| // 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 { | |
| if a[i].freq == a[j].freq { | |
| return a[i].literal < a[j].literal | |
| } | |
| 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) | |
| } |