Depth of a Random Binary Search Tree with Concurrent Insertions

Abstract

Shuffle a deck of n cards numbered 1 through n. Deal out the first c cards into a hand. A player then repeatedly chooses one of the cards from the hand, inserts it into a binary search tree, and then adds the next card from deck to the hand (if the deck is empty). When the player finally runs out of cards, how deep can the search tree be?

This problem is motivated by concurrent insertions by c processes of random keys into a binary search tree, where the order of insertions is controlled by an adversary that can delay individual processes. We show that an adversary that uses any strategy based on comparing keys cannot obtain an expected average depth greater than $O(c + \log n)$. However, the adversary can obtain an expected tree height of $\varOmega (c \log (n/c))$, using a simple strategy of always playing the largest available card.

A Background on Martingales

Here, for the sake of completeness, we present background information about martingales that is used in our paper, following the presentation of Mitzenmacher and Upfal’s textbook [11].

We say that a sequence $Y_0, Y_1, \ldots$ of random variables is a martingale with respect to another sequence $X_0, X_1, \ldots$ of random variables if for all $n\ge 0$

• $Y_n$ is a function of $X_0,X_1,\ldots , X_n$,

• $\hbox {E}\left[ {\,|Y_n|\,}\right] <\infty$, and

• $\hbox {E}\left[ {Y_{n+1}}\;\bigg \vert \;{X_0,X_1,\ldots ,X_n}\right] = Y_n$.

The last property is called the martingale property. When $X_n$ contains all the information in the $X_i$ for $i < n$, we can write it more succinctly as $\hbox {E}\left[ {Y_{n+1}}\;\bigg \vert \;{X_n}\right] = Y_n$.

A random variable $\tau$ that takes values from $\mathbb {N}$ is a stopping time with respect to $X_0,X_1,\ldots$ if, for all $n\ge 0$, the event $\tau = n$ depends only on $X_0,X_1, \ldots ,X_n$.

See [14, Sect. 10.10] for a proof of Doob’s Optional Stopping Theorem, of which the following is a special case.

Theorem 4

If $Y_0,Y_1, \ldots$ is a martingale and $\tau$ is a stopping time, both with respect to $X_0,X_1,\ldots$, and $\tau$ is bounded, then $\hbox {E}\left[ {Y_\tau }\right] =\hbox {E}\left[ {Y_0}\right]$.

For some applications, it makes sense to replace the martingale property with an inequality. A supermartingale is a process defined as above except that $Y_n \ge \hbox {E}\left[ {Y_{n+1}}\;\bigg \vert \;{X_0,\dots ,X_n}\right]$; where a martingale stays the same on average, a supermartingale is non-increasing on average. A straightforward induction shows that supermartingales satisfy $Y_k \ge \hbox {E}\left[ {Y_n}\;\bigg \vert \;{X_0,\dots ,X_k}\right]$ whenever $k \le n$.