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Fixed GA3 again and again

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Claudio Maggioni 2019-05-13 17:16:35 +02:00
parent d37f05664d
commit 59b66ac061
2 changed files with 8 additions and 1 deletions
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GA3/ga3.tex
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@ -639,7 +639,7 @@ FUNCTION JOIN-INTERVALS(X)
c $\gets$ 1 c $\gets$ 1
n $\gets$ 0 n $\gets$ 0
start $\gets$ X[i][1] start $\gets$ X[1][1]
for i from 2 to X.length: for i from 2 to X.length:
c $\gets$ c + X[i][2] c $\gets$ c + X[i][2]
if c == 0: if c == 0:
@ -654,4 +654,11 @@ FUNCTION JOIN-INTERVALS(X)
The complexity of this algorithm is $O(n\log(n))$ since the sorting is in $\Theta(n\log(n))$ and the 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 $\Theta(n)$. union operation afterwards is $\Theta(n)$.
\section{Bonus}
The number of possible trees with $n$ nodes can be defined recursively. The number of trees with 0 or 1 node is clearly 1, since there is no freedom in arranging any remaining elements as children. For $n$ nodes, this number can be defined as the the sum, for each $x, y \in N_0$ s.t. $x + y = n - 1$, of the number of trees with $x$ nodes times the number of trees with $y$ nodes. We consider only $n - 1$ nodes since one node must be the root of the tree.
Using math notation, the number $T_n$ of trees with $n$ nodes can be expressed as:
$$T_n = 1 \hspace{1cm} n = 0 \lor n = 1$$
$$T_n = \sum_{i = 0}^{n-1} C_i C_{n-1-i} \hspace{1cm} n \geq 2$$
\end{document} \end{document}