Geometry of binary search trees
In computer science, one approach to the dynamic optimality problem on online algorithms for binary search trees involves reformulating the problem geometrically, in terms of augmenting a set of points in the plane with as few additional points as possible to avoid rectangles with only two points on their boundary.
Access sequences and competitive ratio
As typically formulated, the online binary search tree problem involves search trees defined over a fixed key set. An access sequence is a sequence ... where each access belongs to the key set.Any particular algorithm for maintaining binary search trees has a cost for each access sequence that models the amount of time it would take to use the structure to search for each of the keys in the access sequence in turn. The cost of a search is modeled by assuming that the search tree algorithm has a single pointer into a binary search tree, which at the start of each search points to the root of the tree. The algorithm may then perform any sequence of the following operations:
- Move the pointer to its left child.
- Move the pointer to its right child.
- Move the pointer to its parent.
- Perform a single tree rotation on the pointer and its parent.
As is standard in competitive analysis, the competitive ratio of an algorithm A is defined to be the maximum, over all access sequences, of the ratio of the cost for A to the best cost that any algorithm could achieve:
The dynamic optimality conjecture states that splay trees have a constant competitive ratio, but this remains unproven. The geometric view of binary search trees provides a different way of understanding the problem that has led to the development of alternative algorithms that could also have a constant competitive ratio.
Translation to a geometric point set
In the geometric view of the online binary search tree problem,an access sequence is mapped to the set of points, where the X-axis represents the key space and the Y-axis represents time; to which a set of touched nodes is added. By touched nodes we mean the following. Consider a BST access algorithm with a single pointer to a node in the tree. At the beginning of an access to a given key, this pointer is initialized to the root of the tree. Whenever the pointer moves to or is initialized to a node, we say that the node is touched.
We represent a BST algorithm for a given input sequence by drawing a point for each item that gets touched.
For example, assume the following BST on 4 nodes is given:
The key set is.
Let 3, 1, 4, 2 be the access sequence.
- In the first access, only the node 3 is touched.
- In the second access, the nodes 3 and 1 are touched.
- In the third access - 3 and 4 are touched.
- In the fourth access, touch 3, then 1, and after that 2.
Arborally satisfied point sets
A point set is said to be arborally satisfied if the following property holds: for anypair of points that do not lie on the same horizontal or vertical line, there exists a third point
which lies in the rectangle spanned by the first two points.
Theorem
A point set containing the points is arborally satisfied if and only if it corresponds to a valid BST for the input sequence.Proof
First, prove that the point set for any valid BST algorithm is arborally satisfied.Consider points and, where is touched at time and is touched at time. Assume by symmetry that and. It needs to be shown that there exists a third point in the rectangle
with corners as and. Also let denote the lowest common ancestor of nodes
and right before time. There are a few cases:
- If, then use the point, since must have been touched if was.
- If, then the point can be used.
- If neither of the above two cases holds, then must be an ancestor of right before time and be an ancestor of right before time. Then at some time, must have been rotated above, so the point can be used.
If a pair of nodes, and, straddle the boundary between the touched and untouched part
of the treap, then if is to be touched sooner than then is an unsatisfied rectangle because the leftmost such point would be the right child of, not.
Corollary
Finding the best BST execution for the input sequence is equivalent to finding the minimum cardinality superset of points that is arborally satisfied. The more general problem of finding the minimum cardinality arborally satisfied superset of a general set of input points, is known to be NP-complete.Greedy algorithm
The following greedy algorithm constructs arborally satisfiable sets:- Sweep the point set with a horizontal line by increasing coordinate.
- At time, place the minimal number of points at to make the point set up to arborally satisfied. This minimal set of points is uniquely defined: for any unsatisfied rectangle formed with in one corner, add the other corner at.