University:
University of California, San Diego
Course:
CSE 100/101 | Data Structures and Algorithms
Academic year:

2023
Views:

187

Pages:

2
Author:

anon

Improving the Analysis of the BuildHeap Procedure: A More Accurate Approach Because the BuildHeap process consists of around n/2 calls to the SiftDown operation, which has a running time of log n, it is sometimes assumed that its execution time is O(n log n). This estimate, nevertheless, might be overly pessimistic. The BuildHeap technique will be further examined and its running time will be more precisely estimated in this article. To do this, we will take into account the fact that if the SiftDown process is invoked for a node near the leaves, sifting it down will take considerably less time than log n. For the reason that it cannot make more swaps than the height of the relevant subtree before it reaches the leaves. In addition, there are a lot of nodes close to the root in a heap. The root is the only node with a height of log n, followed by two nodes with log n - 1, four nodes with log n - 2, and so on, with around n/4 nodes having a height of just 1. Let's estimate the BuildHeap procedure's duration more precisely in light of this. Analyzing the Cost of Sifting Down Let's examine a schematic illustration of a heap to help with this. There is just one node at the top level, and sifting it down incurs a logarithmic cost. There are at most n/2 nodes on the final level, and there is only room for one swap after they are culled. There are at most n/4 nodes on the following level, and there are a maximum of two swaps required to sift them down. We can get the entire running time of the BuildHeap operation by adding up all of these charges. Let's set an upper limit on this sum equal to a similar sum to make it more manageable. The sum is 1/2 + 2/4 + 3/8 + 4/16 + 5/32 +..., where the index of the sum, I can have a maximum value of log n. The sum from I = 1 to infinity of I divided by 2i can serve as the upper boundary of this sum. Although this total is finite in the instance of the BuildHeap procedure's running time, we shall upper-bound it by the infinite sum to simplify the analysis. Estimating the Running Time We will demonstrate that the sum is equal to two. This indicates that the BuildHeap procedure takes O(n + n log n) = O to complete (n log n). This is a considerably more accurate estimation of the BuildHeap procedure's running time than the O estimate that was previously employed (n log n). So by considering the various heap properties as well as the cost of filtering down for nodes at various heights, we have been able to produce a more accurate estimation of the BuildHeap function's execution time.