### pptx

```Lecture 8
Public-Key Encryption I
Stefan Dziembowski
www.dziembowski.net
MIM UW
23.11.12
ver 1.0
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
Public-Key Encryption
Use 2 keys (pk,sk), where
pk is used for encryption,
sk is used for decryption.
m
c := Enc(pk,m)
Dec(sk,c)
Alice
Bob
pk
sk
The RSA trap-door permutation from
the last lecture
N = pq - RSA modulus
e is such that gcd(e, φ(N)) = 1,
d is such that ed = 1 (mod φ(N))
Q: How to
formalize this?
f(x) = xe mod N
A: “RSA
assumption”
easy
ZN
*
• easy
(if you know p,q)
• believed to be hard
(otherwise)
Z N*
A naïve idea
•
•
pk = (N,e) is used for encryption,
sk = (N,d) is used for decryption.
m  ZN*
c := me mod N
Alice
(N,e)
m = cd mod N
Bob
(N,d)
The “handbook RSA encryption”
N = pq - RSA modulus
φ(N) = (p-1)(q-1)
e is such that gcd(e, φ(N)) = 1,
d is such that ed = 1 (mod φ(N))
RSA(e,N) (m) = me mod N,
and RSA(d,N) (c) = cd mod N.
= RSA-1(e,N)
A game
security parameter 1k
(x,e,N)
oracle
choose:
• N = pq where p and q are random
primes such that |p| = |q| = k
• x – a random element of ZN* ,
• e – a random element of Zφ(N)*
outputs
y
We say that the adversary wins if y = RSA-1(e,N) (x) mod N
RSA assumption
No poly-time adversary wins this game with a non-negligible
probability.
RSA assumption – more formally
RSA assumption
For any randomized polynomial time algorithm A we have:
P(ye = x mod N : y := A(x,N,e)) is negligible in k
where N = pq where p and q are random primes such that
|p| = |q| = k, and x is a random element of ZN* ,
and e is random element of Zφ(N)*
The ZN* group is a bit strange
Some elements of
ZN = {0,...,n-1}
are not there but you don’t know which if you don’t know
p and q.
Is it a problem?
No, for 2 reasons:
• it is hard to find an element in ZN* \ ZN (other than 0),
• RSA works also over ZN (“by accident”).
Remember the Chinese Reminder
Theorem?
Example Zpq where p=5, q = 7
x mod 7
x mod 5
0
1
2
3
4
0
0
7
12
17
22
Z35
1
5
1
22
8
29
2
10
16
2
23
9
3
15
31
17
3
24
4
20
11
32
18
4
5
25
26
12
33
19
6
30
6
27
13
34
Z35*
It is hard to find an element in ZN \ ZN*
(other than 0)
Why?
ZN
mod p
Suppose we have found a
non-zero element
x  Z N \ Z N*
mod q
ZN*
For example:
x mod q = 0 and x ≠ 0
Hence gcd(x,N) = q.
So we can factor N.
Example
gcd(15,35) = 5
x mod 7
x mod 5
0
1
2
3
4
0
0
7
12
17
22
Z35
1
5
1
22
8
29
2
10
16
2
23
9
3
15
31
17
3
24
4
20
11
32
18
4
5
25
26
12
33
19
6
30
6
27
13
34
Z35*
RSA works also over ZN
Suppose x is such that
x mod q = 0 and x mod p ≠ 0
We show that
= xed
RSA(N,d)(RSA(N,e)(x)) = x mod N
By CRT it is enough to show that:
this holds because both sides are divisible by q
xed = x mod q
xed = x mod p
Recall that: (p-1)(q-1) | ed - 1
Hence: (p-1) | ed - 1
Therefore: xed-1 = 1 mod p
This implies that xed = x mod p
Question?
Can we use this permutation for encryption?
• yes,
• but not directly...
Problems
RSApk is deterministic, so:
if one encrypts twice the same message then the
ciphertexts are the same
Therefore if the message space M is small, the adversary can
check all possible messages:
given a ciphertext c do:
for every m є M check if RSApk(m) = c
for example if M={yes,no}, then the adversary can decrypt
the message.
RSA has some “algebraic properties”.
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
Algebraic properties of RSA
1. RSA is homomorphic:
RSA(e,N)(m0 · m1) = (m0 · m1)e
= m0e · m1e
= RSA(e,N )(m0) · RSA(e,N )(m1)
By checking if
c0 · c 1 = c
RSA(d,N) (c0) · RSA(d,N)(c1) = RSA (d,N) (c)
2. The Jacobi symbol leaks.
Jacobi Symbol
+1
- 1
for N=pq define JN(x) := Jp(x) · Jq(x)
for any prime p define Jp(x) :=
QR(p)
QR(q)
QR(n)
mod p
if x Є QRp
otherwise
JN(x) :=
+1
-1
-1
+1
mod q
Jacobi symbol can be computed efficiently! (even if p and q are unknown)
Fact: the RSA function “preserves” the
Jacobi symbol
N = pq - RSA modulus
e is such that gcd(e, φ(N)) = 1
JN(x) =
e
JN(x
mod N)
Actually, something even stronger holds:
RSA(N,e) is a permutation on each “quarter” of ZN*
QRp
QRq
In other words:
• m mod p  QRp iff me mod p  QRp
• m mod p  QRq iff me mod q  QRq
QRp
Example Z35*
Calculate RSA(23,35)(m) = m23 mod 35
QR7
mod 5
QR5
1
4
2
3
1
1
29
22
8
2
16
9
2
23
mod 7
4
11
4
32
18
3
31
24
17
3
5
26
19
12
33
6
6
34
27
13
1
4
3
2
1
1
29
8
22
4
11
4
18
32
2
16
9
23
2
5
26
19
33
12
3
31
24
3
17
6
6
34
13
27
How to prove it?
By the CRT and by the fact that p and q are
symmetric it is enough to show that
m is a QRp
iff
me is a QRp
Fact
For an odd e:
me mod p is a QRp
iff
m mod p is a QRp
Proof.
Let g be the generator of Zp*. Let y be such that m=gy.
Recall that x is a QRp iff x is an even power of g
It is easy to see that
gye = gye mod (p-1) (remember that p and e are odd)
(gy)e mod p is an even power of g
iff
gy mod p is an even power of g.
Hence we are done.
Conclusion
The Jacobi symbol “leaks”, i.e.:
from c
one can compute JN(Dec(N,d)(c))
(without knowing the factorization of N)
Is it a big problem?
Depends on the application...
Note
The fact that the Jacobi symbol leaks does not
oracle
(x,e,N)
choose:
• N = pq where p and q are random
primes such that |p| = |q| = k
• x – a random element of ZN* ,
• e – is random element of Zφ(N)*
cannot compute RSA-1(e,N) (x) mod N
but can compute JN(RSA-1(e,N) (x) mod N)
Question: Is RSA secure?
Looks like it has some weaknesses...
Plan:
1. Provide a formal security definition.
2. Modify RSA so that it is secure according to
this definition.
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
A mathematical view
A public-key encryption (PKE) scheme is a triple (Gen, Enc, Dec) of
poly-time algorithms, where
 Gen is a key-generation randomized algorithm that takes as
input a security parameter 1n and outputs a key pair (pk,sk).
 Enc is an encryption algorithm that takes as input the public
key pk and a message m, and outputs a ciphertext c,
 Dec is an decryption algorithm that takes as input the private
key sk and the ciphertext c, and outputs a message m’.
We will sometimes write Encpk(m) and Decsk(c) instead of
Enc(pk,m) and Dec(sk,c).
Correctness
P(Decsk(Encpk(m)) ≠ m) is negligible in n
The security definition
Remember the symmetric-key case?
We considered a chosen-plaintext attack.
How would it look in the case of the public-key
encryption?
A chosen-plaintext attack (CPA)
security parameter
1n
1. selects random
(pk,sk) = Gen(1n)
2. chooses a random b = 0,1
pk
chooses m’1
m’1
c1 = Enc(pk,m’1)
...
chooses m’t
challenge phase:
chooses m0,m1
m’t
ct = Enc(pk,m’t)
m0,m1
c = Enc(pk,mb)
has to guess b
the interaction continues . . .
oracle
This is not
needed.
Why?
Because if
Eve knows pk
she can
compute all
these
ciphertexts
herself!
A simplified view
security parameter
1n
1. selects random
(pk,sk) = Gen(1n)
2. chooses a random b = 0,1
pk
oracle
challenge phase:
chooses m0,m1
m0,m1
c = Enc(pk,mb)
has to guess b
CPA-security
Alternative name: CPA-secure
Security definition:
We say that (Gen,Enc,Dec) has indistinguishable
encryptions under a chosen-plaintext attack (CPA) if any
guesses b correctly
with probability at most 0.5 + ε(n), where ε is negligible.
Is the “handbook RSA” secure?
the “handbook RSA”
N = pq - RSA modulus
e is such that gcd(e,d) = 1, d is such that ed = 1 (mod φ(N))
Enc(N,e)(m) = me mod N, and Dec(d,N)(c) = cd mod N.
Not secure!
In fact:
No deterministic encryption scheme is secure.
How can the adversary win the game?
1. he just chooses any m0,m1 ,
2. computes c0=Enc(pk,m0) himself
3. compares the result.
Moral: encryption has to be randomized.
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
Encoding
Therefore, before encrypting a message we
usually encode it (adding some randomness).
• makes the encryption non-deterministic
• breaks the “algebraic properties” of
encryption.
How is it done in real-life?
PKCS #1: RSA Encryption Standard Version 1.5:
public-key: (N,e)
let k := length on N in bytes.
let D := length of the plaintext
requirement: D ≤ k - 11.
Enc((N,e), m) := xe mod N, where x is equal to:
k bytes
00000000
00000001
r
00000000
(k - D - 3) random non-zero
bytes
m
D bytes
How to encrypt?
m
Encoding(x) :=
00000000
00000001
r
00000000
RSA
RSA(Encoding(x))
m
How to decrypt?
ciphertext y
RSA-1(y)
check if the format agrees....
00000000
00000001
r
00000000
m
output
m
Example
If the adversary can calculate the Jacobi symbol
of
00000000
00000001
r
00000000
m
most probably it doesn’t help him in learning any
Security of the PKCS #1: RSA Encryption
Standard Version 1.5.
It is believed to be CPA-secure.
It has however some weaknesses (it is not “chosenciphertext secure”).
Optimal Asymmetric Encryption Padding (OAEP) is a
more secure encoding.
(we will discuss it later)
The situation
???
the PKCS #1 encryption
scheme is secure
RSA assumption holds
Can we construct a public-key encryption scheme
whose security
can be provably based on the RSA assumption?
yes!
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
Notation
For an integer x we will write
LSB(x)
to denote the least significant bit of x.
LSB(x)
x:
In other words: LSB(x) = x mod 2
Fact (informally)
LSB is the “hardest bit to compute” in RSA.
(it is called a “hard-core bit”).
More precisely:
If you can compute LSB then you can invert RSA.
Note:
In some sense it is a “dual” predicate to Jacobi
symbol...
Recall:
security parameter 1k
(x,e,N)
oracle
choose:
• N = pq where p and q are random
primes such that |p| = |q| = k
• x – a random element of ZN* ,
• e – a random element of Zφ(N)*
outputs
y
We say that the adversary wins if y = RSA-1(e,N) (x) mod N
RSA assumption
No poly-time adversary wins this game with a non-negligible
probability.
Game 2
security parameter 1k
(x,e,N)
oracle
choose:
• N = pq where p and q are random
primes such that |p| = |q| = k
• x – a random element of ZN* ,
• e – is random element of Zφ(N)*
• d – such that ed = 1 mod φ(N)
outputs
b
We say that the adversary wins if b is the least
significant bit of y = RSA-1(e,N) (x) mod N
Theorem
Suppose the RSA assumption holds.
Then every poly-time adversary wins Game 2 with a
probability at most
0.5 + ε(k),
where ε is negligible.
W. Alexi, B. Chor, O. Goldreich, and C.P. Schnorr
On the hardness of the least-signficant bits of the RSA and Rabin functions, 1984
In other words:
The least significant bit is a hard-core bit for RSA.
Proof strategy
Suppose we are given a poly-time adversary
For simplicity suppose
that this happens with
probability 1
that wins Game 2.
(not: 0.5 + something)
that breaks the RSA assumption.
Outline of the construction
(x,e,N)
(x1,e,N)
b1
...
(xt,e,N)
y=xd
bt
Observation
can be used to compute
LSB of xd mod N.
It can also be used to compute (for any c)
LSB of c · xd mod N.
How?
(ce · x, e, N)
outputs
b’ = LSB((ce· x)d)
= LSB (ced · xd )
= LSB (c · xd )
The method
compute:
• LSB(2x)
• LSB(4x)
• LSB(8x)
...
Why is it usefull?
will use
to
Observation
x≤(N-1)/2
1
2x
2
2x mod N
1
...
N-1
...
...
N-1 1
even
= 2x
Moral: x  [1,...,(N-1)/2] iff 2x mod N is even
2N-2
...
= 2x - N
odd
x
x>(N-1)/2
N-1
Remember:
N=pq is odd
(N-1)/2
(N-1)/4
1
x
1
N-1
...
...
even
= 4x
N-1 1
...
N-1 1
= 4x - N
4N-4
...
N-1 1
= 4x – 2N
...
= 4x - 3N
odd
4x
mod N
...
even
4
odd
4x
3(N-1)/4
Moral: x  [1,...,(N-1)/4]  [(N/2)+1,...,3(N-1)/4] iff 4x mod N is even
N-1
(N-1)/8
...
x
...
8x
...
7(N-1)/8
Moral: x  [1,...,(N-1)/8]  [(2N/8)+1,...,3(N-1)/8] 
[4(N/8)+1,...,5(N-1)/8]  [6(N/8)+1,...,7(N-1)/8]
iff 8x mod N is even
= 8x-7N
odd
even
= 8x-5N = 8x-6N
odd
= 8x-4N
even
odd
= 8x-2N = 8x-3N
odd
= 8x-N
even
= 8x
even
8x
mod N
So we can use bisection
1
N-1
calculate
LSB((2·x)d)
calculate
LSB((4·x)d)
calculate
LSB((8·x)d)
calculate
LSB((16·x)d)
...
How to encrypt a one-bit message b?
(N,e) – public key
(N,d) – private key
Enc1(N,e)(b) = (LSB(x)  b, xe mod N),
where x  ZN* is random.
Dec1(N,d)(b’,y) = LSB(yd mod N)  b’
How to encrypt long messages?
Let m=(m1,..., mt)
Use the one-bit scheme bit-by-bit
Enc(N,e)(m1,..., mt) = (Enc1(N,e)(m1),..., (Enc1(N,e)(mt))
Dec(N,d)(c1,..., ct) = (Dec1(N,d)(c1),..., (Dec1(N,d)(ct))
(we omit the security proof)
Conclusion
Security of this scheme is implied by the RSA
assumption.
RSA assumption holds
the public-key encryption
scheme that we just
constructed is secure
The ciphertext is much longer than the plaintext.
It is a rather theoretical construction!
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
question:
The situation
can we construct
PKE based on the
“factoring
assumption”
factoring RSA moduli
is hard
RSA assumption
holds
Yes:
Rabin encryption
public-key encryption
exists
Rabin encrypion
Michael O. Rabin (1931 – )
One of the founding fathers of computer
science.
• introduced non-determinism
• decidability of the monadic second
order logic
• efficient primality testing
• oblivious transfer,
• ....
• introduced by
Michael O. Rabin in 1979
• based on squaring in ZN*
• security equivalent to factoring
Remember squaring modulo N=pq?
ZN*
ZN*
RabinN(x) = x2 mod N
This function “glues” 4 elements together
Example
Z15*:
a
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
a2
0
1
4
3
1
5
6
4
4
9
10
1
12
4
1
A nice thing about Rabin’s function
factoring RSA moduli
is hard
RabinN(x) = x2 mod N
is a one-way function
Finding a square root modulo N is as hard as factoring.
Can we base an encryption scheme on this?
YES!
How to do it?
First step: “make the inversion unique”.
How to do it?
add an encoding (like in the RSA encryption).
In such a way that only 1 out of the 4 square roots “make
sense”.
ZN*
ZN*
In other
words:
make the
set of legal
ciphertexts
“sparse”
f(x) = x2 mod N
Another approach
Such an N is called a
“Blum integer”
Fact
Suppose N=pq where
p = q = 3 (mod 4)
Then the function
RabinN(x) = x2 mod N
is a permutation
RabinN : QRN  QRN
How does it look like?
ZN*
ZN*
RabinN (x) = x2 mod N
QRN
QRN
Rabin restricted to QRN is a
permutation
ZN*
QRN
ZN*
RabinN(x) = x2 mod N
QRN
Rabin restricted to QRN is a permutation - proof
Suppose we have x,y  QRN such that
x2 = y2 mod N

Let i,j N be such that
x2 = y2 mod p
g2i = x mod p and g2j = y mod q

where g is a generator of Zp*
g4i = g4j mod p
and (p-1)/2 ≥ i > j ≥ 0.

g4(i-j) mod p

p-1 | 4(i-j)
= (4k+2)/2

= 2k+1
4k+2 | 4(i-j)

2k+1 | 2(i-j)

2k+1 | i-j

i=j

x = y (mod p)
p = 4k + 3
q = 4k’ + 3
N=pq
By a symmetric argument:
x = y (mod q)
x = y mod N
QED
How to encrypt a one-bit message b?
Fact
The least significant bit is also a hard-core bit for the
Rabin permutation.
a Blum integer
RabinN (x) = x2 mod N
RabinN : QRN  QRN
N – public key
(N,p,q) – private key
Enc1N(b) = (LSB(x)  b, RabinN(x)),
where x  QRN is random.
this can
be computed if one-1 knows p and q
1
Dec
(N,p,q)(b’,y)
= LSB(RabinN (y))  b’
Plan
1. Handbook RSA and its insecurity
a. introduction
b. algebraic properties of RSA
2. Security definitions
3. How to encrypt with RSA
a. a practical construction
b. a theoretical construction based on hardcore bits
4. Rabin encryption
5. Theoretical constructions
Hard-core predicates
A concept of a
hard-core bit
can be generalized to a
hard core predicate.
Definition (informal)
π : {0,1}n  {0,1} is a hard core predicate for a trap-door
permutation f: {0,1}n  {0,1}n if it is impossible to
guess π(f-1(y)) from y
(significantly better than with prob. 0.5)
Example
π : {0,1}n  {0,1} defined as
π(x1,...,xn) = xn
is a hard-core predicate for RSA and Rabin.
π : {0,1}n  {0,1} defined as
π(x) = JN(x)
for sure is not a hard core predicate for RSA.
A fact
Does every trap-door permutation have a hard-core
predicate?
Almost:
Suppose that f is a trap-door permutation.
It can be used to build a trap-door permutation g
that has a hard-core predicate.
How to encrypt with such an g?
public key: description of g
private key: trapdoor t for g
Enc1g(b) = (π(x)  b, g(x)),
where x  ZN* is random.
Dec1t(b’,y) = π(g-1(y))  b’
Is the public-key encryption in
Minicrypt?
As far as we know:
no!
cryptomania
public-key encryption
exists
minicrypt
trap-door permutations
exist
one way functions
exist
©2011 by Stefan Dziembowski. Permission to make digital or hard copies of part or all of
this material is currently granted without fee provided that copies are made only for
personal or classroom use, are not distributed for profit or commercial advantage, and
that new copies bear this notice and the full citation.
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