| Abhay Kumar | a2ae599 | 2025-11-10 14:02:24 +0000 | [diff] [blame^] | 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. |
| 4 | |
| 5 | package flate |
| 6 | |
| 7 | import ( |
| 8 | "math" |
| 9 | "math/bits" |
| 10 | ) |
| 11 | |
| 12 | const ( |
| 13 | maxBitsLimit = 16 |
| 14 | // number of valid literals |
| 15 | literalCount = 286 |
| 16 | ) |
| 17 | |
| 18 | // hcode is a huffman code with a bit code and bit length. |
| 19 | type hcode uint32 |
| 20 | |
| 21 | func (h hcode) len() uint8 { |
| 22 | return uint8(h) |
| 23 | } |
| 24 | |
| 25 | func (h hcode) code64() uint64 { |
| 26 | return uint64(h >> 8) |
| 27 | } |
| 28 | |
| 29 | func (h hcode) zero() bool { |
| 30 | return h == 0 |
| 31 | } |
| 32 | |
| 33 | type huffmanEncoder struct { |
| 34 | codes []hcode |
| 35 | bitCount [17]int32 |
| 36 | |
| 37 | // Allocate a reusable buffer with the longest possible frequency table. |
| 38 | // Possible lengths are codegenCodeCount, offsetCodeCount and literalCount. |
| 39 | // The largest of these is literalCount, so we allocate for that case. |
| 40 | freqcache [literalCount + 1]literalNode |
| 41 | } |
| 42 | |
| 43 | type literalNode struct { |
| 44 | literal uint16 |
| 45 | freq uint16 |
| 46 | } |
| 47 | |
| 48 | // A levelInfo describes the state of the constructed tree for a given depth. |
| 49 | type levelInfo struct { |
| 50 | // Our level. for better printing |
| 51 | level int32 |
| 52 | |
| 53 | // The frequency of the last node at this level |
| 54 | lastFreq int32 |
| 55 | |
| 56 | // The frequency of the next character to add to this level |
| 57 | nextCharFreq int32 |
| 58 | |
| 59 | // The frequency of the next pair (from level below) to add to this level. |
| 60 | // Only valid if the "needed" value of the next lower level is 0. |
| 61 | nextPairFreq int32 |
| 62 | |
| 63 | // The number of chains remaining to generate for this level before moving |
| 64 | // up to the next level |
| 65 | needed int32 |
| 66 | } |
| 67 | |
| 68 | // set sets the code and length of an hcode. |
| 69 | func (h *hcode) set(code uint16, length uint8) { |
| 70 | *h = hcode(length) | (hcode(code) << 8) |
| 71 | } |
| 72 | |
| 73 | func newhcode(code uint16, length uint8) hcode { |
| 74 | return hcode(length) | (hcode(code) << 8) |
| 75 | } |
| 76 | |
| 77 | func reverseBits(number uint16, bitLength byte) uint16 { |
| 78 | return bits.Reverse16(number << ((16 - bitLength) & 15)) |
| 79 | } |
| 80 | |
| 81 | func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxUint16} } |
| 82 | |
| 83 | func newHuffmanEncoder(size int) *huffmanEncoder { |
| 84 | // Make capacity to next power of two. |
| 85 | c := uint(bits.Len32(uint32(size - 1))) |
| 86 | return &huffmanEncoder{codes: make([]hcode, size, 1<<c)} |
| 87 | } |
| 88 | |
| 89 | // Generates a HuffmanCode corresponding to the fixed literal table |
| 90 | func generateFixedLiteralEncoding() *huffmanEncoder { |
| 91 | h := newHuffmanEncoder(literalCount) |
| 92 | codes := h.codes |
| 93 | var ch uint16 |
| 94 | for ch = 0; ch < literalCount; ch++ { |
| 95 | var bits uint16 |
| 96 | var size uint8 |
| 97 | switch { |
| 98 | case ch < 144: |
| 99 | // size 8, 000110000 .. 10111111 |
| 100 | bits = ch + 48 |
| 101 | size = 8 |
| 102 | case ch < 256: |
| 103 | // size 9, 110010000 .. 111111111 |
| 104 | bits = ch + 400 - 144 |
| 105 | size = 9 |
| 106 | case ch < 280: |
| 107 | // size 7, 0000000 .. 0010111 |
| 108 | bits = ch - 256 |
| 109 | size = 7 |
| 110 | default: |
| 111 | // size 8, 11000000 .. 11000111 |
| 112 | bits = ch + 192 - 280 |
| 113 | size = 8 |
| 114 | } |
| 115 | codes[ch] = newhcode(reverseBits(bits, size), size) |
| 116 | } |
| 117 | return h |
| 118 | } |
| 119 | |
| 120 | func generateFixedOffsetEncoding() *huffmanEncoder { |
| 121 | h := newHuffmanEncoder(30) |
| 122 | codes := h.codes |
| 123 | for ch := range codes { |
| 124 | codes[ch] = newhcode(reverseBits(uint16(ch), 5), 5) |
| 125 | } |
| 126 | return h |
| 127 | } |
| 128 | |
| 129 | var fixedLiteralEncoding = generateFixedLiteralEncoding() |
| 130 | var fixedOffsetEncoding = generateFixedOffsetEncoding() |
| 131 | |
| 132 | func (h *huffmanEncoder) bitLength(freq []uint16) int { |
| 133 | var total int |
| 134 | for i, f := range freq { |
| 135 | if f != 0 { |
| 136 | total += int(f) * int(h.codes[i].len()) |
| 137 | } |
| 138 | } |
| 139 | return total |
| 140 | } |
| 141 | |
| 142 | func (h *huffmanEncoder) bitLengthRaw(b []byte) int { |
| 143 | var total int |
| 144 | for _, f := range b { |
| 145 | total += int(h.codes[f].len()) |
| 146 | } |
| 147 | return total |
| 148 | } |
| 149 | |
| 150 | // canReuseBits returns the number of bits or math.MaxInt32 if the encoder cannot be reused. |
| 151 | func (h *huffmanEncoder) canReuseBits(freq []uint16) int { |
| 152 | var total int |
| 153 | for i, f := range freq { |
| 154 | if f != 0 { |
| 155 | code := h.codes[i] |
| 156 | if code.zero() { |
| 157 | return math.MaxInt32 |
| 158 | } |
| 159 | total += int(f) * int(code.len()) |
| 160 | } |
| 161 | } |
| 162 | return total |
| 163 | } |
| 164 | |
| 165 | // Return the number of literals assigned to each bit size in the Huffman encoding |
| 166 | // |
| 167 | // This method is only called when list.length >= 3 |
| 168 | // The cases of 0, 1, and 2 literals are handled by special case code. |
| 169 | // |
| 170 | // list An array of the literals with non-zero frequencies |
| 171 | // |
| 172 | // and their associated frequencies. The array is in order of increasing |
| 173 | // frequency, and has as its last element a special element with frequency |
| 174 | // MaxInt32 |
| 175 | // |
| 176 | // maxBits The maximum number of bits that should be used to encode any literal. |
| 177 | // |
| 178 | // Must be less than 16. |
| 179 | // |
| 180 | // return An integer array in which array[i] indicates the number of literals |
| 181 | // |
| 182 | // that should be encoded in i bits. |
| 183 | func (h *huffmanEncoder) bitCounts(list []literalNode, maxBits int32) []int32 { |
| 184 | if maxBits >= maxBitsLimit { |
| 185 | panic("flate: maxBits too large") |
| 186 | } |
| 187 | n := int32(len(list)) |
| 188 | list = list[0 : n+1] |
| 189 | list[n] = maxNode() |
| 190 | |
| 191 | // The tree can't have greater depth than n - 1, no matter what. This |
| 192 | // saves a little bit of work in some small cases |
| 193 | if maxBits > n-1 { |
| 194 | maxBits = n - 1 |
| 195 | } |
| 196 | |
| 197 | // Create information about each of the levels. |
| 198 | // A bogus "Level 0" whose sole purpose is so that |
| 199 | // level1.prev.needed==0. This makes level1.nextPairFreq |
| 200 | // be a legitimate value that never gets chosen. |
| 201 | var levels [maxBitsLimit]levelInfo |
| 202 | // leafCounts[i] counts the number of literals at the left |
| 203 | // of ancestors of the rightmost node at level i. |
| 204 | // leafCounts[i][j] is the number of literals at the left |
| 205 | // of the level j ancestor. |
| 206 | var leafCounts [maxBitsLimit][maxBitsLimit]int32 |
| 207 | |
| 208 | // Descending to only have 1 bounds check. |
| 209 | l2f := int32(list[2].freq) |
| 210 | l1f := int32(list[1].freq) |
| 211 | l0f := int32(list[0].freq) + int32(list[1].freq) |
| 212 | |
| 213 | for level := int32(1); level <= maxBits; level++ { |
| 214 | // For every level, the first two items are the first two characters. |
| 215 | // We initialize the levels as if we had already figured this out. |
| 216 | levels[level] = levelInfo{ |
| 217 | level: level, |
| 218 | lastFreq: l1f, |
| 219 | nextCharFreq: l2f, |
| 220 | nextPairFreq: l0f, |
| 221 | } |
| 222 | leafCounts[level][level] = 2 |
| 223 | if level == 1 { |
| 224 | levels[level].nextPairFreq = math.MaxInt32 |
| 225 | } |
| 226 | } |
| 227 | |
| 228 | // We need a total of 2*n - 2 items at top level and have already generated 2. |
| 229 | levels[maxBits].needed = 2*n - 4 |
| 230 | |
| 231 | level := uint32(maxBits) |
| 232 | for level < 16 { |
| 233 | l := &levels[level] |
| 234 | if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 { |
| 235 | // We've run out of both leafs and pairs. |
| 236 | // End all calculations for this level. |
| 237 | // To make sure we never come back to this level or any lower level, |
| 238 | // set nextPairFreq impossibly large. |
| 239 | l.needed = 0 |
| 240 | levels[level+1].nextPairFreq = math.MaxInt32 |
| 241 | level++ |
| 242 | continue |
| 243 | } |
| 244 | |
| 245 | prevFreq := l.lastFreq |
| 246 | if l.nextCharFreq < l.nextPairFreq { |
| 247 | // The next item on this row is a leaf node. |
| 248 | n := leafCounts[level][level] + 1 |
| 249 | l.lastFreq = l.nextCharFreq |
| 250 | // Lower leafCounts are the same of the previous node. |
| 251 | leafCounts[level][level] = n |
| 252 | e := list[n] |
| 253 | if e.literal < math.MaxUint16 { |
| 254 | l.nextCharFreq = int32(e.freq) |
| 255 | } else { |
| 256 | l.nextCharFreq = math.MaxInt32 |
| 257 | } |
| 258 | } else { |
| 259 | // The next item on this row is a pair from the previous row. |
| 260 | // nextPairFreq isn't valid until we generate two |
| 261 | // more values in the level below |
| 262 | l.lastFreq = l.nextPairFreq |
| 263 | // Take leaf counts from the lower level, except counts[level] remains the same. |
| 264 | if true { |
| 265 | save := leafCounts[level][level] |
| 266 | leafCounts[level] = leafCounts[level-1] |
| 267 | leafCounts[level][level] = save |
| 268 | } else { |
| 269 | copy(leafCounts[level][:level], leafCounts[level-1][:level]) |
| 270 | } |
| 271 | levels[l.level-1].needed = 2 |
| 272 | } |
| 273 | |
| 274 | if l.needed--; l.needed == 0 { |
| 275 | // We've done everything we need to do for this level. |
| 276 | // Continue calculating one level up. Fill in nextPairFreq |
| 277 | // of that level with the sum of the two nodes we've just calculated on |
| 278 | // this level. |
| 279 | if l.level == maxBits { |
| 280 | // All done! |
| 281 | break |
| 282 | } |
| 283 | levels[l.level+1].nextPairFreq = prevFreq + l.lastFreq |
| 284 | level++ |
| 285 | } else { |
| 286 | // If we stole from below, move down temporarily to replenish it. |
| 287 | for levels[level-1].needed > 0 { |
| 288 | level-- |
| 289 | } |
| 290 | } |
| 291 | } |
| 292 | |
| 293 | // Somethings is wrong if at the end, the top level is null or hasn't used |
| 294 | // all of the leaves. |
| 295 | if leafCounts[maxBits][maxBits] != n { |
| 296 | panic("leafCounts[maxBits][maxBits] != n") |
| 297 | } |
| 298 | |
| 299 | bitCount := h.bitCount[:maxBits+1] |
| 300 | bits := 1 |
| 301 | counts := &leafCounts[maxBits] |
| 302 | for level := maxBits; level > 0; level-- { |
| 303 | // chain.leafCount gives the number of literals requiring at least "bits" |
| 304 | // bits to encode. |
| 305 | bitCount[bits] = counts[level] - counts[level-1] |
| 306 | bits++ |
| 307 | } |
| 308 | return bitCount |
| 309 | } |
| 310 | |
| 311 | // Look at the leaves and assign them a bit count and an encoding as specified |
| 312 | // in RFC 1951 3.2.2 |
| 313 | func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalNode) { |
| 314 | code := uint16(0) |
| 315 | for n, bits := range bitCount { |
| 316 | code <<= 1 |
| 317 | if n == 0 || bits == 0 { |
| 318 | continue |
| 319 | } |
| 320 | // The literals list[len(list)-bits] .. list[len(list)-bits] |
| 321 | // are encoded using "bits" bits, and get the values |
| 322 | // code, code + 1, .... The code values are |
| 323 | // assigned in literal order (not frequency order). |
| 324 | chunk := list[len(list)-int(bits):] |
| 325 | |
| 326 | sortByLiteral(chunk) |
| 327 | for _, node := range chunk { |
| 328 | h.codes[node.literal] = newhcode(reverseBits(code, uint8(n)), uint8(n)) |
| 329 | code++ |
| 330 | } |
| 331 | list = list[0 : len(list)-int(bits)] |
| 332 | } |
| 333 | } |
| 334 | |
| 335 | // Update this Huffman Code object to be the minimum code for the specified frequency count. |
| 336 | // |
| 337 | // freq An array of frequencies, in which frequency[i] gives the frequency of literal i. |
| 338 | // maxBits The maximum number of bits to use for any literal. |
| 339 | func (h *huffmanEncoder) generate(freq []uint16, maxBits int32) { |
| 340 | list := h.freqcache[:len(freq)+1] |
| 341 | codes := h.codes[:len(freq)] |
| 342 | // Number of non-zero literals |
| 343 | count := 0 |
| 344 | // Set list to be the set of all non-zero literals and their frequencies |
| 345 | for i, f := range freq { |
| 346 | if f != 0 { |
| 347 | list[count] = literalNode{uint16(i), f} |
| 348 | count++ |
| 349 | } else { |
| 350 | codes[i] = 0 |
| 351 | } |
| 352 | } |
| 353 | list[count] = literalNode{} |
| 354 | |
| 355 | list = list[:count] |
| 356 | if count <= 2 { |
| 357 | // Handle the small cases here, because they are awkward for the general case code. With |
| 358 | // two or fewer literals, everything has bit length 1. |
| 359 | for i, node := range list { |
| 360 | // "list" is in order of increasing literal value. |
| 361 | h.codes[node.literal].set(uint16(i), 1) |
| 362 | } |
| 363 | return |
| 364 | } |
| 365 | sortByFreq(list) |
| 366 | |
| 367 | // Get the number of literals for each bit count |
| 368 | bitCount := h.bitCounts(list, maxBits) |
| 369 | // And do the assignment |
| 370 | h.assignEncodingAndSize(bitCount, list) |
| 371 | } |
| 372 | |
| 373 | // atLeastOne clamps the result between 1 and 15. |
| 374 | func atLeastOne(v float32) float32 { |
| 375 | if v < 1 { |
| 376 | return 1 |
| 377 | } |
| 378 | if v > 15 { |
| 379 | return 15 |
| 380 | } |
| 381 | return v |
| 382 | } |
| 383 | |
| 384 | func histogram(b []byte, h []uint16) { |
| 385 | if true && len(b) >= 8<<10 { |
| 386 | // Split for bigger inputs |
| 387 | histogramSplit(b, h) |
| 388 | } else { |
| 389 | h = h[:256] |
| 390 | for _, t := range b { |
| 391 | h[t]++ |
| 392 | } |
| 393 | } |
| 394 | } |
| 395 | |
| 396 | func histogramSplit(b []byte, h []uint16) { |
| 397 | // Tested, and slightly faster than 2-way. |
| 398 | // Writing to separate arrays and combining is also slightly slower. |
| 399 | h = h[:256] |
| 400 | for len(b)&3 != 0 { |
| 401 | h[b[0]]++ |
| 402 | b = b[1:] |
| 403 | } |
| 404 | n := len(b) / 4 |
| 405 | x, y, z, w := b[:n], b[n:], b[n+n:], b[n+n+n:] |
| 406 | y, z, w = y[:len(x)], z[:len(x)], w[:len(x)] |
| 407 | for i, t := range x { |
| 408 | v0 := &h[t] |
| 409 | v1 := &h[y[i]] |
| 410 | v3 := &h[w[i]] |
| 411 | v2 := &h[z[i]] |
| 412 | *v0++ |
| 413 | *v1++ |
| 414 | *v2++ |
| 415 | *v3++ |
| 416 | } |
| 417 | } |