Succinct Representation of Data Structures:Partial Sums.

By -

Partial Sums

Let a1, a2,..., an be a sequence of n non-negative k-bit numbers. The partial sums problem maintains the sequence under the following operations:

image

Dietz [13] gave a structure for the partial sum problem that supports sum and update in O(lg n/ lg lg n) time using O(n lg n) bits of extra space, for the case when k = Θ(lg n) and no constraints on δ. The time bounds are optimal due to the lower bound of Fred man and Saks [20]. As the information-theoretic space lower bound is kn bits, this structure uses space within a constant factor of the optimal.

The main idea of this structure is to store the elements at the leaves of a complete tree with branching factor O(lgE n) for some E< 1. The operations are performed by traversing a path from a leaf to the root, querying/updating the nodes along the path.

The searchable partial sums problem is an extension of the partial sums problem that also supports the following operation:

• search(j): find the smallest i such that sum(i) ≥ j.

When k = 1 (i.e., each element is a bit), the special case is commonly known as the dynamic bit vector problem, which maintains a bit vector of length n under rank, select and flip (update) operations.

For the searchable partial sums problem there is a structure that supports all operations in O(lg n/ lg lg n) time, and uses kn + o(kn) bits of space [49]. For k = O(lg lg n), one can also obtain a structure that again takes kn + o(kn) bits and supports sum and search in O(logb n) time and update in O(b) amortized time, for any parameter b ≥ lg n/ lg lg n [30].

For the partial sums problem, one can support the above trade-off for k = O(lg n) [49], and the time bounds can be shown to be optimal [20].

For the dynamic bit vector problem, one can support rank and select in O(logb n) time and flip in O(b) (worst-case) time, for any parameter lg n/ lg lg n ≤ b ≤ n, using o(n) bits of extra space. One can also extend the above trade-off for k = O(lg lg n), using kn + o(kn) bits of space.

See [49] and [30] for details.

Comments

Post a Comment

Popular posts from this blog

0/1 Knapsack Problem Memory function.

By -
Image
0 / 1 Knapsack Problem Memory function Definition: Given a set of n items of known weights w1,…,wn and values v1,…,vn and a knapsack of capacity W, the problem is to find the most valuable subset of the items that fit into the knapsack. Knapsack problem is an OPTIMIZATION PROBLEM Dynamic programming approach to solve knapsack problem Step 1: Identify the smaller sub-problems. If items are labeled 1..n, then a sub-problem would be to find an optimal solution for Sk = {items labeled 1, 2, .. k } Step 2: Recursively define the value of an optimal solution in terms of solutions to smaller problems. Initial conditions: S te p 3: Bottom up computation using iteration Question: Apply bottom-up dynamic programming algorithm to the following instance of the knapsack problem Capacity W= 5 Solution: Using dynamic programming approach, we have: What is the composition of the optimal subset? The composition of the optimal subset if found by tracing back the computations for the...
Read more

Binary Space Partitioning Trees:BSP Tree as a Hierarchy of Regions.

By -
Image
BSP Tree as a Hierarchy of Regions Another way to look at BSP Trees is to focus on the hierarchy of regions created by the recursive partitioning, instead of focusing on the hyperplanes themselves. This view helps us to see more easily how intersections are efficiently computed. The key idea is to think of a BSP Tree region as serving as a bounding volume: each node v corresponds to a convex volume that completely contains all the geometry represented by the sub tree rooted at v . Therefore, if some other geometric entity, such as a point, ray, object, etc., is found to not intersect the bounding volume, then no intersection computations need be performed with any geometry within that volume. Consider as an example a situation in which we are given some test point and we want to find which object if any this point lies in. Initially, we know only that the point lies somewhere space ( Figure 20.12). By comparing the location of the point with respect to the first partitioning hyper pl...
Read more

Drawing Trees:HV-Layout

By -
Image
HV-Layout An hv -layout of a binary tree is an upward (but not strictly upward) straight line orthogonal drawing with edges drawn as rightward horizontal or downward vertical segments. The ”hv” stands for horizontal-vertical. For any vertex v in a binary tree T we either have: 1. A child of v is either • aligned horizontally with v and to the right of v or • vertically aligned below v . 2. The smallest rectangles that cover the area of the subtrees of the children of v in the layout do not intersect. Figure 45.12 shows an example of a hv -layout A hv -layout is generated by applying a dived and conquer approach. The divide step constructs the hv -layout of the left and the right subtrees, while the conquer step either performs a ho rizontal or a ver t i c a l combination . Figure 45.13 shows the two types of com- bination. If the left subtree a node v is placed to the left in a horizontal combination and below in a vertical combination, the layout preserves ...
Read more