I have a 32-bit number that is created by encoding some data, I want to be more confident that the data (a max 32-bit number) is not changed when decoding it, so I am going to add some error detection bits.
I need to keep the data as short as possible, so I can only add a few bits for error detection, in some cases just 1 bit.
I'm looking for an algorithm that detects more bit changes and needs fewer extra bits.
I was thinking of calculating a checksum or CRC and just dropping extra bits or maybe xor the result to make it shorter but I'm not sure if the error detection remains good enough.
Thanks in advance for any help.
A 1-bit CRC, with polynomial x+1 would simply be the parity of your 32 message bits. That will detect any one-bit error in the resulting 32 bits. For a 2-bit CRC, you can use x2+1. You can define a CRC of any length. See Koopman's list for good CRC polynomials for CRCs of degree 3 and higher.
I'm working on a problem out of Cracking The Coding Interview which requires that I swap odd and even bits in an integer with as few instructions as possible (e.g bit 0 and 1 are swapped, bits 2 and 3 are swapped, etc.)
The author's solution revolves around using a mask to grab, in one number, the odd bits, and in another num the even bits, and then shifting them off by 1.
I get her solution, but I don't understand how she grabbed the even/odd bits. She creates two bit masks --both in hex -- for a 32 bit integer. The two are: 0xaaaaaaaa and 0x55555555. I understand she's essentially creating the equivalent of 1010101010... for a 32 bit integer in hexadecimal and then ANDing it with the original num to grab the even/odd bits respectively.
What I don't understand is why she used hex? Why not just code in 10101010101010101010101010101010? Did she use hex to reduce verbosity? And when should you use one over the other?
It's to reduce verbosity. Binary 10101010101010101010101010101010, hexadecimal 0xaaaaaaaa, and decimal 2863311530 all represent exactly the same value; they just use different bases to do so. The only reason to use one or another is for perceived readability.
Most people would clearly not want to use decimal here; it looks like an arbitrary value.
The binary is clear: alternating 1s and 0s, but with so many, it's not obvious that this is a 32-bit value, or that there isn't an adjacent pair of 1s or 0s hiding in the middle somewhere.
The hexadecimal version takes advantage of chunking. Assuming you recognize that 0x0a == 0b1010, you can mentally picture the 8 groups of 1010 in the assumed value.
Another possibility would be octal 25252525252, since... well, maybe not. You can see that something is alternating, but unless you use octal a lot, it's not clear what that alternating pattern in binary is.
Could you please suggest an error detection scheme for detecting
one possible bit flip in the first 32 bytes of a 33-byte message using
no more than 8 bits of additional data?
Could Pearson hashing be a solution?
Detecting a single bit-flip in any message requires only one extra bit, independent of the length of the message: simply xor together all the bits in the message and tack that on the end. If any single bit flips, the parity bit at the end won't match up.
If you're asking to detect which bit flipped, that can't be done, and a simple argument shows it: the extra eight bits can represent up to 256 classes of 32-byte messages, but the zero message and the 256 messages with one on bit each must all be in different classes. Thus, there are 257 messages which must be distinctly classified, and only 256 classes.
You can detect one bit flip with just one extra bit in any length message (as stated by #Daniel Wagner). The parity bit can, simply put, indicate whether the total number of 1-bits is odd or even. Obviously, if the number of bits that are wrong is even, then the parity bit will fail, so you cannot detect 2-bit errors.
Now, for a more accessible understanding of why you can't error-correct 32 bytes (256 bits) with just 8 bits, please read about the Hamming code (like used in ECC memory). Such a scheme uses special error-correcting parity bits (henceforth called "EC parity") that only encode the parity of a subset of the total number of bits. For every 2^m - 1 total bits, you need to use m EC bits. These represent each possible different mask following the pattern "x bits on, x bits off" where x is a power of 2. Thus, the larger the number of bits at once, the better the data/parity bit ratio you get. For example, 7 total bits would allow encoding only 4 data bits after losing 3 EC bits, but 31 total bits can encode 26 data bits after losing 5 EC bits.
Now, to really understand this probably will take an example. Consider the following sets of masks. The first two rows are to be read top down, indicating the bit number (the "Most Significant Byte" I've labeled MSB):
MSB LSB
| |
v v
33222222 22221111 11111100 0000000|0
10987654 32109876 54321098 7654321|0
-------- -------- -------- -------|-
1: 10101010 10101010 10101010 1010101|0
2: 11001100 11001100 11001100 1100110|0
3: 11110000 11110000 11110000 1111000|0
4: 11111111 00000000 11111111 0000000|0
5: 11111111 11111111 00000000 0000000|0
The first thing to notice is that the binary values for 0 to 31 are represented in each column going from right to left (reading the bits in rows 1 through 5). This means that each vertical column is different from each other one (the important part). I put a vertical extra line between bit numbers 0 and 1 for a particular reason: Column 0 is useless because it has no bits set in it.
To perform error-correcting, we will bitwise-AND the received data bits against each EC bit's predefined mask, then compare the resulting parity to the EC bit. For any calculated parities discovered to not match, find the column in which only those bits are set. For example, if error-correcting bits 1, 4, and 5 are wrong when calculated from the received data value, then column #25--containing 1s in only those masks--must be the incorrect bit and can be corrected by flipping it. If only a single error-correcting bit is wrong, then the error is in that error-correcting bit. Here's an analogy to help you understand why this works:
There are 32 identical boxes, with one containing a marble. Your task is to locate the marble using just an old-style scale (the kind with two balanced platforms to compare the weights of different objects) and you are only allowed 5 weighing attempts. The solution is fairly easy: you put 16 boxes on each side of the scale and the heavier side indicates which side the marble is on. Discarding the 16 boxes on the lighter side, you then weigh 8 and 8 boxes keeping the heavier, then 4 and 4, then 2 and 2, and finally locate the marble by comparing the weights of the last 2 boxes 1 to 1: the heaviest box contains the marble. You have completed the task in only 5 weighings of 32, 16, 8, 4, and 2 boxes.
Similarly, our bit patterns have divided up the boxes in 5 different groups. Going backwards, the fifth EC bit determines whether an error is on the left side or the right side. In our scenario with bit #25, it is wrong, so we know that the error bit is on the left side of the group (bits 16-31). In our next mask for EC bit #4 (still stepping backward), we only consider bits 16-31, and we find that the "heavier" side is the left one again, so we have narrowed down the bits 24-31. Following the decision tree downward and cutting the number of possible columns in half each time, by the time we reach EC bit 1 there is only 1 possible bit left--our "marble in a box".
Note: The analogy is useful, though not perfect: 1-bits are not represented by marbles--the erroring bit location is represented by the marble.
Now, some playing around with these masks and thinking how to arrange things will reveal that there is a problem: If we try to make all 31 bits data bits, then we need 5 more bits for EC. But how, then, will we tell if the EC bits themselves are wrong? Just a single EC bit wrong will incorrectly tell us that some data bit needs correction, and we'll wrongly flip that data bit. The EC bits have to somehow encode for themselves! The solution is to position the parity bits inside of the data, in columns from the bit patterns above where only one bit is set. This way, any data bit being wrong will trigger two EC bits to be wrong, making it so that if only one EC bit is wrong, we know it is wrong itself instead of it signifying a data bit is wrong. The columns that satisfy the one-bit condition are 1, 2, 4, 8, and 16. The data bits will be interleaved between these starting at position 2. (Remember, we are not using position 0 as it would never provide any information--none of our EC bits would be set at all).
Finally, adding one more bit for overall parity will allow detecting 2-bit errors and reliably correcting 1-bit errors, as we can then compare the EC bits to it: if the EC bits say something is wrong, but the parity bit says otherwise, we know there are 2 bits wrong and cannot perform correction. We can use the discarded bit #0 as our parity bit! In fact, now we are encoding the following pattern:
0: 11111111 11111111 11111111 11111111
This gives us a final total of 6 Error-Checking and Correcting (ECC) bits. Extending the scheme of using different masks indefinitely looks like this:
32 bits - 6 ECC bits = 26 data
64 bits - 7 ECC bits = 57 data
128 bits - 8 ECC bits = 120 data
256 bits - 9 ECC bits = 247 data
512 bits - 10 ECC bits = 502 data
Now, if we are sure that we only will get a 1-bit error, we can dispense with the #0 parity bit, so we have the following:
31 bits - 5 ECC bits = 26 data
63 bits - 6 ECC bits = 57 data
127 bits - 7 ECC bits = 120 data
255 bits - 8 ECC bits = 247 data
511 bits - 9 ECC bits = 502 data
This is no change because we don't get any more data bits. Oops! 32 bytes (256 bits) as you requested cannot be error-corrected with a single byte, even if we know we can have only a 1-bit error at worst, and we know the ECC bits will be correct (allowing us to move them out of the data region and use them all for data). We need TWO more bits than we have--one must slide up to the next range of 512 bits, then leave out 246 data bits to get our 256 data bits. So that's one more ECC bit AND one more data bit (as we only have 255, exactly what Daniel told you).
Summary:: You need 33 bytes + 1 bit to detect which bit flipped in the first 32 bytes.
Note: if you are going to send 64 bytes, then you're under the 32:1 ratio, as you can error correct that in just 10 bits. But it's that in real world applications, the "frame size" of your ECC can't keep going up indefinitely for a few reasons: 1) The number of bits being worked with at once may be much smaller than the frame size, leading to gross inefficiencies (think ECC RAM). 2) The chance of being able to accurately correct a bit gets less and less, since the larger the frame, the greater the chance it will have more errors, and 2 errors defeats error-correction ability, while 3 or more can defeat even error-detection ability. 3) Once an error is detected, the larger the frame size, the larger the size of the corrupted piece that must be retransmitted.
If you need to use a whole byte instead of a bit, and you only need to detect errors, then the standard solution is to use a cyclic redundancy check (CRC). There are several well-known 8-bit CRCs to choose from.
A typical fast implementation of a CRC uses a table with 256 entries to handle a byte of the message at a time. For the case of an 8 bit CRC this is a special case of Pearson's algorithm.
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.