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CMSC 414 Computer and Network Security Lecture 7 Jonathan Katz

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Review: El Gamal encryption (Some aspects of the actual scheme are simplified) Key generation –Choose a large prime p, and an element g Z p * –Choose random x {0, …, p-2}, set h=g x –The public key is (p, g, h), and the private key is x Encryption –View the message m as an element of Z p * –Choose random r {0, …, p-2} –The ciphertext is (g r, h r m) To decrypt ciphertext (c 1, c 2 ) output c 2 /c 1 x –Correctness?

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Security? Security of El Gamal encryption is based on the decisional Diffie-Hellman assumption Best current algorithm for the decisional Diffie- Hellman problem in Z p * runs in time ≈ exp(|p| 1/3 ) –So if p is a 1024-bit prime, best current attack on El Gamal encryption requires time ≈ 2 60 In other groups, the Diffie-Hellman problem is currently ‘harder’ –E.g., for elliptic curve groups, best current algorithms require time exp(|p|/2) –Can use 120-bit primes to get 2 60 security

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RSA background N=pq, p and q distinct, odd primes (N) = (p-1)(q-1) = |Z N * | –Easy to compute (N) given the factorization of N –Hard to compute (N) without the factorization of N For all x Z N *, it holds that x (N) = 1 mod N If ed=1 mod (N), then for all m: (m e ) d = m mod N I.e., given d, we can compute e th roots

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We have an asymmetry! Let e be relatively prime to (N) –Needed so that ed=1 mod (N) has a solution Given e and the factors of N, can compute d and hence compute e th roots Without the factorization of N, no apparent way to compute e th roots

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Hardness of computing e th roots? The RSA problem: –Given N, e, and c, compute c 1/e mod N If factoring is easy, then the RSA problem is easy We know of no other way to solve the RSA problem besides factoring N –But we do not know how to prove that the RSA problem is as hard as factoring The upshot: we believe factoring is hard, and we believe the RSA problem is hard

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How hard is factoring? Best current algorithms for factoring N=pq a product of two equal-length primes, run in time ≈ exp(|N| 1/3 ) So need |N| ≈ 1024 for reasonable security Currently |N| ≈ 2048 recommended for good security margins

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We have an asymmetry! Given d (which can be computed from e and the factorization of N), possible to compute e th roots Without the factorization of N, no apparent way to compute e th roots Let’s use this to encrypt…

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RSA key generation Generate random p, q of sufficient length Compute N=pq and (N) = (p-1)(q-1) Compute e and d such that ed = 1 mod (N) –e must be relatively prime to (N) –Typical choice: e = 3; other choices possible Public key = (N, e); private key = (N, d)

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“Textbook RSA” encryption Public key (N, e); private key (N, d) To encrypt a message m Z N *, compute c = m e mod N To decrypt a ciphertext c, compute m = c d mod N Correctness… …what about security?

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Textbook RSA is insecure! It is deterministic! Furthermore, it can be shown that the ciphertext leaks specific information about the plaintext

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Padded RSA Introduce randomization… Public key (N, e); private key (N, d) –Say |N| = 1024 bits To encrypt m {0,1} 895, –Choose random r {0,1} 128 –Compute c = (r | m) e mod N Decryption done in the natural way… Essentially this is standardized as PKCS #1 v1.5

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Hybrid encryption Public-key encryption is “slow” Encrypting “block-by-block” would be inefficient for long messages Hybrid encryption gives the functionality of public-key encryption at the (asymptotic) efficiency of private-key encryption!

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Hybrid encryption Enc’ message Enc k “encapsulated key” “encrypted message” ciphertext Enc = public-key encryption scheme Enc’ = private-key encryption scheme pk random!

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Security If public-key component and private-key component are secure against chosen-plaintext attacks, then hybrid encryption is secure against chosen-plaintext attacks

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Extension How should hybrid encryption be done when sending the same message to multiple recipients (e.g., email encryption)?

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Malleability All the public-key encryption schemes we have seen so far are malleable –Given ciphertext c that encrypts (unknown) message m, possible to generate a ciphertext c’ that encrypts a related message m’ In the public-key setting, security against chosen- ciphertext attacks implies non-malleability In many scenarios, malleability/chosen-ciphertext attacks are problematic –E.g., auction example; password example; Bleichenbacher attack…

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Bleichenbacher’s attack RSA PKCS #1 v1.5 is actually defined as: c = (00 || 02 || r || 0 || m) e mod N When decrypting, return an error if formatting is not obeyed This enables a chosen-ciphertext attack that relies only on the ability to detect errors upon decryption

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Bleichenbacher’s attack c = Enc pk (m) c1c1 error/no error c 999 error/no error … If the {c i } are carefully constructed, this information is enough to determine m!

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Malleability All the public-key encryption schemes we have seen so far are malleable –Given a ciphertext c that encrypts an (unknown) message m, possible to generate a ciphertext c’ that encrypts a related message m’ Note: the problem is not integrity (there is no integrity in public-key encryption, anyway), but malleability and/or the ability to conduct a chosen- ciphertext attack

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Malleability in private-key setting Malleability is an issue in the private-key setting as well –Recall that CBC, OFB, CTR mode are all vulnerable to chosen-ciphertext attacks, and are all malleable Authenticated encryption schemes (e.g., “encrypt- then-authenticate”) are secure against chosen- ciphertext attacks (and non-malleable)

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