---
tags: explainers
description: A toy example of data recovery using Vitalik's proposed method for Eth2
image: https://benjaminion.xyz/f/favicon-96x96.png
---
# Data recovery: a toy example
In Ethereum 2.0's data availability model, we have a need to be able to [reconstruct chunks of data](https://hackmd.io/@HWeNw8hNRimMm2m2GH56Cw/r1XzqYIOv#Background-Data-Availability-Sampling) when up to half of that data has been lost or withheld. I [recently implemented](https://github.com/benjaminion/c-kzg/blob/main/src/recover.c) some code to do this, but I always like to understand what's _really_ going on, so I put together this worked example.
This example is going to be super unambitious. We are going to start with two items of data. We will encode these into four items of data, and then throw two of them away. Finally we will recover the original two items.
At this scale, you don't need to be very sophisticated. If our data are $[a,b]$, we can extend it to $[a,b,a+b,a-b]$: now we can lose any two of these values and still recover the original numbers (try it!). But for much larger scales, we need a _general technique_ that can be _efficiently implemented_.
Such a technique is described by Vitalik Buterin in a post [Reed-Solomon erasure code recovery in n\*log^2(n) time with FFTs](https://ethresear.ch/t/reed-solomon-erasure-code-recovery-in-n-log-2-n-time-with-ffts/3039?u=benjaminion). My example here essentially follows that description and notation. It will seem massively over-complex when applied to our meager two data, but remember that the method is both general and efficient.
If you would like to play with it yourself, I've coded up the toy example below [in this repo](https://github.com/benjaminion/data-recovery-toy).
## The basic equipment
The basic items of equipment we shall be using are polynomials, roots of unity, and Fourier transforms. These are all interrelated in very cool ways.
#### Polynomials
If we have four items of data, $[d_0, d_1, d_2, d_3]$, we can make a polynomial out of them like this: $P(x)=d_3x^3+d_2x^2+d_1x+d_0$. The polynomial can then be evaluated for whatever value of $x$ we like.
#### Roots of unity
Powers of primitive roots of unity are the key to making this whole thing work efficiently. They allow us to take lots of shortcuts, and are the points at which we will evaluate our polynomials. If a number is an $n$th root of unity, then raising it to the power of $n$ results in $1$. If it is a _primitive_ $n$th root of unity, then no lower power of $n$ equals $1$.
For our example, we will be using four powers of a primitive fourth root of unity. These are $[1, i, -1, -i]$, where $i$ is a square-root of minus one, an "imaginary" number. The quantity $i$ is a primitive fourth root of unity, and the four powers are generated from from it thus: $[i^0, i^1, i^2, i^3]$. We could have chosen $-i$ as our generator---it is also a primitive fourth root---and ended up with $[1, -i, -1, i]$ which would work just fine, but we will stick with convention. Collectively, to save words, I'll be referring to this list below as "the roots of unity" (which they are, albeit not all primitive), but bear in mind that they are generated from a single primitive root.
#### Fourier transforms
We're going to need to evaluate our polynomial at each of these four roots. Evaluating by hand gives the following relations,
$$
e_0 = P(i^0) = P(1) = d_0+d_2+d_1+d_3 \\
e_1 = P(i^1) = P(i) = d_0-d_2+i(d_1-d_3) \\
e_2 = P(i^2) = P(-1) = d_0+d_2-d_1-d_3 \\
e_3 = P(i^3) = P(-i) = d_0-d_2-i(d_1-d_3)
$$
where I have grouped terms suggestively. The upshot of this is that, if we have the coefficients of a polynomial (the $d_j$) and want to know the polynomial's values at certain roots of unity (the $e_j$), we can just apply a transformation to the coefficients, which is a shortcut when compared to evaluating the polynomial every time.
Magically, it turns out that this is identical to taking the [Fourier transform](https://en.wikipedia.org/wiki/Fourier_transform) of the coefficients, and a very fast way of doing this is known: the [fast Fourier transform](https://en.wikipedia.org/wiki/Fast_Fourier_transform) (FFT).
> Taking the Fourier transform of a polynomial's coefficients gives you its evaluations at a set of roots of unity.
Interestingly, we can do some arithmetic to invert the simultaneous equations above, with the result:
$$
d_0 = \tfrac{1}{4}(e_0+e_2+e_1+e_3) \\
d_1 = \tfrac{1}{4}(e_0-e_2-i(e_1-e_3)) \\
d_2 = \tfrac{1}{4}(e_0+e_2-e_1-e_3) \\
d_3 = \tfrac{1}{4}(e_0-e_2+i(e_1-e_3)) \\
$$
These equations give us a way to reconstruct (interpolate) a polynomial, by calculating its coefficients given only its values at the roots of unity. This is identical to an inverse Fourier transform, which can also be done very efficiently (you can see that it's almost the same as a forward Fourier transform).
> Taking the inverse Fourier transform of a polynomial's evaluations at a set of roots of unity gives you its coefficients.
#### Multiplication and division
Other useful tools are polynomial multiplication and division. Polynomial multiplication is taking, say, $x^2+3x+2$ and multiplying by $x-1$ to get $x^3+2x^2-x-2$. For small polynomials, this is easily done via long multiplication; for large polynomials, we can use fast Fourier transforms for a big speed up.
The [convolution theorem](https://www.theoremoftheday.org/Analysis/Convolution/TotDConvolution.pdf) tells us that, if we have polynomials $P(x)$ and $Q(x)$, then the product $(P*Q)(x)$ is given by ${\tt interpolate({\tt evaluate({\it P})\times{\tt evaluate({\it Q})}})}$ where $\times$ is point-wise multiplication. As we saw above, the interpolate and evaluate steps can be done quickly with fast Fourier transforms. If $P$ and $Q$ are of order $n$ in degree, then the long multiplication method takes $\mathcal{O}(n^2)$ operations, whereas the convolution method takes $\mathcal{O}(n\log{}n)$ operations.
Polynomial division can be done in the same way by replacing pointwise multiplication with pointwise division. Once we have the Fourier transforms, it's easier than implementing [polynomial long division](https://en.wikipedia.org/wiki/Polynomial_long_division).
#### Finite fields
As a final note about the toolkit: if we were doing this "for real", we'd be using a finite field rather than the complex numbers for our coefficients and evaluations. Appropriately constructed finite fields have roots of unity that we can use to construct Fourier transforms (sometimes called [number theoretic transforms](https://en.wikipedia.org/wiki/Discrete_Fourier_transform_(general)#Number-theoretic_transform) in that context), and all the maths carries over. The [full implementation](https://github.com/benjaminion/c-kzg/blob/main/src/recover.c) uses a 256-bit finite field. I've stuck to complex numbers for this example as it seems a little more intuitive to me.
Update: I've added an example using the integers modulo 17 to the [repo](https://github.com/benjaminion/data-recovery-toy). In that field, 4 is a primitive fourth root of unity (we could also use 13), and our roots become [1, 4, 16, 13]. Everything else is unchanged.
## The example
With our toolkit complete, let's attack our toy example.
These are my starting data, the data I wish to make recoverable: $[5,7]$
### Step 1: encode the data
To encode these data we first extend with zeros from a length of two to a length of four. In the general case, we double the length with zeros: it is using this redundancy to extend our domain that is the essence of being able to later recover the data after losing up to half of it.
The padded data:
$$
[5,7,0,0]
$$
We treat this padded data as a polynomial, $D(x)=7x+5$, and encode the data by evaluating the polynomial at the roots of unity. We can do this via our forward FFT.
The encoded data, which is the evaluation of $D(x)$ at the roots of unity:
$$
[12, 5 + 7i, -2, 5 - 7i]
$$
By substituting $d_0=5$, $d_1=7$, and $d_2=d_3=0$, you can see how this relates to the explanation of evaluation above:
$$
e_0 = D(i^0) = D(1) = 5+7 \\
e_1 = D(i^1) = D(i) = 5+7i \\
e_2 = D(i^2) = D(-1) = 5-7 \\
e_3 = D(i^3) = D(-i) = 5-7i
$$
### Step 2: throw away some data!
For this demo, we shall discard two elements of the encoded data. It could be any two, but we will lose indices 1 and 2, and replace them with zeros.
Encoded data with mising elements:
$$
[12, 0, 0, 5 - 7i]
$$
In Vitalik's [article](https://ethresear.ch/t/reed-solomon-erasure-code-recovery-in-n-log-2-n-time-with-ffts/3039?u=benjaminion) This is the evaluation of a polynomial $E(x)$ that we never actually reconstruct.
### Step 3: make the "zero polynomial"
For reasons that will hopefully be clear in a moment, we need to create a polynomial that evaluates to zero at the same places as the missing indices. We'll call this the zero polynomial, $Z(x)$, and in our example it must evaluate to zero at the first and second powers of the root of unity, $x=i^1=i$ and $x=i^2=-1$, corresponding to indices 1 and 2.
Making zero polynomials is conceptually simple: we just multiply up the $(x-x_j)$ for all the $x_j$ values that we want to evaluate to zero. In our case: $Z(x)=(x-i)(x+1)=x^2+(1-i)x-i$. In practice, for large sizes, calculating the zero polynomial ends up being the slowest part of the whole procedure.
$$
Z(x)\mbox{: }[-i,1-i,1,0]
$$
### Step 4: create the polynomial $(E*Z)(x)$
Recalling polynomial multiplication, if we have the evaluation of $E$ (we do), and the evaluation of $Z$ (we already have the two zeros, and we can work out the remainder), then we can interpolate their pointwise multiplications to get the polynomial $(E*Z)(x)$.
Evaluation of $Z$:
$$
[2 - 2i, 0, 0, -2 - 2i]
$$
Pointwise multiplication of the evaluation of $Z$ with the evaluation of $E$:
$$
[24 - 24i, 0, 0, -24 + 4i]
$$
Finally, we interpolate this using the inverse Fourier transform to obtain the polynomial $(E*Z)(x) = 7x^3+(12-7i)x^2+(5-12i)x-5i$. This is represented as,
$$
(E*Z)(x)\mbox{: }[-5i, 5 - 12i, 12 - 7i, 7]
$$
### Step 5: ponder the magic
Now the magic happens. We are trying to reconstruct our original polynomial, $D(x)$, so that we can get our data back. Consider $(E*Z)(x)$ and $(D*Z)(x)$. By construction, both of these polymonials evaluate to zero at the missing indices, 1 ($x=i$) and 2 ($x=-1$). And both evaluate to the same values at the non-missing indices, 0 ($x=1$) and 3 ($x=-i$). Third degree polynomials are fully determined by four values, which means that these two are the same polynomial. Take a moment to convince yourself of this.
As a result of all this, if we can divide the polynomial $(E*Z)(x)=(D*Z)(x)$ by the polyomial $Z(x)$, then we recover $D(x)$ and our work is complete.
### Step 6: divide $(D*Z)(x)$ by $Z(x)$
If only it were that simple! We still have a little hurdle to jump. We would normally do the polynomial division by using the values of the polynomials at the roots of unity. Unfortunately, we know that for indices 1 and 2, both polynomials evaluate to zero because of how we made them, and we have no way to assign values to the $0/0$ result.
#### Step 6a: scaling the polynomials
Thankfully, we can use a little trick to avoid making evaluations at these zero points. The trick is to scale the polynomial by transforming $x\mapsto x/k$ for some constant $k$. (This is called shifting the polynomial in the original article, but it isn't really a shift.)
This is reliable because we know that $Z(x)$ is only zero at two points. As long as our scaled version is not evaluated at those points then we'll not be trying to divide by zero. This works out fine as long as the scale factor, $k$, is not $0$, $1$, or any of our roots.
The shift is easy to implement: we just multiply the polynomial's coefficients by the appropriate power of $k$. Shifting by $k=2$ gives us the following:
$$
\begin{align}
(D*Z)(x/2)\mbox{: } &[-5i, 10 - 24i, 48 - 28i, 56]\\
Z(x/2)\mbox{: } &[-i, 2 - 2i, 4, 0]
\end{align}
$$
#### Step 6b: division via convolution
To do the division, we first form the evaluations of the two polynomials at the roots of unity, using the forward Fourier transform method we are already familiar with.
Evaluation of scaled $D*Z$:
$$
[114 - 57i, -24 - 23i, -18 - 9i, -72 + 69i]
$$
Evaluation of scaled $Z$:
$$
[6 - 3i, -2 + i, 2 + i, -6 - 3i]
$$
Now we need to divide the first by the second, entry by entry. Division of complex numbers is a little non-obvious, but [well-defined](https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex-div/a/dividing-complex-numbers-review):
$$
[19, 5 + 14i, -9, 5 - 14i]
$$
As the last step in the division, we interpolate this back to a polynomial via the inverse Fourier transform:
$$
D(x/2)\mbox{: }[5, 14, 0, 0]
$$
#### Step 6c: unscaling
We're almost there! We just need to unscale $D(x/2)$ by dividing each coefficient by the appropriate power of $k=2$.
$$
D(x)\mbox{: }[5, 7, 0, 0]
$$
### Step 7: profit!
We have recovered the polynomial $D(x)$. And _voila_, our original data is intact there, in the first two places: $[5,7]$.
## Conclusion
As noted at the top, all this palaver is a bit ridiculous for our tiny example with only two data items. However, the real value is that the method described above can scale efficiently to large sizes, using essentially the same toolkit. My [full code](https://github.com/benjaminion/c-kzg/blob/main/src/recover.c) takes about a second to recover 2MB of data when fully half of the encoded data has been lost. About half of that time is spent in generating the zero polynomial - possibly the simplest part conceptually, but apparently hard to do quickly.
As mentioned above, demo code implementing the above toy example is available [on my GitHub](https://github.com/benjaminion/data-recovery-toy).
## Related work
- This toy example [coded up](https://github.com/benjaminion/data-recovery-toy).
- Vitalik: [Reed-Solomon erasure code recovery in n\*log^2(n) time with FFTs](https://ethresear.ch/t/reed-solomon-erasure-code-recovery-in-n-log-2-n-time-with-ffts/3039?u=benjaminion)
- Ethereum [Data Availability Sampling Phase 1 Proposal](https://hackmd.io/@HWeNw8hNRimMm2m2GH56Cw/r1XzqYIOv)
- Dankrad Feist: [polynomial_reconstruction.py](https://github.com/ethereum/research/blob/master/polynomial_reconstruction/polynomial_reconstruction.py). An implementation in Python.
- Protolambda: [go-kzg](https://github.com/protolambda/go-kzg). Includes an implementation in Go, see [recover_from_samples.go](https://github.com/protolambda/go-kzg/blob/master/recover_from_samples.go).
- Me: [c-kzg](https://github.com/benjaminion/c-kzg). Largely follows go-kzg. See [recover.c](https://github.com/benjaminion/c-kzg/blob/main/src/recover.c). Work in progress.
- Vitalik's [artice on fast Fourier transforms](https://vitalik.ca/general/2019/05/12/fft.html) has lots of helpful insights.

[Quick contemporaneous notes by Ben Edgington; fka "Eth2 Implementers' Call"] Agenda: https://github.com/ethereum/pm/issues/667 Livestream: https://youtu.be/KFc1sWYlVZ4 Capella Last week's All Core Devs call recap (of consensus stuff). Danny understands that we will keep withdrawals separately specified from EIP-4844, and slated for Capella. If they do end up getting implemented together, we would not combine the specs, but would just stagger the upgrade. This was confirmed to be the common understanding on this call. testnets / progress updates Running two testnets (no EIP-4844): one post-merge, one pre-merge (to facilitate Prysm). Many participants... (see recording for the details). Full withdrawals working fine.

12/2/2022[Quick contemporaneous notes by Ben Edgington; fka "Eth2 Implementers' Call"] Agenda: https://github.com/ethereum/pm/issues/660 Livestream: https://youtu.be/IK1jNCQz5yk @metachris mev-boost summary and questions Summary document. (Discussion brought forward from later in the agenda.) Release of MEV-Boost 1.4.0 allows setting minimum bid value

11/17/2022[Quick contemporaneous notes by Ben Edgington; fka "Eth2 Implementers' Call"] Agenda: https://github.com/ethereum/pm/issues/630 Livestream: https://youtu.be/GWkhFCvwOT4 Introductory Tim Beiko in the chair this week. Lots on the agenda! Spec, research, etc. Withdrawals

11/3/2022by Ben Edgington 26 August 2022 RSS feed Advertising Back issues 12 August 2022 29 July 2022

8/26/2022
Published on ** HackMD**