// 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 import ( "math" "sort" ) type huffmanEncoder struct { codeBits []uint8 code []uint16 } type literalNode struct { literal uint16 freq int32 } type chain struct { // The sum of the leaves in this tree freq int32 // The number of literals to the left of this item at this level leafCount int32 // The right child of this chain in the previous level. up *chain } type levelInfo struct { // Our level. for better printing level int32 // The most recent chain generated for this level lastChain *chain // The frequency of the next character to add to this level nextCharFreq int32 // The frequency of the next pair (from level below) to add to this level. // Only valid if the "needed" value of the next lower level is 0. nextPairFreq int32 // The number of chains remaining to generate for this level before moving // up to the next level needed int32 // The levelInfo for level+1 up *levelInfo // The levelInfo for level-1 down *levelInfo } func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxInt32} } func newHuffmanEncoder(size int) *huffmanEncoder { return &huffmanEncoder{make([]uint8, size), make([]uint16, size)} } // Generates a HuffmanCode corresponding to the fixed literal table func generateFixedLiteralEncoding() *huffmanEncoder { h := newHuffmanEncoder(maxLit) codeBits := h.codeBits code := h.code var ch uint16 for ch = 0; ch < maxLit; ch++ { var bits uint16 var size uint8 switch { case ch < 144: // size 8, 000110000 .. 10111111 bits = ch + 48 size = 8 break case ch < 256: // size 9, 110010000 .. 111111111 bits = ch + 400 - 144 size = 9 break case ch < 280: // size 7, 0000000 .. 0010111 bits = ch - 256 size = 7 break default: // size 8, 11000000 .. 11000111 bits = ch + 192 - 280 size = 8 } codeBits[ch] = size code[ch] = reverseBits(bits, size) } return h } func generateFixedOffsetEncoding() *huffmanEncoder { h := newHuffmanEncoder(30) codeBits := h.codeBits code := h.code for ch := uint16(0); ch < 30; ch++ { codeBits[ch] = 5 code[ch] = reverseBits(ch, 5) } return h } var fixedLiteralEncoding *huffmanEncoder = generateFixedLiteralEncoding() var fixedOffsetEncoding *huffmanEncoder = generateFixedOffsetEncoding() func (h *huffmanEncoder) bitLength(freq []int32) int64 { var total int64 for i, f := range freq { if f != 0 { total += int64(f) * int64(h.codeBits[i]) } } return total } // Generate elements in the chain using an iterative algorithm. func (h *huffmanEncoder) generateChains(top *levelInfo, list []literalNode) { n := len(list) list = list[0 : n+1] list[n] = maxNode() l := top for { if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 { // We've run out of both leafs and pairs. // End all calculations for this level. // To m sure we never come back to this level or any lower level, // set nextPairFreq impossibly large. l.lastChain = nil l.needed = 0 l = l.up l.nextPairFreq = math.MaxInt32 continue } prevFreq := l.lastChain.freq if l.nextCharFreq < l.nextPairFreq { // The next item on this row is a leaf node. n := l.lastChain.leafCount + 1 l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up} l.nextCharFreq = list[n].freq } else { // The next item on this row is a pair from the previous row. // nextPairFreq isn't valid until we generate two // more values in the level below l.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain} l.down.needed = 2 } if l.needed--; l.needed == 0 { // We've done everything we need to do for this level. // Continue calculating one level up. Fill in nextPairFreq // of that level with the sum of the two nodes we've just calculated on // this level. up := l.up if up == nil { // All done! return } up.nextPairFreq = prevFreq + l.lastChain.freq l = up } else { // If we stole from below, move down temporarily to replenish it. for l.down.needed > 0 { l = l.down } } } } // Return the number of literals assigned to each bit size in the Huffman encoding // // This method is only called when list.length >= 3 // The cases of 0, 1, and 2 literals are handled by special case code. // // list An array of the literals with non-zero frequencies // and their associated frequencies. The array is in order of increasing // frequency, and has as its last element a special element with frequency // MaxInt32 // maxBits The maximum number of bits that should be used to encode any literal. // return An integer array in which array[i] indicates the number of literals // that should be encoded in i bits. func (h *huffmanEncoder) bitCounts(list []literalNode, maxBits int32) []int32 { n := int32(len(list)) list = list[0 : n+1] list[n] = maxNode() // The tree can't have greater depth than n - 1, no matter what. This // saves a little bit of work in some small cases maxBits = minInt32(maxBits, n-1) // Create information about each of the levels. // A bogus "Level 0" whose sole purpose is so that // level1.prev.needed==0. This makes level1.nextPairFreq // be a legitimate value that never gets chosen. top := &levelInfo{needed: 0} chain2 := &chain{list[1].freq, 2, new(chain)} for level := int32(1); level <= maxBits; level++ { // For every level, the first two items are the first two characters. // We initialize the levels as if we had already figured this out. top = &levelInfo{ level: level, lastChain: chain2, nextCharFreq: list[2].freq, nextPairFreq: list[0].freq + list[1].freq, down: top, } top.down.up = top if level == 1 { top.nextPairFreq = math.MaxInt32 } } // We need a total of 2*n - 2 items at top level and have already generated 2. top.needed = 2*n - 4 l := top for { if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 { // We've run out of both leafs and pairs. // End all calculations for this level. // To m sure we never come back to this level or any lower level, // set nextPairFreq impossibly large. l.lastChain = nil l.needed = 0 l = l.up l.nextPairFreq = math.MaxInt32 continue } prevFreq := l.lastChain.freq if l.nextCharFreq < l.nextPairFreq { // The next item on this row is a leaf node. n := l.lastChain.leafCount + 1 l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up} l.nextCharFreq = list[n].freq } else { // The next item on this row is a pair from the previous row. // nextPairFreq isn't valid until we generate two // more values in the level below l.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain} l.down.needed = 2 } if l.needed--; l.needed == 0 { // We've done everything we need to do for this level. // Continue calculating one level up. Fill in nextPairFreq // of that level with the sum of the two nodes we've just calculated on // this level. up := l.up if up == nil { // All done! break } up.nextPairFreq = prevFreq + l.lastChain.freq l = up } else { // If we stole from below, move down temporarily to replenish it. for l.down.needed > 0 { l = l.down } } } // Somethings is wrong if at the end, the top level is null or hasn't used // all of the leaves. if top.lastChain.leafCount != n { panic("top.lastChain.leafCount != n") } bitCount := make([]int32, maxBits+1) bits := 1 for chain := top.lastChain; chain.up != nil; chain = chain.up { // chain.leafCount gives the number of literals requiring at least "bits" // bits to encode. bitCount[bits] = chain.leafCount - chain.up.leafCount bits++ } return bitCount } // Look at the leaves and assign them a bit count and an encoding as specified // in RFC 1951 3.2.2 func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalNode) { code := uint16(0) for n, bits := range bitCount { code <<= 1 if n == 0 || bits == 0 { continue } // The literals list[len(list)-bits] .. list[len(list)-bits] // are encoded using "bits" bits, and get the values // code, code + 1, .... The code values are // assigned in literal order (not frequency order). chunk := list[len(list)-int(bits):] sortByLiteral(chunk) for _, node := range chunk { h.codeBits[node.literal] = uint8(n) h.code[node.literal] = reverseBits(code, uint8(n)) code++ } list = list[0 : len(list)-int(bits)] } } // Update this Huffman Code object to be the minimum code for the specified frequency count. // // freq An array of frequencies, in which frequency[i] gives the frequency of literal i. // maxBits The maximum number of bits to use for any literal. func (h *huffmanEncoder) generate(freq []int32, maxBits int32) { list := make([]literalNode, len(freq)+1) // Number of non-zero literals count := 0 // Set list to be the set of all non-zero literals and their frequencies for i, f := range freq { if f != 0 { list[count] = literalNode{uint16(i), f} count++ } else { h.codeBits[i] = 0 } } // If freq[] is shorter than codeBits[], fill rest of codeBits[] with zeros h.codeBits = h.codeBits[0:len(freq)] list = list[0:count] if count <= 2 { // Handle the small cases here, because they are awkward for the general case code. With // two or fewer literals, everything has bit length 1. for i, node := range list { // "list" is in order of increasing literal value. h.codeBits[node.literal] = 1 h.code[node.literal] = uint16(i) } return } 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 literalNodeSorter struct { a []literalNode less func(i, j int) bool } func (s literalNodeSorter) Len() int { return len(s.a) } func (s literalNodeSorter) Less(i, j int) bool { return s.less(i, j) } func (s literalNodeSorter) Swap(i, j int) { s.a[i], s.a[j] = s.a[j], s.a[i] } func sortByFreq(a []literalNode) { s := &literalNodeSorter{a, func(i, j int) bool { return a[i].freq < a[j].freq }} sort.Sort(s) } func sortByLiteral(a []literalNode) { s := &literalNodeSorter{a, func(i, j int) bool { return a[i].literal < a[j].literal }} sort.Sort(s) }