1 // Copyright 2009 The Go Authors. All rights reserved.
2 // Use of this source code is governed by a BSD-style
3 // license that can be found in the LICENSE file.
12 type huffmanEncoder struct {
17 type literalNode struct {
23 // The sum of the leaves in this tree
26 // The number of literals to the left of this item at this level
29 // The right child of this chain in the previous level.
33 type levelInfo struct {
34 // Our level. for better printing
37 // The most recent chain generated for this level
40 // The frequency of the next character to add to this level
43 // The frequency of the next pair (from level below) to add to this level.
44 // Only valid if the "needed" value of the next lower level is 0.
47 // The number of chains remaining to generate for this level before moving
48 // up to the next level
51 // The levelInfo for level+1
54 // The levelInfo for level-1
58 func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxInt32} }
60 func newHuffmanEncoder(size int) *huffmanEncoder {
61 return &huffmanEncoder{make([]uint8, size), make([]uint16, size)}
64 // Generates a HuffmanCode corresponding to the fixed literal table
65 func generateFixedLiteralEncoding() *huffmanEncoder {
66 h := newHuffmanEncoder(maxLit)
67 codeBits := h.codeBits
70 for ch = 0; ch < maxLit; ch++ {
75 // size 8, 000110000 .. 10111111
80 // size 9, 110010000 .. 111111111
85 // size 7, 0000000 .. 0010111
90 // size 8, 11000000 .. 11000111
95 code[ch] = reverseBits(bits, size)
100 func generateFixedOffsetEncoding() *huffmanEncoder {
101 h := newHuffmanEncoder(30)
102 codeBits := h.codeBits
104 for ch := uint16(0); ch < 30; ch++ {
106 code[ch] = reverseBits(ch, 5)
111 var fixedLiteralEncoding *huffmanEncoder = generateFixedLiteralEncoding()
112 var fixedOffsetEncoding *huffmanEncoder = generateFixedOffsetEncoding()
114 func (h *huffmanEncoder) bitLength(freq []int32) int64 {
116 for i, f := range freq {
118 total += int64(f) * int64(h.codeBits[i])
124 // Generate elements in the chain using an iterative algorithm.
125 func (h *huffmanEncoder) generateChains(top *levelInfo, list []literalNode) {
132 if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 {
133 // We've run out of both leafs and pairs.
134 // End all calculations for this level.
135 // To m sure we never come back to this level or any lower level,
136 // set nextPairFreq impossibly large.
140 l.nextPairFreq = math.MaxInt32
144 prevFreq := l.lastChain.freq
145 if l.nextCharFreq < l.nextPairFreq {
146 // The next item on this row is a leaf node.
147 n := l.lastChain.leafCount + 1
148 l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up}
149 l.nextCharFreq = list[n].freq
151 // The next item on this row is a pair from the previous row.
152 // nextPairFreq isn't valid until we generate two
153 // more values in the level below
154 l.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain}
158 if l.needed--; l.needed == 0 {
159 // We've done everything we need to do for this level.
160 // Continue calculating one level up. Fill in nextPairFreq
161 // of that level with the sum of the two nodes we've just calculated on
168 up.nextPairFreq = prevFreq + l.lastChain.freq
171 // If we stole from below, move down temporarily to replenish it.
172 for l.down.needed > 0 {
179 // Return the number of literals assigned to each bit size in the Huffman encoding
181 // This method is only called when list.length >= 3
182 // The cases of 0, 1, and 2 literals are handled by special case code.
184 // list An array of the literals with non-zero frequencies
185 // and their associated frequencies. The array is in order of increasing
186 // frequency, and has as its last element a special element with frequency
188 // maxBits The maximum number of bits that should be used to encode any literal.
189 // return An integer array in which array[i] indicates the number of literals
190 // that should be encoded in i bits.
191 func (h *huffmanEncoder) bitCounts(list []literalNode, maxBits int32) []int32 {
192 n := int32(len(list))
196 // The tree can't have greater depth than n - 1, no matter what. This
197 // saves a little bit of work in some small cases
198 maxBits = minInt32(maxBits, n-1)
200 // Create information about each of the levels.
201 // A bogus "Level 0" whose sole purpose is so that
202 // level1.prev.needed==0. This makes level1.nextPairFreq
203 // be a legitimate value that never gets chosen.
204 top := &levelInfo{needed: 0}
205 chain2 := &chain{list[1].freq, 2, new(chain)}
206 for level := int32(1); level <= maxBits; level++ {
207 // For every level, the first two items are the first two characters.
208 // We initialize the levels as if we had already figured this out.
212 nextCharFreq: list[2].freq,
213 nextPairFreq: list[0].freq + list[1].freq,
218 top.nextPairFreq = math.MaxInt32
222 // We need a total of 2*n - 2 items at top level and have already generated 2.
227 if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 {
228 // We've run out of both leafs and pairs.
229 // End all calculations for this level.
230 // To m sure we never come back to this level or any lower level,
231 // set nextPairFreq impossibly large.
235 l.nextPairFreq = math.MaxInt32
239 prevFreq := l.lastChain.freq
240 if l.nextCharFreq < l.nextPairFreq {
241 // The next item on this row is a leaf node.
242 n := l.lastChain.leafCount + 1
243 l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up}
244 l.nextCharFreq = list[n].freq
246 // The next item on this row is a pair from the previous row.
247 // nextPairFreq isn't valid until we generate two
248 // more values in the level below
249 l.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain}
253 if l.needed--; l.needed == 0 {
254 // We've done everything we need to do for this level.
255 // Continue calculating one level up. Fill in nextPairFreq
256 // of that level with the sum of the two nodes we've just calculated on
263 up.nextPairFreq = prevFreq + l.lastChain.freq
266 // If we stole from below, move down temporarily to replenish it.
267 for l.down.needed > 0 {
273 // Somethings is wrong if at the end, the top level is null or hasn't used
274 // all of the leaves.
275 if top.lastChain.leafCount != n {
276 panic("top.lastChain.leafCount != n")
279 bitCount := make([]int32, maxBits+1)
281 for chain := top.lastChain; chain.up != nil; chain = chain.up {
282 // chain.leafCount gives the number of literals requiring at least "bits"
284 bitCount[bits] = chain.leafCount - chain.up.leafCount
290 // Look at the leaves and assign them a bit count and an encoding as specified
292 func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalNode) {
294 for n, bits := range bitCount {
296 if n == 0 || bits == 0 {
299 // The literals list[len(list)-bits] .. list[len(list)-bits]
300 // are encoded using "bits" bits, and get the values
301 // code, code + 1, .... The code values are
302 // assigned in literal order (not frequency order).
303 chunk := list[len(list)-int(bits):]
305 for _, node := range chunk {
306 h.codeBits[node.literal] = uint8(n)
307 h.code[node.literal] = reverseBits(code, uint8(n))
310 list = list[0 : len(list)-int(bits)]
314 // Update this Huffman Code object to be the minimum code for the specified frequency count.
316 // freq An array of frequencies, in which frequency[i] gives the frequency of literal i.
317 // maxBits The maximum number of bits to use for any literal.
318 func (h *huffmanEncoder) generate(freq []int32, maxBits int32) {
319 list := make([]literalNode, len(freq)+1)
320 // Number of non-zero literals
322 // Set list to be the set of all non-zero literals and their frequencies
323 for i, f := range freq {
325 list[count] = literalNode{uint16(i), f}
331 // If freq[] is shorter than codeBits[], fill rest of codeBits[] with zeros
332 h.codeBits = h.codeBits[0:len(freq)]
335 // Handle the small cases here, because they are awkward for the general case code. With
336 // two or fewer literals, everything has bit length 1.
337 for i, node := range list {
338 // "list" is in order of increasing literal value.
339 h.codeBits[node.literal] = 1
340 h.code[node.literal] = uint16(i)
346 // Get the number of literals for each bit count
347 bitCount := h.bitCounts(list, maxBits)
348 // And do the assignment
349 h.assignEncodingAndSize(bitCount, list)
352 type literalNodeSorter struct {
354 less func(i, j int) bool
357 func (s literalNodeSorter) Len() int { return len(s.a) }
359 func (s literalNodeSorter) Less(i, j int) bool {
363 func (s literalNodeSorter) Swap(i, j int) { s.a[i], s.a[j] = s.a[j], s.a[i] }
365 func sortByFreq(a []literalNode) {
366 s := &literalNodeSorter{a, func(i, j int) bool {
367 if a[i].freq == a[j].freq {
368 return a[i].literal < a[j].literal
370 return a[i].freq < a[j].freq
375 func sortByLiteral(a []literalNode) {
376 s := &literalNodeSorter{a, func(i, j int) bool { return a[i].literal < a[j].literal }}