Understanding the Hamming 'sphere packing' bound in coding theory
The Hamming bound, or 'sphere-packing bound', is an important result in communications and coding theory. It places an upper limit on the number of distinct codewords that can be encoded using a block of symbols of particular length, to provide a specified error-correcting capability.
The term 'sphere-packing' arises from the traditional formulation of the problem in terms of codewords representing the centres of spheres in multidimensional discrete space. So far as possible, I will avoid using such mind-bending notions in this article. However, it should (I hope) be clear from my explanation how this topological representation arises. In case it isn't, I will explain at the end of the article.
As always, I will start my explanation with a simple, concrete example, and build up incrementally to the more general case that is usually presented in textbooks.
I don't assume any prior knowledge of information theory or coding theory -- only high-school algebra. However, I would expect somebody reading this to have a least a passing familiarity with concepts related to the representation of information.
Stating the problem
Suppose we want to communicate or store information in fixed-size blocks of symbols, in such a way that some errors can be corrected. These errors may arise during a communication process, or just over a period of time. The effect of errors will be to change one symbol into another, and 'error correction' amounts to recovering the original, uncorrupted information.
In many computing applications, the 'symbols' in question will just be the binary numbers 0 and 1 although, of course, many other kinds of symbol exist. In computing, convenient block sizes are multiples of eight binary bits but, in practice, we may get better theoretical performance using coding schemes that do not lend themselves to packing into bytes.
Consider a coding scheme with q symbols (in binary, q=2, of course), in blocks of length n. Each n-symbol block will have some redundancy -- we can't do error correction without it.
The absolute maximum number of symbol patterns that can be encoded is just qn -- this value arises from assuming that each position in the coded block can take any one of the q symbols. An eight-bit binary block, for example, can encode 28=256 different patterns.
However, not every symbol pattern will be a codeword, to be transmitted or stored. For sure, we can encode the numbers 0-255 using an 8-bit binary number, but not with any error-correcting capabilities. Some of the symbol patterns will have to be redundant.
So the question arises: how many distinct codewords can be represented in a block of n symbols, each drawn from an alphabet of q symbols, if we need to be able to correct t errors per block?
This is a surprisingly difficult question to answer, and exact answers are still unknown for some block sizes. The importance of the Hamming bound is that it places an upper limit on the number of codewords for a given block size and alphabet. A coding scheme may not reach that level of performance indicated by the Hamming bound, but it certainly won't exceed it.
Let's start our investigation of the Hamming bound by looking at a particular coding example.
An example: the binary Hamming (7,16,3) code
This particular coding scheme is widely used in tutorial examples, including this one, because it's simple enough that the entire set of codewords can be written out and examined.
The designation "(7,16,3)" indicates that this is a coding scheme with a block size of seven symbols, representing sixteen codewords. The The "3" in the designation refers to the distance of the code, not the number of parity symbols, although this code does, in fact, use three parity bits.
The distance, or Hamming distance, of a coding scheme is the smallest number of symbols that have to be changed by errors, before a codeword cannot be read without ambiguity. The Hamming (7,16,3) code has a distance of three because (as we'll see) it's necessary to corrupt at least three symbols before a codeword is unreadable.
With binary values, the (7,16,3) code has seven bits in each block, of which four bits correspond to information and three to parity. The code can therefore represent 24 = 16 distinct codewords.
Here is one representation of a Hamming binary (7,16,3) code, written out in full. I've numbered the codewords 00-15 but, of course, the codewords don't have to be numbered this way -- they don't even have to represent numbers.
Codeword (7,16,3) Representation -------- ---------------------- 00 0 0 0 0 0 0 0 01 0 0 0 1 0 1 1 02 0 0 1 0 1 1 0 03 0 0 1 1 1 0 1 04 0 1 0 0 1 1 1 05 0 1 0 1 1 0 0 06 0 1 1 0 0 0 1 07 0 1 1 1 0 1 0 08 1 0 0 0 1 0 1 09 1 0 0 1 1 1 0 10 1 0 1 0 0 1 1 11 1 0 1 1 0 0 0 12 1 1 0 0 0 1 0 13 1 1 0 1 0 0 1 14 1 1 1 0 1 0 0 15 1 1 1 1 1 1 1
My grouping puts the four 'data' bits and the three 'parity' bits into separate blocks, just for readability. You'll often see this code written with the parity bits interspersed with the data bits, or with completely different bit patterns, but none of this changes the analysis.
The "distance" of the code should be clear just from inspection. The smallest number of positions in which any two codewords differ is three. For example, codewords 00 and 01 differ in three positions -- 01 has three '1' digits where 00 has '0's. Of course, some pairs of codewords differ more substantially -- 00 and 15 differ in all seven positions. However, 15 only differs in three positions from 14 -- it is the minimum distance that is significant for error correction.
That every codeword differs by exactly the same number of bits from its nearest neighbour is one of the trademarks of a so-called "perfect code" -- every bit is fully utilized.
Could the (7,16,3) code be extended to more than 16 codewords?
The Hamming (7,16,3) code uses seven bits to encode four information bits. On the face of it, that's not very efficient. With a binary representation, seven bits allows for 27=128 symbol patterns, yet only sixteen are used for codewords.
The question naturally arises: can we shuffle the bits around, such that we get more codewords without a longer code, and without loss of error-correction capability?
We can investigate this by considering some of the possible errors that might arise if the real codeword was, for example, 00. This codeword is represented as seven zero bits -- 0000 000. The table below shows some of the one-bit and two-bit errors that might affect the 00 codeword.
Received message Error bits Ambiguous with ---------------- ---------- -------------- 0 0 0 0 0 0 1 1 none 0 0 0 0 0 1 0 1 none 0 0 0 0 0 1 1 2 01 0 0 0 0 1 0 0 1 none 0 0 0 0 1 0 1 2 08 0 0 0 0 1 1 0 2 02 0 0 0 0 1 1 1 2 04 0 0 0 1 0 0 0 1 none 0 0 0 1 0 0 1 2 01 0 0 0 1 0 1 0 2 01 ...
By comparing the received (error-affected) message with the table of correct codewords above, we can see which codewords become equally likely with 00. It's clear that a greater number of errors than one is impossible to correct, even though it can be detected. A distance-3 code allows only one bit to be changed by an error -- more than that, and the received message is ambiguous. This is a general result -- to correct t errors we need a distance of at least 2t + 1.
If we could reduce the distance of the code without sacrificing error performance, then we might be able to redistribute the seven bits so that more (on average) were used for information, and fewer for parity. But that doesn't seem to be possible. So the question whether the (7,16,3) can be extended to handle more than 16 codewords reduces to whether there is a (7,M,3) code, where M > 16. That is, can we achieve a minimum distance between any two codewords of three bits, and still find a way to encode more than 16 codewords?
Codewords as 'pattern cluster centres'
To answer that question, we should think of a particular codeword as being the 'centre' of a cluster of symbol patterns. As we see from the table above, codeword 00 corresponds not only to the pattern 0000 000, but also to patterns 0000 001, 0000 010, and so on. All these patterns are considered to be equivalent to codeword 00, after error correction is taken into account. That is, any of these patterns would be read as codeword 00, as 00 is more similar to the value received than any other potential codeword in the set.
What's important to determine is the size of the cluster associated with each codeword. This size will be the same for each codeword, because the significant factor is the number of bits changed, not the specific values of the bits. In the (7,16,3) example, there are seven different ways a codeword can be changed in one bit, and still not lead to an uncorrectable error. If we add the original codeword -- unaffected by error -- the cluster size is eight patterns per codeword.
Note that this determination depends only on the block size and distance. The cluster size is eight patterns per codeword, regardless of how the block is divided into data and parity bits.
However the code is structured, we can say one thing with certainty: the number of codewords, multiplied by the cluster size, must be less than the total number of symbol patterns. There simply aren't any more symbol patterns than this, regardless of encoding. Because this basic fact will be true for any alphabet of symbols, not just for binary digits, we can write
M c ≤ qn
where M is the number of distinct codewords, c is the cluster size per codeword, q is the number of symbols in the alphabet, and n is the number of symbols per block. For a seven-symbol binary code, with a cluster size of 8 symbols per codeword,
M ≤ qn / c
M ≤ 128 / 8 = 16.
Consequently, there is no way to encode more than sixteen symbols using seven bits, if we need to be able to correct a one-bit error.
Towards the Hamming bound
Although it might not be obvious, we've just worked out the Hamming bound for a seven-bit binary code with one-bit error correction. It's important to understand that the Hamming bound only tells us a maximum number of possible codewords, based on a consideration of how non-code patterns cluster around codeword patterns. It doesn't tell us how to construct a code with particular characteristics, or even if it's theoretically possible to do so.
In the (7,16,3) example we can actually construct the code in a number of different ways, including by simple trial-and-error. So we know the code exists. Moreover, we know that it supports 16 distinct codewords -- the same as the Hamming limit -- simply by counting them. Making this kind of determination with more complicated codes will not be as straightforward.
The Hamming bound for multiple-error-correcting binary codes
The key to determining the Hamming bound will always be calculating the cluster size, that is, the number of symbol patterns per valid codeword. In the case of the single-error-correcting binary code that's easy -- there is exactly one way in which a particular bit can be changed, to change one codeword to a different, erroneous codeword.
But what about multiple error correction? A code that can correct t errors may well have proportionately larger cluster sizes, because there are more ways in which an error pattern can arise. If we need to correct t errors, then symbols cluster around codewords such that up to t bits are changed with respect to the codeword.
With n bits in a block, there are n ways in which a one-bit error can arise. What about a two-bit error? This is more complicated because, for each position in which the first error can occur, there are n-1 positions in which the second error might occur. Moreover, the errors can be introduced in any order.
Finding the number of potential error patterns for a given number of errors is a combinatorial problem. The number of error patterns is equivalent -- in this case -- to the number of unordered selections of two objects that can be made from seven objects. A common way to write this combinatorial function is 7C2, usually read as "seven choose two". In general:
nCm = n! / ( m! (n-m)! )
For the record, 7C2 = 21; that is, there are 21 possible symbol patterns arising from two bits being in error. To get the cluster size we need to add this figure to the number of one bit error patterns, and add one for the original, uncorrupted codeword. In general, for a code that can correct up to t error bits, we need to consider all the symbol patterns that differ in up to and including t places. So the cluster size c is
c = 1 + nC1 + nC2 + ... + nCt
From this cluster size we can work out the Hamming bound, as above.
The general Hamming bound
So far I have considered only binary codes. This makes life a little easier, because there's only one way in which a symbol can be in error -- if it's a binary '1', then an error makes it a '0', and vice versa.
The Hamming bound is not limited to binary codes, though. To handle larger symbol alphabets, we have to think more carefully about all the different ways in which a symbols can be in error.
in the previous section I explained that the number of possible one-bit error patterns was simply n, the block size. If we're using arbitrary symbols, rather than bits, then the number of one-symbol error patterns in n(q-1), where q is the number of symbols in the alphabet. It's q-1, rather than q, because one, and only one, symbol is not an error.
For a two-bit error the number of error patterns was nC2. With two symbols, rather than two bits, we have (q-1)2 symbol patterns for each error. This is a square term because for each of the q-i ways the first symbol can be in error, there are q-i ways the second symbol can be in error. So the total number of two-symbol errors is nC2 (q-1)2
Extending this logic to the case of t errors, the total number of error patterns is nCt (q-1)t.
To find the cluster size we have to add the error patterns corresponding to all the number of errors up to and including t, and the original, error-free, codeword. So we have
c = 1 + n(q-1) + nC2(1-q)2 + ... + nCt(1-q)t
If M is the total number of codewords, then we have the relationship
M [ 1 + n(q-1) + nC2(1-q)2 + ... + nCt(1-q)t ] ≤ qn
which is the way the Hamming bound is typically formulated.
Why is it a 'sphere packing' bound?
In the explanation above, I've used the notion of a 'cluster size', that is, the number of symbol patterns that correspond to each valid codeword. Each codeword is, in a sense, the centre of such a cluster. The larger the number of symbols that is changed in a valid codeword, the "further" the symbol sequence is from that codeword.
Changing the symbol in a specific position in a block can be regarded as a move of one unit in a specific direction. Each position in the block represents a different direction. So, if a valid codeword is regarded as a centre of some kind, each error represents a move from that centre in a particular direction. Unfortunately, the directions have to be orthogonal, and we need to allocate one direction to each position in the code. In the case where n=7, we need to try to visualize a seven-dimensional space.
Moreover, it's an odd kind of space, even for a multidimensional one -- each direction allows for only one position: "correct" or "error". Still, we can kind-of visualize particular codewords as occupying particular regions in this space, with spherical boundaries. So what I've referred to as a "pattern cluster" can alternatively be seen as a region in n-dimensional space that encompasses a particular set of patterns. Each of those patterns "belongs to" the codeword at the centre of the sphere, for the purposes of error correction.
Finding a code with an optimal use of symbols can be regarded as finding a way to fill n-dimensional space with spheres, such that none of the spheres overlap, and there is minimal empty space. This is broadly analogous to finding a cluster size (using my terminology) that gives a codeword set whose size is equal to the Hamming bound. This would be a "perfect" code.
Whether this notion of filling an n-dimensional space with spheres is a helpful one or not is, in the end, a matter for individual discernment. However, for better or worse, it's a widely-used concept in communications engineering.
