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Just because a turing machine prints out 0 and 1 at each step doesn't mean the sequences to factor into the calculation of what to print out next can't be longe

Just because a turing machine prints out 0 and 1 at each step doesn't mean the sequences to factor into the calculation of what to print out next can't be longer binary sequences. Pretty much all the best compression methods are language agnostic and work on bit wise sequences. They also pretty much all predict the next bit and feed that into an alogithmic encoder. Eg. look up dynamic markov coding which is commonly used by Hutter prize winners. The paper is short and readable. They dynamically create a binary tree and binary sequences are seen so if the pattern '01101000 01100101' comes in it walks down the binary tree. It'll probably predict the next bit as '0' as '0110100001100101' just so happened to be a common sequence in English that will likely have a next bit of '0' but the Dynamic Markov coding model has no idea of that. It just has binary sequences of bits and a prediction of the next bit given that. Likewise it can continue reading the bits of the file in and walking down its binary tree where history of the next bit are stored in every node and it see's '111001001011110110100000'. It makes the prediction of the next bit as a likely '1' that it feeds into an arithmetic coder that takes predictions and forms an optimally minimal bitsequence from that. That second binary sequence forms part of 你好. In both cases the turing machine doesn't care about that. It's also just writing 1's and 0's as per a turing machine. Eventually those 1's and 0's form sequences that happen to map to characters in various languages but it doesn't care about that. >The mapping from strings in the binary language to letters in the target language cannot participate in the calculations for the complexity of a string in another language. Because if it did, again, you could make the Kolmogorov complexity of any arbitrary string S you choose to be 0, because you could just say the null output maps to S. One other thing to address here is that Kolmogorov complexity explicitly includes any dictionary you use in it's calculation. A dictionary of the file you wish to compress would just blow out your Kolmogorov complexity to that size exactly. That's why Kolmogorov complexity is an excellent tool. You explicitly cannot cheat in this way.

rhelz 3y ago
> the turing machine doesn't care about that. It's also just writing 1's and 0's as per a turing machine. But a Turing machine does not have to be restricted to just printing out zeros and ones. It can be any finite set of symbols. For example, he Soviets built a computer which used base 3--its symbol set was {-1, 0, 1}. It didn't have bits which could just store "1" or "0", it had trits which could store "-1", "0", or "1". And why the soviets built such a computer is germane to Kolmogorov complexity just because you can make shorter strings in base 3 than you can in base 2. The choice of symbol set absolutely impacts the length of strings and programs, and therefore impacts the Kolmogorov complexity of strings relative to the computer. With this in mind, please consider three Turing Machines: A, B, and C The symbols A prints out on its tape are {"A", "B", "C"} The symbols B prints out on its tape are {"0", "1"} The symbols C prints out on its tape are {"0", "1", "A", "B", "C"} Now consider two strings: "1000001" and "A". Turing machine B could print out "1000001". Turing machine A could print out "A". You might be tempted to equate "1000001" and "A". But consider the same two strings printed out by machine C: "1000001" and "A" Clearly, these are different strings. If C printed out "A", it did not print out "1000001", and vice versa. > Kolmogorov complexity explicitly includes any dictionary you use in it's calculation. Sure, but on a binary turing machine, like Machine B above, that dictionary is not going to be matching between binary strings and, say Roman letters. Its going to be mapping from one binary string to another binary string. Using Machine C, you certainly could write a program which inputed "1000001" and output "A". But you absolutely, positively, cannot write such a program in either machines A or B. Machine A cannot print the string "1000001". And B cannot print the string "A". Different strings, different things.
- AnotherGoodName 3y ago
  All of those examples are exactly equivalent and convertable but where you are confusing yourself is that you're thinking of the differing states of a given n-ary system as having explicit symbols. It's best to think of it as simply numeric. Binary 010 is 2 if i were to write in decimal. Ternary 020 is the number 6 if i were to write it in decimal. Etc. Any actual symbol mapping to the numbers from an n-ary system to actual symbols like you are showing here is actually arbitrary and part of the calculation of space when measuring an algorithms Kolmogov complexity. The program size in kolmogorov complexity includes the dictionary of numerical to symbolic lookup which is what you are getting at here.
  - rhelz 3y ago
    > where you are confusing yourself is that you're thinking of the differing states of a given n-ary system as having explicit symbols. Its best to think of it as simply numeric. But what is "simply numeric"? How do we explain numeric things, and how to compute with them? Uh, we use Turing machines. We don't explain Turing machines using numeric concepts, we explain numeric concepts using Turing machines. We explain what computation is, by describing how a Turing machine can manipulate symbols on its tape. But that's where the analysis stops. The symbols themselves are the simplest entities, there's nothing simpler than they are to explain them with. Nobody who is objecting to me here has ever noticed that the symbols "0" and "1" are atomic, and not further explained. Because that's where they bottom out--explanations have to stop somewhere. But you don't have to bottom out at {0,1}. Trinary computers use a different symbol set, {-1, 0 1}. Saying that "A" can be equated with "1000001" is as silly as saying we could store any of the three values -1, 0 or 1 in a single bit of a binary computer.
    - AnotherGoodName 3y ago
      There just simply is no symbol set for any machine. That's literally an arbitrary mapping. You can literally call the two states whatever you want. You can call one state the contents of file A and another state the contents of file B if you want. But that mapping has a size it takes and kolmogorov complexity very directly says you need to include the size of that mapping. End of story. The reason to think numerically is that there are n^y states for an n-ary machine. You can count them. And you can map each on of those to any other n-ary system. Note that Turing in his paper talked about states. "simply stated that the machine had certain m-configurations (enumerated) and all m-configurations obtainable". There's no binary machine that outputs 0 or 1 directly except by a mapping of their two states to symbols. Kolmogorov knew this and explicitly stated that any dictionary be included as part of the size measurement.

rhelz 3y ago

> the turing machine doesn't care about that. It's also just writing 1's and 0's as per a turing machine. But a Turing machine does not have to be restricted to just printing out zeros and ones. It can be any finite set of symbols. For example, he Soviets built a computer which used base 3--its symbol set was {-1, 0, 1}. It didn't have bits which could just store "1" or "0", it had trits which could store "-1", "0", or "1". And why the soviets built such a computer is germane to Kolmogorov complexity just because you can make shorter strings in base 3 than you can in base 2. The choice of symbol set absolutely impacts the length of strings and programs, and therefore impacts the Kolmogorov complexity of strings relative to the computer. With this in mind, please consider three Turing Machines: A, B, and C The symbols A prints out on its tape are {"A", "B", "C"} The symbols B prints out on its tape are {"0", "1"} The symbols C prints out on its tape are {"0", "1", "A", "B", "C"} Now consider two strings: "1000001" and "A". Turing machine B could print out "1000001". Turing machine A could print out "A". You might be tempted to equate "1000001" and "A". But consider the same two strings printed out by machine C: "1000001" and "A" Clearly, these are different strings. If C printed out "A", it did not print out "1000001", and vice versa. > Kolmogorov complexity explicitly includes any dictionary you use in it's calculation. Sure, but on a binary turing machine, like Machine B above, that dictionary is not going to be matching between binary strings and, say Roman letters. Its going to be mapping from one binary string to another binary string. Using Machine C, you certainly could write a program which inputed "1000001" and output "A". But you absolutely, positively, cannot write such a program in either machines A or B. Machine A cannot print the string "1000001". And B cannot print the string "A". Different strings, different things.

AnotherGoodName 3y ago
All of those examples are exactly equivalent and convertable but where you are confusing yourself is that you're thinking of the differing states of a given n-ary system as having explicit symbols. It's best to think of it as simply numeric. Binary 010 is 2 if i were to write in decimal. Ternary 020 is the number 6 if i were to write it in decimal. Etc. Any actual symbol mapping to the numbers from an n-ary system to actual symbols like you are showing here is actually arbitrary and part of the calculation of space when measuring an algorithms Kolmogov complexity. The program size in kolmogorov complexity includes the dictionary of numerical to symbolic lookup which is what you are getting at here.
- rhelz 3y ago
  > where you are confusing yourself is that you're thinking of the differing states of a given n-ary system as having explicit symbols. Its best to think of it as simply numeric. But what is "simply numeric"? How do we explain numeric things, and how to compute with them? Uh, we use Turing machines. We don't explain Turing machines using numeric concepts, we explain numeric concepts using Turing machines. We explain what computation is, by describing how a Turing machine can manipulate symbols on its tape. But that's where the analysis stops. The symbols themselves are the simplest entities, there's nothing simpler than they are to explain them with. Nobody who is objecting to me here has ever noticed that the symbols "0" and "1" are atomic, and not further explained. Because that's where they bottom out--explanations have to stop somewhere. But you don't have to bottom out at {0,1}. Trinary computers use a different symbol set, {-1, 0 1}. Saying that "A" can be equated with "1000001" is as silly as saying we could store any of the three values -1, 0 or 1 in a single bit of a binary computer.
  - AnotherGoodName 3y ago
    There just simply is no symbol set for any machine. That's literally an arbitrary mapping. You can literally call the two states whatever you want. You can call one state the contents of file A and another state the contents of file B if you want. But that mapping has a size it takes and kolmogorov complexity very directly says you need to include the size of that mapping. End of story. The reason to think numerically is that there are n^y states for an n-ary machine. You can count them. And you can map each on of those to any other n-ary system. Note that Turing in his paper talked about states. "simply stated that the machine had certain m-configurations (enumerated) and all m-configurations obtainable". There's no binary machine that outputs 0 or 1 directly except by a mapping of their two states to symbols. Kolmogorov knew this and explicitly stated that any dictionary be included as part of the size measurement.