Foundations of Cryptography
Lecture 3: One-way on its Iterates, Authentication
Lecturer: Moni Naor
Recap of last week’s lecture
• One-way functions are essential to the two guard
identification problem.
– Important idea: simulation
• Examples of one-way functions
– Subset sum, discrete log, factoring
• Weak one-way functions
– Constructing strong one-way functions from weak one-way
functions
– Important ideas: hardness amplification; reduction
From Weak to Strong One-way
Functions
• Given
– a function f that is guaranteed to be a weak one-way
• Let p(n) be such that Prob[A[f(x)]  f-1(f(x)) ] ≤ 1-1/p(n)
– can we construct a function g that is (strong) one-way?
An instance of a hardness amplification problem
• Simple idea: repetition. For some polynomial q(n) define
g(x1, x2 ,…, xq(n) )=f(x1), f(x2), …, f(xq(n))
• To invert g need to succeed in inverting f in all q(n) places
– If q(n) = p2(n) seems unlikely (1-1/p(n))p2(n) ≈ e-p(n)
– But how to we show? Sequential repetition intuition – not a proof.
Want: Inverting g with low probability implies
inverting f with high probability
• Given a machine B that inverts g want a machine B’
– operating in similar time bounds
– inverts f with high probability
• Idea: given y=f(x) plug it in some place in g and generate the rest
of the locations at random
z=(y, f(x2), …, f(xq(n)))
• Ask machine B to invert g at point z
• Probability of success should be at least (exactly) B’s Probability of
inverting g at a random point
• Running B Once is not enough
• How to amplify?
– Repeat while keeping y fixed
– Put y at random position (or sort the inputs to g )
Proof of Amplification for Repetition of Two
Concentrate on two-repetition: g(x1, x2) = f(x1), f(x2)
• Goal: show that the probability of inverting g is roughly squared the
probability of inverting f
just as would be sequentially
Claim:
Let (n) be a function that for some p(n) satisfies
1/p(n) ≤ (n) ≤ 1-1/p(n)
Let ε(n) be any inverse polynomial function
Suppose that for every polynomial time A and sufficiently large n
Prob[A[f(x)]  f-1(f(x)) ] ≤ (n)
Then: for every polynomial time B and sufficiently large n
Prob[B[g(x1, x2)]  g-1(g(x1, x2)) ] ≤ 2(n) + ε(n)
Proof of Amplification for Two Repetition
Given a better than 2+ε algorithm B for inverting g
construct the following B’ for inverting f:
• B’(y): Inversion algorithm for f
– Repeat t times
•
•
•
•
Choose x’ at random and compute y’=f(x’)
Run B(y,y’).
Check the results
Helpful for
constructive
If correct: Halt with success
– Output failure
algorithm
Inner
loop
Probability of Success
•
Define
S={y=f(x) | Prob[Inner loop successful| y ] > β}
Since the choices of the x’’s are independent of each other
Prob[B’ succeeds|yS] > 1-(1- β)t
Taking t= n/β means that when yS, then almost surely B’
will succeed in inverting it
•
Hence want to show that Prob[yS] > (n)
–
Probability is over the choice of x
The success of B
• Fix the random bits of B.
Define P={(y1, y2)| B succeeds on (y1,y2)}
P= P ⋂ {(y1,y2 )| y1,y2 S}
⋃ P ⋂ {(y1,y2 )| y1 S}
y1
⋃ P ⋂ {(y1,y2 )| y2 S}
Well behaved part
y2
Want to bound P by a square
P
Behaves as in independent choices
S is the only success...
But
Prob[B[y1, y2]  g-1(y1, y2) | y1  S] ≤ β
and similarly
Prob[B[y1, y2]  g-1(y1, y2) | y2  S] ≤ β
So
Prob[(y1, y2)  P and y1,y2  S]
≥ Prob[(y1, y2)  P ] - 2β
≥ 2+
ε - 2β
Setting β =ε/3 we have
Prob[(y1, y2) P and y1,y2  S] ≥ 2 +
ε/3
Contradiction
But
Prob[(y1, y2) P and y1,y2 S]
≤ Prob[y1 S] Prob[y2 S]
= Prob2[y S]
So
Prob[y S] ≥ √(α2+ ε/3) > α
Is there an ultimate one-way function?
aka `universal’
Do not know: P≠NP implies the existence of one-way functions,
Can we show a specific function f so that if some one-way exists, then f
is one-way?
• If f1:{0,1}* → {0,1}* and f2:{0,1}* → {0,1}* are
guaranteed to:
– Be polynomial time computable
– At least one of them is one-way.
then can construct a function g:{0,1}* → {0,1}* which is oneway:
g(x1, x2 ) = (f1(x1),f2 (x2 ))
Robust Combiner
Can generalizes to a 1-out-m combiner
A function that takes that
much time to compute
The Construction
• If a 5n2 time one-way function is guaranteed to exist, can
construct an O(n2 log n) one-way function g:
– Idea: enumerate all Turing Machines and make sure they run at
most 5n2 steps
g(x1, x2 ,…, xlog (n) )=M1(x1), M2(x2), …, Mlog n(xlog (n))
– Eventually: get to the TM of the one-way function
• If a one-way function is guaranteed to exist, then there
exists a 5n2 time one-way:
– Idea: concentrate on the prefix, ignore the rest
n’ where n=p(n’)
Ultimate one-way function conclusions
Original proof due to L. Levin
• Be careful what you wish for
• Problem with resulting one-way function:
– Cannot learn about behavior on large inputs from small inputs
– Whole rational of considering asymptotic results is eroded!
• Construction does not work for non-uniform one-way
functions
• Notion of robust combiner seems fundamental
– See “Robust Combiners for Oblivious Transfer and Other
Primitives” by Harnik, Kilian, Naor, Reingold and Rosen,
Eurocrypt 2005
Distributionally One-Way Functions
• A function f:{0,1}*  {0,1}* is one-way if:
– it is computable in poly-time
– the probability of successfully finding an inverse in poly-time is
negligible (on a random input)
• A function f :{0,1}*  {0,1}* is distributionally one-way if:
– it is computable in poly-time
– No poly-time algorithm can successfully find a random inverse (on a
random input)
• Distribution on inverting algorithm far from uniform on the pre-images
Example: function from two guards problem
Theorem [Impagliazzo Luby 89]: distributionally one-way
functions exist iff one-way functions exists
Identification - many times
• Alice would want to send an `approve’ message
to Bob many times.
• They want to prevent Eve from interfering
– Bob should be sure that Alice indeed approved each
time. How to specify?
Alice
Bob
Eve
Specification of the Problem
Alice and Bob communicate through a channel
Bob has an external counter C (# of times Alice approved)
Eve completely controls the channel
Requirements:
• If Alice wants to approve and Eve does not interfere – Bob increases
the counter C
• The number of times Alice approves is a bound the value of counter C
• If Alice wants to approve and Eve does interfere - no requirements
from the counter C until there is a quiescent period
– A time that Alice wants to approve and Eve does not interfere
Not the only possible specification!
Can mandate that an approval was sent since the last time counter increased
Solution to the many time identification
problem
Let k be an upper bound on the number of
identifications
• If Alice and Bob share in the setup phase k
passwords
y1, y2, … , yk
• Each time Alice want to `approve’ she sends the next
unused password.
– In the ith time: yi
• Bob compare with the next password on the list
Can they do it with sharing less than k passwords?
Compressing the list
Giving y instead of x may
turn out to be dangerous…
– f is a one-way function
– Let k be an upper bound on the number of identifications
• Assume that
• Setup phase: Alice chooses x{0,1}n, defines a sequence
Where yi=f(k-i)(x)
Gives Bob x
y1, y2, … yk=x
• When Alice wants to approve for the ith time:
send the special symbol $ followed by i and yi=f(k-i)(x)
• Bob stores x; Set C Ã 0
• If Bob gets a $ followed by symbols on channel
– denote them (j,z);
– Compare j to C; reject if j ≤ C.
– Check whether z=f(k-j)(x)
• If equal Set C Ã j
Is it secure?
• Need care in choosing f
• Should be difficult to invert any one of the iterated
instances of f
One-way on its iterates
A function f: {0,1}n → {0,1}n is called one-way on its iterates, if
• f is a polynomial-time computable function
• for every probabilistic polynomial-time algorithm A, every polynomial
p(¢), and all sufficiently large n’s: for all k ≤ p(n)
Prob[A[f(k)(x)]  f-1(f (k)(x)) ] ≤ 1/p(n)
Where x is chosen uniformly in {0,1}n and the probability is also over the
internal coin flips of A
• From homework: not all one-way functions are one-way on their iterates
• Every one-way permutation is one-way on its iterates
If you start at a
• Subset sum function one-way on its iterates
– If it one-way then it is one-way of its iterates
random point and
iterate – still random
Example: the squaring function (Rabin)
f(x,N)= (x2 mod N,N)
Quadratic residue mod a prime:
• If s and r satisfy s=r2 mod P then s is called a quadratic residue modulo P
• If P is a prime then:
– s=r2 mod P has exactly two solutions mod P if 0<s<P. Can denote +/-r
– quadratic residues: multiplicative subgroup with (P-1)/2 elements.
– If P=1 mod 4 then -1 is a quadratic residue mod P.
• Both square-roots are either quadratic residues or non residues
– If P=3 mod 4 then -1 is a non-quadratic residue mod P.
• One square-roots is a quadratic residue, the other not.
• Squaring mod P is a permutation on the quadratic residues!
• Computing square-roots: if r=s(p+1)/4 mod P square, then
r2=s(p+1)/2 =s∙s(p-1)/2 = +/- s mod P
•
•
If N=P∙Q then s is a quadratic residue modulo N if and only it is a quadratic
residue for both P and Q
If N=P∙Q where P,Q = 3 mod 4 - called Blum Integers
– Each quadratic residue has 4 square-roots
– Exactly one of which is quadratic residue in itself
– Squaring mod N is a permutation on the quadratic residues!
Finding Square-roots and factoring are equivalent
• If know the factorization of N=P∙Q, then can compute
square-roots
• If there is a procedure that computes square-roots
correctly for non-negligible fraction – can boost it
– Random self reducibility
• If we know (r,t) such that
– s=r2 =t2 mod N
– r =t mod P
– r ≠t mod Q
Then we can factor by computing GCD(t-r,N)
• Homework: show how to use a square-root computing
routine to factor while preserving the probability of success.
A one-way on its iterates function
• To fully specify the function – need a starting
procedure for generating
– N=P∙Q where P,Q=3 mod 4
– Easy to specify given
• deterministic primality testing (even probabilistic is sufficient)
• density of primes
– A quadratic residue mod N
• Easy by generating a random square
• Resulting function – one-way on its iterates
Back to the compressing the list
Giving y instead of x may
turn out to be dangerous…
– f is a one-way function
– Let k be an upper bound on the number of identifications
• Assume that
• Setup phase: Alice chooses x{0,1}n, defines a sequence
Where yi=f(k-i)(x)
Gives Bob x
y1, y2, … yk=x
• When Alice wants to approve for the ith time:
send the special symbol $ followed by i and yi=f(k-i)(x)
• Bob stores x; Set C Ã 0
• If Bob gets a $ followed by symbols on channel
– denote them (j,z);
– Compare j to C; reject if j ≤ C.
– Check whether z=f(k-j)(x)
• If equal Set C Ã j
Security of scheme
Two possible evil actions:
•Substitute a correct value
•Invent a value, forge
If scheme can be broken:
• There is the first time where Eve sent a false value z as yi
By the specification of the protocol:
– If Eve substitutes a value yi which was sent by Alice with her own z
– she is caught
Hence first false z is also an attempt to forge: Alice approved only i-1
times but Eve convinced Bob to accepts i times
If probability of breaking is at least 1/p(n) To forge: Eve must invert yi-1
• There is a j ≤ k where Eve does this with probability at least
1/kp(n)
Important idea: Existence of a large step
…Security of scheme
For this j: can break the (k-j-1)th iterate of f with
probability at least 1/kp(n)
– Given yj-1 = f(k-j-1)(x) compute
y1=f(j-1)(yj-1), y2=f(j-2)(yj-1), …, yj-2=f(yj-1), yj-1
and simulate the adversary for j rounds
Can send the expect values from Alice
– Adversary sees exactly the same distribution as in real life
• Forging at step j must be done by inverting yj-1
• Hence probability adversary succeeds in forgery at step j is at
least 1/kp(n)
From such a success: can invert the (k-j-1)th iterate of f on x
Problems with the scheme
• Need to know an upper bound k on the number of
identifications
• Need to perform work proportional to k before first identification
(what if it flops)
• Total work (in all k sessions) by Alice: O(k2)
– For Bob, if stores last value: O(k)
– If Alice stores all k values yj: total work (in all k sessions) only O(k)
– Homework: how can Alice store O(log k) values and perform
amortized O(log k) work
• More problems:
– need to maintain state, both Alice and Bob (in addition to the counter)
– What happens when there are two verifiers
Possible Pitfalls: why give x and not y
If Bob does not check from scratch
Alternative protocol: Bob knows y0 =f(k)(x)
To verify at step i: Bob computes f(i)(z) and compares to y0
then:
• Eve might substitute yj with a value z which she can invert in
subsequent sessions.
– If possible to find “easy siblings” could be dangerous
– Homework: show that there is a function f that is
• One-way on its iterates
• Given x it is easy to find x' such that f(x)=f(x’) and it is easy to invert
f on x’
Question
• Is it possible to have a protocol based on a function
that it one-way on its iterates without bob
maintaining a state?
Want a scheme with unlimited use
If we have a function that only Alice can compute but
both Bob and Charlie can verify
• Alice can compute for session number i the value
f(i)
• Problem: interleaving of verifiers – can replay
• Solution: challenge response
– Verifier chooses a random nonce r and asks to see
f(r)
To be continued!
The authentication problem
one-time version
• Alice would want to send a message m  {0,1}n
to Bob
• They want to prevent Eve from interfering
– Bob should be sure that the message m’ he receives is
equal to the message m Alice sent
m
Alice
Eve
Bob
Specification of the Problem
Alice and Bob communicate through a channel
Bob has an external register R  N (no message) ⋃ {0,1}n
Eve completely controls the channel
Requirements:
• Completeness: If Alice wants to send m  {0,1}n and Eve
does not interfere – Bob has value m in R
• Soundness: If Alice wants to send m and Eve does interfere
– R is either N or m (but not m’ ≠m )
– If Alice does not want to send a message R is N
Since this is a generalization of the identification problem
– must use shared secrets and probability or complexity
Probabilistic version:
• for any behavior from Eve, for any message m  {0,1}n, the
probability that Bob is in state m’ ≠ m or N is at most ε
Authentication using hash functions
• Suppose that
–
–
–
–
H= {h| h: {0,1}n → {0,1}k } is a family of functions
Alice and Bob share a random function h  H
To authenticate message m  {0,1}n Alice sends (m,h(m))
When receiving (m’,z) Bob computes h(m’) and compares to z
• If equal, moves register R to m’
• If not equal, register R stays in N
• What properties do we require from H
– hard to guess h(m’) - at most ε
• But clearly not sufficient: one-time pad.
– hard to guess h(m’) even after seeing h(m) - at most ε
• Should be true for any m’
– Short representation for h - must have small log|H|
– Easy to compute h(m) given h and m
Universal hash functions
• Given that for hH we have h: {0,1}n → {0,1}k we know that
ε≥2-k
• A family where this is an equality is called universal2
Definition: a family of functions H= {h| h: {0,1}n → {0,1}k } is
called Strongly Universal2 or pair-wise independent if:
– for all m1, m2 {0,1}n and y1, y2 {0,1}k we have
Prob[h(m1) = y1 and h(m2) = y2 ] = 2-2k
Where the probability is over a randomly chosen h H
In particular Prob[h(m2) = y2 | h(m1) = y1 ] = 2-k
Theorem: when a strongly universal2 family is used in the protocol,
Eve’s probability of cheating is at most 2-k
Constructing universal hash functions
The linear polynomial construction:
• fix a finite field F of size at least the message space 2n
– Could be either GF[2n] or GF[P] for some prime P ≥ 2n
• The family H of functions h: F→ F is defined as
H= {ha,b(m) = a∙m + b | a, b  F}
Claim: the family above is strongly universal2
Proof: for every m1≠m2, y1, y2 F there are unique a, b  F such that
a∙m1+b = y1
a∙m2+b = y2
Size: each hH represented by 2n bits
Constructing universal hash functions
The inner product construction:
• fix a finite field F of size at least the target space 2k
•
•
•
– Could be either GF[2k] or GF[P] for some prime P ≥ 2k
Let n= ℓ ∙ k
Treat each message m{0,1}n as an (ℓ +1)-vector over F where the first entry is 1.
Denote by (m0, m1, … ,mℓ)
The family H of functions h: Fℓ → F defined by all (ℓ+1)-vectors a=(a0, a1, … ,aℓ)
H= {ha(m)= ∑i=0ℓ ai ∙mi | a0, a1, … ,aℓ  F}
Claim: the family above is strongly universal2
Proof: for every (m0, m1, … ,ml) , (m’0, m’1, … ,m’l) y1, y2 F there are the same
number of solutions to
∑i=0ℓ ai ∙mi = y1
∑i=0ℓ ai ∙m’i = y2
Size: each hH represented by n+k bits
Lower bound on size of strongly universal
hash functions
Theorem: let H= {h| h: {0,1}n → {0,1} } be a
family of pair-wise independent functions. Then
|H| is Ω(2n)
More precisely, to obtain a d-wise independence
family |H| should be Ω(2n└d/2┘)
Theorem: see
N. Alon and J. Spencer, The Probabilistic Method
Chapter 15 on derandomization, proposition 2.3
An almost perfect solution
By allowing ε to be slightly larger than 2-k we can get much
smaller families
Definition: a family of functions
H= {h| h: {0,1}n → {0,1}k } is called δ-Universal2 if
for all m1, m2 {0,1}n where m1 ≠ m2 we have
Prob[h(m1) = h(m2) ] ≤ δ
Properties:
• Strongly-universal2 implies 2-k -Universal2
• Opposite not true: the function h(x) = x…
An almost perfect solution
Idea: combine
• a family of δ-Universal2 functions H1= {h| {0,1}n → {0,1}k }
with
• a Strongly Universal2 family H2= {h| {0,1}k → {0,1}k }
Consider the family H where each h  H is {0,1}n → {0,1}k and is defined by
h1  H1 and h2  H2
As before Alice sends m, h(m)
h(x) = h2(h1(x))
Claim : probability of cheating is at most δ + 2-k
Proof: when Eve sends m’, y’ we must have m ≠ m‘ but either
– y’ = h(m), which means that Eve succeeds with probability at most δ + 2-k
• Collision in h1 Or in h2
Or
– y’ ≠ h(m) which means that Eve succeeds with probability at most 2-k
• Collision in h2
Size: each hH represented by log |H1 |+ log |H2|
Constructing almost universal hash functions
The polynomial evaluation construction {0,1}n → {0,1}k :
• fix a finite field F of size at least the target space 2k
– Could be either GF[2k] or GF[P] for some prime P ≥ 2k
• Let n= ℓ∙ k
• Treat each (non-zero) message m{0,1}n as a degree (ℓ-1)-polynomial
over F. Denote by Pm.
m
• The family H of functions h: Fℓ → F is defined by all elements in F:
H= {hx (m)= Pm (x)| xF}
Claim: the family above is δ-Universal2 for δ= (ℓ-1)/2k
Proof: the maximum number of points where two different degree (ℓ-1)
polynomials agree is ℓ-1
Size: each hH represented by k bits
Composing universal hash functions
Concatenation
Let H where each h  H is {0,1}n → {0,1}k be a family of δUniversal2 functions
Consider the family H’ where each h’  H’ is {0,1}2n → {0,1}2k and
where
h’(x1 ,x2) = h(x1 ), h(x2)
for some h  H
Claim: the family above is δ-Universal2
Proof: let x1, x2 and x’1, x’2 be a pair of inputs.
• If x1 ≠ x’1 collision must occur in first part h(x1)=h(x’1)
• Else, x2 ≠ x’2 and collision must occur in second part
h(x2)=h(x’2)
In either case the probability is at most δ
Composing universal hash functions
n1
Composition
Let
n2
• H1= {h| h:{0,1}n1 → {0,1}n2 }
with
• H2= {h| h: {0,1}n2 → {0,1}n3 }
n3
be families of δ-Universal2 functions
Consider the family H where each hH is {0,1}n1 → {0,1}n3 is defined by h1  H1 and
h2  H2
h(x) = h2(h1(x))
Claim: the family above is 2δ-Universal2
Proof: the collision must occur either at the first hash function or the second hash function.
Each event happens with probability at most δ and we apply the union bound
The Tree Construction
m
h1
h2
h3
Set n=ℓ∙k. Each hi :{0,1}2k → {0,1}k is chosen independently from a δ-Universal family H.
The result is a family of functions {0,1}n → {0,1}k which is tδ-Universal
t is the number of levels in the tree
Size: t log |H|
Can construct functions from huge domains
Homework
• Given ε,n what is the number of bits needed to
specify an authentication scheme?
• Bonus: Can interaction help?
– Can the number of shared secret bits be smaller than in
a unidirectional scheme
– Can the number of shared bits depend on ε only?
What about the public-key problem?
• Recall: Bob and Charlie share the set-up phase information
• Is it possible to satisfy the requirements:
– Completeness: If Alice wants to send m  {0,1}n and Eve does
not interfere – Bob has value m in R
– Soundness: If Alice wants to send m and Eve and Charlie do
interfere
• R is either N or m (but not m’ ≠m )
• If Alice does not want to send a message R is N
• Who chooses which m Alice will want to approve?
– Adversary does. This is a chosen message attack
• As before: complexity to the rescue