 0
	hi := len(a)
	nRemaining := hi

	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 {
		initRunLen, err := countRunAndMakeAscending(a, lo, hi, lt)
		if err != nil {
			return err
		}

		return binarySort(a, lo, hi, lo+initRunLen, lt)
	}

	/**
	 * 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.
	 */

	ts := newTimSort(a, lt)
	minRun, err := minRunLength(nRemaining)
	if err != nil {
		return
	}
	for {
		// Identify next run
		runLen, err := countRunAndMakeAscending(a, lo, hi, lt)
		if err != nil {
			return err
		}

		// If run is short, extend to min(minRun, nRemaining)
		if runLen < minRun {
			force := minRun
			if nRemaining <= minRun {
				force = nRemaining
			}
			if err = binarySort(a, lo, lo+force, lo+runLen, lt); err != nil {
				return err
			}
			runLen = force
		}

		// Push run onto pending-run stack, and maybe merge
		ts.pushRun(lo, runLen)
		if err = ts.mergeCollapse(); err != nil {
			return err
		}

		// Advance to find next run
		lo += runLen
		nRemaining -= runLen
		if nRemaining == 0 {
			break
		}
	}

	// Merge all remaining runs to complete sort
	if lo != hi {
		return errors.New("lo==hi!")
	}

	if err = ts.mergeForceCollapse(); err != nil {
		return
	}
	if ts.stackSize != 1 {
		return errors.New("ts.stackSize != 1")
	}
	return
}

/**
 * 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
 */
func binarySort(a []interface{}, lo, hi, start int, lt LessThan) (err error) {
	if lo > start || start > hi {
		return errors.New("lo <= start && start <= hi")
	}

	if start == lo {
		start++
	}

	for ; start < hi; start++ {
		pivot := a[start]

		// Set left (and right) to the index where a[start] (pivot) belongs
		left := lo
		right := start

		if left > right {
			return errors.New("left <= right")
		}

		/*
		 * Invariants:
		 *   pivot >= all in [lo, left).
		 *   pivot <  all in [right, start).
		 */
		for left < right {
			mid := (left + right) / 2
			if lt(pivot, a[mid]) {
				right = mid
			} else {
				left = mid + 1
			}
		}

		if left != right {
			return errors.New("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 to make room for pivot.
		 */
		n := start - left // The number of elements to move
		// just an optimization for copy in default case
		if n <= 2 {
			if n == 2 {
				a[left+2] = a[left+1]
			}
			if n > 0 {
				a[left+1] = a[left]
			}
		} else {
			copy(a[left+1:], a[left:left+n])
		}
		a[left] = pivot
	}
	return
}

/**
  * 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
*/
func countRunAndMakeAscending(a []interface{}, lo, hi int, lt LessThan) (int, error) {

	if lo >= hi {
		return 0, errors.New("lo < hi")
	}

	runHi := lo + 1
	if runHi == hi {
		return 1, nil
	}

	// Find end of run, and reverse range if descending
	if lt(a[runHi], a[lo]) { // Descending
		runHi++

		for runHi < hi && lt(a[runHi], a[runHi-1]) {
			runHi++
		}
		reverseRange(a, lo, runHi)
	} else { // Ascending
		for runHi < hi && !lt(a[runHi], a[runHi-1]) {
			runHi++
		}
	}

	return runHi - lo, nil
}

/**
 * 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
 */
func reverseRange(a []interface{}, lo, hi int) {
	hi--
	for lo < hi {
		a[lo], a[hi] = a[hi], a[lo]
		lo++
		hi--
	}
}

/**
 * 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 < c_MIN_MERGE, return n (it's too small to bother with fancy stuff).
 *  Else if n is an exact power of 2, return c_MIN_MERGE/2.
 *  Else return an int k, c_MIN_MERGE/2 <= k <= c_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
 */
func minRunLength(n int) (int, error) {
	if n < 0 {
		return 0, errors.New("n >= 0")
	}
	r := 0 // Becomes 1 if any 1 bits are shifted off
	for n >= _MIN_MERGE {
		r |= (n & 1)
		n >>= 1
	}
	return n + r, nil
}

/**
 * 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
 */
func (self *timSortHandler) pushRun(runBase, runLen int) {
	self.runBase[self.stackSize] = runBase
	self.runLen[self.stackSize] = runLen
	self.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.
 */
func (self *timSortHandler) mergeCollapse() (err error) {
	for self.stackSize > 1 {
		n := self.stackSize - 2
		if n > 0 && self.runLen[n-1] <= self.runLen[n]+self.runLen[n+1] {
			if self.runLen[n-1] < self.runLen[n+1] {
				n--
			}
			if err = self.mergeAt(n); err != nil {
				return
			}
		} else if self.runLen[n] <= self.runLen[n+1] {
			if err = self.mergeAt(n); err != nil {
				return
			}
		} else {
			break // Invariant is established
		}
	}
	return
}

/**
 * Merges all runs on the stack until only one remains.  This method is
 * called once, to complete the sort.
 */
func (self *timSortHandler) mergeForceCollapse() (err error) {
	for self.stackSize > 1 {
		n := self.stackSize - 2
		if n > 0 && self.runLen[n-1] < self.runLen[n+1] {
			n--
		}
		if err = self.mergeAt(n); err != nil {
			return
		}
	}
	return
}

/**
 * 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
 */
func (self *timSortHandler) mergeAt(i int) (err error) {
	if self.stackSize < 2 {
		return errors.New("stackSize >= 2")
	}

	if i < 0 {
		return errors.New(" i >= 0")
	}

	if i != self.stackSize-2 && i != self.stackSize-3 {
		return errors.New("if i == stackSize - 2 || i == stackSize - 3")
	}

	base1 := self.runBase[i]
	len1 := self.runLen[i]
	base2 := self.runBase[i+1]
	len2 := self.runLen[i+1]

	if len1 <= 0 || len2 <= 0 {
		return errors.New("len1 > 0 && len2 > 0")
	}

	if base1+len1 != base2 {
		return errors.New("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.
	 */
	self.runLen[i] = len1 + len2
	if i == self.stackSize-3 {
		self.runBase[i+1] = self.runBase[i+2]
		self.runLen[i+1] = self.runLen[i+2]
	}
	self.stackSize--

	/*
	 * Find where the first element of run2 goes in run1. Prior elements
	 * in run1 can be ignored (because they're already in place).
	 */
	k, err := gallopRight(self.a[base2], self.a, base1, len1, 0, self.lt)
	if err != nil {
		return err
	}
	if k < 0 {
		return errors.New(" 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, err = gallopLeft(self.a[base1+len1-1], self.a, base2, len2, len2-1, self.lt)
	if err != nil {
		return
	}
	if len2 < 0 {
		return errors.New(" len2 >= 0;")
	}
	if len2 == 0 {
		return
	}

	// Merge remaining runs, using tmp array with min(len1, len2) elements
	if len1 <= len2 {
		err = self.mergeLo(base1, len1, base2, len2)
		if err != nil {
			return errors.New(fmt.Sprintf("mergeLo: %v", err))
		}
	} else {
		err = self.mergeHi(base1, len1, base2, len2)
		if err != nil {
			return errors.New(fmt.Sprintf("mergeHi: %v", err))
		}
	}
	return
}

/**
 * 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.
 */
func gallopLeft(key interface{}, a []interface{}, base, len, hint int, c LessThan) (int, error) {
	if len <= 0 || hint < 0 || hint >= len {
		return 0, errors.New(" len > 0 && hint >= 0 && hint < len;")
	}
	lastOfs := 0
	ofs := 1

	if c(a[base+hint], key) {
		// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
		maxOfs := len - hint
		for ofs < maxOfs && c(a[base+hint+ofs], key) {
			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]
		maxOfs := hint + 1
		for ofs < maxOfs && !c(a[base+hint-ofs], key) {
			lastOfs = ofs
			ofs = (ofs << 1) + 1
			if ofs <= 0 { // int overflow
				ofs = maxOfs
			}
		}
		if ofs > maxOfs {
			ofs = maxOfs
		}

		// Make offsets relative to base
		tmp := lastOfs
		lastOfs = hint - ofs
		ofs = hint - tmp
	}

	if -1 > lastOfs || lastOfs >= ofs || ofs > len {
		return 0, errors.New(" -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++
	for lastOfs < ofs {
		m := lastOfs + (ofs-lastOfs)/2

		if c(a[base+m], key) {
			lastOfs = m + 1 // a[base + m] < key
		} else {
			ofs = m // key <= a[base + m]
		}
	}

	if lastOfs != ofs {
		return 0, errors.New(" lastOfs == ofs") // so a[base + ofs - 1] < key <= a[base + ofs]
	}
	return ofs, nil
}

/**
 * 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]
 */
func gallopRight(key interface{}, a []interface{}, base, len, hint int, c LessThan) (int, error) {
	if len <= 0 || hint < 0 || hint >= len {
		return 0, errors.New(" len > 0 && hint >= 0 && hint < len;")
	}

	ofs := 1
	lastOfs := 0
	if c(key, a[base+hint]) {
		// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
		maxOfs := hint + 1
		for ofs < maxOfs && c(key, a[base+hint-ofs]) {
			lastOfs = ofs
			ofs = (ofs << 1) + 1
			if ofs <= 0 { // int overflow
				ofs = maxOfs
			}
		}
		if ofs > maxOfs {
			ofs = maxOfs
		}

		// Make offsets relative to b
		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]
		maxOfs := len - hint
		for ofs < maxOfs && !c(key, a[base+hint+ofs]) {
			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 -1 > lastOfs || lastOfs >= ofs || ofs > len {
		return 0, errors.New("-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++
	for lastOfs < ofs {
		m := lastOfs + (ofs-lastOfs)/2

		if c(key, a[base+m]) {
			ofs = m // key < a[b + m]
		} else {
			lastOfs = m + 1 // a[b + m] <= key
		}
	}
	if lastOfs != ofs {
		return 0, errors.New(" lastOfs == ofs") // so a[b + ofs - 1] <= key < a[b + ofs]
	}
	return ofs, nil
}

/**
 * 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)
 */
func (self *timSortHandler) mergeLo(base1, len1, base2, len2 int) (err error) {
	if len1 <= 0 || len2 <= 0 || base1+len1 != base2 {
		return errors.New(" len1 > 0 && len2 > 0 && base1 + len1 == base2")
	}

	// Copy first run into temp array
	a := self.a // For performance
	tmp := self.ensureCapacity(len1)

	copy(tmp, a[base1:base1+len1])

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

	// Move first element of second run and deal with degenerate cases
	a[dest] = a[cursor2]
	dest++
	cursor2++
	len2--
	if len2 == 0 {
		copy(a[dest:dest+len1], tmp)
		return
	}
	if len1 == 1 {
		copy(a[dest:dest+len2], a[cursor2:cursor2+len2])
		a[dest+len2] = tmp[cursor1] // Last elt of run 1 to end of merge
		return
	}

	lt := self.lt               // Use local variable for performance
	minGallop := self.minGallop //  "    "       "     "      "

outer:
	for {
		count1 := 0 // Number of times in a row that first run won
		count2 := 0 // Number of times in a row that second run won

		/*
		 * Do the straightforward thing until (if ever) one run starts
		 * winning consistently.
		 */
		for {
			if len1 <= 1 || len2 <= 0 {
				return errors.New(" len1 > 1 && len2 > 0")
			}

			if lt(a[cursor2], tmp[cursor1]) {
				a[dest] = a[cursor2]
				dest++
				cursor2++
				count2++
				count1 = 0
				len2--
				if len2 == 0 {
					break outer
				}
			} else {
				a[dest] = tmp[cursor1]
				dest++
				cursor1++
				count1++
				count2 = 0
				len1--
				if len1 == 1 {
					break outer
				}
			}
			if (count1 | count2) >= minGallop {
				break
			}
		}

		/*
		 * 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.
		 */
		for {
			if len1 <= 1 || len2 <= 0 {
				return errors.New("len1 > 1 && len2 > 0")
			}
			count1, err = gallopRight(a[cursor2], tmp, cursor1, len1, 0, lt)
			if err != nil {
				return
			}
			if count1 != 0 {
				copy(a[dest:dest+count1], tmp[cursor1:cursor1+count1])
				dest += count1
				cursor1 += count1
				len1 -= count1
				if len1 <= 1 { // len1 == 1 || len1 == 0
					break outer
				}
			}
			a[dest] = a[cursor2]
			dest++
			cursor2++
			len2--
			if len2 == 0 {
				break outer
			}

			count2, err = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, lt)
			if err != nil {
				return
			}
			if count2 != 0 {
				copy(a[dest:dest+count2], a[cursor2:cursor2+count2])
				dest += count2
				cursor2 += count2
				len2 -= count2
				if len2 == 0 {
					break outer
				}
			}
			a[dest] = tmp[cursor1]
			dest++
			cursor1++
			len1--
			if len1 == 1 {
				break outer
			}
			minGallop--
			if count1 < _MIN_GALLOP && count2 < _MIN_GALLOP {
				break
			}
		}
		if minGallop < 0 {
			minGallop = 0
		}
		minGallop += 2 // Penalize for leaving gallop mode
	} // End of "outer" loop

	if minGallop < 1 {
		minGallop = 1
	}
	self.minGallop = minGallop // Write back to field

	if len1 == 1 {

		if len2 <= 0 {
			return errors.New(" len2 > 0;")
		}
		copy(a[dest:dest+len2], a[cursor2:cursor2+len2])
		a[dest+len2] = tmp[cursor1] //  Last elt of run 1 to end of merge
	} else if len1 == 0 {
		return errors.New("Comparison method violates its general contract!")
	} else {
		if len2 != 0 {
			return errors.New("len2 == 0;")
		}
		if len1 <= 1 {
			return errors.New(" len1 > 1;")
		}

		copy(a[dest:dest+len1], tmp[cursor1:cursor1+len1])
	}
	return
}

/**
 * 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)
 */
func (self *timSortHandler) mergeHi(base1, len1, base2, len2 int) (err error) {
	if len1 <= 0 || len2 <= 0 || base1+len1 != base2 {
		return errors.New("len1 > 0 && len2 > 0 && base1 + len1 == base2;")
	}

	// Copy second run into temp array
	a := self.a // For performance
	tmp := self.ensureCapacity(len2)

	copy(tmp, a[base2:base2+len2])

	cursor1 := base1 + len1 - 1 // Indexes into a
	cursor2 := len2 - 1         // Indexes into tmp array
	dest := base2 + len2 - 1    // Indexes into a

	// Move last element of first run and deal with degenerate cases
	a[dest] = a[cursor1]
	dest--
	cursor1--
	len1--
	if len1 == 0 {
		dest -= len2 - 1
		copy(a[dest:dest+len2], tmp)
		return
	}
	if len2 == 1 {
		dest -= len1 - 1
		cursor1 -= len1 - 1
		copy(a[dest:dest+len1], a[cursor1:cursor1+len1])
		a[dest-1] = tmp[cursor2]
		return
	}

	lt := self.lt               // Use local variable for performance
	minGallop := self.minGallop //  "    "       "     "      "

outer:
	for {
		count1 := 0 // Number of times in a row that first run won
		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.
		 */
		for {
			if len1 <= 0 || len2 <= 1 {
				return errors.New(" len1 > 0 && len2 > 1;")
			}
			if lt(tmp[cursor2], a[cursor1]) {
				a[dest] = a[cursor1]
				dest--
				cursor1--
				count1++
				count2 = 0
				len1--
				if len1 == 0 {
					break outer
				}
			} else {
				a[dest] = tmp[cursor2]
				dest--
				cursor2--
				count2++
				count1 = 0
				len2--
				if len2 == 1 {
					break outer
				}
			}
			if (count1 | count2) >= minGallop {
				break
			}
		}

		/*
		 * 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.
		 */
		for {
			if len1 <= 0 || len2 <= 1 {
				return errors.New(" len1 > 0 && len2 > 1;")
			}
			if gr, err := gallopRight(tmp[cursor2], a, base1, len1, len1-1, lt); err == nil {
				count1 = len1 - gr
			} else {
				return err
			}
			if count1 != 0 {
				dest -= count1
				cursor1 -= count1
				len1 -= count1
				copy(a[dest+1:dest+1+count1], a[cursor1+1:cursor1+1+count1])
				if len1 == 0 {
					break outer
				}
			}
			a[dest] = tmp[cursor2]
			dest--
			cursor2--
			len2--
			if len2 == 1 {
				break outer
			}

			if gl, err := gallopLeft(a[cursor1], tmp, 0, len2, len2-1, lt); err == nil {
				count2 = len2 - gl
			} else {
				return err
			}
			if count2 != 0 {
				dest -= count2
				cursor2 -= count2
				len2 -= count2
				copy(a[dest+1:dest+1+count2], tmp[cursor2+1:cursor2+1+count2])
				if len2 <= 1 { // len2 == 1 || len2 == 0
					break outer
				}
			}
			a[dest] = a[cursor1]
			dest--
			cursor1--
			len1--
			if len1 == 0 {
				break outer
			}
			minGallop--

			if count1 < _MIN_GALLOP && count2 < _MIN_GALLOP {
				break
			}
		}
		if minGallop < 0 {
			minGallop = 0
		}
		minGallop += 2 // Penalize for leaving gallop mode
	} // End of "outer" loop

	if minGallop < 1 {
		minGallop = 1
	}

	self.minGallop = minGallop // Write back to field

	if len2 == 1 {
		if len1 <= 0 {
			return errors.New(" len1 > 0;")
		}
		dest -= len1
		cursor1 -= len1

		copy(a[dest+1:dest+1+len1], a[cursor1+1:cursor1+1+len1])
		a[dest] = tmp[cursor2] // Move first elt of run2 to front of merge
	} else if len2 == 0 {
		return errors.New("Comparison method violates its general contract!")
	} else {
		if len1 != 0 {
			return errors.New("len1 == 0;")
		}

		if len2 <= 0 {
			return errors.New(" len2 > 0;")
		}

		copy(a[dest-(len2-1):dest+1], tmp)
	}
	return
}

/**
 * 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
 */
func (self *timSortHandler) ensureCapacity(minCapacity int) []interface{} {
	if len(self.tmp) < minCapacity {
		// Compute smallest power of 2 > minCapacity
		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 {
			ns := len(self.a) / 2
			if ns < newSize {
				newSize = ns
			}
		}

		self.tmp = make([]interface{}, newSize)
	}

	return self.tmp
}