// 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] }