Complete K-Array Tree MCQ Quiz - Objective Question with Answer for Complete K-Array Tree - Download Free PDF

The correct answer is option 3: To deal with relationships that involve more than two tables

Key Points

• An N-ary association in a database refers to a relationship that involves N entities (or tables), where N ≥ 3.
• It is used when the relationship cannot be accurately or efficiently represented using just binary (2-table) associations.
• Commonly modeled using a relationship table that includes foreign keys referencing each of the N participating entities.

Additional Information

• Option 1: Parent-child relationships are typically represented using recursive relationships or 1:N binary associations.
• Option 2: One-to-many is a 2-entity (binary) relationship, not N-ary.
• Option 4: Inheritance is represented using generalization/specialization hierarchies, not N-ary associations.

Hence, the correct answer is: option 3: To deal with relationships that involve more than two tables

The correct answer is Multilevel sparse indices.

Key Points

• Non-leaf nodes of a B+ tree structure form multilevel sparse indices, which are crucial for its efficiency and performance.
• In a B+ tree, all the actual data records are stored in the leaf nodes, while the non-leaf nodes (or internal nodes) store the keys that act as pointers to guide the search process.
• The non-leaf nodes do not store every key, only enough to direct the search to the appropriate leaf node, making the indices sparse.
• This structure helps in reducing the height of the tree, thus speeding up the search, insert, and delete operations.
• Multilevel indexing means that the tree is structured in multiple levels (root, intermediate levels, and leaf nodes), where each level helps narrow down the search space progressively.
• The B+ tree is a balanced tree, meaning all the leaf nodes are at the same level, which ensures consistent and predictable performance.

Additional Information

• B+ trees are widely used in database indexing and file systems due to their efficiency in handling large amounts of data.
• They are an extension of B-trees, with the primary difference being that in B+ trees, all values are stored at the leaf level, and internal nodes only store keys.
• This structure ensures that sequential traversal of the data can be done efficiently by following the linked leaf nodes.
• Because of the multilevel sparse indexing, B+ trees are well-suited for range queries and ordered traversals.
• B+ trees maintain balance by splitting and merging nodes as necessary during insertions and deletions, ensuring that the tree remains balanced with minimal height.

The correct answer is: 2) Singly Circular Linked List

Explanation:

• In a Singly Circular Linked List, each node has exactly one successor, and the last node’s next pointer points back to the first node, thus ensuring that every node has a successor.

• In a Singly Linked List, the last node's next pointer is null, so it does not have a successor.

• In a Doubly Linked List, nodes have both a successor and a predecessor, but again, the last node’s next is null, so not all nodes have a successor.

• Hence, the only correct option where every node (including the last) has a successor is:

Concept -

No. of total internal nodes = 1 + K + K² + K³ + … + K^N = (K^N -1)/ (K-1) - It constitutes geometric progression.

No. of total nodes = 1 + K + K² + K³ + … + K^N+1 = (K^N+1 -1)/ (K-1)

\(\frac{{{\rm{No}}.{\rm{\;of\;total\;internal\;nodes\;}}}}{{{\rm{No}}.{\rm{\;of\;total\;nodes}}}}\tilde = \;\frac{1}{K}\)

Formula:

L = I (n - 1) + 1

L =number of leaf nodes

I = number of internal nodes

n = n - ary tree

Calculation:
I = 8

n = 5

L = 8(5 - 1) + 1

L = 32 + 1 = 33

Number of leaf nodes = 33

Number of internal nodes = 8

Formula:

L = I (n - 1) + 1

L =number of leaf nodes

I = number of internal nodes

n = n - ary tree

Calculation:
I = 8

n = 5

L = 8(5 - 1) + 1

L = 32 + 1 = 33

Number of leaf nodes = 33

Number of internal nodes = 8

Concept -

No. of total internal nodes = 1 + K + K² + K³ + … + K^N = (K^N -1)/ (K-1) - It constitutes geometric progression.

No. of total nodes = 1 + K + K² + K³ + … + K^N+1 = (K^N+1 -1)/ (K-1)

\(\frac{{{\rm{No}}.{\rm{\;of\;total\;internal\;nodes\;}}}}{{{\rm{No}}.{\rm{\;of\;total\;nodes}}}}\tilde = \;\frac{1}{K}\)

The correct answer is Multilevel sparse indices.

Key Points

• Non-leaf nodes of a B+ tree structure form multilevel sparse indices, which are crucial for its efficiency and performance.
• In a B+ tree, all the actual data records are stored in the leaf nodes, while the non-leaf nodes (or internal nodes) store the keys that act as pointers to guide the search process.
• The non-leaf nodes do not store every key, only enough to direct the search to the appropriate leaf node, making the indices sparse.
• This structure helps in reducing the height of the tree, thus speeding up the search, insert, and delete operations.
• Multilevel indexing means that the tree is structured in multiple levels (root, intermediate levels, and leaf nodes), where each level helps narrow down the search space progressively.
• The B+ tree is a balanced tree, meaning all the leaf nodes are at the same level, which ensures consistent and predictable performance.

Additional Information

• B+ trees are widely used in database indexing and file systems due to their efficiency in handling large amounts of data.
• They are an extension of B-trees, with the primary difference being that in B+ trees, all values are stored at the leaf level, and internal nodes only store keys.
• This structure ensures that sequential traversal of the data can be done efficiently by following the linked leaf nodes.
• Because of the multilevel sparse indexing, B+ trees are well-suited for range queries and ordered traversals.
• B+ trees maintain balance by splitting and merging nodes as necessary during insertions and deletions, ensuring that the tree remains balanced with minimal height.

Formula:

L = I (n - 1) + 1

L =number of leaf nodes

I = number of internal nodes

n = n - ary tree

Calculation:
I = 8

n = 5

L = 8(5 - 1) + 1

L = 32 + 1 = 33

Number of leaf nodes = 33

Number of internal nodes = 8

Concept -

No. of total internal nodes = 1 + K + K² + K³ + … + K^N = (K^N -1)/ (K-1) - It constitutes geometric progression.

No. of total nodes = 1 + K + K² + K³ + … + K^N+1 = (K^N+1 -1)/ (K-1)

\(\frac{{{\rm{No}}.{\rm{\;of\;total\;internal\;nodes\;}}}}{{{\rm{No}}.{\rm{\;of\;total\;nodes}}}}\tilde = \;\frac{1}{K}\)

Consider an example:

n = 7

leaves = 5

Option 1:

\(\frac{n}{2} = \frac{7}{2} = 3.5\)

Option 2:

\(\frac{{\left( {2{\rm{n}}\: + \:1} \right)}}{3} = \frac{{2\: \times \:7\: + \:1}}{3} = 5\;\)

Option 3:

\(\frac{{\left( {n - 1} \right)}}{n} = \frac{{7 - 1}}{7} = \frac{6}{7}\)

Option 4:

\(n - 1 = 7 - 1 = 6\)

The correct answer is: 2) Singly Circular Linked List

Explanation:

• In a Singly Circular Linked List, each node has exactly one successor, and the last node’s next pointer points back to the first node, thus ensuring that every node has a successor.

• In a Singly Linked List, the last node's next pointer is null, so it does not have a successor.

• In a Doubly Linked List, nodes have both a successor and a predecessor, but again, the last node’s next is null, so not all nodes have a successor.

• Hence, the only correct option where every node (including the last) has a successor is:

Data:

number of leaves = L = 10

Formula

If k- ary tree with y internal nodes, then number of leaves

L = (k – 1) × y + 1

Calculation

10 = (k – 1)y + 1

∴y = \(\frac{9}{k -1} \)

Verification

leaves = L = 5

5 = (k - 1)y + 1

y = 4 ÷ (k-1) 

since k = 3

∴ y = 2

The correct answer is option 3: To deal with relationships that involve more than two tables

Key Points

• An N-ary association in a database refers to a relationship that involves N entities (or tables), where N ≥ 3.
• It is used when the relationship cannot be accurately or efficiently represented using just binary (2-table) associations.
• Commonly modeled using a relationship table that includes foreign keys referencing each of the N participating entities.

Additional Information

• Option 1: Parent-child relationships are typically represented using recursive relationships or 1:N binary associations.
• Option 2: One-to-many is a 2-entity (binary) relationship, not N-ary.
• Option 4: Inheritance is represented using generalization/specialization hierarchies, not N-ary associations.

Hence, the correct answer is: option 3: To deal with relationships that involve more than two tables

The correct answer is Multilevel sparse indices.

Key Points

• Non-leaf nodes of a B+ tree structure form multilevel sparse indices, which are crucial for its efficiency and performance.
• In a B+ tree, all the actual data records are stored in the leaf nodes, while the non-leaf nodes (or internal nodes) store the keys that act as pointers to guide the search process.
• The non-leaf nodes do not store every key, only enough to direct the search to the appropriate leaf node, making the indices sparse.
• This structure helps in reducing the height of the tree, thus speeding up the search, insert, and delete operations.
• Multilevel indexing means that the tree is structured in multiple levels (root, intermediate levels, and leaf nodes), where each level helps narrow down the search space progressively.
• The B+ tree is a balanced tree, meaning all the leaf nodes are at the same level, which ensures consistent and predictable performance.

Additional Information

• B+ trees are widely used in database indexing and file systems due to their efficiency in handling large amounts of data.
• They are an extension of B-trees, with the primary difference being that in B+ trees, all values are stored at the leaf level, and internal nodes only store keys.
• This structure ensures that sequential traversal of the data can be done efficiently by following the linked leaf nodes.
• Because of the multilevel sparse indexing, B+ trees are well-suited for range queries and ordered traversals.
• B+ trees maintain balance by splitting and merging nodes as necessary during insertions and deletions, ensuring that the tree remains balanced with minimal height.