115 lines
2.9 KiB
Markdown
115 lines
2.9 KiB
Markdown
# Lecture 5
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Proving that there are one-way functions relies on assumptions.
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Factoring Assumption: $\forall a, \exist \varepsilon (n)$, let $p,q\in prime,p,q<2^n$
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$$
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P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:a(N)\in \{p,q\}]<\varepsilon(n)
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$$
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Evidence: To this point, best known procedure to always factor has run time $O(2^{\sqrt{n}\sqrt{log(n)}})$
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Distribution of prime numbers:
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- We have infinitely many prime
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- Prime Number Theorem $\pi(n)\approx\frac{n}{\ln(n)}$, that means, $\frac{1}{\ln n}$ of all integers are prime.
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We want to (guaranteed to) find prime:
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$\pi(n)>\frac{2^n}{2n}$
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e.g.
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$$
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P[x\gets \{0,1\}^n:x\in prime]\geq {\frac{2^n}{2n}\over 2^n}=\frac{1}{2n}
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$$
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Theorem:
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$$
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f_{mult}:\{0,1\}^{2n}\to \{0,1\}^{2n},f_{mult}(x_1,x_2)=x_1\cdot x_2
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$$
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Idea: There are enough pairs of primes to make this difficult.
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> Reminder: Weak on-way if easy to compute and $\exist p(n)$,
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> $$P[a\ inverts=success]<1-\frac{1}{p(n)}$$
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> $$P[failure]>\frac{1}{p(n)}$$ high enough
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## Prove one-way function (under assumptions)
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To prove $f$ is on-way (under assumption)
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1. Show $\exists p.p.t$ solves $f(x),\forall x$.
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2. Proof by contradiction.
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- For weak: Provide $p(n)$ that we know works.
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- Assume $\exists a$ such that $P[a\ inverts]>1-\frac{1}{p(n)}$
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- For strong: Provide $p(n)$ that we know works.
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- Assume $\exists a$ such that $P[a\ inverts]>\frac{1}{p(n)}$
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Construct p.p.t B
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which uses $a$ to solve a problem, which contradicts assumption or known fact.
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Back to Theorem:
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We will show that $p(n)=8n^2$ works.
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We claim $\forall a$,
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$$
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P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(a(y))=y]<1-\frac{1}{8n^2}
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$$
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For the sake of contradiction, suppose
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$$
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\exists a \textup{ such that} P[success]>1-\frac{1}{8n^2}
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$$
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We will use this $a$ to design p.p.t $B$ which can factor 2 random primes with non-negligible prob.
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```python
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def A(y):
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# the adversary algorithm
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# expecting N to be product of random integer, don't need to be prime
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def is_prime(x):
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# test if x is a prime
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def gen(n):
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# generate number up to n bits
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def B(y):
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# N is the input cipher
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x1,x2=gen(n),gen(n)
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p=x1*x2
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if is_prime(x1) and is_prime(x2):
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return A(p)
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return A(y)
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```
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How often does B succeed/fail?
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B fails to factor $N=p\dot q$, if:
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- $x$ and $y$ are not both prime
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- $P_e=1-P(x\in prime)P(y\in prime)\leq 1-(\frac{1}{2n})^2=1-\frac{1}{4n^2}$
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- if $a$ fails to factor
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- $P_f<\frac{1}{8n^2}$
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So
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$$
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P[B\ fails]\leq P[E\cup F]\leq P[E]+P[F]\leq (1-\frac{1}{4n^2}+\frac{1}{8n^2})=1-\frac{1}{8n^2}
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$$
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So
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$$
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P[B\ succeed]\geq \frac{1}{8n^2}\ (non\ negligible)
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$$
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This contradicting factoring assumption. Therefore, our assumption that $a$ exists was wrong.
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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.
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