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The algorithm to calculate a CRC involves dividing (mod 2) the data by a polynomial, and that, by nature starts at the biggest bit using the basic long division algorithm and works down (unless you're taking the shortcuts and using tables).
Now, the stream I'm dealing with has the requirements that the data is added little endian and the CRC remainder goes at the end, whereas if the CRC was applied and appended; the CRC remainder bits would appear at the leftmost point in the least significant bit given the bitstream is little endian.
So here's the question. We have a little endian stream with the CRC remainder at the "unexpected" end (correct me if I'm wrong please), should the CRC remainder be added big endian at the end of the bytestream, and then the CRC run on the whole bytestream (this is what I expect from the requirements) or something else?
How in industry is this normally done?
Major update for clarity thanks to Mark Adler's highly helpful answer.
I've read a few posts, but seen nothing where there seems to be a little endian bytestream with a CRC in the MSB (rightmost).
The image above should describe what I'm doing. All the bytes are big endian bit order, but the issue is that the requirements state that the bytes should be little endian byte ordered, and then the CRC tacked on the end.
For the bytestream as a sequence of bits to be validated by the CRC remainder being placed at the end, the CRC remainder bytes should be added bit endian, therefore allowing the message as a whole to be validated with the polynomial. However this involves adding bytes to the stream in a mix of endiannesses, which sounds highly ugly and wrong.
I will assume that by "biggest" bit, you mean most significant. First off, you can take either the most-significant bit or the least-significant bit of the first byte as the highest power of x for polynomial division. Both are in common use. There is no "by nature" here. And that has nothing to do with whether tables are used or not. Taking the least-significant bit as the highest power of x, the one you would call "not by nature" is in very common use, due to slightly faster and simpler software implementations as compared to using the most-significant bit.
Second, bit streams are neither "little endian", nor "big endian". Those terms are used for how integers are broken up into a sequence of bytes. That has nothing to do with the interpretation of a stream of bits as a polynomial. The terms you seem to be looking for are "reflected" and "not reflected" bit streams in and CRCs out. "reflected" means that the highest power of x is the least significant bit, and "not reflected" means it is the most significant bit.
If you look at Greg Cook's catalogue of CRCs, you will see as part of each definition refin=false refout=false or refin=true refout=true, meaning that the data coming in is reflected or not, and the CRC coming out is reflected or not, referring to where the highest power of x is found. For the CRC, the entire n-bits is reflected or not. In actual implementations, no bits are flipped for the input data or the output CRC. Instead, the constant CRC polynomial is reflected to match the data and CRC reflections. That is done once as the code is written, never during execution. (There is one outlier CRC in Greg's catalogue, CRC-12/UMTS, that has refin=false refout=true. For that one, the implementation would in fact have to reflect the CRC result every time.)
Given all that, I am left attempting to intepret your question. What do you mean by "the data is added little endian"? Does that mean the CRC is being calculated using the least-significant bit as the highest power of x (the opposite of your "by nature")? What does "the CRC remainder bits would appear at the leftmost point in the least significant bit given the bitstream is little endian" mean? That one is really confusing, since there is no leftmost point of a bit, and I can't tell at all what you're trying to say about the arrangement of the remainder bits.
The only thing I think I understand and can try to answer here is: "How in industry is this normally done?"
Well, as you can tell from the list of over a hundred CRCs, there is little normalcy established. What I can say is that CRCs have a special property that leads to a "natural" (now I can use that word) ordering of the CRC bits and bytes at the end of the stream of bits and bytes that the CRC was calculated on. That property is that if you append it properly, the CRC of the entire message, including the CRC at the end, will always be the same constant, if there are no errors in the message. Now little and big endian are useful terms, but only for the CRC itself, not the bit or byte stream. The proper order is little endian for reflected CRCs and big endian for non-reflected CRCs. (This assumes that the input and output have the same reflection, so this won't work for that one outlier CRC.)
Of course, I have seen many cases where a reflected CRC is used, but is appended to the stream big-endian, and vice versa, in which case this calculation of the CRC on the entire message doesn't work. That's ok, since the alternative way to check the CRC is to simply repeat what was done before transmission, which is to calculate the CRC only on the data portion of the message, then properly assemble the CRC from the bytes that follow it, and compare the two values. That is what would be done for any other hash that doesn't have that elegant mathematical property of CRCs.
I've seen 8-bit, 16-bit, and 32-bit CRCs.
At what point do I need to jump to a wider CRC?
My gut reaction is that it is based on the data length:
1-100 bytes: 8-bit CRC
101 - 1000 bytes: 16-bit CRC
1001 - ??? bytes: 32-bit CRC
EDIT:
Looking at the Wikipedia page about CRC and Lott's answer, here' what we have:
<64 bytes: 8-bit CRC
<16K bytes: 16-bit CRC
<512M bytes: 32-bit CRC
It's not a research topic. It's really well understood: http://en.wikipedia.org/wiki/Cyclic_redundancy_check
The math is pretty simple. An 8-bit CRC boils all messages down to one of 256 values. If your message is more than a few bytes long, the possibility of multiple messages having the same hash value goes up higher and higher.
A 16-bit CRC, similarly, gives you one of the 65,536 available hash values. What are the odds of any two messages having one of these values?
A 32-bit CRC gives you about 4 billion available hash values.
From the wikipedia article: "maximal total blocklength is equal to 2**r − 1". That's in bits. You don't need to do much research to see that 2**9 - 1 is 511 bits. Using CRC-8, multiple messages longer than 64 bytes will have the same CRC checksum value.
The effectiveness of a CRC is dependent on multiple factors. You not only need to select the SIZE of the CRC but also the GENERATING POLYNOMIAL to use. There are complicated and non-intuitive trade-offs depending on:
The expected bit error rate of the channel.
Whether the errors tend to occur in bursts or tend to be spread out (burst is common)
The length of the data to be protected - maximum length, minimum length and distribution.
The paper Cyclic Redundancy Code Polynominal Selection For Embedded Networks, by Philip Koopman and Tridib Chakravarty, publised in the proceedings of the 2004 International Conference on Dependable Systems and Networks gives a very good overview and makes several recomendations. It also provides a bibliography for further understanding.
http://www.ece.cmu.edu/~koopman/roses/dsn04/koopman04_crc_poly_embedded.pdf
The choice of CRC length versus file size is mainly relevant in cases where one is more likely to have an input which differs from the "correct" input by three or fewer bits than to have a one which is massively different. Given two inputs which are massively different, the possibility of a false match will be about 1/256 with most forms of 8-bit check value (including CRC), 1/65536 with most forms of 16-bit check value (including CRC), etc. The advantage of CRC comes from its treatment of inputs which are very similar.
With an 8-bit CRC whose polynomial generates two periods of length 128, the fraction of single, double, or triple bit errors in a packet shorter than that which go undetected won't be 1/256--it will be zero. Likewise with a 16-bit CRC of period 32768, using packets of 32768 bits or less.
If packets are longer than the CRC period, however, then a double-bit error will go undetected if the distance between the erroneous bits is a multiple of the CRC period. While that might not seem like a terribly likely scenario, a CRC8 will be somewhat worse at catching double-bit errors in long packets than at catching "packet is totally scrambled" errors. If double-bit errors are the second most common failure mode (after single-bit errors), that would be bad. If anything that corrupts some data is likely to corrupt a lot of it, however, the inferior behavior of CRCs with double-bit errors may be a non-issue.
I think the size of the CRC has more to do with how unique of a CRC you need instead of of the size of the input data. This is related to the particular usage and number of items on which you're calculating a CRC.
The CRC should be chosen specifically for the length of the messages, it is not just a question of the size of the CRC: http://www.ece.cmu.edu/~koopman/roses/dsn04/koopman04_crc_poly_embedded.pdf
Here is a nice "real world" evaluation of CRC-N
http://www.backplane.com/matt/crc64.html
I use CRC-32 and file-size comparison and have NEVER, in the billions of files checked, run into a matching CRC-32 and File-Size collision. But I know a few exist, when not purposely forced to exist. (Hacked tricks/exploits)
When doing comparison, you should ALSO be checking "data-sizes". You will rarely have a collision of the same data-size, with a matching CRC, within the correct sizes.
Purposely manipulated data, to fake a match, is usually done by adding extra-data until the CRC matches a target. However, that results in a data-size that no-longer matches. Attempting to brute-force, or cycle through random, or sequential data, of the same exact size, would leave a real narrow collision-rate.
You can also have collisions within the data-size, just by the generic limits of the formulas used, and constraints of using bits/bytes and base-ten systems, which depends on floating-point values, which get truncated and clipped.
The point you would want to think about going larger, is when you start to see many collisions which can not be "confirmed" as "originals". (When they both have the same data-size, and (when tested backwards, they have a matching CRC. Reverse/byte or reverse/bits, or bit-offsets)
In any event, it should NEVER be used as the ONLY form of comparison, just for a quick form of comparison, for indexing.
You can use a CRC-8 to index the whole internet, and divide everything into one of N-catagories. You WANT those collisions. Now, with those pre-sorted, you only have to check one of N-directories, looking for "file-size", or "reverse-CRC", or whatever other comparison you can do to that smaller data-set, fast...
Doing a CRC-32 forwards and backwards on the same blob of data is more reliable than using CRC-64 in just one direction. (Or an MD5, for that matter.)
You can detect a single bit error with a CRC in any size packet. Detecting double bit errors or correction of single bit errors is limited to the number of distinct values the CRC can take, so for 8 bits, that would 256; for 16 bits, 65535; etc. 2^n; In practice, though, CRCs actually take on fewer distinct values for single bit errors. For example what I call the 'Y5' polynomial, the 0x5935 polynomial only takes on up to 256 different values before they repeat going back farther, but on the other hand it is able to correct double bit errors that distance, which is 30 bytes plus 2 bytes for errors in the CRC itself.
The number of bits you can correct with forward error correction is also limited by the Hamming Distance of the polynomial. For example, if the Hamming distance is three, you have to flip three bits to change from a set of bits that represents one valid message with matching CRC to another valid message with its own matching CRC. If that is the case, you can correct one bit with confidence. If the Hamming distance were 5, you could correct two bits. But when correcting multiple bits, you are effectively indexing multiple positions, so you need twice as many bits to represent the indexes of two corrected bits rather than one.
With forward error correction, you calculate the CRC on a packet and CRC together, and get a residual value. A good message with zero errors will always have the expected residual value (zero unless there's a nonzero initial value for the CRC register), and each bit position of error has a unique residual value, so use it to identify the position. If you ever get a CRC result with that residual, you know which bit (or bits) to flip to correct the error.
It sounds weird to be going bigger, but that's what I'm trying to do. I want to take the entire sequence of 16-bit integers and hash each one in such a way that it maps to 256-bit space uniformly.
The reason for this is that I'm trying to put a subset of the 16-bit number space into a 256-bit bloom filter, for fast membership testing.
I could use some well-known hashing function on each integer, but I'm looking for an extremely efficient implementation (just a few instructions) so that this runs well in a GPU shader program. I feel like the fact that the hash input is known to be only 16-bits can inform the hash function is designed somehow, but I am failing to see the solution.
Any ideas?
EDITS
Based on the responses, my original question is confusing. Sorry about that. I will try to restate it with a more concrete example:
I have a subset S1 of n numbers from the set S, which is in the range (0, 2^16-1). I need to represent this subset S1 with a 256-bit bloom filter constructed with a single hashing function. The reason for the bloom filter is a space consideration. I've chosen a 256-bit bloom filter because it fits my space requirements, and has a low enough probability of false positives. I'm looking to find a very simple hashing function that can take a number from set S and represent it in 256 bits such that each bit has roughly equal probability of being 1 or 0.
The reason for the requirement of simplicity in the hashing function is that this hashing function is going to have to run thousands of times per pixel, so anywhere where I can trim instructions is a win.
If you multiply (using uint32_t) a 16 bit value by prime (or for that matter any odd number) p between 2^31 and 2^32, then you "probably" smear the results fairly evenly across the 32 bit space. Then you might want to add another prime value, to prevent 0 mapping to 0 (you want each bit to have an equal probability of being 0 or 1, only one input value in 2^256 should have output all zeros, and since there are only 2^16 inputs that means you want none of them to have output all zeros).
So that's how to expand 16 bits to 32 with one operation (plus whatever instructions are needed to load the constant). Use four different values p1 ... p4 to get 256 bits, and run some tests with different p values to find good ones (i.e. those that produce not too many more false positives than what you expect for your Bloom filter given the size of the set you're encoding and assuming an ideal hashing function). For example I'm pretty sure -1 is a bad p-value.
No matter how good the values you'll see some correlations, though: for example as I've described it above the lowest bit of all 4 separate values will be equal, which is a pretty serious dependency. So you probably want a couple more "mixing" operations. For example you might say that each byte of the final output shall be the XOR of two of the bytes of what I've described (and not two least-siginficant bytes!), just to get rid of the simple arithmetic relations.
Unless I've misunderstood the question, though, this is not how a Bloom filter usually works. Usually you want your hash to produce an exact fixed number of set bits for each input, and all the arithmetic to compute the false positive rate relies on this. That's why for a Bloom filter 256 bits in size you'd normally have k 8-bit hashes, not one 256-bit hash. k is normally rather less than half the size of the filter in bits (the optimal value is the number of bits per value in the filter, times ln(2) which is about 0.7). So normally you don't want the probability of each bit being 1 to be anything like as high as 0.5.
The reason is that once you've ORed as few as 4 such 256-bit values together, almost all the bits in your filter are set (15 in 16 of them). So you're looking at a lot of false positives already.
But if you've done the math and you're happy with a single hash function producing a variable number of set bits averaging half of them, then fair enough. Or is the double-occurrence of the number 256 just a coincidence, because k happens to be 32 for the set size you have chosen and you're actually using the 256-bit hash as 32 8-bit hashes?
[Edit: your comment clarifies this, but anyway k should not get so high that you need 256 bits of hash in total. Clearly there's no point in this case using a Bloom filter with more than 16 bits per value (i.e fewer than 16 values), since using the same amount of space you could just list the values, and have a false positive rate of 0. A filter with 16 bits per value gives a false positive rate of something like 1 in 2200. Even there, optimal k is only 23, that is you should set 23 bits in the filter for each value in the set. If you expect the sets to be bigger than 16 values then you want to set fewer bits for each element, and you'll get a higher false positive rate.]
I believe there is some confusion in the question as posed. I will first try to clear up any inconsistencies I've noticed above.
OP originally states that he is trying to map a smaller space into a larger one. If this is truly the case, then the use of the bloom filter algorithm is unnecessary. Instead, as has been suggested in the comments above, the identity function is the only "hash" function necessary to set and test each bit. However, I make the assertion that this is not really what the OP is looking for. If so, then the OP must be storing 2^256 bits in memory (based on how the question is stated) in order for the space of 16-bit integers (i.e. 2^16) to be smaller than his set size; this is an unreasonable amount of memory to be using and is highly unlikely to be the case.
Therefore, I make the assumption that the problem constraints are as follows: we have a 256-bit bit vector in which we want to map the space of 16-bit integers. That is, we have 256 bits available to map 2^16 possible different integers. Thus, we are not actually mapping into a larger space, but, instead, a much smaller space. Similarly, it does appear (again, as previously pointed out in the comments above) that the OP is requesting a single hash function. If this is the case, there is clear misunderstanding about how bloom filters work.
Bloom filters typically use a set of hash independent hash functions to reduce false positives. Without going into too much detail, every input to the bloom filter runs through all n hash functions and then the resulting index in the bit vector is tested for each function. If all indices tested are set to 1, then the value may be in the set (with proper collisions in all n hash functions or overlap, false positives will occur). Moreover, if any of the indices is set to 0, then the value is absolutely not in the set. With this in mind, it is important to notice that an entirely saturated bloom filter has no benefit. That is, every query to the bloom filter will return that the item is in the set.
Hash Function Concerns
Now, back to the OP's original question. It is likely going to be best to use known hashing algorithms (since these are mathematically difficult to write and "rolling your own" typically doesn't end well). If you are worried about efficiency down to clock-cycles, implement the algorithm yourself in the appropriate assembly language for your architecture to reduce running time for each hash function. Remember, algorithmically, hash functions should run in O(1) time, so they should not contribute too much overhead if implemented properly. To start you off, I would recommend considering the modified bernstein hash. I have written a version for your specific case below (mostly for example purposes):
unsigned char modified_bernstein(short key)
{
unsigned ret = key & 0xff;
ret = 33 * ret ^ (key >> 8);
return ret % 256; // Try to do some modulo math to keep it in range
}
The bernstein method I have adapted generally runs as a function of the number of bytes of the input. Since a short type is 2 bytes or 16-bits, I have removed any variables and loops from the algorithm and simply performed some bit twiddling to get at each byte. Finally, an unsigned char can return a value in the range of [0,256) which forces the hash function to return a valid index in the bit vector.
I've seen 8-bit, 16-bit, and 32-bit CRCs.
At what point do I need to jump to a wider CRC?
My gut reaction is that it is based on the data length:
1-100 bytes: 8-bit CRC
101 - 1000 bytes: 16-bit CRC
1001 - ??? bytes: 32-bit CRC
EDIT:
Looking at the Wikipedia page about CRC and Lott's answer, here' what we have:
<64 bytes: 8-bit CRC
<16K bytes: 16-bit CRC
<512M bytes: 32-bit CRC
It's not a research topic. It's really well understood: http://en.wikipedia.org/wiki/Cyclic_redundancy_check
The math is pretty simple. An 8-bit CRC boils all messages down to one of 256 values. If your message is more than a few bytes long, the possibility of multiple messages having the same hash value goes up higher and higher.
A 16-bit CRC, similarly, gives you one of the 65,536 available hash values. What are the odds of any two messages having one of these values?
A 32-bit CRC gives you about 4 billion available hash values.
From the wikipedia article: "maximal total blocklength is equal to 2**r − 1". That's in bits. You don't need to do much research to see that 2**9 - 1 is 511 bits. Using CRC-8, multiple messages longer than 64 bytes will have the same CRC checksum value.
The effectiveness of a CRC is dependent on multiple factors. You not only need to select the SIZE of the CRC but also the GENERATING POLYNOMIAL to use. There are complicated and non-intuitive trade-offs depending on:
The expected bit error rate of the channel.
Whether the errors tend to occur in bursts or tend to be spread out (burst is common)
The length of the data to be protected - maximum length, minimum length and distribution.
The paper Cyclic Redundancy Code Polynominal Selection For Embedded Networks, by Philip Koopman and Tridib Chakravarty, publised in the proceedings of the 2004 International Conference on Dependable Systems and Networks gives a very good overview and makes several recomendations. It also provides a bibliography for further understanding.
http://www.ece.cmu.edu/~koopman/roses/dsn04/koopman04_crc_poly_embedded.pdf
The choice of CRC length versus file size is mainly relevant in cases where one is more likely to have an input which differs from the "correct" input by three or fewer bits than to have a one which is massively different. Given two inputs which are massively different, the possibility of a false match will be about 1/256 with most forms of 8-bit check value (including CRC), 1/65536 with most forms of 16-bit check value (including CRC), etc. The advantage of CRC comes from its treatment of inputs which are very similar.
With an 8-bit CRC whose polynomial generates two periods of length 128, the fraction of single, double, or triple bit errors in a packet shorter than that which go undetected won't be 1/256--it will be zero. Likewise with a 16-bit CRC of period 32768, using packets of 32768 bits or less.
If packets are longer than the CRC period, however, then a double-bit error will go undetected if the distance between the erroneous bits is a multiple of the CRC period. While that might not seem like a terribly likely scenario, a CRC8 will be somewhat worse at catching double-bit errors in long packets than at catching "packet is totally scrambled" errors. If double-bit errors are the second most common failure mode (after single-bit errors), that would be bad. If anything that corrupts some data is likely to corrupt a lot of it, however, the inferior behavior of CRCs with double-bit errors may be a non-issue.
I think the size of the CRC has more to do with how unique of a CRC you need instead of of the size of the input data. This is related to the particular usage and number of items on which you're calculating a CRC.
The CRC should be chosen specifically for the length of the messages, it is not just a question of the size of the CRC: http://www.ece.cmu.edu/~koopman/roses/dsn04/koopman04_crc_poly_embedded.pdf
Here is a nice "real world" evaluation of CRC-N
http://www.backplane.com/matt/crc64.html
I use CRC-32 and file-size comparison and have NEVER, in the billions of files checked, run into a matching CRC-32 and File-Size collision. But I know a few exist, when not purposely forced to exist. (Hacked tricks/exploits)
When doing comparison, you should ALSO be checking "data-sizes". You will rarely have a collision of the same data-size, with a matching CRC, within the correct sizes.
Purposely manipulated data, to fake a match, is usually done by adding extra-data until the CRC matches a target. However, that results in a data-size that no-longer matches. Attempting to brute-force, or cycle through random, or sequential data, of the same exact size, would leave a real narrow collision-rate.
You can also have collisions within the data-size, just by the generic limits of the formulas used, and constraints of using bits/bytes and base-ten systems, which depends on floating-point values, which get truncated and clipped.
The point you would want to think about going larger, is when you start to see many collisions which can not be "confirmed" as "originals". (When they both have the same data-size, and (when tested backwards, they have a matching CRC. Reverse/byte or reverse/bits, or bit-offsets)
In any event, it should NEVER be used as the ONLY form of comparison, just for a quick form of comparison, for indexing.
You can use a CRC-8 to index the whole internet, and divide everything into one of N-catagories. You WANT those collisions. Now, with those pre-sorted, you only have to check one of N-directories, looking for "file-size", or "reverse-CRC", or whatever other comparison you can do to that smaller data-set, fast...
Doing a CRC-32 forwards and backwards on the same blob of data is more reliable than using CRC-64 in just one direction. (Or an MD5, for that matter.)
You can detect a single bit error with a CRC in any size packet. Detecting double bit errors or correction of single bit errors is limited to the number of distinct values the CRC can take, so for 8 bits, that would 256; for 16 bits, 65535; etc. 2^n; In practice, though, CRCs actually take on fewer distinct values for single bit errors. For example what I call the 'Y5' polynomial, the 0x5935 polynomial only takes on up to 256 different values before they repeat going back farther, but on the other hand it is able to correct double bit errors that distance, which is 30 bytes plus 2 bytes for errors in the CRC itself.
The number of bits you can correct with forward error correction is also limited by the Hamming Distance of the polynomial. For example, if the Hamming distance is three, you have to flip three bits to change from a set of bits that represents one valid message with matching CRC to another valid message with its own matching CRC. If that is the case, you can correct one bit with confidence. If the Hamming distance were 5, you could correct two bits. But when correcting multiple bits, you are effectively indexing multiple positions, so you need twice as many bits to represent the indexes of two corrected bits rather than one.
With forward error correction, you calculate the CRC on a packet and CRC together, and get a residual value. A good message with zero errors will always have the expected residual value (zero unless there's a nonzero initial value for the CRC register), and each bit position of error has a unique residual value, so use it to identify the position. If you ever get a CRC result with that residual, you know which bit (or bits) to flip to correct the error.
I've got a large number of integer arrays. Each one has a few thousand integers in it, and each integer is generally the same as the one before it or is different by only a single bit or two. I'd like to shrink each array down as small as possible to reduce my disk IO.
Zlib shrinks it to about 25% of its original size. That's nice, but I don't think its algorithm is particularly well suited for the problem. Does anyone know a compression library or simple algorithm that might perform better for this type of information?
Update: zlib after converting it to an array of xor deltas shrinks it to about 20% of the original size.
If most of the integers really are the same as the previous, and the inter-symbol difference can usually be expressed as a single bit flip, this sounds like a job for XOR.
Take an input stream like:
1101
1101
1110
1110
0110
and output:
1101
0000
0010
0000
1000
a bit of pseudo code
compressed[0] = uncompressed[0]
loop
compressed[i] = uncompressed[i-1] ^ uncompressed[i]
We've now reduced most of the output to 0, even when a high bit is changed. The RLE compression in any other tool you use will have a field day with this. It'll work even better on 32-bit integers, and it can still encode a radically different integer popping up in the stream. You're saved the bother of dealing with bit-packing yourself, as everything remains an int-sized quantity.
When you want to decompress:
uncompressed[0] = compressed[0]
loop
uncompressed[i] = uncompressed[i-1] ^ compressed[i]
This also has the advantage of being a simple algorithm that is going to run really, really fast, since it is just XOR.
Have you considered Run-length encoding?
Or try this: Instead of storing the numbers themselves, you store the differences between the numbers. 1 1 2 2 2 3 5 becomes 1 0 1 0 0 1 2. Now most of the numbers you have to encode are very small. To store a small integer, use an 8-bit integer instead of the 32-bit one you'll encode on most platforms. That's a factor of 4 right there. If you do need to be prepared for bigger gaps than that, designate the high-bit of the 8-bit integer to say "this number requires the next 8 bits as well".
You can combine that with run-length encoding for even better compression ratios, depending on your data.
Neither of these options is particularly hard to implement, and they all run very fast and with very little memory (as opposed to, say, bzip).
You want to preprocess your data -- reversibly transform it to some form that is better-suited to your back-end data compression method, first. The details will depend on both the back-end compression method, and (more critically) on the properties you expect from the data you're compressing.
In your case, zlib is a byte-wise compression method, but your data comes in (32-bit?) integers. You don't need to reimplement zlib yourself, but you do need to read up on how it works, so you can figure out how to present it with easily compressible data, or if it's appropriate for your purposes at all.
Zlib implements a form of Lempel-Ziv coding. JPG and many others use Huffman coding for their backend. Run-length encoding is popular for many ad hoc uses. Etc., etc. ...
Perhaps the answer is to pre-filter the arrays in a way analogous to the Filtering used to create small PNG images. Here are some ideas right off the top of my head. I've not tried these approaches, but if you feel like playing, they could be interesting.
Break your ints up each into 4 bytes, so i0, i1, i2, ..., in becomes b0,0, b0,1, b0,2, b0,3, b1,0, b1,1, b1,2, b1,3, ..., bn,0, bn,1, bn,2, bn,3. Then write out all the bi,0s, followed by the bi,1s, bi,2s, and bi,3s. If most of the time your numbers differ only by a bit or two, you should get nice long runs of repeated bytes, which should compress really nicely using something like Run-length Encoding or zlib. This is my favourite of the methods I present.
If the integers in each array are closely-related to the one before, you could maybe store the original integer, followed by diffs against the previous entry - this should give a smaller set of values to draw from, which typically results in a more compressed form.
If you have various bits differing, you still may have largish differences, but if you're more likely to have large numeric differences that correspond to (usually) one or two bits differing, you may be better off with a scheme where you create ahebyte array - use the first 4 bytes to encode the first integer, and then for each subsequent entry, use 0 or more bytes to indicate which bits should be flipped - storing 0, 1, 2, ..., or 31 in the byte, with a sentinel (say 32) to indicate when you're done. This could result the raw number of bytes needed to represent and integer to something close to 2 on average, which most bytes coming from a limited set (0 - 32). Run that stream through zlib, and maybe you'll be pleasantly surprised.
Did you try bzip2 for this?
http://bzip.org/
It's always worked better than zlib for me.
Since your concern is to reduce disk IO, you'll want to compress each integer array independently, without making reference to other integer arrays.
A common technique for your scenario is to store the differences, since a small number of differences can be encoded with short codewords. It sounds like you need to come up with your own coding scheme for differences, since they are multi-bit differences, perhaps using an 8 bit byte something like this as a starting point:
1 bit to indicate that a complete new integer follows, or that this byte encodes a difference from the last integer,
1 bit to indicate that there are more bytes following, recording more single bit differences for the same integer.
6 bits to record the bit number to switch from your previous integer.
If there are more than 4 bits different, then store the integer.
This scheme might not be appropriate if you also have a lot of completely different codes, since they'll take 5 bytes each now instead of 4.
"Zlib shrinks it by a factor of about 4x." means that a file of 100K now takes up negative 300K; that's pretty impressive by any definition :-). I assume you mean it shrinks it by 75%, i.e., to 1/4 its original size.
One possibility for an optimized compression is as follows (it assumes a 32-bit integer and at most 3 bits changing from element to element).
Output the first integer (32 bits).
Output the number of bit changes (n=0-3, 2 bits).
Output n bit specifiers (0-31, 5 bits each).
Worst case for this compression is 3 bit changes in every integer (2+5+5+5 bits) which will tend towards 17/32 of original size (46.875% compression).
I say "tends towards" since the first integer is always 32 bits but, for any decent sized array, that first integer would be negligable.
Best case is a file of identical integers (no bit changes for every integer, just the 2 zero bits) - this will tend towards 2/32 of original size (93.75% compression).
Where you average 2 bits different per consecutive integer (as you say is your common case), you'll get 2+5+5 bits per integer which will tend towards 12/32 or 62.5% compression.
Your break-even point (if zlib gives 75% compression) is 8 bits per integer which would be
single-bit changes (2+5 = 7 bits) : 80% of the transitions.
double-bit changes (2+5+5 = 12 bits) : 20% of the transitions.
This means your average would have to be 1.2 bit changes per integer to make this worthwhile.
One thing I would suggest looking at is 7zip - this has a very liberal licence and you can link it with your code (I think the source is available as well).
I notice (for my stuff anyway) it performs much better than WinZip on a Windows platform so it may also outperform zlib.