USPS tracking numbers

I noticed the other day that an app on my phone assumed that a long number was a USPS tracking number. I wondered how it decided that and did a little research. I assumed there was some structure to the number, at least a check sum if not more than that.

This turned out to be a deep rabbit hole. USPS and other carriers all have a variety of tracking numbers, either for different kinds of packages or formats that have changed over time. My impression is that much of what is publicly known about these numbers has been reverse engineered, not extracted from documentation. I decided to turn around before I spent any more time looking into this.

Then I took a more empirical approach. What if I change a few digits? That should break the checksum. It seems my app believes a positive integer is a USPS tracking number if and only if the number has 22 digits.

That’s not very clever. Or maybe it is. It’s not very clever at the deepest level. The app apparently does not care about false positives. But that might be a clever choice at a higher level. Simply assuming 22-digit numbers are tracking numbers is a good bet, and this is robust to any changes in how groups of digits are interpreted.

Update: It looks like the software checks whether the first digit is a 9. I can change other digits of a tracking number, but not the first.

Related posts

Checksum polynomials

A large class of checksum algorithms have the following pattern:

  1. Think of the bits in a file as the coefficients in a polynomial P(x).
  2. Divide P(x) by a fixed polynomial Q(x) mod 2 and keep the remainder.
  3. Report the remainder as a sequence of bits.

In practice there’s a little more to the algorithm than this, such as appending the length of the file, but the above pattern is at the heart of the algorithm.

There’s a common misconception that the polynomial Q(x) is irreducible, i.e. cannot be factored. This may or may not be the case.


Perhaps the most common choice of Q is

Q(x) = x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x3 + x2 + x + 1

This polynomial is used in the cksum utility and is part of numerous standards. It’s know as CRC-32 polynomial, though there are other polynomials occasionally used in 32-bit implementations of the CRC algorithm. And it is far from irreducible as the following Mathematica code shows. The command

    Factor[x^32 + x^26 + x^23 + x^22 + x^16 + x^12 + 
           x^11 + x^10 + x^8 +  x^7 + x^5 + x^4 + 
           x^3 + x^2 + x + 1, Modulus -> 2]

shows that Q can be factored as

(1 + x)5 (1 + x + x3 + x4 + x6) (1 + x + x2 + x5 + x6)
(1 + x + x4 + x6 + x7) (1 + x + x4 + x5 + x6 + x7 + x8)

(Mathematica displays polynomials in increasing order of terms.)

Note that the factorization is valid when done over the field with 2 elements, GF(2). Whether a polynomial can be factored, and what the factors are, depends on what field you do your arithmetic in. The polynomial Q(x) above is irreducible as a polynomial with real coefficients. It can be factored working mod 3, for example, but it factors differently mod 3 than it factors mod 2. Here’s the factorization mod 3:

(1 + 2 x2 + 2 x3 + x4 + x5) (2 + x + 2 x2 + x3 + 2 x4 + x6 + x7)
(2 + x + x3 + 2 x7 + x8 + x9 + x10 + 2 x12 + x13 + x15 + 2 x16 + x17 + x18 + x19 + x20)


The polynomial

Q(x) = x64 + x4 + x3 + x + 1

is known as CRC-64, and is part of several standards, including ISO 3309. This polynomial is irreducible mod 2 as the following Mathematica code confirms.

    IrreduciblePolynomialQ[x^64 + x^4 + x^3 + x + 1, Modulus -> 2]

The CRC algorithm uses this polynomial mod 2, but out of curiosity I checked whether it is irreducible in other contexts. The following code tests whether the polynomial is irreducible modulo the first 100 primes.

    Table[IrreduciblePolynomialQ[x^64 + x^4 + x^3 + x + 1, 
        Modulus -> Prime[n]], {n, 1, 100}]

It is irreducible mod p for p = 2, 233, or 383, but not for any other primes up to 541. It’s also irreducible over the real numbers.

Since Q is irreducible mod 2, the check sum essentially views its input P(x) as a member of the finite field GF(264).

Related posts

Luhn checksum algorithm

After writing the previous post on credit card numbers, I intended to link to a previous post that discussed credit card check sums. But I couldn’t find such a post. I’ve written about other kinds of checksums, such as the checksum scheme used in Vehicle Identification Numbers, but apparently I haven’t written about credit card checksums before.

The algorithm used by credit cards, and other identification numbers, is called the Luhn algorithm. I’ll lead up to the algorithm to explain its motivation, going through a thought process Luhn might have gone through in developing his algorithm.

NB: The intermediate steps are not the Luhn algorithm. If you’re just looking for Luhn’s algorithm, skip to the end.

Step 1

A simple way to create a check sum is to add up the digits of a number and tack on the sum mod 10 on the end. For example, this process would replace 1776 with 17761 because

1 + 7 + 7 + 6 = 21

and 21 mod 10 = 1.

This simple checksum will detect if any single digit has been changed. For example, if we receive 17791 we can compute the checksum of 1779 and tell that something is wrong because the last digit should be 4.

Step 2

A common kind of error is transposing consecutive digits, and the scheme above will not detect transpositions: all digits are treated the same way, so scrambling them doesn’t change the checksum.

Luhn’s idea was to double every other digit before taking the sum. This creates an asymmetry between the contribution of consecutive digits to the check sum and makes it possible to detect all transpositions.

Which digits to double

The method described above works whether you start from the left end or the right end, and whether you start doubling with the first or second digit you encounter. But to be compatible with the Luhn algorithm used in practice, start on the right end and double the last digit.

So going back to our little example, 1776 becomes 17764 because

2*6 + 7 + 2*7 + 1 = 34.

If we accidentally transpose the first two digits, 17764 becomes 71764. But the check sum of 7176 is 0. So 71760 would be valid but 71764 is not. We’ve detected an error.

The advantage of starting at the right end is that adding zeros to the front of the number doesn’t change the checksum. So, for example, 01776 becomes 017764.

Step 3

The next refinement is strange. Instead of simply doubling every other digit, we’ll double every other digit and reduce mod 9.

So we could calculate the checksum for 1776 by first doubling every other digit

1776 -> 1, 14, 7, 12

then reducing the doubled digits mod 9, i.e. replacing 14 by 5 and 12 by 3,

1, 14, 7, 12 -> 1, 5, 7, 3

and then finally taking the last digit of the sum.

1 + 5 + 7 + 3 = 16 -> 6.

What does this extra complication buy us? Nothing as far as I can tell. It weakens the algorithm. The algorithm in Step 2 could detect if 09 were transposed to 90; the algorithm here in Step 3 cannot.

Update: Nathan points out in the comments that without the mod 9 step some changes to a single digit could go undetected, namely changing the second of a pair of doubled digits by 5. For example, 33 and 38 would have the same checksum in Step 2 but not in Step 3.

Step 4

The final step is to take the tens compliment of the checksum computed above. This adds nothing to the effectiveness of the algorithm, but it’s what Luhn did.

Luhn algorithm in practice

Let’s describe the position of digits in a number by numbering them from the right end, starting with o: the last digit is in position 0, the one to its left is in position 1, etc.

We can summarize the Luhn algorithm in the following pseudocode.

    sum = 0
    for d in digits:
        if d is in an even position:
            sum = sum + (2*d) mod 9
            sum = sum + d
    return 10 - sum mod 10

The last line isn’t quite correct. If sum mod 10 is 0, then this returns 10 when it should return 0. A clever way around this is to change the last line to

    return (sum*9) mod 10

which will always return a single digit.

NB: In our examples above, the checksum was added to the end. Some credit cards do this, but others put the checksum somewhere in the interior of the number.


The Luhn algorithm was developed in 1954, though it is still widely used. Since that time much more sophisticated error correcting codes have been developed. See the links below for examples.

Tradeoff between alphabet size and word size

Literal alphabets

Natural language alphabets are all within an order of magnitude of the size of the Roman alphabet. The Hebrew alphabet has a few less letters and Russian has a few more. The smallest alphabet I’m aware of is Hawaiian with 13 letters.

Syllabaries are larger than alphabets, but not an order of magnitude larger. For example, Japanese uses 46 symbols and the Cherokee uses 85.

There is a tradeoff between alphabet size and word size that constrains alphabets to be roughly the same size in every culture.

Metaphorical alphabets

Computer science generalizes the term alphabet to mean any fixed collection of symbols and generalizes word to mean any finite sequence of symbols from an alphabet. So, for example, the digits 0 through 9 form an alphabet in this sense, and the words are numbers.

Numbers are sometimes written using smaller or larger alphabets. Binary notation uses an alphabet of only 2 symbols, but numbers require significantly more symbols to express than they do in decimal. Hexadecimal uses an alphabet of 16 symbols and is more compact than decimal notation.


Many design problems can be formulated as a tradeoff between alphabet length and word length. I was working on such a problem recently, which motivated this blog post.

Larger alphabets mean shorter words, but a linear reduction in word size requires an exponential increase in alphabet size. For example, we could write numbers using half as many digits if we thought in base 100 and had symbols for numbers 0 through 99. Or we could think in base 1000, increasing the number of digits by a factor of 100 and reducing the length of number representations by a factor of 3.

Word length is inversely proportional to the logarithm of alphabet size, and so if we have some cost proportional to alphabet size a and another cost proportional to word length, the optimal alphabet size for the sum of these costs would minimize

C1a + C2 / log a.

We could divide this objective function by C1 and set w = C1/C2 to simplify the objective to minimizing

a + w / log a

Setting the derivative with respect to a to zero shows that the optimal alphabet size a is given by solving

a (log a)² = w.

As the weight w given to the cost of word size increases, the optimal alphabet size increases as well. A wide range of weights corresponds to a narrower range of alphabet sizes.

This plot could be misleading. The graph is not showing the cost associated with various alphabet sizes but rather the optimal alphabet size for a given word size cost. The weight w is fixed in any particular application.

An alphabet of size 10 is optimal for a word length cost weight w of about 50. Using w = 50, the following plot shows how alphabet size cost and word size costs compare. A large increase in alphabet size cost only buys a small decrease in word size cost.

Related post: The base with the largest decibel

Morse code numbers and abbreviations

Numbers in Morse code seem a little strange. Here they are:

    | Digit | Code  |
    |     1 | .---- |
    |     2 | ..--- |
    |     3 | ...-- |
    |     4 | ....- |
    |     5 | ..... |
    |     6 | -.... |
    |     7 | --... |
    |     8 | ---.. |
    |     9 | ----. |
    |     0 | ----- |

They’re fairly regular, but not quite. That’s why a couple years ago I thought it would be an interesting exercise to write terse code to encode and decode digits in Morse code. There’s exploitable regularity, but it’s irregular enough to make the exercise challenging.


As with so many things, this scheme makes more sense than it seems at first. When you ask “Why didn’t they just …” there’s often a non-obvious answer.

The letters largely exhausted the possibilities of up to 4 dots and dashes. Some digits would have to take five symbols, and it makes sense that they would all take 5 symbols. But why the ones above? This scheme uses a lot of dashes, and dashes take three times longer to transmit than dots.

A more efficient scheme would be to use binary notation, with dot for 0’s and dash for 1’s. That way the leading symbol would always be a dot and usually the second would be a dot. That’s when encoding digits 0 through 9. As a bonus you could use the same scheme to encode larger numbers in a single Morse code entity.

The problem with this scheme is that Morse code is intended for humans to decode by ear. A binary scheme would be hard to hear. The scheme actually used is easy to hear because you only change from dot to dash at most once. As Morse code entities get longer, the patterns get simpler. Punctuation marks take six or more dots and dashes, but they have simple patterns that are easy to hear.

Code golf

When I posed my coding exercise as a challenge, the winner was Carlos Luna-Mota with the following Python code.

    e=lambda x:S[9-x:14-x]
    d=lambda x:9-S.find(x)

Honorable mention goes to Manuel Eberl with the following code. It only does decoding, but is quite clever and short.

    d=lambda c:hash(c+'BvS+')%10

It only works in Python 2 because it depends on the specific hashing algorithm used in earlier versions of Python.

Cut numbers

If you’re mixing letters and digits, digits have to be five symbols long. But if you know that characters have to be digits in some context, this opens up the possibility of shorter encodings.

The most common abbreviations are T for 0 and N for 9. For example, a signal report is always three digits, and someone may send 5NN rather than 599 because in that context it’s clear that the N’s represent 9s.

When T abbreviates 0 it might be a “long dash,” slightly longer than a dash meant to represent a T. This is not strictly according to Hoyle but sometimes done.

There are more abbreviations, so called cut numbers, though these are much less common and therefore less likely to be understood.

    | Digit | Code  |  T1 | Abbrev | T2 |
    |     1 | .---- |  17 | .-     |  5 |
    |     2 | ..--- |  15 | ..-    |  7 |
    |     3 | ...-- |  13 | ...-   |  9 |
    |     4 | ....- |  11 | ....-  | 11 |
    |     5 | ..... |   9 | .      |  1 |
    |     6 | -.... |  11 | -....  | 11 |
    |     7 | --... |  13 | -...   |  9 |
    |     8 | ---.. |  15 | -..    |  7 |
    |     9 | ----. |  17 | -.     |  5 |
    |     0 | ----- |  19 | -      |  3 |
    | Total |       | 140 |        | 68 |

The space between dots and dashes is equal to one dot, and the length of a dash is the length of three dots. So the time required to send a sequence of dots and dashes equals

2(# dots) + 4(# dashes) – 1

In the table above, T1 is the time to transmit a digit, in units of dots, without abbreviation, and T2 is the time with abbreviation. Both the maximum time and the average time are cut approximately in half. Of course that’s ideal transmission efficiency, not psychological efficiency. If the abbreviations are not understood on the receiving end and the receiver asks for numbers to be repeated, the shortcut turns into a longcut.

Related posts

Frequency shift keying (FSK) spectrum

This post will look encoding digital data as an analog signal using frequency shift keying (FSK), first directly and then with windowing. We’ll look at the spectrum of the encoded signal and show that basic FSK uses much less bandwidth than direct encoding, but more bandwidth than FSK with windowing.

Square waves

The most natural way to encode binary data as an analog signal would be represent 0s and 1s by a sequence of pulses that take on the values 0 and 1.

A problem with this approach is that it would require a lot of bandwidth.

In theory a square wave has infinite bandwidth: its Fourier series has an infinite number of non-zero coefficients. In practice, the bandwidth of a signal is determined by how many Fourier coefficients it has above some threshold. The threshold would depend on context, but let’s say we ignore Fourier components with amplitude smaller than 0.001.

As I wrote about here, the nth Fourier sine series coefficients for a square wave is equal to 4/nπ for odd n. This means we would need on the order of 1,000 terms before the coefficients drop below our threshold.

Frequency shift keying

The rate of convergence of the Fourier series for a function f depends on the smoothness of f. Discontinuities, like the jump in a square wave, correspond to slow convergence, i.e. high bandwidth. We can save bandwidth by encoding our data with smoother functions.

So instead of jumping from 0 to 1, we’ll encode a 0 as a signal of one frequency and a 1 as a signal with another frequency. By changing the frequency after some whole number of periods, the resulting function will be continuous, and so will have smaller bandwidth.

Suppose we have a one second signal f(t) that is made of half a second of a 4 Hz signal and half a second of a 6 Hz signal, possibly encoding a 0 followed by a 1.

What would the Fourier coefficients look like? If we just had a 4 Hz sine wave, the Fourier series would have only one component: the signal itself at 4 Hz. If we just had a 6 Hz sine wave, the only Fourier component would again be the signal itself. There would be no sine components at other frequencies, and no cosine components.

But our signal patched together by concatenating 4 Hz and 6 Hz signals has non-zero cosine terms for every odd n, and these coefficients decay like O(1/n²).

Our Fourier series is

f(t) = 0.25 sin 8πt + 0.25 sin 12πt + 0.0303 cos 2πt + 0.1112 cos 6πt – 0.3151 cos 12πt + 0.1083 cos 10πt + …

We need to go out to 141 Hz before the coefficients drop below 0.001. That’s a lot of coefficients, but it’s an order of magnitude fewer coefficients than we’d need for a square wave.

Pulse shaping

Although our function f is continuous, it is not differentiable. The left-hand derivative at 1/2 is 8π and the right-hand derivative is 12π. If we could replace f with a related function that is differentiable at 1/2, presumably the signal would require less bandwidth.

We could do this by multiplying both halves of our signal by a windowing function. This is called pulse shaping because instead of a simple sine wave, we change the shape of the wave, tapering it at the ends.

Let’s using a cosine window because that’ll be easy; in practice you’d probably use a different window [1].

Now our function is differentiable at 1/2, and its Fourier series converges more quickly. Now we can disregard components above 40 Hz. With a smoother windowing function the windowed function would have more derivatives and we could disregard more of the high frequencies.

Related posts

[1] This kind of window is called a cosine window because you multiply your signal by one lobe of a cosine function, with the peak in the middle of the signal. Since we’re doing this over [0. 1/2] and again over [1/2, 1], we’re actually multiplying by |sin 2πt|.


Varicode is a way of encoding text and control characters into binary using code words of variable length. It was developed as part of the PSK31 protocol for digital communication over amateur radio.

In the spirit of Morse code, it uses short code words for common characters and longer code words for less common characters in the expectation that this will result in shorter encodings.

If you use variable length words, you’ve got to have some way of knowing when one word ends and the next begins. Varicode solves this problem by using only keywords that begin and end with 1 and that do not contain two consecutive zeros. Then 00 is inserted between code words. Since 00 cannot appear inside a code word, these bits unambiguously mark the space between code words.


Varicode is self-synchronizing in the sense that if you jump into a stream of bits produced by Varicode, as soon as you see two zeros, you know you’re at a code word boundary and can start reading from there. You’ve lost any bits that were transmitted before you jumped in, but you can parse everything going forward.

ASCII doesn’t have this problem or this robustness. It doesn’t have the problem of determining code word boundaries because every code word is eight bits long. But then to read a stream of ASCII bits you need to know your position mod 8. If you jump into a stream of bits, you don’t know where the next code word boundary will be, though you may be able to infer it by trying 8 possibilities and seeing which produces the most intelligible results.

Regular expression

Another way to state the rules for forming Varicode code words is to say that 1 is a valid code (the code for a space, ASCII 0x20) and that you can form new codes by prefixing 1 or 10 to a code. In terms of regular expressions, this says a Varicode code word matches


Fibonacci numbers

How many code words are there of length n? Well, there are two ways to make a code word that long: you either put a 1 in front of a code word of length n-1, or you put a 10 in front of a code word of length n-2. So the number of code words of length n equals the number of code words of length n-1 plus the number of code words of length n-2. That is, the number of code words satisfies the same recurrence relation as the Fibonacci numbers.

It’s easy to see that there’s only one code word of length one, and only one code word of length two, so the number of code words of each length satisfies the initial conditions for the Fibonacci sequence as well, and so they are the Fibonacci numbers.


Varicode encodes a lot more than lower case letters—it encodes most ASCII characters— and so it would take some work to discover the relative frequencies of the characters, and this frequency would depend on where Varicode is used. As far as I know Varicode is use primarily (only?) in PSK31, and so the relevant frequency would be the frequencies in messages sent over PSK31, not English more generally.

You can find the code words of each letter here.

To make things easier, let’s suppose messages are limited to lower case letters and spaces, and that the letters follow the same distribution as in English in general.

We can use the letter frequencies here, except these don’t take spaces into account. If we assume words are about 5 letters long, then the probability of a character being a space is 1/6 and the probabilities of the other characters conditional on not being a space are given by the table. This means the letter probabilities need to be multiplied by 5/6.

This gives us a an expected length of 3.89 bits per letter, which is effectively 5.89 bits when you consider the 00 pattern we have to add between letters.

You could represent the 26 letters and a few more characters using 5 bits, but the result would not be self-synchronizing. One way of looking at this to say that the compression provided by variable length encoding nearly pays for the overhead required to make the code self-synchronizing.

Related posts

Q codes in Seveneves

The first time I heard of Q codes was when reading the novel Seveneves by Neal Stephenson. These are three-letter abbreviations using in Morse code that all begin with Q.

Since Q is always followed by U in native English words, Q can be used to begin a sort of escape sequence [1].

There are dozens of Q codes used in amateur radio [2], and more used in other contexts, but there are only 10 Q codes used in Seveneves [3]. All begin with Q, followed by R, S, or T.

Tree[Q, {Tree[R, {A, K, N, S, T}], Tree[S, {B, L, O}], Tree[T, {H, X}]}]

Each Q code can be used both as a question and as an answer or statement. For example, QRS can mean “Would you like me to slow down” or “Please slow down.” I’ll just give the interrogative forms below.

Here are the 10 codes that appear in Stephenson’s novel.

What is your call sign?
Is my signal intelligible?
Is static a problem?
Should I slow down?
Should I stop sending?
Is my signal fading?
Are you still there?
Could you communicate with …?
Where are you?
Will you keep your station open for talking with me?

Related posts

[1] Some Q codes have a U as the second letter. I don’t know why—there are plenty of unused TLAs that begin with Q—but it is what it is.

[2] You can find a list here.

[3] There is one non-standard code in the novel: QET for “not on planet Earth.”

Self-orthogonal vectors and coding

One of the surprising things about linear algebra over a finite field is that a non-zero vector can be orthogonal to itself.

When you take the inner product of a real vector with itself, you get a sum of squares of real numbers. If any element in the sum is positive, the whole sum is positive.

In a finite field, things don’t work that way. A list of non-zero squares can add up to zero.

In the previous post, we looked at the ternary Golay code, an error correcting code that multiplies a vector of data by a rectangular matrix mod 3. That post included the example that

[1, 0, 1, 2, 2]

was encoded as

[1 0 1 2 2 0 1 0 2 0 0].

The output is orthogonal to itself:

1² + 0² + 1² + 2² + 2² + 0² + 1² + 0² + 2² + 0² + 0² ≡ 0 (mod 3).

In fact, every output of the ternary Golay code will be self-orthogonal. To see this, let v be a 1 by 5 row vector. Then its encoding is vG where G is the 5 by 11 matrix in the previous post. The inner product of vG with itself is

vG (vG)T = vGGTvT.

We can easily show that GGT is a zero matrix using Python:

    >>> (G@G.T)%3
    [[0 0 0 0 0]
     [0 0 0 0 0]
     [0 0 0 0 0]
     [0 0 0 0 0]
     [0 0 0 0 0]]

And so

vGGTvT = vOvT = 0.

If a vector is not self-orthogonal, there has been an error. For example, if one of the 0s in our output turned into a 1 or a 2, the inner product of the corrupted vector with itself would be equal to 1 (mod 3) rather than 0.

Self-orthogonality is necessary but not sufficient for an encoded vector to be error-free. Correctly encoded vectors form a 5 dimensional subspace of GF(3)11, namely the row space of G. The set of self-orthogonal vectors in GF(3)11 is a larger space.

A sufficient condition for a row vector w to be the encoding of a data vector v is for


to be the zero vector (carrying out all calculations mod 3).

The ternary Golay code is capable of detecting and correcting corruption in up to two spots in a vector. For example, suppose we take our vector

[1 0 1 2 2 0 1 0 2 0 0]

above and corrupt it by turning the first couple 2’s to 1’s.

[1 0 1 1 1 0 1 0 2 0 0]

We’d still get a self-orthogonal vector, but the product GwT would not be zero.

    >>> v = np.array( [1, 0, 1, 2, 2] )
    >>> w = (v@G)%3
    >>> w[3] = w[4] = 1
    >>> print((G@w)%3)

This prints

    [0 0 0 2 2]

which lets us know that the modified version of w is not a valid encoding, not a part of a row space of G.

If we know (or are willing to assume) that

[1 0 1 1 1 0 1 0 2 0 0]

is the result of no more than two changes applied a valid code word vG, then we can recover vG by looking for the closest vector in the row space of G. Here closeness is measured in Hamming distance: the number of positions in which vectors differ. Every vector in GF(3)11 is within a distance of 2 from a unique code word, and so the code can correct any two errors.

Related posts

Ternary Golay code in Python

Marcel Golay discovered two “perfect” error-correcting codes: one binary and one ternary. These two codes stick out in the classification of perfect codes [1].

The ternary code is a linear code over GF(3), the field with three elements. You can encode a list of 5 base-three digits by multiplying the list as a row vector by the following generator matrix on the right:

\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 1 & 2 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 & 0 & 1 & 2 & 1 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 & 0 & 1 & 2 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 2 & 2 & 1 & 1 \\ \end{pmatrix}

Here the arithmetic is carried out in GF(3): every multiplication and addition is carried out mod 3.

Suppose we want to write this up in Python. How can you tell NumPy that you want to do arithmetic mod 3? I don’t believe you can directly. However, you could just multiply your row vector by the matrix using ordinary arithmetic and reducing the result mod 3 at the end. This gives the same result as if all intermediate operations had been carried out mod 3.

For example, suppose we wanted to encode the list [1, 0, 1, 2, 2].

    import numpy as np

    G = np.array([
      [1, 0, 0, 0, 0, 1, 1, 1, 2, 2, 0], 
      [0, 1, 0, 0, 0, 1, 1, 2, 1, 0, 2], 
      [0, 0, 1, 0, 0, 1, 2, 1, 0, 1, 2], 
      [0, 0, 0, 1, 0, 1, 2, 0, 1, 2, 1], 
      [0, 0, 0, 0, 1, 1, 0, 2, 2, 1, 1], 

    v = np.array( [1, 0, 1, 2, 2] )
    print ((v@G)%3)

The vector-matrix product v@G contains integers that are larger than 3:

    [1 0 1 2 2 6 7 6 8 9 6]

But when we reduce everything mod 3 we get

    [1 0 1 2 2 0 1 0 2 0 0]

This doesn’t mean that linear algebra over a finite field is trivial, though it was trivial in this case.

The same trick would work if we were to work modulo any prime. So the analogous trick would work mod 7, for example.

However, the trick would not work for the field with 8 elements, for example, because arithmetic in this field is not simply arithmetic mod 8. (You can read why here.) So our trick works for GF(p) for any prime p, but not in GF(pk) for any k > 1.

Related posts

[1] From “Sphere Packings, Lattices and Groups” by J. H. Conway and N. J. A. Sloane:

Perfect codes were essentially classified by Tietäväinen and van Lint. The list is as follows.

(i) Certain trivial codes …
(ii) Hamming codes …
(iii) Nonlinear codes with the same parameters as Hamming codes …
(iv) The binary [23, 12, 7] binary Golay code C23 and the ternary [11, 6, 5] Golay code C11.