| // 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) |
| } |
| } |
| |
| 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] |
| } |