How can I provide a proof for this question regarding the Heapify function for Heapsort? Analyzing Heapify(): FormalFixing up relationships between i, 1, and rtakes (1) time• If the heap at i has n elements, how manyelements can the subtrees at l or r have?■Draw it• Answer: 2n/3 (worst case: bottom row 1/2 full)• So time taken by Heapify () is given byT(n)T(2n/3) + (1)provideproof

To prove the complexity analysis of the Heapify function based on the given slide, let us break the…

Answered: Analyzing Heapify(): Formal Fixing up…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Implement a priority queue using a heapordered binary tree, but use a triply linked structure instead of an array. You will needthree links per node: two to traverse down the tree and one to traverse up the tree. Yourimplementation should guarantee logarithmic running time per operation, even if nomaximum priority-queue size is known ahead of time.
From the following sets of linked lists, which set should be chosen as the head, tominimize the time complexity, when we do union operation on them? Why?Set A: 0-3-5-6-8-10Set B: 1-2-4-9
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Construct a priority queue using a heapordered binary tree, but instead of an array, use a triply linked structure. Each node will require three links: two to go down the tree and one to traverse up the tree. Even if no maximum priority-queue size is specified ahead of time, your solution should ensure logarithmic running time per operation.
Par
30. In the implementation for a breadth-first search we studied, a queue was used. The code below replaces the queue with a stack. List the pre-order enumeration that the vertices in the graph below are visited using this modified method, starting from vertex 0. } 3 /** stack-based search */ static void modSearch (Graph G, int start) { Stack S new AStack (G.n()); S.push(start); } 6 G.setMark (start, VISITED); while (S.length) > 0) { } modSearch: intv S.pop(); PreVisit (G, v); for (int w = G.first (v); w < G.n(); w = G.next(v, w)) if (G.getMark (w) == UNVISITED) { G.setMark (w, VISITED); S.push(w);
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