SECONDARY
STORAGE
MANAGEMENT
SECTIONS 13.1 – 13.3
05/11/09 15:18
Sanuja Dabade & Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
• 13.1 The Memory Hierarchy
13.1.1 The Memory Hierarchy
13.1.2 Transfer of Data Between
Levels
o 13.1.3 Volatile and Nonvolatile
Storage
o 13.1.4 Virtual Memory
o
o
• 13.2 Disks
o
o
o
13.2.1 Mechanics of Disks
13.2.2 The Disk Controller
13.2.3 Disk Access Characteristics
05/11/09 15:18
Presentation Outline (con’t)
• 13.3 Accelerating Access to
Secondary Storage
13.3.1 The I/O Model of Computation
13.3.2 Organizing Data by Cylinders
13.3.3 Using Multiple Disks
13.3.4 Mirroring Disks
13.3.5 Disk Scheduling and the
Elevator Algorithm
o 13.3.6 Prefetching and Large-Scale
Buffering
o
o
o
o
o
05/11/09 15:18
13.1.1 Memory Hierarchy
• Several components for data storage having
different data capacities available
• Cost per byte to store data also varies
• Device with smallest capacity offer the fastest
speed with highest cost per bit
05/11/09 15:18
Memory Hierarchy Diagram
Programs,
Main Memory DBMS’s
As Visual Memory
DBMS
Tertiary Storage
Disk File
System
Main Memory
Cache
05/11/09 15:18
13.1.1 Memory Hierarchy
• Cache
Lowest level of the hierarchy
Data items are copies of certain locations of main
memory
o Sometimes, values in cache are changed and
corresponding changes to main memory are
delayed
o Machine looks for instructions as well as data for
those instructions in the cache
o Holds limited amount of data
o
o
05/11/09 15:18
13.1.1 Memory Hierarchy
(con’t)(updated)
• No need to update the data in main memory
immediately in a single processor computer
• In multiple processors data is updated
immediately to main memory….called as write
through
• Data and instructions are moved to cache
from main memory when they are needed by
the processor.
05/11/09 15:18
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
05/11/09 15:18
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
05/11/09 15:18
Tertiary Storage(updated)
• Holds data volumes in terabytes
• Used for databases much larger than what
can be stored on disk
• Higher read/write times than secondary
storage.
• Retrieval takes seconds or minutes.
05/11/09 15:18
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 blocks
• Entire blocks are moved to and from memory
called a buffer
05/11/09 15:18
13.1.2 Transfer of Data Between
level (cont’d)
• 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
• Same idea applies to other hierarchy levels
05/11/09 15:18
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
05/11/09 15:18
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
05/11/09 15:18
13.2.1 Mechanism of Disk
• Mechanisms of Disks
Use of secondary storage is one of the important
characteristic of DBMS
o Consists of 2 moving pieces of a disk
o
 1. disk assembly
 2. head assembly
o
o
o
Disk assembly consists of 1 or more platters
Platters rotate around a central spindle
Bits are stored on upper and lower surfaces of
platters
05/11/09 15:18
13.2.1 Mechanism of
Disk(updated)
• Disk is organized into tracks
The tracks that are at fixed radius from center
among all the surfaces form one cylinder
• Tracks are organized into sectors
• Tracks are the segments of circle separated
by gaps and are used to help identify the
beginnings of sectors.
• The head assembly holds the disk heads.
• A head reads/alters the magnetism passing
under it.
05/11/09 15:18
05/11/09 15:18
13.2.2 Disk Controller(updated)
• One or more disks are controlled by disk
controllers
• Disks controllers are capable of
Controlling the mechanical actuator that moves
the head assembly
o Selecting the sector from among all those in the
cylinder at which heads are positioned
o Transferring bits between desired sector and
main memory
o Possible buffering an entire track in the local
memory of the disk controller.
o
05/11/09 15:18
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’
o The disk controller waits while the first sector of
the block moves under the head. This is a
‘rotational latency’
o 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’
o
05/11/09 15:18
SECONDARY
STORAGE
MANAGEMENT
SECTION 13.3
05/11/09 15:21
Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
• 13.3 Accelerating Access to
Secondary Storage
13.3.1 The I/O Model of Computation
13.3.2 Organizing Data by Cylinders
13.3.3 Using Multiple Disks
13.3.4 Mirroring Disks
13.3.5 Disk Scheduling and the
Elevator Algorithm
o 13.3.6 Prefetching and Large-Scale
Buffering
o
o
o
o
o
05/11/09 15:21
13.3 Accelerating Access to
Secondary Storage
• Several approaches for more-efficiently accessing data
in secondary storage:
o Place blocks that are together in the same cylinder.
o Divide the data among multiple disks.
o Mirror disks.
o Use disk-scheduling algorithms.
o 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.
05/11/09 15:21
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.
o
This 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.
o
o
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.
05/11/09 15:21
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.
o If all blocks are consecutively stored on 1
cylinder:
o
 6.46ms + 8.33ms * 16 = 139ms
(1 average seek) (time per rotation) (# rotations)
05/11/09 15:21
13.3.3 Using Multiple Disks
• If we have n disks, read/write performance will
increase by a factor of n.
• Striping – distributing a relation across multiple
disks following this pattern:
o
o
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.
o
6.46ms + (8.33ms * (16/4)) = 39.8ms
(1 average seek) (time per rotation) (# rotations)
05/11/09 15:21
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.
05/11/09 15:21
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:
o
o
o
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
05/11/09 15:21
13.3.5 Disk Scheduling and the
Elevator Problem (con’t)
Events:
Head starting point
Request data at 8000
Request data at
24000
Request data at
56000
Get data at 8000
Request data at
16000
Get data at 24000
Request data at
64000
Get data at 56000
Request Data at
40000
Get data at 64000
Get data at 40000
Get data at 16000
64000
56000
48000
40000
32000
24000
16000
8000
Current time
13.6
26.9
34.2
45.5
56.8
4.3
10
20
30
0
data
time
8000..
4.3
24000..
24000..
56000..
56000..
13.6
13.6
26.9
26.9
64000..
64000..
34.2
34.2
40000..
40000..
45.5
45.5
16000..
56.8
05/11/09 15:21
13.3.5 Disk Scheduling and the
Elevator Problem (con’t)
Elevator
Algorithm
FIFO
Algorithm
data
data
time
time
8000..
4.3
8000..
4.3
24000..
13.6
24000..
13.6
56000..
26.9
56000..
26.9
64000..
34.2
16000..
42.2
40000..
45.5
64000..
59.5
16000..
56.8
40000..
70.8
05/11/09 15:21
13.3.6 Prefetching and Large-Scale
Buffering(updated)
• 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 is called prefetching or sometimes double
buffering.
05/11/09 15:21
Questions
05/11/09 15:21
• Disk failure ways and their mitigatioPppppn
• Pri
Ways in which disks can fail• Intermittent failure.
• Media Decay.
• Write failure.
• Disk Crash.
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.
• Serious form of failure.
• Bit/Bits are permanently corrupted.
• Impossible to read a sector correctly even after
many trials.
• Stable storage technique for organizing a disk is
used to avoid this failure.
Write failure
• Attempt to write a sector is not possible.
• Attempt to retrieve previously written sector is
unsuccessful.
• Possible reason – power outage while writing of the
sector.
• Stable Storage Technique can be used to avoid this.
Disk Crash
• Most serious form of disk failure.
• Entire disk becomes unreadable, suddenly and
permanently.
• RAID techniques can be used for coping with disk
crashes.
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.
More on Intermittent Failures..
• 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.
Contd..
• 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.
• If it is good sector, then the write was correct
otherwise the write was unsuccessful and must be
repeated.
Checksums.
• Technique used to determine the good/bad status of
a sector.
• Each sector has some additional bits called the
checksum that are set depending on the values of
the data bits in that sector.
• If checksum is not proper on reading, then there is
an error in reading.
Checksums(contd..)
• There is a small chance that the block was not read
correctly even if the checksum is proper.
• The probability of correctness can be increased by
using many checksum bits.
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.
Checksum calculation(contd..)
• 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.
Examples…
• A sequence of bits 01101000 has odd number of
1’s. The parity bit will be 1. So the sequence with the
parity bit will now be 011010001.
• A sequence of bits 11101110 will have an even
parity as it has even number of 1’s. So with the
parity bit 0, the sequence will be 111011100.
Checksum calculation(contd..)
• Any one-bit error in reading or writing the bits results
in a sequence of bits that has odd-parity.
• The disk controller can count the number of 1’s and
can determine if the sector has odd parity in the
presence of an error.
Odds.
• There are chances that more than one bit can be
corrupted and the error can be unnoticed.
• Increasing the number of parity bits can increase the
chances of detecting errors.
• In general, if there are n independent bits as
checksum, the chances of error will be one in 2n.
Stable Storage.
• Checksums can detect the error but cannot correct
it.
• Sometimes we overwrite the previous contents of a
sector and yet cannot read the new contents
correctly.
• To deal with these problems, Stable Storage policy
can be implemented on the disks.
Stable-Storage(contd..)
• 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.
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.
• Also, If the read function returns a good value w for
either XL or XR then it is assumed that w is the true
value of X.
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.
1.Repeat (1) for XR.
Stable-Storage Reading Policy:
• The policy is to alternate trying to read XL and XR
until a good value is returned.
• If a good value is not returned after pre chosen
number of tries, then it is assumed that X is truly
unreadable.
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.
• The probability of both failing is extremely small.
Error-Handling
Capabilities(contd..)
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.
• Half of the sector may be written with part of new
value of X, while the other half remains as it was.
Error-Handling
Capabilities(contd..)
• The possible cases when the system becomes
available:
1.The failure occurred when writing to XL. Then XL is
considered bad. Since XR was never changed, its
status is good. We can make a copy of XR into XL,
which is the old value of X.
2.The failure occurred after XL is written. Then XL will
have the good status and XR which has the old
value of XR has bad status. We can copy the new
value of X to XR from XL.
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.
• 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.
Recovery from Disk
Crashes(contd..)
• 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.
• 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.
Disk Failures
Xiaqing He
ID: 204
Dr. Lin
Content
1)Focus on :
“How to recover from disk crashes”
common term RAID
“redundancy array of independent disks”
2)Several schemes to recover from disk
crashes:
•
•
•
•
Mirroring—RAID level 1;
Parity checks--RAID 4;
Improvement--RAID 5;
RAID 6;
1) Mirroring
• The simplest scheme to recovery from
Disk Crashes
• How does Mirror work?
-- making two or more copied of the data
on different disks
• Benefit:
-- save data in case of one disk will fail;
-- divide data on several disks and let access to
several blocks at once
1) Mirroring (con’t)
• 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.
• Possibility:
Suppose:
• One disk: mean time to failure = 10 years;
• One of the two disk: average of mean time to failure = 5 years;
• The process of replacing the failed disk= 3 hours=1/2920 year;
So:
• the possibility of the mirror disk will fail=1/10 * 1/2,920 =1/29,200;
• The possibility of data loss by mirroring: 1/5 * 1/29,200 = 1/146,000
2)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.
• Example
2)Parity Blocks(con’t)___Example
Data disks:
• Disk1: 11110000
• Disk2: 10101010
• Disk3: 00111000
Redundant disk:
• Disk4: 01100010
2)RAID 4 (con’t)
• Reading
-- Similar with reading blocks from any disk;
• Writing
1)change the data disk;
2)change the corresponding block of the redundant
disk;
• Why?
-- hold the parity checks for the corresponding blocks
of all the data disks
2)RAID 4 (con’t) _ writing
For 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;
2)RAID 4 (con’t) _ writing_Example
•
•
•
•
Data disks:
Disk1: 11110000
Disk2: 10101010 • 01100110
Disk3: 00111000
•
•
•
•
•
to do:
Modulo-2 sum of the old and new version of disk 2: 11001100
So, we need to change the positions 1,2,5,6 of the redundant disk.
Redundant disk:
Disk4: 01100010 • 10101110
2)RAID 4 (con’t) _failure recovery
• Redundant disk crash:
-- swap a new one and recomputed data from all the data disks;
• One of Data disks crash:
-- swap a new one;
-- recomputed data from the other disks including data disks and redundant disk;
• How to recomputed? (same rule, that’s why there will be some
improvement)
-- take modulo-2 sum of all the corresponding bits of all the other disks
3) 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”
So, there is no need to retreat one as redundant disk and others as data
disks
3) RAID 5 (con’t)
• 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 i cylinder
of disk J as redundant if J is the remainder
when I is divided by n+1;
th
• Example;
3) RAID 5 (con’t)_example
N=3;
• The first disk, labeled as 0 : 4,8,12…;
• The second disk, labeled as 1 : 1,5,9…;
• The third disk, labeled as 2 : 2,6,10…;
• ……….
Suppose all the 4 disks are equally likely to be
written, for one of the 4 disks, the possibility of being
written:
• 1/4 + 3 /4 * 1/3 =1/2
• If N=m => 1/m +(m-1)/m * 1/(m-1) = 2/m
4) 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);
• How to set up this 3*7 matrix ?
(why is 3? – there are 3 redundant disks)
1)every column values three 1’s and 0’s except for all
three 0’s;
2) column of the redundant disk has single 1’s;
3) column of the data disk has at least two 1’s;
4) Coping with multiple disk crashes (con’t)
• 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
4) Coping with multiple disk crashes (con’t)
• In those system which has 4 data disks and 3
redundant disk, how they can correct up to 2 disk
crashes?
• Suppose disk a and b failed:
• find some row r (in 3*7 matrix)in which the column
for a and b are different (suppose a is 0’s and b is
1’s);
• Compute the correct b by taking modulo-2 sum of
the corresponding bits from all the other disks
other than b which have 1’s in row r;
• After getting the correct b, Compute the correct a
with all other disks available;
• Example
4) 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
4) Coping with multiple disk crashes
(con’t)_example
First block of all the disks
disk contents
1) 11110000
2) 10101010
3) 00111000
4) 01000001
5) 01100010
6) 00011011
7) 10001001
4) 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
4) Coping with multiple disk crashes
(con’t)_example
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
13.5 Arranging data on disk
Meghna Jain
ID-205
CS257
Prof: Dr. T.Y.Lin
(updated)
Data elements are represented as records, which stores
in consecutive bytes in same same disk block.
Basic layout techniques of storing data :
Fixed-Length Records
Records have fixed-length fields, one for each attribute
of the tuple.
Allocation criteria - data should start at word
boundary.
Fixed Length record header
1. A pointer to record schema.
2. The length of the record.
3. Timestamps to indicate last modified or last read.
4. Pointers to the fields of the records.
Example
CREATE TABLE employee(
name CHAR(30) PRIMARY KEY,
address VARCHAR(255),
gender CHAR(1),
birthdate DATE
);
Data should start at word boundary and contain header and four fields name,
address, gender and birthdate.
Packing Fixed-Length Records into Blocks :
Records are stored in the form of blocks on the disk and they
move into main memory when we need to update or access
them.
A block header is written first, and it is followed by series of
blocks.
Block header contains the following information
:
• Links to one or more blocks that are part of a
network of blocks.
• Information about the role played by this block in such
a network.
• Information about the relation, the tuples in this block
belong to.
• A "directory" giving the offset of each record in the
block.
• Time stamp(s) to indicate time of the block's last
modification and/or access.
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.
Thank You
13.6 REPRESENTING BLOCK AND
RECORD ADDRESSES
Ramya Karri
CS257 Section 2
ID: 206
INTRODUCTION(updated)
• 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
o In Secondary Memory: sequence of bytes describe the
location of the block in the overall system
o
• Sequence of Bytes describe the location of the
block : the device Id for the disk, Cylinder number,
etc.
• The record's address is block address and the
offset of the first byte of the record within the block.
ADDRESSES IN CLIENT-SERVER
SYSTEMS
• 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.
o Logical Addresses: arbitrary string of bytes of some
fixed length
o
• Physical Address bits are used to indicate:
o
o
o
o
o
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
ADDRESSES IN CLIENT-SERVER
SYSTEMS (CONTD..)(updated)
• Logical address is an arbitrary string of bytes of
some fixed length.
• Map Table relates logical addresses to physical
addresses.
Logical
Physical
Logical Address
Physical Address
LOGICAL AND STRUCTURED
ADDRESSES
• Purpose of logical address?
• Gives more flexibility, when we
o
Move the record around within the block
o Move the record to another block
• Gives us an option of deciding what to do when a
record is deleted?
Unused
Recor Recor Recor Recor
d4
d3
d2
d1
Offset table
Header
POINTER SWIZZLING(updated)
• 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:
o database address: address on the disk
o memory address, if the item is in virtual memory
When the item is in main memory, it is more
efficient to use memory address.
A translation table translates database
addresses currently in virtual memory to their
current memory address.
POINTER SWIZZLING (CONTD…)
• Translation Table: Maps database address to
memory address
Dbaddr
Mem-addr
Database address
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
POINTER SWIZZLING (CONTD…)
• Pointer consists of the following two fields
o
o
o
Bit indicating the type of address
Database or memory address
Example 13.17
Disk
Memory
Swizzled
Block 1
Block 1
Unswizzled
Block 2
EXAMPLE 13.7
• Block 1 has a record with pointers to a second
record on the same block and to a record on
another block
• If Block 1 is copied to the memory
The first pointer which points within Block 1 can be
swizzled so it points directly to the memory address of
the target record
o Since Block 2 is not in memory, we cannot swizzle the
second pointer
o
POINTER SWIZZLING (CONTD…)
• Three types of swizzling
o
Automatic Swizzling
 As soon as block is brought into memory,
swizzle all relevant pointers.
o
Swizzling on Demand
 Only swizzle a pointer if and when it is
actually followed.
o
No Swizzling
 Pointers are not swizzled they are accesses
using the database address.
PROGRAMMER CONTROL OF
SWIZZLING(updated)
The programmer calls for the pointers to be swizzled
only as needed.
• Unswizzling
When a block is moved from memory back to disk, all
pointers must go back to database (disk) addresses
o Use translation table again
o Important to have an efficient data structure for the
translation table
o
PINNED RECORDS AND
BLOCKS(updated)
• A block in memory is said to be pinned if it cannot
be written back to disk safely.
• Header of the block has one bit telling whether or
not the block is pinned.
• 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
o Unswizzle those pointers (use translation table to
replace the memory addresses with database (disk)
addresses
o
o
Thank you
Eswara Satya Pavan Rajesh Pinapala
CS 257
ID: 221
Topics
•
•
•
•
•
Records with Variable Length Fields
Records with Repeating Fields
Variable Format Records
Records that do not fit in a block
BLOBS
Example
name
address
gender
birth date
0 30 286 287 297
Fig 1 : Movie star record with four fields
Records with Variable Fields
An effective way to represent variable length
records is as follows
• Fixed length fields are Kept ahead of the
variable length fields
• Record header contains
o Length of the record
o Pointers to the beginning of all variable
length fields except the first one.
Records with Variable Length
Fields
header information
record length
to address
gender birth date
name
address
Figure 2 : A Movie Star record with name and address
implemented as variable length character strings
Records with Repeating Fields
• Records contains variable number of occurrences of a
field F
• All occurrences of field F are grouped together and the
record
header contains a pointer to the first occurrence of field F
• L bytes are devoted to one instance of field F
• Locating an occurrence of field F within the record
o 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
o We stop upon reaching the offset of the field F.
Records with Repeating Fields
other header
information
record length
to address
to movie pointers
name
address
pointers to movies
Figure 3 : A record with a repeating group of references to
movies
Records with Repeating Fields
record header to name length of name
information
to address
length of address
to movie references
number of
references
address
name
Figure 4 : Storing variable-length fields separately from the
record
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
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
A compromise strategy is to allocate a fixed
portion of the record for the repeating fields
• If the number of repeating fields is lesser than
allocated space, then there will be some unused
space
• If the number of repeating fields is greater than
allocated space, then extra fields are stored in a
different location and
• Pointer to that location and count of additional
occurrences is stored in the record
Variable Format Records
• Records that do not have fixed schema
• Variable format records are represented by sequence of
tagged fields
• Each of the tagged fields consist of information
o Attribute or field name
o Type of the field
o Length of the field
o Value of the field
• Why use tagged fields
o Information – Integration applications
o Records with a very flexible schema
Variable Format Records
code for name
code for string type
length
N
S 14
code for restaurant owned
Clint Eastwood R
code for string type
length
S
Fig 5 : A record with tagged fields
16
Hog’s Breath
Inn
Records that do not fit in a block
• When the length of a record is greater than block size
,then
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
Spanned Records
• Spanned records require the following extra
header information
• A bit indicates whether it is fragment or not
• A bit indicates whether it is first or last fragment
of
a record
• Pointers to the next or previous fragment for the
same record
Records that do not fit in a block
block header
record header
record 1
block 1
record
2-a
record
2-b
record 3
block 2
Figure 6 : Storing spanned records across blocks
BLOBS(updated)
• Large binary objects are called BLOBS
e.g. : audio files, video files
Storage of BLOBS – They must be stored
on a sequence of blocks allocated
consecutively on the cylinders of the disk.
They can be striped across several blocks
for faster retrieval.
Retrieval of BLOBS-pass only small fields of
the record first and allow the client to
request blocks of BLOB one at a time.
Record Modifications
Chapter 13
Section 13.8
Neha Samant
CS 257
(Section II) Id 222
5/14/2009
Modification types
• Insertion
• Deletion
• Update
5/14/2009
Insertion(updated)
• Insertion of records without order
Records can be placed in a block with empty space or in a new block.
Insertion of records in fixed order
• Space available in the block
• No space available in the block (outside the block)
Structured address
Pointer to a record from outside the block. It is the block address and the
location of the entry for the record in the offset table.
5/14/2009
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.
Offset
table
header
unused
Record 4
5/14/2009
Record 3
Record 2
Record 1
Insertion in fixed order
No space available within the block (outside the block)
• 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.
o 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.
o
• Create an overflow block
o
o
o
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.
Block B
5/14/2009
Overflow
block for B
Deletion
• Recover space after deletion
o
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.
5/14/2009
Deletion
• Use of tombstone
o
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.
• If pointers go to fixed locations from which the location of the record is found
then we put the tombstone in that fixed location. (See examples)
• Where a tombstone is placed depends on the nature of the record pointers.
• Map table is used to translate logical record address to physical address.
5/14/2009
Deletion
• Use of tombstone
o
If we need to replace records by tombstones, place the bit that serves as the
tombstone at the beginning of the record.
• This bit remains the record location and subsequent bytes can be reused for
another record
Record 1
Record 2
Record 1 can be replaced, but the tombstone remains, record 2 has no
tombstone and can be seen when we follow a pointer to it.
5/14/2009
Update
• Fixed Length update
No effect on storage system as it occupies same space as before
update.
• Variable length update
o Longer length
o Short length
5/14/2009
Update
Variable length update (longer length)
• Stored on the same block:
o
o
Sliding records
Creation of overflow block.
• Stored on another block
o
o
5/14/2009
Move records around that block
Create a new block for storing variable length fields.
Update
Variable length update (Shorter length)
• Same as deletion
o
o
5/14/2009
Recover space
Consolidate space.
BTrees & Bitmap Indexes
14.2
DATABASE SYSTEMS – The Complete
Book
Presented By: Under the supervision of:
Maciej Kicinski Dr.T.Y.Lin
B Trees ►►
Structure(updated)
• A balanced tree, meaning that all paths from the
leaf node have the same length.
They are at least 3 layers – root, intermediate layer and
leaves.
• 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.
Structure
●A
typical node >
● a typical interior
node would have
pointers pointing to
leaves with out
values
● a typical leaf would
have pointers point
to records
N search keys
N+1 pointers.
Structure
●A
typical node >
● a typical interior
node would have
pointers pointing to
leaves with out
values
● a typical leaf would
have pointers point
to records
N search keys
N+1 pointers
Application(updated)
• The (n+1) pointer of the leaf node points to the
next leaf.
• The search key of the Btree is the primary key
for the data file.
• Data file is sorted by its primary key.
• Data file is sorted by an attribute that is not a
key,and this attribute is the search key for the
Btree.
Lookup
If at an interior node, choose the correct pointer to use. This
is done by comparing keys to search value.
Lookup
If at a leaf node, choose the key that matches what
you are looking for and the pointer for that leads
to the data.
Insertion
• When inserting, choose the correct leaf node to
put pointer to data.
• If node is full, create a new node and split keys
between the two.
• 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.
Deletion
Perform lookup to find node to delete and delete it.
If 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.
Efficiency
Btrees allow lookup, insertion, and deletion of
records using very few disk I/Os.
Each level of a Btree would require one read. Then
you would follow the pointer of that to the next or
final read.
Efficiency
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.
References
References
BTrees & Bitmap Indexes
14.7
DATABASE SYSTEMS – The Complete
Book
Presented By: Under the supervision of:
Deepti Kundu Dr.T.Y.Lin
Bitmap Indexes ►►
Definition(updated)
A bitmap index for a field F is a collection of bitvectors of length n, one for each possible value
that may appear in that field F.[1]
The vector for value v has 1 in position i if
the ith record has v in field F, and it has 0
there if not.
What does that mean?
• Assume relation R
with
o
o
o
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…
• A bitmap for attribute B is:
Value
foo
bar
baz
Vector
100100
010010
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?
• 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 little more…
• A bitmap index for attribute A of relation R is:
o A collection of bit-vectors
o The number of bit-vectors = the number of distinct values
of A in R.
o The length of each bit-vector = the cardinality of R.
o 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 Indexes
• Very efficient when used for partial match queries.[3]
• They offer the advantage of buckets [2]
o 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]
Another Example
Multidimensional Array of multiple types
{(5,d),(79,t),(4,d),(79,d),(5,t),(6,a)}
5 = 100010
79 = 010100
4 = 001000
6 = 000001
d = 101100
t = 010010
a = 000001
Example Continued…
{(5,d),(79,t),(4,d),(79,d),(5,t),(6,a)}
Searching for items is easy, just AND together.
To search for (5,d)
5 = 100010
d = 101100
100010 AND 101100 = 100000
The location of the
record has been traced!
Compressed Bitmaps
• Assume:
o The number of records in R are n
o 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.
o Opportunity to encode
• A common encoding approach is called run-length
encoding.[1]
Run-length encoding
• Represents runs
o 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:
o First computing how many bits are needed to represent i, Say k
o Then represent the run by k-1 1’s and a single 0 followed by k bits which
represent i in binary.
o The encoding for i = 1 is 01. k = 1
o 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 Example
• Let us decode the sequence 11101101001011
• Staring at the beginning (left most bit):
o 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
o Second run: 001011
k=1
i=0
o Last run: 1011
k=1
i=3
• Our entire run length is thus 13,0,3, hence our bit-vector is:
0000000000000110001
Managing Bitmap Indexes
1) How do you find a specific bit-vector for a
value efficiently?
2) After selecting results that match, how do you retrieve
the results efficiently?
3) When data is changed, do you you alter bitmap
index?
1) Finding bit vectors
• Think of each bit-vector as a key to a value.[1]
• Any secondary storage technique will be efficient in
retrieving the values.[1]
• Create secondary key with the attribute value as a
search key [3]
o Btree
o Hash
2) Finding Records
• Create secondary key with the record number as a
search key [3]
• Or in other words,
o Once you learn that you need record k, you can create a
secondary index using the kth position as a search key.[1]
3) Handling Modifications
Two things to remember:
Record numbers must remain fixed once
assigned
Changes to data file require changes to
bitmap index
Deletion
 Tombstone replaces deleted record
 Corresponding bit is set to 0
Insertion
 Record assigned the next record number.
 A bit of value 0 or 1 is appended to each bit
vector
 If 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.
References
[1] Database Systems : The Complete Book - Hector Garcia-Molina, Jeffrey D. Ullman,
Jennifer D. Widom
[2] http://en.wikipedia.org/wiki/Bitmap_index#Example
[3] faculty.kfupm.edu.sa/ICS/adam/ICS541/L10-md-bitmap-indexing.ppt
[4] http://csis.bitspilani.ac.in/faculty/goel/Data%20Warehousing/Lecture%20Notes/Lecture%20%239%20%20Bitmap%20Indexes%20in%20DW.doc (- a good doc file to read the concepts of bitmap
indexes)
Concurrency Control
18.1 – 18.2
Chiu Luk
CS257 Database Systems Principles
Spring 2009
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
T1
Read(A); A  A+100
Write(A);
Read(B); B  B+100;
Write(B);
precedes T2
T2
A
25
B
25
125
Read(A);A  A2;
Write(A);
Read(B);B  B2;
Write(B);
125
250
250
250
250
Serial Schedule T2 precedes Tl
T1
Read(A); A  A+100
Write(A);
Read(B); B  B+100;
Write(B);
T2
Read(A);A  A2;
Write(A);
Read(B);B  B2;
Write(B);
A
25
B
25
50
50
150
150
150
150
serializable, but not serial, schedule
T1
Read(A); A  A+100
Write(A);
Read(B); B  B+100;
Write(B);
T2
Read(A);A  A2;
Write(A);
A
25
B
25
125
250
Read(B);B  B2;
Write(B);
125
250
r1(A); w1 (A): r2(A); w2(A); r1 (B); w1 (B); r2(B); w2(B);
250
250
nonserializable schedule
T1
Read(A); A  A+100
Write(A);
Read(B); B  B+100;
Write(B);
T2
Read(A);A  A2;
Write(A);
Read(B);B  B2;
Write(B);
A
25
B
25
125
250
50
150
250 150
schedule that is serializable only because of the detailed
behavior of the transactions
T1
Read(A); A  A+100
Write(A);
Read(B); B  B+100;
Write(B);
•
T2’
Read(A);A  A1;
Write(A);
Read(B);B  B1;
Write(B);
A
25
125
B
25
125
25
125
regardless of the consistent initial state: the final state will be consistent.
125 125
A Notation for Transactions and
Schedules.(added)
• An action is the expression of the form
ri(X) and wi(X) meaning that transaction Ti
reads or writes the database element X.
• A transaction Ti is a sequence of actions
with subscript i.
• A schedule S of a set of Transactions is a
sequence of actions. The actions
appearing in each transaction appear in
the same order in the schedule S.
Non-Conflicting Actions
Two actions are non-conflicting if whenever they
occur consecutively in a schedule, swapping them
does not affect the final state produced by the
schedule. Otherwise, they are conflicting.
Conflicting Actions: General
Rules
• Two actions of the same transaction
conflict:
o r1(A)
w1(B)
• Two actions over the same database
element conflict, if one of them is a write
o r1(A) w2(A)
o w1(A) w2(A)
Conflict actions
• Two or more actions are said to be in conflict if:
o The actions belong to different transactions.
o At least one of the actions is a write operation.
o The actions access the same object (read or write).
• The following set of actions is conflicting:
o T1:R(X), T2:W(X), T3:W(X)
• While the following sets of actions are not:
o T1:R(X), T2:R(X), T3:R(X)
o 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 nonconflicting
swaps.
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 nonconflicting swaps of adjacent actions.
• We shall call a schedule conflict-serializable if it is
conflict-equivalent to a serial schedule.
conflict-serializable
T1
R(A)
W(A)
T2
R(A)
R(B)
W(A)
W(B)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(B)
T2
R(A)
W(A)
W(B)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(A)
T2
R(B)
W(B)
W(A)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(A)
W(B)
T2
Serial
Schedule
R(B)
W(A)
R(B)
W(B)
END of present
References
•
•
•
•
•
Database Systems: The Complete Book (2nd Edition) (Hardcover) by Hector GarciaMolina (Author), Jeffrey D. Ullman (Author), Jennifer Widom (Author) Publisher :
Prenctice Hall.
http://en.wikipedia.org/wiki/Concurrency_control
http://www.utdallas.edu/~mxk055100/db07files/serilizable-defs.ppt
http://en.wikipedia.org/wiki/Schedule_(computer_science)#Serializable
http://www.cs.duke.edu/~shivnath/courses/fall06/Lectures/11_serial.ppt
Concurrency Control
By Donavon Norwood
Ankit Patel
Aniket Mulye
INTRODUCTION
• Enforcing serializability by locks
o
o
o
Locks
Locking scheduler
Two phase locking
• Locking systems with several lock modes
o
o
o
o
Shared and exclusive locks
Compatibility matrices
Upgrading/updating locks
Incrementing locks
Locks(updated)
It works like as follows :
• A request from transaction
• Scheduler checks in the lock table
• Generates a serializable schedule of actions.
A scheduler uses a lock table to help perform its job.
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.
o Unlocking an element is compulsory.
o
• Legality of schedules
o
No two transactions can acquire the lock on same
element without the prior one releasing it.
Locking scheduler(updated)
• Grants lock requests only if it is in a legal
schedule.
• Lock table stores the information about current
locks on the elements.
• The lock table is a relation
Locks(element,transaction), consisting of pairs
(X,T) such that transaction T currently has a lock
on database element X.
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:
o
Obtain all the locks and no unlocks allowed.
• The shrinking phase:
o
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.
o Rigorous two phase locking : Transaction holds
all locks till commit / abort.
o
• 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.
Concurrency Control: 18.4
Locking Systems with Several
Lock Modes
CS257 Spring/2009
Professor: Tsau Lin
Student: Suntorn Sae-Eung
ID: 212
18.4 Locking Systems with
Several Lock Modes
• In 18.3, if a transaction must lock a database
element (X) either reads or writes,
o
No reason why several transactions could not
read X at the same time, as long as none write X
• Introduce locking schemes
o
o
Shared/Read Lock ( For Reading)
Exclusive/Write Lock( For Writing)
18.4.1 Shared & Exclusive
Locks
• Transaction Consistency
o
o
o
Cannot write without Exclusive Lock
Cannot read without holding some lock
Consider lock for writing is “stronger” than for reading
• This basically works on 2 principles
1. A read action can only proceed a shared or an
exclusive lock
2. 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
Lock • R/W • Unlock
• 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
18.4.1 Shared & Exclusive
Locks (cont.)
• Legality of Schedules
o
An 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 lockmanagement policies
Rows correspond to a lock held on an element by
another transaction
o Columns correspond to mode of lock requested.
o Example :
o
Lock requested
Lock in
hold
S
X
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 retry and
succeed
‘B’ is released
• T1 cannot take an exclusive lock on B until all locks on
B are released.
18.4.3 Upgrading Locks (cont.)
• Upgrading can simply cause a “Deadlock”.
o
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.
o An “Update lock” is allowed to grant on X when
there are already shared locks on X.
o Once there is an “Update lock,” it prevents
additional any kinds of lock, and later changes to
a write (exclusive) lock.
o
• Notation: uli (X)
18.4.4 Update locks (cont.)
• Example
18.4.4 Update locks (cont.)
• Compatibility matrix (asymmetric)
Lock requested
Lock in
hold
S
X
U
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),
o
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.)
CASE 1
T1: INC (A,2)
A=7
A=5
T2: INC (A,10)
CASE 2
T2: INC (A,10)
A=17
A=15
T1: INC (A,2)
18.4.5 Increment Locks (cont.)
• What if
T1: INC (A,2)
A=5
A=7
T2: INC (A,10)
A=15
A=5
T2: INC (A,10)
A=15
T1: INC (A,2)
A=5
A=5
A=7
A !=
17
18.4.5 Increment Locks (cont.)
• INC (A, c) –
o
Increment action of writing on database element
A, which is an atomic execution consisting of
1. READ(A,t);
2. t = t+c;
3. WRITE(A,t);
• Notation:
o ili (X)– action of Ti requesting an increment lock on X
o inci (X)– action of Ti increments X by some constant;
don’t care about the value of the constant.
18.4.5 Increment Locks (cont.)
• Example
18.4.5 Increment Locks (cont.)
• Compatibility matrix
Lock requested
Lock in
hold
S
X
I
S
YES
NO
NO
X
NO
NO
NO
I
NO
NO
YES
References
• H. Garcia-Molina, J. Ullman, and J. Widom,
“Database System: The Complete Book,”
second edition: chapter 18.3-18.4, p.897-913,
Prentice Hall, New Jersy, 2008
Concurrency Control
Chapter 18
Section 18.5
Presented by
Khadke, Suvarna
CS 257
(Section II) Id 213
Overview
• Assume knowledge of:
o
o
o
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
o Release locks when the transaction manager tells it
that the transaction will commit or abort
o
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
1. 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
1. 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.
1. 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 Structure
Some 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
locks
2>wait : one transaction
waiting for a lock on A
3>A list :T currently hold
locks on A or Waiting for
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 mode
 S: GM(S) or nothing
 U: GM(S) or nothing
 X: nothing
Handling Unlock Requests
• If the value of waiting is “yes" need to grant one or more locks using following
approaches
o First-Come-First-Served:
o Grant the lock to the longest waiting request.
o No starvation (waiting forever for lock)
o Priority to Shared Locks:
o Grant all S locks waiting, then one U lock.
o Grant X lock if no others waiting
o Priority to Upgrading:
o If there is a U lock waiting to upgrade to an X lock, grant that first.
Reference List
• ULLMAN, J. D., WISDOM J. & HECTOR G., DATABASE SYSTEMS THE
COMPLETE BOOK, 2nd Edition, 2008.
Thank You
Concurrency Control
Managing Hierarchies of Database Elements
(18.6)
Presented by
Ronak Shah
(214)
March 9, 2009
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
elements
o
Determine 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)
o
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 bank
Choosing relations as database elements means we
would have one lock for an entire relation
o If we were dealing with a relation having account
balances, this kind of lock would be very inflexible and
thus provide very little concurrency
o Why? Because balance transactions require exclusive
locks and this would mean only one transaction occurs
for one account at any time
o But as each account is independent of others we could
perform transactions on different accounts
simultaneously
o
…(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 document
o
With 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(updated)
• These are required to manage locks at different
granularities
o
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
• Three levels of database elements:
o
o
o
Relations are the largest lockable elements.
Each relation is composed of blocks are pages
Each block contains one or more tuples.
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:
o
o
o
o
Begin at the root of hierarchy
Request 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 follows
o
o
Select * 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
o
o
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’
o This calculates the total length of all tuples having
attribute1
o Thus, T3 acquires IS for relation and S for targeted
tuples
o
o 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 behavior
o
o
o
o
r3(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
Thank You
SECTION 18.7
THE TREE PROTOCOL
By :
Saloni Tamotia (215)
BASICS
• B-Trees
- Tree data structure that keeps data sorted
- allow searches, insertion, and deletion
- commonly used in database and file systems
• Lock
- Enforce limits on access to resources
- way of enforcing concurrency control
• Lock Granularity
- Level and type of information that lock
protects.
TREE PROTOCOL
• Kind of graph-based protocol
• Alternate 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
DATA
Assumptions:
•
•
•
•
Only one kind of lock
Consistent transactions
Legal schedules
No 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
Timestamps
By :
Rupinder Singh (216)
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.
o Use
the value of system, clock as the
timestamp.
o 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), and
3. 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:
o Most transactions are read-only.
o It is rare that concurrent transaction will try to
read and write the same element.
• In high-conflict situation, locking performs better
than timestamps
By
Swathi Vegesna
217
5/14/2009
At a Glance
•
•
•
•
•
•
•
•
Introduction
Validation based scheduling
Validation based Scheduler
Expected exceptions
Validation rules
Example
Comparisons
Summary
5/14/2009
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)
5/14/2009
Validation based
scheduling(updated)
• Scheduler keeps a record of what the active
transactions are doing.
• A transaction goes through a validation phase before
it starts to write values to database elements.
• The set of read and write elements are compared with
write sets of other active transactions.
• Executes in 3 phases
1.Read- reads from RS( ), computes local address
2.Validate- compares read and write sets
3.Write- writes from WS( )
5/14/2009
Validation based
Scheduler(updated)
• Contains an assumed serial order of transactions.
• Maintains three sets:
o
o
o
o
5/14/2009
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.
Transactions are executed in 3 phases- Read,
Validate and Write.
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.
5/14/2009
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)
5/14/2009
Example
5/14/2009
Solution
• Validation of U:
Nothing to check
• Validation of T:
WS(U) ∩ RS(T)= {D} ∩{A,B}=φ
WS(U) ∩ WS(T)= {D}∩ {A,C}=φ
• Validation of V:
RS(V) ∩ WS(T)= {B}∩{A,C}=φ
WS(V) ∩ WS(T)={D,E}∩ {A,C}=φ
RS(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)
5/14/2009
Comparison
Concurrency
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
5/14/2009
Summary
•
•
•
•
Concurrency control by validation
The three phases
Validation Rules
Comparison
5/14/2009
References
• Database Systems: The Complete Book
5/14/2009
5/14/2009
21.1 Introduction to
Information Integration
CS257 Fan Yang
5/14/2009
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 independently
hard to design a database to support future
use
5/14/2009
University Database
• Registrar: to record student and grade
• Bursar: to record tuition payments by
students
• Human Resources Department: to record
employees
• Other department….
5/14/2009
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
5/14/2009
How to integrate
• Start over
build one database: contains all the legacy
databases; rewrite all the applications
result: painful
• Build a layer of abstraction (middleware)
on top of all the legacy databases
this layer is often defined by a collection of
classes
BUT…
5/14/2009
Heterogeneity Problem(updated)
• What is Heterogeneity Problem – Sources
differ in many ways though they all store
same kinds of data.
Aardvark Automobile Co.
1000 dealers has 1000 databases
to find a model at another dealer
can we use this command:
SELECT * FROM CARS
WHERE MODEL=“A6”;
5/14/2009
Type of Heterogeneity
•
•
•
•
•
•
Communication Heterogeneity
Query-Language Heterogeneity
Schema Heterogeneity
Data type difference
Value Heterogeneity
Semantic Heterogeneity
5/14/2009
Conclusion
• One database system is perfect, but
impossible
• Independent database is inconvenient
• Integrate database
1. start over
2. middleware
• heterogeneity problem
5/14/2009
Thank you very much
5/14/2009
Chapter 21.2
Modes of Information Integration
ID: 219
Name: Qun Yu
Class: CS257 219 Spring 2009
Instructor: Dr. T.Y.Lin
Content Index
21.2 Modes of Information Integration
21.2.1 Federated Database Systems
21.2.2 Data Warehouses
21.2.3 Mediators
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
DB1
DB2
2
Wrappers
2
Wrappers
2
Wrappers
DB3
2
Wrappers
2
Wrappers
2
Wrappers
DB4
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)
wrapper
Dealer-1’s DB
wrapper
Dealer-2’s DB
Example…
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’);
}
}
Dealer 1 queries Dealer 2 for needed cars
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
User
query
result
Warehouse
Combiner
Extractor
Extractor
Source 1
Source 2
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,Options
WHERE Autos.serial=Options.serial AND
option=‘autoTrans’;
INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)
SELECT serialNo,model,color,’no’,’dealer2’
FROM Autos
WHERE NOT EXISTS ( SELECT * FROM serial =Autos.serial
AND option = ‘autoTrans’);
Construct Data Warehouse
There are mainly 3 ways to constructing
the data in the 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.
Construct Data 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.
Construct Data Warehouse
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
Result
User query
Mediator
Query
Result
Result
Wrapper
Query
Result
Source 1
Query
Wrapper
Query
Result
Source 2
Example
Same data sources as the example of data warehouse, the mediator
Integrates 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 AkutoMed
Where color=‘red’
Example
In this simple case, the mediator forwards the same query to each
Of the two wrappers.
Wrapper1: Cars(serialNo, model, color, autoTrans, cdPlayer, …)
SELECT serialNo,model
FROM cars
WHERE color = ‘red’;
Wrapper2: Autos(serial,model,color); Options(serial,option)
SELECT serial, model
FROM Autos
WHERE color=‘red’;
The mediator needs to interprets serial into serialNo, and then
returns the union of these sets of data to user.
Example
There may be different options for the mediator to forward user query,
for example, the user queries if there are a specific model&color car
(i.e. “Gobi”, “blue”).
The mediator decides 2nd query is needed or not based on the result of
1st query. That is, If dealer-1 has the specific car, the mediator doesn’t
have to query dealer-2.
THANK YOU !
Reference:
Database Systems– The complete Book 2nd
Edition, 21.2
notes from
http://infolab.stanford.edu/~ullman/dscb.html
Chapter 21 Information Integration
21.3 Wrappers in Mediator-Based Systems
Presented by: Kai Zhu
Professor: Dr. T.Y. Lin
Class ID: 220
Intro
• Templates for Query patterns
• Wrapper Generator
• Filter
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.
How to design a wrapper?
Classify the possible queries that the
mediator can ask into templates, which are
queries with parameters that represent
constants.
Templates for Query Patterns:
• Use notation T=>S to express the
idea that the template T is turned by
the wrapper into the source query S.
• Example 1
Dealer 1
Cars (serialNo, model, color, autoTrans,
navi,…)
For use by a mediator with schema
AutoMed (serialNo, model, color, autoTrans,
dealer)
• We denote the code representing
that color by the parameter $c, then
the template will be:
SELECT *
FROM AutosMed
WHERE color = ’$c’;
=>
SELECT serialNo, model, color, autoTrans, ’dealer1’
FROM Cars
WHERE color=’$c’;
(Template T => Source query S)
• There will be total 2n templates if we
have the option of specifying n
attributes.
Wrapper Generators
• The wrapper generator creates a
table holds the various query
patterns contained in the templates.
• The source queries that are
associated with each.
A driver is used in each wrapper, the
task of the driver is to:
• Accept a query from the mediator.
• Search the table for a template that
matches the query.
• The source query is sent to the source,
again using a “plug-in” communication
mechanism.
• The response is processed by the wrapper.
Filter
• Have a wrapper filter to supporting
more queries.
• Example 2
• If wrapper is designed with more
complicated template with queries
specify both model and color.
SELECT *
FROM AutosMed
WHERE model = ’$m’ AND color = ’$c’;
=>
SELECT serialNo, model, color, autoTrans, ’dealer1’
FROM Cars
WHERE model = ’$m’ AND color=’$c’;
• Now we suppose the only template
we have is color. However the
wrapper is asked by the Mediator to
find “blue Gobi model car.”
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’;
Thank you
INFORMATION
INTEGRATION
SECTIONS 21.4 – 21.5
05/11/09 15:54
Sanuja Dabade & Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
• 21.4 Capability Based Optimization
o
o
o
o
21.4.1The Problem of Limited Source
Capabilities
21.4.2 A notation for Describing Source
Capabilities
21.4.3 Capability-Based Query-Plan
Selection
21.4.4 Adding Cost-Based Optimization
• 21.5 Optimizing Mediator Queries
o
o
o
o
21.5.1 Simplified Adornment Notation
21.5.2 Obtaining Answers for Subgoals
21.5.3 The Chain Algorithm
21.5.4 Incorporating Union Views at the
05/11/09 15:54
Mediator
21.4 Capability Based
Optimization(updated)
• Introduction
Typical DBMS estimates the cost of each query
plan and picks what it believes to be the best
o Mediator – has knowledge of how long its sources
will take to answer
o Optimization of mediator queries cannot rely on
cost measure alone to select a query plan
o Optimization by mediator follows capability based
optimization
o The central issue is whether the query plan is
executable, if so estimate costs.
o
05/11/09 15:54
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
o
E.g. Select * from books;
05/11/09 15:54
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
o Indexes on large database may make certain
queries feasible, while others are too expensive
to execute
o Security reasons
o
 E.g. Medical database may answer queries about
averages, but won’t disclose details of a particular
patient's information
05/11/09 15:54
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
o u(unspecified) – not permitted to specify a value
for a attribute
o
o
05/11/09 15:54
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
05/11/09 15:54
21.4.2 A notation for Describing
Source Capabilities….(cont’d)
• E.g. Dealer 1 is a source of data in the form:
Cars (serialNo, model, color, autoTrans, navi)
The adornment for this query form is b’uuuu
05/11/09 15:54
21.4.3 Capability-Based QueryPlan 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.
o We can construct no more valid forms of source
queries, yet still cannot answer the mediator
query. It has been an impossible query.
o
05/11/09 15:54
21.4.3 Capability-Based QueryPlan 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
o
o
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
05/11/09 15:54
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
05/11/09 15:54
INFORMATION
INTEGRATION
SECTION 21.5
05/11/09 15:57
Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
• 21.5 Optimizing Mediator Queries
21.5.1 Simplified Adornment Notation
21.5.2 Obtaining Answers for
Subgoals
o 21.5.3 The Chain Algorithm
o 21.5.4 Incorporating Union Views at
the Mediator
o
o
05/11/09 15:57
21.5 Optimizing Mediator
Queries(updated)
• 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.
o The solution may not be optimal.
o The class of queries handled are those involving
joins of relations at the sources and an optional
selection or projection onto output attributes.
o
05/11/09 15:57
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:
o
o
nameadornments(attributes)
where:
 name is the name of the relation
 the number of adornments = the number of attributes
05/11/09 15:57
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.
o
o
o
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:
o
If yi is f, u, or o[S] and xi is either b or f.
05/11/09 15:57
21.5.3 The Chain Algorithm
• Maintains 2 types of information:
o
o
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.
05/11/09 15:57
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.
o
Project X onto its variables that appear in R.
05/11/09 15:57
21.5.3 The Chain Algorithm
(con’t)
1. For each tuple t in the project of X, issue a query to
the source as follows (β is a source adornment).
o 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.
o 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.
o If a component of β is f, and the corresponding
component of α is b, provide a constant value for
source query.
05/11/09 15:57
21.5.3 The Chain Algorithm
(con’t)
• 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].
1.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.
05/11/09 15:57
21.5.3 The Chain Algorithm
(con’t)
1.Replace X with X πs(R), where S is all of the
variables among: a1, a2, …, an.
2.Project out of X all components that
correspond to variables that do not appear in
α
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.
05/11/09 15:57
21.5.3 The Chain Algorithm
Example
• Mediator query:
o
Q: Answer(c) ← Rbf(1,a) AND Sff(a,b) AND Tff(b,c)
• Example:
Relation R S T
Data
w
x
x
y
y
z
1
2
2
4
4
6
1
3
3
5
5
7
1
4
5
8
Adornment bf c’[2,3,5]f bu
05/11/09 15:57
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.
o
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.
05/11/09 15:57
21.5.3 The Chain Algorithm
Example (con’t)
• Project the subgoal’s relation onto its second
component, since only the second component
of R(1,a) is a variable.
a
2
3
4
• This is joined with X, resulting in X equaling
this relation.
• Change adornment on S from ff to bf.
05/11/09 15:57
21.5.3 The Chain Algorithm
Example (con’t)
• Now we resolve Sbf(a,b):
o
o
Project X onto a, resulting in X.
Now, search S for tuples with attribute a
equivalent to attribute a in X.
a
b
2
4
• Join this relation with
3 X,5and remove a
because it doesn’t appear in the head nor
any unresolved subgoal:
b
4
5
05/11/09 15:57
21.5.3 The Chain Algorithm
Example (con’t)
• Now we resolve Tbf(b,c):
b
c
4
6
5
7
5
8
• Join this relation with X and project onto the c
attribute to get the relation for the head.
• Solution is {(6), (7), (8)}.
05/11/09 15:57
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.
05/11/09 15:57
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.
o Less practical because it makes queries harder to
answer and impossible if any source is down.
o
• Best Efforts
We need only 1 source with a matching
adornment to resolve a subgoal.
o Need to modify chain algorithm to revisit each
subgoal when that subgoal has new bound
requirements.
o
05/11/09 15:57
Questions
05/11/09 15:57
LOCAL-AS-VIEW
MEDIATORS
Priya Gangaraju(Class Id:203)
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 queryV1(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 greatgrand 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
o
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 query
Q1: 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.
Information Integration
Entity Resolution – 21.7
Presented By:
Deepti Bhardwaj
Roll No: 223_103
Contents
• 21.7 Entity Resolution
 21.7.1 Deciding Whether Records Represent a
Common Entity
 21.7.2 Merging Similar Records
 21.7.3 Useful Properties of Similarity and Merge
Functions
 21.7.4 The R-Swoosh Algorithm for ICAR Records
 21.7.5 Other Approaches to Entity Resolution
Introduction
• Determining whether two records or tuples
do or do not represent the same person,
organization, place or other entity is called
ENTITY RESOLUTION.
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. Misspellings
2. Variant Names
3. Misunderstanding of Names
Continue: Deciding whether Records
represent a Common Entity
4. Evolution of Values
5. Abbreviations
Thus when deciding whether two records represent
the same entity, we need to look carefully at the
kinds of discrepancies and use the test that
measures the similarity of records.
Deciding Whether Records Represents a
Common Entity - Edit Distance(updated)
• 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.
• Similarity measure can be sum of edit distance or
sum of the squares of the distances.
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 means replacing two records that are similar
enough to merge and replace by one single record
which contain information of both.
• 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 Phone
1. Susan 123 Oak St. 818-555-1234
2. Susan 456 Maple St. 818-555-1234
3. 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- 5555678}
Useful Properties of Similarity and
Merge Functions
The following properties say that the merge operation is
a semi lattice :
1. Idempotence : That is, the merge of a record with
itself should surely be that record.
2. Commutativity : If we merge two records, the order
in which we list them should 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
Clustering: Some time we group the records into
clusters such that members of a cluster are in some
sense similar to each other and members of different
clusters are not similar.
Other Approaches to Entity Resolution Partitioning
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.
Thank You
The Query Compiler
16.1 Parsing and Preprocessing
Meghna Jain(205)
Dr. T.Y. Lin
Presentation Outline
16.1 Parsing and Preprocessing
16.1.1 Syntax Analysis and Parse Tree
16.1.2 A Grammar for Simple Subset of SQL
16.1.3 The Preprocessor
16.1.4 Processing Queries Involving Views
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.
Query
Parser
Preprocessor
Logical Query plan
generator
Query rewrite
Preferred logical query
plan
Form a query to a logical query
plan
Syntax Analysis and Parse
Tree (updated)
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>
An atom has no children but a syntactic category has
and are described by one of the rules of the grammar
for the language.
Grammar for Simple Subset of SQL
<Query> ::= <SFW>
<Query> ::= (<Query>)
<SFW> ::= SELECT <SelList> FROM <FromList> WHERE <Condition>
<SelList> ::= <Attribute>,<SelList>
<SelList> ::= <Attribute>
<FromList> ::= <Relation>, <FromList>
<FromList> ::= <Relation>
<Condition>
<Condition>
<Condition>
<Condition>
::= <Condition> AND <Condition>
::= <Tuple> IN <Query>
::= <Attribute> = <Attribute>
::= <Attribute> LIKE <Pattern>
<Tuple> ::= <Attribute>
Atoms(constants), <syntactic categories>(variable),
::= (can be expressed/defined as)
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%'
);
Another query equivalent
SELECT title
FROM StarsIn, MovieStar
WHERE starName = name AND
birthdate LIKE '%1960%' ;
Parse Tree
<Query>
<SFW>
SELECT <SelList> FROM <FromList> WHERE <Condition>
<Attribute> <RelName> , <FromList> AND
title StarsIn <RelName>
MovieStar
<Query>
<Condition> <Condition>
<Attribute> = <Attribute> <Attribute> LIKE <Pattern>
starName name birthdate ‘%1960’
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 FROMclause 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;
Thank You
16.2 ALGEBRAIC LAWS FOR
IMPROVING QUERY PLANS
Ramya Karri
ID: 206
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.
o
E.g. A + B = B + A - The “+” operator is commutative.
• associativity - operator is independent of operand grouping.
o
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
Associative
RXS=SXR
(R X S) X T = S X (R X T)
R⋈S=S⋈R
(R ⋈ S) ⋈ T= S ⋈ (R ⋈ T)
RS=SR
(R  S)  T = S  (R  T)
R∩S=S∩R
(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
o
o
R={a,a,b,b,b,c}
p1 satisfied by a,b, p2 satisfied by b,c
o σp1vp2 (R) = {a,a,b,b,b,c}
o σp1(R) = {a,a,b,b,b}
o σp2(R) = {b,b,b,c}
o 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
σC(R
σC(R
σC(R
× S) = σC(R) × S
∩ S) = σC(R) ∩ S
⋈ S) = σC(R) ⋈ S
⋈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.
1.Selection can be distributed into joins.
1.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 Involving Duplicate
Elimination(updated)
• 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))
A relation has no duplicates if it has a primary key or it is a
result of grouping or union, intersection or difference.
Laws Involving Duplicate
Elimination
• 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).
Thank you
The Query Compiler
Section 16.3
DATABASE SYSTEMS – The Complete Book
Presented By: Under the supervision of:
Deepti Kundu Dr.T.Y.Lin
Topics to be covered
• From Parse to Logical Query Plans
o Conversion to Relational Algebra
o Removing Subqueries From Conditions
o Improving the Logical Query Plan
o Grouping Associative/ Commutative Operators
16.3 From Parse to Logical Query Plans ►
Review
Query
Parser
Section 16.1
Preprocessor
Logical query
plan generator
Section 16.3
Query Rewriter
Preferred logical query plan
Two steps to turn Parse tree into
Preferred 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.
Reference Relations
• StarsIn (movieTitle, movieYear, starName)
• MovieStar (name, address, gender, birthdate)
Conversion to Relational
Algebra(updated)
• If we have a <Query> with a <Condition> that has no
subqueries, then we may replace the entire construct – the
select-list, from-list, and condition – by a relational-algebra
expression consisting of from bottom to top of:
o The product of all the relations mentioned in the
<FromList> which is the argument of:
o A selection where <Condition> expression in the construct
being replaced which in turn is the argument of:
o A projection with a list of attributes in the <SelList>.
• 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:
o A selection σC, where C is the <Condition> expression in the
construct being replaced, which in turn is the argument of:
o A projection πL , where L is the list of attributes in the <SelList>
o
A query : Example
• SELECT movieTitle
FROM Starsin, MovieStar
WHERE starName = name AND
birthdate LIKE ‘%1960’;
SELECT movieTitle
FROM Starsin, MovieStar
WHERE starName = name AND
birthdate LIKE ‘%1960’;
Translation to an algebraic expression
tree
Removing Subqueries From Conditions
• 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 σ
πmovieTitle
σ
StarsIn
<Condition
>
<Tuple>
<Attribute>
starName
IN
πname
σ birthdate LIKE ‘%1960'
MovieStar
Two argument selection with condition
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 condition
involving IN
σ
R
σC
<Condition>
t
IN
X
S
R
δ
S
The effect
Improving the Logical Query Plan
• Algebraic laws to improve logical query plans:
Selections can be pushed down the expression tree as far
as they can go.
o Similarly, projections can be pushed down the tree, or new
projections can be added.
o Duplicate eliminations can sometimes be removed, or
moved to a more convenient position in the tree.
o Certain selections can be combined with a product below
to turn the pair of operations into an equijoin.
o
Grouping Associative/ Commutative
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 rewriting
Π movieTitle
Starname = name
StarsIn
σbirthdate LIKE ‘%1960’
MovieStar
Final step in producing logical query plan
=>
R
U
U
R
U
S
T
V
W
U
U
S
T
V
W
An 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 m1
WHERE 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
);
SELECT distinct m1.movieTitle, m1,movieYear
FROM StarsIn m1
WHERE 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 to turn
the pair of operations into an equijoin…
Condition pushed up the expression tree…
`
Selections combined…
The Query Compiler
(16.4)
DATABASE SYSTEMS – The Complete Book
Presented By: Under the supervision of:
Maciej Kicinski Dr.T.Y.Lin
Topics to be covered
• From Parse to Logical Query Plans
o
o
o
o
Conversion to Relational Algebra
Removing Subqueries From Conditions
Improving the Logical Query Plan
Grouping Associative/ Commutative Operators
• Estimating the Cost of Operation
o
o
o
o
o
Estimating Sizes of Intermediate Relations
Estimating the Size of a Projection
Estimating the Size of a Selection
Estimating the Size of a Join
Estimating Sizes for Other Operations
16.4 From Estimating the Cost of Operation ►
Estimating the Cost of Operations
• 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:
o
o
o
o
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:
S = σ A=c (R)
Where A is an attribute of R and c is a constant
 The recommended estimate is
T(S) = T(R)/ V(R,A)
Estimating Sizes of Other Operations
(updated)
•
•
•
•
Union – larger plus half the smaller.
Intersection – half the smaller.
Difference – T(R)-T(S)/2.
Duplicate Elimination – smaller of T(R)/2 and the product
of all the V(R,ai)'s.
Grouping and Aggregation – smaller of T(R)/2 and the
product of all the V(R,ai)'s where attribute A ranges over only
the grouping attributes of L.
Estimating the size of a
Join(added)
• Assumptions:
Containment of Value Sets – If R and S are two
relations with a attribute Y and V(R,Y)<=V(S,Y),
then every value of R will be a Y-value of S.
o Preservation of Value Sets – If A is an attribute of
R but not of S, then V(R lX| S, A)=V(R,A).
o
T(R lXl S)=T(R)T(S)/max(V(R,Y),V(S,Y))
Natural Joins with Multiple Join
Attributes.(added)
• The estimate of the size of join R and S is
computed by multiplying T(R) by T(S) and
dividing by the larger of V(R,y) and V(S,y) for
each attribute that is common to R and S.
Joins of Many Relations
(added)
• Start with the product of the number of tuples
in each relation. Then, for each attribute A
appearing at least twice, divide by all but the
least of the V(R,A)'s.
Cost-Based Plan Selection
Choosing an Order for Joins
Chapter 16.5 and16.6 by:Vikas Vittal Rao
ID: 124/227
Chiu Luk
ID: 210
5/14/2009
Agenda
•
•
•
•
•
•
•
Outline
Cost Estimation
Histograms
Computation of statistics
Reducing cost of heuristics
Enumerating Physical Plans
List of other approaches
5/14/2009
Outline(updated)
• Query Optimizer estimates the “cost” of
query evaluation.
• This “cost” is based on number of disk
I/O’s.
• Disk I/O’s influenced by: Logical operators used
 Size of intermediate results
 Physical operators used to implement
logical operators.
 The ordering of similar operations,
especially joins (discussed next in 16.6)
5/14/2009
Cost Estimation(updated)
• Query Optimizer keeps track of certain
parameters like:
 T(R) – number of tuples in a relation R
 V( R , a ) – number of unique values in R for
attribute ‘a’
 B(R) – number of blocks in which R can fit.
The Optimizer also computes a
“histogram” of the above parameters.
Sizes of joins can be estimated more
accurately with the help of a histogram.
5/14/2009
Histograms
Equal Width - divide value range by fixed width w
and keep counts of each width
• Equal Height – similar to equal width, we pick the
lowest value v0, a fraction p and keep count of
values which are at “p from the lowest, 2 p from
the lowest, etc” up to the highest value.
• Most frequent values – list of most frequent values
and the number of their occurances
• Using histograms helps estimate the sizes of joins
more accurately(than previously discussed
methods)
o
5/14/2009
Computation of Statistics
• Statistics are fault-tolerant, good for use in
estimates; e.g. small error does not significantly
change out come.
• Inspect the distribution of values across records
in relation
• Computation on entire relation is expensive;
however, can be computed using smaller
sample size
5/14/2009
Reducing Cost by Heuristics
• Applies for logical query plan
• Estimate cost before and after a transformation
• Only choose/apply transformation when cost
estimations show beneficial
• Example:
o
5/14/2009
Deferring duplicate elimination is better
Enumerating Physical Plans
• Baseline approach (exhaustive): consider all
combinations, pick the smallest cost plan
• Other approaches categorized into:
o
o
5/14/2009
Top-down: compute cost from root, take the best
Bottom-up: compute cost for all combinations for a
sub-expression, select the best, move up until root
evaluated
List of Other Approaches
• Heuristic selection
• Branch-and-bound plan enumeration: keep
record of best cost, skip to next plan if current
plan exceed best known cost
• Hill climbing
• Dynamic programming (PP): keep least cost of
each sub-expression; work bottom-up
• Selinger-style optimization: improved version of
DP; keep others beneficial plans besides least
cost plans
5/14/2009
Choosing an Order for Joins
Chapter 16.6 by:
Chiu Luk
ID: 210
5/14/2009
Introduction
• This section focuses on critical problem in
cost-based optimization:
o 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
5/14/2009
Introduction
5/14/2009
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
o Called the build relation
o Assumed to be smaller
o Stored in main-memory
5/14/2009
Significance of Left and Right Join
Arguments
• The right argument of the join is
o Called the probe relation
o Read a block at a time
o Its tuples are matched with
those of build
relation
• The join algorithms which distinguish
between the arguments are:
o One-pass join
o Nested-loop join
o Index join
5/14/2009
Join Trees
• Order of arguments is important for joining
two relations
• Left argument, since stored in mainmemory, 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
5/14/2009
Join Trees
• Order of arguments is important for joining
two relations
• Left argument, since stored in mainmemory, 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
5/14/2009
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
5/14/2009
Left-Deep Join Trees
• Consider 4 relations. Different ways to join
them are as follows
5/14/2009
(updated)
• In fig (a) all the right children are leaves. This is a
left-deep tree
• In fig (c) all the left children are leaves. This is a
right-deep tree
• Fig (b) is a bushy tree
• Considering left-deep trees is advantageous for
deciding join orders
• Query plans based on left-deep trees and join
implementations are more efficient than the same
algorithms used with non-left deep trees.
5/14/2009
Join order
• Join order selection
o A1 A2 A3 .. An
o Left deep join trees
An
Ai
• Dynamic programming
o
Best plan computed for each subset of relations
 Best
Best
Best
….
Best
5/14/2009
plan (A1, .., An) = min cost plan of(
plan(A2, .., An) A1
plan(A1, A3, .., An) A2
plan(A1, .., An-1)) An
Dynamic Programming to Select a
Join Order and Grouping
• Three choices to pick an order for the join of many
relations are:
o Consider all of the relations
o Consider a subset
o Use a heuristic o pick one
• Dynamic programming is used either to consider all or a
subset
o Construct a table of costs based on relation size
o Remember only the minimum entry which will required
to proceed
5/14/2009
Dynamic Programming to Select a
Join Order and Grouping
5/14/2009
Dynamic Programming to Select a
Join Order and Grouping
5/14/2009
Dynamic Programming to Select a
Join Order and Grouping
5/14/2009
Dynamic Programming to Select a
Join Order and Grouping
5/14/2009
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
o Make
one decision at a time about order of
join and never backtrack on the decisions
once made
5/14/2009
Thank you
5/14/2009
Query Compiler: 16.7 Completing
the Physical Query-Plan
CS257 Spring 2009
Professor Tsau Lin
Student: Suntorn Sae-Eung
ID: 212
Outline
16.7 Completing the Physical-Query-Plan
I. Choosing a Selection Method
II. Choosing a Join Method
III. Pipelining Versus Materialization
IV. Pipelining Unary Operations
V. Pipelining Binary Operations
Before complete Physical-Query-Plan
• A query previously has been
o Parsed and Preprocessed (16.1)
o Converted to Logical Query Plans (16.3)
o Estimated the Costs of Operations (16.4)
o Determined costs by Cost-Based Plan
Selection (16.5)
o 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?
o 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
1. B(R) if R is clustered
2. T(R) if R is not clustered
2. Cost of a plan picking an equality term (e.g. a = 10) w/ index-scan
1. B(R) / V(R, a) clustering index
2. T(R) / V(R, a) nonclustering index
3. Cost of a plan picking an inequality term (e.g. b < 20) w/ index-scan
1. B(R) / 3 clustering index
2. 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 joining
2. Sort-join is preferred when attributes are presorted 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)
o
store (intermediate) result of each operations on disk
• Pipelining (more efficient way)
o
o
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.
In buf
Unary
operation
Out buf
Unary
operation
Out buf
R
In 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
o
o
o
o
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 buffers
2.Organize them for lookup as a hash table
3.Use one buffer left to read each block of U in
turn
4.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
o 10,000 to read U for
one-pass hash join of
(R S) U.
o
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:
o 20,000
to read U and write its tuples into
buckets
o 45,000 for two-pass hash-join R S
o k to write out the buckets of R S
o 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
ok
 49 (case 1)
 Disk I/O’s is 55,000
ok
> 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:
for two-pass hash-join R and S
o k to store R S on disk
o 30,000 + k for two-pass join of U in R S
o 45,000
• The total cost is 75,000+4k.
Example (cont.)
• In summary, costs of physical plan as
function of R S size.
Reference
[1] H. Garcia-Molina, J. Ullman, and J.
Widom, “Database System: The Complete
Book,” second edition: p.897-913, Prentice
Hall, New Jersy, 2008
VI. Notation for Physical Query
Plans
• Several types of operators:
1. Operators for leaves
2. (Physical) operators for Selection
3. (Physical) Sorts Operators
4. 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
o
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.
Notation for Physical Query
Plans (cont.)
1.(Physical) operators for Selection
o
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)
Notation for Physical Query
Plans (cont.)
1.(Physical) Sort Operators
Sorting can occur any point in physical plan, which use a
notation SortScan(R, L).
o It is common to use an explicit operator Sort(L) to sort
relation that is not stored.
o Can apply at the top of physical-query-plan tree if the result
needs to be sorted with ORDER BY clause (г).
o
Notation for Physical Query
Plans (cont.)
1.Other Relational-Algebra Operations
o
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. sortbased, 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
o
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
o
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
o
o
Use Index on condition y = 2 first
Filter 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.
o
o
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.
Summary of Chapter 16
In this part of the presentation I will talk
about the main topics of Chapter 16.
COMPILATION OF
QUERIES
• Compilation means turning a query into a
physical query plan, which can be
implemented by query engine.
• Steps of query compilation :
o
o
o
o
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
VIEW EXPANSION
• A very critical part of query compilation.
• Expands the view references in the query
tree to the actual view.
• Provides opportunities for the query
optimization.
SEMANTIC CHECKING
• Checks the semantics of a SQL query.
• Examines a parse tree.
• Checks :
o
o
o
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 :
o Commutative and associative laws
o Laws involving selection
o Pushing selection
o Laws involving projection
o Laws about joins and products
o Laws involving duplicate eliminations
o Laws 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 :
o
o
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.
o Nested loop and the hash loop joins used.
o Scanning and sorting operations.
o Storing intermediate results.
o
PLAN ENUMERATION
STRATEGIES
• Common approaches for searching the
space for best physical plan .
Dynamic programming : Tabularizing the best plan for each
sub expression
o Selinger style programming : sort-order the results as a part
of table
o Greedy approaches : Making a series of locally optimal
decisions
o Branch-and-bound : Starts with enumerating the worst plans
and reach the best plan
o
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.
Reference
[1] H. Garcia-Molina, J. Ullman, and J. Widom,
“Database System: The Complete Book,”
second edition: p.897-913, Prentice Hall, New
Jersey, 2008
Query Execution
Chapter 15
Section 15.1
Presented by
Khadke, Suvarna
CS 257
(Section II) Id 213
Agenda
•
•
•
•
•
•
•
•
Query Processor and major parts of Query processor
Physical-Query-Plan Operators
Scanning Tables
Basic approaches to locate the tuples of a relation R
Sorting While Scanning Tables
Computation Model for Physical Operator
I/O Cost for Scan Operators
Iterators
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 operations
• Must 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
Operators(updated)
• Physical 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
• The relation is an operand of some other
operation.
Scanning Tables
• One 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 Tables
Basic approaches to locate the tuples of a relation R
• Table Scan
o Relation R is stored in secondary memory with its
tuples arranged in blocks
o It is possible to get the blocks one by one
• Index-Scan
o 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 Tables
• Number of reasons to sort a relation
o Query could include an ORDER BY clause,
requiring that a relation be sorted.
o Algorithms to implement relational algebra
operations requires one or both arguments
to be sorted relations.
o 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
Operator
• Physical-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 query
o Buffer space availability in the main
memory at the time of execution of the query
o Size of input and the size of the output
generated
o 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 Operators
• Many physical operators can be implemented
as an Iterator.
• Three methods forming the iterator for an
operation are:
• 1. Open( ) :
o This method starts the process of getting
tuples
o It initializes any data structures needed to
perform the operation
Iterators for Implementation of
Physical Operators
• 2. GetNext( ):
o Returns the next tuple in the result
o If there are no more tuples to return,
GetNext returns a special value NotFound
• 3. Close( ) :
o Ends the iteration after all tuples
o It calls Close on any arguments of the
operator
Reference
• ULLMAN, J. D., WISDOM J. & HECTOR G., DATABASE
SYSTEMS THE COMPLETE BOOK, 2nd Edition, 2008.
Thank You
Query Execution
One-Pass Algorithms for Database Operations
(15.2)
Presented by
Ronak Shah
(214)
April 22, 2009
Introduction(updated)
• The choice of an algorithm for each operator is an
essential part of the process of transforming a logical
query plan into a physical query plan.
• Main classes of Algorithms:
o
o
o
Sorting-based methods
Hash-based methods
Index-based methods
• Division based on degree difficulty and cost:
1-pass algorithms – read data only once from disk.
2-pass algorithms – reading data a first time and writing computed data to
the disk and reading again for further processing.
o 3 or more pass algorithms
o
o
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-aTime Operations
• Tuple-at-a-time operations are selection and projection
o
o
o
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.
o
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, fillRelation Operations
• Duplicate Elimination
o
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.
o
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.
Managing memory for a one-pass
duplicate-elimination
Duplicate Elimination
• 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.
o 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.
o
• We need a main-memory structure that allows each of
the operations:
Add a new tuple, and
o Tell whether a given tuple is already there
o
Duplicate Elimination (…contd.)
• The different structures that can be used for such main
memory structures are:
o
o
Hash table
Balanced binary search tree
One-Pass Algorithms for Unary, fillRelation Operations
• Grouping
o
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.
Grouping
• The accumulated value is:
For MIN(a) or MAX(a) aggregate, record minimum /maximum value,
respectively.
o For any COUNT aggregation, add 1 for each tuple of group.
o For SUM(a), add value of attribute a to the accumulated sum for its
group.
o AVG(a) is a hard case. We must maintain 2 accumulations: count of no. of
tuples in the group & sum of a-values of these tuples. Each is computed as
we would for a COUNT & SUM aggregation, respectively. After all tuples
of R are seen, take quotient of sum & count to obtain average.
o
One-Pass Algorithms for Binary
Operations
• Binary operations include:
o
o
o
o
o
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.
Bag Difference (…contd.)
• Think of a tuple t with a count of c as c reasons not to
copy t to the output as we read tuples of R.
• Read a tuple t of R; check if t occurs in S. If not, then copy
t to the output. If t does occur in S, then we look at
current count c associated with t. If c = 0, then copy t to
output. If c > 0, do not copy t to output, but decrement c
by 1.
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.)
1. 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.
Thank you
QUERY EXECUTION
15.3
Nested-Loop Joins
By:
Saloni Tamotia (215)
Introduction to Nested-Loop Joins
• Used for relations of any side.
• Not necessary that relation fits in main
memory
• Uses “One-and-a-half” pass method in which
for each variation:
• One argument read just once.
• Other argument read repeatedly.
• Two kinds:
o Tuple-Based Nested Loop Join
o Block-Based Nested Loop Join
ADVANTAGES OF NESTED-LOOP JOIN
• Allows us to avoid storing intermediate relation on
disk.
• Fits in the iterator framework.
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 DO
IF r and s join to make tuple t THEN
output t
R and S are two Relations with r and s as tuples.
carelessness in buffering of blocks causes the use of T(R)T(S)
disk I/O’s
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.
Block-Based Nested-Loop Join Algorithm
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
Block-Based Nested-Loop Join Algorithm
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
Two-Pass Algorithms
Based on Sorting
SECTION 15.4
Rupinder Singh
Two-Pass Algorithms Based on
Sorting(updated)
• 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
• Divide a relation R for which B(R)>M
into chunk of size M, sort them and
process each sorted sublist in main
memory one at a time.
Basic idea
• Step 1: Read M blocks of R into main
memory.
• Step 2:Sort these M blocks in main
memory, using an efficient, mainmemory 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
Sublist
In memory
Waiting on disk
R1
1,2
2,2, 2,5
R2
2,3
4,4, 4,5
After processing
tuple 1 1,1
R3
2,3,5
Sublist
In memory
Waiting on disk
R1
2
2,2, 2,5
2,3
4,4, 4,5
Output: 1 R2
R3
2,3
5
Continue the
same process
with next tuple.
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(updated)
• For bag-union one-pass algorithm is used.
• For set-union
o Step 1:Repeatedly bring M blocks of R into main memory, sort
their tuples, and write the resulting sorted sublist back to disk.
o Step 2:Do the same for S, to create sorted sublists for relation S.
o Step 3:Use one main-memory buffer for each sublist of R and S.
Initialize each with the first block from the corresponding sublist.
o 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(updated)
• 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.
•
• 3(B(R)+B(S)) disk I/O's.
• Approximately B(R)+B(S)<=M*M for the algorithm to work.
A Simple Sort-Based Join
Algorithm
• 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
• 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
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:
o
o
o
o
o
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 Simple Sort-Based Join
Algorithm
• The simple sort-join uses 5(B(R) +
B(S)) disk I/0's.
• It requires B(R) ≤ M² and B(S) ≤ M² to
work.
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.
o Bring the first block of each sublist into a buffer
o 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
o
A More Efficient Sort-Based Join
• The number of disk I/O’s is 3(B(R) +
B(S))
• It requires B(R) + B(S) ≤ M² to work
Summary of Sort-Based
Algorithms
Operators
Approximate
M 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))
By
Swathi Vegesna
5/14/2009
At a glimpse
•
•
•
•
•
•
•
•
Introduction
Partitioning Relations by Hashing
Algorithm for Duplicate Elimination
Grouping and Aggregation
Union, Intersection, and Difference
Hash-Join Algorithm
Sort based Vs Hash based
Summary
5/14/2009
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.
5/14/2009
Partitioning Relations by
Hashing
Algorithm:
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 THEN
BEGIN
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 THEN
write the buffer to disk;
5/14/2009
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
5/14/2009
Requirements
• Number of disk I/O's: 3*B(R)
o B(R) < M(M-1), only then the two-pass, hash-based
algorithm will work
• In order for this to work, we need:
o hash function h evenly distributes the tuples among
the buckets
o each bucket Ri fits in main memory (to allow the onepass algorithm)
o i.e., B(R) ≤ M2
5/14/2009
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
5/14/2009
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
5/14/2009
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
5/14/2009
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
5/14/2009
Summary
•
•
•
•
•
•
Partitioning Relations by Hashing
Algorithm for Duplicate Elimination
Grouping and Aggregation
Union, Intersection, and Difference
Hash-Join Algorithm
Sort based Vs Hash based
5/14/2009
5/14/2009
Index-Based Algorithms
Chapter 15
Section 15.6
Presented by
Fan Yang
CS 257
Class ID218
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(updated)
• For a selection σC(R), suppose C is of the form
a=v, where a is an attribute for which an index
exists.
• A search on the index with value v will return the
pointers to those tuples of R that have a-value v.
These tuples constitute the result of the selection
on R with a=v condition.
• For clustering index R.a:
the number of disk I/O’s will be B(R)/V(R,a)
Index-Based Selection(updated)
• The actual number may be higher:
1. index is not kept entirely in main memory
2. they spread over more blocks
3. may not be packed as tightly as possible into
blocks
When index R.a is non-clustering, then disk I/o's will
be T(R)/V(R,a).
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(updated)
• Natural join R(X, Y) S S(Y, Z), S has an index on the attribute Y.
• Each tuple of R within each block of R is joined with those tuples of S that
have attribute Y retrieved by using index on S.
Number of I/O’s to get R
Clustered: B(R)
Not clustered: T(R)
Number of I/O’s to get tuple t of S
Clustered: 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 blocks
Assume 10 tuples in each block,
so T(R)=10,000 and T(S)=5000
V(S,Y)=100
If R is clustered, and there is a clustering index on Y
for S
the 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
relations
search keys (Y-value) for R: 1,3,4,4,5,6
search keys (Y-value) for S: 2,2,4,6,7,8
Thank You
Chapter 15.7
Buffer Management
ID: 219
Name: Qun Yu
Class: CS257 219 Spring 2009
Instructor: Dr. T.Y.Lin
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 advance
2. 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.
The role of the buffer manager
Read/Writes
Requests
Buffers
Buffer
manager
Figure 1: The role of the buffer manager : responds to requests for
main-memory access to disk blocks
15.7.1 Buffer Management Architecture
Two broad architectures for a buffer manager:
1. The buffer manager controls main memory directly.
o Relational DBMS
2. The buffer manager allocates buffers in virtual memory,
allowing the OS to decide how to use buffers.
o “main-memory” DBMS
o “object-oriented” DBMS
Buffer Pool
Key setting for the Buffer manager to be efficient:
The buffer manager should limit the number of buffers in
use so that they fit in the available main memory, i.e. Don’t
exceed available space.
The number of buffers is a parameter set when the DBMS is
initialized.
No matter which architecture of buffering is used, we simply
assume that there is a fixed-size buffer pool, a set of buffers
available to queries and other database actions.
Buffer Pool
Page Requests from Higher Levels
BUFFER POOL
disk page
free frame
MAIN MEMORY
DISK
DB
choice of frame dictated
by replacement policy
• Data must be in RAM for DBMS to operate on it!
• Buffer Manager hides the fact that not all data is in RAM.
15.7.2 Buffer Management Strategies
Buffer-replacement strategies:
When a buffer is needed for a newly requested
block and the buffer pool is full, what block to
throw out the 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.
o Update the time table for every access.
o 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.
o Keep only the loading time
o 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
0
the buffer with
a 0 flag will be
replaced
0
0
1
1
Start point to search
a 0 flag
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
Buffer-replacement strategy -- Clock
a buffer’s flag set to 1 when:
• a block is read into a buffer
• the contents of the buffer is accessed
a buffer’s flag set to 0 when:
• the buffer manager needs a buffer for a new block,
it looks for the first 0 it can find, rotating clockwise. If
it passes 1’s, it sets them to 0.
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
Problem:
• Physical Operator expected certain number of
buffers M for execution.
• However, the buffer manager may not be able
to guarantee these M buffers are available.
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: LRU
Buffers for R: k
We 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.
Example
CASE 2: YES
• Buffer-replacement strategy: LRU
• Buffers for R: k
• We read the blocks of R in an order that alternates:
first last and then last first.
• 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 bufferreplacement 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.
THANK YOU !
Chapter 15 Query Execution
15.8 Algorithms using more than two
passes
Presented by: Kai Zhu
Professor: Dr. T.Y. Lin
Class ID: 220
Intro
• Why we use more than 2 passes
• Multi-pass Sort-based Algorithms
• Conclusion
Reason that we use more than two
passes:
Two passes are usually enough,
however, for the largest relation, we
use as many passes as necessary.
Multi-pass Sort-based Algorithms
Suppose we have M main-memory
buffers available to sort a relation R,
which we assume is stored clustered.
Then we do the following:
BASIS:
If R fits in M blocks (i.e., B(R)<=M)
1. Read R into main memory.
2. Sort it using any main-memory
sorting algorithm.
3. 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.
If we are not merely sorting R, but
performing a unary operation such as δ
or γ on R.
We can modify the above so that at the
final merge we perform the operation
on the tuples at the front of the sorted
sublists.
That is:
• 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.
Conclusion
The 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.
Thank you