brotli/enc/bit_cost.h
Zoltan Szabadka 667f70adcb Speedups to brotli quality 11.
* Cluster at most 64 histograms at a time in the first
    round of clustering.

  * Use a faster histogram cost estimation function.

  * Don't compute the log2(total) multiple times in the
    block splitter.
2015-06-12 15:29:06 +02:00

135 lines
3.9 KiB
C++

// Copyright 2013 Google Inc. All Rights Reserved.
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
//
// Functions to estimate the bit cost of Huffman trees.
#ifndef BROTLI_ENC_BIT_COST_H_
#define BROTLI_ENC_BIT_COST_H_
#include <stdint.h>
#include "./entropy_encode.h"
#include "./fast_log.h"
namespace brotli {
static inline double BitsEntropy(const int *population, int size) {
int sum = 0;
double retval = 0;
const int *population_end = population + size;
int p;
if (size & 1) {
goto odd_number_of_elements_left;
}
while (population < population_end) {
p = *population++;
sum += p;
retval -= p * FastLog2(p);
odd_number_of_elements_left:
p = *population++;
sum += p;
retval -= p * FastLog2(p);
}
if (sum) retval += sum * FastLog2(sum);
if (retval < sum) {
// At least one bit per literal is needed.
retval = sum;
}
return retval;
}
template<int kSize>
double PopulationCost(const Histogram<kSize>& histogram) {
if (histogram.total_count_ == 0) {
return 12;
}
int count = 0;
for (int i = 0; i < kSize; ++i) {
if (histogram.data_[i] > 0) {
++count;
}
}
if (count == 1) {
return 12;
}
if (count == 2) {
return 20 + histogram.total_count_;
}
double bits = 0;
uint8_t depth[kSize] = { 0 };
if (count <= 4) {
// For very low symbol count we build the Huffman tree.
CreateHuffmanTree(&histogram.data_[0], kSize, 15, depth);
for (int i = 0; i < kSize; ++i) {
bits += histogram.data_[i] * depth[i];
}
return count == 3 ? bits + 28 : bits + 37;
}
// In this loop we compute the entropy of the histogram and simultaneously
// build a simplified histogram of the code length codes where we use the
// zero repeat code 17, but we don't use the non-zero repeat code 16.
int max_depth = 1;
int depth_histo[kCodeLengthCodes] = { 0 };
const double log2total = FastLog2(histogram.total_count_);
for (int i = 0; i < kSize;) {
if (histogram.data_[i] > 0) {
// Compute -log2(P(symbol)) = -log2(count(symbol)/total_count) =
// = log2(total_count) - log2(count(symbol))
double log2p = log2total - FastLog2(histogram.data_[i]);
// Approximate the bit depth by round(-log2(P(symbol)))
int depth = static_cast<int>(log2p + 0.5);
bits += histogram.data_[i] * log2p;
if (depth > max_depth) {
max_depth = depth;
}
++depth_histo[depth];
++i;
} else {
// Compute the run length of zeros and add the appropiate number of 0 and
// 17 code length codes to the code length code histogram.
int reps = 1;
for (int k = i + 1; k < kSize && histogram.data_[k] == 0; ++k) {
++reps;
}
i += reps;
if (i == kSize) {
// Don't add any cost for the last zero run, since these are encoded
// only implicitly.
break;
}
if (reps < 3) {
depth_histo[0] += reps;
} else {
reps -= 2;
while (reps > 0) {
++depth_histo[17];
// Add the 3 extra bits for the 17 code length code.
bits += 3;
reps >>= 3;
}
}
}
}
// Add the estimated encoding cost of the code length code histogram.
bits += 18 + 2 * max_depth;
// Add the entropy of the code length code histogram.
bits += BitsEntropy(depth_histo, kCodeLengthCodes);
return bits;
}
} // namespace brotli
#endif // BROTLI_ENC_BIT_COST_H_