Suppose a B +tree is used for indexing a database file. Consider the following information : size of the search key field = 10 bytes, block size = 1024 bytes, size of the record pointer = 9 bytes, size of the block pointer = 8 bytes. Let K be the order of internal node and L be the order of leaf node of B +tree, then (K, L) = ______.
(57, 53)
Understanding B+ Tree Node Order Calculation
This question asks us to determine the order of the internal nodes (K) and the leaf nodes (L) for a B+ tree index based on given disk block and pointer sizes. The order of a B+ tree node dictates the maximum number of children an internal node can have and the maximum number of data entries a leaf node can hold. We need to figure out how many keys and pointers can fit into a single block of the specified size.
Given Information:
| Parameter | Size |
|---|---|
| Size of search key field | 10 bytes |
| Block size | 1024 bytes |
| Size of record pointer | 9 bytes |
| Size of block pointer | 8 bytes |
Calculating the Order of Internal Nodes (K)
An internal node in a B+ tree with order K typically stores K-1 search key values and K block pointers. These pointers point to child nodes (which can be other internal nodes or leaf nodes). The total space used by these keys and pointers must fit within the block size. We assume the minimum space required for an internal node is when it contains the maximum number of keys and pointers allowed, which is K-1 keys and K pointers.
The space constraint for an internal node can be expressed as:
(Number of keys * Size of key) + (Number of block pointers * Size of block pointer) <= Block size
Substituting the values for an internal node of order K:
(K - 1) * (Size of search key field) + K * (Size of block pointer) <= Block size
Plugging in the given sizes:
\((K - 1) \times 10 \text{ bytes} + K \times 8 \text{ bytes} \leq 1024 \text{ bytes}\)
\(10K - 10 + 8K \leq 1024\)
\(18K - 10 \leq 1024\)
\(18K \leq 1024 + 10\)
\(18K \leq 1034\)
\(K \leq \frac{1034}{18}\)
\(K \leq 57.44...\)
Since the order K must be an integer, the maximum possible integer value for K that satisfies this condition is 57.
Therefore, the order of the internal node K is 57.
Calculating the Order of Leaf Nodes (L)
A leaf node in a B+ tree with order L typically stores L data entries. Each data entry in a leaf node consists of a search key value and a record pointer. Additionally, leaf nodes are often linked together sequentially, requiring space for one block pointer to the next leaf node. The total space used by these entries and the next-pointer must fit within the block size. We assume the minimum space required for a leaf node is when it contains the maximum number of data entries allowed, which is L entries, plus the next-pointer.
The space constraint for a leaf node can be expressed as:
(Number of data entries * (Size of key + Size of record pointer)) + (Size of block pointer for next leaf) <= Block size
Substituting the values for a leaf node storing L data entries:
L * (Size of search key field + Size of record pointer) + (Size of block pointer) <= Block size
Plugging in the given sizes:
\(L \times (10 \text{ bytes} + 9 \text{ bytes}) + 8 \text{ bytes} \leq 1024 \text{ bytes}\)
\(L \times 19 + 8 \leq 1024\)
\(19L + 8 \leq 1024\)
\(19L \leq 1024 - 8\)
\(19L \leq 1016\)
\(L \leq \frac{1016}{19}\)
\(L \leq 53.47...\)
Since the order L must be an integer, the maximum possible integer value for L that satisfies this condition is 53.
Therefore, the order of the leaf node L is 53.
Result
Based on the calculations, the order of the internal node K is 57 and the order of the leaf node L is 53. Thus, (K, L) = (57, 53).
Revision Table: B+ Tree Node Order Calculation
| Node Type | Formula for Order (X) | Calculation | Resulting Order |
|---|---|---|---|
| Internal Node (K) | \((K - 1) \times \text{Key Size} + K \times \text{Block Pointer Size} \leq \text{Block Size}\) | \((K - 1) \times 10 + K \times 8 \leq 1024 \Rightarrow 18K - 10 \leq 1024 \Rightarrow 18K \leq 1034 \Rightarrow K \leq 57.44\dots\) | K = 57 |
| Leaf Node (L) | \(L \times (\text{Key Size} + \text{Record Pointer Size}) + \text{Block Pointer Size} \leq \text{Block Size}\) | \(L \times (10 + 9) + 8 \leq 1024 \Rightarrow 19L + 8 \leq 1024 \Rightarrow 19L \leq 1016 \Rightarrow L \leq 53.47\dots\) | L = 53 |
Additional Information: B+ Tree Structure and Properties
A B+ tree is a self-balancing tree data structure used for indexing in databases and file systems. Its main advantage is its ability to efficiently retrieve records from disk storage. Here are some key properties:
- All data entries (key-pointer pairs pointing to actual data records) are stored only in the leaf nodes.
- Internal nodes store only search key values and block pointers to guide the search towards the correct leaf node.
- Leaf nodes are typically linked together sequentially using pointers, allowing for efficient range queries.
- All leaf nodes are at the same depth, forming a sorted sequence of data entries.
- The root of the tree can be either a leaf or an internal node.
- The order of the tree (often related to K and L) determines the minimum and maximum number of entries (keys/pointers or data entries) a node can hold, except for the root node. This helps keep the tree relatively shallow, minimizing disk I/O for searches.
Calculating the node order is crucial for determining the capacity of the tree and estimating its height, which directly impacts search performance.
Important Questions from Indexing - Teaching
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