CMSC 414
Computer and Network Security
Lecture 6
Jonathan Katz
Diffie-Hellman key exchange
 Before describing the protocol, a brief detour
through number theory…
– Modular arithmetic, Zp, Zp*
– Generators: e.g., 3 is a generator of Z17*, but 2 is not
– The discrete logarithm assumption
The Diffie-Hellman protocol
prime p, element g  Zp*
hA = gx mod p
hB = gy mod p
KAB = (hB)x
KBA = (hA)y
Security?
 Consider security against a passive eavesdropper
– We will cover stronger notions of security for key
exchange in more detail later in the semester
 Under the computational Diffie-Hellman (CDH)
assumption, hard for eavesdropper to compute
KAB = KBA
– Not sufficient for security!
– Can hash the key before using
 Under the decisional Diffie-Hellman (DDH)
assumption, the key KAB looks random to an
eavesdropper
Technical notes
 p and g must be chosen so that the CDH/DDH
assumptions hold
– Need to be chosen with care – in particular, g should be
chosen as a generator of a subgroup of Zp*
– Details in CMSC456
 Can use other groups
– Elliptic curves are also popular
 Modular exponentiation can be done quickly (in
particular, in polynomial time)
– But the naïve algorithm does not work!
Security against active attacks?
 The basic Diffie-Hellman protocol we have shown
is not secure against a ‘man-in-the-middle’ attack
 In fact, impossible to achieve security against such
an attacker unless some information is shared in
advance
– E.g., private-key setting
– Or public-key setting (next)
Public-key cryptography
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
The public-key setting
pk
c = Encpk(m)
pk
c = Encpk(m)
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(n2) 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)
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
• Need stronger assumptions, easier to attack
– 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)
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 (unless
the sender has a public key)
– Can be too inefficient for certain applications
Cryptographic primitives
Private-key setting
Public-key setting
Confidentiality
Private-key
encryption
Public-key
encryption
Integrity
Message
authentication codes
Digital signature
schemes
Public-key encryption
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  Epk(m)
 Decryption algorithm:
– Takes a private key and a ciphertext, and outputs a
message (or perhaps an error); m = Dsk(c)
 Correctness: for all (pk, sk), Dsk(Epk(m)) = m
Security?
 Just as in the case of private-key encryption, but
the attacker gets to see the public key pk
 That is:
– For all m0, m1, no adversary running in time T, given pk
and an encryption of m0 or m1, 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)
 In the public-key setting, security against
ciphertext-only attacks implies security against
chosen-plaintext attacks
El Gamal encryption
 We have already (essentially) seen one encryption
scheme:
Receiver
p, g
Sender
p, g, hA = gx
hA = g.x mod p
c = (KBA m) mod p
hB, c
hB = gy mod p
KAB = (hB)x
KBA = (hA)y
Security
 If the DDH assumption holds, the El Gamal
encryption scheme is secure against chosenplaintext attacks
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  ZN*, it holds that x(N) = 1 mod N
– Proof: take CMSC 456!
 If ed=1 mod (N), then for all x it holds that
(xe)d = x mod N
I.e., this is a way to compute eth roots
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
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
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 clearly holds…
 …what about security?