mirror of
https://codeberg.org/forgejo/forgejo
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594 lines
14 KiB
Go
594 lines
14 KiB
Go
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package brotli
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import "math"
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/* Copyright 2010 Google Inc. All Rights Reserved.
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Distributed under MIT license.
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See file LICENSE for detail or copy at https://opensource.org/licenses/MIT
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*/
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/* Entropy encoding (Huffman) utilities. */
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/* A node of a Huffman tree. */
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type huffmanTree struct {
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total_count_ uint32
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index_left_ int16
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index_right_or_value_ int16
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}
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func initHuffmanTree(self *huffmanTree, count uint32, left int16, right int16) {
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self.total_count_ = count
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self.index_left_ = left
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self.index_right_or_value_ = right
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}
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/* Input size optimized Shell sort. */
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type huffmanTreeComparator func(*huffmanTree, *huffmanTree) bool
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var sortHuffmanTreeItems_gaps = []uint{132, 57, 23, 10, 4, 1}
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func sortHuffmanTreeItems(items []huffmanTree, n uint, comparator huffmanTreeComparator) {
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if n < 13 {
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/* Insertion sort. */
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var i uint
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for i = 1; i < n; i++ {
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var tmp huffmanTree = items[i]
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var k uint = i
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var j uint = i - 1
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for comparator(&tmp, &items[j]) {
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items[k] = items[j]
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k = j
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tmp10 := j
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j--
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if tmp10 == 0 {
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break
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}
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}
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items[k] = tmp
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}
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return
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} else {
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var g int
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if n < 57 {
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g = 2
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} else {
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g = 0
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}
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for ; g < 6; g++ {
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var gap uint = sortHuffmanTreeItems_gaps[g]
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var i uint
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for i = gap; i < n; i++ {
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var j uint = i
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var tmp huffmanTree = items[i]
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for ; j >= gap && comparator(&tmp, &items[j-gap]); j -= gap {
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items[j] = items[j-gap]
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}
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items[j] = tmp
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}
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}
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}
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}
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/* Returns 1 if assignment of depths succeeded, otherwise 0. */
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func setDepth(p0 int, pool []huffmanTree, depth []byte, max_depth int) bool {
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var stack [16]int
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var level int = 0
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var p int = p0
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assert(max_depth <= 15)
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stack[0] = -1
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for {
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if pool[p].index_left_ >= 0 {
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level++
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if level > max_depth {
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return false
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}
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stack[level] = int(pool[p].index_right_or_value_)
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p = int(pool[p].index_left_)
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continue
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} else {
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depth[pool[p].index_right_or_value_] = byte(level)
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}
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for level >= 0 && stack[level] == -1 {
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level--
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}
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if level < 0 {
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return true
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}
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p = stack[level]
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stack[level] = -1
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}
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}
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/* Sort the root nodes, least popular first. */
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func sortHuffmanTree(v0 *huffmanTree, v1 *huffmanTree) bool {
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if v0.total_count_ != v1.total_count_ {
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return v0.total_count_ < v1.total_count_
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}
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return v0.index_right_or_value_ > v1.index_right_or_value_
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}
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/* This function will create a Huffman tree.
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The catch here is that the tree cannot be arbitrarily deep.
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Brotli specifies a maximum depth of 15 bits for "code trees"
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and 7 bits for "code length code trees."
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count_limit is the value that is to be faked as the minimum value
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and this minimum value is raised until the tree matches the
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maximum length requirement.
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This algorithm is not of excellent performance for very long data blocks,
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especially when population counts are longer than 2**tree_limit, but
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we are not planning to use this with extremely long blocks.
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See http://en.wikipedia.org/wiki/Huffman_coding */
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func createHuffmanTree(data []uint32, length uint, tree_limit int, tree []huffmanTree, depth []byte) {
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var count_limit uint32
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var sentinel huffmanTree
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initHuffmanTree(&sentinel, math.MaxUint32, -1, -1)
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/* For block sizes below 64 kB, we never need to do a second iteration
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of this loop. Probably all of our block sizes will be smaller than
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that, so this loop is mostly of academic interest. If we actually
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would need this, we would be better off with the Katajainen algorithm. */
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for count_limit = 1; ; count_limit *= 2 {
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var n uint = 0
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var i uint
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var j uint
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var k uint
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for i = length; i != 0; {
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i--
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if data[i] != 0 {
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var count uint32 = brotli_max_uint32_t(data[i], count_limit)
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initHuffmanTree(&tree[n], count, -1, int16(i))
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n++
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}
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}
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if n == 1 {
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depth[tree[0].index_right_or_value_] = 1 /* Only one element. */
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break
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}
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sortHuffmanTreeItems(tree, n, huffmanTreeComparator(sortHuffmanTree))
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/* The nodes are:
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[0, n): the sorted leaf nodes that we start with.
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[n]: we add a sentinel here.
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[n + 1, 2n): new parent nodes are added here, starting from
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(n+1). These are naturally in ascending order.
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[2n]: we add a sentinel at the end as well.
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There will be (2n+1) elements at the end. */
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tree[n] = sentinel
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tree[n+1] = sentinel
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i = 0 /* Points to the next leaf node. */
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j = n + 1 /* Points to the next non-leaf node. */
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for k = n - 1; k != 0; k-- {
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var left uint
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var right uint
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if tree[i].total_count_ <= tree[j].total_count_ {
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left = i
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i++
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} else {
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left = j
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j++
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}
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if tree[i].total_count_ <= tree[j].total_count_ {
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right = i
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i++
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} else {
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right = j
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j++
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}
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{
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/* The sentinel node becomes the parent node. */
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var j_end uint = 2*n - k
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tree[j_end].total_count_ = tree[left].total_count_ + tree[right].total_count_
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tree[j_end].index_left_ = int16(left)
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tree[j_end].index_right_or_value_ = int16(right)
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/* Add back the last sentinel node. */
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tree[j_end+1] = sentinel
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}
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}
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if setDepth(int(2*n-1), tree[0:], depth, tree_limit) {
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/* We need to pack the Huffman tree in tree_limit bits. If this was not
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successful, add fake entities to the lowest values and retry. */
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break
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}
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}
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}
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func reverse(v []byte, start uint, end uint) {
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end--
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for start < end {
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var tmp byte = v[start]
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v[start] = v[end]
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v[end] = tmp
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start++
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end--
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}
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}
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func writeHuffmanTreeRepetitions(previous_value byte, value byte, repetitions uint, tree_size *uint, tree []byte, extra_bits_data []byte) {
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assert(repetitions > 0)
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if previous_value != value {
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tree[*tree_size] = value
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extra_bits_data[*tree_size] = 0
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(*tree_size)++
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repetitions--
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}
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if repetitions == 7 {
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tree[*tree_size] = value
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extra_bits_data[*tree_size] = 0
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(*tree_size)++
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repetitions--
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}
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if repetitions < 3 {
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var i uint
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for i = 0; i < repetitions; i++ {
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tree[*tree_size] = value
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extra_bits_data[*tree_size] = 0
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(*tree_size)++
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}
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} else {
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var start uint = *tree_size
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repetitions -= 3
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for {
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tree[*tree_size] = repeatPreviousCodeLength
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extra_bits_data[*tree_size] = byte(repetitions & 0x3)
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(*tree_size)++
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repetitions >>= 2
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if repetitions == 0 {
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break
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}
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repetitions--
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}
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reverse(tree, start, *tree_size)
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reverse(extra_bits_data, start, *tree_size)
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}
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}
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func writeHuffmanTreeRepetitionsZeros(repetitions uint, tree_size *uint, tree []byte, extra_bits_data []byte) {
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if repetitions == 11 {
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tree[*tree_size] = 0
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extra_bits_data[*tree_size] = 0
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(*tree_size)++
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repetitions--
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}
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if repetitions < 3 {
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var i uint
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for i = 0; i < repetitions; i++ {
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tree[*tree_size] = 0
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extra_bits_data[*tree_size] = 0
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(*tree_size)++
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}
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} else {
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var start uint = *tree_size
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repetitions -= 3
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for {
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tree[*tree_size] = repeatZeroCodeLength
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extra_bits_data[*tree_size] = byte(repetitions & 0x7)
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(*tree_size)++
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repetitions >>= 3
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if repetitions == 0 {
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break
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}
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repetitions--
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}
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reverse(tree, start, *tree_size)
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reverse(extra_bits_data, start, *tree_size)
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}
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}
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/* Change the population counts in a way that the consequent
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Huffman tree compression, especially its RLE-part will be more
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likely to compress this data more efficiently.
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length contains the size of the histogram.
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counts contains the population counts.
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good_for_rle is a buffer of at least length size */
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func optimizeHuffmanCountsForRLE(length uint, counts []uint32, good_for_rle []byte) {
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var nonzero_count uint = 0
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var stride uint
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var limit uint
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var sum uint
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var streak_limit uint = 1240
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var i uint
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/* Let's make the Huffman code more compatible with RLE encoding. */
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for i = 0; i < length; i++ {
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if counts[i] != 0 {
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nonzero_count++
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}
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}
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if nonzero_count < 16 {
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return
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}
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for length != 0 && counts[length-1] == 0 {
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length--
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}
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if length == 0 {
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return /* All zeros. */
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}
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/* Now counts[0..length - 1] does not have trailing zeros. */
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{
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var nonzeros uint = 0
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var smallest_nonzero uint32 = 1 << 30
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for i = 0; i < length; i++ {
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if counts[i] != 0 {
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nonzeros++
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if smallest_nonzero > counts[i] {
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smallest_nonzero = counts[i]
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}
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}
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}
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if nonzeros < 5 {
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/* Small histogram will model it well. */
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return
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}
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if smallest_nonzero < 4 {
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var zeros uint = length - nonzeros
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if zeros < 6 {
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for i = 1; i < length-1; i++ {
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if counts[i-1] != 0 && counts[i] == 0 && counts[i+1] != 0 {
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counts[i] = 1
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}
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}
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}
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}
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if nonzeros < 28 {
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return
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}
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}
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/* 2) Let's mark all population counts that already can be encoded
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with an RLE code. */
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for i := 0; i < int(length); i++ {
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good_for_rle[i] = 0
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}
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{
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var symbol uint32 = counts[0]
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/* Let's not spoil any of the existing good RLE codes.
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Mark any seq of 0's that is longer as 5 as a good_for_rle.
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Mark any seq of non-0's that is longer as 7 as a good_for_rle. */
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var step uint = 0
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for i = 0; i <= length; i++ {
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if i == length || counts[i] != symbol {
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if (symbol == 0 && step >= 5) || (symbol != 0 && step >= 7) {
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var k uint
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for k = 0; k < step; k++ {
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good_for_rle[i-k-1] = 1
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}
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}
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step = 1
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if i != length {
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symbol = counts[i]
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}
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} else {
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step++
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}
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}
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}
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/* 3) Let's replace those population counts that lead to more RLE codes.
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Math here is in 24.8 fixed point representation. */
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stride = 0
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limit = uint(256*(counts[0]+counts[1]+counts[2])/3 + 420)
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sum = 0
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for i = 0; i <= length; i++ {
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if i == length || good_for_rle[i] != 0 || (i != 0 && good_for_rle[i-1] != 0) || (256*counts[i]-uint32(limit)+uint32(streak_limit)) >= uint32(2*streak_limit) {
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if stride >= 4 || (stride >= 3 && sum == 0) {
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var k uint
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var count uint = (sum + stride/2) / stride
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/* The stride must end, collapse what we have, if we have enough (4). */
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if count == 0 {
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count = 1
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}
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if sum == 0 {
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/* Don't make an all zeros stride to be upgraded to ones. */
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count = 0
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}
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for k = 0; k < stride; k++ {
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/* We don't want to change value at counts[i],
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that is already belonging to the next stride. Thus - 1. */
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counts[i-k-1] = uint32(count)
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}
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}
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stride = 0
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sum = 0
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if i < length-2 {
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/* All interesting strides have a count of at least 4, */
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/* at least when non-zeros. */
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limit = uint(256*(counts[i]+counts[i+1]+counts[i+2])/3 + 420)
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} else if i < length {
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limit = uint(256 * counts[i])
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} else {
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limit = 0
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}
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}
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stride++
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if i != length {
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sum += uint(counts[i])
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if stride >= 4 {
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limit = (256*sum + stride/2) / stride
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}
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if stride == 4 {
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limit += 120
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}
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}
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}
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}
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func decideOverRLEUse(depth []byte, length uint, use_rle_for_non_zero *bool, use_rle_for_zero *bool) {
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var total_reps_zero uint = 0
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var total_reps_non_zero uint = 0
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var count_reps_zero uint = 1
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var count_reps_non_zero uint = 1
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var i uint
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for i = 0; i < length; {
|
||
|
var value byte = depth[i]
|
||
|
var reps uint = 1
|
||
|
var k uint
|
||
|
for k = i + 1; k < length && depth[k] == value; k++ {
|
||
|
reps++
|
||
|
}
|
||
|
|
||
|
if reps >= 3 && value == 0 {
|
||
|
total_reps_zero += reps
|
||
|
count_reps_zero++
|
||
|
}
|
||
|
|
||
|
if reps >= 4 && value != 0 {
|
||
|
total_reps_non_zero += reps
|
||
|
count_reps_non_zero++
|
||
|
}
|
||
|
|
||
|
i += reps
|
||
|
}
|
||
|
|
||
|
*use_rle_for_non_zero = total_reps_non_zero > count_reps_non_zero*2
|
||
|
*use_rle_for_zero = total_reps_zero > count_reps_zero*2
|
||
|
}
|
||
|
|
||
|
/* Write a Huffman tree from bit depths into the bit-stream representation
|
||
|
of a Huffman tree. The generated Huffman tree is to be compressed once
|
||
|
more using a Huffman tree */
|
||
|
func writeHuffmanTree(depth []byte, length uint, tree_size *uint, tree []byte, extra_bits_data []byte) {
|
||
|
var previous_value byte = initialRepeatedCodeLength
|
||
|
var i uint
|
||
|
var use_rle_for_non_zero bool = false
|
||
|
var use_rle_for_zero bool = false
|
||
|
var new_length uint = length
|
||
|
/* Throw away trailing zeros. */
|
||
|
for i = 0; i < length; i++ {
|
||
|
if depth[length-i-1] == 0 {
|
||
|
new_length--
|
||
|
} else {
|
||
|
break
|
||
|
}
|
||
|
}
|
||
|
|
||
|
/* First gather statistics on if it is a good idea to do RLE. */
|
||
|
if length > 50 {
|
||
|
/* Find RLE coding for longer codes.
|
||
|
Shorter codes seem not to benefit from RLE. */
|
||
|
decideOverRLEUse(depth, new_length, &use_rle_for_non_zero, &use_rle_for_zero)
|
||
|
}
|
||
|
|
||
|
/* Actual RLE coding. */
|
||
|
for i = 0; i < new_length; {
|
||
|
var value byte = depth[i]
|
||
|
var reps uint = 1
|
||
|
if (value != 0 && use_rle_for_non_zero) || (value == 0 && use_rle_for_zero) {
|
||
|
var k uint
|
||
|
for k = i + 1; k < new_length && depth[k] == value; k++ {
|
||
|
reps++
|
||
|
}
|
||
|
}
|
||
|
|
||
|
if value == 0 {
|
||
|
writeHuffmanTreeRepetitionsZeros(reps, tree_size, tree, extra_bits_data)
|
||
|
} else {
|
||
|
writeHuffmanTreeRepetitions(previous_value, value, reps, tree_size, tree, extra_bits_data)
|
||
|
previous_value = value
|
||
|
}
|
||
|
|
||
|
i += reps
|
||
|
}
|
||
|
}
|
||
|
|
||
|
var reverseBits_kLut = [16]uint{
|
||
|
0x00,
|
||
|
0x08,
|
||
|
0x04,
|
||
|
0x0C,
|
||
|
0x02,
|
||
|
0x0A,
|
||
|
0x06,
|
||
|
0x0E,
|
||
|
0x01,
|
||
|
0x09,
|
||
|
0x05,
|
||
|
0x0D,
|
||
|
0x03,
|
||
|
0x0B,
|
||
|
0x07,
|
||
|
0x0F,
|
||
|
}
|
||
|
|
||
|
func reverseBits(num_bits uint, bits uint16) uint16 {
|
||
|
var retval uint = reverseBits_kLut[bits&0x0F]
|
||
|
var i uint
|
||
|
for i = 4; i < num_bits; i += 4 {
|
||
|
retval <<= 4
|
||
|
bits = uint16(bits >> 4)
|
||
|
retval |= reverseBits_kLut[bits&0x0F]
|
||
|
}
|
||
|
|
||
|
retval >>= ((0 - num_bits) & 0x03)
|
||
|
return uint16(retval)
|
||
|
}
|
||
|
|
||
|
/* 0..15 are values for bits */
|
||
|
const maxHuffmanBits = 16
|
||
|
|
||
|
/* Get the actual bit values for a tree of bit depths. */
|
||
|
func convertBitDepthsToSymbols(depth []byte, len uint, bits []uint16) {
|
||
|
var bl_count = [maxHuffmanBits]uint16{0}
|
||
|
var next_code [maxHuffmanBits]uint16
|
||
|
var i uint
|
||
|
/* In Brotli, all bit depths are [1..15]
|
||
|
0 bit depth means that the symbol does not exist. */
|
||
|
|
||
|
var code int = 0
|
||
|
for i = 0; i < len; i++ {
|
||
|
bl_count[depth[i]]++
|
||
|
}
|
||
|
|
||
|
bl_count[0] = 0
|
||
|
next_code[0] = 0
|
||
|
for i = 1; i < maxHuffmanBits; i++ {
|
||
|
code = (code + int(bl_count[i-1])) << 1
|
||
|
next_code[i] = uint16(code)
|
||
|
}
|
||
|
|
||
|
for i = 0; i < len; i++ {
|
||
|
if depth[i] != 0 {
|
||
|
bits[i] = reverseBits(uint(depth[i]), next_code[depth[i]])
|
||
|
next_code[depth[i]]++
|
||
|
}
|
||
|
}
|
||
|
}
|