Limit of compression (source coding theorem)

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Discussion Overview

The discussion revolves around the source coding theorem and its implications for data compression. Participants explore whether the theorem's requirement of N*H bits for encoding a message applies universally or if additional information about symbol relationships can lead to more efficient compression methods.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if the source coding theorem's limit of N*H bits applies only when the frequency of symbols is known, suggesting that knowledge of symbol sequences could allow for better compression.
  • Another participant argues that if a symbol x is almost always followed by symbol y, this relationship is already accounted for in the entropy calculation, implying that the entropy is effectively lowered.
  • A further contribution presents a scenario involving joint probabilities and conditional probabilities, indicating that different conditional probabilities can yield the same overall entropy despite differing symbol relationships.
  • One participant acknowledges a misunderstanding regarding the application of the formula, noting that conditional probabilities require a different model for computing the entropy rate.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the source coding theorem and the role of conditional probabilities in determining entropy. The discussion remains unresolved regarding the implications of these relationships for data compression.

Contextual Notes

Limitations include the dependence on the independence of choices in the application of the source coding theorem and the need for different models when considering conditional probabilities.

gop
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Hi

The source coding theorem says that one needs at least N*H bits to encode a message of length N and entropy H. This supposedly is the theoretical limit of data compression.

But is it? Or does it only apply to situations where only the frequency (probability) of a given symbol is known?

For example, If I know that the symbol x is always followed by the symbol y (or that this happens with a very high probability) couldn't I use this to construct a compression algorithm that needs fewer bits than N*H?

thx
 
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gop said:
For example, If I know that the symbol x is always followed by the symbol y (or that this happens with a very high probability) couldn't I use this to construct a compression algorithm that needs fewer bits than N*H?

If x is (almost) always followed by y, then that lowers the entropy. It's already taken into account.
 
But if I have something like this

p(x,y,z)= (1/3,1/3,1/3)

then I have entropy 1.585. But now I could have p(y|x) = 1/2 or p(y|x) = 1 as long as p(y|z)=1-p(y|x) the overall probability p(x,y,z) stays the same. So I have the same entropy. But in the case of p(y|x)=1 I can only use two symbols say p(xy,z) = (1/2,1/2) which has entropy 1.

I guess I'm missing something obvious here but...

thx
 
gop said:
I guess I'm missing something obvious here but...

Yes, your formula doesn't apply unless the choices are made independently.
 
ok I got it if I use conditional probabilities I need to use another model and another way to compute the entropy rate.

thx
 

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