# Difference between revisions of "TADM2E 3.5"

1: A tree with $n$ nodes would have $n-1$ edges, because our internal nodes have both parent and child pointers, we can say that the number of pointers is equal to the double of edges. So, the ratio of data over total should be:
$\frac{4n}{4n + 4*2*(n-1)} = \frac{4}{4 + 8} = \frac{1}{3} (n\to\infty)$
$\frac{4*n}{(4*n + 4*(n-1))} = \frac{4*n}{4 * (n + n -1)} = \frac{n}{2*n - 1}$, this approaches $\frac{1}{2}$ as n gets large