260 lines
7.3 KiB
Go
260 lines
7.3 KiB
Go
// Copyright 2011 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 bzip2
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import "sort"
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// A huffmanTree is a binary tree which is navigated, bit-by-bit to reach a
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// symbol.
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type huffmanTree struct {
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// nodes contains all the non-leaf nodes in the tree. nodes[0] is the
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// root of the tree and nextNode contains the index of the next element
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// of nodes to use when the tree is being constructed.
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nodes []huffmanNode
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nextNode int
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}
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// A huffmanNode is a node in the tree. left and right contain indexes into the
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// nodes slice of the tree. If left or right is invalidNodeValue then the child
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// is a left node and its value is in leftValue/rightValue.
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//
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// The symbols are uint16s because bzip2 encodes not only MTF indexes in the
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// tree, but also two magic values for run-length encoding and an EOF symbol.
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// Thus there are more than 256 possible symbols.
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type huffmanNode struct {
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left, right uint16
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leftValue, rightValue uint16
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}
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// invalidNodeValue is an invalid index which marks a leaf node in the tree.
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const invalidNodeValue = 0xffff
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// Decode reads bits from the given bitReader and navigates the tree until a
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// symbol is found.
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func (t *huffmanTree) Decode(br *bitReader) (v uint16) {
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nodeIndex := uint16(0) // node 0 is the root of the tree.
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for {
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node := &t.nodes[nodeIndex]
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var bit uint16
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if br.bits > 0 {
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// Get next bit - fast path.
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br.bits--
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bit = 0 - (uint16(br.n>>br.bits) & 1)
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} else {
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// Get next bit - slow path.
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// Use ReadBits to retrieve a single bit
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// from the underling io.ByteReader.
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bit = 0 - uint16(br.ReadBits(1))
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}
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// now
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// bit = 0xffff if the next bit was 1
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// bit = 0x0000 if the next bit was 0
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// 1 means left, 0 means right.
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//
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// if bit == 0xffff {
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// nodeIndex = node.left
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// } else {
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// nodeIndex = node.right
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// }
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nodeIndex = (bit & node.left) | (^bit & node.right)
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if nodeIndex == invalidNodeValue {
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// We found a leaf. Use the value of bit to decide
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// whether is a left or a right value.
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return (bit & node.leftValue) | (^bit & node.rightValue)
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}
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}
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}
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// newHuffmanTree builds a Huffman tree from a slice containing the code
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// lengths of each symbol. The maximum code length is 32 bits.
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func newHuffmanTree(lengths []uint8) (huffmanTree, error) {
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// There are many possible trees that assign the same code length to
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// each symbol (consider reflecting a tree down the middle, for
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// example). Since the code length assignments determine the
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// efficiency of the tree, each of these trees is equally good. In
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// order to minimize the amount of information needed to build a tree
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// bzip2 uses a canonical tree so that it can be reconstructed given
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// only the code length assignments.
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if len(lengths) < 2 {
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panic("newHuffmanTree: too few symbols")
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}
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var t huffmanTree
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// First we sort the code length assignments by ascending code length,
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// using the symbol value to break ties.
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pairs := huffmanSymbolLengthPairs(make([]huffmanSymbolLengthPair, len(lengths)))
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for i, length := range lengths {
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pairs[i].value = uint16(i)
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pairs[i].length = length
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}
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sort.Sort(pairs)
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// Now we assign codes to the symbols, starting with the longest code.
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// We keep the codes packed into a uint32, at the most-significant end.
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// So branches are taken from the MSB downwards. This makes it easy to
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// sort them later.
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code := uint32(0)
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length := uint8(32)
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codes := huffmanCodes(make([]huffmanCode, len(lengths)))
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for i := len(pairs) - 1; i >= 0; i-- {
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if length > pairs[i].length {
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length = pairs[i].length
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}
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codes[i].code = code
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codes[i].codeLen = length
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codes[i].value = pairs[i].value
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// We need to 'increment' the code, which means treating |code|
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// like a |length| bit number.
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code += 1 << (32 - length)
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}
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// Now we can sort by the code so that the left half of each branch are
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// grouped together, recursively.
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sort.Sort(codes)
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t.nodes = make([]huffmanNode, len(codes))
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_, err := buildHuffmanNode(&t, codes, 0)
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return t, err
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}
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// huffmanSymbolLengthPair contains a symbol and its code length.
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type huffmanSymbolLengthPair struct {
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value uint16
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length uint8
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}
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// huffmanSymbolLengthPair is used to provide an interface for sorting.
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type huffmanSymbolLengthPairs []huffmanSymbolLengthPair
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func (h huffmanSymbolLengthPairs) Len() int {
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return len(h)
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}
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func (h huffmanSymbolLengthPairs) Less(i, j int) bool {
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if h[i].length < h[j].length {
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return true
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}
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if h[i].length > h[j].length {
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return false
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}
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if h[i].value < h[j].value {
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return true
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}
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return false
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}
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func (h huffmanSymbolLengthPairs) Swap(i, j int) {
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h[i], h[j] = h[j], h[i]
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}
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// huffmanCode contains a symbol, its code and code length.
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type huffmanCode struct {
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code uint32
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codeLen uint8
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value uint16
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}
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// huffmanCodes is used to provide an interface for sorting.
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type huffmanCodes []huffmanCode
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func (n huffmanCodes) Len() int {
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return len(n)
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}
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func (n huffmanCodes) Less(i, j int) bool {
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return n[i].code < n[j].code
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}
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func (n huffmanCodes) Swap(i, j int) {
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n[i], n[j] = n[j], n[i]
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}
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// buildHuffmanNode takes a slice of sorted huffmanCodes and builds a node in
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// the Huffman tree at the given level. It returns the index of the newly
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// constructed node.
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func buildHuffmanNode(t *huffmanTree, codes []huffmanCode, level uint32) (nodeIndex uint16, err error) {
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test := uint32(1) << (31 - level)
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// We have to search the list of codes to find the divide between the left and right sides.
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firstRightIndex := len(codes)
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for i, code := range codes {
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if code.code&test != 0 {
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firstRightIndex = i
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break
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}
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}
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left := codes[:firstRightIndex]
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right := codes[firstRightIndex:]
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if len(left) == 0 || len(right) == 0 {
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// There is a superfluous level in the Huffman tree indicating
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// a bug in the encoder. However, this bug has been observed in
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// the wild so we handle it.
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// If this function was called recursively then we know that
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// len(codes) >= 2 because, otherwise, we would have hit the
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// "leaf node" case, below, and not recursed.
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//
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// However, for the initial call it's possible that len(codes)
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// is zero or one. Both cases are invalid because a zero length
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// tree cannot encode anything and a length-1 tree can only
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// encode EOF and so is superfluous. We reject both.
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if len(codes) < 2 {
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return 0, StructuralError("empty Huffman tree")
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}
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// In this case the recursion doesn't always reduce the length
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// of codes so we need to ensure termination via another
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// mechanism.
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if level == 31 {
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// Since len(codes) >= 2 the only way that the values
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// can match at all 32 bits is if they are equal, which
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// is invalid. This ensures that we never enter
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// infinite recursion.
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return 0, StructuralError("equal symbols in Huffman tree")
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}
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if len(left) == 0 {
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return buildHuffmanNode(t, right, level+1)
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}
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return buildHuffmanNode(t, left, level+1)
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}
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nodeIndex = uint16(t.nextNode)
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node := &t.nodes[t.nextNode]
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t.nextNode++
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if len(left) == 1 {
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// leaf node
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node.left = invalidNodeValue
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node.leftValue = left[0].value
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} else {
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node.left, err = buildHuffmanNode(t, left, level+1)
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}
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if err != nil {
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return
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}
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if len(right) == 1 {
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// leaf node
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node.right = invalidNodeValue
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node.rightValue = right[0].value
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} else {
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node.right, err = buildHuffmanNode(t, right, level+1)
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}
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return
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}
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