Lecture 5

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

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

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.

Theorem:

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 $\forall a$,

For the sake of contradiction, suppose

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

So

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.