129 lines
3.7 KiB
Markdown
129 lines
3.7 KiB
Markdown
# Lecture 4
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## Recap
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Negligible function $\varepsilon(n)$ if $\forall c>0,\exist N$ such that $n>N$, $\varepsilon (n)<\frac{1}{n^c}$
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Ex: $\varepsilon(n)=2^{-n},\varepsilon(n)=\frac{1}{n^{\log (\log n)}}$
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### Strong One-Way Function
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1. $\exists$ a P.P.T. that computes $f(x),\forall x\in\{0,1\}^n$
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2. $\forall a$ adversaries, $\exists \varepsilon(n),\forall n$.
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$$
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P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]<\varepsilon(n)
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$$
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_That is, the probability of success guessing should decreasing as encrypted message increase..._
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To negate statement 2:
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$$
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P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]=\mu_a(n)
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$$
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is a negligible function.
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Negation:
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$\exists a$, $P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]=\mu_a(n)$ is not a negligible function.
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That is, $\exists c>0,\forall N \exists n>N \varepsilon(n)>\frac{1}{n^c}$
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$\mu_a(n)>\frac{1}{n^c}$ for infinitely many $n$. or infinitely often.
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> Keep in mind: $P[success]=\frac{1}{n^c}$, it can try $O(n^c)$ times and have a good chance of succeeding at least once.
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## New materials
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### Week One-Way Function
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$f:\{0,1\}^n\to \{0,1\}^*$
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1. $\exists$ a P.P.T. that computes $f(x),\forall x\in\{0,1\}^n$
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2. $\forall a$ adversaries, $\exists \varepsilon(n),\forall n$.
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$$
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P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]<1-\frac{1}{p(n)}
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$$
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_The probability of success should not be too close to 1_
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### Probability
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### Useful bound $0<p<1$
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$1-p<e^{-p}$
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(most useful when $p$ is small)
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For an experiment has probability $p$ of failure and $1-p$ of success.
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We run experiment $n$ times independently.
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$P[$success all n times$]=(1-p)^n<(e^{-p})^n=e^{-np}$
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Theorem: If there exists a weak one-way function, there there exists a strong one-way function
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In particular, if $f:\{0,1\}^n\to \{0,1\}^*$ is weak one-way function.
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$\exists$ polynomial $q(n)$ such that
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$$
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g(x):\{0,1\}^{nq(n)}\to \{0,1\}^*
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$$
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and for every $n$ bits $x_i$
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$$
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g(x_1,x_2,..,x_{q(n)})=(f(x_1),f(x_2),...,f(x_{q(n)}))
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$$
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is a strong one-way function.
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Proof:
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1. Since $\exist P.P.T.$ that computes $f(x),\forall x$ we use this $q(n)$ polynomial times to compute $g$.
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2. (Idea) $a$ has to succeed in inverting $f$ all $q(n)$ times.
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Since $x$ is a weak one-way, $\exists$ polynomial $p(n)$. $\forall q, P[q$ inverts $f]<1-\frac{1}{p(n)}$ (Here we use $<$ since we can always find a polynomial that works)
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Let $q(n)=np(n)$.
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Then $P[a$ inverting $g]\sim P[a$ inverts $f$ all $q(n)]$ times. $<(1-\frac{1}{p(n)})^{q(n)}=(1-\frac{1}{p(n)})^{np(n)}<(e^{-\frac{1}{p(n)}})^{np(n)}=e^{-n}$ which is negligible function.
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EOP
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_we can always force the adversary to invert the weak one-way function for polynomial time to reach the property of strong one-way function_
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Example: $(1-\frac{1}{n^2})^{n^3}<e^{-n}$
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### Some candidates of one-way function
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#### Multiplication
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$Mult(m_1,m_2)=\begin{cases}
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1,m_1=1 | m_2=1\\
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m_1\cdot m_2
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\end{cases}$
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But we don't want trivial answers like (1,1000000007)
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Idea: Our "secret" is 373 and 481, Eve cna see the product 179413.
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Not strong one-way for all integer inputs because there are trivial answer for $\frac{3}{4}$ of all outputs. `Mult(2,y/2)`
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Factoring Assumption:
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The only way to efficiently factorizing the product of prime is to iterate all the primes.
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In other words:
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$\forall a\exists \varepsilon(n)$ such that $\forall n$. $P[p_1\gets \prod n_j]$
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We'll show this is a weak one-way function under the Factoring Assumption.
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$\forall a,\exists \varepsilon(n)$ such that $\forall n$,
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$$
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P[p_1\gets \Pi_n;p_2\gets \Pi_n;N=p_1\cdot p_2:a(n)=\{p_1,p_2\}]<\varepsilon(n)
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$$
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where $\Pi_n=\{$ all primes $p<2^n\}$ |