Estimating the Security of Ring Learning with Errors (RLWE)

This article was written by my colleague, Cathie Yun. Cathie is an applied cryptographer and security engineer, currently working with me to make fully homomorphic encryption a reality at Google. She’s also done a lot of cool stuff with zero knowledge proofs.


In previous articles, we’ve discussed techniques used in Fully Homomorphic Encryption (FHE) schemes. The basis for many FHE schemes, as well as other privacy-preserving protocols, is the Learning With Errors (LWE) problem. In this article, we’ll talk about how to estimate the security of lattice-based schemes that rely on the hardness of LWE, as well as its widely used variant, Ring LWE (RLWE).

A previous article on modulus switching introduced LWE encryption, but as a refresher:

Reminder of LWE

A literal repetition from the modulus switching article. The LWE encryption scheme I’ll use has the following parameters:

  • A plaintext space $\mathbb{Z}/q\mathbb{Z}$, where $q \geq 2$ is a positive integer. This is the space that the underlying message comes from.
  • An LWE dimension $n \in \mathbb{N}$.
  • A discrete Gaussian error distribution $ D$ with a mean of zero and a fixed standard deviation.

An LWE secret key is defined as a vector in $\{0, 1\}^n$ (uniformly sampled). An LWE ciphertext is defined as a vector $a = (a_1, \dots, a_n)$, sampled uniformly over $(\mathbb{Z} / q\mathbb{Z})^n$, and a scalar $b = \langle a, s \rangle + m + e$, where $e$ is drawn from $D$ and all arithmetic is done modulo $q$. Note that $e$ must be small for the encryption to be valid.

Learning With Errors (LWE) security

Choosing appropriate LWE parameters is a nontrivial challenge when designing and implementing LWE based schemes, because there are conflicting requirements of security, correctness, and performance. Some of the parameters that can be manipulated are the LWE dimension $n$, error distribution $D$ (referred to in the next few sections as $X_e$), secret distribution $X_s$, and plaintext modulus $q$.

Lattice Estimator

Here is where the Lattice Estimator tool comes to our assistance! The lattice estimator is a Sage module written by a group of lattice cryptography researchers which estimates the concrete security of Learning with Errors (LWE) instances.

For a given set of LWE parameters, the Lattice Estimator calculates the cost of all known efficient lattice attacks – for example, the Primal, Dual, and Coded-BKW attacks. It returns the estimated number of “rops” or “ring operations” required to carry out each attack; the attack that is the most efficient is the one that determines the security parameter. The bits of security for the parameter set can be calculated as $\log_2(\text{rops})$ for the most efficient attack.

Running the Lattice Estimator

For example, let’s estimate the security of the security parameters originally published for the popular TFHE scheme:

n = 630
q = 2^32
Xs = UniformMod(2)
Xe = DiscreteGaussian(stddev=2^17)

After installing the Lattice Estimator and sage, we run the following commands in sage:

> from estimator import *
> schemes.TFHE630
LWEParameters(n=630, q=4294967296, Xs=D(σ=0.50, μ=-0.50), Xe=D(σ=131072.00), m=+Infinity, tag='TFHE630')
> _ = LWE.estimate(schemes.TFHE630)
bkw                  :: rop: ≈2^153.1, m: ≈2^139.4, mem: ≈2^132.6, b: 4, t1: 0, t2: 24, ℓ: 3, #cod: 552, #top: 0, #test: 78, tag: coded-bkw
usvp                 :: rop: ≈2^124.5, red: ≈2^124.5, δ: 1.004497, β: 335, d: 1123, tag: usvp
bdd                  :: rop: ≈2^131.0, red: ≈2^115.1, svp: ≈2^131.0, β: 301, η: 393, d: 1095, tag: bdd
bdd_hybrid           :: rop: ≈2^185.3, red: ≈2^115.9, svp: ≈2^185.3, β: 301, η: 588, ζ: 0, |S|: 1, d: 1704, prob: 1, ↻: 1, tag: hybrid
bdd_mitm_hybrid      :: rop: ≈2^265.5, red: ≈2^264.5, svp: ≈2^264.5, β: 301, η: 2, ζ: 215, |S|: ≈2^189.2, d: 1489, prob: ≈2^-146.6, ↻: ≈2^148.8, tag: hybrid
dual                 :: rop: ≈2^128.7, mem: ≈2^72.0, m: 551, β: 346, d: 1181, ↻: 1, tag: dual
dual_hybrid          :: rop: ≈2^119.8, mem: ≈2^115.5, m: 516, β: 314, d: 1096, ↻: 1, ζ: 50, tag: dual_hybrid

In this example, the most efficient attack is the dual_hybrid attack. It uses 2^119.8 ring operations, and so these parameters provide 119.8 bits of security. The reader may notice that the TFHE website claims those parameters give 128 bits of security. This discrepancy is due to the fact that they used an older library (the LWE estimator, which is no longer maintained), which doesn’t take into account the most up-to-date lattice attacks.

For further reading, Benjamin Curtis wrote an article about parameter selection for the CONCRETE implementation of the TFHE scheme. Benjamin Curtis, Martin Albrecht, and other researchers also used the Lattice Estimator to estimate all the LWE and NTRU schemes.

Ring Learning with Errors (RLWE) security

It is often desirable to use Ring LWE instead of LWE, for greater efficiency and smaller key sizes (as Chris Peikert illustrates via meme). We’d like to estimate the security of a Ring LWE scheme, but it wasn’t immediately obvious to us how to do this, since the Lattice Estimator only operates over LWE instances. In order to use the Lattice Estimator for this security estimate, we first needed to do a reduction from the RLWE instance to an LWE instance.

Attempted RLWE to LWE reduction

Given an RLWE instance with $ \text{RLWE_dimension} = k $ and $ \text{poly_log_degree} = N $, we can create a relation that looks like an LWE instance of $ \text{LWE_dimension} = N * k $ with the same security, as long as $N$ is a power of 2 and there are no known attacks that target the ring structure of RLWE that are more efficient than the best LWE attacks. Note: $N$ must be a power of 2 so that $x^N+1$ is a cyclotomic polynomial.

An RLWE encryption has the following form: $ (a_0(x), a_1(x), … a_{k-1}(x), b(x)) $

  •   Public polynomials: $ a_0(x), a_1(x), \dots a_{k-1}(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} (\mathbb{Z}/{q \mathbb{Z}[x]} ) / (x^N + 1)^k$
  •   Secret (binary) polynomials: $ s_0(x), s_1(x), \dots s_{k-1}(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} (\mathbb{B}_N[x])^k$
  •   Error: $ e(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} \chi_e$
  •   RLWE instance: $ b(x) = \sum_{i=0}^{k-1} a_i(x) \cdot s_i(x) + e(x) \in (\mathbb{Z}/{q \mathbb{Z}[x]} ) / (x^N + 1)$

We would like to express this in the form of an LWE encryption. We can make start with the simple case, where $ k=1 $. Therefore, we will only be working with the zero-entry polynomials, $a_0(x)$ and $s_0(x)$. (For simplicity, in the next example you can ignore the zero-subscript and think of them as $a(x)$ and $s(x)$).

Naive reduction for $k=1$ (wrong!)

Naively, if we simply defined the LWE $A$ matrix to be a concatenation of the coefficients of the RLWE polynomial $a(x)$, we get:

$$ A_{\text{LWE}} = ( a_{0, 0}, a_{0, 1}, \dots a_{0, N-1} ) $$

We can do the same for the LWE $s$ vector:

$$ s_{\text{LWE}} = ( s_{0, 0}, s_{0, 1}, \dots s_{0, N-1} ) $$

But this doesn’t give us the value of $b_{LWE}$ for the LWE encryption that we want. In particular, the first entry of $b_{LWE}$, which we can call $b_{\text{LWE}, 0}$, is simply a product of the first entries of $a_0(x)$ and $s_0(x)$:

$$ b_{\text{LWE}, 0} = a_{0, 0} \cdot s_{0, 0} + e_0 $$

However, we want $b_{\text{LWE}, 0}$ to be a sum of the products of all the coefficients of $a_0(x)$ and $s_0(x)$ that give us a zero-degree coefficient mod $x^N + 1$. This modulus is important because it causes the product of high-degree monomials to “wrap around” to smaller degree monomials because of the negacyclic property, such that $x^N \equiv -1 \mod x^N + 1$. So the constant term $b_{\text{LWE}, 0}$ should include all of the following terms:

$$\begin{aligned}
b_{\text{LWE}, 0} = & a_{0, 0} \cdot s_{0, 0} \\
 – & a_{0, 1} \cdot s_{0, N-1} \\
 – & a_{0, 2} \cdot s_{0, N-2} \\
 – & \dots \\
 – & a_{0, N-1} \cdot s_{0, 1}\\
 + & e_0\\
\end{aligned}
$$

Improved reduction for $k=1$

We can achieve the desired value of $b_{\text{LWE}}$ by more strategically forming a matrix $A_{\text{LWE}}$, to reflect the negacyclic property of our polynomials in the RLWE space. We can keep the naive construction for $s_\text{LWE}$.

$$ A_{\text{LWE}} =
\begin{pmatrix}
a_{0, 0}   & -a_{0, N-1} & -a_{0, N-2} & \dots & -a_{0, 1}\\
a_{0, 1}   & a_{0, 0}    & -a_{0, N-1} & \dots & -a_{0, 2}\\
\vdots     & \ddots      &             &       & \vdots   \\
a_{0, N-1} & \dots       &             &       & a_{0, 0} \\
\end{pmatrix}
$$

This definition of $A_\text{LWE}$ gives us the desired value for $b_\text{LWE}$, when $b_{\text{LWE}}$ is interpreted as the coefficients of a polynomial. As an example, we can write out the elements of the first row of $b_\text{LWE}$:

$$
\begin{aligned}
b_{\text{LWE}, 0} = & \sum_{i=0}^{N-1} A_{\text{LWE}, 0, i} \cdot s_{0, i} + e_0 \\
b_{\text{LWE}, 0} = & a_{0, 0} \cdot s_{0, 0} \\
 – & a_{0, 1} \cdot s_{0, N-1} \\
 – & a_{0, 2} \cdot s_{0, N-2} \\
 – & \dots \\
 – & a_{0, N-1} \cdot s_{0, 1}\\
 + & e_0 \\
\end{aligned}
$$

Generalizing for all $k$

In the generalized $k$ case, we have the RLWE equation:

$$ b(x) = a_0(x) \cdot s_0(x) + a_1(x) \cdot s_1(x) \cdot a_{k-1}(x) \cdot s_{k-1}(x) + e(x) $$

We can construct the LWE elements as follows:

$$A_{\text{LWE}} =
\left ( \begin{array}{c|c|c|c}
A_{0, \text{LWE}} & A_{1, \text{LWE}} & \dots & A_{k-1, \text{LWE}} \end{array}
 \right )
$$

where each sub-matrix is the construction from the previous section:

$$ A_{\text{LWE}} =
\begin{pmatrix}
a_{i, 0}   & -a_{i, N-1} & -a_{i, N-2} & \dots & -a_{i, 1}\\
a_{i, 1}   & a_{i, 0}    & -a_{i, N-1} & \dots & -a_{i, 2}\\
\vdots     & \ddots      &             &       & \vdots   \\
a_{i, N-1} & \dots       &             &       & a_{i, 0} \\
\end{pmatrix}
$$

And the secret keys are stacked similarly:

$$ s_{\text{LWE}} = ( s_{0, 0}, s_{0, 1}, \dots s_{0, N-1} \mid s_{1, 0}, s_{1, 1}, \dots s_{1, N-1} \mid \dots ) $$

This is how we can reduce an RLWE instance with RLWE dimension $k$ and polynomial modulus degree $N$, to a relation that looks like an LWE instance of LWE dimension $N * k$.

Caveats and open research

This reduction does not result in a correctly formed LWE instance, since an LWE instance would have a matrix $A$ that is randomly sampled, whereas the reduction results in an matrix $A$ that has cyclic structure, due to the cyclic property of the RLWE instance. This is why I’ve been emphasizing that the reduction produces an instance that looks like LWE. All currently known attacks on RLWE do not take advantage of the structure, but rather directly attack this transformed LWE instance. Whether the additional ring structure can be exploited in the design of more efficient attacks remains an open question in the lattice cryptography research community.

In her PhD thesis, Rachel Player mentions the RLWE to LWE security reduction:

In order to try to pick parameters in Ring-LWE-based schemes (FHE or otherwise) that we hope are sufficiently secure, we can choose parameters such that the underlying Ring-LWE instance should be hard to solve according to known attacks. Each Ring-LWE sample can be used to extract $n$ LWE samples. To the best of our knowledge, the most powerful attacks against $d$-sample Ring-LWE all work by instead attacking the $nd$-sample LWE problem. When estimating the security of a particular set of Ring-LWE parameters we therefore estimate the security of the induced set of LWE parameters.

This indicates that we can do this reduction for certain RLWE instances. However, we must be careful to ensure that the polynomial modulus degree $N$ is a power of two, because otherwise the error distribution “breaks”, as my colleague Baiyu Li explained to me in conversation:

The RLWE problem is typically defined in using the ring of integers of the cyclotomic field $\mathbb{Q}[X]/(f(X))$, where $f(X)$ is a cyclotomic polynomial of degree $k=\phi(N)$ (where $\phi$ is Euler’s totient function), and the error is a spherical Gaussian over the image of the canonical embedding into the complex numbers $\mathbb{C}^k$ (basically the images of primitive roots of unity under $f$). In many cases we set $N$ to be a power of 2, thus $f(X)=X^{N/2}+1$, since the canonical embedding for such $N$ has a nice property that the preimage of the spherical Gaussian error is also a spherical Gaussian over the coefficients of polynomials in $\mathbb{Q}[X]/(f(X))$. So in this case we can sample $k=N/2$ independent Gaussian numbers and use them as the coefficients of the error polynomial $e(x)$. For $N$ not a power of 2, $f(X)$ may have some low degree terms, and in order to get the spherical Gaussian with the same variance $s^2$ in the canonical embedding, we probably need to use a larger variance when sampling the error polynomial coefficients.

The RLWE we frequently use in practice is actually a specialized version called “polynomial LWE”, and instantiated with $N$ = power of 2 and so $f(X)=X^{N/2}+1$. For other parameters the two are not exactly the same. This paper has some explanations: https://eprint.iacr.org/2018/170.pdf

The error distribution “breaks” if $N$ is not a power of 2 due to the fact that the precise form of RLWE is not defined on integer polynomial rings $R = \mathbb{Z}[X]/(f(X))$, but is defined on its dual (or the dual in the underlying number field, which is a fractional ideal of $\mathbb{Q}[X]/(f(x))$), and the noise distribution is on the Minkowski embedding of this dual ring. For non-power of 2 $N$, the product mod $f$ of two small polynomials in $\mathbb{Q}[X]/(f(x))$ may be large, where small/large means their L2 norm on the coefficient vector. This means that in order to sample the required noise distribution, you may need a skewed coefficient distribution. Only when $N$ is a power of 2, the dual of $R$ is a scaling of $R$, and distance in the embedding of $R^{\text{dual}}$ is preserved in $R$, and so we can just sample iid gaussian coefficient to get the required noise.

Because working with a power-of-two RLWE polynomial modulus gives “nice” error behavior, this parameter choice is often recommended and chosen for concrete instantiations of RLWE. For example, the Homomorphic Encryption Standard
recommends and only analyzes the security of parameters for power-of-two cyclotomic fields for use in homomorphic encryption (though future versions of the standard aim to extend the security analysis to generic cyclotomic rings):

We stress that when the error is chosen from sufficiently wide and “well spread” distributions that match the ring at hand, we do not have meaningful attacks on RLWE that are better than LWE attacks, regardless of the ring. For power-of-two cyclotomics, it is sufficient to sample the noise in the polynomial basis, namely choosing the coefficients of the error polynomial $e \in \mathbb{Z}[x] / \phi_k(x)$ independently at random from a very “narrow” distribution.

Existing works analyzing and targeting the ring structure of RLWE include:

It would of course be great to have a definitive answer on whether we can be confident using this RLWE to LWE reduction to estimate the security of RLWE based schemes. In the meantime, we have seen many Fully Homomorphic Encryption (FHE) schemes using this RLWE to LWE reduction, and we hope that this article helps explain how that reduction works and the existing open questions around this approach.

Negacyclic Polynomial Multiplication

In this article I’ll cover three techniques to compute special types of polynomial products that show up in lattice cryptography and fully homomorphic encryption. Namely, the negacyclic polynomial product, which is the product of two polynomials in the quotient ring $\mathbb{Z}[x] / (x^N + 1)$. As a precursor to the negacyclic product, we’ll cover the simpler cyclic product.

All of the Python code written for this article is on GitHub.

The DFT and Cyclic Polynomial Multiplication

A recent program gallery piece showed how single-variable polynomial multiplication could be implemented using the Discrete Fourier Transform (DFT). This boils down to two observations:

  1. The product of two polynomials $f, g$ can be computed via the convolution of the coefficients of $f$ and $g$.
  2. The Convolution Theorem, which says that the Fourier transform of a convolution of two signals $f, g$ is the point-wise product of the Fourier transforms of the two signals. (The same holds for the DFT)

This provides a much faster polynomial product operation than one could implement using the naïve polynomial multiplication algorithm (though see the last section for an implementation anyway). The DFT can be used to speed up large integer multiplication as well.

A caveat with normal polynomial multiplication is that one needs to pad the input coefficient lists with enough zeros so that the convolution doesn’t “wrap around.” That padding results in the output having length at least as large as the sum of the degrees of $f$ and $g$ (see the program gallery piece for more details).

If you don’t pad the polynomials, instead you get what’s called a cyclic polynomial product. More concretely, if the two input polynomials $f, g$ are represented by coefficient lists $(f_0, f_1, \dots, f_{N-1}), (g_0, g_1, \dots, g_{N-1})$ of length $N$ (implying the inputs are degree at most $N-1$, i.e., the lists may end in a tail of zeros), then the Fourier Transform technique computes

\[ f(x) \cdot g(x) \mod (x^N – 1) \]

This modulus is in the sense of a quotient ring $\mathbb{Z}[x] / (x^N – 1)$, where $(x^N – 1)$ denotes the ring ideal generated by $x^N-1$, i.e., all polynomials that are evenly divisible by $x^N – 1$. A particularly important interpretation of this quotient ring is achieved by interpreting the ideal generator $x^N – 1$ as an equation $x^N – 1 = 0$, also known as $x^N = 1$. To get the canonical ring element corresponding to any polynomial $h(x) \in \mathbb{Z}[x]$, you “set” $x^N = 1$ and reduce the polynomial until there are no more terms with degree bigger than $N-1$. For example, if $N=5$ then $x^{10} + x^6 – x^4 + x + 2 = -x^4 + 2x + 3$ (the $x^{10}$ becomes 1, and $x^6 = x$).

To prove the DFT product computes a product in this particular ring, note how the convolution theorem produces the following formula, where $\textup{fprod}(f, g)$ denotes the process of taking the Fourier transform of the two coefficient lists, multiplying them entrywise, and taking a (properly normalized) inverse FFT, and $\textup{fprod}(f, g)(j)$ is the $j$-th coefficient of the output polynomial:

\[ \textup{fprod}(f, g)(j) = \sum_{k=0}^{N-1} f_k g_{j-k \textup{ mod } N} \]

In words, the output polynomial coefficient $j$ equals the sum of all products of pairs of coefficients whose indices sum to $j$ when considered “wrapping around” $N$. Fixing $j=1$ as an example, $\textup{fprod}(f, g)(1) = f_0 g_1 + f_1g_0 + f_2 g_{N-1} + f_3 g_{N-2} + \dots$. This demonstrates the “set $x^N = 1$” interpretation above: the term $f_2 g_{N-1}$ corresponds to the product $f_2x^2 \cdot g_{N-1}x^{N-1}$, which contributes to the $x^1$ term of the polynomial product if and only if $x^{2 + N-1} = x$, if and only if $x^N = 1$.

To achieve this in code, we simply use the version of the code from the program gallery piece, but fix the size of the arrays given to numpy.fft.fft in advance. We will also, for simplicity, assume the $N$ one wishes to use is a power of 2. The resulting code is significantly simpler than the original program gallery code (we omit zero-padding to length $N$ for brevity).

import numpy
from numpy.fft import fft, ifft

def cyclic_polymul(p1, p2, N):
    """Multiply two integer polynomials modulo (x^N - 1).

    p1 and p2 are arrays of coefficients in degree-increasing order.
    """
    assert len(p1) == len(p2) == N
    product = fft(p1) * fft(p2)
    inverted = ifft(product)
    return numpy.round(numpy.real(inverted)).astype(numpy.int32)

As a side note, there’s nothing that stops this from working with polynomials that have real or complex coefficients, but so long as we use small magnitude integer coefficients and round at the end, I don’t have to worry about precision issues (hat tip to Brad Lucier for suggesting an excellent paper by Colin Percival, “Rapid multiplication modulo the sum and difference of highly composite numbers“, which covers these precision issues in detail).

Negacyclic polynomials, DFT with duplication

Now the kind of polynomial quotient ring that shows up in cryptography is critically not $\mathbb{Z}[x]/(x^N-1)$, because that ring has enough easy-to-reason-about structure that it can’t hide secrets. Instead, cryptographers use the ring $\mathbb{Z}[x]/(x^N+1)$ (the minus becomes a plus), which is believed to be more secure for cryptography—although I don’t have a great intuitive grasp on why.

The interpretation is similar here as before, except we “set” $x^N = -1$ instead of $x^N = 1$ in our reductions. Repeating the above example, if $N=5$ then $x^{10} + x^6 – x^4 + x + 2 = -x^4 + 3$ (the $x^{10}$ becomes $(-1)^2 = 1$, and $x^6 = -x$). It’s called negacyclic because as a term $x^k$ passes $k \geq N$, it cycles back to $x^0 = 1$, but with a sign flip.

The negacyclic polynomial multiplication can’t use the DFT without some special hacks. The first and simplest hack is to double the input lists with a negation. That is, starting from $f(x) \in \mathbb{Z}[x]/(x^N+1)$, we can define $f^*(x) = f(x) – x^Nf(x)$ in a different ring $\mathbb{Z}[x]/(x^{2N} – 1)$ (and similarly for $g^*$ and $g$).

Before seeing how this causes the DFT to (almost) compute a negacyclic polynomial product, some math wizardry. The ring $\mathbb{Z}[x]/(x^{2N} – 1)$ is special because it contains our negacyclic ring as a subring. Indeed, because the polynomial $x^{2N} – 1$ factors as $(x^N-1)(x^N+1)$, and because these two factors are coprime in $\mathbb{Z}[x]/(x^{2N} – 1)$, the Chinese remainder theorem (aka Sun-tzu’s theorem) generalizes to polynomial rings and says that any polynomial in $\mathbb{Z}[x]/(x^{2N} – 1)$ is uniquely determined by its remainders when divided by $(x^N-1)$ and $(x^N+1)$. Another way to say it is that the ring $\mathbb{Z}[x]/(x^{2N} – 1)$ factors as a direct product of the two rings $\mathbb{Z}[x]/(x^{N} – 1)$ and $\mathbb{Z}[x]/(x^{N} + 1)$.

Now mapping a polynomial $f(x)$ from the bigger ring $(x^{2N} – 1)$ to the smaller ring $(x^{N}+1)$ involves taking a remainder of $f(x)$ when dividing by $x^{N}+1$ (“setting” $x^N = -1$ and reducing). There are many possible preimage mappings, depending on what your goal is. In this case, we actually intentionally choose a non preimage mapping, because in general to compute a preimage requires solving a system of congruences in the larger polynomial ring. So instead we choose $f(x) \mapsto f^*(x) = f(x) – x^Nf(x) = -f(x)(x^N – 1)$, which maps back down to $2f(x)$ in $\mathbb{Z}[x]/(x^{N} + 1)$. This preimage mapping has a particularly nice structure, in that you build it by repeating the polynomial’s coefficients twice and flipping the sign of the second half. It’s easy to see that the product $f^*(x) g^*(x)$ maps down to $4f(x)g(x)$.

So if we properly account for these extra constant factors floating around, our strategy to perform negacyclic polynomial multiplication is to map $f$ and $g$ up to the larger ring as described, compute their cyclic product (modulo $x^{2N} – 1$) using the FFT, and then the result should be a degree $2N-1$ polynomial which can be reduced with one more modular reduction step to the right degree $N-1$ negacyclic product, i.e., setting $x^N = -1$, which materializes as taking the second half of the coefficients, flipping their signs, and adding them to the corresponding coefficients in the first half.

The code for this is:

def negacyclic_polymul_preimage_and_map_back(p1, p2):
    p1_preprocessed = numpy.concatenate([p1, -p1])
    p2_preprocessed = numpy.concatenate([p2, -p2])
    product = fft(p1_preprocessed) * fft(p2_preprocessed)
    inverted = ifft(product)
    rounded = numpy.round(numpy.real(inverted)).astype(p1.dtype)
    return (rounded[: p1.shape[0]] - rounded[p1.shape[0] :]) // 4

However, this chosen mapping hides another clever trick. The product of the two preimages has enough structure that we can “read” the result off without doing the full “set $x^N = -1$” reduction step. Mapping $f$ and $g$ up to $f^*, g^*$ and taking their product modulo $(x^{2N} – 1)$ gives

\[ \begin{aligned} f^*g^* &= -f(x^N-1) \cdot -g(x^N – 1) \\ &= fg (x^N-1)^2 \\ &= fg(x^{2N} – 2x^N + 1) \\ &= fg(2 – 2x^N) \\ &= 2(fg – x^Nfg) \end{aligned} \]

This has the same syntactical format as the original mapping $f \mapsto f – x^Nf$, with an extra factor of 2, and so its coefficients also have the form “repeat the coefficients and flip the sign of the second half” (times two). We can then do the “inverse mapping” by reading only the first half of the coefficients and dividing by 2.

def negacyclic_polymul_use_special_preimage(p1, p2):
    p1_preprocessed = numpy.concatenate([p1, -p1])
    p2_preprocessed = numpy.concatenate([p2, -p2])
    product = fft(p1_preprocessed) * fft(p2_preprocessed)
    inverted = ifft(product)
    rounded = numpy.round(0.5 * numpy.real(inverted)).astype(p1.dtype)
    return rounded[: p1.shape[0]]

Our chosen mapping $f \mapsto f-x^Nf$ is not particularly special, except that it uses a small number of pre and post-processing operations. For example, if you instead used the mapping $f \mapsto 2f + x^Nf$ (which would map back to $f$ exactly), then the FFT product would result in $5fg + 4x^Nfg$ in the larger ring. You can still read off the coefficients as before, but you’d have to divide by 5 instead of 2 (which, the superstitious would say, is harder). It seems that “double and negate” followed by “halve and take first half” is the least amount of pre/post processing possible.

Negacyclic polynomials with a “twist”

The previous section identified a nice mapping (or embedding) of the input polynomials into a larger ring. But studying that shows some symmetric structure in the FFT output. I.e., the coefficients of $f$ and $g$ are repeated twice, with some scaling factors. It also involves taking an FFT of two $2N$-dimensional vectors when we start from two $N$-dimensional vectors.

This sort of situation should make you think that we can do this more efficiently, either by using a smaller size FFT or by packing some data into the complex part of the input, and indeed we can do both.

[Aside: it’s well known that if all the entries of an FFT input are real, then the result also has symmetry that can be exploted for efficiency by reframing the problem as a size-N/2 FFT in some cases, and just removing half the FFT algorithm’s steps in other cases, see Wikipedia for more]

This technique was explained in Fast multiplication and its applications (pdf link) by Daniel Bernstein, a prominent cryptographer who specializes in cryptography performance, and whose work appears in widely-used standards like TLS, OpenSSH, and he designed a commonly used elliptic curve for cryptography.

[Aside: Bernstein cites this technique as using something called the “Tangent FFT (pdf link).” This is a drop-in FFT replacement he invented that is faster than previous best (split-radix FFT), and Bernstein uses it mainly to give a precise expression for the number of operations required to do the multiplication end to end. We will continue to use the numpy FFT implementation, since in this article I’m just focusing on how to express negacyclic multiplication in terms of the FFT. Also worth noting both the Tangent FFT and “Fast multiplication” papers frame their techniques—including FFT algorithm implementations!—in terms of polynomial ring factorizations and mappings. Be still, my beating cardioid.]

In terms of polynomial mappings, we start from the ring $\mathbb{R}[x] / (x^N + 1)$, where $N$ is a power of 2. We then pick a reversible mapping from $\mathbb{R}[x]/(x^N + 1) \to \mathbb{C}[x]/(x^{N/2} – 1)$ (note the field change from real to complex), apply the FFT to the image of the mapping, and reverse appropriately it at the end.

One such mapping takes two steps, first mapping $\mathbb{R}[x]/(x^N + 1) \to \mathbb{C}[x]/(x^{N/2} – i)$ and then from $\mathbb{C}[x]/(x^{N/2} – i) \to \mathbb{C}[x]/(x^{N/2} – 1)$. The first mapping is as easy as the last section, because $(x^N + 1) = (x^{N/2} + i) (x^{N/2} – i)$, and so we can just set $x^{N/2} = i$ and reduce the polynomial. This as the effect of making the second half of the polynomial’s coefficients become the complex part of the first half of the coefficients.

The second mapping is more nuanced, because we’re not just reducing via factorization. And we can’t just map $i \mapsto 1$ generically, because that would reduce complex numbers down to real values. Instead, we observe that (momentarily using an arbitrary degree $k$ instead of $N/2$), for any polynomial $f \in \mathbb{C}[x]$, the remainder of $f \mod x^k-i$ uniquely determines the remainder of $f \mod x^k – 1$ via the change of variables $x \mapsto \omega_{4k} x$, where $\omega_{4k}$ is a $4k$-th primitive root of unity $\omega_{4k} = e^{\frac{2 \pi i}{4k}}$. Spelling this out in more detail: if $f(x) \in \mathbb{C}[x]$ has remainder $f(x) = g(x) + h(x)(x^k – i)$ for some polynomial $h(x)$, then

\[ \begin{aligned} f(\omega_{4k}x) &= g(\omega_{4k}x) + h(\omega_{4k}x)((\omega_{4k}x)^{k} – i) \\ &= g(\omega_{4k}x) + h(\omega_{4k}x)(e^{\frac{\pi i}{2}} x^k – i) \\ &= g(\omega_{4k}x) + i h(\omega_{4k}x)(x^k – 1) \\ &= g(\omega_{4k}x) \mod (x^k – 1) \end{aligned} \]

Translating this back to $k=N/2$, the mapping from $\mathbb{C}[x]/(x^{N/2} – i) \to \mathbb{C}[x]/(x^{N/2} – 1)$ is $f(x) \mapsto f(\omega_{2N}x)$. And if $f = f_0 + f_1x + \dots + f_{N/2 – 1}x^{N/2 – 1}$, then the mapping involves multiplying each coefficient $f_k$ by $\omega_{2N}^k$.

When you view polynomials as if they were a simple vector of their coefficients, then this operation $f(x) \mapsto f(\omega_{k}x)$ looks like $(a_0, a_1, \dots, a_n) \mapsto (a_0, \omega_{k} a_1, \dots, \omega_k^n a_n)$. Bernstein calls the operation a twist of $\mathbb{C}^n$, which I mused about in this Mathstodon thread.

What’s most important here is that each of these transformations are invertible. The first because the top half coefficients end up in the complex parts of the polynomial, and the second because the mapping $f(x) \mapsto f(\omega_{2N}^{-1}x)$ is an inverse. Together, this makes the preprocessing and postprocessing exact inverses of each other. The code is then

def negacyclic_polymul_complex_twist(p1, p2):
    n = p2.shape[0]
    primitive_root = primitive_nth_root(2 * n)
    root_powers = primitive_root ** numpy.arange(n // 2)

    p1_preprocessed = (p1[: n // 2] + 1j * p1[n // 2 :]) * root_powers
    p2_preprocessed = (p2[: n // 2] + 1j * p2[n // 2 :]) * root_powers

    p1_ft = fft(p1_preprocessed)
    p2_ft = fft(p2_preprocessed)
    prod = p1_ft * p2_ft
    ifft_prod = ifft(prod)
    ifft_rotated = ifft_prod * primitive_root ** numpy.arange(0, -n // 2, -1)

    return numpy.round(
        numpy.concatenate([numpy.real(ifft_rotated), numpy.imag(ifft_rotated)])
    ).astype(p1.dtype)

And so, at the cost of a bit more pre- and postprocessing, we can negacyclically multiply two degree $N-1$ polynomials using an FFT of length $N/2$. In theory, no information is wasted and this is optimal.

And finally, a simple matrix multiplication

The last technique I wanted to share is not based on the FFT, but it’s another method for doing negacyclic polynomial multiplication that has come in handy in situations where I am unable to use FFTs. I call it the Toeplitz method, because one of the polynomials is converted to a Toeplitz matrix. Sometimes I hear it referred to as a circulant matrix technique, but due to the negacyclic sign flip, I don’t think it’s a fully accurate term.

The idea is to put the coefficients of one polynomial $f(x) = f_0 + f_1x + \dots + f_{N-1}x^{N-1}$ into a matrix as follows:

\[ \begin{pmatrix} f_0 & -f_{N-1} & \dots & -f_1 \\ f_1 & f_0 & \dots & -f_2 \\ \vdots & \vdots & \ddots & \vdots \\ f_{N-1} & f_{N-2} & \dots & f_0 \end{pmatrix} \]

The polynomial coefficients are written down in the first column unchanged, then in each subsequent column, the coefficients are cyclically shifted down one, and the term that wraps around the top has its sign flipped. When the second polynomial is treated as a vector of its coefficients, say, $g(x) = g_0 + g_1x + \dots + g_{N-1}x^{N-1}$, then the matrix-vector product computes their negacyclic product (as a vector of coefficients):

\[ \begin{pmatrix} f_0 & -f_{N-1} & \dots & -f_1 \\ f_1 & f_0 & \dots & -f_2 \\ \vdots & \vdots & \ddots & \vdots \\ f_{N-1} & f_{N-2} & \dots & f_0 \end{pmatrix} \begin{pmatrix} g_0 \\ g_1 \\ \vdots \\ g_{N-1} \end{pmatrix} \]

This works because each row $j$ corresponds to one output term $x^j$, and the cyclic shift for that row accounts for the degree-wrapping, with the sign flip accounting for the negacyclic part. (If there were no sign attached, this method could be used to compute a cyclic polynomial product).

The Python code for this is

def cylic_matrix(c: numpy.array) -> numpy.ndarray:
    """Generates a cyclic matrix with each row of the input shifted.

    For input: [1, 2, 3], generates the following matrix:

        [[1 2 3]
         [2 3 1]
         [3 1 2]]
    """
    c = numpy.asarray(c).ravel()
    a, b = numpy.ogrid[0 : len(c), 0 : -len(c) : -1]
    indx = a + b
    return c[indx]


def negacyclic_polymul_toeplitz(p1, p2):
    n = len(p1)

    # Generates a sign matrix with 1s below the diagonal and -1 above.
    up_tri = numpy.tril(numpy.ones((n, n), dtype=int), 0)
    low_tri = numpy.triu(numpy.ones((n, n), dtype=int), 1) * -1
    sign_matrix = up_tri + low_tri

    cyclic_matrix = cylic_matrix(p1)
    toeplitz_p1 = sign_matrix * cyclic_matrix
    return numpy.matmul(toeplitz_p1, p2)

Obviously on most hardware this would be less efficient than an FFT-based method (and there is some relationship between circulant matrices and Fourier Transforms, see Wikipedia). But in some cases—when the polynomials are small, or one of the two polynomials is static, or a particular hardware choice doesn’t handle FFTs with high-precision floats very well, or you want to take advantage of natural parallelism in the matrix-vector product—this method can be useful. It’s also simpler to reason about.

Until next time!

Polynomial Multiplication Using the FFT

Problem: Compute the product of two polynomials efficiently.

Solution:

import numpy
from numpy.fft import fft, ifft


def poly_mul(p1, p2):
    """Multiply two polynomials.

    p1 and p2 are arrays of coefficients in degree-increasing order.
    """
    deg1 = p1.shape[0] - 1
    deg2 = p1.shape[0] - 1
    # Would be 2*(deg1 + deg2) + 1, but the next-power-of-2 handles the +1
    total_num_pts = 2 * (deg1 + deg2)
    next_power_of_2 = 1 << (total_num_pts - 1).bit_length()

    ff_p1 = fft(numpy.pad(p1, (0, next_power_of_2 - p1.shape[0])))
    ff_p2 = fft(numpy.pad(p2, (0, next_power_of_2 - p2.shape[0])))
    product = ff_p1 * ff_p2
    inverted = ifft(product)
    rounded = numpy.round(numpy.real(inverted)).astype(numpy.int32)
    return numpy.trim_zeros(rounded, trim='b')

Discussion: The Fourier Transform has a lot of applications to science, and I’ve covered it on this blog before, see the Signal Processing section of Main Content. But it also has applications to fast computational mathematics.

The naive algorithm for multiplying two polynomials is the “grade-school” algorithm most readers will already be familiar with (see e.g., this page), but for large polynomials that algorithm is slow. It requires $O(n^2)$ arithmetic operations to multiply two polynomials of degree $n$.

This short tip shows a different approach, which is based on the idea of polynomial interpolation. As a side note, I show the basic theory of polynomial interpolation in chapter 2 of my book, A Programmer’s Introduction to Mathematics, along with an application to cryptography called “Secret Sharing.”

The core idea is that given $n+1$ distinct evaluations of a polynomial $p(x)$ (i.e., points $(x, p(x))$ with different $x$ inputs), you can reconstruct the coefficients of $p(x)$ exactly. And if you have two such point sets for two different polynomials $p(x), q(x)$, a valid point set of the product $(pq)(x)$ is the product of the points that have the same $x$ inputs.

\[ \begin{aligned} p(x) &= \{ (x_0, p(x_0)), (x_1, p(x_1)), \dots, (x_n, p(x_n)) \} \\ q(x) &= \{ (x_0, q(x_0)), (x_1, q(x_1)), \dots, (x_n, q(x_n)) \} \\ (pq)(x) &= \{ (x_0, p(x_0)q(x_0)), (x_1, p(x_1)q(x_1)), \dots, (x_n, p(x_n)q(x_n)) \} \end{aligned} \]

The above uses $=$ loosely to represent that the polynomial $p$ can be represented by the point set on the right hand side.

So given two polynomials $p(x), q(x)$ in their coefficient forms, one can first convert them to their point forms, multiply the points, and then reconstruct the resulting product.

The problem is that the two conversions, both to and from the coefficient form, are inefficient for arbitrary choices of points $x_0, \dots, x_n$. The trick comes from choosing special points, in such a way that the intermediate values computed in the conversion steps can be reused. This is where the Fourier Transform comes in: choose $x_0 = \omega_{N}$, the complex-N-th root of unity, and $x_k = \omega_N^k$ as its exponents. $N$ is required to be large enough so that $\omega_N$’s exponents have at least $2n+1$ distinct values required for interpolating a degree-at-most-$2n$ polynomial, and because we’re doing the Fourier Transform, it will naturally be “the next largest power of 2” bigger than the degree of the product polynomial.

Then one has to observe that, by its very formula, the Fourier Transform is exactly the evaluation of a polynomial at the powers of the $N$-th root of unity! In formulas: if $a = (a_0, \dots, a_{n-1})$ is a list of real numbers define $p_a(x) = a_0 + a_1x + \dots + a_{n-1}x^{n-1}$. Then $\mathscr{F}(a)(k)$, the Fourier Transform of $a$ at index $k$, is equal to $p_a(\omega_n^k)$. These notes by Denis Pankratov have more details showing that the Fourier Transform formula is a polynomial evaluation (see Section 3), and this YouTube video by Reducible also has a nice exposition. This interpretation of the FT as polynomial evaluation seems to inspire quite a few additional techniques for computing the Fourier Transform that I plan to write about in the future.

The last step is to reconstruct the product polynomial from the product of the two point sets, but because the Fourier Transform is an invertible function (and linear, too), the inverse Fourier Transform does exactly that: given a list of the $n$ evaluations of a polynomial at $\omega_n^k, k=0, \dots, n-1$, return the coefficients of the polynomial.

This all fits together into the code above:

  1. Pad the input coefficient lists with zeros, so that the lists are a power of 2 and at least 1 more than the degree of the output product polynomial.
  2. Compute the FFT of the two padded polynomials.
  3. Multiply the result pointwise.
  4. Compute the IFFT of the product.
  5. Round the resulting (complex) values back to integers.

Hey, wait a minute! What about precision issues?

They are a problem when you have large numbers or large polynomials, because the intermediate values in steps 2-4 can lose precision due to the floating point math involved. This short note of Richard Fateman discusses some of those issues, and two paths forward include: deal with it somehow, or use an integer-exact analogue called the Number Theoretic Transform (which itself has issues I’ll discuss in a future, longer article).

Postscript: I’m not sure how widely this technique is used. It appears the NTL library uses the polynomial version of Karatsuba multiplication instead (though it implements FFT elsewhere). However, I know for sure that much software involved in doing fully homomorphic encryption rely on the FFT for performance reasons, and the ones that don’t instead use the NTT.

Carnival of Mathematics #209

Welcome to the 209th Carnival of Mathematics!

209 has a few distinctions, including being the smallest number with 6 representations as a sum of 3 positive squares:

$$\begin{aligned}209 &= 1^2 + 8^2 + 12^2 \\ &= 2^2 + 3^2 + 14^2 \\ &= 2^2 + 6^2 + 13^2 \\ &= 3^2 + 10^2 + 10^2 \\ &= 4^2 + 7^2 + 12^2 \\ &= 8^2 + 8^2 + 9^2 \end{aligned}$$

As well as being the 43rd Ulam number, the number of partitions of 16 into relatively prime parts and the number of partitions of 63 into squares.

Be sure to submit fun math you find in October to the next carvinal host!

Math YouTubers

The Heidelberg Laureate forum took place, which featured lectures from renowned mathematicians and computer scientists, like Rob Tarjan and Avi Wigderson on the CS theory side, as well as a panel discussion on post-quantum cryptography with none other than Vint Cerf, Whitfield Diffie, and Adi Shamir. All the videos are on YouTube.

Tom Edgar, who is behind the Mathematical Visual Proofs YouTube channel, published a video (using manim) exploring for which $n$ it is possible to divide a disk into $n$ equal pieces using a straightedge and compass. It was based on a proof from Roger Nelsen’s and Claudi Alsina’s book, “Icons of Mathematics”.

The folks at Ganit Charcha also published a talk “Fascinating Facts About Pi” from a Pi Day 2022 celebration. The video includes a question that was new to me about interpreting subsequences of pi digits as indexes and doing reverse lookups until you find a loop.

Henry Segerman published two nice videos, including one on an illusion of a square and circle in the same shape, and a preview of a genus-2 holonomy maze (Augh, my wallet! I have both of his original holonomy mazes and my houseguests love playing with them!)

Steve Mould published a nice video about the Chladni figures used (or adapted) in the new Lord of the Rings TV series’ title sequence.

The Simons institute has been doing a workshop on graph limits, which aims to cover some of the theory about things like low-rank matrix completion, random graphs, and various models of networks. Their lectures are posted on their YouTube page.

Math Twitter

Peter Rowlett shared a nice activity with his son about distinct colorings of a square divided into four triangular regions.

Krystal Guo showed off her approach to LiveTeX’ing lectures.

Tamás Görbe gave a nice thread about a function that enumerates all rational numbers exactly once.

Every math club leader should be called the Prime Minister.

In doing research for my book, I was writing a chapter on balanced incomplete block designs, and I found a few nice tidbits in threads (thread 1, thread 2). A few here: Latin squares were on Islamic amulets from the 1200’s. The entire back catalog of “The Mathematical Scientist” journal is available on Google Drive, and through it I found an old article describing the very first use of Latin squares for experimental design, in which a man ran an experiment on what crop was best to feed his sheep during the winter months in France in the 1800’s. Finally, I determined that NFL season scheduling is done via integer linear programming.

Math Bloggers

Lúcás Meier published a nice article at the end of August (which I only discovered in September, it counts!) going over the details of his favorite cryptography paper “Unifying Zero-Knowledge Proofs of Knowledge”, by Ueli Maurer, which gives a single zero-knowledge protocol that generalizes Schnorr, Fiat-Shamir, and a few others for proving knowledge of logarithms and roots.

Ralph Levien published a blog post about how to efficiently draw a decent approximation to the curve parallel to a given cubic Bezier curve. He has a previous blog post about fitting cubic Beziers to data, and a variety of other interesting graphics-inspired math articles in between articles about Rust and GPUs.