Added GA3 EX3

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Claudio Maggioni 2019-05-10 12:03:08 +02:00
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\subsection{Point B}
A red-black tree of n distinct elements has an height between $log(n)$
and $2 log(n)$ thanks to the red-black tree invariant. The worst-case insertion
complexity is $log(n)$ since finding the right place to insert is as complex as
A red-black tree of n distinct elements has an height between $\log(n)$
and $2\log(n)$ thanks to the red-black tree invariant. The worst-case insertion
complexity is $\log(n)$ since finding the right place to insert is as complex as
a regular tree (i.e. logarithmic) and the ``fixup'' operation is logarithmic as
well (the tree traversal is logarithmic while operations in each iteration are constant).
In asymptotic terms, the uneven height of leaves in the tree does not make a difference
since it is a constant factor (since the maximum height is $2 log(n)$ and the minimum one is $log(n)$.
since it is a constant factor (since the maximum height is $2\log(n)$ and the minimum one is $\log(n)$.
\section{Exercise 3}
\begin{lstlisting}[caption=Solution for exercise 3, label={lst:ex3}]
FUNCTION JOIN-INTERVALS(X)
if X.length == 0:
return
for i from 1 to X.length:
if i % 2 == 1:
X[i] $\gets$ (X[i], 1)
else:
X[i] $\gets$ (X[i], -1)
$\textit{sort X in place by 1st tuple element in O(n log(n))}$
c $\gets$ 1
n $\gets$ 0
start $\gets$ X[i][1]
for i from 2 to X.length:
c $\gets$ c + X[i][2]
if c == 0:
X[2 * n + 1] $\gets$ start
X[2 * n + 2] $\gets$ X[i][1]
start $\gets$ X[i+1][1]
n $\gets$ n + 1
X.length $\gets$ 2 * n
\end{lstlisting}
\\
The complexity of this algorithm is $O(n\log(n))$ since the sorting is in $\Theta(n\log(n))$ and the
union operation afterwards is $\Theta(n)$.
\end{document}