use quickSort from sort.Sort (#207)

About 5-15% speedup. More on smaller blocks.

Author: luyu6056

flate/huffman_code.go

@@ -7,7 +7,6 @@ package flate
 import (
 	"math"
 	"math/bits"
-	"sort"
 )
 
 const (
@@ -25,8 +24,6 @@ type huffmanEncoder struct {
 	codes     []hcode
 	freqcache []literalNode
 	bitCount  [17]int32
-	lns       byLiteral // stored to avoid repeated allocation in generate
-	lfs       byFreq    // stored to avoid repeated allocation in generate
 }
 
 type literalNode struct {
@@ -270,7 +267,7 @@ func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalN
 		// assigned in literal order (not frequency order).
 		chunk := list[len(list)-int(bits):]
 
-		h.lns.sort(chunk)
+		sortByLiteral(chunk)
 		for _, node := range chunk {
 			h.codes[node.literal] = hcode{code: reverseBits(code, uint8(n)), len: uint16(n)}
 			code++
@@ -315,47 +312,14 @@ func (h *huffmanEncoder) generate(freq []uint16, maxBits int32) {
 		}
 		return
 	}
-	h.lfs.sort(list)
+	sortByFreq(list)
 
 	// Get the number of literals for each bit count
 	bitCount := h.bitCounts(list, maxBits)
 	// And do the assignment
 	h.assignEncodingAndSize(bitCount, list)
 }
 
-type byLiteral []literalNode
-
-func (s *byLiteral) sort(a []literalNode) {
-	*s = byLiteral(a)
-	sort.Sort(s)
-}
-
-func (s byLiteral) Len() int { return len(s) }
-
-func (s byLiteral) Less(i, j int) bool {
-	return s[i].literal < s[j].literal
-}
-
-func (s byLiteral) Swap(i, j int) { s[i], s[j] = s[j], s[i] }
-
-type byFreq []literalNode
-
-func (s *byFreq) sort(a []literalNode) {
-	*s = byFreq(a)
-	sort.Sort(s)
-}
-
-func (s byFreq) Len() int { return len(s) }
-
-func (s byFreq) Less(i, j int) bool {
-	if s[i].freq == s[j].freq {
-		return s[i].literal < s[j].literal
-	}
-	return s[i].freq < s[j].freq
-}
-
-func (s byFreq) Swap(i, j int) { s[i], s[j] = s[j], s[i] }
-
 // histogramSize accumulates a histogram of b in h.
 // An estimated size in bits is returned.
 // Unassigned values are assigned '1' in the histogram.

flate/huffman_sortByFreq.go

@@ -0,0 +1,178 @@
+// Copyright 2009 The Go Authors. All rights reserved.
+// Use of this source code is governed by a BSD-style
+// license that can be found in the LICENSE file.
+
+package flate
+
+// Sort sorts data.
+// It makes one call to data.Len to determine n, and O(n*log(n)) calls to
+// data.Less and data.Swap. The sort is not guaranteed to be stable.
+func sortByFreq(data []literalNode) {
+	n := len(data)
+	quickSortByFreq(data, 0, n, maxDepth(n))
+}
+
+func quickSortByFreq(data []literalNode, a, b, maxDepth int) {
+	for b-a > 12 { // Use ShellSort for slices <= 12 elements
+		if maxDepth == 0 {
+			heapSort(data, a, b)
+			return
+		}
+		maxDepth--
+		mlo, mhi := doPivotByFreq(data, a, b)
+		// Avoiding recursion on the larger subproblem guarantees
+		// a stack depth of at most lg(b-a).
+		if mlo-a < b-mhi {
+			quickSortByFreq(data, a, mlo, maxDepth)
+			a = mhi // i.e., quickSortByFreq(data, mhi, b)
+		} else {
+			quickSortByFreq(data, mhi, b, maxDepth)
+			b = mlo // i.e., quickSortByFreq(data, a, mlo)
+		}
+	}
+	if b-a > 1 {
+		// Do ShellSort pass with gap 6
+		// It could be written in this simplified form cause b-a <= 12
+		for i := a + 6; i < b; i++ {
+			if data[i].freq == data[i-6].freq && data[i].literal < data[i-6].literal || data[i].freq < data[i-6].freq {
+				data[i], data[i-6] = data[i-6], data[i]
+			}
+		}
+		insertionSortByFreq(data, a, b)
+	}
+}
+
+// siftDownByFreq implements the heap property on data[lo, hi).
+// first is an offset into the array where the root of the heap lies.
+func siftDownByFreq(data []literalNode, lo, hi, first int) {
+	root := lo
+	for {
+		child := 2*root + 1
+		if child >= hi {
+			break
+		}
+		if child+1 < hi && (data[first+child].freq == data[first+child+1].freq && data[first+child].literal < data[first+child+1].literal || data[first+child].freq < data[first+child+1].freq) {
+			child++
+		}
+		if data[first+root].freq == data[first+child].freq && data[first+root].literal > data[first+child].literal || data[first+root].freq > data[first+child].freq {
+			return
+		}
+		data[first+root], data[first+child] = data[first+child], data[first+root]
+		root = child
+	}
+}
+func doPivotByFreq(data []literalNode, lo, hi int) (midlo, midhi int) {
+	m := int(uint(lo+hi) >> 1) // Written like this to avoid integer overflow.
+	if hi-lo > 40 {
+		// Tukey's ``Ninther,'' median of three medians of three.
+		s := (hi - lo) / 8
+		medianOfThreeSortByFreq(data, lo, lo+s, lo+2*s)
+		medianOfThreeSortByFreq(data, m, m-s, m+s)
+		medianOfThreeSortByFreq(data, hi-1, hi-1-s, hi-1-2*s)
+	}
+	medianOfThreeSortByFreq(data, lo, m, hi-1)
+
+	// Invariants are:
+	//	data[lo] = pivot (set up by ChoosePivot)
+	//	data[lo < i < a] < pivot
+	//	data[a <= i < b] <= pivot
+	//	data[b <= i < c] unexamined
+	//	data[c <= i < hi-1] > pivot
+	//	data[hi-1] >= pivot
+	pivot := lo
+	a, c := lo+1, hi-1
+
+	for ; a < c && (data[a].freq == data[pivot].freq && data[a].literal < data[pivot].literal || data[a].freq < data[pivot].freq); a++ {
+	}
+	b := a
+	for {
+		for ; b < c && (data[pivot].freq == data[b].freq && data[pivot].literal > data[b].literal || data[pivot].freq > data[b].freq); b++ { // data[b] <= pivot
+		}
+		for ; b < c && (data[pivot].freq == data[c-1].freq && data[pivot].literal < data[c-1].literal || data[pivot].freq < data[c-1].freq); c-- { // data[c-1] > pivot
+		}
+		if b >= c {
+			break
+		}
+		// data[b] > pivot; data[c-1] <= pivot
+		data[b], data[c-1] = data[c-1], data[b]
+		b++
+		c--
+	}
+	// If hi-c<3 then there are duplicates (by property of median of nine).
+	// Let's be a bit more conservative, and set border to 5.
+	protect := hi-c < 5
+	if !protect && hi-c < (hi-lo)/4 {
+		// Lets test some points for equality to pivot
+		dups := 0
+		if data[pivot].freq == data[hi-1].freq && data[pivot].literal > data[hi-1].literal || data[pivot].freq > data[hi-1].freq { // data[hi-1] = pivot
+			data[c], data[hi-1] = data[hi-1], data[c]
+			c++
+			dups++
+		}
+		if data[b-1].freq == data[pivot].freq && data[b-1].literal > data[pivot].literal || data[b-1].freq > data[pivot].freq { // data[b-1] = pivot
+			b--
+			dups++
+		}
+		// m-lo = (hi-lo)/2 > 6
+		// b-lo > (hi-lo)*3/4-1 > 8
+		// ==> m < b ==> data[m] <= pivot
+		if data[m].freq == data[pivot].freq && data[m].literal > data[pivot].literal || data[m].freq > data[pivot].freq { // data[m] = pivot
+			data[m], data[b-1] = data[b-1], data[m]
+			b--
+			dups++
+		}
+		// if at least 2 points are equal to pivot, assume skewed distribution
+		protect = dups > 1
+	}
+	if protect {
+		// Protect against a lot of duplicates
+		// Add invariant:
+		//	data[a <= i < b] unexamined
+		//	data[b <= i < c] = pivot
+		for {
+			for ; a < b && (data[b-1].freq == data[pivot].freq && data[b-1].literal > data[pivot].literal || data[b-1].freq > data[pivot].freq); b-- { // data[b] == pivot
+			}
+			for ; a < b && (data[a].freq == data[pivot].freq && data[a].literal < data[pivot].literal || data[a].freq < data[pivot].freq); a++ { // data[a] < pivot
+			}
+			if a >= b {
+				break
+			}
+			// data[a] == pivot; data[b-1] < pivot
+			data[a], data[b-1] = data[b-1], data[a]
+			a++
+			b--
+		}
+	}
+	// Swap pivot into middle
+	data[pivot], data[b-1] = data[b-1], data[pivot]
+	return b - 1, c
+}
+
+// Insertion sort
+func insertionSortByFreq(data []literalNode, a, b int) {
+	for i := a + 1; i < b; i++ {
+		for j := i; j > a && (data[j].freq == data[j-1].freq && data[j].literal < data[j-1].literal || data[j].freq < data[j-1].freq); j-- {
+			data[j], data[j-1] = data[j-1], data[j]
+		}
+	}
+}
+
+// quickSortByFreq, loosely following Bentley and McIlroy,
+// ``Engineering a Sort Function,'' SP&E November 1993.
+
+// medianOfThreeSortByFreq moves the median of the three values data[m0], data[m1], data[m2] into data[m1].
+func medianOfThreeSortByFreq(data []literalNode, m1, m0, m2 int) {
+	// sort 3 elements
+	if data[m1].freq == data[m0].freq && data[m1].literal < data[m0].literal || data[m1].freq < data[m0].freq {
+		data[m1], data[m0] = data[m0], data[m1]
+	}
+	// data[m0] <= data[m1]
+	if data[m2].freq == data[m1].freq && data[m2].literal < data[m1].literal || data[m2].freq < data[m1].freq {
+		data[m2], data[m1] = data[m1], data[m2]
+		// data[m0] <= data[m2] && data[m1] < data[m2]
+		if data[m1].freq == data[m0].freq && data[m1].literal < data[m0].literal || data[m1].freq < data[m0].freq {
+			data[m1], data[m0] = data[m0], data[m1]
+		}
+	}
+	// now data[m0] <= data[m1] <= data[m2]
+}