Do lossless compression algorithms reduce entropy?












18














According to Wikipedia:




Shannon's entropy measures the information contained in a message as opposed to the portion of the message that is determined (or predictable). Examples of the latter include redundancy in language structure or statistical properties relating to the occurrence frequencies of letter or word pairs, triplets etc.




So entropy is a measure of the amount of information contained in a message. Entropy coders are used to losslessy compress such a message to the minimum number of bits needed to represent it (entropy). To me this looks like a perfect entropy encoder would be all that is needed to losslessy compress a message as much as possible.



Many compression algorithms however use steps before entropy coding to supposedly reduce the entropy of the message.



According to german Wikipedia




Entropiekodierer werden häufig mit anderen Kodierern kombiniert. Dabei dienen vorgeschaltete Verfahren dazu, die Entropie der Daten zu verringern.




In english:




Entropy coders are frequently combined with other encoders. Previous steps serve to reduce the entropy of the data.




i.e. bzip2 uses the Burrows-Wheeler-Transform followed by a Move-To-Front-Transform before applying entropy coding (Huffman coding in this case).



Do these steps really reduce the entropy of the message, which would imply reducing the amount of information contained in the message? This seems contradictory to me, since that would mean that information was lost during compression, preventing lossless decompression. Or do they merely transform the message to improve the efficiency of the entropy coding algorithm? Or does entropy not correspond directly to the amount of information in the message?










share|cite|improve this question







New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Could be a way to estimate the entropy though.
    – pipe
    yesterday
















18














According to Wikipedia:




Shannon's entropy measures the information contained in a message as opposed to the portion of the message that is determined (or predictable). Examples of the latter include redundancy in language structure or statistical properties relating to the occurrence frequencies of letter or word pairs, triplets etc.




So entropy is a measure of the amount of information contained in a message. Entropy coders are used to losslessy compress such a message to the minimum number of bits needed to represent it (entropy). To me this looks like a perfect entropy encoder would be all that is needed to losslessy compress a message as much as possible.



Many compression algorithms however use steps before entropy coding to supposedly reduce the entropy of the message.



According to german Wikipedia




Entropiekodierer werden häufig mit anderen Kodierern kombiniert. Dabei dienen vorgeschaltete Verfahren dazu, die Entropie der Daten zu verringern.




In english:




Entropy coders are frequently combined with other encoders. Previous steps serve to reduce the entropy of the data.




i.e. bzip2 uses the Burrows-Wheeler-Transform followed by a Move-To-Front-Transform before applying entropy coding (Huffman coding in this case).



Do these steps really reduce the entropy of the message, which would imply reducing the amount of information contained in the message? This seems contradictory to me, since that would mean that information was lost during compression, preventing lossless decompression. Or do they merely transform the message to improve the efficiency of the entropy coding algorithm? Or does entropy not correspond directly to the amount of information in the message?










share|cite|improve this question







New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Could be a way to estimate the entropy though.
    – pipe
    yesterday














18












18








18


10





According to Wikipedia:




Shannon's entropy measures the information contained in a message as opposed to the portion of the message that is determined (or predictable). Examples of the latter include redundancy in language structure or statistical properties relating to the occurrence frequencies of letter or word pairs, triplets etc.




So entropy is a measure of the amount of information contained in a message. Entropy coders are used to losslessy compress such a message to the minimum number of bits needed to represent it (entropy). To me this looks like a perfect entropy encoder would be all that is needed to losslessy compress a message as much as possible.



Many compression algorithms however use steps before entropy coding to supposedly reduce the entropy of the message.



According to german Wikipedia




Entropiekodierer werden häufig mit anderen Kodierern kombiniert. Dabei dienen vorgeschaltete Verfahren dazu, die Entropie der Daten zu verringern.




In english:




Entropy coders are frequently combined with other encoders. Previous steps serve to reduce the entropy of the data.




i.e. bzip2 uses the Burrows-Wheeler-Transform followed by a Move-To-Front-Transform before applying entropy coding (Huffman coding in this case).



Do these steps really reduce the entropy of the message, which would imply reducing the amount of information contained in the message? This seems contradictory to me, since that would mean that information was lost during compression, preventing lossless decompression. Or do they merely transform the message to improve the efficiency of the entropy coding algorithm? Or does entropy not correspond directly to the amount of information in the message?










share|cite|improve this question







New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











According to Wikipedia:




Shannon's entropy measures the information contained in a message as opposed to the portion of the message that is determined (or predictable). Examples of the latter include redundancy in language structure or statistical properties relating to the occurrence frequencies of letter or word pairs, triplets etc.




So entropy is a measure of the amount of information contained in a message. Entropy coders are used to losslessy compress such a message to the minimum number of bits needed to represent it (entropy). To me this looks like a perfect entropy encoder would be all that is needed to losslessy compress a message as much as possible.



Many compression algorithms however use steps before entropy coding to supposedly reduce the entropy of the message.



According to german Wikipedia




Entropiekodierer werden häufig mit anderen Kodierern kombiniert. Dabei dienen vorgeschaltete Verfahren dazu, die Entropie der Daten zu verringern.




In english:




Entropy coders are frequently combined with other encoders. Previous steps serve to reduce the entropy of the data.




i.e. bzip2 uses the Burrows-Wheeler-Transform followed by a Move-To-Front-Transform before applying entropy coding (Huffman coding in this case).



Do these steps really reduce the entropy of the message, which would imply reducing the amount of information contained in the message? This seems contradictory to me, since that would mean that information was lost during compression, preventing lossless decompression. Or do they merely transform the message to improve the efficiency of the entropy coding algorithm? Or does entropy not correspond directly to the amount of information in the message?







information-theory data-compression entropy






share|cite|improve this question







New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









robertrobert

9317




9317




New contributor




robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Could be a way to estimate the entropy though.
    – pipe
    yesterday


















  • Could be a way to estimate the entropy though.
    – pipe
    yesterday
















Could be a way to estimate the entropy though.
– pipe
yesterday




Could be a way to estimate the entropy though.
– pipe
yesterday










4 Answers
4






active

oldest

votes


















20














A lot of casual descriptions of entropy are confusing in this way because entropy is not quite as neat and tidy a measure as sometimes presented. In particular, the standard definition of Shannon entropy stipulates that it only applies when, as Wikipedia puts it, "information due to independent events is additive."



In other words, independent events must be statistically independent. If they aren't, then you have to find a representation of the data that defines events in ways that make them truly independent. Otherwise, you will overestimate the entropy.



To put it yet another way, Shannon entropy only applies to true probability distributions, and not to random processes in general. For concrete examples of processes that don't fit the assumptions of Shannon entropy, consider...



Markov processes



A Markov process is a series of events in which the most recent event is sampled from a distribution that depends on one or more previous events. Obviously a huge number of real-world phenomena are better modeled as Markov processes than as discrete, independent probability distributions. For example: the text you're reading right now!



The naively calculated Shannon entropy rate of a Markov process will always be greater than or equal to the true entropy rate of the process. To get the true entropy of the process, you need to take into account the statistical dependence between events. In simple cases, the formula for that looks like this:



$$ H(mathcal{S}) = - sum_i p_i sum_j p_i (j) log p_i (j) $$



This can also be represented like so:



$$ H(Y) = - sum_{ij} mu_i P_{ij} log P_{ij} $$



Again quoting Wikipedia, here "$mu_i$ is the asymptotic distribution of the chain" -- that is, the overall probability that a given event will occur over a long horizon.



This is all a complicated way of saying that even when you can calculate the overall probability of a given event in the long run, certain sequences of events are more likely than others in a Markov process. So for example, the following three strings of English words are increasingly less likely:




  • They ran to the tree

  • The tree ran to they

  • Tree the they to ran


But Shannon entropy will assess all three strings as equally likely. The Markov process entropy takes the difference into account, and as a result, it assigns a lower entropy rate to the process.



Entropy rates are model-dependent



If you zoom way out, here's the big picture: the entropy rate of a given sequence of events from an unknown source is model-dependent. You'll assign a different entropy rate to a particular series of events depending on how you model the process that generated them.



And very frequently, your model of the process isn't going to be quite correct. This isn't a simple or easy to solve problem. In fact, in general, it is impossible to assign a true entropy rate to a sufficiently long and complex sequence of events, if you don't know what the true underlying process is. This is a central result in algorithmic information theory.



What this means in practice is that given an unknown source of sequences of events, different models will yield different entropies, and it's impossible to know which is correct in the long run -- although the one that assigns the lowest entropy is probably the best.






share|cite|improve this answer























  • Thank you very much! This explains perfectly what the mistake in my reasoning was.
    – robert
    19 hours ago



















10














No, if the algorithm is lossless no steps in the compression sequence can reduce its entropy - otherwise it would not be able to be decompressed/decoded. However, the additional entropy may be stored in 'out-of-band' information - such as the list that needs to be maintained in order to decode the move-to-front transform.






share|cite|improve this answer








New contributor




Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
    – robert
    yesterday










  • Indeed, it doesn't (well, depending on the exact meaning of "close").
    – Grimy
    yesterday












  • The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
    – Luke Schwartzkopff
    17 hours ago










  • No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
    – immibis
    15 hours ago










  • Aah, you're right, that wasn't the best example :)
    – Luke Schwartzkopff
    6 hours ago



















3














They reduce the apparent entropy inherent in the structure of the original message. Or in other words they tune the message to make use of the strengths of the next stages of compression.



One simple example would be replacing the name in the end tags of xml with a special symbol. You can perfectly recreate the original xml from that but the compressor doesn't have to include the full name again in that place.



A more real-world example is png compression. It's entropy compressor is DEFLATE, which is a combination of Lempel-Ziff and Huffman. This means that it works best with values and patterns that repeat often. Most adjacent pixels tend to be similar colors. So each row is assigned a filter which turn the original pixel values into a differential encoding. This way the values that end up encoded by DEFLATE are mostly close to 0. In the extreme case this will turn a smooth gradient from all different values into a single value throughout the row which the LZ portion or DEFLATE makes very quick work of.






share|cite|improve this answer























  • Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
    – robert
    yesterday










  • with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
    – ratchet freak
    yesterday










  • But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
    – robert
    yesterday






  • 1




    @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
    – robert
    yesterday






  • 1




    @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
    – kutschkem
    yesterday



















3














Entropy coders do not compress the message to the minimum number of bits needed to represent it. I know it's tempting to think that, but it's not what they do. They aren't magic and they can't achieve that.



Instead, they do something a bit less magical -- but still useful. Suppose for the moment that we knew that each character of the message was chosen independently from some distribution. Then it would be possible to build a lossless compression algorithm that optimally compresses the messages. These algorithms are called entropy encoders.



Now real messages usually don't have that independence property. For instance, if you see a Q, it's likely that the next letter is a U. And so on. It is still possible to apply an entropy encoder algorithm to a real message, where each character isn't chosen independently of the rest. The algorithm will still be lossless, it can still be used for compression, and in practice, it will still often shorten the length of the message. However, it doesn't shorten it to the minimum possible length. It doesn't compress the message to something whose length is equal to the entropy of the message; it compresses it less than that.



Once you realize this property of entropy encoders, then the paradox evaporates.



In general, any lossless step never reduces the entropy of the message. However, it might put the message into a form where some other compression algorithm is more effective, so it might still be useful (on average) in practice.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "419"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    robert is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f102647%2fdo-lossless-compression-algorithms-reduce-entropy%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    20














    A lot of casual descriptions of entropy are confusing in this way because entropy is not quite as neat and tidy a measure as sometimes presented. In particular, the standard definition of Shannon entropy stipulates that it only applies when, as Wikipedia puts it, "information due to independent events is additive."



    In other words, independent events must be statistically independent. If they aren't, then you have to find a representation of the data that defines events in ways that make them truly independent. Otherwise, you will overestimate the entropy.



    To put it yet another way, Shannon entropy only applies to true probability distributions, and not to random processes in general. For concrete examples of processes that don't fit the assumptions of Shannon entropy, consider...



    Markov processes



    A Markov process is a series of events in which the most recent event is sampled from a distribution that depends on one or more previous events. Obviously a huge number of real-world phenomena are better modeled as Markov processes than as discrete, independent probability distributions. For example: the text you're reading right now!



    The naively calculated Shannon entropy rate of a Markov process will always be greater than or equal to the true entropy rate of the process. To get the true entropy of the process, you need to take into account the statistical dependence between events. In simple cases, the formula for that looks like this:



    $$ H(mathcal{S}) = - sum_i p_i sum_j p_i (j) log p_i (j) $$



    This can also be represented like so:



    $$ H(Y) = - sum_{ij} mu_i P_{ij} log P_{ij} $$



    Again quoting Wikipedia, here "$mu_i$ is the asymptotic distribution of the chain" -- that is, the overall probability that a given event will occur over a long horizon.



    This is all a complicated way of saying that even when you can calculate the overall probability of a given event in the long run, certain sequences of events are more likely than others in a Markov process. So for example, the following three strings of English words are increasingly less likely:




    • They ran to the tree

    • The tree ran to they

    • Tree the they to ran


    But Shannon entropy will assess all three strings as equally likely. The Markov process entropy takes the difference into account, and as a result, it assigns a lower entropy rate to the process.



    Entropy rates are model-dependent



    If you zoom way out, here's the big picture: the entropy rate of a given sequence of events from an unknown source is model-dependent. You'll assign a different entropy rate to a particular series of events depending on how you model the process that generated them.



    And very frequently, your model of the process isn't going to be quite correct. This isn't a simple or easy to solve problem. In fact, in general, it is impossible to assign a true entropy rate to a sufficiently long and complex sequence of events, if you don't know what the true underlying process is. This is a central result in algorithmic information theory.



    What this means in practice is that given an unknown source of sequences of events, different models will yield different entropies, and it's impossible to know which is correct in the long run -- although the one that assigns the lowest entropy is probably the best.






    share|cite|improve this answer























    • Thank you very much! This explains perfectly what the mistake in my reasoning was.
      – robert
      19 hours ago
















    20














    A lot of casual descriptions of entropy are confusing in this way because entropy is not quite as neat and tidy a measure as sometimes presented. In particular, the standard definition of Shannon entropy stipulates that it only applies when, as Wikipedia puts it, "information due to independent events is additive."



    In other words, independent events must be statistically independent. If they aren't, then you have to find a representation of the data that defines events in ways that make them truly independent. Otherwise, you will overestimate the entropy.



    To put it yet another way, Shannon entropy only applies to true probability distributions, and not to random processes in general. For concrete examples of processes that don't fit the assumptions of Shannon entropy, consider...



    Markov processes



    A Markov process is a series of events in which the most recent event is sampled from a distribution that depends on one or more previous events. Obviously a huge number of real-world phenomena are better modeled as Markov processes than as discrete, independent probability distributions. For example: the text you're reading right now!



    The naively calculated Shannon entropy rate of a Markov process will always be greater than or equal to the true entropy rate of the process. To get the true entropy of the process, you need to take into account the statistical dependence between events. In simple cases, the formula for that looks like this:



    $$ H(mathcal{S}) = - sum_i p_i sum_j p_i (j) log p_i (j) $$



    This can also be represented like so:



    $$ H(Y) = - sum_{ij} mu_i P_{ij} log P_{ij} $$



    Again quoting Wikipedia, here "$mu_i$ is the asymptotic distribution of the chain" -- that is, the overall probability that a given event will occur over a long horizon.



    This is all a complicated way of saying that even when you can calculate the overall probability of a given event in the long run, certain sequences of events are more likely than others in a Markov process. So for example, the following three strings of English words are increasingly less likely:




    • They ran to the tree

    • The tree ran to they

    • Tree the they to ran


    But Shannon entropy will assess all three strings as equally likely. The Markov process entropy takes the difference into account, and as a result, it assigns a lower entropy rate to the process.



    Entropy rates are model-dependent



    If you zoom way out, here's the big picture: the entropy rate of a given sequence of events from an unknown source is model-dependent. You'll assign a different entropy rate to a particular series of events depending on how you model the process that generated them.



    And very frequently, your model of the process isn't going to be quite correct. This isn't a simple or easy to solve problem. In fact, in general, it is impossible to assign a true entropy rate to a sufficiently long and complex sequence of events, if you don't know what the true underlying process is. This is a central result in algorithmic information theory.



    What this means in practice is that given an unknown source of sequences of events, different models will yield different entropies, and it's impossible to know which is correct in the long run -- although the one that assigns the lowest entropy is probably the best.






    share|cite|improve this answer























    • Thank you very much! This explains perfectly what the mistake in my reasoning was.
      – robert
      19 hours ago














    20












    20








    20






    A lot of casual descriptions of entropy are confusing in this way because entropy is not quite as neat and tidy a measure as sometimes presented. In particular, the standard definition of Shannon entropy stipulates that it only applies when, as Wikipedia puts it, "information due to independent events is additive."



    In other words, independent events must be statistically independent. If they aren't, then you have to find a representation of the data that defines events in ways that make them truly independent. Otherwise, you will overestimate the entropy.



    To put it yet another way, Shannon entropy only applies to true probability distributions, and not to random processes in general. For concrete examples of processes that don't fit the assumptions of Shannon entropy, consider...



    Markov processes



    A Markov process is a series of events in which the most recent event is sampled from a distribution that depends on one or more previous events. Obviously a huge number of real-world phenomena are better modeled as Markov processes than as discrete, independent probability distributions. For example: the text you're reading right now!



    The naively calculated Shannon entropy rate of a Markov process will always be greater than or equal to the true entropy rate of the process. To get the true entropy of the process, you need to take into account the statistical dependence between events. In simple cases, the formula for that looks like this:



    $$ H(mathcal{S}) = - sum_i p_i sum_j p_i (j) log p_i (j) $$



    This can also be represented like so:



    $$ H(Y) = - sum_{ij} mu_i P_{ij} log P_{ij} $$



    Again quoting Wikipedia, here "$mu_i$ is the asymptotic distribution of the chain" -- that is, the overall probability that a given event will occur over a long horizon.



    This is all a complicated way of saying that even when you can calculate the overall probability of a given event in the long run, certain sequences of events are more likely than others in a Markov process. So for example, the following three strings of English words are increasingly less likely:




    • They ran to the tree

    • The tree ran to they

    • Tree the they to ran


    But Shannon entropy will assess all three strings as equally likely. The Markov process entropy takes the difference into account, and as a result, it assigns a lower entropy rate to the process.



    Entropy rates are model-dependent



    If you zoom way out, here's the big picture: the entropy rate of a given sequence of events from an unknown source is model-dependent. You'll assign a different entropy rate to a particular series of events depending on how you model the process that generated them.



    And very frequently, your model of the process isn't going to be quite correct. This isn't a simple or easy to solve problem. In fact, in general, it is impossible to assign a true entropy rate to a sufficiently long and complex sequence of events, if you don't know what the true underlying process is. This is a central result in algorithmic information theory.



    What this means in practice is that given an unknown source of sequences of events, different models will yield different entropies, and it's impossible to know which is correct in the long run -- although the one that assigns the lowest entropy is probably the best.






    share|cite|improve this answer














    A lot of casual descriptions of entropy are confusing in this way because entropy is not quite as neat and tidy a measure as sometimes presented. In particular, the standard definition of Shannon entropy stipulates that it only applies when, as Wikipedia puts it, "information due to independent events is additive."



    In other words, independent events must be statistically independent. If they aren't, then you have to find a representation of the data that defines events in ways that make them truly independent. Otherwise, you will overestimate the entropy.



    To put it yet another way, Shannon entropy only applies to true probability distributions, and not to random processes in general. For concrete examples of processes that don't fit the assumptions of Shannon entropy, consider...



    Markov processes



    A Markov process is a series of events in which the most recent event is sampled from a distribution that depends on one or more previous events. Obviously a huge number of real-world phenomena are better modeled as Markov processes than as discrete, independent probability distributions. For example: the text you're reading right now!



    The naively calculated Shannon entropy rate of a Markov process will always be greater than or equal to the true entropy rate of the process. To get the true entropy of the process, you need to take into account the statistical dependence between events. In simple cases, the formula for that looks like this:



    $$ H(mathcal{S}) = - sum_i p_i sum_j p_i (j) log p_i (j) $$



    This can also be represented like so:



    $$ H(Y) = - sum_{ij} mu_i P_{ij} log P_{ij} $$



    Again quoting Wikipedia, here "$mu_i$ is the asymptotic distribution of the chain" -- that is, the overall probability that a given event will occur over a long horizon.



    This is all a complicated way of saying that even when you can calculate the overall probability of a given event in the long run, certain sequences of events are more likely than others in a Markov process. So for example, the following three strings of English words are increasingly less likely:




    • They ran to the tree

    • The tree ran to they

    • Tree the they to ran


    But Shannon entropy will assess all three strings as equally likely. The Markov process entropy takes the difference into account, and as a result, it assigns a lower entropy rate to the process.



    Entropy rates are model-dependent



    If you zoom way out, here's the big picture: the entropy rate of a given sequence of events from an unknown source is model-dependent. You'll assign a different entropy rate to a particular series of events depending on how you model the process that generated them.



    And very frequently, your model of the process isn't going to be quite correct. This isn't a simple or easy to solve problem. In fact, in general, it is impossible to assign a true entropy rate to a sufficiently long and complex sequence of events, if you don't know what the true underlying process is. This is a central result in algorithmic information theory.



    What this means in practice is that given an unknown source of sequences of events, different models will yield different entropies, and it's impossible to know which is correct in the long run -- although the one that assigns the lowest entropy is probably the best.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 17 hours ago

























    answered 22 hours ago









    senderlesenderle

    37617




    37617












    • Thank you very much! This explains perfectly what the mistake in my reasoning was.
      – robert
      19 hours ago


















    • Thank you very much! This explains perfectly what the mistake in my reasoning was.
      – robert
      19 hours ago
















    Thank you very much! This explains perfectly what the mistake in my reasoning was.
    – robert
    19 hours ago




    Thank you very much! This explains perfectly what the mistake in my reasoning was.
    – robert
    19 hours ago











    10














    No, if the algorithm is lossless no steps in the compression sequence can reduce its entropy - otherwise it would not be able to be decompressed/decoded. However, the additional entropy may be stored in 'out-of-band' information - such as the list that needs to be maintained in order to decode the move-to-front transform.






    share|cite|improve this answer








    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
      – robert
      yesterday










    • Indeed, it doesn't (well, depending on the exact meaning of "close").
      – Grimy
      yesterday












    • The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
      – Luke Schwartzkopff
      17 hours ago










    • No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
      – immibis
      15 hours ago










    • Aah, you're right, that wasn't the best example :)
      – Luke Schwartzkopff
      6 hours ago
















    10














    No, if the algorithm is lossless no steps in the compression sequence can reduce its entropy - otherwise it would not be able to be decompressed/decoded. However, the additional entropy may be stored in 'out-of-band' information - such as the list that needs to be maintained in order to decode the move-to-front transform.






    share|cite|improve this answer








    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
      – robert
      yesterday










    • Indeed, it doesn't (well, depending on the exact meaning of "close").
      – Grimy
      yesterday












    • The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
      – Luke Schwartzkopff
      17 hours ago










    • No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
      – immibis
      15 hours ago










    • Aah, you're right, that wasn't the best example :)
      – Luke Schwartzkopff
      6 hours ago














    10












    10








    10






    No, if the algorithm is lossless no steps in the compression sequence can reduce its entropy - otherwise it would not be able to be decompressed/decoded. However, the additional entropy may be stored in 'out-of-band' information - such as the list that needs to be maintained in order to decode the move-to-front transform.






    share|cite|improve this answer








    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    No, if the algorithm is lossless no steps in the compression sequence can reduce its entropy - otherwise it would not be able to be decompressed/decoded. However, the additional entropy may be stored in 'out-of-band' information - such as the list that needs to be maintained in order to decode the move-to-front transform.







    share|cite|improve this answer








    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered yesterday









    Luke SchwartzkopffLuke Schwartzkopff

    1113




    1113




    New contributor




    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Luke Schwartzkopff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
      – robert
      yesterday










    • Indeed, it doesn't (well, depending on the exact meaning of "close").
      – Grimy
      yesterday












    • The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
      – Luke Schwartzkopff
      17 hours ago










    • No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
      – immibis
      15 hours ago










    • Aah, you're right, that wasn't the best example :)
      – Luke Schwartzkopff
      6 hours ago


















    • So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
      – robert
      yesterday










    • Indeed, it doesn't (well, depending on the exact meaning of "close").
      – Grimy
      yesterday












    • The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
      – Luke Schwartzkopff
      17 hours ago










    • No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
      – immibis
      15 hours ago










    • Aah, you're right, that wasn't the best example :)
      – Luke Schwartzkopff
      6 hours ago
















    So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
    – robert
    yesterday




    So are the additional steps used in compression algorithms before entropy coding just used to allow the entropy coder to come closer to entropy? Does an entropy coder not come close to entropy on its own when applied to an arbitrary message?
    – robert
    yesterday












    Indeed, it doesn't (well, depending on the exact meaning of "close").
    – Grimy
    yesterday






    Indeed, it doesn't (well, depending on the exact meaning of "close").
    – Grimy
    yesterday














    The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
    – Luke Schwartzkopff
    17 hours ago




    The additional steps allow the entropy encoder to maintain the entropy of the original message while reducing superfluous information more effectively than if it were to be applied on its own. Whether you apply the pre-processing or not, entropy will be preserved, but compression would be less effective (you'd end up with a less efficient encoding).
    – Luke Schwartzkopff
    17 hours ago












    No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
    – immibis
    15 hours ago




    No, the move-to-front transform does not output a separate list that must be transferred to the decoder. Unless you mean the initial list.
    – immibis
    15 hours ago












    Aah, you're right, that wasn't the best example :)
    – Luke Schwartzkopff
    6 hours ago




    Aah, you're right, that wasn't the best example :)
    – Luke Schwartzkopff
    6 hours ago











    3














    They reduce the apparent entropy inherent in the structure of the original message. Or in other words they tune the message to make use of the strengths of the next stages of compression.



    One simple example would be replacing the name in the end tags of xml with a special symbol. You can perfectly recreate the original xml from that but the compressor doesn't have to include the full name again in that place.



    A more real-world example is png compression. It's entropy compressor is DEFLATE, which is a combination of Lempel-Ziff and Huffman. This means that it works best with values and patterns that repeat often. Most adjacent pixels tend to be similar colors. So each row is assigned a filter which turn the original pixel values into a differential encoding. This way the values that end up encoded by DEFLATE are mostly close to 0. In the extreme case this will turn a smooth gradient from all different values into a single value throughout the row which the LZ portion or DEFLATE makes very quick work of.






    share|cite|improve this answer























    • Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
      – robert
      yesterday










    • with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
      – ratchet freak
      yesterday










    • But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
      – robert
      yesterday






    • 1




      @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
      – robert
      yesterday






    • 1




      @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
      – kutschkem
      yesterday
















    3














    They reduce the apparent entropy inherent in the structure of the original message. Or in other words they tune the message to make use of the strengths of the next stages of compression.



    One simple example would be replacing the name in the end tags of xml with a special symbol. You can perfectly recreate the original xml from that but the compressor doesn't have to include the full name again in that place.



    A more real-world example is png compression. It's entropy compressor is DEFLATE, which is a combination of Lempel-Ziff and Huffman. This means that it works best with values and patterns that repeat often. Most adjacent pixels tend to be similar colors. So each row is assigned a filter which turn the original pixel values into a differential encoding. This way the values that end up encoded by DEFLATE are mostly close to 0. In the extreme case this will turn a smooth gradient from all different values into a single value throughout the row which the LZ portion or DEFLATE makes very quick work of.






    share|cite|improve this answer























    • Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
      – robert
      yesterday










    • with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
      – ratchet freak
      yesterday










    • But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
      – robert
      yesterday






    • 1




      @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
      – robert
      yesterday






    • 1




      @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
      – kutschkem
      yesterday














    3












    3








    3






    They reduce the apparent entropy inherent in the structure of the original message. Or in other words they tune the message to make use of the strengths of the next stages of compression.



    One simple example would be replacing the name in the end tags of xml with a special symbol. You can perfectly recreate the original xml from that but the compressor doesn't have to include the full name again in that place.



    A more real-world example is png compression. It's entropy compressor is DEFLATE, which is a combination of Lempel-Ziff and Huffman. This means that it works best with values and patterns that repeat often. Most adjacent pixels tend to be similar colors. So each row is assigned a filter which turn the original pixel values into a differential encoding. This way the values that end up encoded by DEFLATE are mostly close to 0. In the extreme case this will turn a smooth gradient from all different values into a single value throughout the row which the LZ portion or DEFLATE makes very quick work of.






    share|cite|improve this answer














    They reduce the apparent entropy inherent in the structure of the original message. Or in other words they tune the message to make use of the strengths of the next stages of compression.



    One simple example would be replacing the name in the end tags of xml with a special symbol. You can perfectly recreate the original xml from that but the compressor doesn't have to include the full name again in that place.



    A more real-world example is png compression. It's entropy compressor is DEFLATE, which is a combination of Lempel-Ziff and Huffman. This means that it works best with values and patterns that repeat often. Most adjacent pixels tend to be similar colors. So each row is assigned a filter which turn the original pixel values into a differential encoding. This way the values that end up encoded by DEFLATE are mostly close to 0. In the extreme case this will turn a smooth gradient from all different values into a single value throughout the row which the LZ portion or DEFLATE makes very quick work of.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    ratchet freakratchet freak

    2,60888




    2,60888












    • Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
      – robert
      yesterday










    • with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
      – ratchet freak
      yesterday










    • But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
      – robert
      yesterday






    • 1




      @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
      – robert
      yesterday






    • 1




      @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
      – kutschkem
      yesterday


















    • Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
      – robert
      yesterday










    • with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
      – ratchet freak
      yesterday










    • But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
      – robert
      yesterday






    • 1




      @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
      – robert
      yesterday






    • 1




      @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
      – kutschkem
      yesterday
















    Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
    – robert
    yesterday




    Does that mean apparent entropy is different from the actual information content of a message? How is that related to the actual entropy of the message?
    – robert
    yesterday












    with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
    – ratchet freak
    yesterday




    with "apparent entropy" I mean the entropy that the entropy encode can compress down to. Different encoder will have different patterns they look for. Huffman does best when the same few symbols are reused often used often, lempel-ziff does best when chunks are repeated, etc.
    – ratchet freak
    yesterday












    But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
    – robert
    yesterday




    But the Lempel-Ziv algorithms are not entropy coding algorithms, right? What I do not understand is why they are used before entropy coders in e.g. LZMA, when the entropy coder on its own could supposedly already compress the message down to its minimum.
    – robert
    yesterday




    1




    1




    @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
    – robert
    yesterday




    @kutschkem Does this mean entropy is not an absolute measure of the information content of a message but is relative to what is defined as a symbol (e.g. a single character is considered a symbol vs. 1 bit being considered a symbol)? I think that would explain where my assumptions were wrong.
    – robert
    yesterday




    1




    1




    @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
    – kutschkem
    yesterday




    @robert ... There is a tradeoff though, which is the "out-of-band" information Luke mentions in his answer, which is generally added by those steps (lookup tables to be able to decode the encoded information). So it makes no sense to define the whole content as one symbol, and encode it as 0 because somewhere the information has to be stored what this 0 encodes.
    – kutschkem
    yesterday











    3














    Entropy coders do not compress the message to the minimum number of bits needed to represent it. I know it's tempting to think that, but it's not what they do. They aren't magic and they can't achieve that.



    Instead, they do something a bit less magical -- but still useful. Suppose for the moment that we knew that each character of the message was chosen independently from some distribution. Then it would be possible to build a lossless compression algorithm that optimally compresses the messages. These algorithms are called entropy encoders.



    Now real messages usually don't have that independence property. For instance, if you see a Q, it's likely that the next letter is a U. And so on. It is still possible to apply an entropy encoder algorithm to a real message, where each character isn't chosen independently of the rest. The algorithm will still be lossless, it can still be used for compression, and in practice, it will still often shorten the length of the message. However, it doesn't shorten it to the minimum possible length. It doesn't compress the message to something whose length is equal to the entropy of the message; it compresses it less than that.



    Once you realize this property of entropy encoders, then the paradox evaporates.



    In general, any lossless step never reduces the entropy of the message. However, it might put the message into a form where some other compression algorithm is more effective, so it might still be useful (on average) in practice.






    share|cite|improve this answer


























      3














      Entropy coders do not compress the message to the minimum number of bits needed to represent it. I know it's tempting to think that, but it's not what they do. They aren't magic and they can't achieve that.



      Instead, they do something a bit less magical -- but still useful. Suppose for the moment that we knew that each character of the message was chosen independently from some distribution. Then it would be possible to build a lossless compression algorithm that optimally compresses the messages. These algorithms are called entropy encoders.



      Now real messages usually don't have that independence property. For instance, if you see a Q, it's likely that the next letter is a U. And so on. It is still possible to apply an entropy encoder algorithm to a real message, where each character isn't chosen independently of the rest. The algorithm will still be lossless, it can still be used for compression, and in practice, it will still often shorten the length of the message. However, it doesn't shorten it to the minimum possible length. It doesn't compress the message to something whose length is equal to the entropy of the message; it compresses it less than that.



      Once you realize this property of entropy encoders, then the paradox evaporates.



      In general, any lossless step never reduces the entropy of the message. However, it might put the message into a form where some other compression algorithm is more effective, so it might still be useful (on average) in practice.






      share|cite|improve this answer
























        3












        3








        3






        Entropy coders do not compress the message to the minimum number of bits needed to represent it. I know it's tempting to think that, but it's not what they do. They aren't magic and they can't achieve that.



        Instead, they do something a bit less magical -- but still useful. Suppose for the moment that we knew that each character of the message was chosen independently from some distribution. Then it would be possible to build a lossless compression algorithm that optimally compresses the messages. These algorithms are called entropy encoders.



        Now real messages usually don't have that independence property. For instance, if you see a Q, it's likely that the next letter is a U. And so on. It is still possible to apply an entropy encoder algorithm to a real message, where each character isn't chosen independently of the rest. The algorithm will still be lossless, it can still be used for compression, and in practice, it will still often shorten the length of the message. However, it doesn't shorten it to the minimum possible length. It doesn't compress the message to something whose length is equal to the entropy of the message; it compresses it less than that.



        Once you realize this property of entropy encoders, then the paradox evaporates.



        In general, any lossless step never reduces the entropy of the message. However, it might put the message into a form where some other compression algorithm is more effective, so it might still be useful (on average) in practice.






        share|cite|improve this answer












        Entropy coders do not compress the message to the minimum number of bits needed to represent it. I know it's tempting to think that, but it's not what they do. They aren't magic and they can't achieve that.



        Instead, they do something a bit less magical -- but still useful. Suppose for the moment that we knew that each character of the message was chosen independently from some distribution. Then it would be possible to build a lossless compression algorithm that optimally compresses the messages. These algorithms are called entropy encoders.



        Now real messages usually don't have that independence property. For instance, if you see a Q, it's likely that the next letter is a U. And so on. It is still possible to apply an entropy encoder algorithm to a real message, where each character isn't chosen independently of the rest. The algorithm will still be lossless, it can still be used for compression, and in practice, it will still often shorten the length of the message. However, it doesn't shorten it to the minimum possible length. It doesn't compress the message to something whose length is equal to the entropy of the message; it compresses it less than that.



        Once you realize this property of entropy encoders, then the paradox evaporates.



        In general, any lossless step never reduces the entropy of the message. However, it might put the message into a form where some other compression algorithm is more effective, so it might still be useful (on average) in practice.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        D.W.D.W.

        98.3k11117276




        98.3k11117276






















            robert is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            robert is a new contributor. Be nice, and check out our Code of Conduct.













            robert is a new contributor. Be nice, and check out our Code of Conduct.












            robert is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to Computer Science Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f102647%2fdo-lossless-compression-algorithms-reduce-entropy%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            What other Star Trek series did the main TNG cast show up in?

            Berlina muro

            Berlina aerponto