Fix typo

Fix typos introduces more

pages/CSE442T/CSE442T_L12.md

@@ -82,7 +82,7 @@ The NBT(Next bit test) is complete.
 
 If $\{X_n\}$ on $\{0,1\}^{l(n)}$ passes NBT, then it's pseudorandom.
 
-Idea of proof: full proof is on the text.
+Ideas of proof: full proof is on the text.
 
 Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$
 
@@ -137,7 +137,7 @@ The other part of proof will be your homework, damn.
 
 If one-way function exists, then Pseudorandom Generator exists.
 
-Idea of proof:
+Ideas of proof:
 
 Let $f:\{0,1\}^n\to \{0,1\}^n$ be a strong one-way permutation (bijection).
 

pages/CSE442T/CSE442T_L13.md

@@ -16,7 +16,7 @@ $$
 Pr[x\gets \{0,1\}^n;y=f(x);A(1^n,y)=h(x)]\leq \frac{1}{2}+\epsilon(n)
 $$
 
-Idea: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function.
+Ideas: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function.
 
 Given $y=f(x)$, it is hard to recover $x$. A cannot produce all of $x$ but can know some bits of $x$.
 
@@ -46,7 +46,7 @@ $\langle x,1^n\rangle=x_1+x_2+\cdots +x_n\mod 2$
 
 $\langle x,0^{n-1}1\rangle=x_ n$
 
-Idea of proof:
+Ideas of proof:
 
 If A could reliably find $\langle x,1^n\rangle$, with $r$ being completely random, then it could find $x$ too often.
 

pages/CSE442T/CSE442T_L14.md

@@ -123,7 +123,7 @@ $Enc_F(m):$ let $r\gets U_n$; output $(r,F(r)\oplus m)$.
 
 $Dec_F(m):$ Given $(r,c)$, output $m=F(r)\oplus c$.
 
-Idea: Adversary sees $r$ but has no idea about $F(r)$. (we choose all outputs at random)
+Ideas: Adversary sees $r$ but has no Ideas about $F(r)$. (we choose all outputs at random)
 
 If we could do this, this is MMS (multi-message secure).
 

pages/CSE442T/CSE442T_L16.md

@@ -77,7 +77,7 @@ With $g^a,g^b$ no one can compute $g^{ab}$.
 
 ### Public key encryption scheme
 
-Idea: The recipient Bob distributes opened Bob-locks
+Ideas: The recipient Bob distributes opened Bob-locks
 
 - Once closed, only Bob can open it.
 

pages/CSE442T/CSE442T_L17.md

@@ -110,7 +110,7 @@ $\{p\gets \tilde{\Pi_n};y\gets Gen_q;a,b,\bold{z}\gets \mathbb{Z}_q:(p,y,y^a,y^b
 
 So DDH assumption implies discrete logarithm assumption.
 
-Idea:
+Ideas:
 
 If one can find $a,b$ from $y^a,y^b$, then one can find $ab$ from $y^{ab}$ and compare to $\bold{z}$ to check whether $y^\bold{z}$ is a valid DDH tuple.
 

pages/CSE442T/CSE442T_L19.md

@@ -28,7 +28,7 @@ This is not more than one-time secure since the adversary can ask oracle for $Si
 
 We will show it is one-time secure
 
-Idea of proof:
+Ideas of proof:
 
 Say their query is $Sign_{sk}(0^n)$ and reveals $pk_0$. 
 

pages/CSE442T/CSE442T_L20.md

@@ -104,7 +104,7 @@ One-time secure:
 
 Then ($Gen',Sign',Ver'$) is one-time secure.
 
-Idea of Proof:
+Ideas of Proof:
 
 If the digital signature scheme ($Gen',Sign',Ver'$) is not one-time secure, then there exists an adversary $\mathcal{A}$ which can ask oracle for one signature on $m_1$ and receive $\sigma_1=Sign'_{sk'}(m_1)=Sign_{sk}(h_i(m_1))$.