CMSC 414Computer and Network Security

Lecture 7

Jonathan Katz

Review: El Gamal encryption (Some aspects of the actual scheme are simplified)

Key generation– Choose a large prime p, and an element g Zp

*

– Choose random x {0, …, p-2}, set h=gx

– The public key is (p, g, h), and the private key is x

Encryption– View the message m as an element of Zp

*

– Choose random r {0, …, p-2}

– The ciphertext is (gr, hr m)

To decrypt ciphertext (c1, c2) output c2/c1x

– Correctness?

Security?

Security of El Gamal encryption is based on the decisional Diffie-Hellman assumption

Best current algorithm for the decisional Diffie-Hellman problem in Zp

* runs in time ≈ exp(|p|1/3)– So if p is a 1024-bit prime, best current attack on El

Gamal encryption requires time ≈ 260

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 260 security

RSA background

N=pq, p and q distinct, odd primes

(N) = (p-1)(q-1) = |ZN*|

– Easy to compute (N) given the factorization of N

– Hard to compute (N) without the factorization of N

For all x ZN*, it holds that x(N) = 1 mod N

If ed=1 mod (N), then for all m: (me)d = m mod NI.e., given d, we can compute eth roots

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 eth roots

Without the factorization of N, no apparent way to compute eth roots

Hardness of computing eth roots?

The RSA problem:– Given N, e, and c, compute c1/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

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

We have an asymmetry!

Given d (which can be computed from e and the factorization of N), possible to compute eth roots

Without the factorization of N, no apparent way to compute eth roots

Let’s use this to encrypt…

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)

“Textbook RSA” encryption

Public key (N, e); private key (N, d)

To encrypt a message m ZN*, compute

c = me mod N

To decrypt a ciphertext c, compute m = cd mod N

Correctness…

…what about security?

Textbook RSA is insecure!

It is deterministic!

Furthermore, it can be shown that the ciphertext leaks specific information about the plaintext

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

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!

Hybrid encryption

Enc’

message

Enck “encapsulated key”

“encrypted message”

ciphertext

Enc = public-key encryption schemeEnc’ = private-key encryption scheme

pk

random!

Security

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

Extension

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

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…

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

Bleichenbacher’s attack

c = Encpk(m)

c1

error/no error

c999

error/no error

…

If the {ci} are carefully constructed,this information is

enough to determine m!

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

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)