Today:
Phase I:
keygen > (pk, sk)
adversary can use encryption oracle
adversary can use decryption oracle
except on the challenge ctexts
adversary produces two message of the
same length m0, m1 and some state s
Phase II:
examiner flips a coin b \in {0,1}
examiner encrypts one of the messages, computes c* = Enc(m_b)
examiner sends c* to the advesary
adversary has to figure out in time polynomial in the security
parameter whether c* is the encryption of m0 or m1
adversary outputs b* and wins if b = b*
This is the strongest notion of security nowadays.
Basic ElGamal does not satisfy this security definition because ciphertexts can be rerandomized and given to the decryption oracle.
What if the decryption function only decrypts messages that it knows it was created by the encryption box, as opposed to other things: like playing with bits, playing with the homomorphic property, etc.
An extension of ElGamal.
We have G_q
a group of prime order q
(remember g^q = 1
).
TODO: Are g1/g2
generators of G_q
?
Keygen(G_q):
g1, g2 <R G_q
x1,x2, y1,y2, z <R Z_q (additive group mod q)
c < g1^x1 * g2^x2
d < g1^y1 * g2^y2
h < g1^z (ElGamallike feature)
pk = (g1, g2, c, d, h)
sk = (x1, x2, y1, y2, z)
Let H
be a hash function which maps G_q
triples to Z_q
Enc(m \in G_q):
r <R Z_q
u1 < g1^r (ElGamallike)
u2 < g2^r (for the checking portion)
e < m * h^r (ElGamallike)
\alpha < H(u1, u2, e)
v < c^r * d^(r*\alpha)
ctext = (u1, u2, e, v)
Decrypt(ctext = u1, u2, e, v):
\alpha = H(u1, u2, e)
[ u1^x1 u2^x2 = g1^rx1 g2^rx2 = c^r ]
[ u1^y1 u2^y2 = d^r ]
[ u1^(x1+y1*\alpha) * u2^(x2+y2*\alpha) =
u1^x1 u2^x2 * u1^(y1*\alpha) * u2^(y2*\alpha) =
u1^x1 u2^x2 * (u1^y1 * u2^y2)^\alpha = c^r * (d^r)^alpha = v ]
checks that u1^(x1+y1*\alpha) * u2^(x2+y2*\alpha) == v
if not equal, then reject
[ need to divide by (invert) h^r to get m out of e ]
[ don't have r, have z => have h = g1^z ]
[ u1 = g1^r, u1^z = (g1^r)^z = h^r ]
return e/u1^z
Theorem: CramerShoup is INDCCA2 secure, assuming:
G_q
(Why not CDH? ElGamal needed
DDH as well so as to maintain indistinguishability under chosenplaintext =>
CramerShoup, based on ElGamal will need it as well)This is really cool because at the time it solved an openproblem of whether INDCCA2secure public key encryption schemes exist.
Z_n* = all numbers < n relatively prime to n
Keygen:
find large *random* primes p, q
let n = pq
\phi(n) = Z_n* = (p1)(q1)
this is unknown to the adversary
knowing \phi(n) <=> knowing factorization of n
e <R Z*_\phi(n)
just means that gcd(e, \phi(n)) = 1
d < e^1 (mod \phi(n))
computed using Extended Euclidian Algorithm
pk < (n, e)
sk < (p, q, d)
Enc(m \in Z_n*):
c = m^e (mod n)
Dec(c \in Z_n*):
m = c^d (mod n)
Note: m
and c
need to be \in Z_n* =>
need to be relatively prime to
n
, so certain messages cannot be encrypted? The only messages that cannot be
encrypted are p
and q
and their multiples. Note that if an adversary wanted
to encrypt such messages, he would literally have the factorization of n
We can prove using the Chinese Remainder Theorem that RSA works even for
m \notin Z_n*
, but m \in Z_n
(i.e. m
does not need to be coprime to n
)
Chinese Remainder Theorem:
n=pq, \forall x,y \in Z_n*, x = y (mod n) <=> x = y (mod p) AND x = y (mod q)
We know, by definition, that e*d = 1 (mod \phi(n)) => e*d = 1 + t*(p1)(q1) =>
e*d = 1 (mod p1) => d = e^1 (mod p1)
We want to show that (m^e)^d = m (mod n), \forall m \in Z_n
Suffices to show that (m^e)^d = m (mod p)
and then use the Chinese Remainder
Theorem.
Case 1: m = 0 (mod p) => 0^ed = 0 (mod p)
q.e.d.
Case 2: m != 0 (mod p), m \in Z_p*
m^(p1) = 1 (mod p) [ element of group raised to group order == identity]
[ or Fermat's theorem ]
m^(ed) = m^(1 + u*(p1)) = m^1 * (m^(p1))^u = m * 1^u = m (mod p)
Relies on assumption that n = pq
is hard to factor when p
and q
are large
primes.
Best factoring algorithms today have running time exp(c*(ln(n))^(1/3) * ln(ln(n))^(2/3))
,
where n
is the number of bits in the N = pq
number, subexponential time. (According to Wikipedia,
it's the number of bits in n
.)
Q: Why is it important to keep \phi(n)
secret?
A: You can factor n
knowing \phi(n)
?
\phi(n) = (p1)(q1) => \phi(n) = pq  p  q + 1 =>
p + q = pq  \phi(n) + 1 => p + q = n  \phi(n) + 1
We expect p and q to be about the same size, so we can
guess them easily? Sure, but we can do better
We have:
(1) n = pq
(2) p + q = n  \phi(n) + 1
=>
q = (n  \phi(n) + 1)  p
=> (substitute in (1))
n = p((n  \phi(n) + 1)  p)
=  p^2 * p((n+1)  \phi(n))
<=>
p^2  p((n+1)  \phi(n)) + n = 0
<=> (2nd degree equation w/ a,b,c coeff.)
p = (b +/ sqrt(b^2  4ac))/(2a)
= ...
Q: If d
is lost, can you factor n
?
A: You can definitely decrypt and sign messages arbitrarily with d
. The
RSA paper in section IX.C says that you can also efficiently factor n
with
knowledge of e
and d
.
Assume G and H are random oracles
 m  0^k1   r 
 
\>  </ 
/ t+k1 / k0
 
\ \
+ <  G  <*
 
\ 
+>  H  > +
 
\ \
  
 x

 RSA 


x^e (mod n)
Decryption:
RSA
, get x
m
and 0^k1
0^k1
are not presentm
Theorem: RSA+OEAP is INDCCA2 secure assuming
G
and H