Simple compression algorithm - how to avoid marker-strings? - algorithm

I was trying to come up with a string compression algorithm for plaintext, e.g.
AAAAAAAABB -> A#8BB
where n symbols y are written out like
y#n
The problem is: what if I need to compress the string "A#8" ? That would confuse the decompression algorithm into thinking that the original input was "AAAAAAAA" instead of just "A#8".
How can I solve this problem? I was thinking of using a "marker" character instead of the #, but what if I wanted the algorithm to work with binary data? There is no marker character that can be used in that case I suppose

A simple solution is escaping: you could represent each # in the source by ##.
Everytime you encounter an #, you look one character ahead and find either a number (repeat previous character) or another # (its literally #).
A variant would be encoding each # as ##1, which would fit nicely into your current scheme and allows encoding n consecutive # as ##n.

Related

Repetition-based, pattern-based data compression algorithm

Suppose I have the following string:
ABCADCADCADABC
I want to compress it by finding repeating substrings.
What's an algorithm that gives the optimal compression?
In the above example it should return
AB*1 CAD*3 ABC*1
For comparison, a greedy algorithm might return
ABC*1 ADC*2 AD*1 ABC*1
Depending on whether you prefer fast and simple or high compression ratio you could take a look into the Lempel-Ziv-Welch (LZW) or Lempel-Ziv-Markov chain (LZMA) algorithms. They both keep dictionaries of recurring strings.
This sounds like a job for suffix arrays/trees!
http://en.wikipedia.org/wiki/Suffix_array
You can use a suffix array built over your string to figure out patterns that repeat. For instance, we can build a suffix array over your example as follows (I'm using $ as always coming after every letter, you can sort it so that $ comes before every letter ... either way will work):
ABCADCADCADABC$
ABC$
ADABC$
ADCADABC$
ADCADCADABC$
BCADCADCADABC$
BC$
CADABC$
CADCADABC$
CADCADCADABC$
C$
DABC$
DCADABC$
DCADCADABC$
$
From this, we can more easily see the common patterns in the string. Using the information in this suffix array representation, we can see that CAD is repeated 3x in a local area, and we'd likely use this as our choice for compression. ADC and DCA and so on are not as attractive because they compress less of the string.
http://en.wikipedia.org/wiki/Suffix_tree
Suffix trees are more efficient ways of doing the same task. Once you wrap your head around how to do something using suffix arrays, it's not too far of a jump to go onto suffix trees. In fact, this is used in popular compression algorithms including LZW 1 and BWT (Bzip) 2.
It may not be practically relevant, but for the particular question you ask there is a dynamic programming solution. If you have computed the optimum way to compress the strings of length 1, 2, 3...n-1 starting from the first character, then you can compute the optimum way to compress the string of length n starting from the first character by looking at the last k characters for each possibility k and seeing if they form a multiple of a simple string. If so, compute the cost of compressing the first n-k characters and then expressing the last k characters using a multiple of a string.
So in your example you would finish up by noticing that ABC was a multiple of itself, and that if you expressed this as ABC*1 you could use the answer you had already worked out for the first 11 characters of AB CAD*3 to produce AB*1 CAD*3 ABC*1
Better still would be:
ABCAD(6,3)(3,11)
where (n,d) is a length and distance back of a match. So (6,3) copies six bytes starting from three bytes back. While that may sound a little odd, by the time it gets three bytes in, the next three bytes it needs have been copied. So CADCAD is appended. The (3,11) causes ABC to be appended.
This is called LZ77 compression. It is what is implemented by zip, gzip, and zlib using the deflate compressed data format. That format not only references previous string matches, but also uses Huffman compression on the literals (e.g. ABCAD) as well as the lengths and distances.

What's the name of this algorithm/routine?

I am writing a utility class which converts strings from one alphabet to another, this is useful in situations where you have a target alphabet you wish to use, with a restriction on the number of characters available. For example, if you can use lower case letters and numbers, but only 12 characters its possible to compress a timestamp from the alphabet 01234567989 -: into abcdefghijklmnopqrstuvwxyz01234567989 so 2010-10-29 13:14:00 might become 5hhyo9v8mk6avy (19 charaters reduced to 16).
The class is designed to convert back and forth between alphabets, and also calculate the longest source string that can safely be stored in a target alphabet given a particular number of characters.
Was thinking of publishing this through Google code, however I'd obviously like other people to find it and use it - hence the question on what this is called. I've had to use this approach in two separate projects, with Bloomberg and a proprietary system, when you need to generate unique file names of a certain length, but want to keep some plaintext, so GUIDs aren't appropriate.
Your examples bear some similarity to a Dictionary coder with a fixed target and source dictionaries. Also worthwhile to look at is Fibonacci coding, which has a fixed target dictionary (of variable-length bits), which is variably targeted.
I think it also depends whether it is very important that your target alphabet has fixed width entries - if you allow for a fixed alphabet with variable length codes, your compression ratio will approach your entropy that much more optimally! If the source alphabet distribution is known in advance, a static Huffman tree could easily be generated.
Here is a simple algorithm:
Consider that you don't have to transmit the alphabet used for encoding. Also, you don't use (and transmit) the probabilities of the input symbols, as in standard compressions, so we just re-encode somehow the data.
In this case we can consider that the input data are in number represented with base equal to the cardinality of the input alphabet. We just have to change its representation to another base, that is a simple task.
EDITED example:
input alpabet: ABC, output alphabet: 0123456789
message ABAC will translate to 0102 in base 3, that is 11 (9 + 2) in base 10.
11 to base 10: 11
We could have a problem decoding it, because we don't know how many 0-es to use at the begining of the decoded result, so we have to use one of the modifications:
1) encode somehow in the stream the size of compressed data.
2) use a dummy 1 at the start of the stream: in this way our example will become:
10102 (base 3) = 81 + 9 + 2 = 92 (base 10).
Now after decoding we just have to ignore the first 1 (this also provides a basic error detection).
The main problem of this approach is that in most cases (GCD == 1) each new encoded character will completely change the output. This will be very inneficient and difficult to implement. We end up with arithmetic coding as the best solution (actually a simplified version of it).
You probably know about Base64 which does the same thing just usually the other way around. Too bad there are way too many Google results on BaseX or BaseN...

decoding algorithm wanted

I receive encoded PDF files regularly. The encoding works like this:
the PDFs can be displayed correctly in Acrobat Reader
select all and copy the test via Acrobat Reader
and paste in a text editor
will show that the content are encoded
so, examples are:
13579 -> 3579;
hello -> jgnnq
it's basically an offset (maybe swap) of ASCII characters.
The question is how can I find the offset automatically when I have access to only a few samples. I cannot be sure whether the encoding offset is changed. All I know is some text will usually (if not always) show up, e.g. "Name:", "Summary:", "Total:", inside the PDF.
Thank you!
edit: thanks for the feedback. I'd try to break the question into smaller questions:
Part 1: How to detect identical part(s) inside string?
You need to brute-force it.
If those patterns are simple like +2 character code like in your examples (which is +2 char codes)
h i j
e f g
l m n
l m n
o p q
1 2 3
3 4 5
5 6 7
7 8 9
9 : ;
You could easily implement like this to check against knowns words
>>> text='jgnnq'
>>> knowns=['hello', '13579']
>>>
>>> for i in range(-5,+5): #check -5 to +5 char code range
... rot=''.join(chr(ord(j)+i) for j in text)
... for x in knowns:
... if x in rot:
... print rot
...
hello
Is the PDF going to contain symbolic (like math or proofs) or natural language text (English, French, etc)?
If the latter, you can use a frequency chart for letters (digraphs, trigraphs and a small dictionary of words if you want to go the distance). I think there are probably a few of these online. Here's a start. And more specifically letter frequencies.
Then, if you're sure it's a Caesar shift, you can grab the first 1000 characters or so and shift them forward by increasing amounts up to (I would guess) 127 or so. Take the resulting texts and calculate how close the frequencies match the average ones you found above. Here is information on that.
The linked letter frequencies page on Wikipedia shows only letters, so you may want to exclude them in your calculation, or better find a chart with them in it. You may also want to transform the entire resulting text into lowercase or uppercase (your preference) to treat letters the same regardless of case.
Edit - saw comment about character swapping
In this case, it's a substitution cipher, which can still be broken automatically, though this time you will probably want to have a digraph chart handy to do extra analysis. This is useful because there will quite possibly be a substitution that is "closer" to average language in terms of letter analysis than the correct one, but comparing digraph frequencies will let you rule it out.
Also, I suggested shifting the characters, then seeing how close the frequencies matched the average language frequencies. You can actually just calculate the frequencies in your ciphertext first, then try to line them up with the good values. I'm not sure which is better.
Hmmm, thats a tough one.
The only thing I can suggest is using a dictionary (along with some substitution cipher algorithms) may help in decoding some of the text.
But I cannot see a solution that will decode everything for you with the scenario you describe.
Why don't you paste some sample input and we can have ago at decoding it.
It's only possible then you have a lot of examples (examples count stops then: possible to get all the combinations or just an linear values dependency or idea of the scenario).
also this question : How would I reverse engineer a cryptographic algorithm? have some advices.
Do the encoded files open correctly in PDF readers other than Acrobat Reader? If so, you could just use a PDF library (e.g. PDF Clown) and use it to programmatically extract the text you need.

Symmetric Bijective String Algorithm?

I'm looking for an algorithm that can do a one-to-one mapping of a string onto another string.
I want an algorithm that given an alphabet I can perform a symmetric mapping function.
For example:
Let's consider that I have the alphabet "A","B","C","D","E","F". I want something like F("ABC") = "CEA" and F("CEA") = "ABC" for every N letter permutation.
Surely, an algorithm like this exists. If you know of an algorithm, please post the name of it and I can research it. If I haven't been clear enough in my request, please let me know.
Thanks in advance.
Edit 1:
I should clarify that I want enough entropy so that F("ABC") would equal "CEA" and F("CEA") = "ABC" but then I do NOT want F("ABD") to equal "CEF". Notice how two input letters stayed the same and the two corresponding output letters stayed the same?
So a Caesar Cipher/ROT13 or shuffling the array would not be sufficient. However, I don't need any "real" security. Just enough entropy for the output of the function to appear random. Weak encryption algorithms welcome.
Just create an array of objects that contain 2 fields -- a letter, and a random number. Sort the array. By the random numbers. This creates a mapping where the i-th letter of the alphabet now maps to the i-th letter in the array.
If simple transposition or substitution isn't quite enough, it sounds like you want to advance to a polyalphabetic cipher. The Vigenère cipher is extremely easy to implement in code, but is still difficult to break without using a computer.
I suggest the following.
Perform a dense coding of the input to positive integers - with an alphabet size of n and string length of m you can code the string into integers between zero and n^m - 1. In your example this would be the range [0,215]. Now perform a fixed involution on the encoded number and decode it again.
Take RC4, settle for some password, and you're done. (Not that this would be very safe.)
Take the set of all permutations of your alphabet, shuffle it, and map the first half of the set onto the second half. Bad for large alphabets, of course. :)
Nah, thought that over, I forgot about character repetitions. Maybe divide the input into chunks without repeating chars and apply my suggestion to all of those chunks.
I would restate your problem thus, and give you a strategy for that restatement:
"A substitution cypher where a change in input leads to a larger change in output".
The blocking of characters is irrelevant-- in the end, it's just mappings between numbers. I'll speak of letters here, but you can extend it to any block of n characters.
One of the easiest routes for this is a rotating substitution based on input. Since you already looked at the Vigenere cipher, it should be easy to understand. Instead of making the key be static, have it be dependent on the previous letter. That is, rotate through substitutions a different amount per each input.
The variable rotation satisfies the condition of making each small change push out to a larger change. Note that the algorithm will only push changes in one direction such that changes towards the end have smaller effects. You could run the algorithm both ways (front-to-back, then back-to-front) so that every letter of cleartext changed has the possibility of changing the entire string.
The internal rotation strategy elides the need for keys, while of course losing of most of the cryptographic security. It makes sense in context, though, as you are aiming for entropy rather than security.
You can solve this problem with Format-preserving encryption.
One Java-Library can be found under https://github.com/EVGStudents/FPE.git. There you can define a Regex and encrypt/decrypt string values matching this regex.

How to elegantly compute the anagram signature of a word in ruby?

Arising out of this question, I'm looking for an elegant (ruby) way to compute the word signature suggested in this answer.
The idea suggested is to sort the letters in the word, and also run length encode repeated letters. So, for example "mississippi" first becomes "iiiimppssss", and then could be further shortened by encoding as "4impp4s".
I'm relatively new to ruby and though I could hack something together, I'm sure this is a one liner for somebody with more experience of ruby. I'd be interested to see people's approaches and improve my ruby knowledge.
edit: to clarify, performance of computing the signature doesn't much matter for my application. I'm looking to compute the signature so I can store it with each word in a large database of words (450K words), then query for words which have the same signature (i.e. all anagrams of a given word, that are actual english words). Hence the focus on space. The 'elegant' part is just to satisfy my curiosity.
The fastest way to create a sorted list of the letters is this:
"mississippi".unpack("c*").sort.pack("c*")
It is quite a bit faster than split('') and join(). For comparison it is also best to pack the array back together into a String, so you dont have to compare arrays.
I'm not much of a Ruby person either, but as I noted on the other comment this seems to work for the algorithm described.
s = "mississippi"
s.split('').sort.join.gsub(/(.)\1{2,}/) { |s| s.length.to_s + s[0,1] }
Of course, you'll want to make sure the word is lowercase, doesn't contain numbers, etc.
As requested, I'll try to explain the code. Please forgive me if I don't get all of the Ruby or reg ex terminology correct, but here goes.
I think the split/sort/join part is pretty straightforward. The interesting part for me starts at the call to gsub. This will replace a substring that matches the regular expression with the return value from the block that follows it. The reg ex finds any character and creates a backreference. That's the "(.)" part. Then, we continue the matching process using the backreference "\1" that evaluates to whatever character was found by the first part of the match. We want that character to be found a minimum of two more times for a total minimum number of occurrences of three. This is done using the quantifier "{2,}".
If a match is found, the matching substring is then passed to the next block of code as an argument thanks to the "|s|" part. Finally, we use the string equivalent of the matching substring's length and append to it whatever character makes up that substring (they should all be the same) and return the concatenated value. The returned value replaces the original matching substring. The whole process continues until nothing is left to match since it's a global substitution on the original string.
I apologize if that's confusing. As is often the case, it's easier for me to visualize the solution than to explain it clearly.
I don't see an elegant solution. You could use the split message to get the characters into an array, but then once you've sorted the list I don't see a nice linear-time concatenate primitive to get back to a string. I'm surprised.
Incidentally, run-length encoding is almost certainly a waste of time. I'd have to see some very impressive measurements before I'd think it worth considering. If you avoid run-length encoding, you can anagrammatize any string, not just a string of letters. And if you know you have only letters and are trying to save space, you can pack them 5 bits to a letter.
---Irma Vep
EDIT: the other poster found join which I missed. Nice.

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