Arithmetic coding increases sequence length

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Giuseppe Esposito
Giuseppe Esposito 2018 年 6 月 25 日
編集済み: Michael Montouchet 2025 年 10 月 2 日
Hi all, I'm using the function "arithenco(seq,counts)" to compress a sequence of 1's,2's,3's and 4's of size 65536. The correspondent counts (number of occurrences for each symbol) is [1991,7759,52117,3669] so the symbol 3 shows an high probability to occur and I would expect a compression gain from the arithmetic code. But this doesn't happen, and the function outputs a code of size 66424 (longer than the original), how is possible? Thank you for the attention.
  1 件のコメント
Michael Montouchet
Michael Montouchet 2025 年 10 月 2 日
編集済み: Michael Montouchet 2025 年 10 月 2 日
To write your sequence of 1, 2, 3, 4 as a binary sequence, you need at least 2 bits per symbol. The space required to write it as a binary sequence is 2 * 65536 bits.
The arithmetic code is made of 0 and 1, so you need 1 bit per symbol. The space required to write it as a binary sequence is 1 * 66424 bits.
So you managed to compress an initial input into a code that is nearly half of the initial size.

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Michael Montouchet
Michael Montouchet 2025 年 10 月 2 日
編集済み: Michael Montouchet 2025 年 10 月 2 日
To write your sequence of 1, 2, 3, 4 as a binary sequence, you need at least 2 bits per symbol. The space required to write it as a binary sequence is 2 * 65536 bits.
The arithmetic code is made of 0 and 1, so you need 1 bit per symbol. The space required to write it as a binary sequence is 1 * 66424 bits.
So you managed to compress an initial input into a code that is nearly half of the initial size.

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