# 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.

```python
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.