spring 2004 ece569 lecture 05.1 ece 569 database system engineering spring 2004 yanyong zhang...

Spring 2004ECE569 Lecture 05.1

ECE 569 Database System Engineering

Spring 2004

Yanyong Zhang www.ece.rutgers.edu/~yyzhang

Course URL www.ece.rutgers.edu/~yyzhang/spring04

http://www.ece.rutgers.edu/~yyzhang

http://www.ece.rutgers.edu/~yyzhang

http://www.ece.rutgers.edu/~yyzhang/spring04


Index “If you don’t find it in the index, look very carefully

through the entire catalog”

An index is a data structure that organizes data records on disk to optimize certain kind of retrieval operations.

A data entry refers to the records stored in an index file. A data entry with search key k, denoted as k*, contains enough information to locate (one or more) data records with search key value k. Three alternatives:1. K* can be an actual data record2. K* is a (k, tid) pair3. K* is a (k, tid-list) pair


Clustered Indexes When a file is organized so that the ordering of

data records is the same as or close to the ordering of data entries in some index, we say that the index is clustered.

Alternative (1) is clustered Alternatives (2) and (3) can be a clustered index only if

the data records are sorted on the search key field => this is expensive => usually they are unclustered.


Index Data Structures Two basic approaches

Hash-based indexing Tree-based indexing

- ISAM tree- B+ tree


Indexed Sequential Access Method (ISAM)

Proposed prior relational database (developed at IBM prior to DB2)

Highly static Each node is a disk page Leaf nodes are first allocated, then index pages, then

overflow pages Once the ISAM file is created, inserts and deletes affect only

the contents of leaf pages.

Index pages

leaf pages

overflow pages

primary pages


ISAM lookup

40

20 33 51 63

10*15* 20*27* 33*37* 40*46* 51*55* 63*97*

Primary leaf pages are allocated sequentially? Is this assumption reasonable? no “next-leaf” pointer is necessary


ISAM insert

40

20 33 51 63

10*15* 20*27* 33*37* 40*46* 51*55* 63*97*

Insert 23, 48, 41, 42

23* 48*41*

42*


ISAM delete Removes the entry

If the page becomes empty If it is overflow page, then delete it If it is primary page, just leave it as a place holder


ISAM discussion Pros

We know that the index nodes will not be changed, so that we don’t need to lock them

Cons Long chains of overflow pages are performance

bottleneck


B+-tree

The tree grows/shrinks dynamically

Root index fits in one page and directs search for records in index below it

B+-tree is balanced, i.e., every path through tree is same length. Reasonably easy to maintain this property

Large fan-out of index nodes result in few levels. Three levels can address 16M pages (256 records / page)

depth

Index entries(to direct search)

data entries


Format of a node Index node

An index node contains m entries, with d m 2d. d is called the order of the tree. The root node is required to have 1 m 2d.

p0 K1 p1 K2 p2 … Km pm

Leaf node (file organization) Leaf nodes contain the data entries. A page contains at most 2e-1 records Records sorted by key value Doubly linked list


Lookup of key K Assume B+-tree is of depth l

Construct path B0B1…Bl-1 where B0 is root node Kj in block Bi-1 covers K and j th block pointer in Bi-1 is Bi.

Example – Find key K = 245168 296

140 220 256 303

120 136 140 151 168 170 190 220 255 256 271 296 299 303 312 318

Path is B0 (168) B2 (220) B7 245 is not in B7 => 245 is not in main file

B0

B1 B2 B3

B4 B5 B6 B7 B8 B9 B10


Lookup of key K

168 296

140 220 256 303

120 136 140 151 168 170 220 255 256 271 296 299 303 312 318

B0

B1 B2 B3

B4 B5 B6 B7 B8 B9 B10

Insert 180

180 190


Insertion of record with key K Follow lookup procedure to find block in which K

belongs. Path is B0B1…Bl-1

If room in Bl-1, then insert there. (Maintain sorted order of Bl-1) Otherwise, allocate B’ and split records evenly between Bl-1 and

B’ Keys in Bl-1 are less than K’ and those in B’ are greater than or

equal to K’- Insert record (K’, B’) in Block Bl-2. (This insertion can also cause

split) Splitting the root (maintain B0 as the root)

- Allocate two new blocks Bl and Br.

- Move half of keys in root (keys smaller than K’) to Bl and the rest (keys greater than or equal to K’) to Br .

- Modify B0 (original root) to contain (Bl , K’, Br)

- Depth is increased to l+1


Delete record with key K lookup K and find path B0B1…Bl-1

Delete record from Bl-1

If Bl-1 now contains fewer than e records If a neighbor B’ has more than e records, divide the

records between Bl-1 and B’ as evenly as possible. Update any ancestors necessary to reflect change.

Otherwise, the records of Bl-1 and one of its neighbors B’ can be combined. B’ is removed and (K’, B’) is removed from parent. This merge can propagate to the root.

If last two children of root are combined, depth is decreased.


Discussion Merge operations have a high performance penalty;

databases tend to grow, so some merges may not be necessary

Remove blocks when empty Treat merge as a maintenance operation, and do it

periodically

What kind of queries can B+-trees help with? Point query Range query

How can you efficiently search a key in a leaf node?


Dense Indices Decouple allocation of tuples from access method

Allocate tuples following a heap organization (good utilization) Access tuples using hashing, B+-tree, etc.

Access methods must be modified slightly B+-trees: Keys adjacent in key space need not be

physically adjacent. Need tuple pointer for each key value (not key range) in leaf nodes. (each tuple can be in a different page)

Hashing: Hash buckets contain key value, tuple pointer pairs.

168 220

140 175 296 303

168 170


Create_table() check for duplicates (consult the dd)

Update dd (tables, cols, etc)

Create a file to store the table (file.c)

How to interface the buffer manager with file.c

Initialize bitmap emptyfix bufferunfix


Data dictionary Needs to be created when the server is up for the

FIRST time

Otherwise, the server just needs to load the dd from disk to memory


Insert a tuple (no long tuples) Check if the table exists

load the bitmap (bufferfix (fileid, pageid)) – is this needed? If not, you may just maintain a page number which can take next tuple in the dd.

load that page into memory

check if that page has enough space

if last page cannot hold the tuple, find a new page

Insert the tuple

bufferunfix


Secondary Indices Primary indices provide access based on primary

key

Secondary indices provide access based on search fields other than primary key

Index can be used to cluster tuples Sparse B-tree can be easily modified

168 220

140 175 296 303

168 170168 168 175 175 175 200180 187 200


Secondary Indexes (cont’d) Non-clustered indexes

Three ways Same as unique attribute

Each attribute value is stored only once

Two-level pointers

168 220

140 175 296 303

168 170 175 200180 187

168 p1 168 p2 168 p3 170 p1

168 p13 p2 p3 170 p11

168 R 170 R

p1 p2 p3 p1


Performance Lookup requires l accesses where l is depth

The depth is directly dependant on the fanout of index nodes

Define (sparse B+-tree) – n = number of records R = number of records / block (max) F = number of index entries / block (max) u = average node occupancy R eff = R u = average number of records / page F eff = F u = average number of index entries / page

n Fl-1eff Reff

logFeff ( n / Reff) + 1 = l


Performance – cont’d Utilization

If nodes are merged as described above u 69% If nodes are removed when empty

- # inserts = # deletes, u 40%- 60% inserts, 40% deletes, u 60%


Example 4000 bytes / block

200 bytes / record

Key requires 20 bytes

Block pointer requires 4 bytes

n = 1000000

What is the depth? ( 4)


Key compression Tree height can be reduced by increasing fan-out

of index nodes

Key compression can increase the number of keys that can be stored in an index node

Suffix compression- Store only enough of the key value to discriminate between

the children of the index node. For the following keys

artful deliver hand

access alert amass artful boom deal Deliver everyone fiddle Hand integral leaf

- Only need to store the following

ar del h

access alert amass artful boom deal Deliver everyone fiddle Hand integral leaf


Key compression (cont’d) Prefix compression

- Rather than storing each key value, store the difference from the previous key value

- Represent keyi as <j, keyi’> where j is the length of the common prefix shared by keyi and keyi-1, and keyi’ is the remainder of keyi after the common prefix is removed

- The following keys (length 36 bytes) – can, cannon, canter, cantor, capacity, capital – can be encoded as (length 27 bytes) - <0, can>, <3, non>, <3, ter>, <4, or>, <2, pacity>, <3, ital>

- How would you decide which of these two techniques to use in a particular situation?

spring 2004 ece569 lecture 05.1 ece 569 database system engineering spring 2004 yanyong zhang...

Documents