B + -tree: In this case, initially it is considered that the B + -tree is empty and the values are inserted in ascending order. It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values. The first two key values (2,3) get added to the root node.

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Answer 1

Question

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

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CPS 2231 Computer Programming Homework #3 Due Date: Posted on Canvas 1. Provide answers to the following Check Point Questions from our textbook (5 points): a. How do you define a class? How do you define a class in Eclipse? b. How do you declare an object's reference variable (Hint: object's reference variable is the name of that object)? c. How do you create an object? d. What are the differences between constructors and regular methods? e. Explain why we need classes and objects in Java programming. 2. Write the Account class. The UML diagram of the class is represented below (10 points): Account id: int = 0 - balance: double = 0 - annualInterestRate: double = 0.02 - dateCreated: java.util.Date + Account() + Account(id: int, balance: double) + getId(): int + setId(newId: int): void + getBalance(): double + setBalance(newBalance: double): void + getAnnualInterestRate(): double + setAnnualInterest Rate (newRate: double): void + toString(): String + getDataCreated(): java.util.Date +…

THIS IS NOT A GRADING ASSIGNMENT: Please only do lab 2.2 (bottom part of the first picture) For that Lab 2.2 do: *Part 1 (do the CODE, that's super important I need it) *Part 2 *Part 3 I also attached Section 2.5.2 which is part of the step 1 so you can read what is it about. Thank you!

THIS IS NOT A GRADING ASSIGNMENT: Please only do lab 2.2 (bottom part of the first picture) For that Lab 2.2 do: *Part 1 *Part 2 *Part 3 I also attached Section 2.5.2 which is part of the step 1 so you can read what is it about. Thank you!

Answer 2

Question

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Answer 6

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

Answer 7

Chapter 14, Problem 3PE

a)

Program Plan Intro

B⁺ -tree:

In this case, initially it is considered that the B⁺-tree is empty and the values are inserted in ascending order.

It is not possible to have a value less that n/2 as the non-root node and in this case the non-root node can only have 2-3 key values.

The first two key values (2,3) get added to the root node.

Answer 8

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

Answer 9

b)

Explanation of Solution

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

Answer 10

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Answer 11

c)

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 14 - Prob. 1PE Ch. 14 - Prob. 2PE Ch. 14 - Prob. 3PE Ch. 14 - Prob. 4PE Ch. 14 - Prob. 5PE Ch. 14 - Prob. 7PE Ch. 14 - Prob. 8PE Ch. 14 - Prob. 9PE Ch. 14 - Prob. 10PE Ch. 14 - Prob. 11PE

Explanation of Solution

Next, when key value 7 is inserted in the tree; Root node now holds key value 7, and its first pointer points to a leaf node. The inserted leaf node contains the key value 2,3, and 5. As more key values are inserted, more leaf nodes are added as per requirement...

Explanation of Solution

The first four key values (2,3,5, and 7) are simply added to the root node. Next, when key value 11 is inserted in the tree; Root node now holds the key value 11, and its first pointer points to a leaf node. The inserted leaf node contains the key values 2,3,5, and 7...

Want to see the full answer?

Chapter 14 Solutions

Next, when key value 7 is inserted in the tree;

Root node now holds key value 7, and its first pointer points to a leaf node.

The inserted leaf node contains the key value 2,3, and 5.

As more key values are inserted, more leaf nodes are added as per requirement...

The first four key values (2,3,5, and 7) are simply added to the root node.

Next, when key value 11 is inserted in the tree;

Root node now holds the key value 11, and its first pointer points to a leaf node.

The inserted leaf node contains the key values 2,3,5, and 7...