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change epsilon expression

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Zheyuan Wu
2024-12-05 21:23:34 -06:00
parent d18e86852c
commit 75ef366b1c
10 changed files with 37 additions and 37 deletions
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pages/CSE442T/CSE442T_L4.md
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@@ -2,16 +2,16 @@
## Recap
Negligible function $\varepsilon(n)$ if $\forall c>0,\exist N$ such that $n>N$, $\varepsilon (n)<\frac{1}{n^c}$
Negligible function $\epsilon(n)$ if $\forall c>0,\exist N$ such that $n>N$, $\epsilon (n)<\frac{1}{n^c}$
Ex: $\varepsilon(n)=2^{-n},\varepsilon(n)=\frac{1}{n^{\log (\log n)}}$
Ex: $\epsilon(n)=2^{-n},\epsilon(n)=\frac{1}{n^{\log (\log n)}}$
### Strong One-Way Function
1. $\exists$ a P.P.T. that computes $f(x),\forall x\in\{0,1\}^n$
2. $\forall a$ adversaries, $\exists \varepsilon(n),\forall n$.
2. $\forall a$ adversaries, $\exists \epsilon(n),\forall n$.
$$
P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]<\varepsilon(n)
P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]<\epsilon(n)
$$
_That is, the probability of success guessing should decreasing as encrypted message increase..._
@@ -28,7 +28,7 @@ Negation:
$\exists a$, $P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]=\mu_a(n)$ is not a negligible function.
That is, $\exists c>0,\forall N \exists n>N \varepsilon(n)>\frac{1}{n^c}$
That is, $\exists c>0,\forall N \exists n>N \epsilon(n)>\frac{1}{n^c}$
$\mu_a(n)>\frac{1}{n^c}$ for infinitely many $n$. or infinitely often.
@@ -41,7 +41,7 @@ $\mu_a(n)>\frac{1}{n^c}$ for infinitely many $n$. or infinitely often.
$f:\{0,1\}^n\to \{0,1\}^*$
1. $\exists$ a P.P.T. that computes $f(x),\forall x\in\{0,1\}^n$
2. $\forall a$ adversaries, $\exists \varepsilon(n),\forall n$.
2. $\forall a$ adversaries, $\exists \epsilon(n),\forall n$.
$$
P[x\gets \{0,1\}^n;y=f(x):f(a(y,1^n))=y]<1-\frac{1}{p(n)}
$$
@@ -116,14 +116,14 @@ The only way to efficiently factorizing the product of prime is to iterate all t
In other words:
$\forall a\exists \varepsilon(n)$ such that $\forall n$. $P[p_1\gets \prod n_j]$
$\forall a\exists \epsilon(n)$ such that $\forall n$. $P[p_1\gets \prod n_j]$
We'll show this is a weak one-way function under the Factoring Assumption.
$\forall a,\exists \varepsilon(n)$ such that $\forall n$,
$\forall a,\exists \epsilon(n)$ such that $\forall n$,
$$
P[p_1\gets \Pi_n;p_2\gets \Pi_n;N=p_1\cdot p_2:a(n)=\{p_1,p_2\}]<\varepsilon(n)
P[p_1\gets \Pi_n;p_2\gets \Pi_n;N=p_1\cdot p_2:a(n)=\{p_1,p_2\}]<\epsilon(n)
$$
where $\Pi_n=\{$ all primes $p<2^n\}$