Download Report

PSEUDO-RANDOM FUNCTIONS
Mihir Bellare
UCSD
1
Recall
We studied security of a block cipher against key recovery.
But we saw that security against key recovery is not sufficient to ensure
that natural usages of a block cipher are secure.
We want to answer the question:
What is a good block cipher?
where “good” means that natural uses of the block cipher are secure.
We could try to define “good” by a list of necessary conditions:
• Key recovery is hard
• Recovery of M from C = EK (M) is hard
• ...
But this is neither necessarily correct nor appealing.
Mihir Bellare
UCSD
2
Turing Intelligence Test
Q: What does it mean for a program to be “intelligent” in the sense of a
human?
Possible answers:
• It can be happy
• It recognizes pictures
• It can multiply
• But only small numbers!
•
•
Clearly, no such list is a satisfactory answer to the question.
Mihir Bellare
UCSD
3
Turing Intelligence Test
Q: What does it mean for a program to be “intelligent” in the sense of a
human?
Turing’s answer: A program is intelligent if its input/output behavior is
indistinguishable from that of a human.
Mihir Bellare
UCSD
4
Turing Intelligence Test
Behind the wall:
• Room 1: The program P
• Room 0: A human
Mihir Bellare
UCSD
5
Turing Intelligence Test
Game:
• Put tester in room 0 and let it interact with object behind wall
• Put tester in rooom 1 and let it interact with object behind wall
• Now ask tester: which room was which?
The measure of “intelligence” of P is the extent to which the tester fails.
Mihir Bellare
UCSD
6
Real versus Ideal
Notion
Intelligence
PRF
Mihir Bellare
Real object
Program
Block cipher
UCSD
Ideal object
Human
?
7
Real versus Ideal
Notion
Intelligence
PRF
Mihir Bellare
Real object
Program
Block cipher
UCSD
Ideal object
Human
Random function
8
Random functions
Game RandR
// here R is a set
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← R
return T[x]
Adversary A
• Make queries to Fn
• Eventually halts with some output
We denote by
h
i
Pr RandA
⇒
d
R
the probability that A outputs d
Mihir Bellare
UCSD
9
Random functions
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y ← Fn(01)
return (y = 000)
h
i
Pr RandA
⇒
true
=
3
{0,1}
Mihir Bellare
UCSD
10
Random functions
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y ← Fn(01)
return (y = 000)
h
i
Pr RandA
⇒
true
= 2−3
3
{0,1}
Mihir Bellare
UCSD
11
Random function
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y1 ← Fn(00)
y2 ← Fn(11)
return (y1 = 010 ∧ y2 = 011)
h
i
Pr RandA
⇒
true
=
3
{0,1}
Mihir Bellare
UCSD
12
Random function
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y1 ← Fn(00)
y2 ← Fn(11)
return (y1 = 010 ∧ y2 = 011)
h
i
Pr RandA
⇒
true
= 2−6
3
{0,1}
Mihir Bellare
UCSD
13
Random function
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y1 ← Fn(00)
y2 ← Fn(11)
return (y1 ⊕ y2 = 101)
h
i
Pr RandA
⇒
true
=
3
{0,1}
Mihir Bellare
UCSD
14
Random function
Game Rand{0,1}3
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}3
return T[x]
adversary A
y1 ← Fn(00)
y2 ← Fn(11)
return (y1 ⊕ y2 = 101)
h
i
Pr RandA
⇒
true
= 2−3
3
{0,1}
Mihir Bellare
UCSD
15
Function families
A family of functions F : Keys(F ) × Dom(F ) → Range(F ) is a
two-argument map. For K ∈ Keys(F ) we let FK : Dom(F ) → Range(F )
be defined by
∀x ∈ Dom(F ) : FK (x) = F (K , x)
Examples:
• DES: Keys(F ) = {0, 1}56 , Dom(F ) = Range(F ) = {0, 1}64
• Any block cipher: Dom(F ) = Range(F ) and each FK is a permutation
Mihir Bellare
UCSD
16
Real versus Ideal
Notion
PRF
Real object
Family of functions
(eg. a block cipher)
Ideal object
Random function
F is a PRF if the input-output behavior of FK looks to a tester like the
input-output behavior of a random function.
Tester does not get the key K !
Mihir Bellare
UCSD
17
Games defining prf advantage of an adversary against F
Let F : Keys(F ) × Dom(F ) → Range(F ) be a family of functions.
Game RealF
Game RandRange(F )
procedure Initialize
$
K ← Keys(F )
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← Range(F )
Return T[x]
procedure Fn(x)
Return FK (x)
Associated to F , A are the probabilities
h
h
i
i
Pr RandA
⇒1
Pr RealA
F ⇒1
Range(F )
that A outputs 1 in each world. The advantage of A is
h
i
h
i
A
A
Advprf
(A)
=
Pr
Real
⇒1
−
Pr
Rand
⇒1
F
Range(F )
F
Mihir Bellare
UCSD
18
PRF advantage
A’s output d
1
0
Intended meaning: I think I am in game
Real
Random
Advprf
F (A) ≈ 1 means A is doing well and F is not prf-secure.
Advprf
F (A) ≈ 0 (or ≤ 0) means A is doing poorly and F resists the attack
A is mounting.
Mihir Bellare
UCSD
19
PRF security
Adversary advantage depends on its
• strategy
• resources: Running time t and number q of oracle queries
Security: F is a (secure) PRF if Advprf
F (A) is “small” for ALL A that use
“practical” amounts of resources.
Example: 80-bit security could mean that for all n = 1, . . . , 80 we have
−n
Advprf
F (A) ≤ 2
for any A with time and number of oracle queries at most 280−n .
Insecurity: F is insecure (not a PRF) if we can specify an A using “few”
resources that achieves “high” advantage.
Mihir Bellare
UCSD
20
Example
Define F : {0, 1}` × {0, 1}` → {0, 1}` by FK (x) = K ⊕ x for all
K , x ∈ {0, 1}` . Is F a secure PRF?
Game RealF
Game Rand{0,1}`
procedure Initialize
$
K ← Keys(F )
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
procedure Fn(x)
Return K ⊕ x
So we are asking: Can we design a low-resource A so that
h
i
h
i
A
A
Advprf
(A)
=
Pr
Real
⇒1
−
Pr
Rand
⇒1
`
F
F
{0,1}
is close to 1?
Mihir Bellare
UCSD
21
Example
Define F : {0, 1}` × {0, 1}` → {0, 1}` by FK (x) = K ⊕ x for all
K , x ∈ {0, 1}` . Is F a secure PRF?
Game RealF
Game Rand{0,1}`
procedure Initialize
$
K ← Keys(F )
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
procedure Fn(x)
Return K ⊕ x
So we are asking: Can we design a low-resource A so that
h
i
h
i
A
A
Advprf
(A)
=
Pr
Real
⇒1
−
Pr
Rand
⇒1
`
F
F
{0,1}
is close to 1?
Exploitable weakness of F : For all K we have
FK (0` ) ⊕ FK (1` ) = (K ⊕ 0` ) ⊕ (K ⊕ 1` ) = 1`
Mihir Bellare
UCSD
22
Example: The adversary
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Mihir Bellare
UCSD
23
Example: Real game analysis
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Game RealF
procedure Initialize
$
K ← Keys(F )
procedure Fn(x)
Return K ⊕ x
h
i
Pr RealA
⇒1
=
F
Mihir Bellare
UCSD
24
Example: Real game analysis
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Game RealF
procedure Initialize
$
K ← Keys(F )
procedure Fn(x)
Return K ⊕ x
h
i
Pr RealA
⇒1
=1
F
because
Fn(0` ) ⊕ Fn(1` ) = FK (0` ) ⊕ FK (1` ) = (K ⊕ 0` ) ⊕ (K ⊕ 1` ) = 1`
Mihir Bellare
UCSD
25
Example: Rand game analysis
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Game Rand{0,1}`
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
h
i
Pr RandA
⇒1
=
`
{0,1}
Mihir Bellare
UCSD
26
Example: Rand game analysis
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Game Rand{0,1}`
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
h
i
h
i
`
`
`
Pr RandA
⇒1
=
Pr
Fn(1
)
⊕
Fn(0
)
=
1
=
`
{0,1}
Mihir Bellare
UCSD
27
Example: Rand game analysis
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Game Rand{0,1}`
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
h
i
h
i
`
`
`
Pr RandA
⇒1
=
Pr
Fn(1
)
⊕
Fn(0
)
=
1
= 2−`
`
{0,1}
because Fn(0` ), Fn(1` ) are random `-bit strings.
Mihir Bellare
UCSD
28
Example: Conclusion
F : {0, 1}` × {0, 1}` → {0, 1}` is defined by FK (x) = K ⊕ x.
adversary A
if Fn(0` ) ⊕ Fn(1` ) = 1` then return 1 else return 0
Then
2−`
1
}|
z h }|
i{ z h
i{
prf
A
A
AdvF (A) = Pr RealF ⇒1 − Pr Rand{0,1}` ⇒1
= 1 − 2−`
and A is efficient .
Conclusion: F is not a secure PRF.
Mihir Bellare
UCSD
29
Exercise
Define the family of functions F : {0, 1}128 × {0, 1}128 → {0, 1}128 by
F (K , M) = AES(M, K ). Assuming AES is a secure PRF, is F a secure
PRF? If so, explain why. If not, present the best attack (with analysis)
that you can.
If you give an attack it should be presented in the format of the above
examples, with an adversary clearly specified in pseudo-code, a claim
about its advantage, and a succinct proof of the claim. Avoid long textual
descriptions.
Mihir Bellare
UCSD
30
Exercise
Let F : {0, 1}k × D → {0, 1}n be a family of functions and A an adversary.
Prove that
Advprf
F (A) 6= 1 .
Mihir Bellare
UCSD
31
Birthday Problem
We have q people 1, . . . , q with birthdays y1 , . . . , yq ∈ {1 . . . , 365}.
Assume each person’s birthday is a random day of the year. Let
C (365, q) = Pr [2 or more persons have same birthday]
= Pr [y1 , . . . , yq are not all different]
• What is the value of C (365, q)?
• How large does q have to be before C (365, q) is at least 1/2?
Mihir Bellare
UCSD
32
Birthday Problem
We have q people 1, . . . , q with birthdays y1 , . . . , yq ∈ {1 . . . , 365}.
Assume each person’s birthday is a random day of the year. Let
C (365, q) = Pr [2 or more persons have same birthday]
= Pr [y1 , . . . , yq are not all different]
• What is the value of C (365, q)?
• How large does q have to be before C (365, q) is at least 1/2?
Naive intuition:
• C (365, q) ≈ q/365
• q has to be around 365
Mihir Bellare
UCSD
33
Birthday Problem
We have q people 1, . . . , q with birthdays y1 , . . . , yq ∈ {1 . . . , 365}.
Assume each person’s birthday is a random day of the year. Let
C (365, q) = Pr [2 or more persons have same birthday]
= Pr [y1 , . . . , yq are not all different]
• What is the value of C (365, q)?
• How large does q have to be before C (365, q) is at least 1/2?
Naive intuition:
• C (365, q) ≈ q/365
• q has to be around 365
The reality
• C (365, q) ≈ q 2 /365
• q has to be only around 23
Mihir Bellare
UCSD
34
Birthday collision bounds
C (365, q) is the probability that some two people have the same birthday
in a room of q people with random birthdays
q
15
18
20
21
23
25
27
30
35
40
50
Mihir Bellare
C (365, q)
0.253
0.347
0.411
0.444
0.507
0.569
0.627
0.706
0.814
0.891
0.970
UCSD
35
Birthday Problem
$
Pick y1 , . . . , yq ← {1, . . . , N} and let
C (N, q) = Pr [y1 , . . . , yq not all distinct]
Birthday setting: N = 365
Mihir Bellare
UCSD
36
Birthday Problem
$
Pick y1 , . . . , yq ← {1, . . . , N} and let
C (N, q) = Pr [y1 , . . . , yq not all distinct]
Birthday setting: N = 365
Fact: C (N, q) ≈
Mihir Bellare
q2
2N
UCSD
37
Birthday collisions formula
$
Let y1 , . . . , yq ← {1, . . . , N}. Then
1 − C (N, q) = Pr [y1 , . . . , yq all distinct]
N − (q − 1)
N −1 N −2
·
· ··· ·
N
N
N
q−1
Y
i
=
1−
N
= 1·
i=1
so
C (N, q) = 1 −
q−1
Y
1−
i=1
Mihir Bellare
UCSD
i
N
38
Birthday bounds
Let
C (N, q) = Pr [y1 , . . . , yq not all distinct]
Fact: Then
q(q − 1)
q(q − 1)
≤ C (N, q) ≤ 0.5 ·
N
N
√
where the lower bound holds for 1 ≤ q ≤ 2N.
0.3 ·
Mihir Bellare
UCSD
39
Block ciphers as PRFs
Let E : {0, 1}k × {0, 1}` → {0, 1}` be a block cipher.
Game RealE
Game Rand{0,1}`
procedure Initialize
$
K ← {0, 1}k
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
procedure Fn(x)
Return EK (x)
Can we design A so that
h
i
h
i
A
A
Advprf
(A)
=
Pr
Real
⇒1
−
Pr
Rand
⇒1
E
E
{0,1}`
is close to 1?
Mihir Bellare
UCSD
40
Block ciphers as PRFs
Defining property of a block cipher: EK is a permutation for every K
So if x1 , . . . , xq are distinct then
• Fn = EK ⇒ Fn(x1 ), . . . , Fn(xq ) distinct
• Fn random ⇒ Fn(x1 ), . . . , Fn(xq ) not necessarily distinct
This leads to the following attack:
adversary A
Let x1 , . . . , xq ∈ {0, 1}` be distinct
for i = 1, . . . , q do yi ← Fn(xi )
if y1 , . . . , yq are all distinct then return 1
else return 0
Mihir Bellare
UCSD
41
Real world analysis
Let E : {0, 1}k × {0, 1}` → {0, 1}` be a block cipher
Game RealE
procedure Initialize
$
K ← {0, 1}k
procedure Fn(x)
Return EK (x)
Then
Mihir Bellare
adversary A
Let x1 , . . . , xq ∈ {0, 1}` be distinct
for i = 1, . . . , q do yi ← Fn(xi )
if y1 , . . . , yq are all distinct
then return 1 else return 0
h
i
Pr RealA
⇒1
=
E
UCSD
42
Real world analysis
Let E : {0, 1}k × {0, 1}` → {0, 1}` be a block cipher
Game RealE
procedure Initialize
$
K ← {0, 1}k
procedure Fn(x)
Return EK (x)
Then
adversary A
Let x1 , . . . , xq ∈ {0, 1}` be distinct
for i = 1, . . . , q do yi ← Fn(xi )
if y1 , . . . , yq are all distinct
then return 1 else return 0
h
i
Pr RealA
⇒1
=1
E
because y1 , . . . , yq will be distinct because EK is a permutation.
Mihir Bellare
UCSD
43
Rand world analysis
Let E : {0, 1}K × {0, 1}` → {0, 1}` be a block cipher
Game Rand{0,1}`
procedure Fn(x)
$
if T[x] = ⊥ then T[x] ← {0, 1}`
Return T[x]
adversary A
Let x1 , . . . , xq ∈ {0, 1}` be distinct
for i = 1, . . . , q do yi ← Fn(xi )
if y1 , . . . , yq are all distinct
then return 1 else return 0
Then
h
i
Pr RandA
⇒1
= Pr [y1 , . . . , yq all distinct] = 1 − C (2` , q)
`
{0,1}
because y1 , . . . , yq are randomly chosen from {0, 1}` .
Mihir Bellare
UCSD
44
Birthday attack on a block cipher
E : {0, 1}k × {0, 1}` → {0, 1}` a block cipher
adversary A
Let x1 , . . . , xq ∈ {0, 1}` be distinct
for i = 1, . . . , q do yi ← Fn(xi )
if y1 , . . . , yq are all distinct then return 1 else return 0
1−C (2` ,q)
1
z h }|
}|
i{ z h
i{
prf
A
A
AdvE (A) = Pr RealE ⇒1 − Pr Rand{0,1}` ⇒1
= C (2` , q) ≥ 0.3 ·
q(q − 1)
2`
so
q ≈ 2`/2 ⇒ Advprf
E (A) ≈ 1 .
Mihir Bellare
UCSD
45
Birthday attack on a block cipher
Conclusion: If E : {0, 1}k × {0, 1}` → {0, 1}` is a block cipher, there is an
attack on it as a PRF that succeeds in about 2`/2 queries.
Depends on block length, not key length!
DES, 2DES, 3DES3
AES
Mihir Bellare
`
64
128
UCSD
2`/2
232
264
Status
Insecure
Secure
46
KR-security versus PRF-security
We have seen two possible metrics of security for a block cipher E
• KR-security: It should be hard to find a key consistent with
input-output examples of a hidden target key.
• PRF-security: It should be hard to distinguish the input-output
behavior of EK from that of a random function.
Fact: PRF-security of E implies
• KR-security of E
• Many other security attributes of E
This is a validation of the choice of PRF security as our main metric.
Mihir Bellare
UCSD
47
If E is PRF-secure then it is KR-secure
Proposition: Let E : {0, 1}k × {0, 1}` → {0, 1}` be a blockcipher. Given a
kr-adversary B making q (distinct!) oracle queries, we can construct a
PRF adversary A making q oracle queries such that
prf
k−q`
Advkr
.
E (B) ≤ AdvE (A) + 2
The running time of A is that of B plus O(q`).
Interpretation:
E is PRF secure
⇒ Advprf
E (A) is small
⇒ Advkr
E (B) is small
⇒ E is KR-secure.
Example: If E = AES and q = 2 then 2k−q` = 2−128 .
Our first example of a reduction and a proof by reduction!
Mihir Bellare
UCSD
48
Game defining kr-advantage of an adversary B against E
Game KRE
procedure Initialize
$
K ← {0, 1}k ; i ← 0
procedure Fn(M)
i ← i + 1; Mi ← M
Ci ← E (K , Mi )
Return Ci
procedure Finalize(K 0 )
win ← true
For j = 1, . . . , i do
If E (K 0 , Mj ) 6= Cj then win ← false
If Mj ∈ {M1 , . . . , Mj−1 } then win ← false
Return win
The kr-advantage of B is
B
Advkr
E (B) = Pr[KRE ⇒ true]
Mihir Bellare
UCSD
49
Games defining prf-advantage of an adversary A against E
Game RealE
Game Rand{0,1}`
procedure Initialize
$
K ← {0, 1}k
procedure Fn(M)
$
if T[M] = ⊥ then T[M] ← {0, 1}`
Return T[M]
procedure Fn(M)
Return E (K , M)
The prf-advantage of A is
h
i
h
i
A
A
Advprf
(A)
=
Pr
Real
⇒1
−
Pr
Rand
⇒1
E
E
{0,1}`
Mihir Bellare
UCSD
50
Proof of Proposition
Given B we build A as follows:
adversary A
i ← 0; d ← 1
K 0 ← B FnKRSim
For j = 1, . . . , i do
If (E (K 0 , Mj ) 6= Cj ) then d ← 0
Return d
subroutine FnKRSim(M)
i ← i + 1; Mi ← M
Ci ← Fn(Mi )
return Ci
A runs B, simulating B’s oracle via a subroutine that in turn invokes A’s
own Fn oracle. When B returns a key K 0 , adversary A returns 1 if K 0 is
consistent with the input-output examples, and 0 otherwise.
Mihir Bellare
UCSD
51
Real game analysis
Game RealE
procedure Initialize
$
K ← {0, 1}k
procedure Fn(M)
Return EK (M)
When A is executed in game RealE , subroutine FnKRSim(M) will return
Fn(M), which equals EK (M).
So B is getting the same responses it would in game KRE .
So K 0 will be consistent with (M1 , C1 ), . . . , (Mq , Cq ) with probability the
kr-advantage of B.
So
h
i
Pr RealA
⇒1
= Advkr
E
E (B) .
Mihir Bellare
UCSD
52
Rand game analysis
Game Rand{0,1}`
procedure Fn(x)
$
if T[M] = ⊥ then T[M] ← {0, 1}`
Return T[M]
When A is executed in game Rand{0,1}` , subroutine FnKRSim(M) will
return Fn(M), which is a random `-bit string.
So B is getting back a sequence of q random, independent `-bit strings.
So K 0 will be consistent with (M1 , C1 ), . . . , (Mq , Cq ) with probability at
most 2k /2q` , because there are 2k choices for K 0 and 2q` choices for
(C1 , . . . , Cq ).
So
h
i
Pr RandA
⇒1
≤ 2k−q` .
`
{0,1}
Mihir Bellare
UCSD
53
Closer look
There is a lot going on in this proof! Look over it slowly, checking each
step. In particular:
So K 0 will be consistent with (M1 , C1 ), . . . , (Mq , Cq ) with probability at most 2k /2q` , because there are 2k choices for K 0 and 2q`
choices for (C1 , . . . , Cq ).
This is subtle because B picks K 0 as a function of C1 , . . . , Cq . The claim
is justified by a counting argument. There are 2q` sequences (C1 , . . . , Cq ),
but for only 2k of them does there even exist a K 0 which is consistent with
(M1 , C1 ), . . . , (Mq , Cq ).
We will do many such proofs, and you will be asked to do them on quizzes,
so spend the time to understand this one! Ask now if you have doubts!
Mihir Bellare
UCSD
54
Our Assumptions
DES, AES are good block ciphers in the sense that they are PRF-secure
up to the inherent limitations of the birthday attack and known
key-recovery attacks.
You can assume this in designs and analyses.
But beware that the future may prove these assumptions wrong!
Mihir Bellare
UCSD
55
Exercise: Setup
Let F : {0, 1}k × {0, 1}l → {0, 1}L be a family of functions where
l, L ≥ 128. Consider the following game G:
procedure Initialize
$
$
K ← {0, 1}k ; b ← {0, 1}
procedure LR(x0 , x1 )
Ret F (K , xb )
procedure Finalize(b 0 )
Ret (b = b 0 )
We define
h
i
B
Advlr
(B)
=
2
·
Pr
G
⇒
true
−1.
F
Let (x01 , x11 ), . . . , (x0q , x1q ) be the queries that B makes to its oracle. (Each
query is a pair of l-bit strings, and there are q queries in all.) We say that
B is legitimate if x01 , . . . , x0q are all distinct, and also x11 , . . . , x1q are all
distinct. We say that F is LR-secure if Advlr
F (B) is “small” for every
legitimate B of “practical” resources.
Mihir Bellare
UCSD
56
Exercise: Questions
1.
Show that the legitimacy condition is necessary for LR-security to be
“interesting” by showing that if F is a block cipher then there is an
efficient, illegitimate B such that Advlr
F (B) = 1.
2.
Let B be a legitimate lr-adversary that makes q oracle queries and
has time-complexity t. Specify a prf-adversary A, also making q
oracle queries and having time-complexity close to t, such that
prf
Advlr
F (B) ≤ 2 · AdvF (A) .
Explain why this reduction shows that if F is a secure PRF then it is
LR-secure.
3.
Is the converse true? Namely, if F is LR-secure, then is it a secure
PRF? Answer YES or NO. If you say YES, justify this via a
reduction, and, if NO, via a counter-example.
Mihir Bellare
UCSD
57