Skip to main content

The (Almost) Secret Algorithm Researchers Used to Break Thousands of RSA Keys

https://algorithmsoup.wordpress.com/2019/01/15/breaking-an-unbreakable-code-part-1-the-hack/



RSA encryption allows for anyone to send me messages that only I can decode. To set this up, I select two large random primes p and q (each of which is hundreds of bits long), and release their product x = p \cdot q online for everyone to see; x is known as my public key. In addition, I pick some number e which shares no factors with p-1 or q-1 and release it online as well.
The beauty of RSA encryption is that using only the information I publicly released, anyone can encode a message they want to send me. But without knowing the values of p and q, nobody but me can decode the message. And even though everyone knows my public key x = p \cdot q, that doesn’t give them any efficient way to find values for p or q. In fact, even factoring a 232-digit number took a group of researchers more than 1,500 years of computing time (distributed among hundreds of computers).
On the surface, RSA encryption seems uncrackable. And it might be too, except for one small problem. Almost everyone uses the same random-prime-number generators.
A few years ago, this gave researchers an idea. Suppose Bob and Alice both post public keys online. But since they both used the same program to generate random prime numbers, there’s a higher-than-random chance that their public keys share a prime factor. Factoring Bob’s or Alice’s public keys individually would be nearly impossible. But finding any common factors between them is much easier. In fact, the time needed to compute the largest common divisor between two numbers is close to proportional to the number of digits in the two numbers. Once I identify the common prime factor between Bob’s and Alice’s keys, I can factor it out to obtain the prime factorization of both keys. In turn, I can decode any messages sent to either Bob or Alice.
Armed with this idea, the researchers scanned the web and collected 6.2 million actual public keys. They then computed the largest common divisor between pairs of keys, cracking a key whenever it shared a prime factor with any other key. All in all, they were able to break 12,934 keys. In other words, if used carelessly, RSA encryption provides less than 99.8\% security.
At first glance this seems like the whole story. Reading through their paper more closely, however, reveals something strange. According to the authors, they were able to run the entire computation in a matter of hours on a single core. But a back-of-the-envelope calculation suggests that it should take years to compute GCD’s between 36 trillion pairs of keys, not hours.
So how did they do it? The authors hint in a footnote that at the heart of their computation is an asymptotically fast algorithm, allowing them to bring the running time of the computation down to nearly linear; but the actual description of the algorithm is kept a secret from the reader, perhaps to guard against malicious use. Within just months of the paper’s publication, though, follow-up papers had already discussed various approaches in detail, both presenting fast algorithms (see this paper and this paper), and even showing how to use GPUs to make the brute-force approach viable (see this paper).
There’s probably a lesson here about not bragging about things if you want them to stay secret. But for this post I’m not interested in lessons. I’m interested in algorithms. And this one turns out to be both relatively simple and quite fun.
Algorithm Prerequisites: Our algorithm will deal with integers having an asymptotically large number of digits. Consequently, we cannot treat addition and multiplication as constant-time operations.
For n-bit integers, addition takes O(n) time. Using long multiplication, multiplication would seem to take O(n^2) time. However, it turns out there is an algorithm which runs in time O(n \log n \log \log n).
Computing the GCD naively using the Euclidean algorithm would take O(n^2 \log n \log \log n) time. Once again, however, researchers have found a better algorithm, running in time O(n \log^2 n \log \log n).
Fortunately, all of these algorithms are already implemented for us in GMP, the C++ big-integer library. For the rest of the post we will use Big-O-Tilde notation, a variant of Big-O notation that ignores logarithmic factors. For example, while GCD-computation takes time O(n \log^2 n \log \log n), in Big-O-Tilde notation we write that it takes time \widetilde{\text{O}}(n).
Transforming the Problem: Denote the set of public RSA keys by k_1, \ldots, k_n, where each key is the product of two large prime numbers (i.e., hundred digits). Note that n is the total number of keys. Rather than computing the GCD of each pair of keys, we can instead compute for each key k_i the GCD of it and the product of all the other keys \prod_{t \neq i} k_t. If a key k_i shares exactly one prime factor with other keys, then this provides that prime factor. If both prime factors of k_i are shared with other keys, however, then the computation will fail to actually extract the individual prime factors. This case is probably rare enough that it’s not worth worrying about.
The Algorithm: The algorithm has a slightly unusual recursive structure in that the recursion occurs in the middle of the algorithm rather than at the end.
At the beginning of the algorithm, all we have is the keys,
k_1,
k_2,
k_3, \cdots
The first step of the algorithm is to pair off the keys and compute their products,
j_1 = k_1 \cdot k_2,
j_2 = k_3 \cdot k_4,
j_3 = k_5 \cdot k_6, \cdots
Next we recurse on the sequence of numbers j_1, \ldots, j_{n / 2}, in order to compute
r_1 = GCD(j_1, \prod_{t \neq 1} j_t),
r_2 = GCD(j_2, \prod_{t \neq 2} j_t),
r_3 = GCD(j_3, \prod_{t \neq 3} j_t), \cdots
Our goal is to compute s_i = GCD(k_i, \prod_{t \neq i} k_t) for each key k_i. The key insight is that when i is odd, s_i can be expressed as
s_i = GCD(k_i, r_{(i + 1) / 2} \cdot k_{i + 1}),
and that when i is even, s_i can be expressed as
s_i = GCD(k_i, r_{i / 2} \cdot k_{i - 1}).
To see why this is the case, one can verify that the term on the right side of the GCD is guaranteed to be a multiple of GCD(k_i, \prod_{t \neq i} k_t), while also being a divisor of \prod_{t \neq i} k_t. This, in turn, implies that the GCD-computation will evaluate to exactly GCD(k_i, \prod_{t \neq i} k_t), as desired.
Computing each of the s_i‘s in terms of the r_i‘s and k_i‘s completes the algorithm.
Bounding the Running Time: Let m denote the total number of bits needed to write down k_1, k_2, \ldots, k_n. Each time the algorithm recurses, the total number of bits in the input being recursed on is guaranteed to be no more than at the previous level of recursion; this is because the new inputs are products of pairs of elements from the old input.
Therefore each of the O(\log n) levels of recursion act on an input of total size O(m) bits. Moreover, the arithmetic operations within each level of recursion take total time at most \tilde{O}(m). Thus the total running time of the algorithm is also \tilde{O}(m) (since the O(\log n) recursion levels can be absorbed into the Big-O-Tilde notation).
If we unwrap the running time into standard Big-O notation, we get
O(m \log^3 m \log \log m).
Is it practical? At first glance, the triple-logarithmic factor might seem like a deal breaker for this algorithm. It turns out the actual performance is pretty reasonable. This paper found that the algorithm takes time roughly 7.65 seconds per thousand keys, meaning it would take a little more than 13 hours to run on 6.2 million keys.
One of the log factors can be removed using a slightly more clever variant of the algorithm, which avoids GCD computations at all but the first level of recursion (See this paper). The improved algorithm takes about 4.5 seconds per thousand keys, resulting in a total running time of about 7.5 hours to handle 6.2 million keys.
So there we go. A computation that should have taken years is reduced to a matter of hours.  And all it took was a bit of clever recursion.

Comments

Popular posts from this blog

The Difference Between LEGO MINDSTORMS EV3 Home Edition (#31313) and LEGO MINDSTORMS Education EV3 (#45544)

http://robotsquare.com/2013/11/25/difference-between-ev3-home-edition-and-education-ev3/ This article covers the difference between the LEGO MINDSTORMS EV3 Home Edition and LEGO MINDSTORMS Education EV3 products. Other articles in the ‘difference between’ series: * The difference and compatibility between EV3 and NXT ( link ) * The difference between NXT Home Edition and NXT Education products ( link ) One robotics platform, two targets The LEGO MINDSTORMS EV3 robotics platform has been developed for two different target audiences. We have home users (children and hobbyists) and educational users (students and teachers). LEGO has designed a base set for each group, as well as several add on sets. There isn’t a clear line between home users and educational users, though. It’s fine to use the Education set at home, and it’s fine to use the Home Edition set at school. This article aims to clarify the differences between the two product lines so you can decide which

Let’s ban PowerPoint in lectures – it makes students more stupid and professors more boring

https://theconversation.com/lets-ban-powerpoint-in-lectures-it-makes-students-more-stupid-and-professors-more-boring-36183 Reading bullet points off a screen doesn't teach anyone anything. Author Bent Meier Sørensen Professor in Philosophy and Business at Copenhagen Business School Disclosure Statement Bent Meier Sørensen does not work for, consult to, own shares in or receive funding from any company or organisation that would benefit from this article, and has no relevant affiliations. The Conversation is funded by CSIRO, Melbourne, Monash, RMIT, UTS, UWA, ACU, ANU, ASB, Baker IDI, Canberra, CDU, Curtin, Deakin, ECU, Flinders, Griffith, the Harry Perkins Institute, JCU, La Trobe, Massey, Murdoch, Newcastle, UQ, QUT, SAHMRI, Swinburne, Sydney, UNDA, UNE, UniSA, UNSW, USC, USQ, UTAS, UWS, VU and Wollongong.

Building a portable GSM BTS using the Nuand bladeRF, Raspberry Pi and YateBTS (The Definitive and Step by Step Guide)

https://blog.strcpy.info/2016/04/21/building-a-portable-gsm-bts-using-bladerf-raspberry-and-yatebts-the-definitive-guide/ Building a portable GSM BTS using the Nuand bladeRF, Raspberry Pi and YateBTS (The Definitive and Step by Step Guide) I was always amazed when I read articles published by some hackers related to GSM technology. H owever , playing with GSM technologies was not cheap until the arrival of Software Defined Radios (SDRs), besides not being something easy to be implemented. A fter reading various articles related to GSM BTS, I noticed that there were a lot of inconsistent and or incomplete information related to the topic. From this, I decided to write this article, detailing and describing step by step the building process of a portable and operational GSM BTS. Before starting with the “hands on”, I would like to thank all the pioneering Hackers and Researchers who started the studies related to previously closed GSM technology. In particul