# On the hardness of finding minimal-weight codewords in a QC-MDPC code

Problem 1  (Knapsack)

We want to maximize the value of a subset of items such that the weight of the set is no larger than weight $W$.

$\sum_{i \in \mathcal{S}} w_i x_i \leq W$
$\max \sum_{i \in \mathcal{S}} v_i x_i$

Let us first study another problem. It is easy to see that this problem is at least as hard as Problem 1.

Problem 2  (Auxiliary problem)

$\max x_1x_2x_3 + x_2x_5 + x_4x_6x_7 + \cdots$
$\|\begin{matrix} x_1 & x_2 & \cdots & x_k \end{matrix} \|_1 = m, x_i \in \{0, 1\}$

We can solve Problem 1 using an oracle for Problem 2. For a knapsack instance with an item of weight $w_i$ and weight $v_i$, we create a $v_i$ terms $\cdot x_1x_2\cdots x_{w_i}$. The $x_i$ must be unique for every item. This is done for every item in the knapsack. So, Problem 2 is NP-hard.

Now, we want to build some theory towards polynomials. Consider the following construction. Assign each $x_i$ with a corresponding monomial $(x - a_i)$. Now assume that we want to represent the term $x_1 x_2 x_3$. Define $p(x) = x^p + 1 \in \mathbb{F}_q$ to be a product $\prod_{i=0}^p(x-a_i)$. $p$ is a prime number.

We map the term $x_1 x_2 x_3$ to the polynomial

$Q_1(x) = p(x) \cdot \left[(x-a_1)(x-a_2)(x-a_3)\right]^{-1}$

This means that $Q_1(x) \cdot (x-a_1)(x-a_2)(x-a_3) = 0 \bmod p(x)$. To consider all terms jointly, pick a number large number $d$. Then, we create a polynomial using the terms $Q_2(x) = p(x) \cdot \left[(x-a_2)(x-a_5)\right]^{-1}$$Q_3(x) = p(x) \cdot \left[(x-a_4)(x-a_6)(x-a_7)\right]^{-1}$ and so on. We assume there are $n$ terms.

Then, the polynomial $Q(x)$ is constructed as $Q(x) = \sum_{i=1}^n x^{di} \cdot Q_i(x)$. Since $d$ was a large number, it means that the terms of $Q_i(x)$ and $Q_{i+1}(x)$ will not interfere.

We now turn this problem into an instance of finding a low-weight codeword in a QC-MDPC code.

Problem 3  (Low-weight codeword of QC-MDPC code)

Let $G$ be the generator of a QC-MDPC code. Find a non-zero codeword with minimal weight, i.e., minimize $\| \mathbf u G \|_1$.

There is a bijective mapping from a quasi-cyclic code to a polynomial, i.e., $G(x)$ generates a code as well. So, $Q(x)$ generates a quasi-cyclic code. Let us first assume we have access to an oracle for Problem 2.

Indeed, a optimal solution to Problem 2 will give rise to a codeword in the corresponding code of low weight. Assume that the optimum is $x_1x_2x_3 + x_2x_5 + x_4x_6x_7 + \cdots = b$. This means that $u(x)$ is of degree $m$ and that $\|u(x) G(x)\|_1 \leq p(n-b)$, i.e., $u(x) G(x)$ has at most $p (n-b)$ non-zero coefficients.

Using an oracle for Problem 3, we find a minimal-weight codeword for the generator $Q(x)$.

# Experiments with index calculus

In index calculus, the main problem is to represent powers of $g$ in a predefined prime-number basis. We are interested in unravel $x$ from $h = g^x\bmod p$.

Normal approach

First, we find some offset $hg^j = g^{x+j}\bmod p$ such that the factorization is $B$-smooth.

Using the prime-number basis $(-1)^{e_1}2^{e_2}\cdots$, generate a corresponding vector

$\mathbf y = \begin{pmatrix}1 & 0 & 3 & 0 & \cdots & 1\end{pmatrix}$

Then, we generate powers $g^1, g^2, ...$ and check if they have the same property. The results are put into a matrix after which one performs Gaussian eliminiation over $\mathbb{Z}_{p-1}$.

$A | \mathbf v^T = \left( \begin{array}{cccccc|cc}0 & 3 & 2 & 2 & \cdots & 1 & \mathbf{13} \\ 1 & 0 & 2 & 1 & \cdots & 0 & \mathbf{112} \\ 0 & 5 & 2 & 0 & \cdots & 0 & \mathbf{127} \\\vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & 1 & 0 & 0 & \cdots & 0 & \mathbf{67114}\end{array}\right)$

$\mathbf xA = \mathbf y$. Then compute $\mathbf x \cdot \mathbf v$ to find $x+j$.

Different approach

Almost like in the normal approach, we find some offset $hg^j = g^{x+j} \bmod p$ such that the factorization of at least fraction of $hg^j \bmod p$ greater than $\sqrt{p}$ is $B$-smooth.

Again, Using the prime-number basis $(-1)^{e_1}2^{e_2}\cdots$, generate a corresponding vector

$\mathbf y | \Delta = \left(\begin{array}{cccccc|c}0 & 2 & 5 & 1 & \cdots & 0 & \Delta\end{array} \right)$

The vector is chosen such that $\Delta$ corresponds to the product of primes not represented in the chosen basis. In other words, $(-1)^0 \cdot 2^2 \cdot 3^5 \cdot 7^1 \cdots > \sqrt{p}$ for the vector above, or equivalently, $\Delta < \sqrt p$.

Again, we generate powers $g^1, g^2, ...$ and check if they have the same property as above and solved as the normal approach.

$A | \mathbf v^T | \mathbf d^T = \left( \begin{array}{cccccc|c|c} 1 & 0 & 3 & 0 & \cdots & 0 & \mathbf{5} & \delta_1 \\ 0 & 3 & 2 & 2 & \cdots & 1 & \mathbf{13}& \delta_2 \\ 0 & 1 & 2 & 1 & \cdots & 0 & \mathbf{65}& \delta_3 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots\\ 0 & 2 & 0 & 0 & \cdots & 0 & \mathbf{13121}& \delta_B \end{array}\right)$

Find $\mathbf xA = \mathbf y$. Then compute $\mathbf x \cdot \mathbf v$ to find $x+j$. There is a catch here. It will be correct if and only if

$\prod \delta_i ^{x_i} \bmod p = \Delta.$

It remains to see if this actually improves over the normal approach.

# Suggestion for proof of retrievability

(Originally posted here)

Scenario

In this scenario, $A$ and $B$ wants to distribute segments of their data (preferably encrypted). We do not care about coverage, we are trying to maximize the amount of remotely stored data. The game is essentially that any peer, will try to minimize their own storage requirements and thus maximize other peers’.

Protocol

Two peers $A$ and $B$ will repeatedly negotiate upon storage space according to the follwing protocol, where $A$ is prover and $B$ verifier. Let $S(B,A)$ be the set of segments owned by $B$ and stored by $A$. Moreover, let $s \in S(B,A)$ be an abritrary segment and let $\text{sig}(s)$ denote the signature of $s$.

1. $B$ picks a random string $r$ of length $n$ bits (equal to the length of the signature)
2. $B$ asks $A$ to send the closests segment from $S(B,A)$, i.e., such that $\| r - \text{sig}(s) \|_1$ is minimized.
3. $B$ verifies the segment $s$ via the signature. Then, using the distance $\| r - \text{sig}(s) \|_1$, $B$ can estimate the number of segments stored by $A$.

Repeat until the variance is low enough. The procedure is performed both ways.

Approximating $\delta$

Intuitively, we can think of $S(B,A)$ as a random (non-linear) code $\mathcal{C}$ over $\mathbb{F}_2^n$ and the task is to find a minimum-weight codeword in $\mathcal{C} + r = \{ c + r | c \in \mathcal C\}$. Then $N$ corresponds to the number of codewords in $\mathcal{C}$.

Assume that A has stored $N$ segments. Without loss of generality, we can assume that $r$ is the all-zero string, hence transforming the problem to finding the minimum weight of $n$ binomial variables. Let $X_1, X_2, \dots, X_N \in \textsf{Bin}(n, \frac12)$ and $Z := \min(X_1, X_2, \dots, X_n)$. Then with high probability and for larger values of $N$,

$\frac n2 - \sqrt{\frac n3 \log N} \geq E(Z) \geq \frac n2 - \sqrt{\frac n2 \log N}.$

The above can be obtained by Gaussian approximation and the conventional expectation of the minimum of Gaussian random variables. We omit the proof here. Of course, this can be pre-computed for reasonable values of $N$.

Assume there exists a peer $A$ with infinite computing power that tries to minimize its required storage for a given minimum distance $d$. Done in an optimal manner, $A$ will pick equidistant points in the space $\mathbb{F}_2^n$. This corresponds to a perfect error-correcting code with minimum distance $d$. The covering-coding bound states that $d\leq n-k$. For a perfect code, which has minimal size, we have $d = n-k$. Therefore, the size of the code is $|\mathcal C| = 2^{n-d}$, which is also the number of elements $A$ needs to store.

A glaring problem is that $A$ cannot arbitrarily pick points (since points must be valid signatures). The best $A$ can do is to decide upon a perfect code $\mathcal{C}$ (or pick one that is closest to the current set of points — which is a very hard problem). Then, $A$ will look for points $v$ and $v'$ that map to the same codeword $c \in \mathcal{C}$ and discard the point that is the farthest away from $c$. For larger $n$ and smaller $d$, it is not realistic to achieve the covering-coding bound since there in reality are not that many shared segments. So, even given infinite computation, $A$ will not be able to sample points until a perfect code is achieved.

The covering-coding bound serves as a crude estimation on how to set the parameters.

Simulating

In Figure $1$, we see that the upper bound for expected minimum distance approximates $N$ for values larger than $2^7$ and $n = 256$.

Assume that $A$ send a signature $s$ along with the segment. Let

$\delta(r,s) = \|r - \text{sig}(s)\|_1 - \frac n2.$

$B$ can now use different bounds at its own choice. Assume that $\delta(r,s) = 100 - \frac{256}{2}.$ Then, using the expression

$N \approx \exp\left(-\frac 2n \cdot \delta(r,s) \cdot \text{abs}\left(\delta(r,s)\right) \right)$

$B$ determines that $A$ most likely has at least $458$ segments with high probability. Using the lower bound, $A$ has at least $9775$ segments.

So, what is all this?

The intention is to create a simplistic protocol for asserting that shared data actually is retrievable, based on a probabilistic model. This could serve as basis for shared storage, in which segments are distributed. Even if the storer does not have access to the segments it has distributed, it can use the protocol to verify that it can retrieve them without actually doing so.

# Writeup for Snurre128

The intended solution for Snurre128 is as follows. We note that the non-linear function is almost linear, the higher-order terms have degree 5.

return v[0] ^ v[1] ^ v[2] ^ v[31] ^ \
v[1]&v[2]&v[3]&v[64]&v[123] ^ \
v[25]&v[31]&v[32]&v[126]


Whenever a higher-order term becomes non-zero, we will get non-linear behaviour. Also note that since there are two of these terms, they will cancel out with some small probability. We write $f = L(x) + P(x) + Q(x)$. Because all indices in the non-linear terms ($\{1,2,3,64,123\} \cap \{25,31,32,126\} = \emptyset$) are disjoint, we get

$\begin{array}{rcl}\mathbb{P}(P(x) + Q(x) = 1) &=& \mathbb{P}(P(x) \neq Q(x))\\ &=& \mathbb{P}(P(x) = 1) + \mathbb{P}(Q(x) = 1) - 2\cdot \mathbb{P}(P(x) = 1 \land Q(x) = 1)\\ &=& 2 \cdot 2^{-5} - 2 \cdot 2^{-10}\\& =& \frac{31}{512} \end{array}$

Let $k$ be the dimension of the state and $n$ be the length of the keystream. Note that there exists a big-ass matrix $\mathbf{G} \in \mathbb{F}_2^{k\times n}$ such that

$x\mathbf{G} + e = v \in \mathbb{F}_2^n,$

where $e$ is governed by the non-linearity of $f$. The expected Hamming weight of $e$ is

$n \cdot \mathbb{P}(P(x) + Q(x) = 1).$

It is possible to solve this using a Walsh transform, but it is rather expensive requiring $\mathcal{O}(k \cdot 2^k)$ computation. Employing BKW will significantly reduce the complexity, by transforming the problem into a problem of smaller dimension at the expense of higher error rate.

Another solution is to treat this as a decoding problem over a random code. There is support for so-called information-set decoding in sage.

from sage.coding.information_set_decoder import LeeBrickellISDAlgorithm


First we generate the generator matrix from the cipher, according to the above.

# define a code from big-ass matrix
C = codes.LinearCode(G.transpose())


The keystream is a vector $\mathbb{F}_2^n$:

v = vector(GF(2), [0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1])


Using Lee-Brickell’s algorithm (a rather simple and not the most efficient algorithm of today), we can now decode:

num_errors = 118
A = LeeBrickellISDAlgorithm(C, (num_errors, num_errors))
A.calibrate()
x = A.decode(v)
# for the record, this is the error vector
e = v + A*x


Once this is complete, we can solve a set of linear equations to unravel the state.

Also consider to read this writeup by @hellman 🙂

# Solving snurre80 @ MidnightsunCTF

Note: I was the author of this challenge. This is my intended solution. Some details have been left out.

Two teams solved this one: TokyoWesterns and LC↯BC, in that chronological order.

class Snurre80:

"""
Snurre80 is a proprietary stream cipher designed for
low-power devices, i.e., in Internet of Things contexts.
More importantly, it offers an incredibly high security
level of 2^80 (who would need more?).

It has the following structure:

+--------+----+------+
|        |    |      |
+---+---+-   -+---+    |
|   |   | ... |   |  c
+-----------------+

Snurre80 is resistant to distinguishing attacks, since
it uses a non-linear(tm) boolean function f as filtering
output function.

Snurre80 is quantum resistant and blockchain ready. It
is the stream cipher of the future. It is 100 % cyber.
We charge a reasonble fee of $0.00001 / encrypted bit. -- The Designers """ def __init__(self, key): self.state = key self.mask = 1284576224436276739441733 self.nbits = self.mask.bit_length()-1 def output(self): var = bin(self.state)[2:].zfill(self.nbits) v = [int(v) for v in var] return v[0] ^ v[1] ^ v[2] ^ v[31] ^ \ v[1]&v[2]&v[3] ^ \ v[25]&v[31]&v[32]&v[33]&v[34] def __str__(self): j = 0 poly = [] x = self.mask while x > 0: if x & 1: poly = ["x^{}".format(j)] + poly x >>= 1 j += 1 return " + ".join(poly) def keystream(self, n): for _ in xrange(n): self.state = (self.state self.nbits if xor != 0: self.state ^= self.mask yield self.output() # Generate a sequence of 800 bits, with a random key. key = int(os.urandom(10).encode('hex'), 16) cipher = Snurre80(key) z = [c for c in cipher.keystream(800)] # Additionally, this may help you solve the CAPTCHA. def solve_proof_of_work(prefix): i = 0 while True: if sha256(prefix+str(i)).digest()[:3] == "\x00\x00\x00": return str(i) i += 1  The feedback polynomial of the stream cipher is $\begin{array}{rcl} P(x) & = & x^{80} + x^{76} + x^{66} + x^{64} + x^{58} + x^{54} + x^{50} + x^{42} \\ & + & x^{38} + x^{34} + x^{24} + x^{22} + x^{18} + x^{10} + x^6 + x^2 + 1 \end{array}$ To create a distinguisher, we may use the feedback polynomial as parity check. Unfortunately, the output is not linear, but we can approximate it as a linear function. However, by the piling-up lemma, the bias quickly becomes small since there are many terms. So, we need to find a low-weight polynomial multiple to avoid getting too small bias: there exists $q(x)$ such that $q(x) P(x) = u(x)$, where $u(x)$ is of low weight. But wait. Every power is even, so it means it is a square $P(x) = p^2(x)$. Turns out we can easily find the square root of it; by simply dividing every power with $2$, we can find $p(x)$ from $P(x)$. $\begin{array}{rcl} p(x) & = & x^{40} + x^{38} + x^{33} + x^{32} + x^{29} + x^{27} + x^{25} + x^{21} \\ & + & x^{19} + x^{17} + x^{12} + x^{11} + x^{9} + x^{5} + x^3 + x + 1 \end{array}$ Now, we can solve the problem of finding a low-weight multiple in a smaller dimension. This can be done in a couple of seconds. In fact, if you stalked my Github repos you will find that I have written code for this exact problem. The idea behind it is to generate a long list of powers $x^i \bmod P(x)$ and use a generalized birthday attack to find $x^{i_1} + x^{i_2} + x^{i_3} + x^{i_4} = 0 \bmod P(x)$. There are several papers on it. This is not as difficult as it may be time consuming to write functional code. Nonethless, one suitable parity check is $x^{17399} + x^{13567} + x^{4098} + 1$ But it will not work for the polynomial $P(x)$? Well, that if no concern. If $q(x) p(x) = u(x)$, then $q^2(x) p^2(x) = q^2(x) P(x) = u^2(x)$. Since $u(x)$ has weight $w$, $u^2(x)$ will have weight $w$.  p = [17399, 13567, 4098] S = sum( z[i + 2*p[0]] ^ z[i + 2*p[1]] ^ z[i + 2*p[2]] ^ z[i] for i in range(0, 1000)) return S < 430 # appropriate magic constant  Running it for every sequence, we can efficiently decide the source. This gives the flag midnight{1z_3z_2_BR34K_W1D_L0W_w31gHTz!!!} # Setting up Google authenticator along SSH auth in macOS This is a guide on how to setup the use of Google Authenticator along with public-key authentication. First, we clone the Google Authenticator PAM module from Github: $ git clone https://github.com/google/google-authenticator-libpam.git

To build it, we need a few packages which are not included by default in macOS.

$brew install autoconf automake libtool To be able to get QR codes without revealing secrets to Google, you can install libqrencode: $ brew install libqrencode

$./bootstrap$ ./configure
$make$ sudo make install

To install the PAM module, invoke

$sudo cp /usr/local/lib/security/pam_google_authenticator.so /usr/lib/pam/ Then, we add the line auth required pam_google_authenticator.so to the configuration file /etc/pam.d/sshd. It will look something like this: # sshd: auth account password session # Not used by me! # auth optional pam_krb5.so use_kcminit # auth optional pam_ntlm.so try_first_pass # auth optional pam_mount.so try_first_pass # auth required pam_opendirectory.so try_first_pass # Relevant methods of authentication: auth required pam_google_authenticator.so account required pam_nologin.so account required pam_sacl.so sacl_service=ssh account required pam_opendirectory.so password required pam_opendirectory.so session required pam_launchd.so session optional pam_mount.so  I have removed the alternative methods of authentication. In /etc/ssh/sshd_config, we set PubkeyAuthentication yes ChallengeResponseAuthentication yes UsePAM yes AuthenticationMethods publickey,keyboard-interactive:pam  This will force the user to prove ownership of a valid SSH-key and along with a verification code from Google Authenticator. Now we are ready to setup the two-factor authentication. Running $ /usr/local/bin/google-authenticator


generate a shared secret and provide you with a QR code to be used with your phone. You can use TOTP or OTP verification codes. I suggest going with TOTP and small time windows.

Another good thing is to add some notifications to your ~/.ssh/rc file (also — if you are in the US, there are a lot of services available for sending SMS, which can be handy, but remember not to send any sensitive data through third parties). For instance:

# Get and sanitize sender address, probably not needed...
ip=echo $SSH_CONNECTION | cut -d " " -f 1 ip=${ip//[^a-zA-Z0-9:.]/}

# Sanitize user, probably not needed...
user=whoami
user=${user//[^a-zA-Z0-9_]/} # Show notification osascript -e 'display notification "SSH connection as '$user' from '$ip'" with title "SSH"' # Send email echo "SSH connection as '$user' from '\$ip'" | sendmail me@mydomain.com


This will give a notification to you directly if you are at the computer being connected to. Additionally, it will send an email to you just in case you are AFK. The regexp are there to mitigate possible command injection, though I doubt it is possible unless there is severe bug in SSHD and the script above is running as a user with higher privileges. On the other hand, santitation is never bad if it negligible in terms of computations and you are doing it right 🙂

Another way of sending notifications is to use a Telegram bot. See below for an example output.

For this purpose, one might use e.g. telegram-send.

# Finding close-prime factorizations

This is a proposal for a simple factoring algorithm which actually performs better than regular attacks based on lattice-type algorithms. Although more efficient algorithms exists, this one is very simple to implement, while the more efficient ones are not. Do not quote me on this, though ;-D

## Finding a good approximation

The first step is to compute an good approximation of $\phi(n)$. Denote this $\phi'$. One such good approximation is $\phi' = n - 2\sqrt{n} + 1$. Let us see why. First, assuming that $n = pq$, note that

$\phi(n) = (p-1)(q-1) = n - (p+q) + 1$

Let us assume there exists a small $\gamma$ such that $p = a + \gamma$ and $q = a - \gamma$. So, we have that

$\delta := \phi' - \phi(n) = 2\sqrt{a^2 - \gamma^2} - 2a = 2a\sqrt{(1 - \gamma^2/a^2)} - 2a.$

Now, let $\beta = \gamma^2/a^2$. Then,

$\delta =2a\left(\sqrt{1 - \beta} - 1\right) = 2a \cdot \left[\beta/2 + \mathcal{O}(\beta^2)\right]$.

Since $\beta$ is assumed to be small ($a >> \gamma$), we can approximate it as

$\delta \approx \gamma^2/a.$

Great! So the running time is bounded by $\mathcal{O}( \gamma^2/a)$ if we use pure brute force. Now, we are going to be just a little more clever than that, aren’t we?

## Searching efficiently

We will now show how to search for a solution faster than brute force. First, we pick an arbitrary invertible element $y$ and compute all powers of $y \bmod n$ up to a certain parameter $b$, i.e.,

$\mathcal B := \{y^0 \bmod n, y^1 \bmod n, \dots , y^b \bmod n\}$

Obviously, this requires $\mathcal O(b)$ time and memory. A reference implementation in Python (with $y = 2$) would look something like.

# create a look-up table
look_up = {}
z = 1
for i in range(0, b + 1):
look_up[z] = i
z = (z * 2) % n


Now, compute

$\mu := (y^{\phi'})^{-1} \bmod n.$

Note that the exponent of $\mu$ is

$-\phi' \bmod \phi(n)$.

We now do a BSGS-type of algorithm (as noted by hellman in the comments, we can use Pollard’s kangaroo algorithm to avoid exhausting precious memory without sacrificing computational effort, up to polynomial factors). Generate all powers $y^{bi} \bmod n$ for  $0 \leq i \leq b$ and multiply with $\mu$ forming $c:= y^{bi} \cdot \mu$. Again, we get an exponent which is

$-\phi' + bi \bmod \phi(n).$

For each such computed, check against $\mathcal{B}$ is it exists. If so, we have found $\phi(n)$.

# check the table
mu = gmpy.invert(pow(2, phi_approx, n), n)
fac = pow(2, b, n)
j = 0

while True:
mu = (mu * fac) % n
j += b
if mu in look_up:
phi = phi_approx + (look_up[mu] - j)
break
if j > b * b:
return


This is $\mathcal O(b)$ in time. Once $\phi(n)$ has been found, we can trivially compute the factors in negligible time as

# compute roots
m = n - phi + 1
roots = (m - gmpy.sqrt(m ** 2 - 4 * n)) / 2, \
(m + gmpy.sqrt(m ** 2 - 4 * n)) / 2


Cool! Since we require that $\gamma^2/a \leq b^2$ for a solution to be found, the running time is $\mathcal O(\gamma/a^{1/2})$ and with equal amount of memory required.

## A real-world example (sort of)

This little finding inspired me to create a CTF challenge. During the SEC-T CTF 2017, one of the challenges (qproximity) could be solved using this method.

Description

We are given a public key and an encrypted message. The key is

-----BEGIN PUBLIC KEY-----
iravDlD4vUlVk9XK79/fwptVzYsjimO42+ZW5VmHF2AUXaPhDC3jBaoNIoa78CXO
ft030bR1S0hGcffcDFMm/tZxwu2/AAXCHoLdjHSwL7gxtXulFxbWoWOdSq+qxtak
zBSZ7R1QlDmbnpwdAgMDEzc=
-----END PUBLIC KEY-----


Extracting the modulus, we find that

$\begin{array}{rl} n = &246264974647736414345408205036830544055349190030438864689361084738619 \\ &430136992438500973098730365134506003143847829773369403632725772343146 \\ &864922044439763512753030199250563829152109289871491767838931495698391 \\ &860322173235862868025625353744920431228772475066920885663471105686331 \\ &5465220853759428826555838536733 \end{array}$

The factors are found as follows

def close_factor(n, b):

# approximate phi
phi_approx = n - 2 * gmpy.sqrt(n) + 1

# create a look-up table
look_up = {}
z = 1
for i in range(0, b + 1):
look_up[z] = i
z = (z * 2) % n

# check the table
mu = gmpy.invert(pow(2, phi_approx, n), n)
fac = pow(2, b, n)
j = 0

while True:
mu = (mu * fac) % n
j += b
if mu in look_up:
phi = phi_approx + (look_up[mu] - j)
break
if j > b * b:
return

m = n - phi + 1
roots = (m - gmpy.sqrt(m ** 2 - 4 * n)) / 2, \
(m + gmpy.sqrt(m ** 2 - 4 * n)) / 2

return roots

n = 2462649746477364143454082050368305440553491900304388646893610847386194301369924385009730987303651345060031438478297733694036327257723431468649220444397635127530301992505638291521092898714917678389314956983918603221732358628680256253537449204312287724750669208856634711056863315465220853759428826555838536733
b = 10000000

close_factor(n, b)


The code runs in matter of seconds and gives the factors

$\begin{array}{rl} p = &156928319511723700336966188419841178779822856669047426731288203806365 \\ &781043486977181195772183437407585036370841822568787188893070958887580 \\ &0968904667752571 \end{array}$

and

$\begin{array}{rl} q = &156928319511723700336966188419841178779822856669047426731288203806365 \\ &838576177312234882203162849666809217683250766146317773098495921211843 \\ &5829588059735623 \end{array}$

Using these, we construct the flag

SECT{w3ll_th4t_wasnt_2_h4rd?}