update notations and fix typos

pages/CSE347/CSE347_L10.md

@@ -96,7 +96,7 @@ We wait for $R$ times and then take the stairs. In worst case, we wait for $R$ t
 
 Competitive ratio = $\frac{2R}{R}=2$.
 
-EOP
+QED
 
 Let's try $R=S-E$ instead.
 
@@ -116,7 +116,7 @@ We wait for $R=S-E$ times and then take the stairs.
 
 Competitive ratio = $\frac{S-E+S}{S}=2-\frac{E}{S}$.
 
-EOP
+QED
 
 What if we wait less time? Let's try $R=S-E-\epsilon$ for some $\epsilon>0$
 
@@ -174,7 +174,7 @@ The optimal offline solution: In each subsequence, must have at least $1$ miss.
 
 So the competitive ratio is at most $k+1$.
 
-EOP
+QED
 
 Using similar analysis, we can show that LRU is $k$ competitive.
 
@@ -184,7 +184,7 @@ Split the sequence into subsequences such that each subsequence LRU has $k$ miss
 
 Argue that OPT has at least $1$ miss in each subsequence.
 
-EOP
+QED
 
 #### Many sensible algorithms are $k$-competitive
 
@@ -210,7 +210,7 @@ So competitive ratio is at most $\frac{ck}{(c-1)k}=\frac{c}{c-1}$.
 
 _Actual competitive ratio is $\sim \frac{c}{c-1+\frac{1}{k}}$._
 
-EOP
+QED
 
 ### Conclusion
 
@@ -297,7 +297,7 @@ 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.
 
-EOP
+QED
 
 MRU is $k$-competitive.
 
@@ -317,4 +317,4 @@ Let's define the random variable $X$ as the number of misses of RAND MRU.
 
 $E[X]\leq 1+\frac{1}{k}$.
 
-EOP
+QED

pages/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 people have tried and could not find a poly-time solution
+A lot of pQEDle have tried and could not find a poly-time solution
 
 ### Example: Vertex Cover (VC)
 

pages/CSE347/CSE347_L7.md

@@ -154,7 +154,7 @@ This is a valid assignment since:
 - We pick either $v_i$ or $\overline{v_i}$
 - For each clause, at least one literal is true
 
-EOP
+QED
 
 Claim 2: If $\Psi$ is satisfiable, then Subset Sum has a solution.
 
@@ -174,7 +174,7 @@ Say $t=\sum$ elements we picked from $S$.
   - If $q_j=2$, then $z_j\in S'$
   - If $q_j=3$, then $y_j\in S'$
 
-EOP
+QED
 
 ### Example 2: 3 Color
 
@@ -228,13 +228,13 @@ For each dangler color is connected to blue, all literals cannot be blue.
 
 ...
 
-EOP
+QED
 
 Direction 2: If $G$ is 3-colorable, then $\Psi$ is satisfiable.
 
 Proof:
 
-EOP
+QED
 
 ### Example 3:Hamiltonian cycle problem (HAMCYCLE)
 

pages/CSE347/CSE347_L8.md

@@ -153,7 +153,7 @@ 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 $\frac{|E|}{2}$.
 
-EOP
+QED
 
 #### Set cover
 
@@ -264,7 +264,7 @@ 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 $(1+\ln n)k$.
 
-EOP
+QED
 
 So the greedy set cover is not too bad...
 
@@ -350,4 +350,4 @@ $$
 
 So the approximation ratio for greedy set cover is $H_d$.
 
-EOP
+QED

pages/CSE347/CSE347_L9.md

@@ -296,7 +296,7 @@ If $c'\geq 8c$, then $T(n)\leq c'n\log n+1$.
 
 $E[T(n)]\leq c'n\log n+1=O(n\log n)$
 
-EOP
+QED
 
 A more elegant proof:
 
@@ -345,5 +345,5 @@ E[X]&=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}\frac{2}{j-i+1}\\
 $$
 
 
-EOP
+QED