fermentord/vendor/github.com/klauspost/compress/flate/huffman_code.go

417 lines
11 KiB
Go

// 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"
"math/bits"
)
const (
maxBitsLimit = 16
// number of valid literals
literalCount = 286
)
// hcode is a huffman code with a bit code and bit length.
type hcode uint32
func (h hcode) len() uint8 {
return uint8(h)
}
func (h hcode) code64() uint64 {
return uint64(h >> 8)
}
func (h hcode) zero() bool {
return h == 0
}
type huffmanEncoder struct {
codes []hcode
bitCount [17]int32
// Allocate a reusable buffer with the longest possible frequency table.
// Possible lengths are codegenCodeCount, offsetCodeCount and literalCount.
// The largest of these is literalCount, so we allocate for that case.
freqcache [literalCount + 1]literalNode
}
type literalNode struct {
literal uint16
freq uint16
}
// A levelInfo describes the state of the constructed tree for a given depth.
type levelInfo struct {
// Our level. for better printing
level int32
// The frequency of the last node at this level
lastFreq int32
// 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
}
// set sets the code and length of an hcode.
func (h *hcode) set(code uint16, length uint8) {
*h = hcode(length) | (hcode(code) << 8)
}
func newhcode(code uint16, length uint8) hcode {
return hcode(length) | (hcode(code) << 8)
}
func reverseBits(number uint16, bitLength byte) uint16 {
return bits.Reverse16(number << ((16 - bitLength) & 15))
}
func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxUint16} }
func newHuffmanEncoder(size int) *huffmanEncoder {
// Make capacity to next power of two.
c := uint(bits.Len32(uint32(size - 1)))
return &huffmanEncoder{codes: make([]hcode, size, 1<<c)}
}
// Generates a HuffmanCode corresponding to the fixed literal table
func generateFixedLiteralEncoding() *huffmanEncoder {
h := newHuffmanEncoder(literalCount)
codes := h.codes
var ch uint16
for ch = 0; ch < literalCount; ch++ {
var bits uint16
var size uint8
switch {
case ch < 144:
// size 8, 000110000 .. 10111111
bits = ch + 48
size = 8
case ch < 256:
// size 9, 110010000 .. 111111111
bits = ch + 400 - 144
size = 9
case ch < 280:
// size 7, 0000000 .. 0010111
bits = ch - 256
size = 7
default:
// size 8, 11000000 .. 11000111
bits = ch + 192 - 280
size = 8
}
codes[ch] = newhcode(reverseBits(bits, size), size)
}
return h
}
func generateFixedOffsetEncoding() *huffmanEncoder {
h := newHuffmanEncoder(30)
codes := h.codes
for ch := range codes {
codes[ch] = newhcode(reverseBits(uint16(ch), 5), 5)
}
return h
}
var fixedLiteralEncoding = generateFixedLiteralEncoding()
var fixedOffsetEncoding = generateFixedOffsetEncoding()
func (h *huffmanEncoder) bitLength(freq []uint16) int {
var total int
for i, f := range freq {
if f != 0 {
total += int(f) * int(h.codes[i].len())
}
}
return total
}
func (h *huffmanEncoder) bitLengthRaw(b []byte) int {
var total int
for _, f := range b {
total += int(h.codes[f].len())
}
return total
}
// canReuseBits returns the number of bits or math.MaxInt32 if the encoder cannot be reused.
func (h *huffmanEncoder) canReuseBits(freq []uint16) int {
var total int
for i, f := range freq {
if f != 0 {
code := h.codes[i]
if code.zero() {
return math.MaxInt32
}
total += int(f) * int(code.len())
}
}
return total
}
// 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.
//
// Must be less than 16.
//
// 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 {
if maxBits >= maxBitsLimit {
panic("flate: maxBits too large")
}
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
if maxBits > n-1 {
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.
var levels [maxBitsLimit]levelInfo
// leafCounts[i] counts the number of literals at the left
// of ancestors of the rightmost node at level i.
// leafCounts[i][j] is the number of literals at the left
// of the level j ancestor.
var leafCounts [maxBitsLimit][maxBitsLimit]int32
// Descending to only have 1 bounds check.
l2f := int32(list[2].freq)
l1f := int32(list[1].freq)
l0f := int32(list[0].freq) + int32(list[1].freq)
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.
levels[level] = levelInfo{
level: level,
lastFreq: l1f,
nextCharFreq: l2f,
nextPairFreq: l0f,
}
leafCounts[level][level] = 2
if level == 1 {
levels[level].nextPairFreq = math.MaxInt32
}
}
// We need a total of 2*n - 2 items at top level and have already generated 2.
levels[maxBits].needed = 2*n - 4
level := uint32(maxBits)
for level < 16 {
l := &levels[level]
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 make sure we never come back to this level or any lower level,
// set nextPairFreq impossibly large.
l.needed = 0
levels[level+1].nextPairFreq = math.MaxInt32
level++
continue
}
prevFreq := l.lastFreq
if l.nextCharFreq < l.nextPairFreq {
// The next item on this row is a leaf node.
n := leafCounts[level][level] + 1
l.lastFreq = l.nextCharFreq
// Lower leafCounts are the same of the previous node.
leafCounts[level][level] = n
e := list[n]
if e.literal < math.MaxUint16 {
l.nextCharFreq = int32(e.freq)
} else {
l.nextCharFreq = math.MaxInt32
}
} 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.lastFreq = l.nextPairFreq
// Take leaf counts from the lower level, except counts[level] remains the same.
if true {
save := leafCounts[level][level]
leafCounts[level] = leafCounts[level-1]
leafCounts[level][level] = save
} else {
copy(leafCounts[level][:level], leafCounts[level-1][:level])
}
levels[l.level-1].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.
if l.level == maxBits {
// All done!
break
}
levels[l.level+1].nextPairFreq = prevFreq + l.lastFreq
level++
} else {
// If we stole from below, move down temporarily to replenish it.
for levels[level-1].needed > 0 {
level--
}
}
}
// Somethings is wrong if at the end, the top level is null or hasn't used
// all of the leaves.
if leafCounts[maxBits][maxBits] != n {
panic("leafCounts[maxBits][maxBits] != n")
}
bitCount := h.bitCount[:maxBits+1]
bits := 1
counts := &leafCounts[maxBits]
for level := maxBits; level > 0; level-- {
// chain.leafCount gives the number of literals requiring at least "bits"
// bits to encode.
bitCount[bits] = counts[level] - counts[level-1]
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.codes[node.literal] = newhcode(reverseBits(code, uint8(n)), 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 []uint16, maxBits int32) {
list := h.freqcache[:len(freq)+1]
codes := h.codes[:len(freq)]
// 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 {
codes[i] = 0
}
}
list[count] = literalNode{}
list = list[: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.codes[node.literal].set(uint16(i), 1)
}
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)
}
// atLeastOne clamps the result between 1 and 15.
func atLeastOne(v float32) float32 {
if v < 1 {
return 1
}
if v > 15 {
return 15
}
return v
}
func histogram(b []byte, h []uint16) {
if true && len(b) >= 8<<10 {
// Split for bigger inputs
histogramSplit(b, h)
} else {
h = h[:256]
for _, t := range b {
h[t]++
}
}
}
func histogramSplit(b []byte, h []uint16) {
// Tested, and slightly faster than 2-way.
// Writing to separate arrays and combining is also slightly slower.
h = h[:256]
for len(b)&3 != 0 {
h[b[0]]++
b = b[1:]
}
n := len(b) / 4
x, y, z, w := b[:n], b[n:], b[n+n:], b[n+n+n:]
y, z, w = y[:len(x)], z[:len(x)], w[:len(x)]
for i, t := range x {
v0 := &h[t]
v1 := &h[y[i]]
v3 := &h[w[i]]
v2 := &h[z[i]]
*v0++
*v1++
*v2++
*v3++
}
}