Doubly-Logarithmic Time

XFastTrie: searching in doubly-logarithmic Time

The performance of the BinaryTrie structure is not very impressive. The number of elements, n, stored in the structure is at most $2^w$, so $\log n \leq w$. In other words, any of the comparison-based SSet are at least as efficient as a BinaryTrie, and are not restricted to only storing integers.

Next, we describe the XFastTrie, which is just a BinaryTrie with $w + 1$ hash tables—one for each level of the trie. These hash tables are used to speed up the find(x) operation to $O(\log w)$ time. Recall that the find(x) operation in a BinaryTrie is almost complete once we reach a node, u, where the search path for x would like to proceed to u.right (or u.left) but u has no right (respectively, left) child. At this point, the search uses u.jump to jump to a leaf, v, of the BinaryTrie and either return v or its successor in the linked list of leaves. An XFastTrie speeds up the search process by using binary search on the levels of the trie to locate the node u.

To use binary search, we need a way to determine if the node u we are looking for is above a particular level, i, or if u is at or below level i. This information is given by the highest-order i bits in the binary representation of x; these bits determine the search path that x takes from the root to level i.

Visual demonstration of the search path

The visual demonstration of the search path is shown below:

As an example, in the above illustration; the last node, u, on the search path for $14$ (whose binary representation is $1110$) is the node labeled $11\star \star$ at level $2$ because there is no node labeled $111\star$ at level $3$. Therefore, we can label each node at level $i$ with an $i$-bit integer.

Then, the node u we are searching for would be at or below the level i if and only if there is a node at level i whose label matches the highest-order i bits of x.

In an XFastTrie, we store, for each $i \in \{0,...,w\}$, all the nodes at level i in a USet, t[i], that is implemented as a hash table. Using this USet allows us to check in constant expected time if there is a node at level i whose label matches the highest-order i bits of x. In fact, we can even find this node using t[i].find (x>>>(w − i)).

The hash tables $t[0],...,t[w]$ allow us to use binary search to find u. Initially, we know that u is at some level i with $0 \leq i < w+1$. We therefore initialize l = 0 and h = w + 1 and repeatedly look at the hash table t[i], where $i = \left \lfloor (l + h)/2\right \rfloor$. If $t[i]$ contains a node whose label matches $x$’s highest-order i bits then we set l = i (u is at or below level i); otherwise, we set h = i (u is above level i). This process terminates when $h − l \leq 1$, in which case we determine that u is at level l. We then complete the find(x) operation using u.jump and the doubly-linked list of leaves.

Each iteration of the while loop in the above method decreases h − l by roughly a factor of two, so this loop finds u after $O(\log w)$ iterations. Each iteration performs a constant amount of work and one find(x) operation in a USet, which takes a constant expected amount of time. The remaining work takes only constant time, so the find(x) method in an XFastTrie takes only $O(\log w)$ expected time.

The add(x) and remove(x) methods for an XFastTrie are almost identical to the same methods in a BinaryTrie. The only modifications are for managing the hash tables $t[0],. . . ,t[w]$ ...