NoteNextra-origin/pages/CSE442T/CSE442T_L5.md

Lecture 5

Proving that there are one-way functions relies on assumptions.

Factoring Assumption: \forall a, \exist \varepsilon (n), let p,q\in prime,p,q<2^n

P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:a(N)\in \{p,q\}]<\varepsilon(n)

Evidence: To this point, best known procedure to always factor has run time O(2^{\sqrt{n}\sqrt{log(n)}})

Distribution of prime numbers:

• We have infinitely many prime
• Prime Number Theorem \pi(n)\approx\frac{n}{\ln(n)}, that means, \frac{1}{\ln n} of all integers are prime.

We want to (guaranteed to) find prime:

\pi(n)>\frac{2^n}{2n}

e.g.

P[x\gets \{0,1\}^n:x\in prime]\geq {\frac{2^n}{2n}\over 2^n}=\frac{1}{2n}

Theorem:

f_{mult}:\{0,1\}^{2n}\to \{0,1\}^{2n},f_{mult}(x_1,x_2)=x_1\cdot x_2

Idea: There are enough pairs of primes to make this difficult.

Reminder: Weak on-way if easy to compute and \exist p(n),

P[a\ inverts=success]<1-\frac{1}{p(n)}

P[failure]>\frac{1}{p(n)} high enough

Prove one-way function (under assumptions)

To prove f is on-way (under assumption)

1. Show \exists p.p.t solves f(x),\forall x.
2. Proof by contradiction.
   • For weak: Provide p(n) that we know works.
     • Assume \exists a such that P[a\ inverts]>1-\frac{1}{p(n)}
   • For strong: Provide p(n) that we know works.
     • Assume \exists a such that P[a\ inverts]>\frac{1}{p(n)}

Construct p.p.t B which uses a to solve a problem, which contradicts assumption or known fact.

Back to Theorem:

We will show that p(n)=8n^2 works.

We claim \forall a,

P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(a(y))=y]<1-\frac{1}{8n^2}

For the sake of contradiction, suppose

\exists a \textup{ such that} P[success]>1-\frac{1}{8n^2}

We will use this a to design p.p.t B which can factor 2 random primes with non-negligible prob.

def A(y):
    # the adversary algorithm
    # expecting N to be product of random integer, don't need to be prime

def is_prime(x):
    # test if x is a prime

def gen(n):
    # generate number up to n bits

def B(y):
    # N is the input cipher
    x1,x2=gen(n),gen(n)
    p=x1*x2
    if is_prime(x1) and is_prime(x2):
        return A(p)
    return A(y)

How often does B succeed/fail?

B fails to factor N=p\dot q, if:

• x and y are not both prime
  • P_e=1-P(x\in prime)P(y\in prime)\leq 1-(\frac{1}{2n})^2=1-\frac{1}{4n^2}
• if a fails to factor
  • P_f<\frac{1}{8n^2}

So

P[B\ fails]\leq P[E\cup F]\leq P[E]+P[F]\leq (1-\frac{1}{4n^2}+\frac{1}{8n^2})=1-\frac{1}{8n^2}

So

P[B\ succeed]\geq \frac{1}{8n^2}\ (non\ negligible)

This contradicting factoring assumption. Therefore, our assumption that a exists was wrong.

Therefore \forall a, P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(a(y))=y]<1-\frac{1}{8n^2} is wrong.