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1 Chapter 17 Disk Storage, Basic File Structures, and Hashing Chapter 18 Index Structures for Files

1 Chapter 17 Disk Storage, Basic File Structures, and Hashing Chapter 18 Index Structures for Files

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Chapter 17 Disk Storage, Basic File Structures,

and Hashing

Chapter 18Index Structures for Files

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Oracle SQL*Loader

• http://www.orafaq.com/faqloadr.htm

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Storage

• Primary storage (main memory)– Can be operated on directly by computer CPU

small, fast

• Secondary storage– http://en.wikipedia.org/wiki/Hard_disk– Can not be operated on directly by computer CPU

– Magnetic disks, optical disks, tapes, etc.

– Larger capacities, inexpensive, slower than main memory

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Storage capacity units

• Kilobytes – 1000 bytes

• Megabytes – 1 million bytes

• Gigabytes (Gbytes) – 1 billion bytes

• Terabytes – 1000 gigabytes

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Memory Hierarchies and Storage Devices

• Primary storage– Cache (static RAM)– most expensive, fast, used

by CPU to speed up execution programshttp://wombat.doc.ic.ac.uk/foldoc/foldoc.cgi?query=cache

- Main memory (dynamic RAM) – work area for CPU

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Secondary storage (Mass storage)

– CD-ROM– Tapes– Disks

Main memory database: entire database is stored in main memory

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File organization• Heap file (unordered file)

place new records in no order at the end of the file

• Sorted file ( sequential file) keeps the records ordered by the value of a particular

file

• Hashed fileUses hash function applied to a field (hash key) to

determine a record’s placement on disk

• B-trees, B+ trees – use tree structure

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Binary codes

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Tracks

The part of a disk which passes under one read/write head while the head is stationary. The number of tracks on a disk surface therefore corresponds to the number of different radial positions of the head(s). The collection of all tracks on all surfaces at a given radial position is known

a cylinder and each track is divided into sectors.

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Cylinder

• The set of tracks on a multi-headed disk that may be accessed without head movement. That is, the collection of disk tracks which are the same distance from the spindle about which the disks rotate.

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Sector

• one sector lies within a continuous range of rotational angle of the disk

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Data transfer between main memory and disks

(in blocks)

Hardware Address of a block– Surface number– Track number– Block number

• Time requires– Seek time– Rotational delay time (latency)– Block transfer time

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Hashing techniques

Static hashing – hash address space is fixed

Extendible hashing

Linear hashing

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Hashing algorithm

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Hash Table (Wikipedia) http://en.wikipedia.org/wiki/Hash_table

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• A search tree of order p is a tree such that each node contains at most p - 1 search values and p pointers in the order < P1, K1, P2, K2, ..., Pq-1, Kq-1, Pq

>, where q 1 p; each Pi is a pointer to a child node

(or a null pointer); and each Ki is a search value

from some ordered set of values.

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B tree of order p

1. Each internal node in the B-tree is of the form

<P1, <K1, Pr1> , P2, <K2, Pr2> , ..., <Kq-1,Prq-1> , Pq> where q 1 p. Each Pi is a tree pointer—a pointer to another node in the B-tree. Each Pri is a data pointer —a pointer to the record whose search key field value is equal to Ki (or to the data file block containing that record).

2. Within each node, K1 <K2 < ... < Kq-1.

3. For all search key field values X in the subtree pointed at by P i (the ith subtree, see Figure 06.10a), we have:

Ki-1 < X < Ki for 1 < i < q; X < Ki for i = 1; and Ki-1 < X for i = q. 4. Each node has at most p tree pointers. 5. Each node, except the root and leaf nodes, has at least (p/2) tree pointers. The root

node has at least two tree pointers unless it is the only node in the tree. 6. A node with q tree pointers, q 1 p, has q - 1 search key field values (and hence has q -

1 data pointers). 7. All leaf nodes are at the same level. Leaf nodes have the same structure as internal

nodes except that all of their tree pointers Pi are null.

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EXAMPLE 5: Suppose that the search field of Example 4 is a nonordering key field, and we construct a B-tree on this field. Assume that each node of the B-tree is 69 percent full. Each node, on the average, will have p * 0.69 = 23 * 0.69 or approximately 16 pointers and, hence, 15 search key field values. The average fan-out fo =16. We can start at the root and see how many values and pointers can exist, on the average, at each subsequent level:

Root: 1 node 15 entries 16 pointers

Level 1: 16 nodes 240 entries 256 pointers

Level 2: 256 nodes 3840 entries 4096 pointers

Level 3: 4096 nodes 61,440 entries 65536 pointers

Level 4: 65536 nodes 983,040 entries

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B+ TreesThe structure of the internal nodes of a B+-

tree of order p is as follows: 1. Each internal node is of the form <P1, K1, P2, K2, ..., Pq-1, Kq-1, Pq>

where q 1 p and each Pi is a tree pointer.

2. Within each internal node, K1 < K2 < ... <Kq-1.

3. For all search field values X in the subtree pointed at by Pi, we have

Ki-1 < X 1 Ki for 1 < i < q; X 1 Ki for i = 1; and Ki-1 < X for i = q.

4. Each internal node has at most p tree pointers.

5. Each internal node, except the root, has at least (p/2) tree pointers. The root node has at least two tree pointers if it is an internal node.

6. An internal node with q pointers, q 1 p, has q - 1 search field values.

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The structure of the leaf nodes of a B+-tree of order p (Figure 14.11b) is as follows:

1. Each leaf node is of the form <<K1, Pr1> , <K2, Pr2>, ..., <Kq-1, Prq-1>, Pnext>

Where q 1 p, each Pri is a data pointer, and Pnext points to the next leaf

node of the B+-tree.

2. Within each leaf node, K1 < K2 < ... < Kq-1, q 1 p.

3. Each Pri is a data pointer that points to the record whose search field

value is Ki or to a file block containing the record (or to a block of

record pointers that point to records whose search field value is Ki if

the search field is not a key).

4. Each leaf node has at least (p/2) values.

5. All leaf nodes are at the same level.

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EXAMPLE 7: Suppose that we construct a B+-tree on the field of Example 6. To calculate the approximate number of entries of the B+-tree, we assume that each node is 69 percent full. On the average, each internal node will have 34 * 0.69 or approximately 23 pointers, and hence 22 values. Each leaf node, on the average, will hold 0.69 * pleaf = 0.69 * 31 or

approximately 21 data record pointers. A B+-tree will have the following average number of entries at each level:

Root: 1 node 22 entries 23 pointers

Level 1: 23 nodes 506 entries 529 pointers

Level 2: 529 nodes 11,638 entries 12,167 pointers

Level 3: 12,167nodes 255,507entries 279,841 pointers

Level 4: 279,841 nodes 5,876,661entries

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