| /* NOLINT(build/header_guard) */ |
| /* Copyright 2013 Google Inc. All Rights Reserved. |
| |
| Distributed under MIT license. |
| See file LICENSE for detail or copy at https://opensource.org/licenses/MIT |
| */ |
| |
| /* template parameters: FN */ |
| |
| #define HistogramType FN(Histogram) |
| |
| double FN(BrotliPopulationCost)(const HistogramType* histogram) { |
| static const double kOneSymbolHistogramCost = 12; |
| static const double kTwoSymbolHistogramCost = 20; |
| static const double kThreeSymbolHistogramCost = 28; |
| static const double kFourSymbolHistogramCost = 37; |
| const size_t data_size = FN(HistogramDataSize)(); |
| int count = 0; |
| size_t s[5]; |
| double bits = 0.0; |
| size_t i; |
| if (histogram->total_count_ == 0) { |
| return kOneSymbolHistogramCost; |
| } |
| for (i = 0; i < data_size; ++i) { |
| if (histogram->data_[i] > 0) { |
| s[count] = i; |
| ++count; |
| if (count > 4) break; |
| } |
| } |
| if (count == 1) { |
| return kOneSymbolHistogramCost; |
| } |
| if (count == 2) { |
| return (kTwoSymbolHistogramCost + (double)histogram->total_count_); |
| } |
| if (count == 3) { |
| const uint32_t histo0 = histogram->data_[s[0]]; |
| const uint32_t histo1 = histogram->data_[s[1]]; |
| const uint32_t histo2 = histogram->data_[s[2]]; |
| const uint32_t histomax = |
| BROTLI_MAX(uint32_t, histo0, BROTLI_MAX(uint32_t, histo1, histo2)); |
| return (kThreeSymbolHistogramCost + |
| 2 * (histo0 + histo1 + histo2) - histomax); |
| } |
| if (count == 4) { |
| uint32_t histo[4]; |
| uint32_t h23; |
| uint32_t histomax; |
| for (i = 0; i < 4; ++i) { |
| histo[i] = histogram->data_[s[i]]; |
| } |
| /* Sort */ |
| for (i = 0; i < 4; ++i) { |
| size_t j; |
| for (j = i + 1; j < 4; ++j) { |
| if (histo[j] > histo[i]) { |
| BROTLI_SWAP(uint32_t, histo, j, i); |
| } |
| } |
| } |
| h23 = histo[2] + histo[3]; |
| histomax = BROTLI_MAX(uint32_t, h23, histo[0]); |
| return (kFourSymbolHistogramCost + |
| 3 * h23 + 2 * (histo[0] + histo[1]) - histomax); |
| } |
| |
| { |
| /* 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. */ |
| size_t max_depth = 1; |
| uint32_t depth_histo[BROTLI_CODE_LENGTH_CODES] = { 0 }; |
| const double log2total = FastLog2(histogram->total_count_); |
| for (i = 0; i < data_size;) { |
| 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))) */ |
| size_t depth = (size_t)(log2p + 0.5); |
| bits += histogram->data_[i] * log2p; |
| if (depth > 15) { |
| depth = 15; |
| } |
| if (depth > max_depth) { |
| max_depth = depth; |
| } |
| ++depth_histo[depth]; |
| ++i; |
| } else { |
| /* Compute the run length of zeros and add the appropriate number of 0 |
| and 17 code length codes to the code length code histogram. */ |
| uint32_t reps = 1; |
| size_t k; |
| for (k = i + 1; k < data_size && histogram->data_[k] == 0; ++k) { |
| ++reps; |
| } |
| i += reps; |
| if (i == data_size) { |
| /* 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[BROTLI_REPEAT_ZERO_CODE_LENGTH]; |
| /* 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 += (double)(18 + 2 * max_depth); |
| /* Add the entropy of the code length code histogram. */ |
| bits += BitsEntropy(depth_histo, BROTLI_CODE_LENGTH_CODES); |
| } |
| return bits; |
| } |
| |
| #undef HistogramType |
