Merge from mainline (167278:168000).

[official-gcc/graphite-test-results.git] / libgo / go / compress / flate / huffman_code.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"
        "sort"
)

type huffmanEncoder struct {
        codeBits []uint8
        code     []uint16
}

type literalNode struct {
        literal uint16
        freq    int32
}

type chain struct {
        // The sum of the leaves in this tree
        freq int32

        // The number of literals to the left of this item at this level
        leafCount int32

        // The right child of this chain in the previous level.
        up *chain
}

type levelInfo struct {
        // Our level.  for better printing
        level int32

        // The most recent chain generated for this level
        lastChain *chain

        // 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

        // The levelInfo for level+1
        up *levelInfo

        // The levelInfo for level-1
        down *levelInfo
}

func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxInt32} }

func newHuffmanEncoder(size int) *huffmanEncoder {
        return &huffmanEncoder{make([]uint8, size), make([]uint16, size)}
}

// Generates a HuffmanCode corresponding to the fixed literal table
func generateFixedLiteralEncoding() *huffmanEncoder {
        h := newHuffmanEncoder(maxLit)
        codeBits := h.codeBits
        code := h.code
        var ch uint16
        for ch = 0; ch < maxLit; ch++ {
                var bits uint16
                var size uint8
                switch {
                case ch < 144:
                        // size 8, 000110000  .. 10111111
                        bits = ch + 48
                        size = 8
                        break
                case ch < 256:
                        // size 9, 110010000 .. 111111111
                        bits = ch + 400 - 144
                        size = 9
                        break
                case ch < 280:
                        // size 7, 0000000 .. 0010111
                        bits = ch - 256
                        size = 7
                        break
                default:
                        // size 8, 11000000 .. 11000111
                        bits = ch + 192 - 280
                        size = 8
                }
                codeBits[ch] = size
                code[ch] = reverseBits(bits, size)
        }
        return h
}

func generateFixedOffsetEncoding() *huffmanEncoder {
        h := newHuffmanEncoder(30)
        codeBits := h.codeBits
        code := h.code
        for ch := uint16(0); ch < 30; ch++ {
                codeBits[ch] = 5
                code[ch] = reverseBits(ch, 5)
        }
        return h
}

var fixedLiteralEncoding *huffmanEncoder = generateFixedLiteralEncoding()
var fixedOffsetEncoding *huffmanEncoder = generateFixedOffsetEncoding()

func (h *huffmanEncoder) bitLength(freq []int32) int64 {
        var total int64
        for i, f := range freq {
                if f != 0 {
                        total += int64(f) * int64(h.codeBits[i])
                }
        }
        return total
}

// Generate elements in the chain using an iterative algorithm.
func (h *huffmanEncoder) generateChains(top *levelInfo, list []literalNode) {
        n := len(list)
        list = list[0 : n+1]
        list[n] = maxNode()

        l := top
        for {
                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 m sure we never come back to this level or any lower level,
                        // set nextPairFreq impossibly large.
                        l.lastChain = nil
                        l.needed = 0
                        l = l.up
                        l.nextPairFreq = math.MaxInt32
                        continue
                }

                prevFreq := l.lastChain.freq
                if l.nextCharFreq < l.nextPairFreq {
                        // The next item on this row is a leaf node.
                        n := l.lastChain.leafCount + 1
                        l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up}
                        l.nextCharFreq = list[n].freq
                } 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.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain}
                        l.down.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.
                        up := l.up
                        if up == nil {
                                // All done!
                                return
                        }
                        up.nextPairFreq = prevFreq + l.lastChain.freq
                        l = up
                } else {
                        // If we stole from below, move down temporarily to replenish it.
                        for l.down.needed > 0 {
                                l = l.down
                        }
                }
        }
}

// 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.
// 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 {
        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
        maxBits = minInt32(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.
        top := &levelInfo{needed: 0}
        chain2 := &chain{list[1].freq, 2, new(chain)}
        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.
                top = &levelInfo{
                        level:        level,
                        lastChain:    chain2,
                        nextCharFreq: list[2].freq,
                        nextPairFreq: list[0].freq + list[1].freq,
                        down:         top,
                }
                top.down.up = top
                if level == 1 {
                        top.nextPairFreq = math.MaxInt32
                }
        }

        // We need a total of 2*n - 2 items at top level and have already generated 2.
        top.needed = 2*n - 4

        l := top
        for {
                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 m sure we never come back to this level or any lower level,
                        // set nextPairFreq impossibly large.
                        l.lastChain = nil
                        l.needed = 0
                        l = l.up
                        l.nextPairFreq = math.MaxInt32
                        continue
                }

                prevFreq := l.lastChain.freq
                if l.nextCharFreq < l.nextPairFreq {
                        // The next item on this row is a leaf node.
                        n := l.lastChain.leafCount + 1
                        l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up}
                        l.nextCharFreq = list[n].freq
                } 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.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain}
                        l.down.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.
                        up := l.up
                        if up == nil {
                                // All done!
                                break
                        }
                        up.nextPairFreq = prevFreq + l.lastChain.freq
                        l = up
                } else {
                        // If we stole from below, move down temporarily to replenish it.
                        for l.down.needed > 0 {
                                l = l.down
                        }
                }
        }

        // Somethings is wrong if at the end, the top level is null or hasn't used
        // all of the leaves.
        if top.lastChain.leafCount != n {
                panic("top.lastChain.leafCount != n")
        }

        bitCount := make([]int32, maxBits+1)
        bits := 1
        for chain := top.lastChain; chain.up != nil; chain = chain.up {
                // chain.leafCount gives the number of literals requiring at least "bits"
                // bits to encode.
                bitCount[bits] = chain.leafCount - chain.up.leafCount
                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.codeBits[node.literal] = uint8(n)
                        h.code[node.literal] = reverseBits(code, 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 []int32, maxBits int32) {
        list := make([]literalNode, len(freq)+1)
        // 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 {
                        h.codeBits[i] = 0
                }
        }
        // If freq[] is shorter than codeBits[], fill rest of codeBits[] with zeros
        h.codeBits = h.codeBits[0:len(freq)]
        list = list[0: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.codeBits[node.literal] = 1
                        h.code[node.literal] = uint16(i)
                }
                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)
}

type literalNodeSorter struct {
        a    []literalNode
        less func(i, j int) bool
}

func (s literalNodeSorter) Len() int { return len(s.a) }

func (s literalNodeSorter) Less(i, j int) bool {
        return s.less(i, j)
}

func (s literalNodeSorter) Swap(i, j int) { s.a[i], s.a[j] = s.a[j], s.a[i] }

func sortByFreq(a []literalNode) {
        s := &literalNodeSorter{a, func(i, j int) bool { return a[i].freq < a[j].freq }}
        sort.Sort(s)
}

func sortByLiteral(a []literalNode) {
        s := &literalNodeSorter{a, func(i, j int) bool { return a[i].literal < a[j].literal }}
        sort.Sort(s)
}