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// 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]
}
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