latex fix

Author: pl3onasm

IP-Finals/2017resit/problem1.md

@@ -1,8 +1,8 @@
-$\color{cadetblue}{\text{\huge Problem 1}}$
+$\huge\color{cadetblue}{\text{Problem 1}}$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.1: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.1: }}\space{\color{olive}\text{b}}$
 
 ```java
 // x + 2 * y = A ∧ x + y = 2 * A
@@ -21,7 +21,7 @@ $\quad \lbrace \space y = A \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.2: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.2: }}\space{\color{olive}\text{b}}$
 
 ```java
 // 2 * x + y = A + B ∧ x + y = 2 * B
@@ -55,7 +55,7 @@ $\quad \lbrace \space x = A \space \land \space y = B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.3: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.3: }}\space{\color{olive}\text{c}}$
 
 ```java
 // x + A = B ∧ y + B = 2 * A
@@ -79,7 +79,7 @@ $\quad \lbrace \space x = 2 * A - B \space \land \space y = 2 * B - 3 * A \space
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.4: }}\space{\color{olive}\text{\Large a}}$
+$\Large{\color{darkkhaki}\text{Prob 1.4: }}\space{\color{olive}\text{a}}$
 
 ```java
 // x = A + 2 * B ∧ y = A + B
@@ -104,7 +104,7 @@ $\quad \lbrace \space x = A + B \space \land \space y = B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.5: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.5: }}\space{\color{olive}\text{c}}$
 
 ```java
 // x = A ∧ y = 2 * B
@@ -126,7 +126,7 @@ $\quad \lbrace \space x = 2 * B \space \land \space y = A \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.6: }}\space{\color{olive}\text{\Large a}}$
+$\Large{\color{darkkhaki}\text{Prob 1.6: }}\space{\color{olive}\text{a}}$
 
 ```java
 // x + z = A ∧ y + z = A - B

IP-Finals/2017resit/problem2.md

@@ -1,8 +1,8 @@
-$\color{cadetblue}{\text{\huge Problem 2}}$
+$\huge\color{cadetblue}{\text{Problem 2}}$
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.1: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.1: }}{\color{darkseagreen}{{\space \mathcal{O}(N)}}}$  
 
 <br/>
 
@@ -15,7 +15,7 @@ for (int i = N; i > 0; i /= 2) {
 }
 ```
 
-The outer loop runs $\log(N)$ times, since $i$ starts at $N$ and is then halved after each iteration. The inner loop runs $i$ times for each value of $i$, which is $\frac{N}{2^k}$ for $k \in \lbrace 0, 1, \dots, \lceil \log(N) \rceil \rbrace$. The total number of iterations of the inner loop is therefore given by:
+The outer loop runs $\log(N)$ times, since $i$ starts at $N$ and is then halved after each iteration. The inner loop runs $i$ times for each value of $i$, which is $N/2^k$ for $k \in \lbrace 0, 1, \dots, \lceil \log(N) \rceil \rbrace$. The total number of iterations of the inner loop is therefore given by:
 
 $$
 \begin{align*}
@@ -27,7 +27,7 @@ $$
 \end{align*}
 $$
 
-In $\color{peru}{(1)}$ we use the formula for the sum of a geometric series, with $a = \frac{1}{2}$ and $n = \lceil \log(N) \rceil$:
+In $\color{peru}{(1)}$ we use the formula for the sum of a geometric series, with $a = 1/2$ and $n = \lceil \log(N) \rceil$:
 
 $$
 \sum_{k=0}^n a^k = \frac{a^{n+1} - 1}{a - 1} \quad \text{for} \quad a \neq 1
@@ -41,7 +41,7 @@ Note that we could have run an argument for a complexity in $\mathcal{O}(N \log(
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.2: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N \log (N))}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.2: }}{\color{darkseagreen}{{\space \mathcal{O}(N \log (N))}}}$  
 
 <br/>
 
@@ -72,7 +72,7 @@ Therefore, the fragment's time complexity is in $\mathcal{O}(N\log(N))$.
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.3: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N^2)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.3: }}{\color{darkseagreen}{{\space \mathcal{O}(N^2)}}}$  
 
 <br/>
 
@@ -85,7 +85,7 @@ for (int i = N; i > 0; i--) {
 }
 ```
 
-The outer loop runs $N$ times, and the inner loop runs $\frac {i}{2}$ times for each value of $i$. The total number of iterations of the inner loop is therefore given by:
+The outer loop runs $N$ times, and the inner loop runs $i/2$ times for each value of $i$. The total number of iterations of the inner loop is therefore given by:
 
 $$
 \begin{align*}
@@ -102,7 +102,7 @@ Therefore, the fragment's time complexity is in $\mathcal{O}(N^2)$.
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.4: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(\sqrt N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.4: }}{\color{darkseagreen}{{\space \mathcal{O}(\sqrt N)}}}$  
 
 <br/>
 
@@ -120,7 +120,7 @@ To be fair, though, the program fragment will actually result in undefined behav
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.5: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(\log (N))}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.5: }}{\color{darkseagreen}{{\space \mathcal{O}(\log (N))}}}$  
 
 <br/>
 
@@ -139,7 +139,7 @@ The variable $i$ is initialized to $N$, and is either decremented by $1$ or divi
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.6: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.6: }}{\color{darkseagreen}{{\space \mathcal{O}(N)}}}$  
 
 <br/>
 

IP-Finals/2018/problem1.md

@@ -1,8 +1,8 @@
-$\color{cadetblue}{\text{\huge Problem 1}}$
+$\huge\color{cadetblue}{\text{Problem 1}}$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.1: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.1: }}\space{\color{olive}\text{c}}$
 
 ```java
 // 10 < 2 * x + 6 * y < 20
@@ -23,7 +23,7 @@ $\quad \lbrace \space 10 \space < x \space < 15 \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.2: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.2: }}\space{\color{olive}\text{b}}$
 
 ```java
 // x = A + B ∧ y = A - B
@@ -49,7 +49,7 @@ $\quad \lbrace \space x = A \space \land \space y = B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.3: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.3: }}\space{\color{olive}\text{c}}$
 
 ```java
 // x = A ∧ y = B
@@ -73,7 +73,7 @@ $\quad \lbrace \space x - 2 * B = A \space \land \space y = B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.4: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.4: }}\space{\color{olive}\text{b}}$
 
 ```java
 // x = A ∧ y = B
@@ -104,7 +104,7 @@ $\quad \lbrace \space x = 6 * A + 4 * B \space \land \space y = 2 * A + 2 * B \s
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.5: }}\space{\color{olive}\text{\Large a}}$
+$\Large{\color{darkkhaki}\text{Prob 1.5: }}\space{\color{olive}\text{a}}$
 
 ```java
 // x = A ∧ y = B
@@ -126,7 +126,7 @@ $\quad \lbrace \space x = B \space \land \space y = A \space \land \space z = A
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.6: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.6: }}\space{\color{olive}\text{b}}$
 
 ```java
 // y = A ∧ x = z = B

IP-Finals/2018/problem2.md

@@ -1,8 +1,8 @@
-$\color{cadetblue}{\text{\huge Problem 2}}$
+$\huge\color{cadetblue}{\text{Problem 2}}$
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.1: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N^2)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.1: }}{\color{darkseagreen}{{\space \mathcal{O}(N^2)}}}$  
 
 <br/>
 
@@ -17,14 +17,14 @@ while (j > 5) {
 }
 ```
 
-The first loop runs $N$ times, computing the sum of the first $N$ integers, so that after the loop has terminated, we have $j = \frac{N(N+1)}{2}$ by Gauss' formula. The second loop iterates a total number of $j - 5$ times. It is quadratic in $N$, since $j$ is decremented by $1$ at each iteration.  
+The first loop runs $N$ times, computing the sum of the first $N$ integers, so that after the loop has terminated, we have $j = N(N+1)/2$ by Gauss' formula. The second loop iterates a total number of $j - 5$ times. It is quadratic in $N$, since $j$ is decremented by $1$ at each iteration.  
 The loops are not nested, so the most expensive one determines the fragment's time complexity, which is therefore in $\mathcal{O}(N^2)$.
 
 <br/>
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.2: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(\log (N))}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.2: }}{\color{darkseagreen}{{\space \mathcal{O}(\log (N))}}}$  
 
 <br/>
 
@@ -42,7 +42,7 @@ The variable $i$ is doubled at each iteration, so that the loop runs $\log(N^2)
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.3: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.3: }}{\color{darkseagreen}{{\space \mathcal{O}(N)}}}$  
 
 <br/>
 
@@ -53,13 +53,13 @@ for (int i = 42; i < 7*N; i += 3) {
 }
 ```
 
-The variable $s$ does not affect the complexity of the fragment. The loop runs $\frac{7N - 42}{3}$ $= \frac{7}{3}N - 14$ times. The fragment's time complexity is therefore in $\mathcal{O}(N)$.
+The variable $s$ does not affect the complexity of the fragment. The loop runs $(7N - 42)/3$ $= 7/3N - 14$ times. The fragment's time complexity is therefore in $\mathcal{O}(N)$.
 
 <br/>
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.4: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(\sqrt N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.4: }}{\color{darkseagreen}{{\space \mathcal{O}(\sqrt N)}}}$  
 
 <br/>
 
@@ -81,7 +81,7 @@ The first loop ends when $s = i^2 \geq N$, so that $i \geq \sqrt{N}$. In the sec
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.5: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N \log (N))}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.5: }}{\color{darkseagreen}{{\space \mathcal{O}(N \log (N))}}}$  
 
 <br/>
 
@@ -94,7 +94,7 @@ for (int i = N; i > 0; i = i/2) {
 }
 ```
 
-The index $i$ is initialized to $N$ and is halved at the end of each iteration of the outer loop, which therefore runs a total number of $\log(N)$ times. The inner loop runs $N - i$ times for each value of $i$, which is $\frac{N}{2^k}$ for the $k$-th iteration of the outer loop, where $k \in \lbrace 1, \dots, \lfloor \log(N) \rfloor \rbrace$.[^1]
+The index $i$ is initialized to $N$ and is halved at the end of each iteration of the outer loop, which therefore runs a total number of $\log(N)$ times. The inner loop runs $N - i$ times for each value of $i$, which is $N/2^k$ for the $k$-th iteration of the outer loop, where $k \in \lbrace 1, \dots, \lfloor \log(N) \rfloor \rbrace$.[^1]
 The total number of iterations of the inner loop is therefore given by:
 
 $$
@@ -117,7 +117,7 @@ From the above, we conclude that the fragment's time complexity is in $\mathcal{
 
 ----------------------
 
-${\color{rosybrown}\text{\Large Prob 2.6: }}{\color{darkseagreen}{{\Large \space \mathcal{O}(N)}}}$  
+$\Large{\color{rosybrown}\text{Prob 2.6: }}{\color{darkseagreen}{{\space \mathcal{O}(N)}}}$  
 
 <br/>
 
@@ -131,6 +131,6 @@ for (i = 0; i*i < j; i++) {
 }
 ```
 
-The first loop runs $N$ times, computing the sum of the first $N - 1$ integers, so that in the end $j = \frac{N(N-1)}{2}$ by Gauss' formula. The second loop runs about $\sqrt{j}$ times, which is also linear in $N$. The loops are not nested, so the fragment's time complexity is therefore in $\mathcal{O}(N)$.
+The first loop runs $N$ times, computing the sum of the first $N - 1$ integers, so that in the end $j = N(N-1)/2$ by Gauss' formula. The second loop runs about $\sqrt{j}$ times, which is also linear in $N$. The loops are not nested, so the fragment's time complexity is therefore in $\mathcal{O}(N)$.
 
 <br/>

IP-Finals/2018resit/problem1.md

@@ -1,8 +1,8 @@
-$\color{cadetblue}{\text{\huge Problem 1}}$
+$\huge\color{cadetblue}{\text{Problem 1}}$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.1: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.1: }}\space{\color{olive}\text{b}}$
 
 ```java
 // x = y + 7
@@ -22,7 +22,7 @@ $\quad \lbrace \space x = 11 \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.2: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.2: }}\space{\color{olive}\text{c}}$
 
 ```java
 // x = A ∧ y = B
@@ -49,7 +49,7 @@ $\quad \lbrace \space x = A - B \space \land \space y = A + B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.3: }}\space{\color{olive}\text{\Large b}}$
+$\Large{\color{darkkhaki}\text{Prob 1.3: }}\space{\color{olive}\text{b}}$
 
 ```java
 // x = A ∧ y = B
@@ -72,7 +72,7 @@ $\quad \lbrace \space x - y = A - B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.4: }}\space{\color{olive}\text{\Large a}}$
+$\Large{\color{darkkhaki}\text{Prob 1.4: }}\space{\color{olive}\text{a}}$
 
 ```java
 // x = A ∧ y = B
@@ -97,7 +97,7 @@ $\quad \lbrace \space x = B \space \land \space y = A - B \space \rbrace$
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.5: }}\space{\color{olive}\text{\Large a}}$
+$\Large{\color{darkkhaki}\text{Prob 1.5: }}\space{\color{olive}\text{a}}$
 
 ```java
 // x = A ∧ y = B
@@ -124,7 +124,7 @@ $\quad \lbrace \space x = 3 * A + B \space \land \space y = A + B \space \land \
 
 ---------------
 
-${\color{darkkhaki}\text{\Large Prob 1.6: }}\space{\color{olive}\text{\Large c}}$
+$\Large{\color{darkkhaki}\text{Prob 1.6: }}\space{\color{olive}\text{c}}$
 
 ```java
 // x = A + B ∧ y = A - B

IP-Finals/2018resit/problem2.md

@@ -14,7 +14,7 @@ while (i > 1) {
 }
 ```
 
-The variable $i$ is halved in each loop iteration, so that after $k$ iterations $i = \frac{N}{2^k}$. The loop terminates when $i \leq 1$, which happens when $k \geq \log(N)$. The loop therefore runs in $\mathcal{O}(\log(N))$ time.
+The variable $i$ is halved in each loop iteration, so that after $k$ iterations $i = N/2^k$. The loop terminates when $i \leq 1$, which happens when $k \geq \log(N)$. The loop therefore runs in $\mathcal{O}(\log(N))$ time.
 
 <br/>
 
@@ -31,7 +31,7 @@ for (int i = 0; i < N; i += 10) {
 }
 ```
 
-Since the loop variable $i$ is incremented by 10 in each iteration, the loop runs $\frac{N}{10}$ times. The fragment's time complexity is therefore in $\mathcal{O}(N)$.
+Since the loop variable $i$ is incremented by 10 in each iteration, the loop runs $N/10$ times. The fragment's time complexity is therefore in $\mathcal{O}(N)$.
 
 <br/>
 
@@ -49,7 +49,7 @@ while (s < N) {
 }
 ```
 
-The variable $s$ is incremented by $i$ in each iteration, while $i$ just counts the number of iterations. After $i$ iterations, $s = \frac{i(i+1)}{2}$ by Gauss' formula. The loop terminates when:
+The variable $s$ is incremented by $i$ in each iteration, while $i$ just counts the number of iterations. After $i$ iterations, $s = i(i+1)/2$ by Gauss' formula. The loop terminates when:
 
 $$
 \begin{align*}