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I have a guess about the problem above. Suppose I have a binary search tree $T$ initially empty. Suppose I drawn $x_1,\ldots,x_k$ (from some real interval $[a,b]$) keys and I want to insert the keys in such order in $T$, what's the expected depth/height of this tree?

My attempt is the following, suppose $x_1$ has been assigned as root, I define a random variable $L$ as

$$ L = L(x_2,\ldots,x_k | x_1) = \# \left\{x_i : x_i < x_1 ,2 \leq i \leq k\right\} = \sum_{j=2}^k u(x_1 - x_j) $$

This function $L$ models how many elements would go in the left subtree rooted at $x_1$, here $u(x)$ is defined as

$$ u(x) = \begin{cases} 1 & x > 0 \\ 0 & otherwise \end{cases}, $$

we also have

$$ R = k - 1 -L $$

the number of elements at the right of $x_1$. Now the expected value for $L$ is

$$ \bar{L} = \int_{[a,b]^{k-1}} L(x_2,\ldots,x_k | x_1) p(x_2,\ldots, x_k) dx_2 \ldots dx_k = \\ \int_{[a,b]^{k-1}} \sum_{j=2}^k u(x_1 - x_j) p(x_2,\ldots, x_k) dx_2 \ldots dx_k = \\ \int_{[a,b]^{k-1}} \sum_{j=2}^k u(x_1 - x_j) p(x_j) \prod_{l=2,l\neq j}^{k}p(x_l) dx_2 \ldots dx_k = \sum_{j=2}^k \int_{[a,b]} u(x_1 - x_j) p(x_j) dx_j = \\ (k-1) \frac{x_1-a}{b-a} $$

We then have

$$\bar{L} = (k-1) \frac{x_1-a}{b-a} = f(x_1)$$

If we now compute the expected value for $f$ we will get the on average how many elements will go on the left subtree rooted at $x_1$, regardless of $x_1$. This is easier to compute and we get a value I would expect indeed

$$ \bar{f} = \frac{k-1}{2} $$

We know now that given random values they will be equally distributed, we can write down the recurrence

$$ h(k) = 1 + h \left(\frac{k-1}{2} \right) $$

which models the average height for $k$ random elements, clearly $h$ is monotonic and we can write the inequality

$$ h(k) \leq 1 + h(k/2), $$

and therefore we end up with the solution

$$ h(k) = O(\log k) $$

Assuming the analysis is correct the questions are:

  1. I'm only analyzing with random inputs, what about if now I'll do deletion, how would the probabilistic model change? By deletion mean pick $y_1,\ldots, y_m \in \left\{x_1,\ldots x_k \right\}$ random and delete those, I expect a similar result, but I don't have a clue on the probabilistic model.
  2. Is there an alternative proof maybe more combinatorial?

Thank you

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    $\begingroup$ Unfortunately, it is generally not the case that $\mathbb{E}[\phi(X)] = \phi(\mathbb{E}[X])$. Therefore your recursive formula for $h$ doesn't quite hold. Perhaps you can get a similar inequality by convexity, if you can show that $h$ is convex. $\endgroup$ – Yuval Filmus Jul 22 '18 at 21:12
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    $\begingroup$ The height of a random binary search tree is a classical topic, and it is easy to find material on it, in various levels of detail. $\endgroup$ – Yuval Filmus Jul 22 '18 at 21:12
  • $\begingroup$ Which formula doesn't hold? $\endgroup$ – user8469759 Jul 22 '18 at 21:23
  • $\begingroup$ $h(k) = 1 + h(\frac{k-1}{2})$. $\endgroup$ – Yuval Filmus Jul 22 '18 at 21:55
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    $\begingroup$ Possible duplicate of Proof that a randomly built binary search tree has logarithmic height $\endgroup$ – Yuval Filmus Jul 24 '18 at 6:58

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