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I have to deal with sequences of a lot of small numbers, about a million, and I have to put as many as possible (more is better) in 4KB. Obviously that's just too little space to put all of them. Also, while this is a specific scenario, I'd love an answer as general as possible.
The numbers don't follow any pattern, but here is what a small script has to say about their distribution:
407037 times 1
165000 times 2
85389 times 3
52257 times 4
34749 times 5
23567 times 6
15892 times 7
11183 times 8
7636 times 9
5402 times 10
3851 times 11
2664 times 12
2023 times 13
1547 times 14
1113 times 15
... many more lines ...
1 times 62
62 is the biggest number I have, so let's set the maximum number we care about at 64. If the method is easily adaptable to accommodate for bigger max numbers, that would be better.
Here is a sample of the numbers:
20
1
1
1
13
1
5
1
15
1
3
4
3
2
2
A naive way to do this would just be to use 6 bits per number, but I think we can do better.
EDIT: adding a bit of info following discussion in comments.
I also have 2KB of ram and a dozen cycles on a microprocessor to decode each number. I need to store, sequentially, from the first number, as many numbers as I can.
EDIT: see graybeard's comment and my followup too.
The correct way to do this would be Rangecoding, Huffman or Shannon-Fano which you can see in any of the digital-communication blogs over the net, so I'm not exactly explaining you these.
I can suggest you a custom method, which is really simple and you can compare it with other methods if you can use this to store more numbers or not.
I see that there are no 0's in your script. So just decrease each number by 1 (while decoding, add 1 to decoded result). Use either 4 or 7 bits to encode numbers. All numbers up-to 8 can be represented in 3-bits. If the number is n <= 8, set the 1st bit as 0 and next 3 bits can represent the number. Else, if the number is n > 8, set 1st bit as 1 and represent the number as 6 bits from there.
Though in Huffman or Shannon-Fano, few of the representations can be as long as over 20 bits.
For provide correct answer, need to know - is decoder size also limited, or there is not limit for decodes size?
If no limit for decoder (just limit for data), I suggest you to use rangecoder, or Huffman coding. Rangecoder has better compression, but extensive arithmetic operation usage.
However, both decoders uses memory for a code, and for statistical tables. So, perhaps, better answer to create something more easy (custom compressor), but with simple and compact code and without any tables. As easy, code-compact, I can propose the run-1 algorithm. This algorithm is not very efficient for your data (rangecoder or Huffman better), but has trivial compact decoder without any tables.
Idea - sequence can contain zero or more bit_1, and use bit_0 as symbol separator. For example, if we would like encode with run-1 the sequence:
1, 1, 2, 1, 5
There will be bit sequence:
0-0-10-0-11110
There, you needed just count number of sequenced bit_1, add 1, and return value as decoded number.
Maybe slightly better than straight Huffman can be attempted by combining with run-length coding.
If you count the successive identical elements, you can rewrite your sequence as a pairs of (value, count). Every such pair appears with some probability and you can use Huffman coding on these. (I don't mean to code the values and the counts separately, but the pairs as a whole).
Your sample yields
(20, 1), (1, 3), (13 1), (1, 1), (5, 1), (1, 1), (15, 1), (3, 1), (4, 1), (3, 1), (5, 2)
The singletons will be (practically) coded as before, and there are more opportunities for compression of longer runs.
You can limit the maximum count(s) that are supported; if the actual count exceeds the limit, it is no big deal to insert several pairs.
The very first step is to compute an histogram of the count values to see if there are enough repetitions for this approach to be worth.
Alternatively, you can try Huffman coding on the deltas (signed differences between successive values). If there are many repetitions, the frequency of 0 will be much higher, increasing the entropy. Obviously, run-length coding of the deltas is also possible.
I took the distribution you listed, and tried an exponential fit. The result was decently good:
More importantly, the fit was reasonably close to p(x) ~= 2^-x. This suggests a very simple coding, known as "unary coding": to encode the number k, output k-1 zeroes, followed by a 1. If your numbers exactly fit the p(x) ~= 2^-x distribution, that would give you an expected code length of 2 bits. Since your numbers appear to be heavier-tailed than that (otherwise it would be vanishingly unlikely to see a 62 in only a million numbers), you won't quite achieve that. Still, given the simplicity of the coding and the ease of decoding (twelve cycles should be sufficient), you should consider trying it out.
You might also look into other universal codes, such as Elias Delta. Golomb coding would be optimal, but decoding it is an involved process.
[Background Story]
I am working with a 5 year old user identification system, and I am trying to add IDs to the database. The problem I have is that the system that reads the ID numbers requires some sort of checksum, and no-one working here now has ever worked with it, so no-one knows how it works.
I have access to the list of existing IDs, which already have correct checksums. Also, as the checksum only has 16 possible values, I can create any ID I want and run it through the authentication system up to 16 times until I get the correct checksum (but this is quite time consuming)
[Question]
What methods can I use to help guess the checksum algorithm of used for some data?
I have tried a few simple methods such as XORing and summing, but these have not worked.
So my question is: if I have data (in hexadecimal) like this:
data checksum
00029921 1
00013481 B
00026001 3
00004541 8
What methods can I use work out what sort of checksum is used?
i.e. should I try sequential numbers such as 00029921,00029922,00029923,... or 00029911,00029921,00029931,... If I do this what patterns should I look for in the changing checksum?
Similarly, would comparing swapped digits tell me anything useful about the checksum?
i.e. 00013481 and 00031481
Is there anything else that could tell me something useful? What about inverting one bit, or maybe one hex digit?
I am assuming that this will be a common checksum algorithm, but I don't know where to start in testing it.
I have read the following links, but I am not sure if I can apply any of this to my case, as I don't think mine is a CRC.
stackoverflow.com/questions/149617/how-could-i-guess-a-checksum-algorithm
stackoverflow.com/questions/2896753/find-the-algorithm-that-generates-the-checksum
cosc.canterbury.ac.nz/greg.ewing/essays/CRC-Reverse-Engineering.html
[ANSWER]
I have now downloaded a much larger list of data, and it turned out to be simpler than I was expecting, but for completeness, here is what I did.
data:
00024901 A
00024911 B
00024921 C
00024931 D
00042811 A
00042871 0
00042881 1
00042891 2
00042901 A
00042921 C
00042961 0
00042971 1
00042981 2
00043021 4
00043031 5
00043041 6
00043051 7
00043061 8
00043071 9
00043081 A
00043101 3
00043111 4
00043121 5
00043141 7
00043151 8
00043161 9
00043171 A
00044291 E
From these, I could see that when just one value was increased by a value, the checksum was also increased by the same value as in:
00024901 A
00024911 B
Also, two digits swapped did not change the checksum:
00024901 A
00042901 A
This means that the polynomial value (for these two positions at least) must be the same
Finally, the checksum for 00000000 was A, so I calculated the sum of digits plus A mod 16:
( (Σxi) +0xA )mod16
And this matched for all the values I had. Just to check that there was nothing sneaky going on with the first 3 digits that never changed in my data, I made up and tested some numbers as Eric suggested, and those all worked with this too!
Many checksums I've seen use simple weighted values based on the position of the digits. For example, if the weights are 3,5,7 the checksum might be 3*c[0] + 5*c[1] + 7*c[2], then mod 10 for the result. (In your case, mod 16, since you have 4 bit checksum)
To check if this might be the case, I suggest that you feed some simple values into your system to get an answer:
1000000 = ?
0100000 = ?
0010000 = ?
... etc. If there are simple weights based on position, this may reveal it. Even if the algorithm is something different, feeding in nice, simple values and looking for patterns may be enlightening. As Matti suggested, you/we will likely need to see more samples before decoding the pattern.
I've got a special need and the most important concerns are:
in-memory
very low memory footprint
speed
Here's my "problem": I need to store, in-memory, a huge number of very sparse bit arrays. Those bitsets are "append only" and are to be used mostly for intersections. By huge, I mean as high as 200 000 bit arrays.
The range shall be between [0...16 000 000] for each bitset.
I ran some pre-test with "only" 10 673 bit arrays containing some actual data I've got and got the following results:
1% of the bit arrays ( 106 bit arrays) Hamming weight: at most 1 bit set
5% of the bit arrays ( 534 bit arrays) Hamming weight: at most 4 bits set
10% of the bit arrays ( 1068 bit arrays) Hamming weight: at most 8 bits set
15% of the bit arrays ( 1603 bit arrays) Hamming weight: at most 12 bits set
20% of the bit arrays ( 2137 bit arrays) Hamming weight: at most 17 bits set
25% of the bit arrays ( 2671 bit arrays) Hamming weight: at most 22 bits set
30% of the bit arrays ( 3206 bit arrays) Hamming weight: at most 28 bits set
35% of the bit arrays ( 3740 bit arrays) Hamming weight: at most 35 bits set
40% of the bit arrays ( 4274 bit arrays) Hamming weight: at most 44 bits set
45% of the bit arrays ( 4809 bit arrays) Hamming weight: at most 55 bits set
50% of the bit arrays ( 5343 bit arrays) Hamming weight: at most 67 bits set
55% of the bit arrays ( 5877 bit arrays) Hamming weight: at most 83 bits set
60% of the bit arrays ( 6412 bit arrays) Hamming weight: at most 103 bits set
65% of the bit arrays ( 6946 bit arrays) Hamming weight: at most 128 bits set
70% of the bit arrays ( 7480 bit arrays) Hamming weight: at most 161 bits set
75% of the bit arrays ( 8015 bit arrays) Hamming weight: at most 206 bits set
80% of the bit arrays ( 8549 bit arrays) Hamming weight: at most 275 bits set
85% of the bit arrays ( 9083 bit arrays) Hamming weight: at most 395 bits set
90% of the bit arrays ( 9618 bit arrays) Hamming weight: at most 640 bits set
95% of the bit arrays (10152 bit arrays) Hamming weight: at most 1453 bits set
96% of the bit arrays (10259 bit arrays) Hamming weight: at most 1843 bits set
97% of the bit arrays (10366 bit arrays) Hamming weight: at most 2601 bits set
98% of the bit arrays (10473 bit arrays) Hamming weight: at most 3544 bits set
99% of the bit arrays (10580 bit arrays) Hamming weight: at most 4992 bits set
100% of the bit arrays (10687 bit arrays) Hamming weight: at most 53153 bits set
Seen the numbers involved, I obviously need to use compressed bit arrays and that is not an issue: it shall stay easy to deal with seen that the bit arrays are "append only".
The bit array bits that are on are kinda grouped, but not totally. So you'll tend to have several bits on in the same area (but usually not one after another, making RLE kinda not great for bits that are on).
My question is what kind of compression to use?
Now I don't know if I should put my first approach here or in an answer to my own question.
Basically I imagined a "worst case" scenario using a very dumb encoding:
1 bit: if on, the following 5 bits determine how many bits are needed to compute the 'skip', if off, optimization: the following 5 bits determine how many bits are too be taken literally (that is 'on' or 'off', without skipping) [this would only be switched to when determined to be more efficient than the other representation, so when it kicks in, it shall always be an optimization (size-wise)]
5 bits: how many bits we can skip before the next bit on
x bits: skip
Here's an example: a bit array has 3 bit set, the first bit being at 3 098 137, the second at 3 098 141 and the third at 3 098 143.
+-- now we won't skip
|
| +-- 3 because we need 3 bits to store "6" (from 3 098 138 to 3 098 143)
| | +--- 3 098 141 is on
22 3 098 137 | 3 | +- 3 098 143 is on
1 10110 1011110100011000011001 0 00011 000101 etc.
First bit on tells we're going to skip bits.
5 next bits (always 5) tells how many bits we need to tell how many bits we'll skip
22 bits telling to skip to 3 098 137
one bit off telling now we're not skipping bits
5 next bits (always 5) tells how many bits we'll read "as is"
6 bits: off, off, off, on, off, on meaning 3 098 141 and 3 098 143 are on
etc.
Seen the amazing sparsity of these bit arrays, this seems quite size-efficient.
So using that encoding, I took my sample data and computed a "worst case" scenario (I haven't written the algo yet, I'd rather have a few from here inputs first): basically I considered that not only the "size optimization" would never kick in and, also, that the 5 bits would always be set to their maximum value (24 bits), which of course cannot happen.
I did it just to have a very crude approximation of what the "worst of the worst" case could be.
I was very pleasantly surprised:
Worst case scenario:
108 913 290 bits needed for the 10 687 very sparse bit arrays
12.9 MB (13 295 KB)
The data being actual data and all the data being similar, I know that, if worse comes to worse, I could store my 200 000 bit arrays in about 240 MB, which is fine.
I'm pretty sure the actual encoding will comes to way less than that but as I haven't actually written it yet, I can only (very easily) compute the "worst case" which is why I only show that one.
Any hints / ideas as to how to make this more size-efficient (remembering these are super-sparse bit arrays, that there shall be hundreds thousands of them, that they must be in memory, and that they shall be "append only")?
About my 'append-only' case
Basically I've got one growing "expanse" (the range, but "expanse" is the actual term as I understand it) and a lot of bit arrays that have a few bit sets. When the range goes from, say, 0 to 1 000 000, all the bit arrays goes from 0 to 1 000 000 to. When the range grows to 1 000 001, then all the bit arrays are growing too, all by one bit. But most of these bit arrays will have a '0' appended at their end, while about 4 to 8 of the bit arrays will have a '1' appended at their end. However I cannot predict in advance which of the bit arrays will have a 0 or a 1 appended.
So I've got a lot of bit arrays that have all the same size, that are all very sparse (< 0.5% of their bits set) and that are all "growing" as the range growth (so they're all always growing at the same rate).
Judy arrays are great. But I read about them a few years ago and that stuff was "above my head". Judy arrays are a C-only 20KLOC lib and I'm definitely not re-implementing that. But they're amazing.
So I guess I need to add I'd like all this to stay relatively simple, which is not that far-fetched seen the special "append only" property of my very sparse bit arrays.
You didn't say what programming language you want to use. It sounds like you don't want Judy because it's "C-only"... if you are using C# then you could use my Compact Patricia Trie instead. Is is almost 4500 LOC (commented) and uses similar ideas to Judy, but the size and speed of each trie are not ideal due to limitations of .NET. It is not optimized for computing intersections either, but such an algorithm could be added. The article about CP Tries does not emphasize this point, but it can store sets (sparse bit arrays) much more compactly than dictionaries (the graphs in the article show the size and speed of dictionaries, not sets).
The best case is a dense cluster of bits. With 50% occupancy (every other bit set), it requires less than 8 bits per key (less than 4 bits per integer). (correction: less than 8 bits, not more.)
If you only need an approximate representation of the data, use a Bloom filter.
By the way, what do you mean by "append only"? Does it mean that you only add keys, or that each key you add is greater than the keys you added before?
Update: Since you are only adding larger keys, you should probably design a special algorithm just for your case. IMO, when designing a custom algorithm, you should make it as simple as possible. So here's my idea, which assumes the keys of different bitsets are uncorrelated (therefore there is no benefit of attempting to compress data between different bitsets):
A bitset is represented by a sorted array of 32-bit slots. Because it's sorted, you can use binary search to find keys. Each slot consists of a 24-bit "prefix" and 8 bits of "flags". Each slot represents a region of 8 keys. The "flags" tell you which of the 8 keys in the region are present in the bitset, and the "prefix" tells you which region we're talking about, by specifying bits 3 to 26 of the key. For example, if the following bits are "1" in the bitset:
1, 3, 4, 1094, 8001, 8002, 8007, 8009
...then the bitset is represented by an array of 4 slots (16 bytes):
Prefix: 0, 136, 1000, 1001
Flags: 0x15, 0x40, 0x86, 0x02
The first slot represents 1, 3, 4 (notice that bits 1, 3 and 4 are set in the number 0x15); the second slot represents 1094 (136 * 8 + 6); the third slot represents 8001, 8002, and 8007; the fourth slot represents 8009. Does this make sense?
I don't know if this is as compact as your idea. But I think you'll get faster queries and faster modifications, and it will be fairly easy to implement.
You may use binary tree for bit array.
Say, you have array with range of [M..N].
Store it in such a manner:
Choose some number encoding for [0...ram size], like Fibonacci, Golomb or Rice code (you may choose most suitable representation after profiling your program with actual data).
If array is empty (have no bits set), store it as number 0.
If array is full (have all bits set), store it as number 1.
Else split it in two parts: A in [M..(M+N)/2-1] and B in [(M+N)/2..N]
Generate representations of P0 and P1 using this algorithm recursively.
Get length of P0 (in bits or other units length may be whole number of) and store it as a number (you may need to add 1 if length may be 1, e.g. you store 0 as single bit 0).
Store P0 then P1.
In this case, if limits are common, operations of intersection an union are trivial recursions:
Intersection:
If array A is empty, store 0.
If array A is full, store copy of B
Else split arrays, make intersections of both halves, store length of first half, then both halves.
This algorithm may deal with bits (if you need them to be most compact) and bytes/words (if bit operations are so slow).
Also you may add specical encodings for arrays with single bit set, all arrays with size less than some limit (8 elements for example) to decrease level of recursion.
Drawback is that without some hacks adding/removing element to/from array is complex operation (as complex as intersection/union operations).
For example, array with single 0xAB bit set should be stored in array of 0..0xFF as (pseudocode for):
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1, EMPTY, 13, 1, EMPTY, 9, 1, EMPTY, 5, 1, EMPTY, 1, EMPTY, FULL, EMPTY, EMPTY, EMPTY
| AA | AB |
|A8..A9| AA .. AB |
| A8 .. AB |AC..AF|
|A0..A7| A8 .. AF |
| A0 .. AF |B0..BF|
|80..9F| A0 .. BF |
| 80 .. BF |C0..FF|
| 0..7F| 80 .. FF |
EMPTY and FULL are codes for empty and full arrays, numbers are lengths in elements (should be replaced with actual lengthts in bytes, bits or so)
Ff you do not need fast single bit check, you may use most simple approach:
Just store distances between set bits using codes: fibonacci, rice, golomb, levenshtein, elias etc. or invent another one.
Note, that in order to get minimal code length, you should use code with code lengths as close as possible to -log p/log 2, where p is probability of that code. You may use huffman code for that.
For example, use elias gamma code, so array like this:
0 1 0000 1 1 000 1 0 1 000000000000000000 1 000000000000000000
2 5 1 4 2 19 18 (distance)
Should be encoded as:
010 00101 1 00100 010 000010011 000010010
2 5 1 4 2 19 18 (distance code explained)
And mostly compact for array with uniform bits distribution would be arithmetic encoding, but it is very CPU time counsumpting. Becaouse you'll have to read and write such arrays bit by bit with no fast skipping available.
You may look into compressed bitmaps. A common strategy is to use word-aligned run-length encoding.
C++ implementation:
https://github.com/lemire/EWAHBoolArray
Java implementation:
https://github.com/lemire/javaewah
Reference:
Daniel Lemire, Owen Kaser, Kamel Aouiche, Sorting improves word-aligned bitmap indexes. Data & Knowledge Engineering 69 (1), pages 3-28, 2010.
http://arxiv.org/abs/0901.3751
Even if they aren't exactly what you're looking for, it's worth checking out Judy trees. Judy is a heavily optimized library for ordered maps, and one configuration is specifically designed as a bitset rather than a map. I don't think intersection is one of the operations natively optimized for, though...
The general idea is to use a tree with a fixed number of address bits per level, and take advantage of the sparseness at each level. This results in quite good compression even in the worst case, and fast query performance as well. I believe an intersection operation would be relatively straightforward and potentially very fast.
At any rate, it's always a good idea to steal from the best!
Considering you are going to do a bunch of intersection tests anyway, maybe you should try storing all of the bitvectors in parallel. One sparse, 16M entry list. Each entry in that list contains a list of which of the 200k input bitvectors has a '1' at that location. It looks like you expect to have only about 5 bits set per input vector, or 1M total entries? Taking a straw-man linked list implementation for the toplevel and the buckets, and a worst case of no intersections at all (thus 1M buckets with 1 element each) you could store it all in 32MB.
You might be interested in Binary Decision Diagrams (BDD), and more precisely Zero-suppressed Binary Decision Diagram (ZBDD).
They are used to represent sets in a compressed way. Unlike other compressed forms, operations (such as set intersections, or insertions of elements - your "append only" thing?) work directly on the compressed form.
Is it mathematically feasible to encode and initial 4 byte message into 8 bytes and if one of the 8 bytes is completely dropped and another is wrong to reconstruct the initial 4 byte message? There would be no way to retransmit nor would the location of the dropped byte be known.
If one uses Reed Solomon error correction with 4 "parity" bytes tacked on to the end of the 4 "data" bytes, such as DDDDPPPP, and you end up with DDDEPPP (where E is an error) and a parity byte has been dropped, I don't believe there's a way to reconstruct the initial message (although correct me if I am wrong)...
What about multiplying (or performing another mathematical operation) the initial 4 byte message by a constant, then utilizing properties of an inverse mathematical operation to determine what byte was dropped. Or, impose some constraints on the structure of the message so every other byte needs to be odd and the others need to be even.
Alternatively, instead of bytes, it could also be 4 decimal digits encoded in some fashion into 8 decimal digits where errors could be detected & corrected under the same circumstances mentioned above - no retransmission and the location of the dropped byte is not known.
I'm looking for any crazy ideas anyone might have... Any ideas out there?
EDIT:
It may be a bit contrived, but the situation that I'm trying to solve is one where you have, let's say, a faulty printer that prints out important numbers onto a form, which are then mailed off to a processing firm which uses OCR to read the forms. The OCR isn't going to be perfect, but it should get close with only digits to read. The faulty printer could be a bigger problem, where it may drop a whole number, but there's no way of knowing which one it'll drop, but they will always come out in the correct order, there won't be any digits swapped.
The form could be altered so that it always prints a space between the initial four numbers and the error correction numbers, ie 1234 5678, so that one would know whether a 1234 initial digit was dropped or a 5678 error correction digit was dropped, if that makes the problem easier to solve. I'm thinking somewhat similar to how they verify credit card numbers via algorithm, but in four digit chunks.
Hopefully, that provides some clarification as to what I'm looking for...
In the absence of "nice" algebraic structure, I suspect that it's going to be hard to find a concise scheme that gets you all the way to 10**4 codewords, since information-theoretically, there isn't a lot of slack. (The one below can use GF(5) for 5**5 = 3125.) Fortunately, the problem is small enough that you could try Shannon's greedy code-construction method (find a codeword that doesn't conflict with one already chosen, add it to the set).
Encode up to 35 bits as a quartic polynomial f over GF(128). Evaluate the polynomial at eight predetermined points x0,...,x7 and encode as 0f(x0) 1f(x1) 0f(x2) 1f(x3) 0f(x4) 1f(x5) 0f(x6) 1f(x7), where the alternating zeros and ones are stored in the MSB.
When decoding, first look at the MSBs. If the MSB doesn't match the index mod 2, then that byte is corrupt and/or it's been shifted left by a deletion. Assume it's good and shift it back to the right (possibly accumulating multiple different possible values at a point). Now we have at least seven evaluations of a quartic polynomial f at known points, of which at most one is corrupt. We can now try all possibilities for the corruption.
EDIT: bmm6o has advanced the claim that the second part of my solution is incorrect. I disagree.
Let's review the possibilities for the case where the MSBs are 0101101. Suppose X is the array of bytes sent and Y is the array of bytes received. On one hand, Y[0], Y[1], Y[2], Y[3] have correct MSBs and are presumed to be X[0], X[1], X[2], X[3]. On the other hand, Y[4], Y[5], Y[6] have incorrect MSBs and are presumed to be X[5], X[6], X[7].
If X[4] is dropped, then we have seven correct evaluations of f.
If X[3] is dropped and X[4] is corrupted, then we have an incorrect evaluation at 3, and six correct evaluations.
If X[5] is dropped and X[4] is corrupted, then we have an incorrect evaluation at 5, and six correct evaluations.
There are more possibilities besides these, but we never have fewer than six correct evaluations, which suffices to recover f.
I think you would need to study what erasure codes might offer you. I don't know any bounds myself, but maybe some kind of MDS code might achieve this.
EDIT: After a quick search I found RSCode library and in the example it says that
In general, with E errors, and K erasures, you will need
* 2E + K bytes of parity to be able to correct the codeword
* back to recover the original message data.
So looks like Reed-Solomon code is indeed the answer and you may actually get recovery from one erasure and one error in 8,4 code.
Parity codes work as long as two different data bytes aren't affected by error or loss and as long as error isn't equal to any data byte while a parity byte is lost, imho.
Error correcting codes can in general handle erasures, but in the literature the position of the erasure is assumed known. In most cases, the erasure will be introduced by the demodulator when there is low confidence that the correct data can be retrieved from the channel. For instance, if the signal is not clearly 0 or 1, the device can indicate that the data was lost, rather than risking the introduction of an error. Since an erasure is essentially an error with a known position, they are much easier to fix.
I'm not sure what your situation is where you can lose a single value and you can still be confident that the remaining values are delivered in the correct order, but it's not a situation classical coding theory addresses.
What algorithmist is suggesting above is this: If you can restrict yourself to just 7 bits of information, you can fill the 8th bit of each byte with alternating 0 and 1, which will allow you to know the placement of the missing byte. That is, put a 0 in the high bit of bytes 0, 2, 4, 6 and a 1 in the high bits of the others. On the receiving end, if you only receive 7 bytes, the missing one will have been dropped from between bytes whose high bits match. Unfortunately, that's not quite right: if the erasure and the error are adjacent, you can't know immediately which byte was dropped. E.g., high bits 0101101 could result from dropping the 4th byte, or from an error in the 4th byte and dropping the 3rd, or from an error in the 4th byte and dropping the 5th.
You could use the linear code:
1 0 0 0 0 1 1 1
0 1 0 0 1 0 1 1
0 0 1 0 1 1 0 1
0 0 0 1 1 1 1 0
(i.e. you'll send data like (a, b, c, d, b+c+d, a+c+d, a+b+d, a+b+c) (where addition is implemented with XOR, since a,b,c,d are elements of GF(128))). It's a linear code with distance 4, so it can correct a single-byte error. You can decode with syndrome decoding, and since the code is self-dual, the matrix H will be the same as above.
In the case where there's a dropped byte, you can use the technique above to determine which one it is. Once you've determined that, you're essentially decoding a different code - the "punctured" code created by dropping that given byte. Since the punctured code is still linear, you can use syndrome decoding to determine the error. You would have to calculate the parity-check matrix for each of the shortened codes, but you can do this ahead of time. The shortened code has distance 3, so it can correct any single-byte errors.
In the case of decimal digits, assuming one goes with first digit odd, second digit even, third digit odd, etc - with two digits, you get 00-99, which can be represented in 3 odd/even/odd digits (125 total combinations) - 00 = 101, 01 = 103, 20 = 181, 99 = 789, etc. So one encodes two sets of decimal digits into 6 total digits, then the last two digits signify things about the first sets of 2 digits or a checksum of some sort... The next to last digit, I suppose, could be some sort of odd/even indicator on each of the initial 2 digit initial messages (1 = even first 2 digits, 3 = odd first two digits) and follow the pattern of being odd. Then, the last digit could be the one's place of a sum of the individual digits, that way if a digit was missing, it would be immediately apparent and could be corrected assuming the last digit was correct. Although, it would throw things off if one of the last two digits were dropped...
It looks to be theoretically possible if we assume 1 bit error in wrong byte. We need 3 bits to identify dropped byte and 3 bits to identify wrong byte and 3 bits to identify wrong bit. We have 3 times that many extra bits.
But if we need to identify any number of bits error in wrong byte, it comes to 30 bits. Even that looks to be possible with 32 bits, although 32 is a bit too close for my comfort.
But I don't know hot to encode to get that. Try turbocode?
Actually, as Krystian said, when you correct a RS code, both the message AND the "parity" bytes will be corrected, as long as you have v+2e < (n-k) where v is the number of erasures (you know the position) and e is the number of errors. This means that if you only have errors, you can correct up to (n-k)/2 errors, or (n-k-1) erasures (about the double of the number of errors), or a mix of both (see Blahut's article: Transform techniques for error control codes and A universal Reed-Solomon decoder).
What's even nicer is that you can check that the correction was successful: by checking that the syndrome polynomial only contains 0 coefficients, you know that the message+parity bytes are both correct. You can do that before to check if the message needs any correction, and also you can do the check after the decoding to check that both the message and the parity bytes were completely repaired.
The bound v+2e < (n-k) is optimal, you cannot do better (that's why Reed-Solomon is called an optimal error correction code). In fact it's possible to go beyond this limit using bruteforce approaches, up to a certain point (you can gain 1 or 2 more symbols for each 8 symbols) using list decoding, but it's still a domain in its infancy, I don't know of any practical implementation that works.
I seek an algorithm that will let me represent an incoming sequence of bits as letters ('a' .. 'z' ), in a minimal matter such that the stream of bits can be regenerated from the letters, without ever holding the entire sequence in memory.
That is, given an external bit source (each read returns a practically random bit), and user input of a number of bits, I would like to print out the minimal number of characters that can represent those bits.
Ideally there should be a parameterization - how much memory versus maximum bits before some waste is necessary.
Efficiency Goal - The same number of characters as the base-26 representation of the bits.
Non-solutions:
If sufficient storage was present, store the entire sequence and use a big-integer MOD 26 operation.
Convert every 9 bits to 2 characters - This seems suboptimal, wasting 25% of information capacity of the letters output.
If you assign a different number of bits per letter, you should be able to exactly encode the bits in the twenty-six letters allowed without wasting any bits. (This is a lot like a Huffman code, only with a pre-built balanced tree.)
To encode bits into letters: Accumulate bits until you match exactly one of the bit codes in the lookup table. Output that letter, clear the bit buffer, and keep going.
To decode letters into bits: For each letter, output the bit sequence in the table.
Implementing in code is left as an exercise to the reader. (Or to me, if I get bored later.)
a 0000
b 0001
c 0010
d 0011
e 0100
f 0101
g 01100
h 01101
i 01110
j 01111
k 10000
l 10001
m 10010
n 10011
o 10100
p 10101
q 10110
r 10111
s 11000
t 11001
u 11010
v 11011
w 11100
x 11101
y 11110
z 11111
Convert each block of 47 bits to a base 26 number of 10 digits. This gives you more than 99.99% efficiency.
This method, as well as others like Huffman, needs a padding mechanism to support variable-length input. This introduces some inefficiency which is less significant with longer inputs.
At the end of the bit stream, append an extra 1 bit. This must be done in all cases, even when the length of the bit stream is a multiple of 47. Any high-order letters of "zero" value can be skipped in the last block of encoded output.
When decoding the letters, a truncated final block can be filled out with "zero" letters and converted to a 47-bit base 2 representation. The final 1 bit is not data, but marks the end of the bit stream.
Could Huffman coding be what you're looking for? It's a compression algorithm, which pretty much represents any information with a minimum of wasted bits.
Zero waste would be log_2(26) bits per letter. As pointed out earlier, you can get to 4.7 by reading 47 bits and converting them to 10 letters. However, you can get to 4.67 by converting every 14 bits into 3 characters. This has the advantage that it fits into an integer. If you have storage space and run time is important, you can create a lookup table with 17,576 entries mapping the possible 14 bits into 3 letters. Otherwise, you can do mod and div operations to compute the 3 letters.
number of letters number of bits bits/letter
1 4 4
2 9 4.5
3 14 4.67
4 18 4.5
5 23 4.6
6 28 4.67
7 32 4.57
8 37 4.63
9 42 4.67
10 47 4.7
Any solution you use is going to be space-inefficient because 26 is not a power of 2. As far as an algorithm goes, I'd rather use a lookup table than an on-the-fly calculation for each series of 9 bits. Your lookup table would 512 entries long.
If you want the binary footprint of each letter to have the same size, the optimal solution would be given by Arithmetic Encoding. However, it will not reach your goal of a mean representation of 4.5 bits/char. Given 26 different characters (not including space etc) 4.7 would be the best you can reach without using variable-length encoding (Huffman, for instance. See Jaegers's answer) or other compression algoritms.
A suboptimal, although simpler, solution could be to find a feasible number of characters to fit into a big integer. For instance, if you form a 32-bit integer out of every 6 charachter chunk (which is possible as 26^6 < 2^32), you use 5.33 bits/char. You can actually even fit 13 letters into a 64 bit integer (4.92 bits/char). This is quite close to the optimal solution, and still rather easy to implement. Using bigger ints than 64 bits can be tricky due to missing native support in many progamming languages.
If you want even better compression rates for text, you should definitely also look into dictionary-based compression algorithms, such as LZW or Deflate.