D ATABASE S YSTEM P RINCIPLES Ankit Patel 112_226

DATABASE SYSTEM PRINCIPLESAnkit Patel

112_226

CHAPTER 13

SECTION 13.1-13.3

13.1.1 MEMORY HIERARCHY

Several components for data storage having different data capacities available

Cost per byte to store data also varies in different storages.

Device with smallest capacity offer the fastest speed with highest cost per bit

MEMORY HIERARCHY DIAGRAM

Programs, DBMS

Main Memory DBMS’s

Main Memory

Cache

As Visual Memory Disk File System

Tertiary Storage

13.1.1 MEMORY HIERARCHY

Cache Lowest level of the hierarchy Data items are copies of certain locations of

main memory Sometimes, values in cache are changed and

corresponding changes to main memory are delayed

Machine looks for instructions as well as data for those instructions in the cache

limited amount of data capacity

MAIN MEMORY

Everything happens in the computer i.e. instruction execution, data manipulation, as working on information that is resident in main memory

Main memories are random access….one can obtain any byte in the same amount of time

Cost per data storage in main memory is the highest.

Main memory is volatile.

SECONDARY STORAGE

Used to store data and programs when they are not being processed

More permanent than main memory, as data and programs are retained when the power is turned off

E.g. magnetic disks, hard disks Main memory uses secondary memory

sometimes

TERTIARY STORAGE

Holds data volumes in terabytes Used for databases much larger than what

can be stored on disk This type of storage is the cheapest It is a comprehensive computer storage

system that is usually very slow, so it is usually used to archive data that is not accessed frequently.

13.1.2 TRANSFER OF DATA BETWEEN LEVELS Data moves between adjacent levels of the

hierarchy At the secondary or tertiary levels accessing the

desired data or finding the desired place to store the data takes a lot of time

Disk is organized into bocks A key technique for speeding up database

operations is to arrange the data so that when one piece of data block is needed it is likely that other data on the same block will be needed at the same time

Entire blocks are moved to and from memory called a buffer.

At the secondary or tertiary levels accessing the desired data or finding the desired place to store the data takes a lot of time

13.1.3 VOLATILE AND NON VOLATILE STORAGE

A volatile device forgets what data is stored on it after power off

Non volatile holds data for longer period even when device is turned off

All the secondary and tertiary devices are non volatile and main memory is volatile

Secondary and tertiary devices are non volatile

Main memory is volatile

13.1.4 VIRTUAL MEMORY

Typical software executes in virtual memory Address space is typically 32 bit or 232 bytes or 4GB Transfer between memory and disk is in terms of

blocks technique make programming of large applications

easier and use real physical memory (e.g. RAM) more efficiently

13.2.1 MECHANISM OF DISK

Mechanisms of Disks Use of secondary storage is one of the important

characteristic of DBMS Consists of 2 moving pieces of a disk

1. disk assembly 2. head assembly

Disk assembly consists of 1 or more platters Platters rotate around a central spindleBits are stored on upper and lower

surfaces of platters Bits are stored on upper and lower surfaces of

platters

13.2.1 MECHANISM OF DISK

Disk is organized into tracks The track that are at fixed radius from center

form one cylinder Tracks are organized into sectors Tracks are the segments of circle separated

by gap The track that are at fixed radius from center

form one cylinder

13.2.2 DISK CONTROLLER

One or more disks are controlled by disk controllers

Disks controllers are capable of Controlling the mechanical actuator that moves

the head assemblyControlling the mechanical actuator that

moves the head assembly Transferring bits between desired sector and main memory

Possible buffering an entire track

13.2.3 DISK ACCESS CHARACTERISTICS

Accessing (reading/writing) a block requires 3 steps Disk controller positions the head assembly at

the cylinder containing the track on which the block is located. It is a ‘seek time’

The disk controller waits while the first sector of the block moves under the head. This is a ‘rotational latency’

All the sectors and the gaps between them pass the head, while disk controller reads or writes data in these sectors. This is a ‘transfer time’

13.3 ACCELERATING ACCESS TO SECONDARY STORAGE

Several approaches for more-efficiently accessing data in secondary storage: Place blocks that are together in the same cylinder. Divide the data among multiple disks. Mirror disks. Use disk-scheduling algorithms. Prefetch blocks into main memory.

Scheduling Latency – added delay in accessing data caused by a disk scheduling algorithm.

Throughput – the number of disk accesses per second that the system can accommodate.

13.3.1 THE I/O MODEL OF COMPUTATION

The number of block accesses (Disk I/O’s) is a good time approximation for the algorithm. Disk I/o’s proportional to time taken, so should

be minimized. Ex 13.3: You want to have an index on R to

identify the block on which the desired tuple appears, but not where on the block it resides. For Megatron 747 (M747) example, it takes 11ms to

read a 16k block. A standard microprocessor can execute millions of

instruction in 11ms, making any delay in searching for the desired tuple negligible.

delay in searching for the desired tuple is negligible.

13.3.2 ORGANIZING DATA BY CYLINDERS

If we read all blocks on a single track or cylinder consecutively, then we can neglect all but first seek time and first rotational latency.

Ex 13.4: We request 1024 blocks of M747. If data is randomly distributed, average latency is

10.76ms by Ex 13.2, making total latency 11s. If all blocks are consecutively stored on 1 cylinder:

6.46ms + 8.33ms * 16 = 139ms

(1 average seek) (time per rotation) (# rotations)

13.3.3 USING MULTIPLE DISKS If we have n disks, read/write performance will

increase by a factor of n. Number of disks is proportional to the factor by

which performance is performance will increase by improved

Striping – distributing a relation across multiple disks following this pattern: Data on disk R1: R1, R1+n, R1+2n,… Data on disk R2: R2, R2+n, R2+2n,… … Data on disk Rn: Rn, Rn+n, Rn+2n, …

Ex 13.5: We request 1024 blocks with n = 4. 6.46ms + (8.33ms * (16/4)) = 39.8ms

(1 average seek) (time per rotation) (# rotations)

13.3.4 MIRRORING DISKS

Mirroring Disks – having 2 or more disks hold identical copied of data.

Benefit 1: If n disks are mirrors of each other, the system can survive a crash by n-1 disks.

Benefit 2: If we have n disks, read performance increases by a factor of n.

Performance increases further by having the controller select the disk which has its head closest to desired data block for each read.

Performance and efficiency increases

13.3.5 DISK SCHEDULING AND THE ELEVATOR PROBLEM

Disk controller will run this algorithm to select which of several requests to process first.

Pseudo code: requests[] // array of all non-processed data requests upon receiving new data request:

requests[].add(new request) while(requests[] is not empty)

move head to next location if(head location is at data in requests[])

retrieve data remove data from requests[]

if(head reaches end) reverse head direction


(CON’T)Events:

Head starting point

Request data at 8000



Get data at 8000Request data at

16000Get data at 24000Request data at

64000Get data at 56000Request Data at

40000Get data at 64000Get data at 40000Get data at 16000

data time

Current time

Current time

0

Current time

4.3

Current time

10

Current time

13.6

Current time

20

Current time

26.9

Current time

30

Current time

34.2

Current time

45.5

Current time

56.8

800016000240003200040000480005600064000

data time

8000.. 4.3

data time

8000.. 4.3

24000.. 13.6

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

64000.. 34.2

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

64000.. 34.2

40000.. 45.5

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

64000.. 34.2

40000.. 45.5

16000.. 56.8


(CON’T)

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

64000.. 34.2

40000.. 45.5

16000.. 56.8

data time

8000.. 4.3

24000.. 13.6

56000.. 26.9

16000.. 42.2

64000.. 59.5

40000.. 70.8

Elevator Algorithm

FIFOAlgorithm

13.3.6 PREFETCHING AND LARGE-SCALE BUFFERING

If at the application level, we can predict the order blocks will be requested, we can load them into main memory before they are needed. This even reduces the cost and even save the time

13.2.DISK FAILURES

INTERMITTENT FAILURES. Read or write operation on a sector successful not on first try, but

after repeated tries.

The most common form of failure.

Parity checks can be used to detect this kind of failure.

MEDIA DECAY.

Impossible to read a sector correctly even after many trials.

Stable storage technique for organizing a disk is used to avoid this failure.

Serious form of failure.

Bit/Bits are permanently corrupted.

WRITE FAILURE Attempt to retrieve previously written sector is

unsuccessful. Stable Storage Technique can be used to avoid this. Possible reason – power outage while writing of the

sector. Attempt to write a sector is not possible.

DISK CRASH

• RAID techniques can be used for coping with disk crashes.

• Most serious form of disk failure.• Entire disk becomes unreadable, suddenly and

permanently.

MORE ON INTERMITTENT FAILURES…

When we try to read a sector, but the correct content of that sector is not delivered to the disk controller.

If the controller has a way to tell that the sector is good or bad (checksums), it can then reissue the read request when bad data is read.

The controller can attempt to write a sector, but the contents of the sector are not what was intended.

The only way to check this is to let the disk go around again read the sector.

One way to perform the check is to read the sector and compare it with the sector we intend to write.

If it is good sector, then the write was correct otherwise the write was unsuccessful and must be repeated.

Instead of performing the complete comparison at the disk controller, simpler way is to read the sector and see if a good sector was read.

CHECKSUMS. Technique used to determine the good/bad

status of a sector. There is a small chance that the block was

not read correctly even if the checksum is proper.

Each sector has some additional bits called the checksum that are set depending on the values of the data bits in that sector.

The probability of correctness can be increased by using many checksum bits.

If checksum is not proper on reading, then there is an error in reading.

CHECKSUM CALCULATION. Checksum is based on the parity of all bits in the sector.

If there are odd number of 1’s among a collection of bits, the bits are said to have odd parity. A parity bit ‘1’ is added.

If there are even number of 1’s then the collection of bits is said to have even parity. A parity bit ‘0’ is added.

The number of 1’s among a collection of bits and their parity bit is always even.

During a write operation, the disk controller calculates the parity bit and append it to the sequence of bits written in the sector.

Every sector will have a even parity.

STABLE STORAGE.

Sectors are paired and each pair represents one sector-contents X.

The left copy of the sector may be represented as XL and XR as the right copy.

Checksums can detect the error but cannot correct it.

Sometimes we overwrite the previous contents of a sector and yet cannot read the new conents correctly.

To deal with these problems, Stable Storage policy can be implemented on the disks.

ASSUMPTIONS. We assume that copies are written with

sufficient number of parity bits to decrease the chance of bad sector looks good when the parity checks are considered.

STABLE -STORAGE WRITING POLICY:

1. Write the value of X into XL. Check the value has status “good”; i.e., the parity-check bits are correct in the written copy. If not repeat write. If after a set number of write attempts, we have not successfully written X in XL, assume that there is a media failure in this sector. A fix-up such as substituting a spare sector for XL must be adopted.

2. Repeat (1) for XR.

STABLE-STORAGE READING POLICY:

If a good value is not returned after pre chosen number of tries, then it is assumed that X is truly unreadable.

The policy is to alternate trying to read XL and XR until a good value is returned.

ERROR-HANDLING CAPABILITIES:

Media failures:• If after storing X in sectors XL and XR, one of

them undergoes media failure and becomes permanently unreadable, we can read from the second one.

• If both the sectors have failed to read, then sector X cannot be read.

Write Failure:• When writing X, if there is a system

failure(like power shortage), the X in the main memory is lost and the copy of X being written will be erroneous.

RECOVERY FROM DISK CRASHES. To reduce the data loss by Dish crashes, schemes which

involve redundancy, extending the idea of parity checks or duplicate sectors can be applied.

The term used for these strategies is RAID or Redundant Arrays of Independent Disks.

When there is a disk crash of either of the disks, then the other disks can be used to restore the failed disk to avoid a permanent information loss.

In general, if the mean time to failure of disks is n years, then in any given year, 1/nth of the surviving disks fail.

Each of the RAID schemes has data disks and redundant disks.

Data disks are one or more disks that hold the data.

Redundant disks are one or more disks that hold information that is completely determined by the contents of the data disks.

SECTION 13.4

MIRRORING

The simplest scheme to recovery from Disk Crashes

Benefit:

-- save data in case of one disk will fail;

-- divide data on several disks and let access to several blocks at once

For mirroring, when the data can be lost?

-- the only way data can be lost if there is a second (mirror/redundant) disk crash while the first (data) disk crash is being repaired.

PARITY BLOCKS

why changes? -- disadvantages of Mirroring: uses so many redundant

disks What’s new? -- RAID level 4: uses only one redundant disk

How this one redundant disk works? -- modulo-2 sum; -- the jth bit of the redundant disk is the modulo-2 sum

of the jth bits of all the data disks.

Data disks:Disk1: 11110000Disk2: 10101010Disk3: 00111000

Redundant disk:Disk4: 01100010

RAID 4 (CON’T) _ WRITINGFor a total N data disks:1) naïve way: read N data disks and compute the modulo-2 sum

of the corresponding blocks; rewrite the redundant disk according to modulo-2

sum of the data disks;

2) better way: Take modulo-2 sum of the old and new version of

the data block which was rewritten; Change the position of the redundant disk which

was 1’s in the modulo-2 sum;

Data disks:Disk1: 11110000Disk2: 10101010 01100110Disk3: 00111000

Redundant disk:Disk4: 01100010 10101110

AN IMPROVEMENT: RAID 5 Why need a improvement? -- Shortcoming of RAID level 4: suffers from a bottleneck defect

(when updating data disk need to read and write the redundant disk);

Principle of RAID level 5 (RAID 5): -- treat each disk as the redundant disk for some of the blocks;

Why it is feasible?The rule of failure recovery for redundant disk and data disk is the

same:

“take modulo-2 sum of all the corresponding bits of all the other disks”

How to recognize which blocks of each disk treat this disk as redundant disk?

-- if there are n+1 disks which were labeled from 0 to N, then we can treat the ith cylinder of disk J as redundant if J is the remainder when I is divided by n+1;

So, there is no need to retreat one as redundant disk and others as data disks

COPING WITH MULTIPLE DISK CRASHES RAID 6 – deal with any number of disk crashes if using

enough redundant disks Example a system of seven disks ( four data disks_numer

1-4 and 3 redundant disks_ number 5-7); Reading: read form the data disks and ignore the redundant

disk

Writing: Change the data disk change the corresponding bits of all the redundant

disks

COPING WITH MULTIPLE DISK CRASHES (CON’T)_EXAMPLE

3*7 matrix data disk

redundant disk disk number 1 2 3 4 5 6

7

1 1 1 0 1 0 0

1 1 0 1 0 1 0

1 0 1 1 0 0 1

COPING WITH MULTIPLE DISK CRASHES (CON’T)_EXAMPLE

Two disks crashes; disk contents 1) 11110000 2) ????????? 3) 00111000 4) 01000001 5) ????????? 6) 00011011 7) 10001001

In that 3*7 matrix, find in row 2, disk 2 and 5 have different value and disk 2’s value is 1 and 5’s value is 0.

so: compute the first block of disk 2 by modulo-2 sum of all the corresponding bits of disk 1,4,6;

then compute the first block of disk 2 by modulo-2 sum of all the corresponding bits of disk 1,2,3;

1) 11110000 2) ????????? => 00001111 3) 00111000 4) 01000001 5) ????????? => 01100010 6) 00011011 7) 10001001

SECTION 13.5

Data elements are represented as records, which stores Data elements are represented as records, which stores in consecutive bytes in same same disk block. in consecutive bytes in same same disk block.

Basic layout techniques of storing data : Basic layout techniques of storing data :

Fixed-Length RecordsFixed-Length Records

Allocation criteria - data should start at word boundary.Allocation criteria - data should start at word boundary.

Fixed Length record header Fixed Length record header

1. A pointer to record schema.1. A pointer to record schema.2. The length of the record.2. The length of the record.

3. Timestamps to indicate last modified or last read.3. Timestamps to indicate last modified or last read.

EXAMPLE EXAMPLE

CREATE TABLE employee(CREATE TABLE employee(

name CHAR(30) PRIMARY KEY,name CHAR(30) PRIMARY KEY,

address VARCHAR(255),address VARCHAR(255),

gender CHAR(1),gender CHAR(1),

birthdate DATEbirthdate DATE

););

Data should start at word boundary and contain header and Data should start at word boundary and contain header and four fields name, address, gender and birthdate.four fields name, address, gender and birthdate.

PACKING FIXED-LENGTH RECORDS INTO PACKING FIXED-LENGTH RECORDS INTO BLOCKSBLOCKS : :

Records are stored in the form of blocks on the disk Records are stored in the form of blocks on the disk and they move into main memory when we need to and they move into main memory when we need to update or access them.update or access them.

A block header is written first, and it is followed by A block header is written first, and it is followed by series of blocks.series of blocks.

EXAMPLE

Along with the header we can pack as many record as we can

in one block as shown in the figure and remaining space will

be unused.

SECTION 13.6

INTRODUCTION

Address of a block and Record In Main Memory

Address of the block is the virtual memory address of the first byte

Address of the record within the block is the virtual memory address of the first byte of the record

In Secondary Memory: sequence of bytes describe the location of the block in the overall system

Sequence of Bytes describe the location of the block : the device Id for the disk, Cylinder number, etc.

ADDRESSES IN CLIENT-SERVER SYSTEMS

Physical Address bits are used to indicate: Host to which the storage is attached Identifier for the disk Number of the cylinder Number of the track Offset of the beginning of the record

The addresses in address space are represented in two ways Physical Addresses: byte strings that determine

the place within the secondary storage system where the record can be found.

Logical Addresses: arbitrary string of bytes of some fixed length

Map Table relates logical addresses to physical addresses.

Logical Physical

Logical Address

Physical Address

ADDRESSES IN CLIENT-SERVER SYSTEMS

LOGICAL AND STRUCTURED ADDRESSES

Purpose of logical address? Gives more flexibility, when we

Move the record around within the block Move the record to another block

Gives us an option of deciding what to do when a record is deleted?

Record 4

Record 3

Record 2

Record 1

HeaderOffset table

Unused

POINTER SWIZZLING

Having pointers is common in an object-relational database systems

Important to learn about the management of pointers

Every data item (block, record, etc.) has two addresses:database address: address on the diskmemory address, if the item is in virtual

memory

POINTER SWIZZLING (CONTD…)

Translation Table: Maps database address to memory address

All addressable items in the database have entries in the map table, while only those items currently in memory are mentioned in the translation table

Dbaddr Mem-addr

Database address

Memory Address

POINTER SWIZZLING (CONTD…) Pointer consists of the following two fields

Bit indicating the type of address Database or memory address Example 13.17

Disk

Block 2

Block 1

Memory

Swizzled

Unswizzled

Block 1

POINTER SWIZZLING (CONTD…) Three types of swizzling

Automatic Swizzling As soon as block is brought into memory, swizzle all

relevant pointers. No Swizzling

Pointers are not swizzled they are accesses using the database address.

Swizzling on Demand Only swizzle a pointer if and when it is actually followed.PROGRAMMER CONTROL OF

SWIZZLING

Unswizzling When a block is moved from memory back to

disk, all pointers must go back to database (disk) addresses

Use translation table again Important to have an efficient data structure for

the translation table

PINNED RECORDS AND BLOCKS

A block in memory is said to be pinned if it cannot be written back to disk safely.

Unswizzle those pointers (use translation table to replace the memory addresses with database (disk) addresses

If block B1 has swizzled pointer to an item in block B2, then B2 is pinned

Unpin a block, we must unswizzle any pointers to it Keep in the translation table the places in memory

holding swizzled pointers to that item

SECTION 13.7

RECORDS WITH VARIABLE FIELDS

An effective way to represent variable length records is as followsRecord header contains

• Length of the record• Pointers to the beginning of all variable length fields except the first one.

Fixed length fields are Kept ahead of the variable length fields

RECORDS WITH REPEATING FIELDSRecords contains variable number of occurrences of a field F

All occurrences of field F are grouped together and the record Locating an occurrence of field F within the record

• Add to the offset for the field F which are the integer multiples of L starting with 0 , L ,2L,3L and so on to locate•We stop upon reaching the offset of the field F.

header contains a pointer to the first occurrence of field F

L bytes are devoted to one instance of field F

Figure 4 : Storing variable-length fields separately from the record

address

name

record header information

length of nameto address

length of address

to name

to movie references number of

references

Records with Repeating Fields

Advantage

Keeping the record itself fixed length allows record to be searched more efficiently, minimizes the overhead in the block headers, and allows records to be moved within or among the blocks with minimum effort.

Disadvantage

Storing variable length components on another block increases the number of disk I/O’s needed to examine all components of a record.

Records with Repeating Fields

VARIABLE FORMAT RECORDS

Records that do not have fixed schema

Variable format records are represented by sequence of tagged fields

Why use tagged fields• Information – Integration applications• Records with a very flexible schema

Each of the tagged fields consist of information• Attribute or field name• Type of the field• Length of the field• Value of the field

VARIABLE FORMAT RECORDS

Fig 5 : A record with tagged fields

N 16

S S14

Clint Eastwood

Hog’s Breath InnR

code for name code for restaurant ownedcode for string

typecode for string type

lengthlength

RECORDS THAT DO NOT FIT IN A BLOCK

When the length of a record is greater than block size , then record is divided and placed into two or more blocks

Portion of the record in each block is referred to as a RECORD FRAGMENT

Record with two or more fragments is called SPANNED RECORD

Record that do not cross a block boundary is called UNSPANNED RECORD

RECORDS THAT DO NOT FIT IN A BLOCK

Figure 6 : Storing spanned records across blocks

record 1 record 3 record 2 - a

record 2 - b

block header

record header

block 1 block 2

SECTION 13.8

INSERTION

Insertion of records without order Records can be placed in a block with empty space or in a new

block.

Structured address Pointer to a record from outside the block.

Insertion of records in fixed order Space available in the block No space available in the block (outside the block)

69

INSERTION IN FIXED ORDER

Space available within the block Use of an offset table in the header of each block with pointers to

the location of each record in the block. The records are slid within the block and the pointers in the offset

table are adjusted.

70

Record 2Record 3Record 4

header unused

Offset table

Record 1

INSERTION IN FIXED ORDER

71

No space available within the block (outside the block) Create an overflow block

• Records can be stored in overflow block.• Each block has place for a pointer to an overflow block in its header.• The overflow block can point to a second overflow block as shown below.

Find space on a “nearby” block.• In case of no space available on a block, look at the following block in sorted

order of blocks.• If space is available in that block ,move the highest records of first block 1 to

block 2 and slide the records around on both blocks.

Block B

Overflow block for B

DELETION

Recover space after deletion When using an offset table, the records can be slid around the

block so there will be an unused region in the center that can be recovered.

In case we cannot slide records, an available space list can be maintained in the block header.

The list head goes in the block header and available regions hold the links in the list.

72

Use of tombstone The tombstone is placed in a record in order to avoid pointers to

the deleted record to point to new records.

The tombstone is permanent until the entire database is reconstructed.

UPDATE

Fixed Length update No effect on storage system as it occupies same space as

before update.

Variable length update Longer length Short length

73

Variable length update (longer length)Stored on the same block:

Sliding recordsCreation of overflow block.

Stored on another blockMove records around that blockCreate a new block for storing variable length fields.

CHAPTER 14

SECTION 14.2

STRUCTURESTRUCTURE A balanced tree, meaning that all paths from the leaf node have the

same length. There is a parameter n associated with each Btree block. Each block

will have space for n searchkeys and n+1 pointers. The root may have only 1 parameter, but all other

blocks most be at least half full.● A typical node >● a typical interior node would havepointers pointing toleaves with outvalues● a typical leaf wouldhave pointers pointto recordsN search keysN+1 pointers

APPLICATIONAPPLICATION The search key of the Btree is the primary key for the

data file. Data file is sorted by an attribute that is not a key,and

this attribute is the search key for the Btree. Data file is sorted by its primary key.

LOOKUPLOOKUP

If at an interior node, choose the correct pointer to use. This is done by comparing keys to search value.

INSERTIONINSERTION When inserting, choose the correct leaf node to put

pointer to data. Recursively move up, if cannot create new pointer to

new node because full, create new node. This would end with creating a new root node, if

the current root was full. If node is full, create a new node and split keys

between the two.

DELETIONDELETIONIf node is no longer half full, perform join on adjacent node and recursively delete up, or key move if that node is full and recursively change pointer up.

EFFICIENCYEFFICIENCY

Three levels are sufficient for Btrees. Having each block have 255 pointers, 255^3 is about 16.6 million.

You can even reduce disk I/Os by keeping a level of a Btree in main memory. Keeping the first block with 255 pointers would reduce the reads to 2, and even possible to keep the next 255 pointers in memory to reduce reads

to 1.

SECTION 14.7

DEFINITIONDEFINITION

A bitmap index for a field F is a collection of bit-vectors of length n, one for each possible value that may appear in that field F.[1]

WHAT DOES THAT MEAN?WHAT DOES THAT MEAN? Assume relation R with

2 attributes A and B.Attribute A is of type

Integer and B is of type String.

6 records, numbered 1 through 6 as shown.

A B

1 30 foo

2 30 bar

3 40 baz

4 50 foo

5 40 bar

6 30 baz

EXAMPLE CONTINUED…EXAMPLE CONTINUED…

A bitmap for attribute B is:A bitmap for attribute B is:

Value Vector

foo 100100

bar 010010

baz 001001

A B

1 30 foo

2 30 bar

3 40 baz

4 50 foo

5 40 bar

6 30 baz

WHERE DO WE REACH?WHERE DO WE REACH?

A bitmap index is a special kind of database index that uses bitmaps.[2]

Bitmap indexes have traditionally been considered to work well for data such as gender, which has a small number of distinct values, e.g., male and female, but many occurrences of those values.[2]

A bitmap index for attribute A of relation R is: A collection of bit-vectors The number of bit-vectors = the number of distinct values of A in R. The length of each bit-vector = the cardinality of R. The bit-vector for value v has 1 in position i, if the ith record has v in

attribute A, and it has 0 there if not.[3] Records are allocated permanent numbers.[3] There is a mapping between record numbers and record

addresses.[3]

MOTIVATION FOR BITMAP INDEXESMOTIVATION FOR BITMAP INDEXES

Very efficient when used for partial match queries.[3]

They offer the advantage of buckets [2] Where we find tuples with several specified attributes without first retrieving all

the record that matched in each of the attributes.

They can also help answer range queries [3]

COMPRESSED BITMAPSCOMPRESSED BITMAPS Assume:

The number of records in R are n Attribute A has m distinct values in R

The size of a bitmap index on attribute A is m*n. If m is large, then the number of 1’s will be around 1/m.

Opportunity to encode A common encoding approach is called run-length encoding.[1]

RUN-LENGTH ENCODING Represents runs

A run is a sequence of i 0’s followed by a 1, by some suitable binary encoding of the integer i.

A run of i 0’s followed by a 1 is encoded by: First computing how many bits are needed to represent i, Say k Then represent the run by k-1 1’s and a single 0 followed by k

bits which represent i in binary. The encoding for i = 1 is 01. k = 1 The encoding for i = 0 is 00. k = 1

We concatenate the codes for each run together, and the sequence of bits is the encoding of the entire bit-vector

UNDERSTANDING WITH AN EXAMPLEUNDERSTANDING WITH AN EXAMPLE Let us decode the sequence 11101101001011 Staring at the beginning (left most bit):

First run: The first 0 is at position 4, so k = 4. The next 4 bits are 1101, so we know that the first integer is i = 13

Second run: 001011 k = 1 i = 0

Last run: 1011 k = 1 i = 3Our entire run length is thus 13,0,3, hence our bit-vector

is: 0000000000000110001

DeletionTombstone replaces deleted recordCorresponding bit is set to 0

InsertionRecord assigned the next record number. A bit of value 0 or 1 is appended to each

bit vectorIf new record contains a new value of the

attribute, add one bit-vector.

Modification

Change the bit corresponding to the old value of the modified record to 0

Change the bit corresponding to the new value of the modified record to 1

If the new value is a new value of A, then insert a new bit-vector.

CHAPTER 15

SECTION 15.1

WHAT IS A QUERY PROCESSOR

Group of components of a DBMS that converts a user queries and data-modification commands into a sequence of database operations

It also executes those operationsMust supply detail regarding how the

query is to be executed

MAJOR PARTS OF QUERY PROCESSOR

Query Execution:The algorithms that manipulate the data of the database.

Focus on the operations of extended relational algebra.

OUTLINE OF QUERY COMPILATION

Query compilation Parsing : A parse tree for the

query is constructed Query Rewrite : The parse

tree is converted to an initial query plan and transformed into logical query plan (less time)

Physical Plan Generation : Logical Q Plan is converted into physical query plan by selecting algorithms and order of execution of these operator.

PHYSICAL-QUERY-PLAN OPERATORSPhysical operators are

implementations of the operator of relational algebra.

They can also be use in non relational algebra operators like “scan” which scans tables, that is, bring each tuple of some relation into main memory

SCANNING TABLESOne of the basic thing we can do in

a Physical query plan is to read the entire contents of a relation R.

Variation of this operator involves simple predicate, read only those tuples of the relation R that satisfy the predicate.

SCANNING TABLESBasic approaches to locate the tuples of a

relation R Table Scan

Relation R is stored in secondary memory with its tuples arranged in blocks

It is possible to get the blocks one by one Index-Scan

If there is an index on any attribute of Relation R, we can use this index to get all the tuples of Relation R

SORTING WHILE SCANNING TABLESNumber of reasons to sort a relation

Query could include an ORDER BY clause, requiring that a relation be sorted.

Algorithms to implement relational algebra operations requires one or both arguments to be sorted relations.

Physical-query-plan operator sort-scan takes a relation R, attributes on which the sort is to be made, and produces R in that sorted order

COMPUTATION MODEL FOR PHYSICAL OPERATORPhysical-Plan Operator should be

selected wisely which is essential for good Query Processor .

For “cost” of each operator is estimated by number of disk I/O’s for an operation.

The total cost of operation depends on the size of the answer, and includes the final write back cost to the total cost of the query.

PARAMETERS FOR MEASURING COSTS

Parameters that affect the performance of a queryBuffer space availability in the

main memory at the time of execution of the query

Size of input and the size of the output generated

The size of memory block on the disk and the size in the main memory also affects the performance

PARAMETERS FOR MEASURING COSTS B: The number of blocks are needed to hold

all tuples of relation R. Also denoted as B(R) T:The number of tuples in relationR. Also denoted as T(R)

V: The number of distinct values that appear in a column of a relation R

V(R, a)- is the number of distinct values of column for a in relation R

I/O COST FOR SCAN OPERATORS

If relation R is clustered, then the number of disk I/O for the table-scan operator is = ~B disk I/O’s

If relation R is not clustered, then the number of required disk I/O generally is much higher

A index on a relation R occupies many fewer than B(R) blocks

That means a scan of the entire relation R which takes at least B disk I/O’s will require more I/O’s than the entire index

ITERATORS FOR IMPLEMENTATION OF PHYSICAL OPERATORSMany physical operators can be

implemented as an Iterator.Three methods forming the iterator

for an operation are:1. Open( ) :

This method starts the process of getting tuples

It initializes any data structures needed to perform the operation

ITERATORS FOR IMPLEMENTATION OF PHYSICAL OPERATORS

2. GetNext( ): Returns the next tuple in the resultIf there are no more tuples to return,

GetNext returns a special value NotFound

3. Close( ) : Ends the iteration after all tuplesIt calls Close on any arguments of

the operator

SECTION 15.2

ONE-PASS ALGORITHM METHODS

Tuple-at-a-time, unary operations: (selection & projection)

Full-relation, unary operations

Full-relation, binary operations (set & bag versions of union)

ONE-PASS ALGORITHMS FOR TUPLE-AT-A-TIME OPERATIONS

Tuple-at-a-time operations are selection and projection read the blocks of R one at a time into an input

buffer perform the operation on each tuple move the selected tuples or the projected tuples

to the output buffer

The disk I/O requirement for this process depends only on how the argument relation R is provided. If R is initially on disk, then the cost is whatever it

takes to perform a table-scan or index-scan of R.

A SELECTION OR PROJECTION BEING PERFORMED ON A RELATION R

ONE-PASS ALGORITHMS FOR UNARY, FILL-RELATION OPERATIONS

Duplicate Elimination To eliminate duplicates, we can read each block

of R one at a time, but for each tuple we need to make a decision as to whether:

1. It is the first time we have seen this tuple, in which case we copy it to the output, or

2. We have seen the tuple before, in which case we must not output this tuple.

One memory buffer holds one block of R's tuples, and the remaining M - 1 buffers can be used to hold a single copy of every tuple.

DUPLICATE ELIMINATION

We need a main-memory structure that allows each of the operations: Add a new tuple, and Tell whether a given tuple is already there

When a new tuple from R is considered, we compare it with all tuples seen so far if it is not equal: we copy both to the output

and add it to the in-memory list of tuples we have seen.

if there are n tuples in main memory: each new tuple takes processor time proportional to n, so the complete operation takes processor time proportional to n2.

ONE-PASS ALGORITHMS FOR UNARY, FILL-RELATION OPERATIONS

Grouping The grouping operation gives us zero or more

grouping attributes and presumably one or more aggregated attributes

If we create in main memory one entry for each group then we can scan the tuples of R, one block at a time.

The entry for a group consists of values for the grouping attributes and an accumulated value or values for each aggregation.

ONE-PASS ALGORITHMS FOR BINARY OPERATIONS

Binary operations include: Union Intersection Difference Product Join

SET UNION

We read S into M - 1 buffers of main memory and build a search structure where the search key is the entire tuple.

All these tuples are also copied to the output.

Read each block of R into the Mth buffer, one at a time.

For each tuple t of R, see if t is in S, and if not, we copy t to the output. If t is also in S, we skip t.

SET INTERSECTION

Read S into M - 1 buffers and build a search structure with full tuples as the search key.

Read each block of R, and for each tuple t of R, see if t is also in S. If so, copy t to the output, and if not, ignore t.

SET DIFFERENCE Read S into M - 1 buffers and build a search structure

with full tuples as the search key.

To compute R -s S, read each block of R and examine each tuple t on that block. If t is in S, then ignore t; if it is not in S then copy t to the output.

To compute S -s R, read the blocks of R and examine each tuple t in turn. If t is in S, then delete t from the copy of S in main memory, while if t is not in S do nothing.

After considering each tuple of R, copy to the output those tuples of S that remain.

BAG INTERSECTION Read S into M - 1 buffers.

Multiple copies of a tuple t are not stored individually. Rather store 1 copy of t & associate with it a count equal to no. of times t occurs.

Next, read each block of R, & for each tuple t of R see whether t occurs in S. If not ignore t; it cannot appear in the intersection. If t appears in S, & count associated with t is (+)ve, then output t & decrement count by 1. If t appears in S, but count has reached 0, then do not output t; we have already produced as many copies of t in output as there were copies in S.

BAG DIFFERENCE

To compute S -B R, read tuples of S into main memory & count no. of occurrences of each distinct tuple.

Then read R; check each tuple t to see whether t occurs in S, and if so, decrement its associated count. At the end, copy to output each tuple in main memory whose count is positive, & no. of times we copy it equals that count.

To compute R -B S, read tuples of S into main memory & count no. of occurrences of distinct tuples.

PRODUCT Read S into M - 1 buffers of main memory

Then read each block of R, and for each tuple t of R concatenate t with each tuple of S in main memory.

Output each concatenated tuple as it is formed.

This algorithm may take a considerable amount of processor time per tuple of R, because each such tuple must be matched with M - 1 blocks full of tuples. However, output size is also large, & time/output tuple is small.

NATURAL JOIN Convention: R(X, Y) is being joined with S(Y,

Z), where Y represents all the attributes that R and S have in common, X is all attributes of R that are not in the schema of S, & Z is all attributes of S that are not in the schema of R. Assume that S is the smaller relation.

To compute the natural join, do the following:

1. Read all tuples of S & form them into a main-memory search structure. Hash table or balanced tree are good e.g. of such structures. Use M - 1 blocks of memory for this purpose.

NATURAL JOIN (…CONTD.)

2. Read each block of R into 1 remaining main-memory buffer. For each tuple t of R, find tuples of S that agree with t on all attributes of Y, using the search structure. For each matching tuple of S, form a tuple by joining it with t, & move resulting tuple to output.

SECTION 15.3

TUPLE-BASED NESTED-LOOP JOIN

Simplest variation of the nested-loop join

Loop ranges over individual tuples

TUPLE-BASED NESTED-LOOP JOIN

Algorithm to compute the Join R(X,Y) | | S(Y,Z) FOR each tuple s in S DO

FOR each tuple r in R DOIF r and s join to make tuple t THENoutput t

carelessness in buffering of blocks causes the use of T(R)T(S) disk I/O’s

R and S are two Relations with r and s as tuples.

IMPROVEMENT & MODIFICATION

To decrease the cost Method 1: Use algorithm for Index-Based joins

We find tuple of R that matches given tuple of S

We need not to read entire relation R

Method 2: Use algorithm for Block-Based joins Tuples of R & S are divided into blocks Uses enough memory to store blocks in

order to reduce the number of disk I/O’s.

BLOCK-BASED NESTED-LOOP JOIN ALGORITHM

Access to arguments is organized by block.While reading tuples of inner relation we

use less number of I/O’s disk.

Using enough space in main memory to store tuples of relation of the outer loop.Allows to join each tuple of the inner

relation with as many tuples as possible.


ALGORITHM:

FOR each chunk of M-1 blocks of S DO

FOR each block b of R DO FOR each tuple t of b DO find the tuples of S in memory that join with t output the join of t with each of these tuples


Assumptions:

B(S) ≤ B(R)B(S) > M

This means that the neither relation fits in the entire main memory.

ANALYSIS OF NESTED-LOOP JOIN

Number of disk I/O’s:

[B(S)/(M-1)]*(M-1 +B(R))

or

B(S) + [B(S)B(R)/(M-1)]

or approximately B(S)*B(R)/M

SECTION 15.4

TWO-PASS ALGORITHMS BASED ON SORTING

Two-pass Algorithms: where data from the operand relations is read into main memory, processed in some way, written out to disk again, and then reread from disk to complete the operation

BASIC IDEA Step 1: Read M blocks of R into main memory. Step 2:Sort these M blocks in main memory, using an

efficient, main-memory sorting algorithm. so we expect that the time to sort will not exceed the disk 1/0 time for step (1).

Step 3: Write the sorted list into M blocks of disk.

DUPLICATE ELIMINATION USING SORTING Δ(R) First we sort the tuples of R in sublists Then we use the available main memory to hold one

block from each sorted sublist Then we repeatedly copy one to the output and ignore

all tuples identical to it. The total cost of this algorithm is 3B(R) This algorithm requires only √B(R)blocks of main

memory, rather than B(R) blocks(one-pass algorithm).

EXAMPLE

Suppose that tuples are integers, and only two tuples fit on a block. Also, M = 3 and the relation R consists of 17 tuples:

2,5,2,1,2,2,4,5,4,3,4,2,1,5,2,1,3 After first-pass

Sublists Elements

R1 1,2,2,2,2,5

R2 2,3,4,4,4,5

R3 1,1,2,3,5

EXAMPLE Second pass

After processing tuple 1

Output: 1Continue the same process with next tuple.

Sublist In memory Waiting on disk

R1 1,2 2,2, 2,5

R2 2,3 4,4, 4,5

R3 1,1 2,3,5

Sublist In memory Waiting on disk

R1 2 2,2, 2,5

R2 2,3 4,4, 4,5

R3 2,3 5

GROUPING AND AGGREGATION USING SORTING Γ(R) Two-pass algorithm for grouping and aggregation is

quite similar to the previous algorithm. Step 1:Read the tuples of R into memory, M blocks at a

time. Sort each M blocks, using the grouping attributes of L as the sort key. Write each sorted sublist to disk.

Step 2:Use one main-memory buffer for each sublist, and initially load the first block of each sublist into its buffer.

Step 3:Repeatedly find the least value of the sort key (grouping attributes) present among the first available tuples in the buffers.

This algorithm takes 3B(R) disk 1/0's, and will work as long as B(R) < M².

A SORT-BASED UNION ALGORITHM

For bag-union one-pass algorithm is used. For set-union

Step 1:Repeatedly bring M blocks of R into main memory, sort their tuples, and write the resulting sorted sublist back to disk.

Step 2:Do the same for S, to create sorted sublists for relation S. Step 3:Use one main-memory buffer for each sublist of R and S.

Initialize each with the first block from the corresponding sublist. Step 4:Repeatedly find the first remaining tuple t among all the

buffers. Copy t to the output. and remove from the buffers all

copies of t (if R and S are sets there should be at most two copies)

This algorithm takes 3(B(R)+B(S)) disk 1/0's, and will work as long as B(R)+B(S) < M².

SORT-BASED INTERSECTION AND DIFFERENCE

For both set version and bag version, the algorithm is same as that of set-union except that the way we handle the copies of a tuple t at the fronts of the sorted sublists.

For set intersection, output t if it appears in both R and S.

For bag intersection, output t the minimum of the number of times it appears in R and in S.

For set difference, R-S, output t if and only if it appears in R but not in S.

For bag difference, R-S, output t the number of times it appears in R minus the number of times it appears in S.

A SIMPLE SORT-BASED JOIN ALGORITHM

To avoid facing this situation, are can try to reduce main-memory use for other aspects of the algorithm, and thus make available a large number of buffers to hold the tuples with a given join-attribute value

When taking a join, the number of tuples from the two relations that share a common value of the join attribute(s), and therefore need to be in main memory simultaneously, can exceed what fits in memory

A SIMPLE SORT-BASED JOIN ALGORITHM Given relations R(X, Y) and S(Y, Z) to join, and given M blocks

of main memory for buffers. Step 1:Sort R and S, using a two-phase, multiway merge sort,

with Y as the sort key. Step 2:Merge the sorted R and S. The following steps are

done repeatedly: Find the least value y of the join attributes Y that is currently at the

front of the blocks for R and S. If y does not appear at the front of the other relation, then remove

the tuple(s) with sort key y. Otherwise, identify all the tuples from both relations having sort

key y. Output all the tuples that can be formed by joining tuples from R

and S with a common Y-value y. If either relation has no more unconsidered tuples in main

memory.,reload the buffer for that relation.

A MORE EFFICIENT SORT-BASED JOIN If we do not have to worry about very large numbers of

tuples with a common value for the join attribute(s), then we can save two disk 1/0's per block by combining the second phase of the sorts with the join itself

To compute R(X, Y) S(Y, Z) using M►◄ main-memory buffers Create sorted sublists of size M, using Y as the sort key, for both

R and S. Bring the first block of each sublist into a buffer Repeatedly find the least Y-value y among the first available

tuples of all the sublists. Identify all the tuples of both relations that have Y-value y. Output the join of all tuples from R with all tuples from S that share this common Y-value

SUMMARY OF SORT-BASED ALGORITHMS

Operators ApproximateM required

Disk I/O

γ,δ √B 3B

U,∩,− √(B(R) + B(S)) 3(B(R) + B(S))

►◄ √(max(B(R),B(S))) 5(B(R) + B(S))

►◄(more efficient) √(B(R) + B(S)) 3(B(R) + B(S))

SECTION 15.5

INTRODUCTION

Hashing is done if the data is too big to store in main memory buffers. Hash all the tuples of the argument(s) using an

appropriate hash key. For all the common operations, there is a way to

select the hash key so all the tuples that need to be considered together when we perform the operation have the same hash value.

This reduces the size of the operand(s) by a factor equal to the number of buckets.

PARTITIONING RELATIONS BY HASHINGAlgorithm:

initialize M-1 buckets using M-1 empty buffers;FOR each block b of relation R DO BEGIN

read block b into the Mth buffer;FOR each tuple t in b DO BEGIN

IF the buffer for bucket h(t) has no room for t THENBEGIN

copy the buffer t o disk;initialize a new empty block in that buffer;

END; copy t to the buffer for bucket h(t);END ;

END ;FOR each bucket DO

IF the buffer for this bucket is not empty THENwrite the buffer to disk;

DUPLICATE ELIMINATION For the operation δ(R) hash R to M-1 Buckets.(Note that two copies of the same tuple t will

hash to the same bucket) Do duplicate elimination on each bucket Ri

independently, using one-pass algorithm The result is the union of δ(Ri), where Ri is

the portion of R that hashes to the ith bucket

REQUIREMENTS

Number of disk I/O's: 3*B(R)B(R) < M(M-1), only then the two-pass, hash-based

algorithm will work In order for this to work, we need: hash function h evenly distributes the tuples among

the buckets each bucket Ri fits in main memory (to allow the

one-pass algorithm) i.e., B(R) ≤ M2

GROUPING AND AGGREGATION Hash all the tuples of relation R to M-1

buckets, using a hash function that depends only on the grouping attributes(Note: all tuples in the same group end up in the same bucket)

Use the one-pass algorithm to process each bucket independently

Uses 3*B(R) disk I/O's, requires B(R) ≤ M2

UNION, INTERSECTION, AND DIFFERENCE

For binary operation we use the same hash function to hash tuples of both arguments.

R U S we hash both R and S to M-1 R ∩ S we hash both R and S to 2(M-1) R-S we hash both R and S to 2(M-1) Requires 3(B(R)+B(S)) disk I/O’s. Two pass hash based algorithm requires

min(B(R)+B(S))≤ M2

HASH-JOIN ALGORITHM

Use same hash function for both relations; hash function should depend only on the join attributes

Hash R to M-1 buckets R1, R2, …, RM-1

Hash S to M-1 buckets S1, S2, …, SM-1

Do one-pass join of Ri and Si, for all i 3*(B(R) + B(S)) disk I/O's; min(B(R),B(S)) ≤

M2

SORT BASED VS HASH BASED For binary operations, hash-based only limits

size to min of arguments, not sum

Sort-based can produce output in sorted order, which can be helpful

Hash-based depends on buckets being of equal size

Sort-based algorithms can experience reduced rotational latency or seek time

SECTION 15.6

CLUSTERING AND NONCLUSTERING INDEXES

Clustered Relation: Tuples are packed into roughly as few blocks as can possibly hold those tuples

Clustering indexes: Indexes on attributes that all the tuples with a fixed value for the search key of this index appear on roughly as few blocks as can hold them

CLUSTERING AND NONCLUSTERING INDEXES

A relation that isn’t clustered cannot have a clustering index

A clustered relation can have nonclustering indexes

INDEX-BASED SELECTION For a selection σC(R), suppose C is of the

form a=v, where a is an attribute

For clustering index R.a: the number of disk I/O’s will be B(R)/V(R,a)

INDEX-BASED SELECTION

The actual number may be higher:1. index is not kept entirely in main

memory2. they spread over more blocks3. may not be packed as tightly as

possible into blocks

EXAMPLE

B(R)=1000, T(R)=20,000 number of I/O’s required:

1. clustered, not index 1000 2. not clustered, not index 20,000 3. If V(R,a)=100, index is clustering 10 4. If V(R,a)=10, index is nonclustering 2,000

JOINING BY USING AN INDEX

Natural join R(X, Y) S S(Y, Z)

Number of I/O’s to get RClustered: B(R)Not clustered: T(R)

Number of I/O’s to get tuple t of SClustered: T(R)B(S)/V(S,Y)Not clustered: T(R)T(S)/V(S,Y)

EXAMPLE

R(X,Y): 1000 blocks S(Y,Z)=500 blocksAssume 10 tuples in each block, so T(R)=10,000 and T(S)=5000V(S,Y)=100If R is clustered, and there is a clustering index on Y for Sthe number of I/O’s for R is: 1000 the number of I/O’s for S is10,000*500/100=50,000

JOINS USING A SORTED INDEX

Natural join R(X, Y) S (Y, Z) with index on Y for either R or S

Extreme case: Zig-zag join Example:

relation R(X,Y) and R(Y,Z) with index on Y for both relationssearch keys (Y-value) for R: 1,3,4,4,5,6search keys (Y-value) for S: 2,2,4,6,7,8

SECTION 15.7

WHAT DOES A BUFFER MANAGER DO?

Assume there are M of main-memory buffers needed for the operators on relations to store needed data.

In practice: 1) rarely allocated in advance2) the value of M may vary depending on system

conditions Therefore, buffer manager is used to allow processes

to get the memory they need, while minimizing the delay and unclassifiable requests.

Buffer manager

Buffers

RequestsRead/Writes

Figure 1: The role of the buffer manager : responds to requests for main-memory access to disk blocks

The role of the buffer manager

15.7.1 Buffer Management Architecture

Two broad architectures for a buffer manager:

1) The buffer manager controls main memory directly. • Relational DBMS

2) The buffer manager allocates buffers in virtual memory, allowing the OS to decide how to use buffers. • “main-memory” DBMS • “object-oriented” DBMS

Data must be in RAM for DBMS to operate on it! Buffer Manager hides the fact that not all data is in RAM.

DB

MAIN MEMORY

DISK

disk page

free frame

Page Requests from Higher Levels

BUFFER POOL

choice of frame dictatedby replacement policy

Buffer Pool

Buffer-replacement strategy -- LRU

Least-Recently Used (LRU):

To throw out the block that has not been read or written for the longest time.

• Requires more maintenance but it is effective. • Update the time table for every access.• Least-Recently Used blocks are usually less likely to

be accessed sooner than other blocks.

Buffer-replacement strategy -- FIFO

First-In-First-Out (FIFO):

The buffer that has been occupied the longest by the same block is emptied and used for the new block.

• Requires less maintenance but it can make more mistakes.• Keep only the loading time• The oldest block doesn’t mean it is less likely to be

accessed. Example: the root block of a B-tree index

Buffer-replacement strategy – “Clock”

The “Clock” Algorithm (“Second Chance”)

Think of the 8 buffers as arranged in a circle, shown as Figure 3

Flag 0 and 1:

buffers with a 0 flag are ok to sent their contents back to disk, i.e. ok to be replaced

buffers with a 1 flag are not ok to be replaced

Buffer-replacement strategy – “Clock”

0

0

1

1

1

0

0

0

Start point to search a 0 flag

the buffer with a 0 flag will be replaced

The flag will be set to 0

By next time the hand reaches it, if the content of this buffer is not accessed, i.e. flag=0, this buffer will be replaced.That’s “Second Chance”.

Figure 3: the clock algorithm

System Control helps Buffer-replacement strategy

System Control

The query processor or other components of a DBMS can give advice to the buffer manager in order to avoid some of the mistakes that would occur with a strict policy such as LRU,FIFO or Clock.

For example:

A “pinned” block means it can’t be moved to disk without first modifying certain other blocks that point to it.

In FIFO, use “pinned” to force root of a B-tree to remain in memory at all times.

15.7.3 The Relationship Between Physical Operator Selection and Buffer Management

Questions:

Can the algorithm adapt to changes of M, the number of main-memory buffers available?

When available buffers are less than M, and some blocks have to be put in disk instead of in memory.

How the buffer-replacement strategy impact the performance (i.e. the number of additional I/O’s)?

Example

FOR each chunk of M-1 blocks of S DO BEGIN

read these blocks into main-memory buffers;

organize their tuples into a search structure whose

search key is the common attributes of R and S;

FOR each block b of R DO BEGIN

read b into main memory;

FOR each tuple t of b DO BEGIN

find the tuples of S in main memory that

join with t ;

output the join of t with each of these tuples;

END ;

END ;

END ;

Figure 15.8: The nested-loop join algorithm

Example

The outer loop number (M-1) depends on the average number of buffers are available at each iteration.

The outer loop use M-1 buffers and 1 is reserved for a block of R, the relation of the inner loop.

If we pin the M-1 blocks we use for S on one iteration of the outer loop, we shall not lose their buffers during the round.

Also, more buffers may become available and then we could keep more than one block of R in memory.

Will these extra buffers improve the running time?

Example

CASE1: NO

Buffer-replacement strategy: LRUBuffers for R: kWe read each block of R in order into buffers.By end of the iteration of the outer loop, the last k blocks of R are in buffers.However, next iteration will start from the beginning of R again. Therefore, the k buffers for R will need to be replaced.

CASE 2: YES

Buffer-replacement strategy: LRUBuffers for R: kWe read the blocks of R in an order that alternates: firstlast and then lastfirst.In this way, we save k disk I/Os on each iteration of the outer loop except the first iteration.

Other Algorithms and M buffers

Other Algorithms also are impact by M and the buffer-replacement strategy. Sort-based algorithm

If M shrinks, we can change the size of a sublist.

Unexpected result: too many sublists to allocate each sublist a buffer. Hash-based algorithm

If M shrinks, we can reduce the number of buckets, as long as the buckets still can fit in M buffers.

SECTION 15.8

Multi-pass Sort-based AlgorithmsSuppose we have M main-memory buffers available to sort a relation R, which we assume is stored clustered.

If R fits in M blocks (i.e., B(R)<=M) If R fits in M blocks (i.e., B(R)<=M)

1.1. Read R into main memory.Read R into main memory.

2.2. Sort it using any main-memory Sort it using any main-memory sorting algorithm.sorting algorithm.

3.3. Write the sorted relation to disk.Write the sorted relation to disk.

INDUCTION:If R does not fit into main memory.

1. Partition the blocks holding R into M groups, which we shall call R1, R2, R3…

2. Recursively sort Ri for each i=1,2,3…M.3. Merge the M sorted sublists.

For a δ, output one copy of each distinct tuple, and skip over copies of the tuple.

For a γ, sort on the grouping attributes only, and combine the tuples with a given value of these grouping attributes.

ConclusionThe two pass algorithms based on sorting or hashing have natural recursive analogs that take three or more passes and will work for larger amounts of data.

CHAPTER 16

SECTION 16.1

Query compilation is divided into three steps

1. Parsing: Parse SQL query into parser tree.

2. Logical query plan: Transforms parse tree into expression tree of relational algebra.

3.Physical query plan: Transforms logical query plan into physical query plan.

. Operation performed

. Order of operation

. Algorithm used

. The way in which stored data is obtained and passed from one

operation to another.

Parser

Preprocessor

Logical Query plan generator

Query rewrite

Preferred logical query plan

Query

Form a query to a logical query plan

Syntax Analysis and Parse Tree Parser takes the sql query and convert it to parse tree. Nodes of parse tree: 1. Atoms: known as Lexical elements such as key words, constants, parentheses, operators, and other schema elements.

2. Syntactic categories: Subparts that plays a similar role in a query as <Query> ,

<Condition>

QUERY AND PARSE T REE

StarsIn(title,year,starName)

MovieStar(name,address,gender,birthdate)

Query: Give titles of movies that have at least one star

born in 1960

SELECT title FROM StarsIn WHERE starName IN (

SELECT name FROM MovieStar WHERE birthdate LIKE '%1960%'

);

Parse Tree<Query>

<SFW>

SELECT <SelList> FROM <FromList> WHERE <Condition>

<Attribute> <RelName> , <FromList> AND

title StarsIn <RelName>

<Condition> <Condition>

<Attribute> = <Attribute> <Attribute> LIKE <Pattern>

starName name birthdate ‘%1960’

MovieStar <Query>

THE PREPROCESSOR

Functions of Preprocessor . If a relation used in the query is virtual view then each

use of this relation in the form-list must replace by parser tree that describe the view.

. It is also responsible for semantic checking 1. Checks relation uses : Every relation mentioned in

FROM-

clause must be a relation or a view in current schema. 2. Check and resolve attribute uses: Every attribute

mentioned

in SELECT or WHERE clause must be an attribute of same relation in the current scope.

3. Check types: All attributes must be of a type appropriate to

their uses.

StarsIn(title,year,starName)

MovieStar(name,address,gender,birthdate)

Query: Give titles of movies that have at least one star

born in 1960

SELECT title FROM StarsIn WHERE starName IN (

SELECT name FROM MovieStar WHERE birthdate LIKE '%1960%'

);

PREPROCESSING QUERIES INVOLVING VIEWS

When an operand in a query is a virtual view, the preprocessor needs to replace the operand by a piece of parse tree that represents how the view is constructed from base table.

Base Table: Movies( title, year, length, genre, studioname, producerC#)

View definition : CREATE VIEW ParamountMovies AS

SELECT title, year FROM movies

WHERE studioName = 'Paramount';

Example based on view:

SELECT title FROM ParamountMovies WHERE year = 1979;

SECTION 16.2

OPTIMIZING THE LOGICAL QUERY PLAN The translation rules converting a parse tree

to a logical query tree do not always produce the best logical query tree.

It is often possible to optimize the logical query tree by applying relational algebra laws to convert the original tree into a more efficient logical query tree.

Optimizing a logical query tree using relational algebra laws is called heuristic optimization

RELATIONAL ALGEBRA LAWS

These laws often involve the properties of:

commutativity - operator can be applied to operands independent of order.

E.g. A + B = B + A - The “+” operator is commutative.

associativity - operator is independent of operand grouping.

E.g. A + (B + C) = (A + B) + C - The “+” operator is associative.

ASSOCIATIVE AND COMMUTATIVE OPERATORS

The relational algebra operators of cross-product (×), join (⋈), union, and intersection are all associative and commutative.

Commutative

R X S = S X R

R ⋈ S = S ⋈ R

R S = S R

R ∩ S = S ∩ R

Associative

(R X S) X T = S X (R X T)

(R ⋈ S) ⋈ T= S ⋈ (R ⋈ T)

(R S) T = S (R T)

(R ∩ S) ∩ T = S ∩ (R ∩ T)

LAWS INVOLVING SELECTION

Complex selections involving AND or OR can be broken into two or more selections: (splitting laws)

σC1 AND C2 (R) = σC1( σC2 (R))σC1 OR C2 (R) = ( σC1 (R) ) S ( σC2 (R) )

Example R={a,a,b,b,b,c} p1 satisfied by a,b, p2 satisfied by b,c σp1vp2 (R) = {a,a,b,b,b,c} σp1(R) = {a,a,b,b,b} σp2(R) = {b,b,b,c} σp1 (R) U σp2 (R) = {a,a,b,b,b,c}

LAWS INVOLVING SELECTION (CONTD..)

Selection is pushed through both arguments for union:

σC(R S) = σC(R) σC(S)

Selection is pushed to the first argument and optionally the second for difference:

σC(R - S) = σC(R) - S

σC(R - S) = σC(R) - σC(S)

LAWS INVOLVING SELECTION (CONTD..)

All other operators require selection to be pushed to only one of the arguments.

For joins, may not be able to push selection to both if argument does not have attributes selection requires.

σC(R × S) = σC(R) × S

σC(R ∩ S) = σC(R) ∩ S

σC(R ⋈ S) = σC(R) ⋈ S

σC(R ⋈D S) = σC(R) ⋈D S

LAWS INVOLVING SELECTION (CONTD..) Example Consider relations R(a,b) and S(b,c) and the

expression

σ (a=1 OR a=3) AND b<c (R ⋈S)σ a=1 OR a=3(σ b<c (R ⋈S))σ a=1 OR a=3(R ⋈ σ b<c (S))σ a=1 OR a=3(R) ⋈ σ b<c (S)

LAWS INVOLVING PROJECTION

Like selections, it is also possible to push projections down the logical query tree. However, the performance gained is less than selections because projections just reduce the number of attributes instead of reducing the number of tuples.

LAWS INVOLVING PROJECTION

Laws for pushing projections with joins:

πL(R × S) = πL(πM(R) × πN(S))

πL(R ⋈ S) = πL((πM(R) ⋈ πN(S))

πL(R ⋈D S) = πL((πM(R) ⋈D πN(S))

LAWS INVOLVING PROJECTION Laws for pushing projections with set

operations.

Projection can be performed entirely before union.

πL(R UB S) = πL(R) UB πL(S)

Projection can be pushed below selection as long as we also keep all attributes needed for the selection (M = L attr(C)).

πL ( σC (R)) = πL( σC (πM(R)))

LAWS INVOLVING JOIN

We have previously seen these important rules about joins:

1. Joins are commutative and associative.

2. Selection can be distributed into joins.

3. Projection can be distributed into joins.

LAWS INVOLVING DUPLICATE ELIMINATION

The duplicate elimination operator (δ) can be pushed through many operators.

R has two copies of tuples t, S has one copy of t,

δ (RUS)=one copy of t δ (R) U δ (S)=two copies of t


Laws for pushing duplicate elimination operator (δ):

δ(R × S) = δ(R) × δ(S)

δ(R S) = δ(R) δ(S)

δ(R D S) = δ(R) D δ(S)

δ( σC(R) = σC(δ(R))


The duplicate elimination operator (δ) can also be pushed through bag intersection, but not across union, difference, or projection in general.

δ(R ∩ S) = δ(R) ∩ δ(S)

LAWS INVOLVING GROUPING

The grouping operator (γ) laws depend on the aggregate operators used.

There is one general rule, however, that grouping subsumes duplicate elimination:

δ(γL(R)) = γL(R)

The reason is that some aggregate functions are unaffected by duplicates (MIN and MAX) while other functions are (SUM, COUNT, and AVG).

SECTION 16.3

TWO STEPS TO TURN PARSE TREE INTO TWO STEPS TO TURN PARSE TREE INTO PREFERRED LOGICAL QUERY PLANPREFERRED LOGICAL QUERY PLAN

Replace the nodes and structures of the parse tree, in appropriate groups, by an operator or operators of relational algebra.

Take the relational algebra expression and turn it into an expression that we expect can be converted to the most efficient physical query plan.

REVIEWREVIEW

Query

Preferred logical query plan

Parser

Preprocessor

Logical query plan generator

Query Rewriter

Section 16.1

Section 16.3

The relational-algebra expression consists of the following from bottom to top:The products of all the relations mentioned in the

<FromList>, which Is the argument of:A selection σC, where C is the <Condition> expression

in the construct being replaced, which in turn is the argument of:

A projection πL , where L is the list of attributes in the <SelList>

A QUERY : EXAMPLEA QUERY : EXAMPLE

SELECT movieTitle

FROM Starsin, MovieStar

WHERE starName = name AND

birthdate LIKE ‘%1960’;

SELECT MOVIETITLESELECT MOVIETITLEFROM STARSIN, MOVIESTARFROM STARSIN, MOVIESTARWHERE STARNAME = NAME AND WHERE STARNAME = NAME AND BIRTHDATE LIKE ‘%1960’; BIRTHDATE LIKE ‘%1960’;

TRANSLATION TO AN ALGEBRAIC TRANSLATION TO AN ALGEBRAIC EXPRESSION TREEEXPRESSION TREE

REMOVING SUBQUERIES FROM REMOVING SUBQUERIES FROM CONDITIONSCONDITIONS

For parse trees with a <Condition> that has a subquery Intermediate operator – two argument selection It is intermediate in between the syntactic categories of

the parse tree and the relational-algebra operators that apply to relations.

USING A TWO-ARGUMENT USING A TWO-ARGUMENT ΣΣ

πmovieTitle

σ

StarsIn <Condition>

MovieStar

IN πname<Tuple>

starName

σ birthdate LIKE ‘%1960'

<Attribute>

TWO ARGUMENT SELECTION WITH TWO ARGUMENT SELECTION WITH CONDITION INVOLVING INCONDITION INVOLVING IN Now say we have, two arguments – some relation and

the second argument is a <Condition> of the form t IN S.

‘t’ – tuple composed of some attributes of R ‘S’ – uncorrelated subquery

Steps to be followed:1. Replace the <Condition> by the tree that is the expression for

S ( δ is used to remove duplicates)2. Replace the two-argument selection by a one-argument

selection σC.

3. Give σC an argument that is the product of R and S.

TWO ARGUMENT SELECTION WITH TWO ARGUMENT SELECTION WITH CONDITION INVOLVING INCONDITION INVOLVING IN

σ

R <Condition>

t IN S

σC

X

R δ

S

THE EFFECTTHE EFFECT

IMPROVING THE LOGICAL QUERY IMPROVING THE LOGICAL QUERY PLANPLAN

Algebraic laws to improve logical query plans: Selections can be pushed down the expression tree as far as

they can go. Similarly, projections can be pushed down the tree, or new

projections can be added. Duplicate eliminations can sometimes be removed, or moved

to a more convenient position in the tree. Certain selections can be combined with a product below to

turn the pair of operations into an equijoin.

GROUPING ASSOCIATIVE/ GROUPING ASSOCIATIVE/ COMMUTATIVE OPERATORSCOMMUTATIVE OPERATORS An operator that is associative and commutative

operators may be though of as having any number of operands.

We need to reorder these operands so that the multiway join is executed as sequence of binary joins.

Its more time consuming to execute them in the order suggested by parse tree.

For each portion of subtree that consists of nodes with the same associative and commutative operator (natural join, union, and intersection), we group the nodes with these operators into a single node with many children.

THE EFFECT OF QUERY REWRITINGTHE EFFECT OF QUERY REWRITING

Π movieTitle

Starname = name

StarsIn σbirthdate LIKE ‘%1960’

MovieStar

FINAL STEP IN PRODUCING LOGICAL FINAL STEP IN PRODUCING LOGICAL QUERY PLANQUERY PLAN

=>

U

U

U

W

R

S T

VU

U V W

R S T

AN EXAMPLE TO SUMMARIZEAN EXAMPLE TO SUMMARIZE “find movies where the average age of the stars was at

most 40 when the movie was made” SELECT distinct m1.movieTitle, m1,movieYear

FROM StarsIn m1WHERE m1.movieYear – 40 <= (

SELECT AVG (birthdate)FROM StartsIn m2, MovieStar sWHERE m2.starName = s.name AND

m1.movieTitle = m2.movieTitle ANDm1.movieYear = m2.movieyear

);

SELECT distinct m1.movieTitle, m1,movieYearFROM StarsIn m1WHERE m1.movieYear – 40 <= ( SELECT AVG (birthdate) FROM StartsIn m2, MovieStar s WHERE m2.starName = s.name AND m1.movieTitle = m2.movieTitle AND m1.movieYear = m2.movieyear );

SELECTIONS COMBINED WITH A PRODUCT SELECTIONS COMBINED WITH A PRODUCT TO TURN THE PAIR OF OPERATIONS INTO TO TURN THE PAIR OF OPERATIONS INTO AN EQUIJOIN…AN EQUIJOIN…

CONDITION PUSHED UP THE EXPRESSION TREE…

`

SELECTIONS COMBINED…SELECTIONS COMBINED…

SECTION 16.4

ESTIMATING THE COST OF ESTIMATING THE COST OF OPERATIONSOPERATIONS

After getting to the logical query plan, we turn it into physical plan.

Consider all the possible physical plan and estimate their costs – this evaluation is known as cost-based enumeration.

The one with least estimated cost is the one selected to be passed to the query-execution engine.

SELECTION FOR EACH PHYSICAL PLAN

We select for each physical plan:An order and grouping for associative-and-

commutative operations like joins, unions, and intersections.

An algorithm for each operator in the logical plan, for instance, deciding whether a nested-loop join or hash-join should be used.

Additional operators – scanning, sorting etc. – that are needed for the physical plan but that were not present explicitly in the logical plan.

The way in which the arguments are passed from on operator to the next.

ESTIMATING SIZES OF INTERMEDIATE RELATIONS

1. Give accurate estimates.

2. Are easy to compute.

3. Are logically consistent; that is, the size estimate for an intermediate relation should not depend on how that relation is computed.

ESTIMATING THE SIZE OF A PROJECTION

We should treat a classical, duplicate-eliminating projection as a bag-projection.

The size of the result can be computed exactly. There may be reduction in size (due to

eliminated components) or increase in size (due to new components created as combination of attributes).

ESTIMATING THE SIZE OF A SELECTION

While performing selection, we may reduce the number of tuples but the sizes of tuple remain same.

Size can be computed as:

Where A is an attribute of R and c is a constant

The recommended estimate isT(S) = T(R)/ V(R,A)

S = σ A=c (R)

SECTION 16.6

INTRODUCTION

This section focuses on critical problem in cost-based optimization: Selecting order for natural join of three or more

relations Compared to other binary operations, joins take

more time and therefore need effective optimization techniques

SIGNIFICANCE OF LEFT AND RIGHT JOIN ARGUMENTS

The argument relations in joins determine the cost of the join

The left argument of the join is Called the build relation Assumed to be smaller Stored in main-memory

SIGNIFICANCE OF LEFT AND RIGHT JOIN ARGUMENTS

The right argument of the join is Called the probe relation Read a block at a time Its tuples are matched with those of build relation

The join algorithms which distinguish between the arguments are: One-pass join Nested-loop join Index join

JOIN TREES

Order of arguments is important for joining two relations

Left argument, since stored in main-memory, should be smaller

With two relations only two choices of join tree With more than two relations, there are n! ways

to order the arguments and therefore n! join trees, where n is the no. of relations

JOIN TREES

Total # of tree shapes T(n) for n relations given by recurrence:

T(1) = 1 T(2) = 1 T(3) = 2 T(4) = 5 … etc

LEFT-DEEP JOIN TREES

Consider 4 relations. Different ways to join them are as follows

JOIN ORDER Join order selection

A1 A2 A3 .. An Left deep join trees

Dynamic programming Best plan computed for each subset of relations

Best plan (A1, .., An) = min cost plan of( Best plan(A2, .., An) A1 Best plan(A1, A3, .., An) A2 …. Best plan(A1, .., An-1)) An

Ai

An

DYNAMIC PROGRAMMING TO SELECT A JOIN ORDER AND GROUPING

Three choices to pick an order for the join of many relations are:Consider all of the relationsConsider a subsetUse a heuristic o pick one

Dynamic programming is used either to consider all or a subsetConstruct a table of costs based on relation

sizeRemember only the minimum entry which will

required to proceed





A GREEDY ALGORITHM FOR SELECTING A JOIN ORDER

It is expensive to use an exhaustive method like dynamic programming

Better approach is to use a join-order heuristic for the query optimization

Greedy algorithm is an example of that Make one decision at a time about order of join and

never backtrack on the decisions once made

SECTION 16.7

BEFORE COMPLETE PHYSICAL-QUERY-PLAN

A query previously has been Parsed and Preprocessed (16.1) Converted to Logical Query Plans (16.3) Estimated the Costs of Operations (16.4) Determined costs by Cost-Based Plan Selection

(16.5) Weighed costs of join operations by choosing an

Order for Joins

16.7 COMPLETING THE PHYSICAL-QUERY-PLAN

3 topics related to turning LP into a complete physical plan

1. Choosing of physical implementations such as Selection and Join methods

2. Decisions regarding to intermediate results (Materialized or Pipelined)

3. Notation for physical-query-plan operators

I. CHOOSING A SELECTION METHOD (A) Algorithms for each selection operators

1. Can we use an created index on an attribute? If yes, index-scan. Otherwise table-scan)2. After retrieve all condition-satisfied tuples in (1),

then filter them with the rest selection conditions

CHOOSING A SELECTION METHOD(A) (CONT.) Recall Cost of query = # disk I/O’s How costs for various plans are estimated from σC(R) operation

1. Cost of table-scan algorithm

a) B(R) if R is clusteredb) T(R) if R is not clustered

2. Cost of a plan picking an equality term (e.g. a = 10) w/ index-scan

a) B(R) / V(R, a) clustering indexb) T(R) / V(R, a) nonclustering index

3. Cost of a plan picking an inequality term (e.g. b < 20) w/ index-scan

a) B(R) / 3 clustering indexb) T(R) / 3 nonclustering index

EXAMPLE

Selection: σx=1 AND y=2 AND z<5 (R)

- Where paremeters of R(x, y, z) are :

T(R)=5000, B(R)=200,

V(R,x)=100, and V(R, y)=500

- Relation R is clustered- x, y have nonclustering indexes, only index

on z is clustering.

EXAMPLE (CONT.)Selection options:1. Table-scan filter x, y, z. Cost is B(R) = 200

since R is clustered.2. Use index on x =1 filter on y, z. Cost is 50

since T(R) / V(R, x) is (5000/100) = 50 tuples, index is not clustering.

3. Use index on y =2 filter on x, z. Cost is 10 since T(R) / V(R, y) is (5000/500) = 10 tuples using nonclustering index.

4. Index-scan on clustering index w/ z < 5 filter x ,y. Cost is about B(R)/3 = 67

EXAMPLE (CONT.) Costs

option 1 = 200 option 2 = 50 option 3 = 10 option 4 = 67

The lowest Cost is option 3. Therefore, the preferred physical

plan 1. retrieves all tuples with y = 2 2. then filters for the rest two conditions (x,

z).

II. CHOOSING A JOIN METHOD

Determine costs associated with each join algorithms: 1. One-pass join, and nested-loop join

devotes enough buffer to joining2. Sort-join is preferred when attributes are

pre-sorted or two or more join on the same attribute such as

(R(a, b) S(a, c)) T(a, d)

- where sorting R and S on a will produce result of R S to be sorted on a and used directly in next join

CHOOSING A JOIN METHOD (CONT.)

3. Index-join for a join with high chance of using

index created on the join attribute such as R(a, b) S(b, c)

4. Hashing join is the best choice for unsorted or non-indexing relations which needs multipass join.

III. PIPELINING VERSUS MATERIALIZATION

Materialization (naïve way) store (intermediate) result of each operations on

disk

Pipelining (more efficient way)

Interleave the execution of several operations, the

tuples produced by one operation are passed

directly to the operations that used it

store (intermediate) result of each operations on

buffer, which is implemented on main memory

IV. PIPELINING UNARY OPERATIONS

Unary = a-tuple-at-a-time or full relation selection and projection are the best

candidates for pipelining.

R

In buf Unaryoperation

Out buf

In buf Unaryoperation

Out buf

M-1 buffers

PIPELINING UNARY OPERATIONS (CONT.) Pipelining Unary Operations are implemented by iterators

V. PIPELINING BINARY OPERATIONS

Binary operations : , , - , , x The results of binary operations can also be

pipelined. Use one buffer to pass result to its consumer,

one block at a time. The extended example shows tradeoffs and

opportunities

EXAMPLE

Consider physical query plan for the expression

(R(w, x) S(x, y)) U(y, z) Assumption

R occupies 5,000 blocks, S and U each 10,000 blocks.

The intermediate result R S occupies k blocks for some k.

Both joins will be implemented as hash-joins, either one-pass or two-pass depending on k

There are 101 buffers available.

EXAMPLE (CONT.)

First consider join R S, neither relations fits in buffers Needs two-pass hash-join to partition R into 100 buckets (maximum possible) each bucket has 50

blocks The 2nd pass hash-join uses 51 buffers,

leaving the rest 50 buffers for joining result of R S with U.

EXAMPLE (CONT.)

Case 1: suppose k 49, the result of R S occupies at most 49 blocks.

Steps 1. Pipeline in R S into 49 buffers2. Organize them for lookup as a hash table3. Use one buffer left to read each block of U in

turn4. Execute the second join as one-pass join.

EXAMPLE (CONT.) The total number of I/O’s is 55,000 45,000 for two-pass hash

join of R and S 10,000 to read U for one-

pass hash join of (R S) U.

EXAMPLE (CONT.)

Case 2: suppose k > 49 but < 5,000, we can still pipeline, but need another strategy which intermediate results join with U in a 50-bucket, two-pass hash-join. Steps are:

1. Before start on R S, we hash U into 50 buckets of 200 blocks each.

2. Perform two-pass hash join of R and U using 51 buffers as case 1, and placing results in 50 remaining buffers to form 50 buckets for the join of R S with U.

3. Finally, join R S with U bucket by bucket.

EXAMPLE (CONT.)

The number of disk I/O’s is: 20,000 to read U and write its tuples into buckets 45,000 for two-pass hash-join R S k to write out the buckets of R S k+10,000 to read the buckets of R S and U in

the final join The total cost is 75,000+2k.

EXAMPLE (CONT.)

Compare Increasing I/O’s between case 1 and case 2 k 49 (case 1)

Disk I/O’s is 55,000 k > 50 5000 (case 2)

k=50 , I/O’s is 75,000+(2*50) = 75,100 k=51 , I/O’s is 75,000+(2*51) = 75,102 k=52 , I/O’s is 75,000+(2*52) = 75,104

Notice: I/O’s discretely grows as k increases from 49 50.

EXAMPLE (CONT.)

Case 3: k > 5,000, we cannot perform two-pass join in 50 buffers available if

result of R S is pipelined. Steps are

1. Compute R S using two-pass join and store the result on disk.

2. Join result on (1) with U, using two-pass join.

EXAMPLE (CONT.)

The number of disk I/O’s is: 45,000 for two-pass hash-join R and S

k to store R S on disk 30,000 + k for two-pass join of U in R S

The total cost is 75,000+4k.

SECTION 16.7-16.8

VI. NOTATION FOR PHYSICAL QUERY PLANS

Several types of operators: 1. Operators for leaves2. (Physical) operators for Selection3. (Physical) Sorts Operators4. Other Relational-Algebra Operations

In practice, each DBMS uses its own internal notation for physical query plan.

NOTATION FOR PHYSICAL QUERY PLANS (CONT.)

1. Operator for leaves A leaf operand is replaced in LQP tree

TableScan(R) : read all blocks

SortScan(R, L) : read in order according to L

IndexScan(R, C): scan index attribute A by

condition C of form Aθc.

IndexScan(R, A) : scan index attribute R.A.

This behaves like TableScan but more efficient if R is not clustered.


2. (Physical) operators for Selection Logical operator σC(R) is often combined with

access methods. If σC(R) is replaced by Filter(C), and there is no index on

R or an attribute on condition C Use TableScan or SortScan(R, L) to access R

If condition C Aθc AND D for condition D, and there is an index on R.A, then we may Use operator IndexScan(R, Aθc) to access R and

Use Filter(D) in place of the selection σC(R)


3. (Physical) Sort Operators Sorting can occur any point in physical plan,

which use a notation SortScan(R, L). It is common to use an explicit operator Sort(L)

to sort relation that is not stored. Can apply at the top of physical-query-plan

tree if the result needs to be sorted with ORDER BY clause (г).


4. Other Relational-Algebra Operations Descriptive text definitions and signs to

elaborate Operations performed e.g. Join or grouping. Necessary parameters e.g. theta-join or list of

elements in a grouping. A general strategy for the algorithm e.g. sort-

based, hashed based, or index-based. A decision about number of passed to be used e.g.

one-pass, two-pass or multipass. An anticipated number of buffers the operations

will required.

NOTATION FOR PHYSICAL QUERY PLANS (CONT.) Example of a physical-query-plan

A physical-query-plan in example 16.36 for the case k > 5000 TableScan Two-pass hash join Materialize (double line) Store operator

NOTATION FOR PHYSICAL QUERY PLANS (CONT.) Another example

A physical-query-plan in example 16.36 for the case k < 49 TableScan (2) Two-pass hash join Pipelining Different buffers needs Store operator

NOTATION FOR PHYSICAL QUERY PLANS (CONT.) A physical-query-plan in example 16.35

Use Index on condition y = 2 firstFilter with the rest condition later on.

VII. ORDERING OF PHYSICAL OPERATIONS

The PQP is represented as a tree structure implied order of operations.

Still, the order of evaluation of interior nodes may not always be clear.

Iterators are used in pipeline manner Overlapped time of various nodes will make

“ordering” no sense.

ORDERING OF PHYSICAL OPERATIONS (CONT.) 3 rules summarize the ordering of events

in a PQP tree:1. Break the tree into sub-trees at each edge

that represent materialization. Execute one subtree at a time.

2. Order the execution of the subtree Bottom-top Left-to-right

3. All nodes of each sub-tree are executed simultaneously.

COMPILATION OF QUERIES

Compilation means turning a query into a physical query plan, which can be implemented by query engine.

Steps of query compilation : Parsing Semantic checking Selection of the preferred logical query plan Generating the best physical plan

THE PARSER

The first step of SQL query processing. Generates a parse tree Nodes in the parse tree corresponds to the SQL

constructs Similar to the compiler of a programming

language

SEMANTIC CHECKING

Checks the semantics of a SQL query. Examines a parse tree. Checks :

Attributes Relation names Types

Resolves attribute references.

CONVERSION TO A LOGICAL QUERY PLAN

Converts a semantically parsed tree to a algebraic expression.

Conversion is straightforward but sub queries need to be optimized.

Two argument selection approach can be used.

ALGEBRAIC TRANSFORMATION

Many different ways to transform a logical query plan to an actual plan using algebraic transformations.

The laws used for this transformation :Commutative and associative laws Laws involving selectionPushing selectionLaws involving projectionLaws about joins and productsLaws involving duplicate eliminationsLaws involving grouping and aggregation

ESTIMATING SIZES OF RELATIONS

True running time is taken into consideration when selecting the best logical plan.

Two factors the affects the most in estimating the sizes of relation : Size of relations ( No. of tuples ) No. of distinct values for each attribute of each

relation Histograms are used by some systems.

COST BASED OPTIMIZING

Best physical query plan represents the least costly plan.

Factors that decide the cost of a query plan : Order and grouping operations like joins, unions and

intersections. Nested loop and the hash loop joins used. Scanning and sorting operations. Storing intermediate results.

PLAN ENUMERATION STRATEGIES

Common approaches for searching the space for best physical plan . Dynamic programming : Tabularizing the best plan for

each sub expression Selinger style programming : sort-order the results as

a part of table Greedy approaches : Making a series of locally

optimal decisions Branch-and-bound : Starts with enumerating the worst

plans and reach the best plan

LEFT-DEEP JOIN TREES

Left – Deep Join Trees are the binary trees with a single spine down the left edge and with leaves as right children.

This strategy reduces the number of plans to be considered for the best physical plan.

Restrict the search to Left – Deep Join Trees when picking a grouping and order for the join of several relations.

PHYSICAL PLANS FOR SELECTION

Breaking a selection into an index-scan of relation, followed by a filter operation.

The filter then examines the tuples retrieved by the index-scan.

Allows only those to pass which meet the portions of selection condition.

PIPELINING VERSUS MATERIALIZING

This flow of data between the operators can be controlled to implement “ Pipelining “ .

The intermediate results should be removed from main memory to save space for other operators.

This techniques can implemented using “ materialization “ .

Both the pipelining and the materialization should be considered by the physical query plan generator.

An operator always consumes the result of other operator and is passed through the main memory.

CHAPTER 18

SECTION 18.1-18.2

CONCURRENCY CONTROL Concurrency control in database management

systems (DBMS) ensures that database transactions are performed concurrently without the concurrency violating the data integrity of a database.

Executed transactions should follow the ACID rules. The DBMS must guarantee that only serializable (unless Serializability is intentionally relaxed), recoverable schedules are generated.

It also guarantees that no effect of committed transactions is lost, and no effect of aborted (rolled back) transactions remains in the related database.

TRANSACTION ACID RULES

Atomicity - Either the effects of all or none of its operations remain when a transaction is completed - in other words, to the outside world the transaction appears to be indivisible, atomic.

Consistency - Every transaction must leave the database in a consistent state. Isolation - Transactions cannot interfere with each other. Providing isolation is the main goal of concurrency control.

Durability - Successful transactions must persist through crashes.

SERIAL AND SERIALIZABLE SCHEDULES

In the field of databases, a schedule is a list of actions, (i.e. reading, writing, aborting, committing), from a set of transactions. In this example, Schedule D is the set of 3 transactions T1, T2, T3. The schedule describes the actions of the transactions as seen by the DBMS. T1 Reads and writes to object X, and then T2 Reads and writes to object Y, and finally T3 Reads and writes to object Z. This is an example of a serial schedule, because the actions of the 3 transactions are not interleaved.

SERIAL AND SERIALIZABLE SCHEDULES A schedule that is equivalent to a serial schedule has the serializability

property. In schedule E, the order in which the actions of the transactions are executed

is not the same as in D, but in the end, E gives the same result as D.

SERIAL SCHEDULE TI PRECEDES T2T1 T2Read(A); A A+100Write(A);Read(B); B B+100;Write(B);

Read(A);A A2;Write(A);

Read(B);B B2;Write(B);

A B25 25

125

125

250

250250 250

SERIAL SCHEDULE T2 PRECEDES TL

T1 T2



Read(A); A A+100Write(A);Read(B); B B+100;Write(B);

A B25 25

50

50

150

150150 150

SERIALIZABLE, BUT NOT SERIAL, SCHEDULE

T1 T2Read(A); A A+100Write(A);


Read(B); B B+100;Write(B);


A B25 25

125

250

125

250250 250

r1(A); w1 (A): r2(A); w2(A); r1 (B); w1 (B); r2(B); w2(B);

NONSERIALIZABLE SCHEDULET1 T2Read(A); A A+100Write(A);




A B25 25

125

250

50

150250 150

SCHEDULE THAT IS SERIALIZABLE ONLY BECAUSE OF THE DETAILED BEHAVIOR OF THE TRANSACTIONS

T1 T2’Read(A); A A+100Write(A);




regardless of the consistent initial state: the final state will be consistent.

A B25 25

125

125

25

125125 125

NON-CONFLICTING ACTIONS

Two actions are non-conflicting if whenever theyoccur consecutively in a schedule, swapping themdoes not affect the final state produced by theschedule. Otherwise, they are conflicting.

CONFLICTING ACTIONS: GENERAL RULES Two actions of the same transaction conflict:

r1(A) w1(B) Two actions over the same database element

conflict, if one of them is a write r1(A) w2(A) w1(A) w2(A)

CONFLICT ACTIONS Two or more actions are said to be in conflict if:

The actions belong to different transactions. At least one of the actions is a write operation. The actions access the same object (read or write).

The following set of actions is conflicting: T1:R(X), T2:W(X), T3:W(X)

While the following sets of actions are not: T1:R(X), T2:R(X), T3:R(X) T1:R(X), T2:W(Y), T3:R(X)

CONFLICT SERIALIZABLE

We may take any schedule and make as many nonconflicting swaps as we wish.

With the goal of turning the schedule into a serial schedule.

If we can do so, then the original schedule is serializable, because its effect on the database state remains the same as we perform each of the nonconflictingswaps.

CONFLICT SERIALIZABLE A schedule is said to be conflict-serializable when the schedule is conflict-

equivalent to one or more serial schedules. Another definition for conflict-serializability is that a schedule is conflict-

serializable if and only if there exists an acyclic precedence graph/serializability graph for the schedule.

Which is conflict-equivalent to the serial schedule <T1,T2>, but not <T2,T1>.

CONFLICT EQUIVALENT / CONFLICT-SERIALIZABLE

Let Ai and Aj are consecutive non-conflicting actions that belongs to different transactions. We can swap Ai and Aj without changing the result.

Two schedules are conflict equivalent if they can be turned one into the other by a sequence of non-conflicting swaps of adjacent actions.

We shall call a schedule conflict-serializable if it is conflict-equivalent to a serial schedule.

CONFLICT-SERIALIZABLE

T1 T2

R(A)

W(A)

R(A)

R(B)

W(B)

W(A)

R(B)

W(B)

CONFLICT-SERIALIZABLE

T1 T2

R(A)

W(A)

R(A)

W(B)

R(B)

W(A)

R(B)

W(B)

SerialSchedule

SECTION 18.3-18.4

INTRODUCTION

Enforcing serializability by locks Locks Locking scheduler Two phase locking

Locking systems with several lock modes Shared and exclusive locks Compatibility matrices Upgrading/updating locks Incrementing locks

LOCKS It works like as follows :

A request from transaction Scheduler checks in the lock table Generates a serializable schedule of actions.

CONSISTENCY OF TRANSACTIONS

Actions and locks must relate each other Transactions can only read & write only if has a

lock and has not released the lock. Unlocking an element is compulsory.

Legality of schedules No two transactions can aquire the lock on same

element without the prior one releasing it.

LOCKING SCHEDULER Grants lock requests only if it is in a legal

schedule. Lock table stores the information about

current locks on the elements.

THE LOCKING SCHEDULER (CONTD.) A legal schedule of consistent transactions but

unfortunately it is not a serializable.

LOCKING SCHEDULE (CONTD.)

The locking scheduler delays requests that would result in an illegal schedule.

TWO-PHASE LOCKING

Guarantees a legal schedule of consistent transactions is conflict-serializable.

All lock requests proceed all unlock requests. The growing phase:

Obtain all the locks and no unlocks allowed. The shrinking phase:

Release all the locks and no locks allowed.

WORKING OF TWO-PHASE LOCKING

Assures serializability. Two protocols for 2PL:

Strict two phase locking : Transaction holds all its exclusive locks till commit / abort.

Rigorous two phase locking : Transaction holds all locks till commit / abort.

Possible to find a transaction Tj that has a 2PL and a schedule S for Ti ( non 2PL ) and Tj that is not conflict serializable.

FAILURE OF 2PL.

2PL fails to provide security against deadlocks.

SHARED & EXCLUSIVE LOCKS

Consistency of TransactionsCannot write without Exclusive LockCannot read without holding some lock

This basically works on 2 principlesA read action can only proceed a shared or

an exclusive lockA write lock can only proceed a exclusice

lock All locks need to be unlocked before

commit

SHARED AND EXCLUSIVE LOCKS (CONT.) Two-phase locking of transactions

Must precede unlocking Legality of Schedules

An element may be locked exclusively by one transaction or by several in shared mode, but not both.

COMPATIBILITY MATRICES

Has a row and column for each lock mode. Rows correspond to a lock held on an element by

another transaction Columns correspond to mode of lock requested. Example :

LOCK REQUESTED

S X

LOCK S YES NO

HOLD X NO NO

UPGRADING LOCKS

Suppose a transaction wants to read as well as write : It aquires a shared lock on the element Performs the calculations on the element And when its ready to write, It is granted a

exclusive lock. Transactions with unpredicted read write

locks can use UPGRADING LOCKS.

UPGRADING LOCKS (CONT.)

Indiscriminating use of upgrading produces a deadlock.

Example : Both the transactions want to upgrade on the same element

UPDATE LOCKS

Solves the deadlock occurring in upgrade lock method.

A transaction in an update lock can read but cant write.

Update lock can later be converted to exclusive lock.

An update lock can only be given if the element has shared locks.

UPDATE LOCKS (CONT.)

An update lock is like a shared lock when you are requesting it and is like a exclusive lock when you have it.

Compatibility matrix :

S X U

S YES NO YES

X NO NO NO

U NO NO NO

INCREMENT LOCKS

Used for incrementing & decrementing stored values.

E.g. - Transfer money from one bank to another, Ticket selling transactions in which number seats are decremented after each transaction.

INCREMENT LOCK (CONT.)

A increment lock does not enable read or write locks on element.

Any number of transaction can hold increment lock on element

Shared and exclusive locks can not be granted if an increment lock is granted on element

S X I

S YES NO NO

X NO NO NO

I NO NO YES

SECTION 18.4

18.4 LOCKING SYSTEMS WITH SEVERAL LOCK MODES

In 18.3, if a transaction must lock a database element (X) either reads or writes, No reason why several transactions could not read X

at the same time, as long as none write X

Introduce locking schemes Shared/Read Lock ( For Reading) Exclusive/Write Lock( For Writing)

18.4.1 SHARED & EXCLUSIVE LOCKS

Transaction Consistency Cannot write without Exclusive LockCannot read without holding some lock

Consider lock for writing is “stronger” than for reading

This basically works on 2 principles1. A read action can only proceed a shared or an

exclusive lock2. A write lock can only proceed a exclusive lock

All locks need to be unlocked before commit

18.4.1 SHARED & EXCLUSIVE LOCKS (CONT.)

Two-phase locking (2PL) of transactions

Ti

Notation:

sli (X)– Ti requests shared lock on DB element X

xli (X)– Ti requests exclusive lock on DB element X

ui (X)– Ti relinquishes whatever lock on X

Lock R/W Unlock

18.4.1 SHARED & EXCLUSIVE LOCKS (CONT.)

Legality of SchedulesAn element may be locked by: one write

transaction or by several read transactions shared mode, but not both

18.4.2 COMPATIBILITY MATRICES

A convenient way to describe lock-management policies Rows correspond to a lock held on an element by

another transaction Columns correspond to mode of lock requested. Example :

Lock requested

S X

Lock inhold

S YES NO

X NO NO

18.4.3 UPGRADING LOCKS

A transaction (T) taking a shared lock is friendly toward other transaction.

When T wants to read and write a new

value X,

1. T takes a shared lock on X.

2. performs operations on X (may spend long time)

3. When T is ready to write a new value, “Upgrade” shared lock to exclusive lock on X.

18.4.3 UPGRADING LOCKS (CONT.)

Observe the example

T1 cannot take an exclusive lock on B until all locks on

B are released.

‘B’ is releasedT1 retry and

succeed

18.4.3 UPGRADING LOCKS (CONT.)

Upgrading can simply cause a “Deadlock”. Both the transactions want to upgrade on the

same element

Both transactions will wait forever !!

18.4.4 UPDATE LOCKS

The third lock mode resolving the deadlock problem, which rules are

Only “Update lock” can be upgraded to a write (exclusive) lock later.

An “Update lock” is allowed to grant on X when there are already shared locks on X.

Once there is an “Update lock,” it prevents additional any kinds of lock, and later changes to a write (exclusive) lock.

Notation: uli (X)

18.4.4 UPDATE LOCKS (CONT.)

Example

18.4.4 UPDATE LOCKS (CONT.)

• Compatibility matrix (asymmetric)

Lock requestedS X U

Lock inhold

S YES NO YES

X NO NO NO

U NO NO NO

18.4.5 INCREMENT LOCKS

A useful lock for transactions which increase/decrease value. e.g. money transfer between two bank

accounts. If 2 transactions (T1, T2) add constants

to the same database element (X), It doesn’t matter which goes first, but no

reads are allowed in between transaction processing

Let see on following exhibits

18.4.5 INCREMENT LOCKS (CONT.)

A=5

A=15

A=17

A=7T1: INC (A,2)

T1: INC (A,2)

T2: INC (A,10)

T2: INC (A,10)

CASE 1

CASE 2


What if

A=5

A=15

A=15

A=7T1: INC (A,2)

T1: INC (A,2)

T2: INC (A,10)

T2: INC (A,10)

A=5 A=7

A != 17

A=5

A=5

18.4.5 INCREMENT LOCKS (CONT.) INC (A, c) –

Increment action of writing on database element A, which is an atomic execution consisting of1. READ(A,t);2. t = t+c;3. WRITE(A,t);

Notation: ili (X)– action of Ti requesting an increment

lock on X inci (X)– action of Ti increments X by some

constant; don’t care about the value of the constant.


Example


• Compatibility matrix

Lock requestedS X I

Lock inhold

S YES NO NO

X NO NO NO

I NO NO YES

CONCURRENCY CONTROL

Chapter 18

Section 18.5

Presented by Khadke, Suvarna

CS 257 (Section II) Id 213

SECTION 18.5

OVERVIEW

Assume knowledge of: Lock Two phase lock Lock modes: shared, exclusive, update

A simple scheduler architecture based on following principle : Insert lock actions into the stream of reads,

writes, and other actions Release locks when the transaction manager

tells it that the transaction will commit or abort

SCHEDULER THAT INSERTS LOCK ACTIONS INTO THE TRANSACTIONS REQUEST STREAM

SCHEDULER THAT INSERTS LOCK ACTIONS

If transaction is delayed, waiting for a lock, Scheduler performs following actions

Part I: Takes the stream of requests generated by the transaction & insert appropriate lock modes to db operations (read, write, or update)

Part II: Take actions (a lock or db operation) from Part I and executes it.

Determine the transaction (T) that action belongs and status of T (delayed or not). If T is not delayed then

1. Database access action is transmitted to the database and executed

SCHEDULER THAT INSERTS LOCK ACTIONS

2. If lock action is received by PartII, it checks the L Table whether lock can be granted or not

i> Granted, the L Table is modified to include granted lock

ii>Not G. then update L Table about requested lock then PartII delays transaction T

3. When a T = commits or aborts, PartI is notified by the transaction manager and releases all locks.

If any transactions are waiting for locks PartI notifies PartII.

3. Part II when notified about the lock on some DB element, determines next transaction T’ to get lock to continue.

THE LOCK TABLE

A relation that associates database elements with locking information about that element

Implemented with a hash table using database elements as the hash key

Size is proportional to the number of lock elements only, not to the size of the entire database

DB element A

Lock information for A

LOCK TABLE ENTRIES STRUCTURESome Sort of information found in Lock Table entry 1>Group modes-S: only shared locks are held-X: one exclusive lock and no other locks- U: one update lock and one or more shared locks2>wait : one transaction waiting for a lock on A3>A list : T currently hold locks on A or Waiting for lock on A

HANDLING LOCK REQUESTS

Suppose transaction T requests a lock on A If there is no lock table entry for A, then

there are no locks on A, so create the entry and grant the lock request

If the lock table entry for A exists, use the group mode to guide the decision about the lock request

HANDLING LOCK REQUESTS If group mode is U (update) or X (exclusive)No other lock can be granted

Deny the lock request by T Place an entry on the list saying T requests a

lock And Wait? = ‘yes’

If group mode is S (shared)Another shared or update lock can be

granted Grant request for an S or U lock Create entry for T on the list with Wait? = ‘no’ Change group mode to U if the new lock is an

update lock

HANDLING UNLOCK REQUESTS

Now suppose transaction T unlocks A Delete T’s entry on the list for A If T’s lock is not the same as the group mode,

no need to change group mode Otherwise check entire list for new group

modeS: GM(S) or nothingU: GM(S) or nothingX: nothing

HANDLING UNLOCK REQUESTS

If the value of waiting is “yes" need to grant one or more locks using following approachesFirst-Come-First-Served: Grant the lock to the longest waiting request. No starvation (waiting forever for lock)Priority to Shared Locks: Grant all S locks waiting, then one U lock. Grant X lock if no others waitingPriority to Upgrading: If there is a U lock waiting to upgrade to an X lock, grant that first.

SECTION 18.6

MANAGING HIERARCHIES OF DATABASE ELEMENTS

Two problems that arise with locks when there is a tree structure to the data are:

When the tree structure is a hierarchy of lockable elementsDetermine how locks are granted for both

large elements (relations) and smaller elements (blocks containing tuples or individual tuples)

When the data itself is organized as a tree (B-tree indexes)This will be discussed in the next section

LOCKS WITH MULTIPLE GRANULARITY

A database element can be a relation, block or a tuple

Different systems use different database elements to determine the size of the lock

Thus some may require small database elements such as tuples or blocks and others may require large elements such as relations

EXAMPLE OF MULTIPLE GRANULARITY LOCKS

Consider a database for a bankChoosing relations as database elements means

we would have one lock for an entire relation If we were dealing with a relation having account

balances, this kind of lock would be very inflexible and thus provide very little concurrency

Why? Because balance transactions require exclusive locks and this would mean only one transaction occurs for one account at any time

But as each account is independent of others we could perform transactions on different accounts simultaneously

…(CONTD.)

Thus it makes sense to have block element for the lock so that two accounts on different blocks can be updated simultaneously

Another example is that of a documentWith similar arguments as above, we see

that it is better to have large element (a complete document) as the lock in this case

WARNING (INTENTION) LOCKS

These are required to manage locks at different granularities In the bank example, if the a shared lock is

obtained for the relation while there are exclusive locks on individual tuples, unserializable behavior occurs

The rules for managing locks on hierarchy of database elements constitute the warning protocol

DATABASE ELEMENTS ORGANIZED IN HIERARCHY

RULES OF WARNING PROTOCOL These involve both ordinary (S and X) and

warning (IS and IX) locks The rules are:

Begin at the root of hierarchyRequest the S/X lock if we are at the

desired element If the desired element id further down the

hierarchy, place a warning lock (IS if S and IX if X)

When the warning lock is granted, we proceed to the child node and repeat the above steps until desired node is reached

COMPATIBILITY MATRIX FOR SHARED, EXCLUSIVE AND INTENTION LOCKS

IS IX S X

IS Yes Yes Yes No

IX Yes Yes No No

S Yes No Yes No

X No No No No

• The above matrix applies only to locks held by other transactions

GROUP MODES OF INTENTION LOCKS An element can request S and IX locks at the

same time if they are in the same transaction (to read entire element and then modify sub elements)

This can be considered as another lock mode, SIX, having restrictions of both the locks i.e. No for all except IS

SIX serves as the group mode

EXAMPLE

Consider a transaction T1 as followsSelect * from table where attribute1 =

‘abc’Here, IS lock is first acquired on the entire

relation; then moving to individual tuples (attribute = ‘abc’), S lock in acquired on each of them

Consider another transaction T2

Update table set attribute2 = ‘def’ where attribute1 = ‘ghi’

Here, it requires an IX lock on relation and since T1’s IS lock is compatible, IX is granted

On reaching the desired tuple (ghi), as there is no lock, it gets X too

If T2 was updating the same tuple as T1, it would have to wait until T1 released its S lock

PHANTOMS AND HANDLING INSERTIONS CORRECTLY

This arises when transactions create new sub elements of lockable elements

Since we can lock only existing elements the new elements fail to be locked

The problem created in this situation is explained in the following example

EXAMPLE

Consider a transaction T3

Select sum(length) from table where attribute1 = ‘abc’

This calculates the total length of all tuples having attribute1

Thus, T3 acquires IS for relation and S for targeted tuples

Now, if another transaction T4 inserts a new tuple having attribute1 = ‘abc’, the result of T3 becomes incorrect

EXAMPLE (…CONTD.) This is not a concurrency problem since the

serial order (T3, T4) is maintained But if both T3 and T4 were to write an element

X, it could lead to unserializable behaviorr3(t1);r3(t2);w4(t3);w4(X);w3(L);w3(X)r3 and w3 are read and write operations by

T3 and w4 are the write operations by T4 and L is the total length calculated by T3 (t1 + t2)

At the end, we have result of T3 as sum of lengths of t1 and t2 and X has value written by T3

This is not right; if value of X is considered to be that written by T3 then for the schedule to be serializable, the sum of lengths of t1, t2 and t3 should be considered

EXAMPLE (…CONTD.)

Else if the sum is total length of t1 and t2 then for the schedule to be serializable, X should have value written by T4

This problem arises since the relation has a phantom tuple (the new inserted tuple), which should have been locked but wasn’t since it didn’t exist at the time locks were taken

The occurrence of phantoms can be avoided if all insertion and deletion transactions are treated as write operations on the whole relation

SECTION 18.7

BASICSB-Trees

- Tree data structure that keeps data sorted

- allow searches, insertion, and deletion

- commonly used in database and file systemsLock

- Enforce limits on access to resources

- way of enforcing concurrency controlLock Granularity

- Level and type of information that lock

protects.

TREE PROTOCOLKind of graph-based protocolAlternate to Two-Phased

Locking (2PL) database elements are disjoint

pieces of data Nodes of the tree DO NOT

form a hierarchy based on containment

Way to get to the node is through its parent

Example: B-Tree

ADVANTAGES OF TREE PROTOCOL

Unlocking takes less time as compared to 2PL

Freedom from deadlocks

18.7.1 MOTIVATION FOR

TREE-BASED LOCKING

Consider B-Tree Index, treating individual nodes as lockable database elements.

Concurrent use of B-Tree is not possible with standard set of locks and 2PL.

Therefore, a protocol is needed which can assure serializability by allowing access to the elements all the way at the bottom of the tree even if the 2PL is violated.

18.7.1 MOTIVATION FOR TREE-BASED LOCKING (CONT.)

Reason for : “Concurrent use of B-Tree is not possible with standard set of locks and 2PL.”

every transaction must begin with locking the root node

2PL transactions can not unlock the root until all the required locks are acquired.

18.7.2 ACCESSING TREE STRUCTURED

DATAAssumptions:

Only one kind of lockConsistent transactionsLegal schedulesNo 2PL requirement on transaction

18.7.2 RULES FOR ACCESSING TREE STRUCTURED DATA

RULES:

First lock can be at any node.Subsequent locks may be acquired only after

parent node has a lock.Nodes may be unlocked any time.No relocking of the nodes even if the node’s

parent is still locked

18.7.3 WHY TREE PROTOCOL WORKS?

Tree protocol implies a serial order on transactions in the schedule.

Order of precedence:Ti < s Tj

If Ti locks the root before Tj, then Ti locks every node in common with Tj before Tj.

ORDER OF PRECEDENCE

SECTION 18.8

WHAT IS TIMESTAMPING?

Scheduler assign each transaction T a unique number, it’s timestamp TS(T).

Timestamps must be issued in ascending order, at the time when a transaction first notifies the scheduler that it is beginning.

TIMESTAMP TS(T)

Two methods of generating Timestamps.Use the value of system, clock as

the timestamp.Use a logical counter that is

incremented after a new timestamp has been assigned.

Scheduler maintains a table of currently active transactions and their timestamps irrespective of the method used

TIMESTAMPS FOR DATABASE ELEMENT X AND COMMIT BIT

RT(X):- The read time of X, which is the highest timestamp of transaction that has read X.

WT(X):- The write time of X, which is the highest timestamp of transaction that has write X.

C(X):- The commit bit for X, which is true if and only if the most recent transaction to write X has already committed.

PHYSICALLY UNREALIZABLE BEHAVIOR

Read too late: A transaction U that started after

transaction T, but wrote a value for X before T reads X.

U writes X

T reads X

T start U start

PHYSICALLY UNREALIZABLE BEHAVIOR

Write too late A transaction U that started after T,

but read X before T got a chance to write X.

U reads X

T writes X

T start U start

Figure: Transaction T tries to write too late

DIRTY READ

It is possible that after T reads the value of X written by U, transaction U will abort.

U writes X

T reads X

U start T start U aborts

T could perform a dirty read if it reads X when shown

RULES FOR TIMESTAMPS-BASED SCHEDULING

1. Scheduler receives a request rT(X)a) If TS(T) ≥ WT(X), the read is physically realizable.

1. If C(X) is true, grant the request, if TS(T) > RT(X), set RT(X) := TS(T); otherwise do not change RT(X).

2. If C(X) is false, delay T until C(X) becomes true or transaction that wrote X aborts.

b) If TS(T) < WT(X), the read is physically unrealizable. Rollback T.

RULES FOR TIMESTAMPS-BASED SCHEDULING (CONT.)

2. Scheduler receives a request WT(X).a) if TS(T) ≥ RT(X) and TS(T) ≥ WT(X), write is physically realizable and must be performed.

1. Write the new value for X,2. Set WT(X) := TS(T), and3. Set C(X) := false.

b) if TS(T) ≥ RT(X) but TS(T) < WT(X), then the write is physically realizable, but there is already a later values in X.

a. If C(X) is true, then the previous writers of X is committed, and ignore the write by T.

b. If C(X) is false, we must delay T.c) if TS(T) < RT(X), then the write is physically unrealizable, and T must be rolled back.

RULES FOR TIMESTAMPS-BASED SCHEDULING (CONT.)

3. Scheduler receives a request to commit T. It must find all the database elements X written by T and set C(X) := true. If any transactions are waiting for X to be committed, these transactions are allowed to proceed.

4. Scheduler receives a request to abort T or decides to rollback T, then any transaction that was waiting on an element X that T wrote must repeat its attempt to read or write.

MULTIVERSION TIMESTAMPS

Multiversion schemes keep old versions of data item to increase concurrency.

Each successful write results in the creation of a new version of the data item written.

Use timestamps to label versions. When a read(X) operation is issued,

select an appropriate version of X based on the timestamp of the transaction, and return the value of the selected version.

TIMESTAMPS AND LOCKING

Generally, timestamping performs better than locking in situations where:Most transactions are read-only.It is rare that concurrent transaction

will try to read and write the same element.

In high-conflict situation, locking performs better than timestamps

SECTION 18.9

INTRODUCTION

What is optimistic concurrency control?(assumes no unserializable behavior will occur) Timestamp- based scheduling and

Validation-based scheduling(allows T to access data without locks)

VALIDATION BASED SCHEDULING

Scheduler keeps a record of what the active transactions are doing.

Executes in 3 phases1. Read- reads from RS( ), computes local address

2. Validate- compares read and write sets

3. Write- writes from WS( )

VALIDATION BASED SCHEDULER

Contains an assumed serial order of transactions.

Maintains three sets: START( ): set of T’s started but not completed

validation. VAL( ): set of T’s validated but not finished the

writing phase. FIN( ): set of T’s that have finished.

EXPECTED EXCEPTIONS

1. Suppose there is a transaction U, such that: U is in VAL or FIN; that is, U has validated, FIN(U)>START(T); that is, U did not finish before T started RS(T) ∩WS(T) ≠φ; let it contain database element X.

2. Suppose there is transaction U, such that:• U is in VAL; U has successfully validated.•FIN(U)>VAL(T); U did not finish before T entered its validation phase.•WS(T) ∩ WS(U) ≠φ; let x be in both write sets.

VALIDATION RULES

Check that RS(T) ∩ WS(U)= φ for any previously validated U that did not finish before T has started i.e. FIN(U)>START(T).

Check that WS(T) ∩ WS(U)= φ for any previously validated U that did not finish before T is validated i.e. FIN(U)>VAL(T)

EXAMPLE

SOLUTION Validation of U:

Nothing to check Validation of T:

WS(U) ∩ RS(T)= {D} ∩{A,B}=φWS(U) ∩ WS(T)= {D}∩ {A,C}=φ

Validation of V:RS(V) ∩ WS(T)= {B}∩{A,C}=φWS(V) ∩ WS(T)={D,E}∩ {A,C}=φRS(V) ∩ WS(U)={B} ∩{D}=φ

Validation of W:RS(W) ∩ WS(T)= {A,D}∩{A,C}={A}WS(W) ∩ WS(V)= {A,D}∩{D,E}={D}WS(W) ∩ WS(V)= {A,C}∩{D,E}=φ (W is not

validated)

COMPARISONConcurrency control Mechanisms

Storage Utilization Delays

Locks Space in the lock table is proportional to the number of database elements locked.

Delays transactions but avoids rollbacks

Timestamps Space is needed for read and write times with every database element, neither or not it is currently accessed.

Do not delay the transactions but cause them to rollback unless Interface is low

Validation Space is used for timestamps and read or write sets for each currently active transaction, plus a few more transactions that finished after some currently active transaction began.

Do not delay the transactions but cause them to rollback unless interface is low

CHAPTER 21

SECTION 21.1

NEED FOR INFORMATION INTEGRATION

All the data in the world could put in a single database (ideal database system)

In the real world (impossible for a single database):databases are created independentlyhard to design a database to support future use

INCONVENIENT

Record grades for students who pay tuition Want to swim in SJSU aquatic center for free

in summer vacation?(all the cases above cannot achieve the function by a single database)

Solution: one database

HOW TO INTEGRATE

Start overbuild one database: contains all the legacy databases; rewrite all the applicationsresult: painful

Build a layer of abstraction (middleware)on top of all the legacy databasesthis layer is often defined by a collection of classes

BUT…

HETEROGENEITY PROBLEM

What is Heterogeneity ProblemAardvark Automobile Co. 1000 dealers has 1000 databasesto find a model at another dealercan we use this command:

SELECT * FROM CARS WHERE MODEL=“A6”;

TYPE OF HETEROGENEITY

Communication Heterogeneity Query-Language Heterogeneity Schema Heterogeneity Data type difference Value Heterogeneity Semantic Heterogeneity

CONCLUSION

One database system is perfect, but impossible

Independent database is inconvenient Integrate database

1. start over2. middleware

heterogeneity problem

CHAPTER 21.2MODES OF INFORMATION

INTEGRATION

ID: 219Name: Qun YuClass: CS257 219 Spring 2009Instructor: Dr. T.Y.Lin

SECTION 21.2

FEDERATIONS

The simplest architecture for integrating several DBs

One to one connections between all pairs of DBs

n DBs talk to each other, n(n-1) wrappers are needed

Good when communications between DBs are limited

WRAPPER

Wrapper : a software translates incoming queries and outgoing answers. In a result, it allows information sources to conform to some shared schema.

FEDERATIONS DIAGRAM

DB2DB1

DB3 DB4

2 Wrappers

2 Wrappers

2 Wrappers

2 Wrappers

2 Wrappers 2 Wrappers

A federated collection of 4 DBs needs 12 components to translate queries from one to another.

EXAMPLE

Car dealers want to share their inventory. Each dealer queries the other’s DB to find the needed car.

Dealer-1’s DB relation: NeededCars(model,color,autoTrans)

Dealer-2’s DB relation: Auto(Serial, model, color)

Options(serial,option)

Dealer-1’s DB Dealer-2’s DB

wrapper

wrapper

EXAMPLE…

Dealer 1 queries Dealer 2 for needed cars

For(each tuple(:m,:c,:a) in NeededCars){

if(:a=TRUE){/* automatic transmission wanted */

SELECT serial

FROM Autos, Options

WHERE Autos.serial = Options.serial AND Options.option = ‘autoTrans’

AND Autos.model = :m AND Autos.color =:c;

}

Else{/* automatic transmission not wanted */

SELECT serial

FROM Auto

WHERE Autos.model = :m AND

Autos.color = :c AND

NOT EXISTS( SELECT * FROM Options WHERE serial = Autos.serial

AND option=‘autoTrans’);

}

}

DATA WAREHOUSE

Sources are translated from their local schema to a global schema and copied to a central DB.

User transparent: user uses Data Warehouse just like an ordinary DB

User is not allowed to update Data Warehouse

WAREHOUSE DIAGRAM

Warehouse

Extractor Extractor

Source 1 Source 2

User query

result

Combiner

EXAMPLE

Construct a data warehouse from sources DB of 2 car dealers:

Dealer-1’s schema: Cars(serialNo, model,color,autoTrans,cdPlayer,…)Dealer-2’s schema: Auto(serial,model,color)

Options(serial,option)

Warehouse’s schema: AutoWhse(serialNo,model,color,autoTrans,dealer)

Extractor --- Query to extract data from Dealer-1’s data:

INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)SELECT serialNo,model,color,autoTrans,’dealer1’ From Cars;

EXAMPLE

Extractor --- Query to extract data from Dealer-2’s data:

INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)SELECT serialNo,model,color,’yes’,’dealer2’ FROM Autos,OptionsWHERE Autos.serial=Options.serial AND

option=‘autoTrans’;

INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)SELECT serialNo,model,color,’no’,’dealer2’ FROM AutosWHERE NOT EXISTS ( SELECT * FROM serial =Autos.serial AND option = ‘autoTrans’);

CONSTRUCT DATA WAREHOUSE

1) Periodically reconstructed from the current data in the sources, once a night or at even longer intervals.

Advantages:

simple algorithms.

Disadvantages:

1) need to shut down the warehouse;

2) data can become out of date.

There are mainly 3 ways to constructing the data in the warehouse:


2) Updated periodically based on the changes(i.e. each night) of the sources.

Advantages:

involve smaller amounts of data. (important when warehouse is large and needs to be modified in a short period)

Disadvantages:

1) the process to calculate changes to the warehouse is complex.

2) data can become out of date.


3) Changed immediately, in response to each change or a small set of changes at one or more of the sources.

Advantages:data won’t become out of date. Disadvantages: requires too much communication, therefore, it is generally too expensive.

(practical for warehouses whose underlying sources changes slowly.)

MEDIATORS

Virtual warehouse, which supports a virtual view or a collection of views, that integrates several sources.

Mediator doesn’t store any data. Mediators’ tasks: 1)receive user’s query, 2)send queries to wrappers, 3)combine results from wrappers, 4)send the final result to user.

A MEDIATOR DIAGRAM

Mediator

Wrapper Wrapper

Source 1 Source 2

User query

Query

Query

QueryQuery

Result

Result

Result

Result

Result

EXAMPLE

Same data sources as the example of data warehouse, the mediatorIntegrates the same two dealers’ source into a view with schema:

AutoMed(serialNo,model,color,autoTrans,dealer)

When the user have a query:

SELECT sericalNo, model FROM AkutoMedWhere color=‘red’

EXAMPLE

In this simple case, the mediator forwards the same query to eachOf the two wrappers.

Wrapper1: Cars(serialNo, model, color, autoTrans, cdPlayer, …)SELECT serialNo,model FROM carsWHERE color = ‘red’;

Wrapper2: Autos(serial,model,color); Options(serial,option)SELECT serial, modelFROM AutosWHERE color=‘red’;

The mediator needs to interprets serial into serialNo, and then returns the union of these sets of data to user.

SECTION 21.3

WRAPPERS IN MEDIATOR-BASED SYSTEMS

More complicated than that in most data

warehouse system.Able to accept a variety of queries from the

mediator and translate them to the terms of the source.

Communicate the result to the mediator.

Templates for Query Patterns:Templates for Query Patterns:

Use notation T=>S to express the idea Use notation T=>S to express the idea that the template T is turned by the that the template T is turned by the wrapper into the source query S.wrapper into the source query S.

Example 1

Dealer 1Cars (serialNo, model, color,

autoTrans, navi,…)

For use by a mediator with schemaAutoMed (serialNo, model, color,

autoTrans, dealer)

We denote the code representing that color by the parameter $c, then the template will be:

SELECT *FROM AutosMedWHERE color = ’$c’;

=>SELECT serialNo, model, color, autoTrans, ’dealer1’FROM CarsWHERE color=’$c’;

(Template T => Source query S)

There will be total 2n templates if we have the option of specifying n attributes.

WRAPPER GENERATORS

The wrapper generator creates a table holds the various query patterns contained in the templates.

The source queries that are associated with each.

A driver is used in each wrapper, the task of the driver is to:

Accept a query from the mediator.Search the table for a template that

matches the query.The source query is sent to the source,

again using a “plug-in” communication mechanism.

The response is processed by the wrapper.

Filter Have a wrapper filter to supporting more

queries.

Example 2 If wrapper is designed with more complicated

template with queries specify both model and color.

SELECT *FROM AutosMedWHERE model = ’$m’ AND color = ’$c’;

=>SELECT serialNo, model, color, autoTrans, ’dealer1’

FROM CarsWHERE model = ’$m’ AND color=’$c’;

Solution:

1. Use template with $c=‘blue’ find all blue cars and store them in a temporary relation:

TemAutos (serialNo, model, color, autoTrans, dealer)

2.The wrapper then return to the mediator the desired set of automobiles by excuting the local query:SELECT*

FROM TemAutos

WHERE model= ’Gobi’;

SECTION 21.4-21.5

21.4 CAPABILITY BASED OPTIMIZATION

Introduction Typical DBMS estimates the cost of each query

plan and picks what it believes to be the best Mediator – has knowledge of how long its sources

will take to answer Optimization of mediator queries cannot rely on

cost measure alone to select a query plan Optimization by mediator follows capability

based optimization

21.4.1 THE PROBLEM OF LIMITED SOURCE CAPABILITIES

Many sources have only Web Based interfaces

Web sources usually allow querying through a query form

E.g. Amazon.com interface allows us to query about books in many different ways.

But we cannot ask questions that are too general E.g. Select * from books;

21.4.1 THE PROBLEM OF LIMITED SOURCE CAPABILITIES (CON’T)

Reasons why a source may limit the ways in which queries can be asked Earliest database did not use relational DBMS

that supports SQL queries Indexes on large database may make certain

queries feasible, while others are too expensive to execute

Security reasons E.g. Medical database may answer queries about

averages, but won’t disclose details of a particular patient's information

21.4.2 A NOTATION FOR DESCRIBING SOURCE CAPABILITIES

For relational data, the legal forms of queries are described by adornments

Adornments – Sequences of codes that represent the requirements for the attributes of the relation, in their standard order f(free) – attribute can be specified or not b(bound) – must specify a value for an attribute

but any value is allowed u(unspecified) – not permitted to specify a value

for a attribute

21.4.2 A NOTATION FOR DESCRIBING SOURCE CAPABILITIES….

(CONT’D)

c[S](choice from set S) means that a value must be specified and value must be from finite set S.

o[S](optional from set S) means either do not specify a value or we specify a value from finite set S

A prime (f’) specifies that an attribute is not a part of the output of the query

A capabilities specification is a set of adornments A query must match one of the adornments in its

capabilities specification

21.4.3 CAPABILITY-BASED QUERY-PLAN SELECTION

Given a query at the mediator, a capability based query optimizer first considers what queries it can ask at the sources to help answer the query

The process is repeated until: Enough queries are asked at the sources to

resolve all the conditions of the mediator query and therefore query is answered. Such a plan is called feasible.

We can construct no more valid forms of source queries, yet still cannot answer the mediator query. It has been an impossible query.

21.4.3 CAPABILITY-BASED QUERY-PLAN SELECTION (CONT’D)

The simplest form of mediator query where we need to apply the above strategy is join relations

E.g we have sources for dealer 2 Autos(serial, model, color) Options(serial, option)

Suppose that ubf is the sole adornment for Auto and Options have two adornments, bu and uc[autoTrans, navi]

Query is – find the serial numbers and colors of Gobi models with a navigation system

21.4.4 ADDING COST-BASED OPTIMIZATION

Mediator’s Query optimizer is not done when the capabilities of the sources are examined

Having found feasible plans, it must choose among them

Making an intelligent, cost based query optimization requires that the mediator knows a great deal about the costs of queries involved

Sources are independent of the mediator, so it is difficult to estimate the cost

21.5 OPTIMIZING MEDIATOR QUERIES

Chain algorithm – a greed algorithm that finds a way to answer the query by sending a sequence of requests to its sources. Will always find a solution assuming at least one

solution exists. The solution may not be optimal.

21.5.1 SIMPLIFIED ADORNMENT NOTATION

A query at the mediator is limited to b (bound) and f (free) adornments.

We use the following convention for describing adornments: nameadornments(attributes) where:

name is the name of the relation the number of adornments = the number of attributes

21.5.2 OBTAINING ANSWERS FOR SUBGOALS

Rules for subgoals and sources: Suppose we have the following subgoal:

Rx1x2…xn(a1, a2, …, an),

and source adornments for R are: y1y2…yn. If yi is b or c[S], then xi = b. If xi = f, then yi is not output restricted.

The adornment on the subgoal matches the adornment at the source: If yi is f, u, or o[S] and xi is either b or f.

21.5.3 THE CHAIN ALGORITHM

Maintains 2 types of information: An adornment for each subgoal. A relation X that is the join of the relations for all

the subgoals that have been resolved. Initially, the adornment for a subgoal is b iff

the mediator query provides a constant binding for the corresponding argument of that subgoal.

Initially, X is a relation over no attributes, containing just an empty tuple.

21.5.3 THE CHAIN ALGORITHM (CON’T)

First, initialize adornments of subgoals and X. Then, repeatedly select a subgoal that can

be resolved. Let Rα(a1, a2, …, an) be the subgoal:

1. Wherever α has a b, we shall find the argument in R is a constant, or a variable in the schema of R. Project X onto its variables that appear in R.


2. For each tuple t in the project of X, issue a query to the source as follows (β is a source adornment).

If a component of β is b, then the corresponding component of α is b, and we can use the corresponding component of t for source query.

If a component of β is c[S], and the corresponding component of t is in S, then the corresponding component of α is b, and we can use the corresponding component of t for the source query.

If a component of β is f, and the corresponding component of α is b, provide a constant value for source query.


If a component of β is u, then provide no binding for this component in the source query.

If a component of β is o[S], and the corresponding component of α is f, then treat it as if it was a f.

If a component of β is o[S], and the corresponding component of α is b, then treat it as if it was c[S].

3. Every variable among a1, a2, …, an is now bound. For each remaining unresolved subgoal, change its adornment so any position holding one of these variables is b.


4. Replace X with X πs(R), where S is all of the variables among: a1, a2, …, an.

5. Project out of X all components that correspond to variables that do not appear in the head or in any unresolved subgoal.

If every subgoal is resolved, then X is the answer.

If every subgoal is not resolved, then the algorithm fails.

α

21.5.3 THE CHAIN ALGORITHM EXAMPLE

Mediator query: Q: Answer(c) ← Rbf(1,a) AND Sff(a,b) AND Tff(b,c)

Example:

Relation R S TData

Adornment bf c’[2,3,5]f bu

w x

1 2

1 3

1 4

x y

2 4

3 5

y z

4 6

5 7

5 8

21.5.3 THE CHAIN ALGORITHM EXAMPLE (CON’T)

Initially, the adornments on the subgoals are the same as Q, and X contains an empty tuple. S and T cannot be resolved because they each

have ff adornments, but the sources have either a b or c.

R(1,a) can be resolved because its adornments are matched by the source’s adornments.

Send R(w,x) with w=1 to get the tables on the previous page.


Project the subgoal’s relation onto its second component, since only the second component of R(1,a) is a variable.

This is joined with X, resulting in X equaling this relation.

Change adornment on S from ff to bf.

a

2

3

4


Now we resolve Sbf(a,b): Project X onto a, resulting in X. Now, search S for tuples with attribute a equivalent to

attribute a in X.

Join this relation with X, and remove a because it doesn’t appear in the head nor any unresolved subgoal:

a b

2 4

3 5

b

4

5


Now we resolve Tbf(b,c):

Join this relation with X and project onto the c attribute to get the relation for the head.

Solution is {(6), (7), (8)}.

b c

4 6

5 7

5 8

21.5.4 INCORPORATING UNION VIEWS AT THE MEDIATOR

This implementation of the Chain Algorithm does not consider that several sources can contribute tuples to a relation.

If specific sources have tuples to contribute that other sources may not have, it adds complexity.

To resolve this, we can consult all sources, or make best efforts to return all the answers.

21.5.4 INCORPORATING UNION VIEWS AT THE MEDIATOR (CON’T)

Consulting All Sources We can only resolve a subgoal when each source for

its relation has an adornment matched by the current adornment of the subgoal.

Less practical because it makes queries harder to answer and impossible if any source is down.

Best Efforts We need only 1 source with a matching adornment to

resolve a subgoal. Need to modify chain algorithm to revisit each

subgoal when that subgoal has new bound requirements.

SECTION 21.6

LOCAL-AS-VIEW MEDIATORS.

In a LAV mediator, global predicates defined are not views of the source data.

for each source, expressions are defined, involving the global predicates that describe the tuples that the source is able to produce.

Queries are answered at the mediator by discovering all possible ways to construct the query using the views provided by the source.

MOTIVATION FOR LAV MEDIATORS

Sometimes the the relationship between what the mediator should provide and what the sources provide is more subtle.

For example, consider the predicate Par(c, p) meaning that p is a parent of c which represents the set of all child parent facts that could ever exist.

The sources will provide information about whatever child-parent facts they know.

MOTIVATION(CONTD..)

There can be sources which may provide child-grandparent facts but not child- parent facts at all.

This source can never be used to answer the child-parent query under GAV mediators.

LAV mediators allow to say that a certain source provides grand parent facts.

They help discover how and when to use that source in a given query.

TERMINOLOGY FOR LAV MEDIATION.

The queries at the mediator and the queries that describe the source will be single Datalog rules.

A query that is a single Datalog rule is often called a conjunctive query.

The global predicates of the LAV mediator are used as the subgoals of mediator queries.

There are conjunctive queries that define views.

CONTD..

Their heads each have a unique view predicate that is the name of a view.

Each view definition has a body consisting of global predicates and is associated with a particular source.

It is assumed that each view can be constructed with an all-free adornment.

EXAMPLE..

Consider global predicate Par(c, p) meaning that p is a parent of c.

One source produces parent facts. Its view is defined by the conjunctive query-

V1(c, p) Par(c, p) Another source produces some grand parents

facts. Then its conjunctive query will be – V2(c, g) Par(c, p) AND Par(p, g)

EXAMPLE CONTD..

The query at the mediator will ask for great-grand parent facts that can be obtained from the sources. The mediator query is –

Q(w, z) Par(w, x) AND Par(x, y) AND Par(y, z)

One solution can be using the parent predicate(V1) directly three times.

Q(w, z) V1(w, x) AND V1 (x, y) AND V1(y, z)

EXAMPLE CONTD..

Another solution can be to use V1(parent facts) and V2(grandparent facts).

Q(w, z) V1(w, x) AND V2(x, z)

Or Q(w, z) V2(w, y) AND V1(y, z)

EXPANDING SOLUTIONS.

Consider a query Q, a solution S that has a body whose subgoals are views and each view V is defined by a conjunctive query with that view as the head.

The body of V’s conjunctive query can be substituted for a subgoal in S that uses the predicate V to have a body consisting of only global predicates.

EXPANSION ALGORITHM

A solution S has a subgoal V(a1, a2,…an) where ai’s can be any variables or constants.

The view V can be of the form V(b1, b2,….bn) B

Where B represents the entire body. V(a1, a2, … an) can be replaced in solution S

by a version of body B that has all the subgoals of B with variables possibly altered.

EXPANSION ALGORITHM CONTD..

The rules for altering the variables of B are:1. First identify the local variables B, variables

that appear in the body but not in the head.

2. If there are any local variables of B that appear in B or in S, replace each one by a distinct new variable that appears nowhere in the rule for V or in S.

3. In the body B, replace each bi by ai for i = 1,2…n.

EXAMPLE.

Consider the view definitions, V1(c, p) Par(c, p)

V2(c, g) Par(c, p) AND Par(p, g) One of the proposed solutions S is Q(w, z) V1(w, x) AND V2(x, z)

The first subgoal with predicate V1 in the solution can be expanded as Par(w, x) as there are no local variables.

EXAMPLE CONTD.

The V2 subgoal has a local variable p which doesn’t appear in S nor it has been used as a local variable in another substitution. So p can be left as it is.

Only x and z are to be substituted for variables c and g.

The Solution S now will be Q(w, z) Par(w, x) AND Par(x, p) AND Par(p,z)

CONTAINMENT OF CONJUNCTIVE QUERIES

A containment mapping from Q to E is a function т from the variables of Q to the variables and constants of E, such that:

1. If x is the ith argument of the head of Q, then т(x) is the ith argument of the head of E.

2. Add to т the rule that т(c)=c for any constant c. If P(x1,x2,… xn) is a subgoal of Q, then P(т(x1), т(x2),… т(xn)) is a subgoal of E.

EXAMPLE.

Consider two Conjunctive queries:Q1: H(x, y) A(x, z) and B(z, y)

Q2: H(a, b) A(a, c) AND B(d, b) AND A(a, d) When we apply the substitution, Т(x) = a, Т(y) = b, Т(z) = d, the head of Q1

becomes H(a, b) which is the head of Q2.

So,there is a containment mapping from Q1 to Q2.

EXAMPLE CONTD..

The first subgoal of Q1 becomes A(a, d) which is the third subgoal of Q2.

The second subgoal of Q1 becomes the second subgoal of Q2.

There is also a containment mapping from Q2 to Q1 so the two conjunctive queries are equivalent.

WHY THE CONTAINMENT-MAPPING TEST WORKS

Suppose there is a containment mapping т from Q1 to Q2.

When Q2 is applied to the database, we look for substitutions σ for all the variables of Q2.

The substitution for the head becomes a tuple t that is returned by Q2.

If we compose т and then σ, we have a mapping from the variables of Q1 to tuples of the database that produces the same tuple t for the head of Q1.

FINDING SOLUTIONS TO A MEDIATOR QUERY

There can be infinite number of solutions built from the views using any number of subgoals and variables.

LMSS Theorem can limit the search which states that• If a query Q has n subgoals, then any answer

produced by any solution is also produced by a solution that has at most n subgoals.

If the conjunctive query that defines a view V has in its body a predicate P that doesn’t appear in the body of the mediator query, then we need not consider any solution that uses V.

EXAMPLE.

Recall the queryQ1: Q(w, z) Par(w, x) AND Par(x, y) AND Par(y, z) This query has three subgoals, so we don’t

have to look at solutions with more than three subgoals.

WHY THE LMSS THEOREM HOLDS

Suppose we have a query Q with n subgoals and there is a solution S with more than n subgoals.

The expansion E of S must be contained in Query Q, which means that there is a containment mapping from Q to E.

We remove from S all subgoals whose expansion was not the target of one of Q’s subgoals under the containment mapping.

CONTD..

We would have a new conjunctive query S’ with at most n subgoals.

If E’ is the expansion of S’ then, E’ is a subset of Q.

S is a subset of S’ as there is an identity mapping.

Thus S need not be among the solutions to query Q.

SECTION 21.7

INTRODUCTION

ENTITY RESOLUTION: Entity resolution is a problem that arises in many information integration scenarios.

It refers to determining whether two records or tuples do or do not represent the same person, organization, place or other entity.

DECIDING WHETHER RECORDS REPRESENT A COMMON ENTITY

Two records represent the same individual if the two records have similar values for each of the fields associated with those records.

It is not sufficient that the values of corresponding fields be identical because of following reasons:

1. Misspellings2. Variant Names3. Misunderstanding of Names

CONTINUE: DECIDING WHETHER RECORDS REPRESENT A COMMON ENTITY

4. Evolution of Values5. Abbreviations

Thus when deciding whether two records representthe same entity, we need to look carefully at thekinds of discrepancies and use the test thatmeasures the similarity of records.

DECIDING WHETHER RECORDS REPRESENTS A COMMON ENTITY - EDIT DISTANCE

First approach to measure the similarity of records is Edit Distance.

Values that are strings can be compared by counting the number of insertions and deletions of characters it takes to turn one string into another.

So the records represent the same entity if their similarity measure is below a given threshold.

DECIDING WHETHER RECORDS REPRESENTS A COMMON ENTITY - NORMALIZATION

To normalize records by replacing certain substrings by others. For instance: we can use the table of abbreviations and replace abbreviations by what they normally stand for.

Once normalize we can use the edit distance to measure the difference between normalized values in the fields.

MERGING SIMILAR RECORDS

Merging refers to removal of redundant data in two records.

There are many merge rules:1. Set the field in which the records disagree to the

empty string.2. (i) Merge by taking the union of the values in each

field(ii) Declare two records similar if at least two of the three fields have a nonempty intersection.

CONTINUE: MERGING SIMILAR RECORDS

Name Address Phone1. Susan 123 Oak St. 818-555-1234 2. Susan 456 Maple St. 818-555-12343. Susan 456 Maple St. 213-555-5678

After Merging

Name Address Phone(1-2-3) Susan {123 Oak St.,456 Maple St} {818-555-

1234, 213- 555-5678}

USEFUL PROPERTIES OF SIMILARITY AND MERGE FUNCTIONS

The following properties say that merge operation is a semi lattice:

1. Idempotence: Merge of a record with itself yeilds the same record.

2. Commutativity: Order of merged records does not matter

3. Associativity : The order in which we group records for a merger should not matter.

CONTINUE: USEFUL PROPERTIES OF SIMILARITY AND MERGE FUNCTIONS

There are some other properties that we expect similarity relationship to have:

• Idempotence for similarity: A record is always similar to itself

• Commutativity of similarity: In deciding whether two records are similar it does not matter in which order we list them

• Representability: If r is similar to some other record s, but s is instead merged with some other record t, then r remains similar to the merger of s and t and can be merged with that record.

R-SWOOSH ALGORITHM FOR ICAR RECORDS

Input: A set of records I, similarity function and a merge function.

Output: A set of merged records O. Method:

O:= emptyset; WHILE I is not empty DO BEGIN

Let r be any record in I; Find, if possible, some record s in O that is similar to r; IF no record s exists THEN

move r from I to O ELSE BEGIN

delete r from I;

delete s from O;add the merger of r

and s to I; END; END;

OTHER APPROACHES TO ENTITY RESOLUTION

The other approaches to entity resolution are :

Non- ICAR Datasets Clustering Partitioning

OTHER APPROACHES TO ENTITY RESOLUTION - NON ICAR DATASETS

Non ICAR Datasets : We can define a dominance relation r<=s that means record s contains all the information contained in record r.

If so, then we can eliminate record r from further consideration.

OTHER APPROACHES TO ENTITY RESOLUTION – CLUSTERING & PARTITIONING

Clustering: Clustering refers to creating clusters for members that are similar to each other

Partitioning: We can group the records, perhaps several times, into groups that are likely to contain similar records and look only within each group for pairs of similar records.

REFERENCES H. Garcia-Molina, J. Ullman, and J. Widom, “Database

System: The Complete Book,” second edition: p.897-913, Prentice Hall, New Jersey, 2008

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D ATABASE S YSTEM P RINCIPLES Ankit Patel 112_226