# Tag Info

72

In the first decade of the 21th century, and counting, on a given $\text{year}$, no RSA key bigger than $(\text{year} - 2000) \cdot 32 + 512$ bits has been openly factored other than by exploitation of a flaw of the key generator (a pitfall observed in poorly implemented devices including Smart Cards). This linear estimate of academic factoring progress ...

26

Let's assume for an instant that you could build a large table of all primes. Then... what ? How would you use it ? What would you look up ? If you "just" scan the table and try to divide the number to factor by each prime, then this is known as trial division; there is no need to store the primes (they can be regenerated on-the-fly; that's the division ...

20

Those appear to be based on the complexity of the General Number Field Sieve, one of the fastest (if not the fastest) classical factoring algorithms. I confirmed this in Mathematica. Here is the complexity for the GNFS (source): $$\exp\left( \left(\sqrt[3]{\frac{64}{9}} + o(1)\right)(\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}}\right)$$ where $n$ is a ...

19

You might want to look at NIST SP800-57, section 5.2. As of 2011, new RSA keys generated by unclassified applications used by the U.S. Federal Government, should have a moduli of at least bit size 2048, equivalent to 112 bits of security. If you are not asking on behalf of the U.S. Federal Government, or a supplier of unclassified software applications to ...

19

No, it is not at all feasible to build an index of prime factors to break RSA. Even if we consider 384-bit RSA, which was in use but breakable two decades ago, the index would need to include a sizable portion of the 160 to 192-bit primes, so that the smallest factor of the modulus has a chance to be in the index. Per the Prime number theorem there are in ...

16

The three main general-purpose algorithms for factorization are the quadratic sieve (QS), the elliptic curve method (ECM) and the number field sieve (NFS). On Complexity The running time of these algorithms is expressed with the L-notation: $L_n[a,c]$ means that the asymptotic complexity of factoring a number $n$ is $O(e^{(c+o(1))(\log n)^a(\log \log n)})^{... 14 If$p=2q+1$is a safe prime (that is,$q$is a prime as well), then$p-1=2q$has exactly two prime factors:$2$and$q=(p-1)/2$. 12 The generic discrete logarithm problem is this: Given a group$(G, ·)$with generator$g$and$y \in G$, find$x \in \mathbb N$such that$y = g^x$. The "classic" discrete logarithm problem (the one used in "classic" DSA and ECDSA) is this with some subgroup of the (multiplicative) group of a prime field, i.e.$(\mathbb Z/p \mathbb Z)^*$: Given a ... 11 RSA themselves (the company) say this in their RSA Factoring Challenge FAQ: Why is the RSA Factoring Challenge no longer active? Various cryptographic challenges — including the RSA Factoring Challenge — served in the early days of commercial cryptography to measure the state of progress in practical cryptanalysis and reward researchers for the new ... 10 Yes there are other hard problems you can base asymmetric cryptography on. Lattices. The NTRU systems is based on the shortest vector problem in ideal lattices. Lattice-based cryptography is of much interest these days for two reasons: (1) unlike factorization and discrete logarithms, there isn't an efficient algorithm for breaking these problems on a (... 10 If the RSA keys were generated randomly, then it is inconceivable that two different devices would happen to pick the same key. Taking 2048 bit RSA keys as an example, there are approximately$2^{1014}$1024 bit primes; if we consider them pairwise (and realise that about half the pairs yield a 2047 bit number), that means there are about$2^{2026}$RSA ... 10 Breaking such a scheme is easy. Suppose Alice wants to transmit a message$M$to Bob. First thing, Alice picks an integer$R_a$and sends the cipher text$C_a = M \times R_a$to Bob. Bob then picks another integer$R_b$and transmits$C_b = C_a \times R_b$back to Alice. Alice calculates$D_a = \frac{C_b}{R_a}$and sends$D_a$to Bob. Bob calculates$D_b = ...

9

There is an asymptotic formula for the General Number Field Sieve for factoring big integers. This is the most efficient known algorithm for breaking RSA keys which are longer than 400 bits or so (since the current world record is 768 bits, a 400-bit RSA key is quite weak). For discrete logarithm (to break DH), the best known algorithm is also known as "...

9

I don't believe a lower bound has ever been proven for the "fewest" number of bits needed. Coppersmith showed, however, how given either the $n/4$ least or $n/4$ most significant bits of $p$ where $n$ is the size of the modulus $N=pq$, $N$ can be efficiently factored. Additionally, given the $n/4$ least significant bits of $d$, one can reconstruct $d$ (and ...

9

Actually, the problem is that the above quote uses the term "discrete log" in a way that's different from what you're thinking of. When someone uses the term "discrete log", they can mean two things: A discrete log in the group $Z^*_p$; that is, given $p$, $g$ and $g^x \bmod p$, recover $x$ A discrete log in some other group; that is, given a group $G$, a ...

9

No, there is no specific vulnerability associated to choosing $p$ and $q$ with size differing by $i$ bits (or $2\cdot i$ bits as in the statement) for small $i$. However, if $i$ gets too big: That improves the odds that ECM will manage to factor $n$ for some fixed size of $n$, and at some point ECM will become the best algorithm; this is the case if $i$ is ...

9

No, it is not collision-free. All possible sequences of 0's produce the same output: 0 --> (2 ** 0) = 1 00 --> (2 ** 0) * (3 ** 0) = 1 000 --> (2 ** 0) * (3 ** 0) * (5 ** 0) = 1 0000 --> (2 ** 0) * (3 ** 0) * (5 ** 0) * (7 ** 0) = 1 In fact, it can be seen that $f(s) = f(s||0)$, for every bit-string $s$. This could be easily solved by ...

8

"Discrete logarithm" is a wide class. Originally, this means that we work in a finite field (e.g. integers modulo a big prime) and, given $g$, $p$ and $g^x \bmod p$, it is computationally difficult to recover $x$ (it becomes impossible with today's technology once $p$ is big enough). At some point, someone noticed that discrete logarithm was a special case ...

8

Let me try a simple explanation of NFS. I will necessarily skip lots of details, but I hope you will get the main ideas. The number field sieve algorithm (NFS) is a member of a large family: index calculus algorithms. All algorithms in the family, which can be used for factoring and discrete logarithms in finite fields, share a common structure: ...

8

It's probably best to understand Lenstra's Elliptic Curve factorization algorithm by way of contrast with its predecessors, the Pollard's p-1 method, the Williams' p+1 method and the Cyclotomic Polynomial method of Bach and Shallit. These are all Algebraic-group factorisation algorithms which require you to select a stage 1 bound $B_1$ and stage 2 bound $... 8 No, it not possible to attack RSA (and practical modulus size) with a WalkSat derivative, as far as we know, or using the algorithm in the question. Problem with that algorithm is: in order to have a sizable/constant rate of success as$n$increases, we have to repeat steps 2 and 3 not the stated$t\cdot m^2$times, but rather$t\cdot 2^m$times. That's ... 8 Does the factorization of N somehow help me? It sure does. I think I could compute the logarithm modulo each prime and then combine it, but do not know how exactly. Seems similar like problems for Chinese remainder theorem but I cannot find the way how to do it. You're real close; you do recombine them using the Chinese Remainder Theorem; however the ... 7 There are close to$2.27\cdot10^{613}$primes smaller than$2^{2048}$. There are close to$9.4\cdot10^{79}$atoms in the observable universe. Even if you could compute all those prime numbers, you wouldn't have any place to store them... 7 For numbers over about 115 (decimal) digits, the best algorithm currently known in the General Number Field Sieve (GNFS – sometimes just called the Number Field Sieve, though there's also a Special Number Field Sieve for factoring numbers of a special form). The GNFS, unfortunately, is an exceedingly complex algorithm, and I don't know of any online ... 7 We are dealing here with two assumptions and we always refer to$N=pq$the product of two$n$bit primes$p$and$q$. The one is factoring and the second is the so called RSA assumption which is a formalization of "inverting the RSA function is hard". These two assumption can be described as follows: Factoring assumption: There exists no probabilistic ... 7 Yes, they are (deterministically) equivalent. The original RSA paper (Section IX.C), working off Miller's results (Theorem 3), showed how knowing the secret exponent$d$was probabilistically equivalent to factoring$n$. Later, using more advanced techniques, Coron and May showed how to deterministically reduce finding$d$to factoring$n$. 7 Loosely define the RSA problem as solving for$x$the equation$c=x^e\bmod N$, with$x$random in$\{0,1,\dots,N-1\}$(or equivalently$c$random in this set), and$(N,e)$properly chosen. The best method we know for tackling that problem is factoring$N$, but we have no proof that there is no substantially better method. We do not know if the RSA problem ... 6 Pure nonsense. For choosing the random$\Delta$between$\sqrt{\min(N, Ň)}$and$\sqrt{\max(N, Ň)}$there are too many possibilities for it to work. For example whenever the first and last digits of$N$differ, you get something like$\frac{1}{10} \cdot \sqrt N$possibilities (the exact formula doesn't matter). So you can replace the first formula$gcd[N, (...

6

Yes, using Miller-Rabin with a random witness does give a practical factoring method. When you run the Miller-Rabin algorithm, it can end in one of three ways: The final value is not 1; this case causes Miller-Rabin to output "Composite" An intermediate value was not 1 or N-1, but the next value was 1; this causes Miller-Rabin to output "Composite" The ...

6

Yes, what you have is enough to recover p in both cases. In the first case you just need to write the proper equation to be then solved thanks to Coppersmith method to find small roots of univariate polynomial. As explained by Alexander May in pages 40 and 41 of his thesis what you ask is always doable if the unknown bits are consecutive (and you have at ...

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