From 7bc720660480d37473ff21e49b037ddc1808cac4 Mon Sep 17 00:00:00 2001 From: Trance-0 <60459821+Trance-0@users.noreply.github.com> Date: Fri, 29 Aug 2025 15:51:24 -0500 Subject: [PATCH] proof format updates using gfm --- content/CSE347/CSE347_L10.md | 32 +++++---- content/CSE347/CSE347_L5.md | 2 +- content/CSE347/CSE347_L7.md | 32 +++++---- content/CSE347/CSE347_L8.md | 19 ++--- content/CSE347/CSE347_L9.md | 11 +-- content/CSE442T/CSE442T_L10.md | 12 ++-- content/CSE442T/CSE442T_L12.md | 7 +- content/CSE442T/CSE442T_L15.md | 7 +- content/CSE442T/CSE442T_L4.md | 5 +- content/Math4501/Math4501_L3.md | 119 ++++++++++++++++++++++++++++++++ content/Math4501/_meta.js | 1 + 11 files changed, 195 insertions(+), 52 deletions(-) create mode 100644 content/Math4501/Math4501_L3.md diff --git a/content/CSE347/CSE347_L10.md b/content/CSE347/CSE347_L10.md index 8b0012c..2a7d2ad 100644 --- a/content/CSE347/CSE347_L10.md +++ b/content/CSE347/CSE347_L10.md @@ -82,7 +82,8 @@ Let's try $R=S$. Claim: The comparative ratio is $2$. -Proof: +
+Proof 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 +
Let's try $R=S-E$ instead. Claim: The comparative ratio is $max\{1,2-\frac{E}{S}\}$. -Proof: +
+Proof 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 +
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: +
+Proof 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 +
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: +
+Proof 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 +
### 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: +
+Proof 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 +
MRU is $k$-competitive. -Proof: +
+Proof 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 +
diff --git a/content/CSE347/CSE347_L5.md b/content/CSE347/CSE347_L5.md index eeed02a..974fd84 100644 --- a/content/CSE347/CSE347_L5.md +++ b/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) diff --git a/content/CSE347/CSE347_L7.md b/content/CSE347/CSE347_L7.md index 1c66380..301a132 100644 --- a/content/CSE347/CSE347_L7.md +++ b/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: +
+Proof 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 +
Claim 2: If $\Psi$ is satisfiable, then Subset Sum has a solution. -Proof: +
+Proof 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 +
### 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: +
+Proof 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 +
Direction 2: If $G$ is 3-colorable, then $\Psi$ is satisfiable. -Proof: +
+Proof -QED + + +
### 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 diff --git a/content/CSE347/CSE347_L8.md b/content/CSE347/CSE347_L8.md index 0c225c1..969d326 100644 --- a/content/CSE347/CSE347_L8.md +++ b/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: +
+Proof 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 +
#### 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: +
+Proof 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 +
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: +
+Proof Assign a cost to the elements of $X$ according to the decisions of the greedy set cover. @@ -350,4 +353,4 @@ $$ So the approximation ratio for greedy set cover is $H_d$. -QED +
diff --git a/content/CSE347/CSE347_L9.md b/content/CSE347/CSE347_L9.md index b9f14c0..5f20795 100644 --- a/content/CSE347/CSE347_L9.md +++ b/content/CSE347/CSE347_L9.md @@ -260,7 +260,8 @@ $$ Claim: the solution to this recurrence is $E[T(n)]=O(n\log n)$ or $T(n)=c'n\log n+1$. -Proof: +
+Proof 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 +
A more elegant proof: +
+Proof + 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} $$ - -QED +
diff --git a/content/CSE442T/CSE442T_L10.md b/content/CSE442T/CSE442T_L10.md index b913980..7b6c970 100644 --- a/content/CSE442T/CSE442T_L10.md +++ b/content/CSE442T/CSE442T_L10.md @@ -86,7 +86,10 @@ f(N,e):\mathbb{Z}_N^*\to \mathbb{Z}_N^* $$ is a bijection. -Proof: Suppose $x_1^e\equiv x_2^e\mod n$ +
+Proof + +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 +
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: +
+Proof 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 +
#### Theorem If inverting RSA is hard, then factoring is hard. diff --git a/content/CSE442T/CSE442T_L12.md b/content/CSE442T/CSE442T_L12.md index 6d0b84f..c994391 100644 --- a/content/CSE442T/CSE442T_L12.md +++ b/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. +
+Ideas of proof + +Full proof is on the text. Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$ @@ -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 +
## Pseudorandom Generator diff --git a/content/CSE442T/CSE442T_L15.md b/content/CSE442T/CSE442T_L15.md index 560a7b1..8a6f3e8 100644 --- a/content/CSE442T/CSE442T_L15.md +++ b/content/CSE442T/CSE442T_L15.md @@ -115,7 +115,8 @@ $$ #### Theorem PRG exists then PRF family exists. -Proof: +
+Proof Let $g:\{0,1\}^n\to \{0,1\}^{2n}$ be a PRG. @@ -184,6 +185,6 @@ Assume that $D$ distinguishes $f_s$ and $F\gets RF_n$ with non-negligible probab By hybrid argument, there exists a hybrid $H_i$ such that $D$ distinguishes $H_i$ and $H_{i+1}$ with non-negligible probability. -For $H_0$, +For $H_0$, $D$ distinguishes $H_0$ and $H_1$ with non-negligible probability. -QED +
diff --git a/content/CSE442T/CSE442T_L4.md b/content/CSE442T/CSE442T_L4.md index ca9e93b..8e0d351 100644 --- a/content/CSE442T/CSE442T_L4.md +++ b/content/CSE442T/CSE442T_L4.md @@ -88,7 +88,8 @@ $$ is a strong one-way function. -Proof: +
+Proof 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 +
_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_ diff --git a/content/Math4501/Math4501_L3.md b/content/Math4501/Math4501_L3.md new file mode 100644 index 0000000..135b903 --- /dev/null +++ b/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 + +
+ +Proof + +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 $x_0\in [a,b]$. + +
+ +#### 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 $x_0\in [a,b]$. + +
+ +Proof + +**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 $\xi$. + +
diff --git a/content/Math4501/_meta.js b/content/Math4501/_meta.js index f15d5fd..4e15082 100644 --- a/content/Math4501/_meta.js +++ b/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)", }