Augustow Noclegi Staszica 1800petmeds

### Augustow Noclegi Staszica 1800petmeds

To widzi wszystko najlepsze tej pięknej przyjaźni muzykę bezbarwną stronę path from the root For example, suppose that we are given the alphabet with each symbol having probability 0. We start with each symbol a one-node tree: .1 .1 .1 .1 .1 0 2 4 6 8 Because each small tree has the same probability, we pick any two and combine them: .2 .1 .1 .1 .1 .1 1 3 5 7 9 Continuing, .2 .2 .1 .1 .1 .1 .1 0 2 4 6 8 At this point, 8 and 9 have the two lowest probabilities we have to choose those: .2 .2 .2 .1 .1 .1 .1 .1 1 3 5 7 9 all of the trees have probability .2 we choose any pair of them: .4 .2 .2 .1 .1 .1 .1 .1 0 2 4 6 8 We choose any two of the three remaining trees with probability .2 .4 .2 .2 .1 .1 .1 .1 .1 0 2 4 6 8 the two smallest probabilities are .2 and one of the .4 .4 .2 .2 .2 .1 .1 .1 .1 .1 1 3 5 7 9 the two smallest are .4 and .6. After this step, the tree is finished. We can label the branches 0 for left and 1 for right, although the choice is arbitrary. 1 0 .6 1 .4 1 1 .2 .2 .1 .1 .1 .1 .1 0 2 4 6 8 From this tree we construct the code: Symbol Code 0 1 2 3 4 5 6 7 8 9 A code be static or dynamic. A static code is computed by the compressor and transmitted to the decompresser as part of the compressed data. A dynamic code is computed by the compressor and periodically updated, but not transmitted. Instead, the decompresser reconstructs the code using exactly the same algorithm using the previously decoded data to estimate the probabilities. Neither method compresses better because any space saved by not transmitting the model is paid back by having less data with which to estimate probabilities. Huffman codes are typically static, mainly for speed. The compressor only needs to compute the code once, using the entire input to compute probabilities. To transmit a Huffman table, it is only necessary to send the size of each symbol, for example: Both the compressor and decompresser would then assign codes by starting with the shortest symbols, counting up from 0, and appending a 0 bit whenever the code gets longer. This would result the following different but equally effective code: Symbol Size Code 0 000 3 2 010 3 8 100 3 4 1100 4 6 1110 4 For file compression, Huffman coded data still needs to be packed into bytes. JPEG packs bits MSB to LSB order. For example, the codes 00001 would be packed as 00001001....... The deflate format used zip, gzip, and png files packs bits LSB to MSB order, as if each byte is written backward, i.e. 10010000. One other complication is that the last byte has to be padded such a way that it is not interpreted as a Huffman code. JPEG does this by not assigning any symbol to a code of all 1 bits, and then padding the last byte with 1 bits. Deflate handles this by reserving a code to indicate the end of the data. This tells the decoder not to decode the remaining bits of the last byte. Huffman coding has the drawback that code lengths must be a whole number of bits. This effectively constrains the model to probabilities that are powers of 1. The size penalty for modeling errors is roughly proportional to the square of the error. For example, a 10% error results a 1% size penalty. The penalty can be large for small codes. For example, the only possible ways to Huffman code a binary alphabet is to code each bit as itself resulting no compression. Arithmetic coding also called range coding, does not suffer from this difficulty. Let P be a model, meaning that for any string x, P is the probability of that string. Let P be the sum of the probabilities of all strings lexicographically less than x. Let P P P. Then the arithmetic code for a string x is the shortest binary number y such that P y prediction and count int context 0; bytewise order n context Compress byte c MSB to LSB order void compress context CONTEXT_SIZE; Decompress and return a byte int decompress c 256; decoded byte context CONTEXT_SIZE; return c; Update the model void update The compress as a binary number 1 followed by up to 7 earlier bits. For example, if c 00011100, then bit_context takes the 8 successive values 1, 100, 10001, 1000111. decompress and adjusts the prediction inverse proportion to the count. The count is incremented up to a maximum value. At this point, the model switches from stationary to