5 data storage_and_indexing

Prof P Sreenivasa Kumar
Department of CS&E, IITM
1
File Organization and Indexing
The data of a RDB is ultimately stored in disk files
Disk space management:
Should Operating System services be used ?
Should RDBMS manage the disk space by itself ?
2nd option is preferred as RDBMS requires complete
control over when a block or page in main memory buffer
is written to the disk.
This is important for recovering data when system
crash occurs

2
Structure of Disks
Disk
several platters stacked on
a rotating spindle
one read / write head per surface
for fast access
platter has several tracks
• ~10,000 per inch
each track - several sectors
each sector - blocks
unit of data transfer - block
cylinder i - track i on all platters
Platters
Read/write head
}
track
sector
Speed:
7000 to
10000 rpm

3
Data Transfer from Disk
Address of a block: Surface No, Cylinder No, Block No
Data transfer:
Move the r/w head to the appropriate track
• time needed - seek time – ~ 12 to 14 ms
Wait for the appropriate block to come under r/w head
• time needed - rotational delay - ~3 to 4ms (avg)
Access time: Seek time + rotational delay
Blocks on the same cylinder - roughly close to each other
- access time-wise
- cylinder i, cylinder (i + 1), cylinder (i + 2) etc.

4
Data Records and Files
Fixed length record type: each field is of fixed length
• in a file of these type of records, the record number can be
used to locate a specific record
• the number of records, the length of each field are available
in file header
Variable length record type:
• arise due to missing fields, repeating fields, variable length
fields
• special separator symbols are used to indicate the field
boundaries and record boundaries
• the number of records, the separator symbols used are
recorded in the file header

5
Packing Records into Blocks
Record length much less than block size
• The usual case
• Blocking factor b = B/r B - block size (bytes)
r - record length (bytes)
- maximum no. of records that can be stored in a block
Record length greater than block size
• spanned organization is used
Record
1 1 2 2 3 3
File blocks:
sequence of blocks containing all the records of the file

6
Mapping File Blocks onto the Disk Blocks
Contiguous allocation
• Consecutive file blocks are stored in consecutive disk blocks
• Pros: File scanning can be done fast using double buffering
Cons: Expanding the file by including a new block in the middle
of the sequence - difficult
Linked allocation
• each file block is assigned to some disk block
• each disk block has a pointer to next block of the sequence
• file expansion is easy; but scanning is slow
Mixed allocation

7
Operations on Files
Insertion of a new record: may involve searching for appropriate
location for the new record
Deletion of a record: locating a record –may involve search;
delete the record –may involve movement of other records
Update a record field/fields: equivalent to delete and insert
Search for a record: given value of a key field / non-key field
Range search: given range values for a key / non-key field
How successfully we can carry out these operations
depends on the organization of the file and the availability
of indexes

8
Primary File Organization
The logical policy / method used for placing records into file blocks
Example: Student file - organized to have students records sorted
in increasing order of the “rollNo” values
Goal: To ensure that operations performed frequently on the file
execute fast
• conflicting demands may be there
• example: on student file, access based on rollNo and also
access based on name may both be frequent
• we choose to make rollNo access fast
• For making name access fast, additional access structures
are needed.
- more details later

9
Different File Organization Methods
We will discuss Heap files, Sorted files and Hashed files
Heap file:
Records are appended to the file as they are inserted
Simplest organization
Insertion - Read the last file block, append the record and
write back the block - easy
Locating a record given values for any attribute
• requires scanning the entire file – very costly
Heap files are often used only along with other access structures.

10
Sorted files / Sequential files (1/2)
Ordering field: The field whose values are used for sorting the
records in the data file
Ordering key field: An ordering field that is also a key
Sorted file / Sequential file:
Data file whose records are arranged such that the values of the
ordering field are in ascending order
Locating a record given the value X of the ordering field:
Binary search can be performed
Address of the nth file block can be obtained from
the file header
O(log N) disk accesses to get the required block- efficient
Range search is also efficient

11
Sorted files / Sequential files (2/2)
Inserting a new record:
Ordering gets affected
• costly as all blocks following the block in which insertion is
performed may have to be modified
Hence not done directly in the file
• all inserted records are kept in an auxiliary file
• periodically file is reorganized - auxiliary file and main file
are merged
• locating record
• carried out first on auxiliary file and then the main file.
Deleting a record
• deletion markers are used.

12
Hashed Files
Very useful file organization, if quick access to the data record is
needed given the value of a single attribute.
Hashing field: The attribute on which quick access is needed and
on which hashing is performed
Data file: organized as a buckets with numbers 0,1, …, (M − 1)
(bucket - a block or a few consecutive blocks)
Hash function h: maps the values from the domain of the hashing
attribute to bucket numbers

13
Inserting Records into a Hashed File
Insertion: for the given record R,
apply h on the value of hashing
attribute to get the bucket number r.
If there is space in bucket r,
place R there else place R in the
overflow chain of bucket r.
The overflow chains of all the
buckets are maintained in the
overflow buckets.
0
1
2
M-1
Main buckets
Overflow
buckets
Overflow
chain

14
Deleting Records from a Hashed File
Deletion: Locate the record R to be
deleted by applying h.
Remove R from its bucket/overflow
chain. If possible, bring a record from
the overflow chain into the bucket
0
1
2
M-1
Main buckets
Overflow
buckets
Overflow
chain
Search: Given the hash filed value
k, compute r = h(k). Get the bucket
r and search for the record. If not
found, search the overflow chain
of bucket r.

15
Performance of Static Hashing
Static hashing:
The hashing method discussed so far
The number of main buckets is fixed
Locating a record given the value of the hashing attribute
most often – one block access
Capacity of the hash file C = r * M records
(r - no. of records per bucket, M - no. of main buckets)
Disadvantage with static hashing:
If actual records in the file is much less than C
• wastage of disk space
If actual records in the file is much more than C
• long overflow chains – degraded performance

16
Hashing for Dynamic File Organization
Dynamic files
files where record insertions and deletion take place frequently
the file keeps growing and also shrinking
Hashing for dynamic file organization
Bucket numbers are integers
The binary representation of bucket numbers
Exploited cleverly to devise dynamic hashing schemes
Two schemes
• Extendible hashing
• Linear hashing

17
The k-bit sequence corresponding to a record R:
Apply hashing function to the value of the hashing field of R
to get the bucket number r
Convert r into its binary representation to get the bit sequence
Take the trailing k bits
Extendible Hashing (1/2)

18
All records with 3-bit
Sequence ‘111’
Extendible Hashing (2/2)
The # of
trailing
bits used in
the directory
Global depth d=3
000
001
010
011
100
101
110
111
Directory
2
3
3
2
3
3
Local depth
All records with 2-bit
Sequence ‘01’
The number of bits in the
common suffix of bit
sequences corresponding to
the records in the bucket
Locating a record
Match the d-bit sequence with an entry in the directory and go to
the corresponding bucket to find the record

19
Insertion in Extendible Hashing Scheme (1/2)
2 - bit sequence for the record to be inserted: 00
full
00
01
10
11
1
2
2
b0
b1
b2
d=2
b0 Full: Bucket b0 is split
All records whose 2-bit sequence is ‘10’ are
sent to a new bucket b3. Others are retained in b0
Directory is modified.
b0 Not full: New record is placed in b0. No changes in the directory.
00
01
10
11
d=2
all local
depth = 2
b0
b3
b2
b1

20
Insertion in Extendible Hashing Scheme (2/2)
2 - bit sequence for the record to be inserted: 10
00
01
10
11
d=2
full
b0
b1
b2
b3
all local
depth = 2
000
001
010
011
100
101
110
111
d=3
2
2
3
3
2
b0
b1
b3
b2
b4
b3 not full: new record placed in b3. No changes.
b3 full : b3 is split, directory is doubled, all records with 3-bit
sequence 110 sent to b4. Others in b3.
In general, if the local depth of the bucket to be split is equal to the
global depth, directory is doubled

21
Deletion in Extendible Hashing Scheme
00
01
10
11
d=2
b0
b1
b2
b3
all local
depth = 2
000
001
010
011
100
101
110
111
d=3
2
2
3
3
2
b0
b1
b3
b2
b4
Matching pair of data buckets:
k-bit sequences have a common k-1 bit suffix, e.g, b3 & b4
Due to deletions, if a pair of matching data buckets
-- become less than half full – try to merge them into one bucket
If the local depth of all buckets is one less than the global depth
-- reduce the directory to half its size

22
Extendible Hashing Example
Bucket capacity – 2 Initial buckets = 1
Insert 45,22
0
0
45
22
22
12
45
1
1
22
12
45
1
1
11
1
1
0
1
1
0
Global
depth
Local depth
Insert 12
Insert 11
Bucket overflows
local depth = global depth
⇒ Directory doubles and split image
is created
45 101101
22 10110
12 1100
11 1011

23
Insert 15
2
00
01
10
11
45
12
2
2
2
00
01
10
11
45
2
22
12
1
2
11
15
10
22
15
11
2
2
Insert 10
Overflow occurs.
Global depth = local depth
Directory doubles and split occurs
Overflows occurs.
Since local depth < global depth
Split image is created
Directory is not doubled
45 101101
22 10110
12 1100
11 1011
15 1111
10 1010

24
Linear Hashing
Does not require a separate directory structure
Uses a family of hash functions h0, h1, h2,….
• the range of hi is double the range of hi-1
• hi(x) = x mod 2iM
M - the initial no. of buckets
(Assume that the hashing field is an integer)
Initial hash functions
h0(x) = x mod M
h1(x) = x mod 2M

25
Insertion (1/3)
Initially the structure has M main buckets
( 0 ,…, M-1 ) and a few overflow buckets
To insert a record with hash field value x,
place the record in bucket ho(x)
When the first overflow in any bucket occurs:
Say, overflow occurred in bucket s
Insert the record in the overflow chain of bucket s
Create a new bucket M
Split the bucket 0 by using h1
Some records stay in bucket 0 and
some go to bucket M.
.
.
0
1
2
M-1
M
Overflow
buckets
Split image
of bucket 0

26
Insertion (2/3)
On first overflow,
irrespective of where it occurs, bucket 0 is split
On subsequent overflows
buckets 1, 2, 3, … are split in that order
(This why the scheme is called linear hashing)
N: the next bucket to be split
After M overflows,
all the original M buckets are split.
We switch to hash functions h1, h2
and set N = 0.
ho h1 hi
h1 h2 hi+1
… …
.
.
.
0
1
2
M-1
Split
images
M
M+1
.
.

27
Nature of Hash Functions
hi(x) = x mod 2iM. Let M' = 2iM
Note that if hi(x) = k then x = M'r + k, k < M'
and hi+1(x) = (M'r + k) mod 2M' = k or M' + k
M'– the current number of original buckets.
Since,
r – even – (M'2s + k) mod 2M' = k
r – odd – ( M'(2s + 1) + k ) mod 2M' = M' + k

28
Insertion (3/3)
Say the hash functions in use are hi, hi+1
To insert record with hash field value x,
Compute hi(x)
if hi(x) < N, the original bucket is already split
place the record in bucket hi+1(x)
else place the record in bucket hi(x)

29
Linear Hashing Example
Initial Buckets = 1 Bucket capacity = 2 records
N
0
Hash functions
h0 = x mod 1
h1 = x mod 2
Split pointer
Insert 12, 11
N
0 12
11
N
0 12
14
1 11
h0 = x mod 2
h1 = x mod 4
Insert 14
B0 overflows
Bucket pointed by
N is split
Hash functions are
changed

30
Insert 13
N
0 12
14
1 11
N
0 12
1 11
h0 = x mod 2
h1 = x mod 4
13
9
142
Insert 9
B1 overflows
B0 is split using h1
and split image
is created
N
0 12
1 11
13
9
142
Insert 10
h1 is
applied here
10
Insert 18
overflow at B2
split B1
h0 = x mod 4
h1 = x mod 8
0
1
2
3
12
9
13
14
10
11
18
N
13

31
Index Structures
Index: A disk data structure
– enables efficient retrieval of a record
given the value (s) of certain attributes
– indexing attributes
Primary Index:
Index built on ordering key field of a file
Clustering Index:
Index built on ordering non-key field of a file
Secondary Index:
Index built on any non-ordering field of a file

32
Primary Index
Can be built on ordered / sorted files
Index attribute – ordering key field (OKF)
Index Entry:
Index file: ordered file (sorted on OKF)
size-no. of blocks in the data file
Index file blocking factor BFi = B/(V +P)
(B-block size, V-OKF size, P-block pointer size)
- generally more than data file blocking factor
No of Index file blocks bi = b/BFi
(b - no. of data file blocks)
value of OKF for
the first record of
a block Bj
disk address
of Bj
101
121
129
240
.
.
.
.
101
104
121
123
129
130
240
244
.
.
.
.
0
1
2
b
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Ordering key
(RollNo) Data
file

33
Record Access Using Primary Index
Given Ordering key field (OKF) value: x
Carry out binary search on the index file
m – value of OKF for the first record in the middle block k of
the index file
x < m: do binary search on blocks 0 – (k −1) of index file
x ≥ m: if there is an index entry in block k with OKF value x,
use the corresponding block pointer,
get the data file block and
search for the data record with OKF value x
else do binary search on blocks k +1,…, bi of index file
Maximum block accesses required: ⌈log2
bi⌉

34
An Example
Data file:
No. of blocks b = 9500
Block size B = 4KB
OKF length V = 15 bytes
Block pointer length p = 6 bytes
Index file
No. of records ri = 9500
Size of entry V + P = 21 bytes
Blocking factor BFi = 4096/21 = 195
No. of blocks bi = ri/BFi = 49
Max No. of block accesses for getting record
using the primary index 1 + log2
bi = 7
Max No. of block accesses for getting record
without using primary index log2
b = 14

35
Making the Index Multi-level
49 entries
9500
entries
Second level
index
1 block
First level
index
49 blocks
data file
9500 blocks
Index file – itself an ordered file
– another level of index can be built
Multilevel Index –
Successive levels of indices are built till the last level has one block
height – no. of levels
block accesses: height + 1
(no binary search required)
For the example data file:
No of block accesses required with
multi-level primary index: 3
without any index: 14
.
.
.
.
.
.
.
.
.
.

36
Range Search, Insertion and Deletion
Range search on the ordering key field:
Get records with OKF value between x1 and x2 (inclusive)
Use the index to locate the record with OKF value x1 and read
succeeding records till OKF value exceeds x2.
Very efficient
Insertion: Data file – keep 25% of space in each block free
-- to take care of future insertions
index doesn't get changed
-- or use overflow chains for blocks that overflow
Deletion: Handle using deletion markers so that index doesn’t get
affected
Basically, avoid changes to index

37
Clustering Index
Built on ordered files where ordering field is not a key
Index attribute: ordering field (OF)
Index entry:
Index file: Ordered file (sorted on OF)
size – no. of distinct values of OF
Distinct value Vi
of the OF
address of the first
block that has a record with OF value Vi

38
Secondary Index
Built on any non-ordering field (NOF) of a data file.
Case I: NOF is also a key (Secondary key)
Case II: NOF is not a key: two options
(1)
(2)
Remarks:
(1) index entry – variable length record
(2) index entry – fixed length – One more level of indirection
value of the NOF Vi pointer to the record with Vi as the NOF value
value of the NOF Vi
value of the NOF Vi
pointer(s) to the record(s) with Vi as the NOF value
pointer to a block that has pointer(s) to the record(s)
with Vi as the NOF value

39
Secondary Index (key)
Can be built on ordered and also other type of files
Index attribute: non-ordering key field
Index entry:
Index file: ordered file (sorted on NOF values)
No. of entries – same as the no. of records in the data file
Index file blocking factor Bfi = B/(V+Pr)
(B: block size, V: length of the NOF,
Pr: length of a record pointer)
Index file blocks = ⎡r/Bfi⎤
(r – no. of records in the data file)
value of the NOF Vi pointer to the record with Vi as the NOF value

40
An Example
Data file:
No. of records r = 90,000 Block size B = 4KB
Record length R = 100 bytes BF = 4096/100 = 40,
b = 90000/40 = 2250
NOF length V = 15 bytes length of a record pointer Pr = 7 bytes
Index file :
No. of records ri = 90,000 record length = V + Pr = 22 bytes
BFi = 4096/22 = 186 No. of blocks bi = 90000/186 = 484
Max no. of block accesses to get a record
using the secondary index 1 + log2
bi = 10
Avg no. of block accesses to get a record
without using the secondary index b/2 = 1125
A very significant improvement

41
Multi-level Secondary Indexes
Secondary indexes can also be converted to multi-level indexes
First level index
– as many entries as there are records in the data file
First level index is an ordered file
so, in the second level index, the number of entries will be
equal to the number of blocks in the first level index
rather than the number of records
Similarly in other higher levels

42
Making the Secondary Index Multi-level
484 entries
90000
entries
Second level
index
3 blocks
First level
index
484 blocks
data file
90000
records
Multilevel Index –
Successive levels of indices are built
till the last level has one block
height – no. of levels
block accesses: height + 1
For the example data file:
No of block accesses required:
multi-level index: 4
single level index: 10
.
.
.
.
.
.
.
.
.
.
3 entries
1 block
2250
blocks

43
Index Sequential Access Method (ISAM) Files
ISAM files –
Ordered files with a multilevel primary/clustering index
Insertions:
Handled using overflow chains at data file blocks
Deletions:
Handled using deletion markers
Most suitable for files that are relatively static
If the files are dynamic, we need to go for dynamic multi-level
index structures based on B+- trees

44
B+
- trees
Balanced search trees
• all leaves are at the same level
Leaf node entries point to the actual data records
• all leaf nodes are linked up as a list
Internal node entries carry only index information
In B-trees, internal nodes carry data records also
The fan-out in B-trees is less
Makes sure that blocks are always at least half filled
Supports both random and sequential access of records

45
Order
Order (m) of an Internal Node
• Order of an internal node is the maximum number of tree
pointers held in it.
• Maximum of (m-1) keys can be present in an internal node
Order (mleaf) of a Leaf Node
• Order of a leaf node is the maximum number of record
pointers held in it. It is equal to the number of keys in a
leaf node.

46
Internal Nodes
An internal node of a B+
- tree of order m:
It contains at least pointers, except when it is the root node
It contains at most m pointers.
If it has P1, P2, …, Pj pointers with
K1 < K2 < K3 … < Kj-1 as keys, where ≤ j ≤ m, then
• P1 points to the subtree with records having key value x ≤ K1
• Pi (1 < i < j) points to the subtree with records having
key value x such that Ki-1 < x ≤ Ki
• Pj points to records with key value x > Kj-1
2
m
2
m

47
Internal Node Structure
2
m
≤ j ≤ m
P1 K1 P2 Pi PjK2 KiKi-1 Kj-1… … …
x ≤ K1 Ki-1 < x ≤ Ki Kj-1 < x
2 5 12
x ≤ 2
2 < x ≤ 5 5 < x ≤ 12
x > 12
Example
-

48
Leaf Node Structure
Structure of leaf node of B+
- of order mleaf :
It contains one block pointer P to point to next leaf node
At least record pointers and key values
At most mleaf record pointers and key values
If a node has keys K1 < K2 < … < Kj with Pr1, Pr2… Prj as record
pointers and P as block pointer, then
Pri points to record with Ki as the search field value, 1 ≤ i ≤ j
P points to next leaf block
K1 K2 KjPr1 Pr2 Pj P… …
leafm
2
leafm
2
……

49
Order Calculation
Block size: B, Size of Indexing field: V
Size of block pointer: P, Size of record pointer: Pr
Order of Internal node (m):
As there can be at most m block pointers and (m-1) keys
(m*P) + ((m-1) * V) ≤ B
m can be calculated by solving the above equation.
Order of leaf node:
As there can be at most mleaf record pointers and keys
with one block pointer in a leaf node,
mleaf can be calculated by solving
(mleaf * (Pr + V)) + P ≤ B

50
Example Order Calculation
Given B = 512 bytes V = 8 bytes
P = 6 bytes Pr = 7 bytes. Then
Internal node order m = ?
m * P + ((m-1) *V) ≤ B
m * 6 + ((m-1) *8) ≤ 512
14m ≤ 520
m ≤ 37
Leaf order mleaf = ?
mleaf (Pr + V) + P ≤ 512
mleaf (7 + 8) + 6 ≤ 512
15mleaf ≤ 506
mleaf ≤ 33

51
Example B+
- tree
m = 3 mleaf = 2
3
2
7
4 9
1 2 3 4 6 7 8 9 12 15
- - -
- - ^

52
Insertion into B+
- trees
1. Every node is inserted at leaf level
If leaf node overflows, then
• Node is split at j =
• First j entries are kept in original node
• Entities from j+1 are moved to new node
• jth key value is replicated in the parent of the leaf.
If Internal node overflows
• Node is split at j =
• Values and pointers up to Pj are kept in original node
• jth key value is moved to parent of the internal node
• Pj+1 to the rest of entries are moved to new node.
leaf(m 1)
2
+
(m 1)
2
+

53
Example of Insertions
m = 3 mleaf = 2
Insert 20, 11
11 20 11 14
14
20^
^
^
-
-
Insert 14
Overflow. leaf is split
at j = = 2
14 is replicated to upper level
leaf(m 1)
2
+
Insert 25
14
1411 20 25
14
1411 20 25 30
^- 25
^
Inserted at
leaf level
Insert 30
Overflow.
split at 25.
25 is moved
up
1
2
3 4

54
Insert 12 Overflow at leaf level.
- Split at leaf level,
- Triggers overflow at internal node
- Split occurs at internal node
14
30
12 25
11 12 14 2520
.5
-
- -
- -
^ ^
Internal node split
at j =
split at 14 and 14 is
moved up
2
m

55
Insert 22
14
12 22 25
11 12 14 20 22 25 30
-
-
-
^
^
Insert 23, 24
14 24
12 22 25
11 12 14 20 22 23 24 25 30
- ^
6
7

56
Deletion in B+
- trees
Delete the entry from the leaf node
Delete the entry if it is present in Internal node and replace with
the entry to its left in that position.
If underflow occurs after deletion
• Distribute the entries from left sibling
if not possible – Distribute the entries from right sibling
if not possible – Merge the node with left and right sibling

57
Example
14 24
12 22 25
11 12 14 20 22 23 24 25 30
14 24
12 22 25
11 12 14 22 23 24 25 30
Delete 20
Removed entry
from leaf here

58
14 24
12 23 25
11 12 14 23 24 25 30
14
12 23
11 12 14 23 25 30
Delete 24
Delete 22
25
Entry 22 is removed
from leaf and
internal node
Entries from right
sibling are
distributed to left

59
12
11 23
11 12 23 25 30
Delete 14
25
Delete 12
23 25
2311 25 30
Level drop has occurred

60
Advantages of B+
- trees:
1) Any record can be fetched in equal number of disk accesses.
2) Range queries can be performed easily as leaves are linked up
3) Height of the tree is less as only keys are used for indexing
4) Supports both random and sequential access.
Disadvantages of B+
- trees:
Insert and delete operations are complicated
Root node becomes a hotspot

5 data storage_and_indexing

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