## Archive for June, 2017

### Hamming distances

Monday, June 5th, 2017 Nigel Jones

My guess is that most readers of this blog have at least a vague idea of what “Hamming distance” is.  At the most abstract level it is a measure of how different two equal length “strings” are. In the case where the “strings” are actually binary integers, then the the Hamming distance is simply the number of bit positions that differ. Thus if we are comparing 8-bit integers, then the values 0x00 and 0x01 have a Hamming distance of 1, whereas 0xFF and 0x00 have a Hamming distance of 8.

So what you ask? Well over the years I’ve noticed a recurring “problem” in the industry. The scenario is the following. A design calls for some sort of distributed processing architecture, often with a master processor and a number of slaves. These slaves communicate with the master over some sort of serial bus, and each slave has a unique address. In just about every application I have ever seen, the slave addresses are assigned sequential addresses starting at 1 (0 is usually reserved for the master or for broadcasts). So what’s wrong with this you ask? Well, if you do this you are increasing the probability that the wrong processor will be addressed. For example if two processors have addresses of 0x02 and 0x03, then a single bit flip in the LSB will cause the wrong processor to be addressed. Conversely if the two addresses are e.g. 0x55 and 0xAA then you’ll need 8 bit flips to cause the wrong processor to be addressed. In short, you can minimize the probability of incorrect addressing by maximizing the Hamming distance of the addresses.

Now before you rush to the comment section to tell me that you always protect your address bytes with a block check character or a CRC, go and read this article I wrote a few years back (or more to the point read the referenced article by Phil Koopman). Unless you have chosen the optimal CRC, the chances are you aren’t getting the level of protection you think you are. Even if you are using a good CRC, employing the concept of maximizing the Hamming distance essentially comes for free and so why not use it?

I’ll assume for the sake of argument that I’ve convinced you that this is a good idea. Now if you have just two processors on the bus, then it’s easy to choose addresses that give a Hamming distance of 8. For example if the first processor has an address of A, then just assign the second processor an address of A XOR 0xFF. E.g. 0x55 and 0xAA.

However, what happens when you have three processors on the bus? In this case you want to maximize the Hamming distance between all three. The questions now become:

1. What is the maximum Hamming distance obtainable when choosing three unique values out of an 8 bit byte?
2. What are example values I can use?

Now I’m sure that some mathematician (Hamming?) has worked out a generalized solution to this problem. In my case I wrote a quick program to explore the solution space and discovered that the maximum Hamming distance achievable between three 8-bit values is 5. An example is {0xFF, 0xE0, 0x1C}. To see that this is the case: 0xFF XOR 0xE0 = 0x1F (1.e. five ones); 0xFF XOR 0x1C = 0xE3 (i.e. five ones); 0xE0 XOR 0x1C = 0xFC (i.e. 6 ones).  Note that the fact that the last pair yields a Hamming distance of 6 is nice, but does not alter the fact that the minimum of the maximum Hamming distances is 5.  FWIW my program tells me that there are something like 143,360 different combinations that will yield a Hamming distance of 5 for this particular set of requirements.

What about when you have four devices on the bus? In this case the minimum of the maximum achievable Hamming distance drops to 4. An example is {0xFF, 0xF0, 0xCC, 0x0F}.  For five devices the minimum of the maximum Hamming distance is also 4. An example is {0xFF, 0xF0, 0xCC, 0xC3, 0xAA}

So to summarize. With an 8 bit address field, the maximum Hamming distance achievable by device count is:

• Two devices. Maximum Hamming distance = 8.
• Three devices. Maximum Hamming distance = 5.
• Four devices. Maximum Hamming distance = 4.
• Five devices. Maximum Hamming distance = 4.

Beyond 5 devices I’ll need a better algorithm than what I’m using to find a solution. However given that in most of my work I rarely go beyond 4 devices I’ve never felt the need to put the effort in to find out. If anyone out there has a closed form solution or an efficient algorithm for determining any arbitrary solution (e.g. I have 13 devices sitting on a CANBus which uses 11-bit identifiers.  What addresses should I assign to maximize the Hamming distance), then I’d be very interested in hearing about it.

Note that I have concentrated here on address bytes. The same principle applies to any other parameter over which you have control. Typical examples are command bytes and also enumerated values. In short, if a typical message that you send looks something like this: <STX> 01 00 00 02 00 [BCC] <ETX>, then try applying the Hamming distance principles such that a typical message looks more like <STX> FC 73 00 42 D9 [BCC] <ETX>. Not only will you get more protection, you’ll also find that the messages are easier to interpret on a protocol analyzer since you aren’t dealing with a sea of zeros and ones.

Finally, as in most things in engineering, there is another way of looking at the problem. See this article I wrote a few years back.