proof format updates using gfm

content/CSE347/CSE347_L10.md

@@ -82,7 +82,8 @@ Let's try $R=S$.
 
 Claim: The comparative ratio is $2$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Case 1: The optimal offline solution takes the elevator, then $T+E\leq S$.
 
@@ -96,13 +97,14 @@ We wait for $R$ times and then take the stairs. In worst case, we wait for $R$ t
 
 Competitive ratio = $\frac{2R}{R}=2$.
 
-QED
+</details>
 
 Let's try $R=S-E$ instead.
 
 Claim: The comparative ratio is $max\{1,2-\frac{E}{S}\}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Case 1: The optimal offline solution takes the elevator, then $T+E\leq S$.
 
@@ -116,7 +118,7 @@ We wait for $R=S-E$ times and then take the stairs.
 
 Competitive ratio = $\frac{S-E+S}{S}=2-\frac{E}{S}$.
 
-QED
+</details>
 
 What if we wait less time? Let's try $R=S-E-\epsilon$ for some $\epsilon>0$
 
@@ -162,7 +164,8 @@ Cache: $D A C$, the evict $D$ for $B$. 1 miss.
 
 Claim: LRU is $k+1$-competitive.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Split the sequence into subsequences such that each subsequence contains $k+1$ distinct blocks.
 
@@ -174,7 +177,7 @@ The optimal offline solution: In each subsequence, must have at least $1$ miss.
 
 So the competitive ratio is at most $k+1$.
 
-QED
+</details>
 
 Using similar analysis, we can show that LRU is $k$ competitive.
 
@@ -184,8 +187,6 @@ Split the sequence into subsequences such that each subsequence LRU has $k$ miss
 
 Argue that OPT has at least $1$ miss in each subsequence.
 
-QED
-
 #### Many sensible algorithms are $k$-competitive
 
 **Lower Bound**: No deterministic online algorithm is better than $k$-competitive.
@@ -196,7 +197,8 @@ QED
 
 Say $c=2$. LRU cache has twice as much as cache. LRU is $2$-competitive.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 LRU has cache of size $2k$.
 
@@ -210,7 +212,7 @@ So competitive ratio is at most $\frac{ck}{(c-1)k}=\frac{c}{c-1}$.
 
 _Actual competitive ratio is $\sim \frac{c}{c-1+\frac{1}{k}}$._
 
-QED
+</details>
 
 ### Conclusion
 
@@ -273,7 +275,8 @@ Claim: RAND is $k$-competitive.
 2. There exists $k$ pages each of which is in the cache with probability $1-\frac{1}{k}$
 3. All other pages are in the cache with probability $0$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 By induction.
 
@@ -297,11 +300,12 @@ Let $P$ be a page in the cache with probability $1-\frac{1}{k}$.
 
 With probability $\frac{1}{k}$, $P$ is not in the cache and RAND evicts $P'$ in the cache and brings $P$ to the cache.
 
-QED
+</details>
 
 MRU is $k$-competitive.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Case 1: Access MRU page.
 
@@ -317,4 +321,4 @@ Let's define the random variable $X$ as the number of misses of RAND MRU.
 
 $E[X]\leq 1+\frac{1}{k}$.
 
-QED
+</details>

content/CSE347/CSE347_L5.md

@@ -161,7 +161,7 @@ $ISET(G,k)$  returns true if $G$ contains an independent set of size $\geq k$, a
 
 Algorithm? NO! We think that this is a hard problem.
 
-A lot of pQEDle have tried and could not find a poly-time solution
+A lot of people have tried and could not find a poly-time solution
 
 ### Example: Vertex Cover (VC)
 

content/CSE347/CSE347_L7.md

@@ -142,7 +142,8 @@ How many digits are in each integer?
 
 Claim 1: If Subset Sum has a solution, then $\Psi$ is satisfiable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Say $S'$ is a solution to Subset Sum. Then there exists a subset $S' \subseteq S$ such that $\sum_{a_i\in S'} a_i = t$. Here is an assignment of truth values to variables in $\Psi$ that satisfies $\Psi$:
 
@@ -154,11 +155,12 @@ This is a valid assignment since:
 - We pick either $v_i$ or $\overline{v_i}$
 - For each clause, at least one literal is true
 
-QED
+</details>
 
 Claim 2: If $\Psi$ is satisfiable, then Subset Sum has a solution.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 If $A$ is a satisfiable assignment for $\Psi$, then we can construct a subset $S'$ of $S$ such that $\sum_{a_i\in S'} a_i = t$.
 
@@ -174,7 +176,7 @@ Say $t=\sum$ elements we picked from $S$.
   - If $q_j=2$, then $z_j\in S'$
   - If $q_j=3$, then $y_j\in S'$
 
-QED
+</details>
 
 ### Example 2: 3 Color
 
@@ -210,15 +212,16 @@ Key for dangler design:
 
 Connect to all $v_i$ with true to the same color. and connect to all $v_i$ with false to another color.
 
-'''
-TODO: Add dangler design image here.
-'''
+> [!TIP]
+>
+> TODO: Add dangler design image here.
 
 #### Proof of reduction for 3-Color
 
 Direction 1: If $\Psi$ is satisfiable, then $G$ is 3-colorable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Say $\Psi$ is satisfiable. Then $v_i$ and $\overline{v_i}$ are in different colors.
 
@@ -228,13 +231,16 @@ For each dangler color is connected to blue, all literals cannot be blue.
 
 ...
 
-QED
+</details>
 
 Direction 2: If $G$ is 3-colorable, then $\Psi$ is satisfiable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
-QED
+
+
+</details>
 
 ### Example 3:Hamiltonian cycle problem (HAMCYCLE)
 
@@ -242,9 +248,7 @@ Input: $G(V,E)$
 
 Output: Does $G$ have a Hamiltonian cycle? (A cycle that visits each vertex exactly once.)
 
-Proof is too hard.
-
-but it is an existing NP-complete problem.
+Proof is too hard. But it is an existing NP-complete problem.
 
 ## On lecture
 

content/CSE347/CSE347_L8.md

@@ -139,7 +139,8 @@ We could first upper bound the size of the optimal cut is at most $|E|$.
 
 We will then prove that solution we found is at least half of the optimal cut $\frac{|E|}{2}$ for any graph $G$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 When we terminate, no vertex could be moved
 
@@ -153,7 +154,7 @@ Summing over all vertices, the total number of crossing edges is at least $\frac
 
 So the total number of non-crossing edges is at most $\frac{|E|}{2}$.
 
-QED
+</details>
 
 #### Set cover
 
@@ -226,9 +227,10 @@ Need to prove its:
 
 We claim that the size of the set cover found is at most $H_n\log n$ times the size of the optimal set cover.
 
-###### First bound:
+Proof of first bound:
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 If the optimal picks $k$ sets, then the size of the set cover found is at most $(1+\log n)k$ sets.
 
@@ -264,15 +266,16 @@ So $n(1-\frac{1}{k})^{|C|-1}=1$, $|C|\leq 1+k\ln n$.
 
 So the size of the set cover found is at most $(1+\ln n)k$.
 
-QED
+</details>
 
 So the greedy set cover is not too bad...
 
-###### Second bound:
+Proof of second bound:
 
 Greedy set cover is a $H_d$-approximation algorithm of set cover.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Assign a cost to the elements of $X$ according to the decisions of the greedy set cover.
 
@@ -350,4 +353,4 @@ $$
 
 So the approximation ratio for greedy set cover is $H_d$.
 
-QED
+</details>

content/CSE347/CSE347_L9.md

@@ -260,7 +260,8 @@ $$
 
 Claim: the solution to this recurrence is $E[T(n)]=O(n\log n)$ or $T(n)=c'n\log n+1$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We prove by induction.
 
@@ -296,10 +297,13 @@ If $c'\geq 8c$, then $T(n)\leq c'n\log n+1$.
 
 $E[T(n)]\leq c'n\log n+1=O(n\log n)$
 
-QED
+</details>
 
 A more elegant proof:
 
+<details>
+<summary>Proof</summary>
+
 Let $X_{ij}$ be an indicator random variable that is $1$ if element of rank $i$ is compared to element of rank $j$.
 
 Running time: $$X=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}X_{ij}$$
@@ -344,6 +348,5 @@ E[X]&=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}\frac{2}{j-i+1}\\
 \end{aligned}
 $$
 
-
-QED
+</details>
 

content/CSE442T/CSE442T_L10.md

@@ -86,7 +86,10 @@ f(N,e):\mathbb{Z}_N^*\to \mathbb{Z}_N^*
 $$
 is a bijection.
 
-Proof: Suppose $x_1^e\equiv x_2^e\mod n$
+<details>
+<summary>Proof</summary>
+
+Suppose $x_1^e\equiv x_2^e\mod n$
 
 Then let $d=e^{-1}\mod \phi(N)$ (exists b/c $e\in\phi(N)^*$)
 
@@ -98,13 +101,14 @@ $x_1\equiv x_2\mod N$
 
 So it's one-to-one.
 
-QED
+</details>
 
 Let $y\in \mathbb{Z}_N^*$, letting $x=y^d\mod N$, where $d\equiv e^{-1}\mod \phi(N)$
 
 $x^e\equiv (y^d)^e \equiv y\mod n$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 It's easy to sample from $I$:
 
@@ -130,7 +134,7 @@ By RSA assumption
 
 The second equality follows because for any finite $D$ and bijection $f:D\to D$, sampling $y\in D$ directly is equivalent to sampling $x\gets D$, then computing $y=f(x)$.
 
-QED
+</details>
 
 #### Theorem If inverting RSA is hard, then factoring is hard.
 

content/CSE442T/CSE442T_L12.md

@@ -82,7 +82,10 @@ The NBT(Next bit test) is complete.
 
 If $\{X_n\}$ on $\{0,1\}^{l(n)}$ passes NBT, then it's pseudorandom.
 
-Ideas of proof: full proof is on the text.
+<details>
+<summary>Ideas of proof</summary>
+
+Full proof is on the text.
 
 Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$
 
@@ -119,7 +122,7 @@ $\mathcal{D}$ can distinguish $x_{i+1}$ from a truly random $U_{i+1}$, knowing t
 
 So $\mathcal{D}$ can predict $x_{i+1}$ from $x_1\dots x_i$ (contradicting with that $X$ passes NBT)
 
-QED
+</details>
 
 ## Pseudorandom Generator
 

content/CSE442T/CSE442T_L15.md

@@ -115,7 +115,8 @@ $$
 
 #### Theorem PRG exists then PRF family exists.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $g:\{0,1\}^n\to \{0,1\}^{2n}$ be a PRG.
 
@@ -184,6 +185,6 @@ Assume that $D$ distinguishes $f_s$ and $F\gets RF_n$ with non-negligible probab
 
 By hybrid argument, there exists a hybrid $H_i$ such that $D$ distinguishes $H_i$ and $H_{i+1}$ with non-negligible probability.
 
-For $H_0$, 
+For $H_0$, $D$ distinguishes $H_0$ and $H_1$ with non-negligible probability.
 
-QED
+</details>

content/CSE442T/CSE442T_L4.md

@@ -88,7 +88,8 @@ $$
 
 is a strong one-way function.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 1. Since $\exist P.P.T.$ that computes $f(x),\forall x$ we use this $q(n)$ polynomial times to compute $g$.
 2. (Idea) $a$ has to succeed in inverting $f$ all $q(n)$ times.
@@ -98,7 +99,7 @@ Proof:
 
     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.
 
-QED
+</details>
 
 _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_
 

content/Math4501/Math4501_L3.md

@@ -0,0 +1,119 @@
+# Math4501 Lecture 3
+
+## Review from last lecture
+
+### Bisection method for finding root
+
+P1. Let $f$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $f(\xi)=0$.
+
+P2. Let $g$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $g(\xi)=\xi$.
+
+#### Theorem 1:
+
+solution to P1 exists if $f(a)f(b)<0$.
+
+#### Theorem 2:
+
+Fixed point theorem:
+
+If solution to P2 exists, then $g:[a,b]\to [a,b]$.
+
+#### Bijection method
+
+Obtain two sequence $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$.
+
+Initially, we set $a_0=a$ and $b_0=b$.
+
+If $|a_n-b_n|<2^{-n}|a_0-b_0|$, then $a_n$ and $b_n$ are Cauchy sequence. So their limit exists.
+
+$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\xi$.
+
+### Simple iteration method
+
+#### Definition of Simple Iteration
+
+Given $x_0\in [a,b]$, we define a sequence $(x_n)_{n=0}^\infty$ by
+
+$$
+x_{n+1}=g(x_n)
+$$
+
+If a simple iteration converges, the it converges to a fixed point of
+
+<details>
+
+<summary>Proof</summary>
+
+Let $c\coloneqq \lim_{n\to\infty} x_n=g(c)$.
+
+$g:[a,b]\to \mathbb{R}$ is continuous if and only if $g$ is continuous at $x_0\in [a,b]$.
+
+</details>
+
+#### Definition of Lipschitz continuous
+
+A function $g:[a,b]\to \mathbb{R}$ is Lipschitz continuous if for all $x,y\in [a,b]$, there exists a constant $L>0$ such that
+
+$$
+|g(x)-g(y)|\leq L|x-y|
+$$
+
+for some $L>0$.
+
+> [!NOTE]
+> Lipschitz continuous is a stronger condition than continuous.
+>
+> If a function is Lipschitz continuous, then it is continuous.
+>
+> However, the converse is not true.
+
+#### Definition of contraction mapping
+
+A function $g:[a,b]\to \mathbb{R}$ is a contraction mapping if it is Lipschitz continuous with $L<1$.
+
+#### Theorem of simple iteration
+
+Let $g:[a,b]\to [a,b]$ is a contraction mapping (Lipschitz continuous with $L<1$, with pointwise ontinuous $g$).
+
+Then
+
+- $g$ has a unique fixed point $\xi \in [a,b]$
+- Simple iteration $x_{n+1}=g(x_n)$ converges to $\xi$ for any $x_0\in [a,b]$.
+
+<details>
+
+<summary>Proof</summary>
+
+**Uniqueness**:
+
+Suppose $\xi_1$ and $\xi_2$ are two fixed points of $g$.
+
+Then $|x_1-x_2|=|g(\xi_1)-g(\xi_2)|\leq L|\xi_1-\xi_2|$.
+
+Thus, $(1-L)|\xi_1-\xi_2|\leq 0$, which implies $|\xi_1-\xi_2|=0$.
+
+A more general result:
+
+Brouwer's fixed point theorem
+
+**Convergence**:
+
+Let $\xi\in [a,b]$ be the unique fixed point of $g$.
+
+Then,
+
+$$
+\begin{aligned}
+|x_n-\xi|&=|g(x_{n-1})-g(\xi)|\\
+&\leq L|x_{n-1}-\xi|\\
+&=L|g(x_{n-2})-g(\xi)|\\
+&\leq L^2|x_{n-2}-\xi|\\
+&\vdots\\
+&\leq L^n|x_0-\xi|
+$$
+
+Thus, we can always find $N$ such that $L^N|x_0-\xi|<\epsilon$ for any $\epsilon>0$. Choose $N=\log(\frac{\epsilon}{|x_0-\xi|})/\log(L)$.
+
+Therefore, $x_n$ converges to $\xi$.
+
+</details>

content/Math4501/_meta.js

@@ -5,4 +5,5 @@ export default {
     },
     Math4501_L1: "Numerical Applied Mathematics (Lecture 1)",
     Math4501_L2: "Numerical Applied Mathematics (Lecture 2)",
+    Math4501_L3: "Numerical Applied Mathematics (Lecture 3)",
 }