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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]
+}