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If I read a book I can see with my inner eye (or ear) see the plot developing. With a random string of letters, with random spacings, this isn't the case. Then why contains the former less information than the latter?

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  • $\begingroup$ Related: physics.stackexchange.com/q/263197/44126 $\endgroup$ – rob Jan 26 '17 at 8:55
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    $\begingroup$ Mathematically, you have to look at the distribution of all possible books (entropy is a function defined on statistical distributions, not statistical samples). Physically/practically, try compressing it in a .zip file! A written word text document will be compressed a lot. The random character file will barely be compressed at all. Hence the latter has more information, in the sense that it takes more bits to represent. $\endgroup$ – user12029 Jan 26 '17 at 8:57
  • $\begingroup$ Google "Kolmogorov complexity" (or "algorithmic complexity", and note that G.Chaitin has written lots about this) for much additional info along the lines described here. And note that @NeuroFuzzy seems to have, very logically, misread your question which explicitly asks why the non-random string contains more info. And actually, you're wrong, the random string contains more, like everybody said. But the combination (maybe tensor product) of the string's info with info already in your mind is what counts here. And the non-random string, acting as "input", generates much more "output". $\endgroup$ – John Forkosh Jan 26 '17 at 10:37
  • $\begingroup$ 1. Your title is the exact opposite of the question in the body: "[...]less information in a written book" vs "[...]the former [book] more information than the letter [random string]". Please fix this. 2. How is this a physics question? It appears purely information-theoretic in nature, with no direct physical connection. 3. You need to carefully distinguish between the colloquial meaning of "information" and the technical meaning of information given by measures like Shannon entropy or Kolmogorov complexity. Please give the definition of "information" you are using here. $\endgroup$ – ACuriousMind Jan 26 '17 at 14:15
  • $\begingroup$ @ACuriousMind Hmm ... understanding that entropy is a feature of the (possibly constrained) ensemble of possible microstates representing a macrostate rather than of the specific microstate in front of you at this moment is as important in statistical physics as in information theory. Admittedly the OP hadn't framed in that fashion, but if I was to answer this question it is the approach I would take. $\endgroup$ – dmckee Jan 26 '17 at 17:50
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I ran out of room before completing my comment above, which I cut-and-pasted below, and then continued below that...

Google "Kolmogorov complexity" (or "algorithmic complexity", and note that G.Chaitin has written lots about this) for much additional info along the lines described here. And note that @NeuroFuzzy seems to have, very logically, misread your question which explicitly asks why the non-random string contains more info. And actually, you're wrong, the random string contains more, like everybody said. But the combination (maybe tensor product) of the string's info with info already in your mind is what counts here. And the non-random string, acting as "input", generates much more "output"

...to continue, the preceding input$\to$output analogy is a bit too simplistic, which is why I need more words here. A more accurate analogy is between a Turing machine state and a "brain state". An input string, random or non-random, that you read, takes your brain word-by-word from some initial state to some final state. Reading random words/strings doesn't take your brain much of anywhere, semantically speaking. That is, the "meaning" of your final brain state contains no more information than your initial state. On the other hand, a non-random string takes your brain to a final state with additional semantic meaning.

"Meaning" seems vague, but is actually mathematically precise in Domain Theory, where Denotational Semantics provides what's called a semantic function from syntax (our random or non-random strings) to semantics. And domain elements are characterized by a poset-like ordering that measures information content. But you'd pretty much need a textbook's worth of discussion for a reasonably complete picture. If interested, maybe try https://en.wikipedia.org/wiki/Denotational_semantics for a start.

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Firstly, the concept of information vs. noise boils down to the concept of known coding vs. unknown (or useless) coding of information. So-called 'noise' is just information that we don't know the coding for, don't care, or both--like information on the particular configuration of the groups of particles (temperature) comprising a transmission line. Normally we'd call that 'noise'. Whether a given piece of 'noise' information is even theoretically decodable (has a meaning that can be resolved, whether we want to know it or not), is the subject of another conversation.

Now I might decide the letter 'W' represents the entire works of Tolstoy. But somewhere the works must be recorded for that to have meaning. It must be in the original data (as a stack of books), the compression dictionary, in our brains, or combinations thereof. For example, I might see the uncompressed word 'tomorrow' on a page of a book. Most of that word's information must be in my brain in the form of a description of that set of letters and a detailed description of what that word means. So 'information' is a measure of the total data needed to describe a concept, not just the encrypted symbol of that concept.

Your list of random letters might have been generated by a pseudo-random algorithm given certain input. So if you know that algorithm (i.e. its in your compression dictionary), then you could resolve the input of that algorithm that generated that 'noise', and thus its no longer 'noise' in the theoretical sense.

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Because you cannot compress the string of random letters, whereas you can compress words from a specific language. Example: axbheodpiurt hspe dirbe siwoebx , versus , one car one man one universe. The first string you cannot write it as anything but the sequence of about 30 letters and spaces. The second string you can write it as 1 car 1 man 1 universe for a total of about 23 letters and spaces. 30 is more than 23. Now of course there has to be an index saying one=1, but in a book there will be most likely a lot of "one" words , thereby reducing "one" to "1" everytime "one" is used.

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  • $\begingroup$ @NickUshor-So by assigning a number to each word of the English language you need lesser numbers than letters? $\endgroup$ – descheleschilder Jan 27 '17 at 1:13
  • $\begingroup$ Again an example: "Maryelizabeth had a lamb and maryelizabeths lamb talked to mary elizabeth, but maryelixabeth didnt understand it, only maryelisabeths mother understood it." Thats about 150 characters. This sentence can be compressed to: " maryelizabeth=* . *had a lamb and *s lamb talked to *, but * didnt understand it, only *s mother understood it" Thats about 110 characters. This is the basis of information compression , I just compressed 150 down to 110. Therefore the original sentence had only 110 units of information so to speak. Now write 150 random characters and try to compress it. $\endgroup$ – Nick Ushor Jan 27 '17 at 10:58
  • $\begingroup$ @NickUsher-The random characters can't be comprised. So it's becáuse the strings of characters have a meaning, they possess less information? I understand, but it sounds counterintuitive. $\endgroup$ – descheleschilder Jan 28 '17 at 15:09
  • $\begingroup$ @descheleschilder Its because some of the strings of characters repeat at regular intervals. If you have a paper that has written on it maryelisabeth 100 times, for a total of 1300 characters, you can just write " maryelisabeth written 100 times" and you will have about 25 characters that have as much information. $\endgroup$ – Nick Ushor Jan 28 '17 at 16:40
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You might think the written book has more information than the string of letters and spaces. But that's only because you happen to have personal opinions on what is useful information (words) and what is useless information (random letters and spaces). But say you don't have a preference for dictionary words over random words, and look at it from a different point of view. Imagine you were trying to memorize a book vs a string of letters and spaces. You'd have an easier time with the book, because it is made up of fewer arrangement of letters. It is in this sense the book is considered to have less information. The book is easier to store (remember) than the random strings of letters.

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The English language contains about 170 000 words. Some words occur more frequently than others. So if we assign the number 0 to the spaces between words (which doesn't reduce the information) the number 1 to the most frequently used words, the number 2 to the second most frequently used ones, etc. until we arrive at the less frequently used word to which we assign the number 170 000, we can replace the string of words by a string of decimal digits (with a space between them) varying from 0 to 170 000, which obviously contains lesser characters (decimal digits) than the number of alphabetic characters in the original text. For simplicity, I omitted the grammar (syntax) of the words, which gives us a bigger number of words (I am, you are, he/she is, etc.)

Of course, you need a "dictionary", in which you can see what number to assign to each word, and because of that, the text will not be easy to read. But the number of characters is reduced, and so is the information.

The meaning of the words (for example the associations I have in my mind when reading the word "woman") is not information inherent in the written text, but (literally) contextual information, which doesn't count as an objective measure.

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protected by Qmechanic Jan 26 '17 at 10:35

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