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CMSC 414 Computer and Network Security Lecture 6 Jonathan Katz

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Administrative announcements Midterm I –March 6 GRACE accounts set up Read the essays and papers linked from the course webpage –Will discuss next Tuesday and/or Thursday

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Public-key cryptography

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The public-key setting A party (Alice) generates a public key along with a matching secret key (aka private key) The public key is widely distributed, and is assumed to be known to anyone (Bob) who wants to communicate with Alice –We will discuss later how this can be ensured Alice’s public key is also known to the attacker! Alice’s secret key remains secret Bob may or may not have a public key of his own

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The public-key setting c = Enc pk (m) pk c = Enc pk (m) pk

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Private- vs. public-key I Disadvantages of private-key cryptography –Need to securely share keys What if this is not possible? Need to know in advance the parties with whom you will communicate Can be difficult to distribute/manage keys in a large organization –O(n 2 ) keys needed for person-to-person communication in an n-party network All these keys need to be stored securely –Inapplicable in open systems (think: e-commerce)

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Private- vs. public-key II Why study private-key at all? –Private-key is orders of magnitude more efficient –Private-key still has domains of applicability Military settings, disk encryption, … –Public-key crypto is “harder” to get right Needs stronger assumptions, more math –Can combine private-key primitives with public-key techniques to get the best of both (for encryption) Still need to understand the private-key setting! –Can distribute keys using trusted entities (KDCs)

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Private- vs. public-key III Public-key cryptography is not a cure-all –Still requires secure distribution of public keys May (sometimes) be just as hard as sharing a key Technically speaking, requires only an authenticated channel instead of an authenticated + private channel –Not clear with whom you are communicating (for public-key encryption) –Can be too inefficient for certain applications

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Cryptographic primitives Private-key settingPublic-key setting Confidentiality Private-key encryption Public-key encryption Integrity Message authentication codes Digital signature schemes

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Public-key encryption

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Functional definition Key generation algorithm: randomized algorithm that outputs (pk, sk) Encryption algorithm: –Takes a public key and a message (plaintext), and outputs a ciphertext; c E pk (m) Decryption algorithm: –Takes a private key and a ciphertext, and outputs a message (or perhaps an error); m = D sk (c) Correctness: for all (pk, sk), D sk (E pk (m)) = m

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Security? Just as in the case of private-key encryption, but the attacker gets to see the public key pk That is: –For all m 0, m 1, no adversary running in time T, given pk and an encryption of m 0 or m 1 can determine the encrypted message with probability better than 1/2 + Public-key encryption must be randomized (even to achieve security against ciphertext-only attacks) Security against ciphertext-only attacks implies security against chosen-plaintext attacks

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El Gamal encryption We have already (essentially) seen one encryption scheme: p, g h A = g x mod p h B = g y mod p K AB = (h B ) x K BA = (h A ) y p, g, h A = g x ReceiverSender c = (K BA. m) mod p h B, c

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Security If the DDH assumption holds, the El Gamal encryption scheme is secure against chosen- plaintext attacks

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RSA: background N=pq, p and q distinct, odd primes (N) = (p-1)(q-1) –Easy to compute (N) given the factorization of N –Hard to compute (N) without the factorization of N Fact: for all x Z N *, it holds that x (N) = 1 mod N –Proof: take CMSC 456! If ed=1 mod (N), then for all x it holds that (x e ) d = x mod N

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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) We have an asymmetry! –Given c=x e mod N, receiver can compute x=c d mod N –No apparent way for anyone else to recover x

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Hardness of the RSA problem? The RSA problem: –Compute x given N, e, and x 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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“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 clearly holds… …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 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 idea has been standardized as RSA 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 pk “encapsulated key” “encrypted message” ciphertext Enc = public-key encryption scheme Enc’ = private-key encryption scheme

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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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Malleability/chosen-ciphertext security All the public-key encryption schemes we have seen so far are malleable –Given a ciphertext c that encrypts an (unknown) message m, may be possible to generate a ciphertext c’ that encrypts a related message m’ In many scenarios, this is problematic –E.g., auction example; password example Note: the problem is not integrity (there is no integrity in public-key encryption, anyway), but malleability

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