413 lines
11 KiB
Go
413 lines
11 KiB
Go
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// Copyright 2009 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package flate
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import (
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"math"
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"math/bits"
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)
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const (
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maxBitsLimit = 16
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// number of valid literals
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literalCount = 286
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)
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// hcode is a huffman code with a bit code and bit length.
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type hcode uint32
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func (h hcode) len() uint8 {
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return uint8(h)
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}
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func (h hcode) code64() uint64 {
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return uint64(h >> 8)
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}
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func (h hcode) zero() bool {
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return h == 0
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}
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type huffmanEncoder struct {
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codes []hcode
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bitCount [17]int32
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// Allocate a reusable buffer with the longest possible frequency table.
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// Possible lengths are codegenCodeCount, offsetCodeCount and literalCount.
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// The largest of these is literalCount, so we allocate for that case.
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freqcache [literalCount + 1]literalNode
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}
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type literalNode struct {
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literal uint16
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freq uint16
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}
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// A levelInfo describes the state of the constructed tree for a given depth.
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type levelInfo struct {
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// Our level. for better printing
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level int32
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// The frequency of the last node at this level
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lastFreq int32
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// The frequency of the next character to add to this level
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nextCharFreq int32
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// The frequency of the next pair (from level below) to add to this level.
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// Only valid if the "needed" value of the next lower level is 0.
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nextPairFreq int32
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// The number of chains remaining to generate for this level before moving
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// up to the next level
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needed int32
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}
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// set sets the code and length of an hcode.
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func (h *hcode) set(code uint16, length uint8) {
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*h = hcode(length) | (hcode(code) << 8)
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}
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func newhcode(code uint16, length uint8) hcode {
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return hcode(length) | (hcode(code) << 8)
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}
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func reverseBits(number uint16, bitLength byte) uint16 {
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return bits.Reverse16(number << ((16 - bitLength) & 15))
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}
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func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxUint16} }
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func newHuffmanEncoder(size int) *huffmanEncoder {
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// Make capacity to next power of two.
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c := uint(bits.Len32(uint32(size - 1)))
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return &huffmanEncoder{codes: make([]hcode, size, 1<<c)}
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}
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// Generates a HuffmanCode corresponding to the fixed literal table
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func generateFixedLiteralEncoding() *huffmanEncoder {
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h := newHuffmanEncoder(literalCount)
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codes := h.codes
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var ch uint16
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for ch = 0; ch < literalCount; ch++ {
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var bits uint16
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var size uint8
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switch {
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case ch < 144:
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// size 8, 000110000 .. 10111111
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bits = ch + 48
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size = 8
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case ch < 256:
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// size 9, 110010000 .. 111111111
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bits = ch + 400 - 144
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size = 9
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case ch < 280:
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// size 7, 0000000 .. 0010111
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bits = ch - 256
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size = 7
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default:
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// size 8, 11000000 .. 11000111
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bits = ch + 192 - 280
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size = 8
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}
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codes[ch] = newhcode(reverseBits(bits, size), size)
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}
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return h
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}
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func generateFixedOffsetEncoding() *huffmanEncoder {
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h := newHuffmanEncoder(30)
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codes := h.codes
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for ch := range codes {
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codes[ch] = newhcode(reverseBits(uint16(ch), 5), 5)
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}
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return h
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}
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var fixedLiteralEncoding = generateFixedLiteralEncoding()
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var fixedOffsetEncoding = generateFixedOffsetEncoding()
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func (h *huffmanEncoder) bitLength(freq []uint16) int {
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var total int
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for i, f := range freq {
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if f != 0 {
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total += int(f) * int(h.codes[i].len())
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}
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}
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return total
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}
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func (h *huffmanEncoder) bitLengthRaw(b []byte) int {
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var total int
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for _, f := range b {
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total += int(h.codes[f].len())
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}
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return total
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}
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// canReuseBits returns the number of bits or math.MaxInt32 if the encoder cannot be reused.
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func (h *huffmanEncoder) canReuseBits(freq []uint16) int {
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var total int
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for i, f := range freq {
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if f != 0 {
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code := h.codes[i]
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if code.zero() {
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return math.MaxInt32
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}
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total += int(f) * int(code.len())
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}
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}
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return total
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}
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// Return the number of literals assigned to each bit size in the Huffman encoding
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//
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// This method is only called when list.length >= 3
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// The cases of 0, 1, and 2 literals are handled by special case code.
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//
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// list An array of the literals with non-zero frequencies
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// and their associated frequencies. The array is in order of increasing
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// frequency, and has as its last element a special element with frequency
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// MaxInt32
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// maxBits The maximum number of bits that should be used to encode any literal.
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// Must be less than 16.
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// return An integer array in which array[i] indicates the number of literals
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// that should be encoded in i bits.
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func (h *huffmanEncoder) bitCounts(list []literalNode, maxBits int32) []int32 {
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if maxBits >= maxBitsLimit {
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panic("flate: maxBits too large")
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}
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n := int32(len(list))
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list = list[0 : n+1]
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list[n] = maxNode()
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// The tree can't have greater depth than n - 1, no matter what. This
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// saves a little bit of work in some small cases
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if maxBits > n-1 {
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maxBits = n - 1
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}
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// Create information about each of the levels.
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// A bogus "Level 0" whose sole purpose is so that
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// level1.prev.needed==0. This makes level1.nextPairFreq
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// be a legitimate value that never gets chosen.
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var levels [maxBitsLimit]levelInfo
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// leafCounts[i] counts the number of literals at the left
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// of ancestors of the rightmost node at level i.
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// leafCounts[i][j] is the number of literals at the left
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// of the level j ancestor.
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var leafCounts [maxBitsLimit][maxBitsLimit]int32
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// Descending to only have 1 bounds check.
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l2f := int32(list[2].freq)
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l1f := int32(list[1].freq)
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l0f := int32(list[0].freq) + int32(list[1].freq)
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for level := int32(1); level <= maxBits; level++ {
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// For every level, the first two items are the first two characters.
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// We initialize the levels as if we had already figured this out.
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levels[level] = levelInfo{
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level: level,
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lastFreq: l1f,
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nextCharFreq: l2f,
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nextPairFreq: l0f,
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}
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leafCounts[level][level] = 2
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if level == 1 {
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levels[level].nextPairFreq = math.MaxInt32
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}
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}
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// We need a total of 2*n - 2 items at top level and have already generated 2.
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levels[maxBits].needed = 2*n - 4
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level := uint32(maxBits)
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for level < 16 {
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l := &levels[level]
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if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 {
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// We've run out of both leafs and pairs.
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// End all calculations for this level.
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// To make sure we never come back to this level or any lower level,
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// set nextPairFreq impossibly large.
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l.needed = 0
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levels[level+1].nextPairFreq = math.MaxInt32
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level++
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continue
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}
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prevFreq := l.lastFreq
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if l.nextCharFreq < l.nextPairFreq {
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// The next item on this row is a leaf node.
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n := leafCounts[level][level] + 1
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l.lastFreq = l.nextCharFreq
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// Lower leafCounts are the same of the previous node.
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leafCounts[level][level] = n
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e := list[n]
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if e.literal < math.MaxUint16 {
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l.nextCharFreq = int32(e.freq)
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} else {
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l.nextCharFreq = math.MaxInt32
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}
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} else {
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// The next item on this row is a pair from the previous row.
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// nextPairFreq isn't valid until we generate two
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// more values in the level below
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l.lastFreq = l.nextPairFreq
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// Take leaf counts from the lower level, except counts[level] remains the same.
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if true {
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save := leafCounts[level][level]
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leafCounts[level] = leafCounts[level-1]
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leafCounts[level][level] = save
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} else {
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copy(leafCounts[level][:level], leafCounts[level-1][:level])
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}
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levels[l.level-1].needed = 2
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}
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if l.needed--; l.needed == 0 {
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// We've done everything we need to do for this level.
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// Continue calculating one level up. Fill in nextPairFreq
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// of that level with the sum of the two nodes we've just calculated on
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// this level.
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if l.level == maxBits {
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// All done!
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break
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}
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levels[l.level+1].nextPairFreq = prevFreq + l.lastFreq
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level++
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} else {
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// If we stole from below, move down temporarily to replenish it.
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for levels[level-1].needed > 0 {
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level--
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}
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}
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}
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// Somethings is wrong if at the end, the top level is null or hasn't used
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// all of the leaves.
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if leafCounts[maxBits][maxBits] != n {
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panic("leafCounts[maxBits][maxBits] != n")
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}
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bitCount := h.bitCount[:maxBits+1]
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bits := 1
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counts := &leafCounts[maxBits]
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for level := maxBits; level > 0; level-- {
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// chain.leafCount gives the number of literals requiring at least "bits"
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// bits to encode.
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bitCount[bits] = counts[level] - counts[level-1]
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bits++
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}
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return bitCount
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}
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// Look at the leaves and assign them a bit count and an encoding as specified
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// in RFC 1951 3.2.2
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func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalNode) {
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code := uint16(0)
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for n, bits := range bitCount {
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code <<= 1
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if n == 0 || bits == 0 {
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continue
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}
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// The literals list[len(list)-bits] .. list[len(list)-bits]
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// are encoded using "bits" bits, and get the values
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// code, code + 1, .... The code values are
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// assigned in literal order (not frequency order).
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chunk := list[len(list)-int(bits):]
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sortByLiteral(chunk)
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for _, node := range chunk {
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h.codes[node.literal] = newhcode(reverseBits(code, uint8(n)), uint8(n))
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code++
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}
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list = list[0 : len(list)-int(bits)]
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}
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}
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// Update this Huffman Code object to be the minimum code for the specified frequency count.
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//
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// freq An array of frequencies, in which frequency[i] gives the frequency of literal i.
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// maxBits The maximum number of bits to use for any literal.
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func (h *huffmanEncoder) generate(freq []uint16, maxBits int32) {
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list := h.freqcache[:len(freq)+1]
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codes := h.codes[:len(freq)]
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// Number of non-zero literals
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count := 0
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// Set list to be the set of all non-zero literals and their frequencies
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for i, f := range freq {
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if f != 0 {
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list[count] = literalNode{uint16(i), f}
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count++
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} else {
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codes[i] = 0
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}
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}
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list[count] = literalNode{}
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list = list[:count]
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if count <= 2 {
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// Handle the small cases here, because they are awkward for the general case code. With
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// two or fewer literals, everything has bit length 1.
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for i, node := range list {
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// "list" is in order of increasing literal value.
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h.codes[node.literal].set(uint16(i), 1)
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}
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return
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}
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sortByFreq(list)
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// Get the number of literals for each bit count
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bitCount := h.bitCounts(list, maxBits)
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// And do the assignment
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h.assignEncodingAndSize(bitCount, list)
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}
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// atLeastOne clamps the result between 1 and 15.
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func atLeastOne(v float32) float32 {
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if v < 1 {
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return 1
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}
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if v > 15 {
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return 15
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}
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return v
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}
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func histogram(b []byte, h []uint16) {
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if true && len(b) >= 8<<10 {
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// Split for bigger inputs
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histogramSplit(b, h)
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} else {
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h = h[:256]
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for _, t := range b {
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h[t]++
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}
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}
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}
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func histogramSplit(b []byte, h []uint16) {
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// Tested, and slightly faster than 2-way.
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// Writing to separate arrays and combining is also slightly slower.
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h = h[:256]
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for len(b)&3 != 0 {
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h[b[0]]++
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b = b[1:]
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}
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n := len(b) / 4
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x, y, z, w := b[:n], b[n:], b[n+n:], b[n+n+n:]
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y, z, w = y[:len(x)], z[:len(x)], w[:len(x)]
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for i, t := range x {
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v0 := &h[t]
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v1 := &h[y[i]]
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v3 := &h[w[i]]
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v2 := &h[z[i]]
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*v0++
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*v1++
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*v2++
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*v3++
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}
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}
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