New answers tagged trees
2
Assume $n>1$. The solution to $k^k=n$ is $k=e^{W(\log n)}$, where $W(\cdot)$ is the Lambert W function.
Another way to express the solution is
$$k = \dfrac{\log n}{\log\dfrac{\log n}{\log\dfrac{\log n}{\log\dfrac{\log n}{\cdots}}}}.$$
In other words, $k$ is the limit of $k_0=\log n$, $k_1=\dfrac{\log n}{\log k_1}$, $\cdots$, $k_{t+1}=\dfrac{\log n}{\...
2
If a binary tree has height $h$ then it has at most 1 node at depth 0, at most 2 nodes at depth 1, ..., at most $2^{h-1}$ nodes at depth $h-1$ (the maximal depth), and so at most $1+2+\cdots+2^{h-1} = 2^h-1$ nodes in total.
You can also prove this by induction. When $h=1$, the tree consists only of a root, so at most $1=2^h-1$ vertices. Given a tree of ...
0
Hint:
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