Learning to SQI

Implementing SQISign in SageMath

Maria Corte-Real Santos, Giacomo Pope

Estimating Bounds for KLPT

The KLPT algorithm takes an input left-$\OO_0$ ideal $I$ and computes an equivalent ideal $J$ with norm dividing some smooth integer $\TT$. At a high level, this is accomplished with the following steps:

  1. Compute an ideal $L \sim I$ with prime norm: $n(L) = N$.
  2. Compute $\gamma \in \BB$ with reduced norm: $\Nrd(\gamma) = L_1 N$ with $L_1 \mathrel{|} \TT $.
  3. Compute $\mu \in \BB$ such that $\gamma \cdot \mu \in L$.
  4. Compute $\nu \in \BB$ with $\Nrd(\nu) = L_2$ with $L_1 L_2 \mathrel{|} \TT $, using $\mu$.
  5. Compute $\beta = \gamma \cdot \nu \in L$ with reduced norm $\Nrd(\beta) = L_1 L_2 N$.
  6. Compute $J = \chi_L(\beta)$ with $n(J) = L_1 L_2$ such that $J \sim L \sim I$.

In this discussion, we will not detail how all these steps work, but rather how we choose the bounds we place on deriving the values $L_1$ and $L_2$ such that we can heuristically rely on the algorithm to return $J$ with high probability.

Applying KLPT to SQISign, we want $L_1 L_2$ to be as small as possible to ensure that each step is efficient. Most importantly, we need this to divide the available torsion on the curve so we do not have to work with extension fields. For SigningKLPT(), $L_1 L_2 = \ell^e$ is fixed from the parameter set, for security considerations, but keeping the output norm small is important, as this controls the degree of the response isogeny $\sigma$, and this computation is by far the most expensive in the SQISign protocol.

Searching for small primes

Although it is maybe not obvious from the steps outlined above, making $n(J)$ small is all wrapped up in finding the smallest possible value of $n(L) = N$.

The trick to solving Step 1 is computing the Minkowski reduced basis of the input ideal: $I = \langle \beta_1, \beta_2, \beta_3, \beta_4 \rangle$ and taking random linear combinations of these basis elements until we find some $\alpha \in I$ with norm $\Nrd(\alpha) = n(I) \cdot N$. We can then compute the equivalent ideal $L \sim I = \chi_I(\alpha)$ with norm $n(L) = N$.

Note: for random input, we would expect all elements of the reduced basis to have reduced norm $\Nrd(\beta_i) \simeq \sqrt{p}$, and so we expect the norm of $L$ to be $n(L) = N \simeq \sqrt{p}$. However, in SQISign, we occasionally have that ideals have particularly small norm. When this happens, the reduced basis is orthogonal with two elements of exceptionally small reduced norm. In these cases, it can be impossible to find prime norm ideals with $n(L) < p$. This is discussed in more detail on the page Equivalent Prime Norm Ideals.

For this discussion, let us write $p \simeq 2^{2\lambda}$ and assume we have obtained some ideal $L$ with prime norm $N \simeq 2^{\lambda + \delta}$, where $\delta$ is some small positive value. The question is, how large can $\delta$ be before we are likely to find a solution? We wish to pick $\delta$ small enough that we have a small output for $n(J)$, but large enough that we are overwhelmingly likely to find solutions for the equivalent prime norm ideal and hence for the KLPT algorithm.

Finding a quaternion algebra element with given norm

The next step is to find $\gamma$ with norm $M = L_1 \cdot N$. This is accomplished with RepresentIntegerHeuristic() and requires that the norm $M > p$. To find $\gamma$, we have to be able to find a solution $M’ = x^2 + y^2$ using Cornacchia’s algorithm. On random input, we expect this to take roughly $\log(p)$ attempts, and so a good rule of thumb is to ensure that $M > p\log(p) \simeq 2^{2\lambda + \epsilon}$. Assuming that $N \simeq 2^{\lambda + \delta}$ we then have a bound for $L_1 \simeq 2^{\lambda + \epsilon - \delta}$ where we expect $\epsilon \simeq \log\log (p)$.

Practically, we do this by taking the input target norm $\TT$ and factoring it. We then randomly take factors from $\TT$ for use in $L_1$ until $M$ is large enough such that we expect to find a solution for $\gamma$. We then update $\TT^\prime = \TT / L_1$ for the following steps.

Finding a solution to the strong approximation

Step four is where $N$ comes back into focus. Our goal is to find some $\nu$ with norm $L_2$ such that $L_1 L_2$ divides the target norm $\TT$. From the estimates above, we then have a maximal size for $L_2$ given $L_1$ computed earlier:

$$ L_{2}^{\text{Max}} = \TT / L_1 \simeq \TT \cdot 2^{\delta - \lambda - \epsilon}. $$

Using the lattice method described in Strong Approximation Lattice Trick, we know that an approximate minimum bound for the reduced norm is $\Nrd(\nu) \simeq pN^3$. This means practically we require $\Nrd(\nu) = L_2 \geq pN^3$ for there to be a reasonable chance at finding a close vector.

Combining these two bounds, we must have that:

$$ L_{2}^{\text{Max}} \simeq \TT \cdot 2^{\delta - \lambda - \epsilon} \geq pN^3 \simeq 2^{5\lambda + 3\delta}, $$

which allow us to represent our heuristic bounds in terms of the target norm $\TT$:

$$ \TT > 2^{6\lambda + 2\delta + \epsilon} \simeq p^3 \cdot 2^{2\delta + \epsilon}. $$

Concrete bounds for SQISign

When running the KLPT algorithm in SQISign, we use it to find ideals with norm that divides the available odd torsion. Further tricks described on the page Computing Isogenies from Ideals allow us to square this such that $\TT = T^2$. We can then estimate our bounds by considering:

$$ \frac{T^2}{p^3} > 2^{2\delta + \epsilon}, $$

where both $p$ and $T$ are fixed constants for the protocol. For the SQISign prime $p_{6983}$, we have that $p \simeq 2^{256}$ and $T \simeq 2^{428}$. Using the bounds above, we have that $T^2 p^{-3} \simeq 2^{89}$. Allowing $\epsilon = \log\log(p) \simeq 8$ for solving $\textsf{RepresentInteger}_{\OO_0}$ we find the maximum possible value which allows solutions would be $\delta = 40$.

Practically, we find that allowing $\delta \simeq \epsilon \simeq \log\log(p)$ seems to work with good probability, however fine-tuning these constants will require more testing. Allowing $\delta$ to be smaller than its maximum makes finding the prime norm a little harder, but makes finding a solution to the strong approximation a little easier. As we randomly sample $L_i$, the bounds input into the strong approximation are more fuzzy, so it appears helpful to save some bits by shrinking $\delta$.

As described on the page Equivalent Prime Norm Ideals, there are a few additional tricks which help keep the size of $N$ small, so setting $\delta \simeq \log\log(p)$ seems fairly robust.

Setting the signing length of SQISign

For the SQISign identification protocol to be zero-knowledge, the output of response() must have fixed length $\ell^e$.

The output of response() is a degree $\ell^e$ isogeny, which is derived from a cyclic ideal $J$ of norm $n(J) = \ell^e$. The ideal $J$ is found by running SigningKLPT(), a generalisation of the KLPT algorithm which allows as input an ideal $I$ with generic left order $\OO$ providing that a connecting ideal $I_\tau$ with left order $\OO_0$ and right order $\OO$ is known.

We decompose $e$ as $e_1 + e_2$, where $L_1 = \ell^{e_1}$ and $L_2 = \ell^{e_2}$. The goal is to minimise $e$ for efficiency of the protocol, while allowing it to be large enough that heuristically we always find an equivalent ideal from SigningKLPT().

Setting $e_1$: In SQISign, we have a connecting ideal $I_\tau$ of prime norm, so we need not find an equivalent prime norm ideal. Following the efficient keygen() from appendix D of the SQISign paper, the norm is also particularly small: $n(I_\tau) \simeq p^{1/4}$. For the input ideal $I$, we will need an equivalent prime norm ideal $L \sim I$ which we expect to have norm $n(L) \simeq \sqrt{p}$. We can set $e_1$ such that $L_1 \cdot n(L) > p$, and it very likely that we find $\gamma \in \BB$ with norm $L_1 \cdot n(L)$.

Setting $e_2$: As we want $e$ to be fixed, we then just allow $e_2 = e - e_1$, but how small can we make $L_2$ while still finding solutions? The size of $L_2$ becomes important when we attempt to solve a generalisation of the strong approximation where the modulus is a composite integer: $N = n(L) \cdot n(I_\tau)$.

Using the lattice trick described in Strong Approximation Lattice Trick, we expect to find solutions of size $L_2 > pN^3$. Using the estimates above, we have $L_1 > p^{1/2}$ and $L_2 > p^{13/4}$, so we should have $e > \frac{15}{4} \log_\ell(p)$. For the 256-bit prime used in SQISign this would restrict us to $e > 960$ for a theoretical bound and $e = 1000$ is a comfortable heuristic bound.

In practice, we set this bound for $e$ in setup.py as a global variable. Then, when running SigningKLPT(), we set a sensible value for $e_1$ once we find $n(L)$ for $L \sim I$. Finally, we set $e_2 = e - e_1$ and attempt to solve the strong approximation for fixed $L_2$. If successful, we find an ideal $J$ of norm $\ell^e$ equivalent to the input ideal $I$.

Fixing the signing length

In fact, there is a little additional complication in setting $e_2$. When we compute an ideal using the KLPT algorithm, the output ideal is not necessarily cyclic. So although we may have some ideal $J$ such that $n(J) = \ell^e$, once we scale the ideal to ensure $J$ is cyclic (this is necessary such that the corresponding isogeny is cyclic), we may find $J’$ has norm $n(J’) = \ell^{e-\epsilon_0}$, for some small $\epsilon_0$.

To deal with this, SigningKLPT() attempts to “overshoot” $L_2$ and pick some $\epsilon_0$ so we have $e_2 = e - e_1 + \epsilon_0$. The output from the KLPT algorithm with then be some ideal of norm $n(J) = \ell^{e + \epsilon_0}$ and we hope that the cyclic ideal $n(J’)$ has norm exactly $\ell^e$.

Note: we were unable to derive a way to ensure that the output had the correct norm. So ultimately we guess $\epsilon_0$ as best as we can, and if $J’$ has the wrong norm, we just run SigningKLPT() again. This does not add too much overhead to the computation time, mainly because of how expensive the isogenies are currently.

Our current technique is to take

$$ \epsilon_0 = 2 \left\lfloor \frac{\log\log(p)}{4} \right\rfloor + \epsilon(\gamma). $$

Where $\epsilon(\gamma)$ is a dynamic correction we make depending on the derived $\gamma \in \BB$ with $\Nrd(\gamma) = L_1 n(L)$.

First, we write $\gamma$ in the basis of $\OO_0$ and compute $g$, the $\gcd$ of its coefficients. We then compute the $\epsilon(\gamma) = \gcd(g, L_1)$. This might be easier to follow in code, so we include the estimate below:

def quaternion_change_basis(γ, O):
    Computes the coefficients of a quaternion γ
    in the basis of a given order O
    O_matrix = Matrix([b.coefficient_tuple() for b in O.basis()])
    γ_vector = vector(γ.coefficient_tuple())
    γO_coeffs = γ_vector * O_matrix.inverse()

    assert γ == sum([a * b for a, b in zip(γO_coeffs, O.basis())])
    return γO_coeffs

def quarternion_basis_gcd(γ, O):
    Computes the gcd of the coefficients of a
    quaternion γ in the basis of a given order O
    γO_coeffs = quaternion_change_basis(γ, O)
    return gcd(γO_coeffs)

def derive_L2_SigningKLPT(γ, L1, e1):
    Given L1 = l^e1 and γ try and compute L2
    so that the output of SigningKLPT has norm
    exactly 2^e
    g = quarternion_basis_gcd(γ, O0)
    extra = 2 * (floor(loglogp / 4) + ZZ(gcd(g, L1).valuation(l)))
    e2 = e - e1 + extra
    return l**e2

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