
diff -r a20bf59d1775 -r 862ab5c7c6df tex/theory1_ex6.tex
--- a/tex/theory1_ex6.tex	Sun Jun 26 19:46:12 2011 +0200
+++ b/tex/theory1_ex6.tex	Mon Jun 27 01:01:33 2011 +0200
@@ -18,58 +18,82 @@
 \maketitle
 
 \section*{Exercise 6.1}
-\[\Phi(T) = |2 \cdot num - size|\]
-\[a_i = \Phi_i - \Phi_{i-1} + t_i\]
-\[a_i = |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\]
+\begin{align*}
+\Phi(T) &= |2 \cdot num - size|\\
+a_i &= \Phi_i - \Phi_{i-1} + t_i\\
+a_i &= |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\\
+\end{align*}
 We calculate the amortized costs for the two distinct cases.
 
 \subsection*{case 1 (no contraction required)}
 Since there was no contraction we know that $s_i = s_{i-1}$ and it follows that $\frac{1}{3}s_i \leq k_i \leq s_i - 1$. Since we have deleted an element we know that $k_i = k_{i-1} - 1$. Because we have only removed an element it follows that $t_i = 1$. We reduce the formula using these substitutions.
-\[a_i = |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\]
-\[= |2k_i - s_i| - |2k_i + 2 - s_i| + 1\]
+\begin{align*}
+a_i &= |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\\
+&= |2k_i - s_i| - |2k_i + 2 - s_i| + 1\\
+\end{align*}
 Now we look for the minimum and maximum value of the above formula. We choose the obvious values for $k_i$ which are within valid range.
-\[(k_i = \frac{s_i}{2}) \implies a_i = |s_i - s_i| - |s_i - s_i + 2| + 1\]
-\[= 0 - 2 + 1 = -1\]
-
-\[(k_i = \frac{s_i}{2} - 1) \implies a_i = |s_i - 2 - s_i| - |s_i - 2 - s_i + 2| + 1\]
-\[= 2 - 0 + 1 = 3\]
-
-\[\implies -1 \leq a_i \leq 3\]
+\begin{align*}
+(k_i = \frac{s_i}{2}) \implies a_i &= |s_i - s_i| - |s_i - s_i + 2| + 1\\
+&= 0 - 2 + 1 = -1\\
+(k_i = \frac{s_i}{2} - 1) \implies a_i &= |s_i - 2 - s_i| - |s_i - 2 - s_i + 2| + 1\\
+&= 2 - 0 + 1 = 3\\
+\implies &-1 \leq a_i \leq 3\\
+\end{align*}
+\newpage
 \subsection*{case 2 (contraction required)}
 Since there was a contraction we know that $s_i = \frac{2}{3}s_{i-1}$ and it follows that $k_i = \frac{1}{3}s_{i-1} - 1$. Since we have deleted an element we know that $k_i = k_{i-1} - 1$. Because we have removed an element and additionally copied all previously contained elements it follows that $t_i = \frac{1}{3}s_{i-1} + 1$. We reduce the formula again using these substitutions.
-\[a_i = |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\]
-\[= |2k_i - \frac{2}{3}s_{i-1}| - |2k_i + 2 - s_{i-1}| + t_i\]
-\[= |\frac{2}{3}s_{i-1} - 2 - \frac{2}{3}s_{i-1}| - |\frac{2}{3}s_{i-1} - 2 + 2 - s_{i-1}| + \frac{1}{3}s_{i-1} + 1\]
-\[= |-2| - |-\frac{1}{3}s_{i-1}| + \frac{1}{3}s_{i-1} + 1\]
-\[(\text{since } s_{i-1} > 0) \implies a_i = 2 - \frac{1}{3}s_{i-1} + \frac{1}{3}s_{i-1} + 1\]
-\[= 2 + 1 = 3 \leq 3\]
+\begin{align*}
+a_i &= |2k_i - s_i| - |2k_{i-1} - s_{i-1}| + t_i\\
+&= |2k_i - \frac{2}{3}s_{i-1}| - |2k_i + 2 - s_{i-1}| + t_i\\
+&= |\frac{2}{3}s_{i-1} - 2 - \frac{2}{3}s_{i-1}| - |\frac{2}{3}s_{i-1} - 2 + 2 - s_{i-1}| + \frac{1}{3}s_{i-1} + 1\\
+&= |-2| - |-\frac{1}{3}s_{i-1}| + \frac{1}{3}s_{i-1} + 1\\
+(\text{since } s_{i-1} > 0) \implies a_i &= 2 - \frac{1}{3}s_{i-1} + \frac{1}{3}s_{i-1} + 1\\
+&= 2 + 1 = 3 \leq 3\\
+\end{align*}
 By calculating the amortized costs for both Table-Delete cases, we have have shown that the amortized costs are bounded by constant 3. \qed
 
 \section*{Exercise 6.2}
-\[\alpha = \frac{k}{s}\]
-\[\Phi(T) = 
+\begin{align*}
+\alpha &= \frac{k}{s}\\
+\Phi(T) &= 
 \left\{ 
 \begin{array}{l l} 
   2k-s & \text{if $\alpha \geq \frac{1}{2}$}\\
   \frac{s}{2}-k & \text{if $\alpha < \frac{1}{2}$}\\
-\end{array} \right.\]
+\end{array} \right.\\
+\end{align*}
 
 We show $\sum a_i \geq \sum t_i$.
-\[a_i = \Phi_i - \Phi_{i-1} + t_i\]
-\[\implies \sum_{i=1}^{n}a_i = \sum_{i=1}^{n}\Phi_i - \Phi_{i-1} + t_i\]
-\[= \sum_{i=1}^{n}\Phi_i - \sum_{i=0}^{n-1}\Phi_i + \sum_{i=1}^{n}t_i\]
-\[= \Phi_n - \Phi_0 + \sum_{i=1}^{n}t_i\]
-
+\begin{align*}
+a_i &= \Phi_i - \Phi_{i-1} + t_i\\
+\implies \sum_{i=1}^{n}a_i &= \sum_{i=1}^{n}\Phi_i - \Phi_{i-1} + t_i\\
+&= \sum_{i=1}^{n}\Phi_i - \sum_{i=0}^{n-1}\Phi_i + \sum_{i=1}^{n}t_i\\
+&= \Phi_n - \Phi_0 + \sum_{i=1}^{n}t_i\\
+\end{align*}
 By definition we know that $\Phi_0 = -1$ and since $2k-s \geq 0$ for $\alpha \geq \frac{1}{2}$ and $\frac{s}{2}-k \geq 0$ for $\alpha < \frac{1}{2}$ it follows that $\Phi_n - \Phi_0 >= 0$.
 \[\implies \Phi_n - \Phi_0 + \sum_{i=1}^{n}t_i \geq \sum t_i\]\qed
 
 \section*{Exercise 6.3}
-We show
-\[\sum_{k=2}^{n-1}k\,lg\,k \leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2\]
-\[\sum_{k=2}^{n-1}k\,lg\,k = \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k\,lg\,k\ + \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\,lg\,k\]
-\[\leq \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k\,lg\,\lceil \frac{n}{2} \rceil + \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\,lg\,n\]
-\[= lg\,\lceil \frac{n}{2} \rceil \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\]
-\[= (lg\,n - lg\,2) \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\]
-\[= lg\,n \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k - \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\]
-\[= lg\,n \frac{(n-1)(n-2)}{2} - \]
+We show $\sum_{k=2}^{n-1}k\,lg\,k \leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2$.
+\begin{align*}
+\sum_{k=2}^{n-1}k\,lg\,k &\leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2\\
+\sum_{k=2}^{n-1}k\,lg\,k &= \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k\,lg\,k\ + \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\,lg\,k\\
+\sum_{k=2}^{n-1}k\,lg\,k &\leq \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k\,lg\,\lceil\frac{n}{2}\rceil + \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\,lg\,n\\
+&= lg\,\lceil \frac{n}{2} \rceil \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\\
+&= (lg\,n - lg\,2) \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\\
+&= lg\,n \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k - \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k + lg\,n \sum_{k=\lceil \frac{n}{2} \rceil}^{n-1}k\\
+&= lg\,n \sum_{k=2}^{n-1}k - \sum_{k=2}^{\lceil \frac{n}{2} \rceil - 1}k\\
+&\leq lg\,n \frac{(n-1)(n-2)}{2} - \frac{(\frac{n}{2} - 1)(\frac{n}{2} - 2)}{2}\\
+&= \frac{1}{2}lg\,n (n^2 - 3n + 2) - \frac{1}{2}(\frac{n^2}{4} - n - \frac{n}{2} + 2)\\
+&= \frac{1}{2}n^2\,lg\,n - \frac{3}{2}n\,lg\,n + lg\,n - \frac{1}{2}(\frac{n^2}{4} - n - \frac{n}{2} + 2)\\
+&= \frac{1}{2}n^2\,lg\,n - \frac{3}{2}n\,lg\,n + lg\,n - \frac{1}{8}n^2 + \frac{n}{2} + \frac{n}{4} - 1\\
+&= \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2 + lg\,n - \frac{3}{2}n\,lg\,n + \frac{3}{4}n - 1\\
+&\leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2 + lg\,n - \frac{3}{2}n + \frac{3}{4}n - 1\\
+&= \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2 + lg\,n - \frac{3}{4}n - 1\\
+\implies \sum_{k=2}^{n-1}k\,lg\,k &\leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2 + lg\,n - \frac{3}{4}n - 1 \leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2\\
+\implies &\frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2 + lg\,n - \frac{3}{4}n - 1 \leq \frac{1}{2}n^2\,lg\,n - \frac{1}{8}n^2\\
+\iff &lg\,n - \frac{3}{4}n - 1 \leq 0\\
+\text{since } \forall{n \in \mathbb{N}}:\, lg\,n - \frac{3}{4}n \leq 0 &\implies \, lg\,n - \frac{3}{4}n - 1 \leq -1 \leq 0\\
+\end{align*}
+\qed
 \end{document}




