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This scheme works well for incrementally encoding a message. Enough accuracy is retained during the double-precision integer calculations to ensure that the message is accurately encoded. But there is potential for a loss of precision under certain circumstances.

If the encoded word has a string of 0s or 9s in it, the high and low values will slowly converge on a value, but they may not see their most significant digits match immediately. High may be 700004, and low may be 699995. At this point, the calculated range will be only a single digit long, which means the output word will not have enough precision to be accurately encoded. Worse, after a few more iterations, high could be 70000, and low could be 69999.

At this point, the values are permanently stuck. The range between high and low has become so small that any iteration through another symbol will leave high and low at their same values. But since the most significant digits of both words are not equal, the algorithm can’t output the digit and shift. It seems to have reached an impasse.

You can defeat this underflow problem by preventing things from ever getting that bad. The original algorithm said something like, “If the most significant digit of high and low match, shift it out.” If the two digits don’t match, but are now on adjacent numbers, a second test needs to be applied. If high and low are one apart, we test to see if the second most significant digit in high is a 0 and if the second digit in low is a 9. If so, it means we are on the road to underflow and need to take action.

Head off underflow with a slightly different shift operation. Instead of shifting the most significant digit out of the word, delete the second digits from high and low and shift the rest of the digits left to fill the space. The most significant digit stays in place. Then set an underflow counter to remember that we threw away a digit and aren’t quite sure whether it was going to be a 0 or a 9. This process is shown here:

Before | After | |
---|---|---|

High: | 40344 | 43449 |

Low: | 39810 | 38100 |

Underflow: | 0 | 1 |

After every recalculation, check for underflow digits again if the most significant digit don’t match. If underflow digits are present, we shift them out and increment the counter.

When the most significant digits do finally converge to a single value, output that value. Then output the underflow digits previously discarded. The underflow digits will all be 9s or 0s, depending on whether high and low converged on the higher or lower value. In the C implementation of this algorithm, the underflow counter keeps track of how many ones or zeros to output.

In the “ideal” decoding process, we had the entire input number to work with, and the algorithm had to do things like “divide the encoded number by the symbol probability.” In practice, we can’t perform an operation like that on a number that could be billions of bytes long. As in the encoding process, however, the decoder can operate using 16- and 32-bit integers for calculations.

Instead of using just two numbers, high and low, the decoder has to use three numbers. The first two, high and low, correspond exactly to the high and low values maintained by the encoder. The third number, code, contains the current bits being read in from the input bit stream. The code value always falls between the high and low values. As they come closer and closer to it, new shift operations will take place, and high and low will move back away from code.

The high and low values in the decoder will be updated after every symbol, just as they were in the encoder, and they should have exactly the same values. By performing the same comparison test on the upper digit of high and low, the decoder knows when it is time to shift a new digit into the incoming code. The same underflow tests are performed as well.

In the ideal algorithm, it was possible to determine what the current encoded symbol was just by finding the symbol whose probabilities enclosed the present value of the code. In the integer math algorithm, things are somewhat more complicated. In this case, the probability scale is determined by the difference between high and low. So instead of the range being between .0 and 1.0, the range will be between two positive 16-bit integer counts. Where the present code value falls along that range determines current probability. Divide (value - low) by (high - low + 1) to get the actual probability for the present symbol.

It is not immediately obvious why this encoding process is an improvement over Huffman coding. It becomes clear when we examine a case in which the probabilities are a little different. If we have to encode the stream “AAAAAAA,” and the probability of A is known to be .9, there is a 90 percent chance that any incoming character will be the letter A. We set up our probability table so that A occupies the .0 to . 9 range, and the end-of-message symbol occupies the .9 to 1 range. The encoding process is shown next:

New Character | Low value | High value |
---|---|---|

0.0 | 1.0 | |

A | 0.0 | 0.9 |

A | 0.0 | 0.81 |

A | 0.0 | 0.729 |

A | 0.0 | 0.6561 |

A | 0.0 | 0.59049 |

A | 0.0 | 0.531441 |

A | 0.0 | 0.4782969 |

END | 0.43046721 | 0.4782969 |

Now that we know the range of high and low values, all that remains is to pick a number to encode this message. The number .45 will make this message uniquely decode to “AAAAAAA.” Those two decimal digits take slightly less than seven bits to specify, which means we have encoded eight symbols in less than eight bits! An optimal Huffman message would have taken a minimum of nine bits.

To take this point to an even further extreme, I set up a test that had only two symbols. In it, 0 had a 16382/16383 probability, and an end-of-file symbol had a 1/16383 probability. I then created a file filled with 100,000 0s. After compression using arithmetic coding, the output file was only three bytes long! The minimum size using Huffman coding would have been 12,501 bytes. This is obviously a contrived example, but it shows that arithmetic coding compresses data at rates much better than one bit per byte when the symbol probabilities are right.

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