change epsilon expression

pages/CSE442T/CSE442T_L10.md

@@ -32,7 +32,7 @@ $$
 Let $e$ be the exponents
 
 $$
-P[p,q\gets \Pi_n;N\gets p\cdot q;e\gets \mathbb{Z}_{\phi(N)}^*;y\gets \mathbb{N}_n;x\gets \mathcal{A}(N,e,y);x^e=y\mod N]<\varepsilon(n)
+P[p,q\gets \Pi_n;N\gets p\cdot q;e\gets \mathbb{Z}_{\phi(N)}^*;y\gets \mathbb{N}_n;x\gets \mathcal{A}(N,e,y);x^e=y\mod N]<\epsilon(n)
 $$
 
 #### Theorem RSA Algorithm
@@ -190,7 +190,7 @@ $\mathcal{F}=\{f_i:D_i\to R_i\}_{i\in I}$
 2. $(i,t)\gets Gen(1^n)$ efficient. ($i\in I$ paired with $t$), $t$ is the "trapdoor info"
 3. $\forall i,D_i$ can be sampled efficiently.
 4. $\forall i,\forall x,f_i(x)$ can be computed in polynomial time.
-5. $P[(i,t)\gets Gen(1^n);y\gets R_i:f_i(\mathcal{A}(1^n,i,y))=y]<\varepsilon(n)$ (note: $\mathcal{A}$ is not given $t$)
+5. $P[(i,t)\gets Gen(1^n);y\gets R_i:f_i(\mathcal{A}(1^n,i,y))=y]<\epsilon(n)$ (note: $\mathcal{A}$ is not given $t$)
 6. (trapdoor) There is a p.p.t. $B$ such that given $i,y,t$, B always finds x such that $f_i(x)=y$. $t$ is the "trapdoor info"
 
 #### Theorem RSA is a trapdoor

pages/CSE442T/CSE442T_L11.md

@@ -45,7 +45,7 @@ Let $\{X_n\}_n$ and $\{Y_n\}_n$ be probability ensembles (separate of dist over
 $\{X_n\}_n$ and $\{Y_n\}_n$ are computationally **in-distinguishable** if for all non-uniform p.p.t adversary $D$ ("distinguishers")
 
 $$
-|P[x\gets X_n:D(x)=1]-P[y\gets Y_n:D(y)=1]|<\varepsilon(n)
+|P[x\gets X_n:D(x)=1]-P[y\gets Y_n:D(y)=1]|<\epsilon(n)
 $$
 
 this basically means that the probability of finding any pattern in the two array is negligible.

pages/CSE442T/CSE442T_L12.md

@@ -9,7 +9,7 @@ $$
 $$
 
 - If $\mu(n)\geq \frac{1}{p(n)}\gets poly(n)$ for infinitely many n, then $\{X_n\}$ and $\{Y_n\}$ are distinguishable.
-- Otherwise, indistinguishable ($|diff|<\varepsilon(n)$)
+- Otherwise, indistinguishable ($|diff|<\epsilon(n)$)
 
 Property: Closed under efficient procedures.
 
@@ -58,7 +58,7 @@ $$
 
 ### Next bit test (NBT)
 
-We say $\{X_n\}$ passes the next bit test if $\forall i\in\{0,1,...,l(n)-1\}$ on $\{0,1\}^{l(n)}$ and for all adversaries $\mathcal{A}:P[t\gets X_n:\mathcal{A}(t_1,t_2,...,t_i)=t_{i+1}]\leq \frac{1}{2}+\varepsilon(n)$ (given first $i$ bit, the probability of successfully predicts $i+1$ th bit is almost random $\frac{1}{2}$)
+We say $\{X_n\}$ passes the next bit test if $\forall i\in\{0,1,...,l(n)-1\}$ on $\{0,1\}^{l(n)}$ and for all adversaries $\mathcal{A}:P[t\gets X_n:\mathcal{A}(t_1,t_2,...,t_i)=t_{i+1}]\leq \frac{1}{2}+\epsilon(n)$ (given first $i$ bit, the probability of successfully predicts $i+1$ th bit is almost random $\frac{1}{2}$)
 
 Note that for any $\mathcal{A}$, and any $i$,
 
@@ -71,7 +71,7 @@ If $\{X_n\}\approx\{U_{l(n)}\}$ (pseudorandom), then $X_n$ must pass NBT for all
 Otherwise $\exists \mathcal{A},i$ where for infinitely many $n$,
 
 $$
-P[t\gets X_n:\mathcal{A}(t_1,t_2,...,t_i)=t_{i+1}]\leq \frac{1}{2}+\varepsilon(n)
+P[t\gets X_n:\mathcal{A}(t_1,t_2,...,t_i)=t_{i+1}]\leq \frac{1}{2}+\epsilon(n)
 $$
 
 We can build a distinguisher $D$ from $\mathcal{A}$.
@@ -147,6 +147,6 @@ $f(x)||x$
 
 Not all bits of $x$ would be hard to predict.
 
-**Hard-core bit:** One bit of information about $x$ which is hard to determine from $f(x)$. $P[$ success $]\leq \frac{1}{2}+\varepsilon(n)$
+**Hard-core bit:** One bit of information about $x$ which is hard to determine from $f(x)$. $P[$ success $]\leq \frac{1}{2}+\epsilon(n)$
 
 Depends on $f(x)$

pages/CSE442T/CSE442T_L3.md

@@ -50,13 +50,13 @@ $P_k[Dec_k(Enc_k(m))=m]=1$
 
 ## Negligible function
 
-$\varepsilon:\mathbb{N}\to \mathbb{R}$ is a negligible function if $\forall c>0$, $\exists N\in\mathbb{N}$ such that $\forall n\geq N, \varepsilon(n)<\frac{1}{n^c}$
+$\epsilon:\mathbb{N}\to \mathbb{R}$ is a negligible function if $\forall c>0$, $\exists N\in\mathbb{N}$ such that $\forall n\geq N, \epsilon(n)<\frac{1}{n^c}$
 
-Idea: for any polynomial, even $n^{100}$, in the long run $\varepsilon(n)\leq \frac{1}{n^{100}}$
+Idea: for any polynomial, even $n^{100}$, in the long run $\epsilon(n)\leq \frac{1}{n^{100}}$
 
-Example: $\varepsilon (n)=\frac{1}{2^n}$, $\varepsilon (n)=\frac{1}{n^{\log (n)}}$
+Example: $\epsilon (n)=\frac{1}{2^n}$, $\epsilon (n)=\frac{1}{n^{\log (n)}}$
 
-Non-example: $\varepsilon (n)=O(\frac{1}{n^c})\forall c$
+Non-example: $\epsilon (n)=O(\frac{1}{n^c})\forall c$
 
 ## One-way function
 
@@ -74,10 +74,10 @@ $$
 f:\{0,1\}^n\to \{0,1\}^*(n\to \infty)
 $$
 
-There is a negligible function $\varepsilon (n)$ such that for any adversary $a$ (n.u.p.p.t)
+There is a negligible function $\epsilon (n)$ such that for any adversary $a$ (n.u.p.p.t)
 
 $$
-P[x\gets\{0,1\}^n;y=f(x):f(a(y))=y,a(y)=x']\leq\varepsilon(n)
+P[x\gets\{0,1\}^n;y=f(x):f(a(y))=y,a(y)=x']\leq\epsilon(n)
 $$
 
 _Probability of guessing correct message is negligible_
@@ -95,7 +95,7 @@ Example: Suppose $f$ is one-to-one, then $a$ must find our $x$, $P[x'=x]=\frac{1
 
 Why do we allow $a$ to get a different $x'$?
 
-> Suppose the definition is $P[x\gets\{0,1\}^n;y=f(x):a(y)=x]\neq\varepsilon(n)$, then a trivial function $f(x)=x$ would also satisfy the definition.
+> Suppose the definition is $P[x\gets\{0,1\}^n;y=f(x):a(y)=x]\neq\epsilon(n)$, then a trivial function $f(x)=x$ would also satisfy the definition.
 
 To be technically fair, $a(y)=a(y,1^n)$, size of input $\approx n$, let them use $poly(n)$ operations.
 

pages/CSE442T/CSE442T_L4.md

@@ -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 $p<2^n\}$

pages/CSE442T/CSE442T_L5.md

@@ -2,10 +2,10 @@
 
 Proving that there are one-way functions relies on assumptions.
 
-Factoring Assumption: $\forall a, \exist \varepsilon (n)$, let $p,q\in prime,p,q<2^n$
+Factoring Assumption: $\forall a, \exist \epsilon (n)$, let $p,q\in prime,p,q<2^n$
 
 $$
-P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:a(N)\in \{p,q\}]<\varepsilon(n)
+P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:a(N)\in \{p,q\}]<\epsilon(n)
 $$
 
 Evidence: To this point, best known procedure to always factor has run time $O(2^{\sqrt{n}\sqrt{log(n)}})$

pages/CSE442T/CSE442T_L6.md

@@ -87,8 +87,8 @@ $F=\{f_i:D_i\to R_i\},i\in I$, $I$ is the index set.
 1. We can effectively choose $i\gets I$ using $Gen$.
 2. $\forall i$ we ca efficiently sample $x\gets D_i$.
 3. $\forall i\forall x\in D_i,f_i(x)$ is efficiently computable
-4. For any n.u.p.p.t $a$, $\exists$ negligible function $\varepsilon (n)$.
-    $P[i\gets Gen(1^n);x\gets D_i;y=f_i(x):f(a(y,i,1^n))=y]\leq \varepsilon(n)$
+4. For any n.u.p.p.t $a$, $\exists$ negligible function $\epsilon (n)$.
+    $P[i\gets Gen(1^n);x\gets D_i;y=f_i(x):f(a(y,i,1^n))=y]\leq \epsilon(n)$
 
 #### Theorem
 
@@ -107,7 +107,7 @@ Algorithm for sampling a random prime $p\gets \Pi_n$
    - Deterministic poly-time procedure
    - In practice, a much faster randomized procedure (Miller-Rabin) used
 
-        $P[x\cancel{\in} prime|test\ said\ x\ prime]<\varepsilon(n)$
+        $P[x\cancel{\in} prime|test\ said\ x\ prime]<\epsilon(n)$
 
 3. If not, repeat. Do this for polynomial number of times
 

pages/CSE442T/CSE442T_L9.md

@@ -108,7 +108,7 @@ Denote safe prime as $\tilde{\Pi}_n=\{p\in \Pi_n:q=\frac{p-1}{2}\in \Pi_{n-1}\}$
 Then
 
 $$
-P\left[p\gets \tilde{\Pi_n};a\gets\mathbb{Z}_p^*;g=a^2\neq 1;x\gets \mathbb{Z}_q;y=g^x\mod p:\mathcal{A}(y)=x\right]\leq \varepsilon(n)
+P\left[p\gets \tilde{\Pi_n};a\gets\mathbb{Z}_p^*;g=a^2\neq 1;x\gets \mathbb{Z}_q;y=g^x\mod p:\mathcal{A}(y)=x\right]\leq \epsilon(n)
 $$
 
 $p\gets \tilde{\Pi_n};a\gets\mathbb{Z}_p^*;g=a^2\neq 1$ is the function condition when we do the encryption on cyclic groups.

pages/CSE442T/Exam_reviews/CSE442T_E1.md

@@ -82,7 +82,7 @@ $$
 
 #### Definition 27.2 Negligible function
 
-A function $\varepsilon(n)$ is negligible if for every $c$. there exists some $n_0$ such that for all $n>n_0$, $\epsilon (n)\leq \frac{1}{n^c}$.
+A function $\epsilon(n)$ is negligible if for every $c$. there exists some $n_0$ such that for all $n>n_0$, $\epsilon (n)\leq \frac{1}{n^c}$.
 
 #### Definition 27.3 Strong One-Way Function
 

pages/Math4111/Math4111_L13.md

@@ -106,19 +106,19 @@ To avoid confusion with sets, we use $(p_n)_{n=1}^\infty$ or $(p_n)$
 
 Let $(X,d)$ be a metric space. Let $(p_n)$ be a sequence in $X$.
 
-Let $p\in X$. We say $(p_x)$ **converges** to $p$ if $\forall \varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\varepsilon$. ($p_n\in B_\varepsilon (p)$)
+Let $p\in X$. We say $(p_x)$ **converges** to $p$ if $\forall \epsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\epsilon$. ($p_n\in B_\epsilon (p)$)
 
 Notation $\lim_{n\to \infty} p_n=p$, $p_n\to p$
 
 We say $(p_n)$ converges if $\exists p\in X$ such that $p_n\to p$.
 
-i.e. $\exists p\in X$ such that $\forall\varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N,d(p_n,p)<\varepsilon$
+i.e. $\exists p\in X$ such that $\forall\epsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N,d(p_n,p)<\epsilon$
 
 We say $(p_n)$ **diverges** if $(p_n)$ doesn't converge.
 
 i.e. $\forall p\in X$, $p_n\cancel{\to} p$
 
-i.e. $\forall p\in X$ such that $\exists \varepsilon>0,\forall N\in\mathbb{N}$ such that $\exists n\geq N,d(p_n,p)\geq\varepsilon$
+i.e. $\forall p\in X$ such that $\exists \epsilon>0,\forall N\in\mathbb{N}$ such that $\exists n\geq N,d(p_n,p)\geq\epsilon$
 
 #### Definition 3.2
 
@@ -128,16 +128,16 @@ Example:
 
 $X=\mathbb{C}$, $s_n=\frac{1}{n}$
 
-Then $s_n\to 0$ i.e. $\forall \varepsilon>0 \exists N\in \mathbb{N}$ such that $\forall n\geq N$, $|s_n-0|<\varepsilon$.
+Then $s_n\to 0$ i.e. $\forall \epsilon>0 \exists N\in \mathbb{N}$ such that $\forall n\geq N$, $|s_n-0|<\epsilon$.
 
 Proof:
 
-Let $\varepsilon >0$ (arbitrary)
+Let $\epsilon >0$ (arbitrary)
 
-Let $N\in \mathbb{N}$ be greater than $\frac{1}{\varepsilon}$ (by Archimedean property) e.g. $N=\frac{1}{\varepsilon}+1$ (we choose $N$)
+Let $N\in \mathbb{N}$ be greater than $\frac{1}{\epsilon}$ (by Archimedean property) e.g. $N=\frac{1}{\epsilon}+1$ (we choose $N$)
 
 Let $n\geq N$ (arbitrary)
 
-Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \varepsilon$
+Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
 
 EOP