Cryptography

1. Cryptography Lecture 11Stefan Dziembowskiwww.dziembowski.net stefan@dziembowski.net

2. Plan • Definition of CCA-security • Construction of the CCA-secure schemes • in the private-key settings, • in the public-key settings.

3. Problem How to encode a message m before encrypting it (with RSA, for example)? m decode(v) x := encode(m) v := zd mod N xe mod N z

4. Remember the chosen-plaintext attack? security parameter 1n • selects random(pk,sk) = Gen(1n) • chooses a random b = 0,1 pk oracle challenge phase: chooses m0,m1 m0,m1 c = Enc(pk,mb) has to guess b

5. The PKCS #1 v1.5 encoding We observed that encoding has to be randomized. This is the encoding that we presented: k bytes (k - D - 3) random bytes D bytes

6. Today’s lecture The PKCS #1 v1.5encoding looks ad-hoc... Today we present a more “scientific” encoding. For this, we are going to use a stronger security definition.

7. Chosen Ciphertext Attack (CCA) The adversary may also choose a ciphertext and learn the corresponding plaintext. Does it make sense? • Aren’t we too paranoid? • How to formalize it?

8. Aren’t we too paranoid? No! Bleichenbacher [1998] showed a “practical” chosen ciphertext attack on encoding proposed for the PKCS #1 v.2 standard. [see also: Bleichenbacher, D., Kaliski B., Staddon J., "Recent results on PKCS #1: RSA encryption standard", RSA Laboratories' bulletin #7,ftp://ftp.rsasecurity.com/pub/pdfs/bulletn7.pdf ] Why is Blaichenbacher’s attack practical? Because it assumes that the adversary can get only one bit of information about the plaintext...

9. PKCS #1: RSA Encryption Standard Version 2 public-key: (N,e) let k := length on N in bytes. let D := length of the plaintext requirement: D ≤ k - 11. Enc((N,e), m) := xe mod N, where x is equal to k bytes (k - D - 3) random bytes D bytes

10. Bleichenbacher’s attack – the scenario pk = (N,d) sk = (N,d) c c1 computes x = c1d mod N and checks if x is a correct PKCS #1 v2 encoding yes/no . . . Goal: compute cd mod N ck yes/no Bleichenbacher [1998]: There exists a successful attack that requires k = 220 questions for N = 1024.

11. So, chosen ciphertext attacks are practical! In Bleichenbacher’s attack the adversary could obtain just one bit of information. Conservative approach: assume that he can get the entire plaintext.

12. Idea • Provide a security definition that “covers” this type of an attack. • Propose a scheme that is “provably secure” according to this definition. This will lead to an encoding that is less ad-hoc than PKCS #1 v1.5

13. CCA - security It makes sense to consider CCA-security in • private-key settings • public-key settings more interesting

14. Decryption oracle To define the CCA-security we consider a decryption oracle. c1 sk Dsk(c1) c2 Dsk(c2) Dsk(ci) := error if ci cannot be decrypted . . . ck Dsk(ck)

15. Decryption/encryption oracle Two types of queries: Decrypt ci (sk,pk) Dsk(ci) Encrypt mi Epk(mi) We assume that also CPA is allowed.

16. CCA-security– the game in the private-key settings: pk = sk security parameter 1n • selects random(pk,sk) = Gen(1n) • chooses a random b = 0,1 pk (in the public-key settings) CCA-attack oracle challenge phase: chooses m0,m1 m0,m1 c = Enc(pk,mb) |m0| = |m1| CCA-attack Here Eve cannot ask for decryptionof c. has to guess b

17. CCA-security Alternative name:CCA-secure Security definition: We say that (Gen,Enc,Dec)hasindistinguishable encryptions under a chosen-ciphertext attack (CCA)if any randomized polynomial timeadversary guesses b correctly with probability at most 0.5 +ε(n), whereεis negligible.

18. CCA in practice (1/2) Some actions of the receiver may depend on the decrypted message. For example, the receiver may communicate an error if the message “looks strange”. (like in the Bleichenbacher’s attack)

19. CCA in practice (2/2) m := “Some top-secret information” Alice Alice: c := E(pk,m) replies (in an encrypted way) to Alice quoting m (pk,sk) Alice Eve: c’ replies quoting m’=D(sk,c’) wants to decrypt c Why Eve cannot just set c’ := c ? Because Bob would get suspicious (why message from Eve has Alice’s name inside?)

20. CCA in the private-key settings CCA-security in the private-key settings can be achieved by adding authentication. How to combine authentication with encryption? We already considered this problem some time ago.

21. Authentication and Encryption Options: • Encrypt-and-authenticate: c ← Enck1(m) and t← Mack2 (m) • Authenticate-then-encrypt: t← Mack2 (m) and c ← Enck1(m||t) • Encrypt-then-authenticate: c ← Enck1(m) and t← Mack2 (c) wrong better the best

22. A CCA-secure encryption scheme (Gen,Enc,Dec) – a CPA-secure encryption scheme (GenMAC,Tag,Vrfy) – a MAC. Create a new encryption scheme (Gen’,Enc’,Dec’) where: • Gen’(1n) := (Gen(1n),GenMAC(1n)), • Enc’((k0,k1),m) := (Enc(k0,m), Tag(k1,Enc(k0,m))) • Dec’((k0,k1),m) := decrypt and verify thetag

23. Why is it secure? Intuition An adversary cannot create a new valid pair (Enc(k0,m), Tag(k1,Enc(k0,m))) without knowing k1. So he will always receive an error message from the oracle (unless he replays the ciphertexts that he already received from the oracle – but this gives him no extra information)

24. Is authenticate-then-encryptsecure? Authenticate-then-encrypt: t← Mack2 (m) and c ← Enck1(m||t) Not always! There exists (artificial) counter-examples...

25. The first counter-example Authenticate-then-encrypt: t← Mack2 (m) and c ← Enck1(m||t) Suppose the encryption scheme adds a random bit at the end of the ciphertext. B Enck1(m || Mack2 (m)) Then Enck1(m || Mack2 (m)) neg B is a different ciphertext and the adversary is allowed to ask the oracle to decrypt it. This example is really artificial. There exist better ones...

26. The second counter-example Consider the following transformation T : {0,1}* → {0,1}* defined on every (x1,x2,...,xn) as T(x1,x2,...,xn) = (U(x1),U(x2),...,U(xn)), where • U(0) = 00 • U(1) = 01, or 10, randomly. This transformation is of course invertible. Example: T

27. Remember the stream ciphers? IV s G(IV,s) m xor IV G(IV,s) xor m Enc(s,m)

28. Our new (artificial) encryption scheme To encrypt a message m do the following: IV s G(IV,s) T(m) xor IV G(IV,s) xor T(m) Enc(s,m) If Enc is CPA-secure then also this scheme is CPA-secure.

29. CCA-security? Suppose we use this encryption scheme with the authenticate-then-encryptmethod: t← Mack2 (m) and c ← Enck1(m||t) IV s G(IV,s) T(m, Mack2 (m)) xor IV G(IV,s) xor T(m, Mack2 (m)) Enc(s,m)

30. How does the ciphertext look? pad1 pad2 T(m) T(Mack2 (m)) xor T(m) xor pad1 T(Mack2 (m)) xor pad2

31. The attack The adversary that wants to decrypt the first bit of C1 C2 can modify the ciphertext by flipping the first two bits: C1 C2 1 1 xor C’ X X C2 If the first two bits are 01 or 10 then the corresponding plaintext doesn’t change. If the first two bits are 00 then the plaintext changes and the tag becomes invalid!

32. The chosen-ciphertext attack(just based on the error messages) The adversary is given cand wants to learn the first bit of the corresponding plaintext. Let c’be the ciphertext c with the first two bits flipped. The adversary sends c’ to the oracle. If the oracle answers “error” then the adversary knows that the first bit was 0. The same can be done for any other bit.

33. These examples are artificial It is likely that for many “normal” schemes this combination is secure. However, these examples show that the authenticate-then-encrypt method cannot be proven secure... (from the standard assumptions)

34. How does it look in the public-key settings There are many constructions of a CCA-secure public-key encryption scheme. Probably the most famous is the one of Cramer and Shoup: [Ronald Cramer and Victor Shoup: "A practical public key cryptosystem provably secure against adaptive chosen ciphertext attack." 1998]. It is based on hardness of discrete logarithm and is quite efficient. Still, many practitioners prefer more efficient schemes (with a weaker security proof).

35. Plan We present two CCA-secure schemes based on RSA. • efficient and simple, • even more efficient and a bit less simple.

36. Firstattempt Idea: take the symmetric-key CCA-secure scheme (Enc’,Dec’) and use something similar to hybrid encryption. public key: (N,e) private key: (N,d) Enc((N,e),m) := (re mod N, Enc’(r,m)) Dec((N,d),(c0,c1)) := Dec’(c0d mod N, c1) r is random

37. Problem Enc((N,e),m) := (re mod N, Enc’(r,m)) |N| is normally much larger than the length of a key for symmetric encryption. Typically |N| = 1024 and length of the key is 128. First idea: truncate. But is it secure? It may be the case that • RSA is hard to invert, but • 128 first bits are easy to compute...

38. Idea Instead of truncating – hash! t – length of the symmetric key H : {0,1}* → {0,1}t – a hash function Enc((N,e),m) := (re mod N, Enc’(H(r),m)) Dec((N,d),(c0,c1)) := Dec’((H(c0)d mod N, c1) But can we prove anything about it? depends...

39. Which properties should H have? If we just assume that H is collision-resistant we cannot prove anything... We have to assume that H “outputs random values on different inputs”. This can be formalized by modeling H as random oracle. This is also called a Random Oracle Model. And it is controversial.

40. The Random Oracle Model (ROM) In the proofs we model the hash function as a random oracle. real protocol: in the proof: SHA1 oracle Ω

41. The oracle Ω Everybody (including the adversary) can query the oracle: oracle Ω has a random function H : {0,1}* → {0,1}n x H(x) In the proof:every call to the hash function is replaced with the query to the oracle Ω

42. Problems with the ROM This model is too strong. Random Oracle cannot be implemented in real-life. Moreover, there are examples of protocols that are secure in ROM, butthey are not secure if the random oracle is replaced with any hash function. [The Random-Oracle Model, Revisited. R. Canetti, O. Goldreich and S. Halevi. J. ACM 51(4): 557-594 (2004).]

43. Security proof – the intuition H – a hash function Enc((N,e),m) := (re mod N, Enc’(H(r), m)) Why is this scheme secure in the random oracle model? Because, as long as the adversary did not query the oracle on r, the value of H(r) is completely random. To learn r the adversary would need to compute it from re mod N, so he would need to invert RSA. So (with a very high probability) from the point of view of the adversary H(r)is random. Therefore the CCA security of (Enc,Dec) follows from the CCA-security of (Enc’,Dec’).

44. Disadvantages of this method Enc((N,e),m) := (re mod N, Enc’(H(r), m)) The ciphertext is longer than the plaintext. This is especially important if the message is short. Therefore in practice another method is used: Optimal Asymmetric Encryption Padding (OAEP).

45. Optimal Asymmetric Encryption Padding (OAEP) – the history • Introduced in:[M. Bellare, P. Rogaway. Optimal Asymmetric Encryption -- How to encrypt with RSA. Eurocrypt '94] • An error in the security proof was spoted in[V. Shup. OAEP Reconsidered. Crypto ’01] • This error was repaired in[E. Fujisaki, T. Okamoto, D. Pointcheval, and J. Stern. RSA-OAEP is secure under the RSA assumption. Crypto ’01] It is now a part of a PKCS#1 v. 2.0 standard.

46. OAEP G,H – hash functions OAEP(m) := n/4 n/4 n/4 m 000...0 random r G H X Y

47. How to invert? check if Z =000...0 m Z G H X Y

48. RSA-OAEP private key: (N,d) public key: (N,e) Enc((N,e),m) = (OAEP(m))e mod N Dec((N,e),m) = (OAEP-1(m))d mod N

49. Security – the intuition 1. OAEP is hard to invert if you don’t know X and Y completely. m 000...0 random r G H X Y

50. Why? Assume G and H are random oracles... m Z G H X Y