A Gentle Introduction to Erasure Codes Fred Akalin November 30, 2017

1. Overview This article explains Reed-Solomon erasure codes and the problems they solve in gory detail, with the aim of providing enough background to understand how the PAR1 and PAR2 file formats work, the details of which will be covered in future articles. I’m assuming that the reader is familiar with programming, but has not had much exposure to coding theory or linear algebra. Thus, I’ll review the basics and treat the results we need as a “black box”, stating them and moving on. However, I’ll give self-contained proofs of those results in a companion article. So let’s start with the problem we’re trying to solve! Let’s say you have \(n\) files of roughly the same size, and you want to guard against \(m\) of them being lost or corrupted. To do so, you generate \(m\) parity files ahead of time, and if in the future you lose up to \(m\) of the data files, you can use an equal number of parity files to recover the lost data files. cashcat0.jpg cashcat1.jpg cashcat2.jpg \(\xmapsto{\mathtt{GenerateParityFiles}}\) cashcats.p00 cashcats.p01 Figure 1 Using parity codes to protect against the loss or corruption of up to two images (out of three) of cashcats. \(\xmapsto{\mathtt{ReconstructDataFiles}}\) Figure 2 With a corrupted and a missing file, recovering the original cashcat images using the parity files from Figure 1. Note that this works even if you lose some of the parity files also; as long as you have \(n\) files, whether they be data or parity files, you’ll be able to recover the original \(n\) data files. Compare making \(n\) parity files with simply making a copy of the \(n\) data files (for \(n > 1\)). In the latter case, if you lose both a data file and its copy, that data file becomes unrecoverable! So parity files take the same amount of space but provide superior recovery capabilities. Now we can reduce the problem above to a byte-level problem as follows. Have ComputeParityFiles pad all the data files so they’re the same size, and then for each byte position i call a function ComputeParityBytes on the i th byte of each data file, and store the results into the i th byte of each parity file. Also take a checksum or hash of each data file and store those (along with the original data file sizes) with the parity files. Then, ReconstructDataFiles can detect corrupted files using the checksums/hashes and treat them as missing, and then for each byte position i it can call a function ReconstructDataBytes on the i th byte of each good data and parity file to recover the i th byte of the corrupted/missing data files. A byte error where we know the position of the dropped/corrupted byte is called an erasure. Then, the pair of functions ComputeParityBytes and ReconstructDataBytes which behave as described above implements what is called an optimal erasure code; it’s an erasure code because it guards only against byte erasures, and not more general errors where we don’t know which data bytes have been corrupted, and it’s optimal because in general you need at least \(n\) known bytes to recover the \(n\) data bytes, and that bound is achieved. ComputeParityBytes<n, m>(data: byte[n]) -> (parity: byte[m]) that computes \(m\) parity bytes given \(n\) data bytes, and a function ReconstructDataBytes<n, m>(partialData: (byte?)[n], partialParity: (byte?)[m]) -> ((data: byte[n]) | Error) that takes in a partial list of data and parity bytes from an earlier call to ComputeParity , and returns the full list of data bytes if there are at least \(n\) known data or parity bytes (i.e., there are no more than \(m\) omitted data or parity bytes), and an error otherwise. In detail, an optimal erasure code is composed of some set of possible \((n, m)\) pairs, and for each possible pair, a functionthat computes \(m\) parity bytes given \(n\) data bytes, and a functionthat takes in a partial list of data and parity bytes from an earlier call to, and returns the full list of data bytes if there are at least \(n\) known data or parity bytes (i.e., there are no more than \(m\) omitted data or parity bytes), and an error otherwise. (In the above pseudocode, I’m using T[n] to mean an array of n objects of type T , and byte? to mean byte | None . Also, I’ll omit the -Bytes<n, m> suffix from now on.) By the end of this article, we’ll find out exactly how the following example works: Example 1: ComputeParity and ReconstructData ComputeParity Let d = [ da, db, 0d ] be the input data bytes and let m = 2 be the desired parity byte count. Then the output parity bytes are p = [ 52 , 0c ].

Let d partial = [ ??, db, ?? ] be the input partial data bytes and p partial = [ 52, 0c ] be the input partial parity bytes. Then the output data bytes are d = [ da , db , 0d ].

2. Erasure codes for \(m = 1\) ComputeParitySum(data: byte[n]) { return [data[0] + … + data[n-1]] } where we consider byte to be an unsigned type such that addition and subtraction wrap around, i.e. byte arithmetic is done modulo \(256\). Then also define ReconstructDataSum(partialData: (byte?)[n], partialParity: (byte?)[1]) { if there is more than one entry of partialData or partialParity set to None { return Error } else if partialData has no entry set to None { return partialData } i := partialData.firstIndexOf(None); partialSum = partialData[0] + … + partialData[i-1] + partialData[i+1] + … + partialData[n-1] return partialData[0:i] ++ [partialParity[0] - partialSum] ++ partialData[i+1:n] } where a[i:j] means the subarray of a starting at i and ending (without inclusion) at j , and ++ is array concatenation. The simplest erasure codes are when \(m = 1\). For example, definewhere we considerto be an unsigned type such that addition and subtraction wrap around, i.e. byte arithmetic is done modulo \(256\). Then also definewheremeans the subarray ofstarting atand ending (without inclusion) at, andis array concatenation. This simple erasure code uses the fact that if you have the sum of a list of numbers, then you can recover a missing number by subtracting the sum of the other numbers from the total sum, and also that this works even if you do the arithmetic modulo \(256\). ^ , or \(\oplus\)) instead of arithmetic modulo \(256\). Define ComputeParityXor(data: byte[n]) { return [data[0] ⊕ … ⊕ data[n-1]] } and ReconstructDataXor(partialData: (byte?)[n], partialParity: (byte?)[1]) { if there is more than one entry of partialData or partialParity set to None { return Error } else if partialData has no entry set to None { return partialData } i := partialData.firstIndexOf(None); partialXor = partialData[0] ⊕ … ⊕ partialData[i-1] ⊕ partialData[i⊕1] ⊕ … ⊕ partialData[n-1] return partialData[0:i] ++ [partialParity[0] ⊕ partialXor] ++ partialData[i+1:n] } Another erasure code for \(m = 1\) uses bitwise exclusive or (denoted as xor,, or \(\oplus\)) instead of arithmetic modulo \(256\). Defineand This relies on the fact that \(a \oplus a = 0\), so given the xor of a list of bytes, you can recover a missing byte by xoring with all the known bytes.

3. Erasure codes for \(m = 2\) (almost) Now coming up with an erasure code for \(m = 2\) is more involved, but we can get an inkling of how it could work by letting \(n = 3\) for simplicity, and also letting the output of ComputeParity be non-negative integers, instead of just bytes (i.e., less than \(256\)). In that case, we can consider parity numbers that are weighted sums of the data bytes. For example, like in the \(m = 1\) case, we can have the first parity number be \[ p_0 = d_0 + d_1 + d_2\text{,} \] (using \(d_i\) for data bytes and \(p_i\) for parity numbers) but for the second parity number, we can pick different weights, say \[ p_1 = 1 \cdot d_0 + 2 \cdot d_1 + 3 \cdot d_2\text{.} \] We want to make sure that the weights for the second parity number are “sufficiently different” from that of the first parity number, which we’ll clarify later, but for example note that setting \[ p_1 = 2 \cdot d_0 + 2 \cdot d_1 + 2 \cdot d_2 \] can’t add any new information, since then \(p_1\) will always be equal to \(2 \cdot p_0\). ComputeParity function looks like ComputeParityWeighted(data: byte[3]) { return [ int(data[0]) + int(data[1]) + int(data[2]), int(data[0]) + 2 * int(data[1]) + 3 * int(data[2]), ] } So then ourfunction looks like ReconstructData , if we have two missing data bytes, say \(d_i\) and \(d_j\) for \(i < j\), and \(p_0\) and \(p_1\), we can rearrange the equations \[ \begin{aligned} p_0 &= d_0 + d_1 + d_2 \\ p_1 &= 1 \cdot d_0 + 2 \cdot d_1 + 3 \cdot d_2 \end{aligned} \] to get all the unknowns on the left side, letting \(d_k\) be the known data byte: \[ \begin{aligned} d_i + d_j &= X = p_0 - d_k \\ (i+1) \cdot d_i + (j+1) \cdot d_j &= Y = p_1 - (k + 1) \cdot d_k\text{.} \end{aligned} \] We can then multiply the first equation by \(i + 1\) and subtract it from the second to cancel the \(d_i\) term and get \[ d_j = (Y - (i + 1) \cdot X) / (j - i)\text{,} \] and then we can use the first equation to solve for \(d_i\): \[ d_i = X - d_j = ((j + 1) \cdot X - Y) / (j - i)\text{.} \] Thus with these equations, we can implement ReconstructData : ReconstructDataWeighted(partialData: (byte?)[3], partialParity: (int?)[2]) { Handle all cases except when there are exactly two entries set to none in partialData. [i, j] := indices of the unknown data bytes k := index of the known data byte X := partialParity[0] - partialData[k] Y := partialParity[1] - (k + 1) * partialData[k]; d_i := ((j + 1) * X - Y) / (j - i) d_j := (Y - (i + 1) * X) / (j - i) return an array with d_i, d_j, and d[k] in the right order } (Generalizing this to larger values of \(n\) is straightforward; \(p_0\) will still have a weight of \(1\) for each data byte, and \(p_1\) will have a weight of \(i + 1\) for \(d_i\). \(X\) and \(Y\) will then have terms for all known bytes, and everything else proceeds the same after that.) As for, if we have two missing data bytes, say \(d_i\) and \(d_j\) for \(i < j\), and \(p_0\) and \(p_1\), we can rearrange the equations \[ \begin{aligned} p_0 &= d_0 + d_1 + d_2 \\ p_1 &= 1 \cdot d_0 + 2 \cdot d_1 + 3 \cdot d_2 \end{aligned} \] to get all the unknowns on the left side, letting \(d_k\) be the known data byte: \[ \begin{aligned} d_i + d_j &= X = p_0 - d_k \\ (i+1) \cdot d_i + (j+1) \cdot d_j &= Y = p_1 - (k + 1) \cdot d_k\text{.} \end{aligned} \] We can then multiply the first equation by \(i + 1\) and subtract it from the second to cancel the \(d_i\) term and get \[ d_j = (Y - (i + 1) \cdot X) / (j - i)\text{,} \] and then we can use the first equation to solve for \(d_i\): \[ d_i = X - d_j = ((j + 1) \cdot X - Y) / (j - i)\text{.} \] Thus with these equations, we can implement(Generalizing this to larger values of \(n\) is straightforward; \(p_0\) will still have a weight of \(1\) for each data byte, and \(p_1\) will have a weight of \(i + 1\) for \(d_i\). \(X\) and \(Y\) will then have terms for all known bytes, and everything else proceeds the same after that.) Now what goes wrong if we just try to do everything modulo \(256\)? The most obvious difference from the \(m = 1\) case is that solving for \(d_i\) or \(d_j\) involves division, which works fine for non-negative integers as long as there’s no remainder, but it is not immediately clear how division can make sense modulo \(256\). One possible way to define division modulo \(256\) would be to first define the multiplicative inverse modulo \(256\) of an integer \(0 \le x \lt 256\) as the integer \(0 \le y \lt 256\) such that \((x \cdot y) \bmod 256 = 1\), if it exists, and then define division by \(x\) modulo \(256\) to be multiplication by \(y\) modulo \(256\). But this immediately runs into problems; \(2\) has no multiplicative inverse modulo \(256\), and the same holds for any even number, so reconstruction will fail if, for example, we have the first and third data bytes missing, since then we’d be trying to divide by \(j - i = 2\). But for now, let’s leave aside the problem of generating parity bytes instead of parity numbers, and instead focus on how we can generalize the above for larger values of \(m\). To do so, we need to first review some linear algebra.

4. Just enough linear algebra to get by[1] In our \(n = 3, m = 2\) example in the previous section, the equations for the parity numbers have the form \[ p = a_0 \cdot d_0 + a_1 \cdot d_1 + a_2 \cdot d_2 \] for constants \(a_0\), \(a_1\), and \(a_2\). We call such a weighted sum of the \(d_i\)s a linear combination of the \(d_i\)s, and we write this in a tabular form \[ p = \begin{pmatrix} a_0 & a_1 & a_2 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{,} \] where we define the multiplication of a row vector and a column vector by the equation above, generalized in the straightforward manner for any \(n\). Then since we have two parity numbers \(p_0\) and \(p_1\), each a linear combination of the \(d_i\)s, i.e. \[ \begin{aligned} p_0 &= a_{00} \cdot d_0 + a_{01} \cdot d_1 + a_{02} \cdot d_2 \\ p_1 &= a_{10} \cdot d_0 + a_{11} \cdot d_1 + a_{12} \cdot d_2\text{,} \end{aligned} \] we can write this in a single tabular form as \[ \begin{bmatrix} p_0 \\ p_1 \end{bmatrix} = \begin{pmatrix} a_{00} & a_{01} & a_{02} \\ a_{10} & a_{11} & a_{12} \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{,} \] where we define the multiplication of a matrix and a column vector by the equations above. Now if we restrict parity numbers to be linear combinations of the data bytes, then we can identify a function ComputeParity using some set of weights with the matrix formed from that set of weights as above. This holds in general: if a function is defined as a list of linear combinations of its inputs, then it can be represented using a matrix as above, and we call it a linear function. Then we have a correspondence between linear functions that take \(n\) numbers to \(m\) numbers and matrices with \(m\) rows and \(n\) columns, which are denoted as \(m \times n\) matrices. As the first example of this correspondence, note that we denote the elements of the matrix above as \(a_{ij}\), where the first index is the row index and the second index is the column index. Looking back to the parity equations, we also see that the first index corresponds to the output arguments of ComputeParity , and the second index corresponds to the input arguments.[2] The usefulness of the correspondence between linear functions and matrices is that we can store and manipulate a linear function by storing and manipulating its corresponding matrix of weights, which you wouldn’t be able to easily do for functions in general. For example, as we’ll see below, we’ll be able to compute the inverse of a linear function by matrix operations, which will be useful for ReconstructData . Deleting a row of a matrix corresponds to deleting an output of a linear function.

Swapping two rows of a matrix corresponds to swapping two outputs of a linear function.

Appending a row to a matrix corresponds to adding an output to a linear function. In general, matrix row operations correspond to manipulations of a linear function’s outputs. First, let’s examine some simple matrix operations and their effects on the corresponding linear function:In general, matrix row operations correspond to manipulations of a linear function’s outputs. matrixMultiply(f: Matrix, g: Matrix) { if (f.columns != g.rows) { return Error } h := new Matrix(f.rows, g.columns) for i := 0 to f.rows - 1 { for j := 0 to g.columns - 1 { t := 0 for k := 0 to f.columns - 1 { t += f[i, k] * g[k, j] } h[i, j] = t } } return h } You can convince yourself that the above formula and code is correct by trying to compose some small linear functions by hand. An important operation on functions is composition: if \(f\) takes \(k\) inputs to \(m\) outputs, and \(g\) takes \(m\) inputs to \(n\) outputs, then we can define \((g \circ f)(x_0, \dotsc, x_k) = g(f(x_0, \dotsc, x_k))\) which takes \(k\) inputs to \(n\) outputs. It turns out that the composition of two linear functions is again a linear function, and so there must be an operation which takes the corresponding \(m \times k\) matrix \(F\) and the \(n \times m\) matrix \(G\) and yields a \(n \times k\) matrix. This important operation, the bane of high-schoolers everywhere, is called matrix multiplication , denoted by \(F \cdot G\). If \(H = F \cdot G\), then the explicit formula for its elements is \[ h_{ij} = \sum_{k=0}^{m-1} f_{ik} \cdot g_{kj}\text{,} \] which corresponds to the following code:You can convince yourself that the above formula and code is correct by trying to compose some small linear functions by hand. A useful property of matrix multiplication is that it’s a generalization of the product of a row vector and a column vector, and the product of a matrix and a column vector as we defined above. I would be remiss if I didn’t talk about the consequences of defining matrix multiplication as the matrix of the composition of the corresponding linear functions. First, this immediately implies that you can only multiply matrices if the left matrix has the same number of rows as the number of columns of the right matrix, which corresponds to the fact that you can only compose functions if the left function takes the same number of inputs as the number of outputs of the right function. Furthermore, even if you have two \(n \times n\) matrices \(F\) and \(G\), unlike numbers, it is not true that \(F \cdot G = G \cdot F\), which corresponds to the fact that in general, for two functions that take \(n\) inputs to \(n\) outputs, it is not true that \(f \circ g = g \circ f\). If you learned matrix multiplication just from the formula above, then these facts are much less obvious! Finally, an important function is the identity function \(\mathrm{Id}_n\), which return its \(n\) inputs as its outputs. It corresponds to the identity matrix \[ I_n = \begin{pmatrix} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & 1 \end{pmatrix}\text{.} \] For a linear function \(f\) that takes \(n\) inputs to \(n\) outputs, if there is a function \(g\) such that \(f \circ g = \mathrm{Id}_n\), then we call \(g\) the inverse of \(f\), and denote it as \(f^{-1}\). (It is also true that \(f^{-1} \circ f = \mathrm{Id}_n\), i.e. \((f^{-1})^{-1} = f\).) Not all linear functions taking \(n\) inputs to \(n\) outputs have inverses, but if the inverse exists, it is also linear (and unique, which is why we call it the inverse). Therefore, we can define the inverse of an \(n \times n\) (or square) matrix \(M\) as the unique matrix \(M^{-1}\) such that \(M \cdot M^{-1} = M^{-1} \cdot M = I_n\), if it exists; also, if \(M\) has an inverse, we say that \(M\) is invertible. Example 2: The matrix/linear function correspondence f(x: rational[2]) { return [ 1 * x[0] + 2 * x[1], 3 * x[0] + 4 * x[1], ] } where rational is an arbitrary-precision rational number type. Let \[ M = \begin{pmatrix} 1 & 2 \\ 3 & 4\end{pmatrix}\text{.} \] This corresponds to the linear functionwhereis an arbitrary-precision rational number type. g(y: rational[2]) { return [ -2 * x[0] + 1 * x[1], (3/2) * x[0] + (-1/2) * x[1], ] } so g is the inverse function of f . Indeed, f([5, 6]) is [17, 39] and g([17, 39]) is [5, 6] . \(M\) is invertible with inverse \[ M^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2\end{pmatrix}\text{.} \] This corresponds to the linear functionsois the inverse function of. Indeed,isandis So now we’ve reduced the problem of finding the inverse of a linear function taking \(n\) inputs to \(n\) outputs to finding the inverse of an \(n \times n\) matrix. Before we tackle the question of computing those inverses, let’s first recast our problem in the language of linear algebra and see why we need to find the inverse of a linear function.

5. Erasure codes in general ComputeParityWeighted(data: byte[3]) { return [ int(data[0]) + int(data[1]) + int(data[2]), int(data[0]) + 2 * int(data[1]) + 3 * int(data[2]), ] } which therefore corresponds to the parity matrix \[ P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix}\text{.} \] So in mathematical notation, ComputeParityWeighted looks like: \[ \begin{bmatrix} p_0 \\ p_1 \end{bmatrix} = \mathtt{ComputeParityWeighted}(d_0, d_1, d_2) = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{.} \] So, revisiting our \(n = 3, m = 2\) erasure code from above, we have the linear functionwhich therefore corresponds to the parity matrix \[ P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix}\text{.} \] So in mathematical notation,looks like: \[ \begin{bmatrix} p_0 \\ p_1 \end{bmatrix} = \mathtt{ComputeParityWeighted}(d_0, d_1, d_2) = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{.} \] So let’s now reimplement ReconstructDataWeighted using linear algebra. First, append the rows of \(P\) to the identity matrix \(I_3\) to get the matrix equation \[ \begin{bmatrix} d_0 \\ d_1 \\ d_2 \\ p_0 \\ p_1 \end{bmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{,} \] which corresponds to a linear function that returns the input data bytes in addition to computing the parity numbers. Now let’s say we lose the data bytes \(d_0\) and \(d_2\). Then let’s remove the rows corresponding to those bytes: \[ \begin{bmatrix} \xcancel{d_0} \\ d_1 \\ \xcancel{d_2} \\ p_0 \\ p_1 \end{bmatrix} = \begin{pmatrix} \xcancel{1} & \xcancel{0} & \xcancel{0} \\ 0 & 1 & 0 \\ \xcancel{0} & \xcancel{0} & \xcancel{1} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{,} \] which turns into \[ \begin{bmatrix} d_1 \\ p_0 \\ p_1 \end{bmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix}\text{,} \] which corresponds to a linear function that maps the input data bytes to the non-lost data bytes and the parity bytes. This is the inverse of the function we want, so we want to invert the \(3 \times 3\) matrix above, which we’ll call \(M\). That inverse is \[ M^{-1} = \begin{pmatrix} -1/2 & 3/2 & -1/2 \\ 1 & 0 & 0 \\ -1/2 & -1/2 & 1/2 \end{pmatrix}\text{.} \] Multiplying both sides above by \(M^{-1}\), we get \[ \begin{bmatrix} d_0 \\ d_1 \\ d_2 \end{bmatrix} = \begin{pmatrix} -1/2 & 3/2 & -1/2 \\ 1 & 0 & 0 \\ -1/2 & -1/2 & 1/2 \end{pmatrix} \cdot \begin{bmatrix} d_1 \\ p_0 \\ p_1 \end{bmatrix}\text{,} \] which is exactly what we want: the original data bytes in terms of the known data bytes and the parity numbers![3] Comparing this equation to the one we manually computed previously, they don’t look immediately similar, but some rearrangement will reveal that they indeed compute the same thing. As a sanity check, notice that the second row of \(M^{-1}\) means that the first input argument is mapped unchanged to the second output argument, which is exactly what we want for the known byte \(d_1\). Now what does this look like in general, i.e. for arbitrary \(n\) and \(m\)? First, we have to generate an \(m \times n\) parity matrix \(P\) whose rows have to be “sufficiently different” from each other, which we still have to clarify. Then ComputeParity just multiplies \(P\) by \([d]\), the column matrix of input bytes, like so: \[ \begin{bmatrix} p_0 \\ \vdots \\ p_{m-1} \end{bmatrix} = \mathtt{ComputeParity}(d_0, \dotsc, d_{n-1}) = \begin{pmatrix} p_0 \\ \vdots \\ p_{m-1} \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ \vdots \\ d_{n-1} \end{bmatrix}\text{,} \] where the \(p_i\) are the rows of \(P\). As for ReconstructData , we first append the rows of \(P\) to \(I_n\), whose rows we’ll denote as \(e_i\): \[ \begin{bmatrix} d_0 \\ \vdots \\ d_{n-1} \\ p_0 \\ \vdots \\ p_{m-1} \end{bmatrix} = \begin{pmatrix} e_0 \\ \vdots \\ e_{n-1} \\ p_0 \\ \vdots \\ p_{m-1} \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ \vdots \\ d_{n-1} \end{bmatrix}\text{.} \] Now assume that the indices of the missing \(k \le m\) data bytes are \(i_0, \dotsc, i_{k-1}\). Then we remove the rows corresponding to the missing data bytes, and keep some \(k\) parity rows, e.g. \(p_0\) to \(p_{k-1}\). This yields the equation \[ \begin{bmatrix} d_{j_0} \\ \vdots \\ d_{j_{n-k-1}} \\ p_0 \\ \vdots \\ p_{k-1} \end{bmatrix} = \begin{pmatrix} e_{j_0} \\ \vdots \\ e_{j_{n-k-1}} \\ p_0 \\ \vdots \\ p_{k-1} \end{pmatrix} \cdot \begin{bmatrix} d_0 \\ \vdots \\ d_{n-1} \end{bmatrix}\text{,} \] where \(j_0, \dotsc, j_{m-k-1}\) are the indices of the present \(n - k\) data bytes. Call that \(n \times n\) matrix \(M\), and compute its inverse \(M^{-1}\). If \(P\) was chosen correctly, \(M^{-1}\) should always exist, so if the inverse computation fails, raise an error. Therefore, ReconstructData just multiplies \(M^{-1}\) by the column matrix of present data bytes and chosen parity numbers: \[ \begin{bmatrix} d_0 \\ \vdots \\ d_{n-1} \end{bmatrix} = \mathtt{ReconstructData}(d_{j_0}, \dotsc, d_{j_{n-k-1}}, p_0, \dotsc, p_{k-1}) = M^{-1} \cdot \begin{bmatrix} d_{j_0} \\ \vdots \\ d_{j_{n-k-1}} \\ p_0 \\ \vdots \\ p_{k-1} \end{bmatrix}\text{.} \] As an optimization, some rows of \(M^{-1}\) correspond to just shuffling around the known data bytes \(d_{j_*}\), so we can just remove those rows, compute the missing data bytes, and do the shuffling ourselves. ComputeParity and ReconstructData , but we still have missing pieces. In particular, How do we compute matrix inverses? How do we generate “optimal” parity matrices so that \(M^{-1}\) always exists? How do we compute parity bytes instead of parity numbers? So we now have outlines of implementations of bothand, but we still have missing pieces. In particular, So first, let’s see how to compute matrix inverses using row reduction.

6. Finding matrix inverses using row reduction We developed the theory of matrices by identifying them with linear functions of numbers. To show how to find matrix inverses, we have to look at them in a slightly different way by identifying matrix equations with systems of linear equations of numbers. For example, consider the matrix equation \[ M \cdot x = y\text{,} \] where \[ M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\text{,} \quad x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \text{,} \quad \text{and } y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\text{.} \] This expands to \[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\text{,} \] or \[ \begin{aligned} y_1 &= 1 \cdot x_1 + 2 \cdot x_2 \\ y_2 &= 3 \cdot x_1 + 4 \cdot x_2\text{,} \end{aligned} \] which is a linear system of equations of numbers. Letting \(M\) be any matrix, and \(x\) and \(y\) be appropriately-sized column matrices of variables, we can see that the matrix equation \(M \cdot x = y\) is shorthand for a system of linear equations of numbers. If we could find \(M^{-1}\), we could solve the matrix equation easily by multiplying both sides by it: \[ \begin{aligned} M^{-1} \cdot (M \cdot x) &= M^{-1} \cdot y \\ x &= M^{-1} \cdot y\text{,} \end{aligned} \] and therefore solve the linear system for \(x\) in terms of \(y\). Conversely, if we were able to solve the linear system for \(x\), we’d then be able to read off \(M^{-1}\) from the new linear system. swapping two equations,

multiplying an equation by a number,

adding one equation to another, possibly multiplying the equation by a number before adding. But how do we solve a linear system? From the theory of linear systems of equations, we have a few tools at our disposal: All of these are valid transformations because they don’t change the solution set of the linear system. For example, in the equation above, the first step would be to subtract \(3\) times the first equation from the second equation to yield \[ \begin{aligned} y_1 &= x_1 + 2 \cdot x_2 \\ y_2 - 3 \cdot y_1 &= -2 \cdot x_2\text{.} \end{aligned} \] Then, add the second equation back to the first equation: \[ \begin{aligned} y_2 - 2 \cdot y_1 &= x_1 \\ y_2 - 3 \cdot y_1 &= -2 \cdot x_2\text{.} \end{aligned} \] Finally, divide the second equation by \(-2\): \[ \begin{aligned} y_2 - 2 \cdot y_1 &= x_1 \\ (3/2) \cdot y_1 - (1/2) \cdot y_2 &= x_2\text{.} \end{aligned} \] This is equivalent to the matrix equation \[ \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} \cdot \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\text{,} \] so \[ M^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}\text{.} \] So how do we translate the above process to an algorithm operating on matrices? First, express our matrix equation in a slightly different form: \[ M \cdot x = I \cdot y\text{.} \] Using the example above, this is \[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\text{.} \] Then, you can see that the first step above corresponds to subtracting \(-3\) times the first row from the second row to yield: \[ \begin{pmatrix} 1 & 2 \\ 0 & -2 \end{pmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix} \cdot \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\text{.} \] We don’t even need to keep writing the \(x\) and \(y\) column matrices; we can just write the “augmented” matrix. \[ A = \left( \hskip -5pt \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array} \hskip -5pt \right) \] and operate on it. swapping two equations corresponds to swapping two rows;

multiplying an equation by a number corresponds to multiplying a row by a number; and

adding an equation to another, possibly multiplying the equation by a number before adding, corresponds to adding a row to another row, possibly multiplying the row by a number before adding. Then, the goal is to use these row operations to transform the initial augmented matrix, where the right side looks like the identity matrix, into one where the left side looks like the identity matrix. Then, translating the augmented matrix back into a matrix equation, that would give \(M^{-1}\) on the right side.[4] Thus, the operations listed above on linear systems have corresponding operations on augmented matrices:Then, the goal is to use these row operations to transform the initial augmented matrix, where the right side looks like the identity matrix, into one where the left side looks like the identity matrix. Then, translating the augmented matrix back into a matrix equation, that would give \(M^{-1}\) on the right side. Turn the left side of \(A\) into a unit upper triangular matrix, which means that all the elements on the main diagonal are \(1\), and all elements below the main diagonal are \(0\), i.e. that \(a_{ii} = 1\) for all \(i\), and \(a_{ij} = 0\) for all \(j > i\). Then turn the left side of \(A\) into the identity matrix. This algorithm is called For each column \(i\) of the left side in ascending order: If \(a_{ii}\) is zero, look at the rows below the \(i\)th row for a row \(j > i\) such that \(a_{ji}

e 0\), and swap rows \(i\) and \(j\). If no such row exists, return an error, as that means that \(A\) is non-invertible. Divide the \(i\)th row by \(a_{ii}\), so that \(a_{ii} = 1\). For each row \(j > i\), subtract \(a_{ji}\) times the \(i\)th row from it, which will set \(a_{ji}\) to \(0\). The second step can be similarly broken down: For each column \(i\) of the left side, in order: For each row \(j < i\), subtract \(a_{ji}\) times the \(i\)th row from it, which will set \(a_{ji}\) to \(0\). When doing this by hand, one usually works with the linear system itself, trying to see which variables can be easily eliminated so as to minimize arithmetic. However, to translate this to an algorithm, we’re more interested in a systematic way of doing this. Fortunately, there’s an easy two-step process:This algorithm is called row reduction . The first step can be further broken down:The second step can be similarly broken down: Note that we’re assuming that all arithmetic is exact, i.e. we use a arbitrary-precision rational number type. If we use floating point numbers, we’d have to worry a lot more about the order in which we do operations and numerical stability. Example 3: Matrix inversion via row reduction Let / 0 2 2 \ M = | 3 4 5 | \ 6 6 7 /. The initial augmented matrix A is / 0 2 2 | 1 0 0 \ | 3 4 5 | 0 1 0 | \ 6 6 7 | 0 0 1 /. We need A 00 to be non-zero, so swap rows 0 and 1 : / 0 2 2 | 1 0 0 \ / 3 4 5 | 0 1 0 \ | 3 4 5 | 0 1 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /. We need A 00 to be 1, so divide row 0 by 3: / 3 4 5 | 0 1 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /. We need A 20 to be 0, so subtract row 0 scaled by 6 from row 2 : / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 0 -2 -3 | 0 -2 1 /. We need A 11 to be 1, so divide row 1 by 2: / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 2 2 | 1 0 0 | --> | 0 1 1 | 1/2 0 0 | \ 0 -2 -3 | 0 -2 1 / \ 0 -2 -3 | 0 -2 1 /. We need A 21 to be 0, so subtract row 1 scaled by −2 from row 2 : / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 1 1 | 1/2 0 0 | --> | 0 1 1 | 1/2 0 0 | \ 0 -2 -3 | 0 -2 1 / \ 0 0 -1 | 1 -2 1 /. We need A 22 to be 1, so divide row 2 by −1, which makes the left side of A a unit upper triangular matrix: / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 1 1 | 1/2 0 0 | --> | 0 1 1 | 1/2 0 0 | \ 0 0 -1 | 1 -2 1 / \ 0 0 1 | -1 2 -1 /. We need A 12 to be 0, so subtract row 2 from row 1 : / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \ | 0 1 1 | 1/2 0 0 | --> | 0 1 0 | 3/2 -2 1 | \ 0 0 1 | -1 2 -1 / \ 0 0 1 | -1 2 -1 /. We need A 02 to be 0, so subtract row 2 scaled by 5/3 from row 0 : / 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 0 | 5/3 -3 5/3 \ | 0 1 0 | 3/2 -2 1 | --> | 0 1 0 | 3/2 -2 1 | \ 0 0 1 | -1 2 -1 / \ 0 0 1 | -1 2 -1 /. We need A 01 to be 0, so subtract row 1 scaled by 4/3 from row 0 , which makes the left side of A the identity matrix: / 1 4/3 0 | 5/3 -3 5/3 \ / 1 0 0 | -1/3 -1/3 1/3 \ | 0 1 0 | 3/2 -2 1 | --> | 0 1 0 | 3/2 -2 1 | \ 0 0 1 | -1 2 -1 / \ 0 0 1 | -1 2 -1 /. Since the left side of A is the identity matrix, the right side of A is M -1. Therefore, / -1/3 -1/3 1/3 \ M^{-1} = | 3/2 -2 1 | \ -1 2 -1 /. LetThe initial augmented matrixisWe needto be non-zero, so swap rowsandWe needto be 1, so divide rowby 3:We needto be 0, so subtract rowscaled by 6 from rowWe needto be 1, so divide rowby 2:We needto be 0, so subtract rowscaled by −2 from rowWe needto be 1, so divide rowby −1, which makes the left side ofa unit upper triangular matrix:We needto be 0, so subtract rowfrom rowWe needto be 0, so subtract rowscaled by 5/3 from rowWe needto be 0, so subtract rowscaled by 4/3 from row, which makes the left side ofthe identity matrix:Since the left side ofis the identity matrix, the right side ofis. Therefore, Now notice one thing: if \(M\) has a row that is proportional to another row, then row reduction would eventually zero out one of the rows, causing the algorithm to fail, and signaling that \(M\) is non-invertible. In fact, a stronger statement is true: \(M\) has a row that can be expressed as a linear combination of other rows of \(M\) exactly when \(M\) is non-invertible. Informally, this means that the linear function corresponding to \(M\) has one of its outputs redundant with the other outputs, so it is essentially a a linear function taking \(n\) inputs to fewer than \(n\) outputs, and such functions aren’t invertible. This gets us a partial explanation for what “sufficiently different” means for our parity functions. If one parity function is a linear combination of other parity functions, then it is redundant, and therefore not “sufficiently different”. Therefore, we want our parity matrix \(P\) to be such that no row can be expressed as a linear combination of other rows. However, this criterion for \(P\) isn’t quite enough to guarantee that all possible matrices \(M\) computed as part of ReconstructData are invertible. For example, this criterion holds for the identity matrix \(I_n\), but if \(n > 1\) and you pick \(I_n\) as the parity matrix for \(n = m\), you can certainly end up with a constructed matrix \(M\) with repeated rows, since you’re starting by appending another copy of \(I_n\) on top of \(P = I_n\)! This explains in a different way why simply making a copy of the original data files makes for a poor erasure code, unless of course you only have one data file. We’re led to our next topic: what makes a parity matrix “optimal”?

7. Optimal parity matrices Recall from above that we form the square matrix \[ M = \begin{pmatrix} e_{j_0} \\ \vdots \\ e_{j_{n-k-1}} \\ p_0 \\ \vdots \\ p_{k-1} \end{pmatrix} \] by prepending some rows of the identity matrix to the first \(k\) rows of the parity matrix. We can generalize this a bit more, since we don’t have to take the first \(k\) rows, but instead can take any \(k\) rows of the parity matrix, whose indices we denote here as \(l_0, \dotsc, l_{k-1}\): \[ M = \begin{pmatrix} e_{j_0} \\ \vdots \\ e_{j_{n-k-1}} \\ p_{l_0} \\ \vdots \\ p_{l_{k-1}} \end{pmatrix}\text{.} \] So we want to construct \(P\) so that any such square matrix \(M\) formed from the rows of \(P\) is invertible. Therefore, we call a parity matrix \(P\) optimal if it satisfies this criterion. Theorem 1 .) A parity matrix \(P\) is optimal exactly when any non-empty square submatrix of \(P\) is invertible.[5] .) A parity matrix \(P\) is optimal exactly when any non-empty square submatrix of \(P\) is invertible. [6] However, it is possible to have a matrix \(P\) where no row can be expressed as a linear combination of other rows, but which is not optimal. We’ve already seen an example above: \(I_n\) for \(n \gt 1\), and indeed, \[ I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\text{,} \] has the \(1 \times 1\) non-invertible submatrix \(\begin{pmatrix} 0 \end{pmatrix}\). Fortunately, there is a simpler criterion for optimal parity matrices. First, define a submatrix of a matrix \(P\) to be a matrix that you get by deleting any number of rows or columns, and call a matrix non-empty if it has at least one row and one column. Then:Note that this criterion is stronger than the one in the previous section, where we want a parity matrix \(P\) to have no row that can be expressed as a linear combination of other rows. That is, if any non-empty square submatrix of \(P\) is invertible, that means that no row can be expressed as a linear combination of other rows.However, it is possible to have a matrix \(P\) where no row can be expressed as a linear combination of other rows, but which is not optimal. We’ve already seen an example above: \(I_n\) for \(n \gt 1\), and indeed, \[ I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\text{,} \] has the \(1 \times 1\) non-invertible submatrix \(\begin{pmatrix} 0 \end{pmatrix}\). Example 4: A optimal parity matrix for \(m = 2\) Recall the parity matrix \[ P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{pmatrix} \] that we were using for our \(n = 3, m = 2\) example. For any \(n\), this matrix looks like \[ P = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 2 & \cdots & n-1 \end{pmatrix}\text{.} \] A \(1 \times 1\) matrix is invertible exactly when its single element is non-zero, so any \(1 \times 1\) submatrix of \(P\) is invertible. Any \(2 \times 2\) submatrix of \(P\) looks like \[ A = \begin{pmatrix} 1 & 1 \\ a & b \end{pmatrix} \] for \(a

e b\), which, using the formula for inverses of \(2 \times 2\) matrices, has inverse \[ A^{-1} = \begin{pmatrix} b/(b-a) & -1/(b-a) \\ -a/(b-a) & 1/(b-a) \end{pmatrix}\text{.} \] These are all the possible square submatrices of \(P\), so therefore this \(P\) is a optimal parity matrix for \(m = 2\). Then, finally, we now have a complete definition of what makes a list of parity functions “sufficiently different”; it is exactly when the corresponding parity matrix is optimal as we’ve defined it above. Now this leads us to the question: how do we find such optimal matrices? Fortunately, there’s a whole class of matrices that are optimal: the Cauchy matrices. Let \(a_0, \dotsc, a_{m+n-1}\) be a sequence of distinct integers, meaning that no two \(a_i\) are equal, and let \(x_0, \dotsc, x_{m-1}\) be the first \(m\) integers of \(a_i\) with \(y_0, \dotsc, y_{n-1}\) the remaining integers. Then form the \(m \times n\) matrix \(A\) by setting its elements according to: \[ a_{ij} = \frac{1}{x_i - y_j}\text{,} \] which is always defined since the denominator is never zero, by the distinctness of the \(a_i\). Then \(A\) is a Cauchy matrix. ( Theorem 2 .) Any non-empty square Cauchy matrix is invertible. Combining this with the simple fact that any submatrix of a Cauchy matrix is also a Cauchy matrix, we get: ( Corollary 1 .) Any non-empty square submatrix of a Cauchy matrix is invertible, and thus any Cauchy parity matrix is optimal. What makes Cauchy matrices useful is the following theorem:Combining this with the simple fact that any submatrix of a Cauchy matrix is also a Cauchy matrix, we get: Example 5: Cauchy matrices Let x = [ 1, 2, 3 ] and y = [ -1, 4, 0 ]. Then, the Cauchy matrix constructed from x and y is / 1/2 -1/3 1 \ | 1/3 -1/2 1/2 | \ 1/4 -1 1/3 /, which has inverse / -36/5 96/5 -36/5 \ | -3/10 9/5 -9/5 | \ 9/2 -9 3 /. LetandThen, the Cauchy matrix constructed fromandiswhich has inverse Therefore, to generate a optimal parity matrix for any \((n, m)\), all we need to do is to generate an \(m \times n\) Cauchy matrix. We can pick any sequence of distinct \(m + n\) integers, so for simplicity let’s just use \[ x_i = n + i \quad \text{and} \quad y_i = i\text{.} \] Example 6: Cauchy parity matrices for \(m = 2\) For \(n = 3, m = 2\), we have the sequences \[ x_0 = 3, x_1 = 4 \quad \text{and} \quad y_0 = 0, y_1 = 1, y_2 = 2\text{,} \] so the corresponding Cauchy parity matrix is \[ P = \begin{pmatrix} 1/3 & 1/2 & 1 \\ 1/4 & 1/3 & 1/2 \end{pmatrix}\text{.} \] Similarly, for any \(n\), \[ P = \begin{pmatrix} 1/n & \cdots & 1/2 & 1 \\ 1/{n + 1} & \cdots & 1/3 & 1/2 \end{pmatrix}\text{.} \] All entries of \(P\) are non-zero, so any \(1 \times 1\) submatrix of \(P\) is invertible. Any \(2 \times 2\) submatrix of \(P\) looks like \[ A = \begin{pmatrix} 1/a & 1/b \\ 1/(a+1) & 1/(b+1) \end{pmatrix} \] for \(a

e b\), which, using the formula for inverses of \(2 \times 2\) matrices, has inverse \[ A^{-1} = \begin{pmatrix} \frac{ab(a+1)}{b-a} & -\frac{a(a+1)(b+1)}{b-a} \\ -\frac{ab(b+1)}{b-a} & \frac{b(a+1)(b+1)}{b-a} \end{pmatrix}\text{.} \] These are all the possible square submatrices of \(P\), so therefore this \(P\) is a optimal parity matrix for \(m = 2\). Note that our first parity matrix for \(n = 3, m = 2\) isn’t a Cauchy matrix, since no Cauchy matrix can have repeating elements in a single row. That means that there are other possible optimal parity matrices that aren’t Cauchy matrices.[7] Also, our previous parity matrices had integers, and Cauchy matrices have rational numbers (i.e., fractions). This means that our parity numbers are now fractions. This isn’t a serious difference, though, since we’d have to deal with fractions when computing matrix inverses anyway. You could also change a parity matrix with fractions into one without by simply multiplying the entire matrix by some non-zero number that gets rid of all the fractions, which doesn’t change the optimality of the matrix. For example, we can multiply \[ \begin{pmatrix} 1/3 & 1/2 & 1 \\ 1/4 & 1/3 & 1/2 \end{pmatrix} \] by \(12\) to get the equivalent parity matrix \[ \begin{pmatrix} 4 & 6 & 12 \\ 3 & 4 & 6 \end{pmatrix}\text{.} \] Now our only remaining missing piece is this: how do we compute parity bytes instead of parity numbers? Answering this would render the above discussion moot. However, to do so, we first have to take another look at how we’re doing linear algebra.

8. Linear algebra over fields We ultimately want our parity numbers to be parity bytes, which means that we want to work with matrices of bytes instead of matrices of rational numbers. In order to do that, we need to define an interface for matrix elements that preserves the operations and properties we care about, and then we have to figure out how to implement that interface using bytes. Looking at the rule for matrix multiplication, we need to be able to add and multiply matrix elements. Looking at how we do matrix inversion, we also need to be able to subtract and divide matrix elements. Finally, there are certain properties that hold for rational numbers that we implicitly assume when doing matrix operations, but that we have to make explicit for matrix elements. interface Field<T> { static Zero: T, One: T plus(b: T): T negate(): T times(b: T): T reciprocate(): T equals(b: T): bool minus(b: T) = this.plus(b.negate()) dividedBy(b: T) = this.times(b.reciprocate()) } This leads us to the concept of a field, which essentially defines the interface that matrix elements should implement. Here it is: We need to be able to add and multiply field elements, which we’ll denote generically by \(\oplus\) and \(\otimes\). We also need to be able to take the negation (additive inverse) of an element \(x\), which we’ll denote by \(-x\), and the reciprocal (multiplicative inverse) of a non-zero element \(x\), which we’ll denote by \(x^{-1}\). Then we can define subtraction of field elements to be \[ a \ominus b = a \oplus -b \] and division of field elements to be \[ a \cldiv b = a \otimes b^{-1}\text{,} \] when \(b

e 0\). Field must satisfy further properties, which are copied from the number laws you learn in school: Identities: \(a \oplus 0 = a \otimes 1 = a\).

Inverses: \(a \oplus -a = 0\), and for \(a

e 0\), \(a \otimes a^{-1} = 1\).

Associativity: \((a \oplus b) \oplus c = a \oplus (b \oplus c)\), and \((a \otimes b) \otimes c = a \otimes (b \otimes c)\).

Commutativity: \(a \oplus b = b \oplus a\), and \(a \otimes b = b \otimes a\).

Distributivity: \(a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)\). Of the above, guaranteeing the existence of reciprocals of non-zero elements is usually the non-trivial part. Now the rational numbers satisfy all of the above, since \[ (p/q)^{-1} = q/p\text{,} \] so we say that they form a field. However, the integers do not form a field, since for example \(2\) has no integer reciprocal; only \(1\) and \(-1\) have integer reciprocals. Furthermore, as we saw above, the integers modulo \(256\), i.e. the numbers from \(0\) to \(255\) with standard arithmetic operations modulo \(256\), do not form a field, as we saw earlier, since \((2 \cdot b) \bmod 256

e 1\) for any \(b\). Also, an implementation ofmust satisfy further properties, which are copied from the number laws you learn in school:Of the above, guaranteeing the existence of reciprocals of non-zero elements is usually the non-trivial part. Now the rational numbers satisfy all of the above, since \[ (p/q)^{-1} = q/p\text{,} \] so we say that they form a field. However, the integers do not form a field, since for example \(2\) has no integer reciprocal; only \(1\) and \(-1\) have integer reciprocals. Furthermore, as we saw above, the integers modulo \(256\), i.e. the numbers from \(0\) to \(255\) with standard arithmetic operations modulo \(256\), do not form a field, as we saw earlier, since \((2 \cdot b) \bmod 256

e 1\) for any \(b\). ( Theorem 3 .) Given a prime number \(p\), for every integer \(0 \lt a \lt p\), there is exactly one \(0 \lt b \lt p\) such that \((a \cdot b) \bmod p = 1\). There are clever ways to find multiplcative inverses mod \(p\), but since \(257\) is so small, we can just brute-force it. So an implementation would look like: class Field257Element : implements Field<Field257Element> { plus(b) { return (this + b) % 257 } negate() { return (257 - this) } times(b) { return (this * b) % 257 } reciprocate() { if (this == 0) { return Error } for i := 0 to 256 { if (this.times(b) == 1) { return i; } } return Error } ... } However, we can construct a field with \(257\) elements, using the fact that \(257\) is a prime number, and the following theorem:There are clever ways to find multiplcative inverses mod \(p\), but since \(257\) is so small, we can just brute-force it. So an implementation would look like: Example 7: Field with 257 elements Denote operations on the field with 257 elements by a 257 subscript, and let a = 23 and b = 54. Then a + 257 b = (23 + 54) mod 257 = 77 ;

− 257 b = (257 − 54) mod 257 = 203 ;

a − 257 b = a + 257 − 257 b = (23 + 203) mod 257 = 226 ;

a × 257 b = (23 × 54) mod 257 = 214 ;

54 × 257 119 = 1, so b -1 257 = 119 ;

so a ÷ 257 b = a × 257 b -1 257 = (23 × 119) mod 257 = 167 , and indeed b × 257 ( a ÷ 257 b ) = (54 × 167) mod 257 = 23 = a . Denote operations on the field with 257 elements by asubscript, and letandThen Field257Element instead of a rational number type when implementing ComputeParity and ReconstructData , and if we’ve abstracted our Matrix type correctly, almost everything should just work. However, there is one thing we need to check: Are Cauchy parity matrices still optimal if we use fields other than the rational numbers? Fortunately, the answer is yes: ( Theorem 1, general version .) A parity matrix \(P\) over any field is optimal exactly when any non-empty square submatrix of \(P\) is invertible. ( Theorem 2, general version .) Any non-empty square Cauchy matrix over any field is invertible. ( Corollary 1, general version .) Any square submatrix of a Cauchy matrix over any field is invertible, and thus any Cauchy parity matrix over any field is optimal. However, note that to construct an \(m \times n\) Cauchy matrix, we need \(m + n\) distinct elements. So if we’re working with a field with \(257\) elements, then this imposes the condition that \(m + n \le 257\), i.e. using a finite field limits the number of data bytes and parity numbers you can have. So this gets us closer, since we can useinstead of a rational number type when implementingand, and if we’ve abstracted ourtype correctly, almost everything should just work. However, there is one thing we need to check: Are Cauchy parity matrices still optimal if we use fields other than the rational numbers? Fortunately, the answer is yes:However, note that to construct an \(m \times n\) Cauchy matrix, we need \(m + n\) distinct elements. So if we’re working with a field with \(257\) elements, then this imposes the condition that \(m + n \le 257\), i.e. using a finite field limits the number of data bytes and parity numbers you can have. Now the question remains: can we construct a field with \(256\) elements? As we saw above, we can’t do so the same way as we constructed the field with \(257\) elements. In fact, we need to start with defining different arithmetic operations on the integers. This brings us to the topic of binary carry-less arithmetic.

9. Binary carry-less arithmetic The basic idea with binary carry-less (which I’ll henceforth shorten to “carry-less”) arithmetic is to express all integers in binary, then perform all arithmetic operations using binary arithmetic, except ignoring all the carries.[8] How does this work with addition? Let’s denote binary carry-less add as \(\clplus\),[9] and let’s see how it behaves on single binary digits: \[ \begin{aligned} 0 \clplus 0 &= 0 \\ 0 \clplus 1 &= 1 \\ 1 \clplus 0 &= 1 \\ 1 \clplus 1 &= 0\text{.} \end{aligned} \] This is just the exclusive or operation on bits, so if we do carry-less addition on any two integers, it turns out to be nothing but xor! Since xor can also be denoted by \(\clplus\), we can conveniently think of \(\clplus\) as meaning both carry-less addition and xor. Example 8: Carry-less addition Let a = 23 and b = 54. Then, with carry-less arithmetic, a = 23 = 10111b ^ b = 54 = 110110b ------- 100001b so a ⊕ b = 100001 b = 33 . LetandThen, with carry-less arithmetic,so What about subtraction? Recall that \((a \clplus b) \clplus b = a\) for any \(a\) and \(b\). Therefore, every element \(b\) is its own (carry-less binary) additive inverse, which means that \(a \clminus b = a \clplus b\), i.e. carry-less subtraction is also just xor. Carry-less multiplication isn’t as simple, but recall that binary multiplication is just adding shifted copies of \(a\) based on which bits are set in \(b\) (or vice versa). To do carry-less multiplication, just ignore carries when adding the shifted copies again, i.e. xor shifted copies instead of adding them. Example 9: Carry-less multiplication Let a = 23 and b = 54. Then, with carry-less arithmetic, a = 23 = 10111b ^* b = 54 = 110110b ------------ 10111 ^ 10111 ^ 10111 ^ 10111 ------------ 1111100010b so a ⊗ b = 1111100010 b = 994 . LetandThen, with carry-less arithmetic,so Finally, we can define carry-less division with remainder. Binary division with remainder is subtracting shifted copies of \(b\) from \(a\) until you get a remainder less than the divisor; then carry-less binary division with remainder is xor-ing shifted copies of \(b\) with \(a\) until you get a remainder. However, there’s a subtlety; with carry-less arithmetic, it’s not enough to stop when the remainder (for that step) is less than the divisor, because if the highest set bit of the remainder is the same as the highest set bit of the divisor, you can still xor with the divisor one more time to yield a smaller number (which then becomes the final remainder). Consider the example below, where we’re dividing \(55\) by \(19\). The first remainder is \(17\), which is less than \(19\), but still shares the same highest set bit, so we can xor one more time with \(19\) to get the remainder \(2\). Example 10: Carry-less division Let a = 55 and b = 19. Then, with carry-less arithmetic, 11b -------- b = 19 = 10011b )110111b = 55 = a ^ 10011 ----- 10001 ^ 10011 ----- 10b so a ⨸ b = 11 b = 3 with remainder 10 b = 2 . LetandThen, with carry-less arithmetic,sowith remainder This leads to an interesting difference between the carry-less modulo operation and the standard modulo operation. If you mod by a number \(n\), you get \(n\) possible remainders, from \(0\) to \(n - 1\). However, if you clmod (carry-less mod) by a number \(2^k \le n \lt 2^{k+1}\), you get \(2^k\) possible remainders, from \(0\) to \(2^k-1\), since those are the numbers whose highest set bit is lower than the highest set bit of \(n\). In particular, if you clmod by a number \(256 \le n < 512\), you always get \(256\) possible remainders. This is very close to what we want—now the hope is to find some \(256 \le n < 512\) so that doing binary carry-less arithmetic clmod \(n\) yields a field, which will then be a field with \(256\) elements!

10. The finite field with \(256\) elements Since there are only a few numbers between \(256\) and \(512\), we can just try each one of them to see if clmod-ing by one of them yields a field. However, with a bit of math, we can gain more insight into which numbers will work. Recall the situation with the standard arithmetic operations: arithmetic mod \(p\) yields a field exactly when \(p\) is prime.[10] But recall the definition of a prime number: it is an integer greater than \(1\) whose positive divisors are only itself and \(1\). Stated another way, a prime number is an integer \(p \gt 1\) that cannot be expressed as \(p = a \cdot b\), for \(a, b \gt 1\). Thus, the concept of a prime number is determined by the multiplication operation, and therefore we can define a “carry-less” prime number to be an integer \( p \gt 1\) that cannot be expressed as \(p = a \clmul b\), for \(a, b \gt 1\).[11] ( Theorem 4 .) Given a carry-less prime number \(2^k \lt p \le 2^{k+1}\), for every integer \(0 \lt a \lt 2^k\), there is a exactly one \(0 \lt b \lt 2^k\) such that \((a \clmul b) \bclmod p = 1\). Now we just need to find a carry-less prime number \(256 \le p < 512\). However, the set of prime numbers and the set of carry-less prime numbers are not necessarily related, so for example, even though \(257\) is a prime number, it is not a carry-less prime number. The only question remaining is whether there is an equivalent of Theorem 3 for carry-less arithmetic. And indeed there is:Now we just need to find a carry-less prime number \(256 \le p < 512\). However, the set of prime numbers and the set of carry-less prime numbers are not necessarily related, so for example, even though \(257\) is a prime number, it is not a carry-less prime number. It is easy enough to test each number \(256 \le n < 512\) for carry-less primality though; doing so, we find the lowest one, \(283\).[12] class Field256Element : implements Field<Field256Element> { plus(b) { return this ^ b } negate() { return b } times(b) { return clmod(clmul(this, b), 283) } reciprocate() { if (this == 0) { return Error } for i := 0 to 255 { if (this.times(b) == 1) { return i; } } return Error } ... } Similarly to how we find reciprocals mod \(257\), we brute-force finding reciprocals clmod \(283\) also. So finally, we have a field with \(256\) elements: the integers with binary carry-less arithmetic clmod \(283\). An implementation would look like:Similarly to how we find reciprocals mod \(257\), we brute-force finding reciprocals clmod \(283\) also. Example 11: Field with 256 elements Denote operations on the field with 256 elements by a 256 subscript, and let a = 23 and b = 54. Then a ⊕ 256 b = 23 ⊕ 54 = 33 ;

⊖ 256 b = b = 54 ;

a ⊖ 256 b = a ⊕ 256 ⊖ 256 b = a ⊕ 256 b = 33 ;

a ⊗ 256 b = (23 ⊗ 54) clmod 283 = 207 ;

54 ⊗ 256 102 = 1, so b -1 256 = 102 ;

so a ø 256 b = a ⊗ 256 b -1 256 = (23 ⊗ 102) clmod 283 = 19 , and indeed b ⊗ 256 ( a ø 256 b ) = (54 × 19) clmod 283 = 23 = a . Denote operations on the field with 256 elements by asubscript, and letandThen

11. The full algorithm Now we have all the pieces we need to construct erasure codes for any \((n, m)\) such that \(m + n \le 256\). First, we can compute an \(m \times n\) Cauchy parity matrix over the field with \(256\) elements. (Recall that this needs \(m + n\) distinct field elements, which is what imposes the condition \(m + n \le 256\).) Example 12: Cauchy matrices in general Working over the field with 256 elements, let x = [ 1, 2, 3 ] and y = [ 4, 5, 6 ]. Then, the Cauchy matrix constructed from x and y is / 82 203 209 \ | 123 209 203 | \ 209 123 82 /, which has inverse / 130 31 176 \ | 252 219 31 | \ 108 252 130 /. Working over the field with 256 elements, letandThen, the Cauchy matrix constructed fromandiswhich has inverse Then we can implement matrix multiplication over arbitrary fields, and thus we can implement ComputeParity . Example 13: ComputeParity in detail Let d = [ da, db, 0d ] be the input data bytes and let m = 2 be the desired parity byte count. Then, with the input byte count n = 3, the m × n Cauchy parity matrix computed using x i = n + i and y i = i is / f6 8d 01 \ \ cb 52 7b /. Therefore, the parity bytes are computed as _ _ _ _ / f6 8d 01 \ | da | | 52 | \ cb 52 7b / * | db | = |_ 0c _|, |_ 0d _| and thus the output parity bytes are p = [ 52 , 0c ]. Letbe the input data bytes and letbe the desired parity byte count. Then, with the input byte counttheCauchy parity matrix computed usingandisTherefore, the parity bytes are computed asand thus the output parity bytes are Then we can implement matrix inversion using row reduction over arbitrary fields. Example 14: Matrix inversion via row reduction in general Working over the field with 256 elements, let / 0 2 2 \ M = | 3 4 5 | \ 6 6 7 /. The initial augmented matrix A is / 0 2 2 | 1 0 0 \ | 3 4 5 | 0 1 0 | \ 6 6 7 | 0 0 1 /. We need A 00 to be non-zero, so swap rows 0 and 1 : / 0 2 2 | 1 0 0 \ / 3 4 5 | 0 1 0 \ | 3 4 5 | 0 1 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /. We need A 00 to be 1, so divide row 0 by 3: / 3 4 5 | 0 1 0 \ / 1 245 3 | 0 246 0 \ | 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /. We need A 20 to be 0, so subtract row 0 scaled by 6 from row 2 : / 1 245 3 | 0 246 0 \ / 1 245 3 | 0 246 0 \ | 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 | \ 6 6 7 | 0 0 1 / \ 0 14 13 | 0 2 1 /. We need A 11 to be 1, so divide row 1 by 2: / 1 245 3 | 0 246 0 \ / 1 245 3 | 0 246 0 \ | 0 2 2 | 1 0 0 | --> | 0 1 1 | 141 0 0 | \ 0 14 13 | 0 2 1 / \ 0 14 13 | 0 2 1 /. We need A 21 to be 0, so subtract row 1 scaled by 14 from row 2 : / 1 245 3 | 0 246 0 \ / 1 245 3 | 0 246 0 \ | 0 1 1 | 141 0 0 | --> | 0 1 1 | 141 0 0 | \ 0 14 13 | 0 2 1 / \ 0 0 3 | 7 2 1 /. We need A 22 to be 1, so divide row 2 by 3, which makes the left side of A a unit upper triangular matrix: / 1 245 3 | 0 246 0 \ / 1 245 3 | 0 246 0 \ | 0 1 1 | 141 0 0 | --> | 0 1 1 | 141 0 0 | \ 0 0 3 | 7 2 1 / \ 0 0 1 | 244 247 246 /. We need A 12 to be 0, so subtract row 2 from row 1 : / 1 245 3 | 0 246 0 \ / 1 245 3 | 0 246 0 \ | 0 1 1 | 141 0 0 | --> | 0 1 0 | 121 247 246 | \ 0 0 1 | 244 247 246 / \ 0 0 1 | 244 247 246 /. We need A 02 to be 0, so subtract row 2 scaled by 3 from row 0 : / 1 245 3 | 0 246 0 \ / 1 245 0 | 7 244 1 \ | 0 1 0 | 121 247 246 | --> | 0 1 0 | 121 247 246 | \ 0 0 1 | 244 247 246 / \ 0 0 1 | 244 247 246 /. We need A 01 to be 0, so subtract row 1 scaled by 245 from row 0 , which makes the left side of A the identity matrix: / 1 245 0 | 7 244 1 \ / 1 0 0 | 82 82 82 \ | 0 1 0 | 121 247 246 | --> | 0 1 0 | 121 247 246 | \ 0 0 1 | 244 247 246 / \ 0 0 1 | 244 247 246 /. Since the left side of A is the identity matrix, the right side of A is M -1. Therefore, / 82 82 82 \ M^{-1} = | 121 247 246 | \ 244 247 246 /. Working over the field with 256 elements, letThe initial augmented matrixisWe needto be non-zero, so swap rowsandWe needto be 1, so divide rowby 3:We needto be 0, so subtract rowscaled by 6 from rowWe needto be 1, so divide rowby 2:We needto be 0, so subtract rowscaled by 14 from rowWe needto be 1, so divide rowby 3, which makes the left side ofa unit upper triangular matrix:We needto be 0, so subtract rowfrom rowWe needto be 0, so subtract rowscaled by 3 from rowWe needto be 0, so subtract rowscaled by 245 from row, which makes the left side ofthe identity matrix:Since the left side ofis the identity matrix, the right side ofis. Therefore, Finally, we can use that to implement ReconstructData . Example 15: ReconstructData in detail Let d partial = [ ??, db, ?? ] be the input partial data bytes and p partial = [ 52, 0c ] be the input partial parity bytes. Then, with the data byte count n = 3 and the parity byte count m = 2, and appending the rows of the m × n Cauchy parity matrix to the n × n identity matrix, we get / X01X X00X X00X \ | 00 01 00 | | X00X X00X X01X | | f6 8d 01 | \ cb 52 7b /, where the rows corresponding to the unknown data and parity bytes are crossed out. Taking the first n rows that aren’t crossed out, we get the square matrix / 00 01 00 \ | f6 8d 01 | \ cb 52 7b / which has inverse / 01 d0 d6 \ | 01 00 00 | \ 7b b8 bb /. Therefore, the data bytes are reconstructed from the first n known data and parity bytes as _ _ _ _ / 01 d0 d6 \ | db | | da | | 01 00 00 | * | 52 | = | db | \ 7b b8 bb / |_ 0c _| |_ 0d _|, and thus the output data bytes are d = [ da , db , 0d ]. Letbe the input partial data bytes andbe the input partial parity bytes. Then, with the data byte countand the parity byte countand appending the rows of theCauchy parity matrix to theidentity matrix, we getwhere the rows corresponding to the unknown data and parity bytes are crossed out. Taking the firstrows that aren’t crossed out, we get the square matrixwhich has inverseTherefore, the data bytes are reconstructed from the firstknown data and parity bytes asand thus the output data bytes are And we’re done!

Thanks to Ying-zong Huang, Ryan Hitchman, Charles Ellis, and Josh Gao for comments/corrections/discussion.

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