This approach easily captures to symmetric and public key cryptography, hash function, stream cipher and the like. The difficult part is finding a good encoding that a modern SAT solver can actually crack.

Encode the input/output relation of a cryptographic algorithm into a propositional formula.In this way, the bit values of the plaintext, the ciphertext, and possibly the key (if used) are encoded into propositional variables and the formula is true if and only if the ciphertext corresponds to the plaintext encrypted with the key.Use a SAT solver to find out the solution.

We invented logical cryptanalysis in 1999 (See also the full list of papers farther below)

- Massacci, Fabio, and Laura Marraro. "Logical cryptanalysis as a SAT problem." Journal of Automated Reasoning 24.1-2 (2000). (PDF)

However, some researchers started using it for finding collisions on the MD4 and MD5 hash functions

- Mironov, Ilya, and Lintao Zhang. "Applications of SAT solvers to cryptanalysis of hash functions." In International Conference on Theory and Applications of Satisfiability Testing, 2006. (PDF).

- Marc Stevens, Elie Bursztein, Pierre Karpman, Ange Albertini, Yarik Markov. The first collision for full SHA-1 2017.

[...] but all these attempts led to results that not only were unsatisfactory but that even threatened the feasibility of the second near-collision attack.[...]Our final solution was to encode this problem into a satisfiability (SAT) problem and use a SAT solver to find a drop-in replacement differential path over the first eight steps that is solvable.

More specifically, we adapted the SHA-1 SAT system generator from Nossum.

- Vegard Nossum. "SAT-based preimage attacks on SHA-1", Master Thesis, Institutt for informatikk, University of Oslo, 2012.

If you are interested in the history of the field you might continue reading here.

- Symmetric crypto: namely the encoding of the (former) Data Encryption Standard
- Asymmetric crypto: namely the generation of signatures of RSA with a fixed public exponent 3 (recommended in a former version of the PKCS signature standard).

- F. MASSACCI Using walk-SAT and rel-sat for cryptographic key search. In
*Proceedings of the 16th International Joint Conference on Artificial Intelligence (IJCAI’99)*(1999), T. Dean, Ed., Morgan Kaufmann.

http://www.ing.unitn.it/~massacci/papers/mass-99-IJCAI.pdf - F. MASSACCI and L. MARRARO. Logical cryptanalysis as a SAT-problem: Encoding and analysis of the U.SS. Data Encryption Standard.
*Journal of Automated Reasoning 24(1-2)*, (2000). (Preliminary versions in Third LICS Workshop on Formal Methods and Security Protocols (July 1999), NMR-2000, SAT-2000: Highlights of Satisfiability Research at the Year 2000)

http://www.ing.unitn.it/~massacci/papers/mass-marr-00-JAR.pdf - C. FIORINI, E. MARTINELLI, and F. MASSACCI. How to fake an RSA signature by encoding modular root finding as a SAT problem.
*Discrete Applied Mathematics 130(2)*, (2003). (Preliminary version at MFPS-99)

http://www.ing.unitn.it/~massacci/papers/fior-mart-mass-03-DAM.pdf

If you are not familiar on the work of DES, you can find background information in any good handbook of cryptography. If you are more interested just in the encoding, then our papers give you enough background information on the problem. The main thing you need to know is that this cipher works by ``repeating'' an operation (called round) a suitable number of times. The task of a round is to mix and combine the input bits with the secret key. Loosely speaking, the higher the number of rounds, the better. For instance, in the case of DES, after eight rounds all input bits affect all output bits.

Thus, from the view point of the propositional encoding (for key search), the higher the number of rounds, the harder the formula. In contrast, the higher the number of plaintext and ciphertext blocks which are conjoned the more constrained is the search: after conjoining a suitable number of pairs there is only one model.

The problems can be of two kind:

**Key Search with a known or unknown plaintext**- The formula encode the input/output relation of one or more blocks of ciphertexts and known plaintext. If you find a model the first N bits will give you the value of the secret key used for encryption
**Verification of cryptographic properties**- The formula encodes some good property of a cryptographic algorithm (e.g. there is no universal key which can decrypt any message, or there are no two keys that can produce the same plaintext/ciphetext pair). You must prove that the formula is valid.

Sami Liedes sliedes AT cc dot hut dot fi a student from Ilkka Niemela HUT TCSlaboratory has provided a patch that also fixes the scripts (patch of version 0.9).

If you are not familiar with RSA, you can find background information in any good handbook of cryptography. If you are more interested just in the encoding, then our papers give you enough background information on the problem. The main thing you need to know is that this cipher works by suitably generating a triple of numbers $n,d,e$ such that the following equation holds

The pair (E,N) is the public key and the pair (D,N) is the private signing key. Basically to sign a message M≤N you elevate M to the D-th power (modulo N) and that is the signature F (from the italian Firma). when you receive the message signature pair (M,F) one simply checks that the e-th power of f is equal to m (modulo n). A PKCS standard recommended 3 as a possible value for E.

Thus, from the view point of the propositional encoding, we have encoded a relation

where M is given and E=3. So any propositional solution of the encoding yields a fake (but valid) signature of M. Mathematically speaking this is called a cubic-residuosity problem.

The problems can be of two kind:

**signature falsification**- When the triple (E,D,N) is generated according the RSA algorithm there is always a solution to the equation given above and so we have the problem of finding the solution to problem whose satisfiability status is known (yes)
**solution of cubic-residuosity**- If the triple (E,D,N) is arbitrary the equation below may not have a solution for all values of M. Then the problem may be satisfiable (M is a cubic residue modulo N) or unsatisfiable (M is not a cubic residue). For example for N=35, M=2 is not a cubic residue, whereas M=6 is a cubic residue. Smodels takes few seconds to prove or disprove the cubic residuosity of all numbers from M<35.