# Tag Info

13

Decrypt the ciphertext with every possible key and store the result: $2^{56}$ decryptions. Now encrypt the (known) plaintext of the ciphertext with every possible key: $2^{56}$ encryptions. Now you have to check every entry, which is in both lists and try it with another plaintext-ciphertext pair. If you can successfully decrypt that, you are very likely to ...

11

Most of these algorithms (i.e. the block ciphers DES, Triple DES, AES, Blowfish) are normally only working on a fixed block size, and take approximately the same time independently of input, thus they are $O(1)$. If you put them into a mode of operation to encrypt longer messages, you usually get an $O(m)$ complexity, where $m$ is the message size, as you ...

11

A proof of P = NP would prove that one-way functions do not exist. That in turn would imply, that almost no secure cryptographic primitives can exist according to the accepted definitions of security. (No symmetric encryption, no MACs, no pseudorandom generators, no signature schemes, ...) However, it would just mean that no scheme can be provably secure. ...

10

As you probably know $f(\lambda)=O(\lambda^4)$ means that $|f|$ asymptotically upper bounded by some constant times $\lambda^4$. The notation $f(\lambda)=\Omega(\lambda^4)$ corresponds to an asymptotic lower-bound. Now, the $\tilde O$ and $\tilde \Omega$ are closely related notations, where we not only ignore constants but also values which are polynomial ...

10

There is no direct inference from $P = NP$ or $P \neq NP$ to security or insecurity of any particular encryption algorithm. As far as practical consequences are concerned, the "$P = NP$" problem is severely overhyped. If $P = NP$ then any problem for which a solution can be verified in polynomial time can also be solved in polynomial time. "Polynomial time" ...

10

Do the post-quantum ciphers also automag/tically address the 1st problem? Not really, however to explore that in any detail, we need to explore what the 1st problem is. If $P=NP$ is proven true, what does that practically mean? Well, it might have absolutely no practical ramifications, or it might mean that virtually all known cryptographical systems can ...

9

Perfect secrecy is achievable in a few cases, such as one-time pads, and, well, that's pretty much it. Most cryptographic protocols are vulnerable to an all-powerful, all-knowing attacker. If you do not put any restriction on what the attacker can do, then Guess the key. Profit. breaks almost any cryptography, as does Wave a magic wand. Profit. So at ...

8

There are techniques for doing online surveys on sensitive subjects. They don't follow the approach you outlined, but here's a sketch of how they work. Suppose we want to survey people to determine how many people have ever seriously considered suicide (say), but we suspect many people might be unwilling to answer honestly because of the stigma associated ...

8

The last major effort I know of for cracking keys was the Distributed.net effort. You can find the project page at http://www.distributed.net/RC5/en. In 2002, they cracked a 64-bit RC5 key using at total of 331,252 computers over 1,757 days. Their maximum throughput was "equivalent to 32,504 800MHz Apple PowerBook G4 laptops or 45,998 2GHz AMD Athlon XP ...

8

Let $n = \lceil \log q \rceil$ (with "$\log$" being the base-2 logarithm, so $n$ is the size, in bits, of $q$). If $q$ is a prime integer (i.e. $\mathbb{F}_q$ is the field of integers modulo $q$), then classical implementations will have cost $O(n)$ for addition and subtraction, $O(n^2)$ for multiplications and divisions. The cost of multiplications can be ...

8

Presumably, it's because they rounded it down to a nice round number of bits. Nobody's going to use an 86.76611925028119 bit key in practice, but an 80-bit key is plausible. Besides, the 86.whatever bit symmetric key length is only approximate, anyway: even using the GNFS, implementation details could easily swing it several bits either way, and of course, ...

8

What Dan Boneh says is not a formal definition as you want it. Let me quote Rogaway on this: In cryptographic practice, a collision-resistant hash-function (also called a collision-free or collision-intractable hash-function) maps arbitrary-length strings to fixed-length ones; it’s an algorithm $H:\{0,1\}^*\rightarrow \{0,1\}^n$ for some fixed $n$...

8

I know an algorithm that runs in polynomial time would be able to break an RSA key pair "quickly". But how quickly is "quickly"? No way to say, it might be microseconds, and it might be large multiplies of the age of the universe. When we say that an algorithm runs in polynomial time, we're not saying anything about how fast the algorithm runs given any ...

8

Just looking for a Turing machine vs circuit is a bit misleading. The important distinction is uniform (complexity class BPP) vs non-uniform (complexity class P/poly) adversaries. You can characterize P/poly in terms of circuit families, but also in terms of Turing machines with arbitrary "advice strings." In fact, the latter is the more traditional ...

7

Well, one assumption you appear to be making is that, with 2DES, there will be approximately $2^{56}$ possible key matches. Actually, there are an expected $2^{48}$ possible key matches; here's why: Let us assume we're running the meet-in-the-middle attack on 2DES, and consider an arbitrary incorrect encryption trial (that is, we try an encryption key that ...

7

When $n$ is prime, solving for $e$-th roots modulo $n$ is easy, since it suffices to compute $d = e^{-1} \pmod {n-1}$ and then $s = m^d \pmod n$. If $n$ is not prime, but is instead a RSA modulus (a composite integer that is the product of two big primes), then the problem becomes apparently hard (in the sense that we don't have a clue how to do it ...

7

Actually, it's not true that public key encryption is based on Discrete Log; the ones in common use (DH, ECDH, ECDSA) are (and even RSA can be viewed as "based on Discrete Log", at least from the standpoint of "if you can solve the Discrete Log modulo a composite, you can break RSA"). However, we do have a number of public key systems (NTRU, McEliece) which ...

7

This begs the question, why would you in any real-world circumstance wish to reduce the difficulty for an attacker to break your cryptosystem? To answer your question practically, the only reasonable way I can think of to accomplish this is to simply reduce the entropy in the key. At 100%, all 128 bits of the key are used. At 50%, 64 bits of the key are ...

7

Is AES solvable in this way? In other words, will the algorithm eventually complete, producing the correct key? Almost yes. It will produce some correct key — there might be more than one. (It should quite plausibly be unique given "enough" plaintext-ciphertext samples, but this need not be the case in general.) Generally, computing the key in a known-...

6

Well, if $q$ is a prime (and not $p^n$ with $n>1$), then addition, subtraction and multiplication can be performed by doing the traditional operations modulo $q$, that is: $a +_{\mathbb{F}_q}b \equiv (a+b) \bmod q$ $a -_{\mathbb{F}_q}b \equiv (a-b) \bmod q$ $a \times_{\mathbb{F}_q}b \equiv (a\times b) \bmod q$ As such, addition and subtraction can be ...

6

I'll expand on the comment I left on my answer. The purpose of Part 2 of NIST SP 800-57 is to "[provide] guidance on policy and security planning requirements for U.S. government agencies". Keeping that in mind, the table on page 64, i.e. the table from whence the numbers in that question came, includes more than just RSA key sizes. Namely, it includes some ...

6

First, it's not said that AES is unbreakable, merely that none of the currently known attacks reduce the computational cost to a point where it's feasible. The current best attack on AES-128 takes 2^126.1 operations, if we had a computer (or cluster) several million times more efficient than any current computer and could operate at the thermodynamic ...

5

I've previously answered this question over at How will security need to be changed if P=NP? (on our sister site, the IT Security Stack Exchange). In addition, see the answers to What would be the scenario if P = NP for RSA algorithm? for still more on the subject. The short answer is that a proof that P=NP doesn't necessarily mean that all cryptography is ...

5

I don't know the general answer, however, it appears that Baby-step Giant-step is able to give you the solution in $O(\sqrt{in})$ time (where $n$ is the size of $G$); this is $O(\sqrt{i})$ times longer than it takes the same algorithm to solve a single discrete-log problem. The first observation is that if you know the group order $n$ and a group generator $... 5 On the third case, I have a comment. The third oracle may help the adversary using Cheon's algorithm for the DL problem. Let$q$be a prime order of the subgroup$\mathbb{G}$of$(\mathbb{Z}/p\mathbb{Z})^{\times}$. In the third case, the adversary has an oracle$a \mapsto a^k$for any$a$. Hence, it can obtain$g^{k^i}$from$g^{k^{i-1}}$and so on. When$...

5

The Handbook on Applied Cryptography (link to the pdf version is on Alfred's webpage) has some of the known techniques to do finite field arithematic. If you are doing arithmetic to implement Elliptic Curve Cryptography (note the comment made by Paulo), then there are methods that depends on whether you are doing it in Jacobian or Projective plane (inverse ...

5

Summary. The short answer is: Cryptography would be insecure. Any encryption you can do with a non-deterministic algorithm, can be broken (in approximately the same running time) by another non-deterministic algorithm. Non-determinism is extremely powerful. If you give everyone access to non-determinism, then secure encryption becomes impossible: the ...

5

How on earth did you arrive at that formula? You can break a Caesar cipher by calculating the result of applying all of the $n-1$ (i.e., 25) possible shifts to the ciphertext and picking the one that makes sense. The computational complexity is just $\mathcal{O}(n)$. If you want to automate the process based on frequency analysis, the correlation step ...

5

If one-way functions exist, then there is a distribution over graphs (or SAT formulas, or ...) having the property you're asking for. In short, just put the OWF through the Cook-Levin reduction. In a little more detail, Cook-Levin transforms the NP witness-finding question "what is a preimage of $y = f(x)$?" (for random unknown $x$) into the NP witness-...

5

Is the running times of corresponding steps true? No. Step 3 of the dealer has to be executed $n$ times (once for each party) with each execution taking $O(t)$ time. So it must be $O(t\cdot n)$. Step 4 of the dealer needs $O(n)$ to distribute each share to every party. I count $O(t\cdot n)$ as the overall time complexity for the dealer. Of course, you ...

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