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>