Let's assume that a binary heap is represented using a binary tree such that each node may have a left child node and a right child node. For this type of representation, we can still label the nodes of the tree in the same way as we label the nodes for an array representation. That is, the root node has a label 1. In general, for a node with label i, its left child node will have a label 2i and the right child has a label 2i+1. For any i with 1 <= I <= n , Terry says that the following easy algorithm will walk you from the root node to the node with label i: First find the binary representation P of i. Start with the rightmost bit (least significant bit) of P, walk down from the root as follow: For a 0 bit, walk to the left child, for a 1 bit walk to the right child. At the end, you’ll reach the node with label i. Which of the following is the most appropriate? A. Terry’s algorithm is wrong and not fixable. B. Terry’s algorithm is right. C. Terry's algorithm can be fixed as follows: Walk from leftmost bit (most significant bit) of P. D. Terry's algorithm can be fixed as follows: Ignore the leftmost bit (the most significant bit) of P and Walk from the second leftmost (second most significant) bit of P.

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Let's assume that a binary heap is represented using a binary tree such that each node may have a left child node and a right child node. For this type of representation, we can still label the nodes of the tree in the same way as we label the nodes for an array representation. That is, the root node has a label 1. In general, for a node with label i, its left child node will have a label 2i and the right child has a label 2i+1. For any i with 1 <= I <= n , Terry says that the following easy algorithm will walk you from the root node to the node with label i:

First find the binary representation P of i.
Start with the rightmost bit (least significant bit) of P, walk down from the root as follow:
- For a 0 bit, walk to the left child, for a 1 bit walk to the right child.
- At the end, you’ll reach the node with label i.

Which of the following is the most appropriate?

A. Terry’s algorithm is wrong and not fixable.

B. Terry’s algorithm is right.

C. Terry's algorithm can be fixed as follows: Walk from leftmost bit (most significant bit) of P.

D. Terry's algorithm can be fixed as follows: Ignore the leftmost bit (the most significant bit) of P and Walk from the second leftmost (second most significant) bit of P.

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