Relational Algebra SQL (Structured Query Language)

Report
Basic Concepts of
Relational Databases
R1 sid
Example Instances
A table or relation stores records (or
tuples) that have the same attributes
22
58
S1 sid
Order of records is not important
conceptually
22
31
58
“Sailors” and “Reserves” relations for our
examples.
Questions: Can we have two records
with the same sid in the same sailor
(e.g., S1 or S2) table?
S2 sid
28
31
44
58
Can the same sailor reserve the same
boat many times?
2
bid
day
101 10/10/96
103 11/12/96
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Relational Algebra
Basic operations:
Selection (cR )
Selects the subset of rows that satisfy a condition c from R.
Projection (listR ) Keeps only the attributes in list from R.
Set-difference ( R1 – R2) Finds tuples in R1, but not in R2.
Union (R1  R2) Finds tuples that belong to R1 or R2.
Cross-product (R1 x R2) Allows us to combine two relations.
Renaming (p Rnew Rold) Allows us to rename a relation Rold to Rnew.
Additional operations:
Intersection, join, division: Not essential, but (very!) useful.
Since each operation returns a relation, operations can be composed! (Algebra is
“closed”.)
3
Projection
Deletes attributes that are not in projection list.
Schema of result contains exactly the fields in
the projection list, with the same names that
they had in the (only) input relation.
The formal projection operator has to eliminate
duplicates!
Note: real systems typically don’t do
duplicate elimination unless the user
explicitly asks for it.
sname
rating
yuppy
lubber
guppy
rusty
9
8
5
10
 sname,rating(S2)
age
35.0
55.5
 age(S2)
4
Selection
Selects rows that satisfy selection
condition.
sid sname rating age
28 yuppy 9
35.0
58 rusty
10
35.0
 rating 8(S2)
No duplicates in result! (Why?)
Schema of result identical to
schema of (only) input relation.
Result relation can be the input for
another relational algebra
operation! (Operator composition.)
sname rating
yuppy 9
rusty
10
 sname,rating( rating 8(S2))
5
Union, Intersection, Set-Difference
sid sname rating age
All of these operations take two input
relations, which must be unioncompatible:
•Same
number of fields.
•Corresponding
fields have the
same type.
22
31
58
44
28
dustin
lubber
rusty
guppy
yuppy
7
8
10
5
9
45.0
55.5
35.0
35.0
35.0
S1 S2
sid sname
22 dustin
sid sname
31 lubber
58 rusty
rating age
7
45.0
rating age
8
55.5
10
35.0
S1 S2
S1 S2
6
Cross (Cartesian)-Product, X
Each row of S is paired with each row of R.
Result schema has one field per field of S and R, with field names
`inherited’ if possible.
Conflict: Both S and R have a field called sid.
(sid) sname rating age
(sid) bid day
22
dustin
7
45.0
22
101 10/10/96
22
dustin
7
45.0
58
103 11/12/96
31
lubber
8
55.5
22
101 10/10/96
31
lubber
8
55.5
58
103 11/12/96
58
rusty
10
35.0
22
101 10/10/96
58
rusty
10
35.0
58
103 11/12/96
7
Joins
Condition Join:
R  c S   c ( R  S)
(sid) sname rating age
22
dustin 7
45.0
31
lubber 8
55.5
S
(sid) bid
58
103
58
103
S.sid R.sid
day
11/12/96
11/12/96
R
Result schema same as that of cross-product.
Fewer tuples than cross-product, might be able to compute more
efficiently
Sometimes called a theta-join.
8
Joins
Equi-Join: A special case of condition join where the condition c contains
only equalities.
S
sid
22
58
sname
dustin
rusty
sid
R
rating age
7
45.0
10
35.0
bid
101
103
day
10/10/96
11/12/96
Result schema similar to cross-product, but only one copy of fields for
which equality is specified.
Natural Join: Equijoin on all common fields.
9
Basic SQL Query
SELECT
FROM
WHERE
[DISTINCT] target-list
relation-list
qualification
target-list A list of attributes of relations in relation-list
relation-list A list of relation names (possibly with a range-variable after each
name).
qualification Comparisons (Attr op const or Attr1 op Attr2, where op is one of
<,>.=,,,) combined using AND, OR and NOT.
DISTINCT is an optional keyword indicating that the answer should not contain
duplicates. Default is that duplicates are not eliminated!
10
Conceptual Evaluation Strategy
Semantics of an SQL query defined in terms of the following conceptual
evaluation strategy:
Compute the cross-product of relation-list.
Discard resulting tuples if they fail qualifications.
Delete attributes that are not in target-list.
If DISTINCT is specified, eliminate duplicate rows.
This strategy is probably the least efficient way to compute a query! An
optimizer will find more efficient strategies to compute the same
answers.
11
Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
(sid) sname rating age
(sid) bid day
22 dustin
7
45.0
22
101 10/10/96
22 dustin
7
45.0
58
103 11/12/96
31 lubber
8
55.5
22
101 10/10/96
31 lubber
8
55.5
58
103 11/12/96
58 rusty
10
35.0
22
101 10/10/96
58 rusty
10
35.0
58
103 11/12/96
12
A Note on Range Variables
Really needed only if the same relation appears twice in the FROM
clause. The previous query can also be written as:
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND bid=103
OR
SELECT sname
FROM Sailors, Reserves
WHERE Sailors.sid=Reserves.sid
AND bid=103
13
Set Operations - Union
Find sid’s of sailors who’ve reserved a red or a green boat
UNION: Can be used to compute the
union of any two union-compatible sets
of tuples (which are themselves the
result of SQL queries).
If we replace OR by AND in the first
version, what do we get?
Do we need the Sailors table?
Also available: EXCEPT (corresponds to
the set difference operation of relational
algebra)
What do we get if we replace UNION by
EXCEPT?
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.color=‘red’ OR B.color=‘green’)
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
UNION
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘green’
14
Aggregate Operators
Significant extension of relational algebra.
SELECT COUNT (*)
FROM Sailors S
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
single column
SELECT S.sname
FROM Sailors S
WHERE S.rating= (SELECT MAX(S2.rating)
FROM Sailors S2)
SELECT COUNT (DISTINCT S.rating)
FROM Sailors S
WHERE S.sname=‘Bob’
15
SELECT AVG ( DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
GROUP BY and HAVING
So far, we’ve applied aggregate operators to all (qualifying) tuples.
Sometimes, we want to apply them to each of several groups of
tuples.
Consider: Find the age of the youngest sailor for each rating level.
In general, we don’t know how many rating levels exist, and what the rating
values for these levels are!
Suppose we know that rating values go from 1 to 10; we can write 10
queries that look like this (!):
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
For i = 1, 2, ... , 10:
16
Queries With GROUP BY and HAVING
SELECT
FROM
WHERE
GROUP BY
HAVING
[DISTINCT] target-list
relation-list
qualification
grouping-list
group-qualification
The target-list contains (i) an attribute list (ii) terms with aggregate operations
(e.g., MIN (S.age)).
The attribute list (i) must be a subset of grouping-list. Intuitively, each answer
tuple corresponds to a group, and these attributes must have a single value per
group. (A group is a set of tuples that have the same value for all attributes in
grouping-list.)
17
For each red boat, find the number of
reservations for this boat
SELECT B.bid, COUNT (*) AS scount
FROM Boats B, Reserves R
WHERE R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid
Grouping over a join of two relations.
We cannot remove B.color=‘red’ from the WHERE clause and add a
HAVING clause with this condition. Only columns that appear in
the Group-By can appear in HAVING, unless they are
arguments of an aggregate operator in HAVING.
18
Basic Concepts of Indexing
Indexing mechanisms used to speed up access to desired data.
E.g., author catalog in library
Search Key - attribute to set of attributes used to look up records
in a file.
An index file consists of records (called index entries) of the form
search-key
pointer
Index files are typically much smaller than the original file
Two basic kinds of indices:
Ordered indices: search keys are stored in sorted order
Hash indices: search keys are distributed uniformly across
“buckets” using a “hash function”.
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Ordered Indices
In an ordered index, index entries are stored sorted on the search
key value. E.g., author catalog in library.
Primary index: in a sequentially ordered file, the index whose
search key specifies the sequential order of the file.
Also called clustering index
The search key of a primary index is usually but not necessarily the
primary key
Secondary index: an index whose search key specifies an order
different from the sequential order of the file. Also called
non-clustering index.
Index-sequential file: ordered sequential file with a primary index
(also called ISAM - indexed sequential access method).
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Dense Index Files
Dense index — Index record appears for every search-key value in
the file (in this example we have clustering index).
21
Sparse Index Files
Sparse Index: contains index records for only some search-key
values.
Applicable when records are sequentially ordered on search-key (i.e.,
only applicable to clustering indexes)
To locate a record with search-key value K we:
Find index record with largest search-key value <= K
Search file sequentially starting at the record to which the index record
points
Less space and less maintenance overhead for insertions and
deletions.
Generally slower than dense index for locating records.
Good tradeoff: sparse index with an index entry for every block in
file, corresponding to least search-key value in the block.
22
Example of Sparse Index Files
23
Multilevel Index
If primary index does not fit in memory, access becomes expensive.
To reduce number of disk accesses to index records, treat primary
index kept on disk as a sequential file and construct a sparse
index on it.
outer index – a sparse index of primary index
inner index – the primary index file
If even outer index is too large to fit in main memory, yet another
level of index can be created, and so on.
Indices at all levels must be updated on insertion or deletion from
the file.
24
Multilevel Index (Cont.)
25
B+-Tree Index
A B+-tree is a rooted tree satisfying the following properties:
All paths from root to leaf are of the same length (i.e.,
balanced tree)
Each node has between n/2 and n pointers. Each leaf node
stores between (n–1)/2 and n–1 values.
n is called fanout (it corresponds to the maximum number of
pointers/children). The value (n-1)/2 is called order (it
corresponds to the minimum number of values).
Special cases:
If the root is not a leaf, it has at least 2 children.
If the root is a leaf (that is, there are no other nodes in the tree), it
can have between 0 and (n–1) values.
26
Example of clustering (primary) B+-tree on
candidate key
41
45 51
11 21 30
1
3
11 13 15
21 23
30 33
41 43
45 47
51 53
FILE W ITH REC O RDS
record with
search key 1
record with
search key 3
This example corresponds to dense B+-tree index:
Every search key value appears in a leaf node
You may also have sparse B+-tree, e.g., entries in leaf nodes correspond to pages
27
Example of non-clustering (secondary) B+tree on candidate key
41
45 51
11 21 30
1
3
11 13 15
21 23
30 33
41 43
45 47
51 53
FILE W ITH REC O RDS
record wit h
record wit h
search key 11 search key 3
record wit h
search key 1
Should be alw ays dense
28
Example of clustering B+-tree on noncandidate key
41
45 51
11 21 30
1
3
11 13 15
21 23
30 33
41 43
45 47
FILE W ITH REC O RDS
records wit h
search key 1
record with
search key 3
29
51 53
Example of non-clustering B+-tree on noncandidate key
41
45 51
11 21 30
1
3
11 13 15
point ers to
records wit h
search key 1
21 23
30 33
41 43
45 47
point ers to
records wit h
search key 3
FILE W ITH REC O RDS
30
51 53
Inserting a Data Entry into a B+ Tree
Find correct leaf L.
Put data entry onto L.
If L has enough space, done!
Else, must split L (into L and a new node L2)
Redistribute entries evenly, copy up middle key.
Insert index entry pointing to L2 into parent of L.
This can happen recursively
To split index node, redistribute entries evenly, but push up middle key.
(Contrast with leaf splits.)
Splits “grow” tree; root split increases height.
Tree growth: gets wider or one level taller at top.
31
Deleting a Data Entry from a B+ Tree
Start at root, find leaf L where entry belongs.
Remove the entry.
If L is at least half-full, done!
If L less that half-full,
Try to re-distribute, borrowing from sibling (adjacent node to the
right).
If re-distribution fails, merge L and sibling.
If merge occurred, must delete entry (pointing to L or sibling) from parent
of L.
Merge could propagate to root, decreasing height.
32
B+-tree Update Examples
Consider the B+-tree below with order 2 (each node except for the root
must contain at least two search key values – and 3 pointers). Show the
tree that would result after successively applying each of the following
operations.
41
45 51
11 21 30
1
3
11 13 15
21 23
30 33
41 43
45 47
Remove 1
51 53
Remove 41
41
45 51
13 21 30
3
11
13 15
21 23
30 33
41 43
33
45 47
51 53
B+-tree Update Examples (cont)
After removing 41
30
43 51
13 21
3
11
13 15
21 23
30 33
43 45 47
Remove 3
51 53
Insert 41
21 30 43 51
11 13 15
21 23
30 33
43 45 47
34
51 53
B+-tree Update Examples (cont)
After inserting 41
21 30 43 51
3
11 13 15
21 23
30 33 41
51 53
43 45 47
Insert 1
30
43 51
13 21
1
3
11
13 15
21 23
30 33 41
35
43 45 47
51 53
Bulk Loading of a B+ Tree
If we have a large collection of records, and we want to create a B+ tree
on some field, doing so by repeatedly inserting records is very slow.
Bulk Loading can be done much more efficiently.
Initialization: Sort all data entries (using external sorting), insert pointer
to first (leaf) page in a new (root) page.
Root
3* 4*
Sorted pages of data entries; not yet in B+ tree
6* 9*
10* 11*
12* 13* 20* 22* 23* 31* 35* 36*
36
38* 41* 44*
Bulk Loading (Cont.)
Root
Index entries for leaf pages
always entered into right-most
index page just above leaf
level. When this fills up, it
splits. (Split may go up rightmost path to the root.)
3*
10
20
Data entry pages
6
4*
6* 9*
12
23
35
not yet in B+ tree
10* 11* 12* 13* 20*22* 23* 31* 35* 36* 38*41* 44*
Much faster than repeated
inserts!
Root
20
10
6
3* 4*
6* 9*
12
Data entry pages
not yet in B+ tree
35
23
38
10* 11* 12* 13* 20*22* 23* 31* 35* 36* 38*41* 44*
37
Hash Indices
Hashing can be used not only for file organization, but also for
index-structure creation.
A hash index organizes the search keys, with their associated
record pointers, into a hash file structure.
Strictly speaking, hash indices are always secondary indices
if the file itself is organized using hashing, a separate primary hash
index on it using the same search-key is unnecessary.
38
Example of Hash Index
39
Hash Functions
In the worst case, the hash function maps all search-key values to
the same bucket; this makes access time proportional to the
number of search-key values in the file.
Ideal hash function is random, so each bucket will have the same
number of records assigned to it irrespective of the actual
distribution of search-key values in the file.
Typical hash functions perform computation on the internal binary
representation of the search-key.
For example, for a string search-key, the binary representations of all
the characters in the string could be added and the sum modulo the
number of buckets could be returned.
40
External Sorting (disk-resident files)
Merging Sorted Files with 3 pages of main memory buffer
sorted file 2
merged file
(1,…)
(2,…)
(1,…)
(5,…)
(4,…)
(2,…)
(7,…)
(6,…)
(4,…)
(11,…)
(9,…)
(5,…)
(12,…)
(10,…)
(6,…)
(15,…)
(14,…)
(7,…)
(20,…)
(17,…)
(9,…)
(21,…)
(18,…)
(11,…)
sorted file 1
…
41
External Sorting (disk-resident files)
sorted file 1
sorted file 2
merged file
(1,…)
(2,…)
(1,…)
(5,…)
(4,…)
(2,…)
(7,…)
(6,…)
(4,…)
(11,…)
(9,…)
(5,…)
bring next page of file 1
write page
to disk
bring next page of file 2
(6,…)
(12,…)
(10,…)
(15,…)
(14,…)
(9,…)
(20,…)
(17,…)
(10,…)
(21,…)
(18,…)
(11,…)
(7,…)
42
write page
to disk
External Sorting (disk-resident files)
Continuing the previous example:
Question: I assumed that each file is already sorted. If the file is not
sorted, how do I sort it (using only my 3 buffer pages?)
Answer: Each file in the example is only 2 pages. Therefore, I can
bring the entire file in memory, and sort it using any main-memory
algorithm.
Question: The previous example assumes two separate files. How do I
apply this idea to sort a single file?
Answer: You can split the file in two parts and merge them as if they
were separate files.
Question: Can I do better if I have M>3 main memory pages.
Answer: Yes, instead of 2 you can merge up to M-1 files (because you
need 1 page for writing the output).
43
Sorting a file that has 108 pages, using only 5 pages
of main memory.
1
2
1
2
5
108
PASS 0
5
6
7
sorted run 1
PASS 1
10
......
sorted run 2
run 1
run 2
run 3
run 4
sorted run 22
run 21
......
4x5=20
pages
sorted run 6
run 1 run 2 run 3 run 4 run 5 run 6
80
pages
28
pages
sorted run 1
PASS 3
run 22
8
pages
sorted run 1
PASS 2
106 107 108
sorted run 2
run 1 run 2
108
pages
sorted file
44
Basic Steps in Query Processing
Parsing and translation
translate the query into its internal form. This is then translated into relational
algebra.
Parser checks syntax, verifies relations
Evaluation
The query-execution engine takes a query-evaluation plan, executes that plan,
and returns the answers to the query.
45
Basic Steps in Query Processing :
Optimization
A relational algebra expression may have many equivalent expressions
E.g., balance2500(balance(account)) is equivalent to
balance(balance2500(account))
Each relational algebra operation can be evaluated using one of several
different algorithms
Correspondingly, a relational-algebra expression can be evaluated in many
ways.
Annotated expression specifying detailed evaluation strategy is called
an evaluation-plan.
E.g., can use an index on balance to find accounts with balance < 2500,
or can perform complete relation scan and discard accounts with balance 
2500
46
Selection Operation
File scan – search algorithms that locate and retrieve records that
fulfill a selection condition.
Algorithm A1 (linear search). Scan each file block and test all
records to see whether they satisfy the selection condition.
Cost estimate (number of disk blocks scanned) = br
br denotes number of blocks containing records from relation r
If selection is on a key attribute, cost = (br /2)
stop on finding record
Linear search can be applied regardless of
selection condition or
ordering of records in the file, or
availability of indices
47
Selection Operation (Cont.)
A2 (binary search). Applicable if selection is an equality
comparison on the attribute on which file is ordered.
Assume that the blocks of a relation are stored contiguously
Cost estimate (number of disk blocks to be scanned):
log2(br) — cost of locating the first tuple by a binary search on
the blocks
Plus number of blocks containing records that satisfy selection
condition
48
Selections Using Indices
Index scan – search algorithms that use an index
selection condition must be on search-key of index.
A3 (primary index on candidate key, equality). Retrieve a single record that
satisfies the corresponding equality condition
Cost = HTi + 1 (HTi is the height of the tree index)
A4 (primary index on nonkey, equality) Retrieve multiple records.
Records will be on consecutive blocks
Cost = HTi + number of blocks containing retrieved records
A5 (equality on search-key of secondary index).
Retrieve a single record if the search-key is a candidate key
Cost = HTi + 1 (if hash index HTi = 1 or HTi = 1.2 if we assume that there
exist overflow buckets)
Retrieve multiple records if search-key is not a candidate key
Cost = HTi + number of records retrieved
Can be very expensive!
each record may be on a different block
one block access for each retrieved record
49
Selections Involving Comparisons
Can implement selections of the form AV (r) or A  V(r) by using
a linear file scan or binary search,
or by using indices in the following ways:
A6 (primary index, comparison). (Relation is sorted on A)
For A  V(r) use index to find first tuple  v and scan relation
sequentially from there
For AV (r) just scan relation sequentially till first tuple > v; do not use
index
A7 (secondary index, comparison).
For A  V(r) use index to find first index entry  v and scan index
sequentially from there, to find pointers to records.
For AV (r) just scan leaf pages of index finding pointers to records, till
first entry > v
In either case, retrieve records that are pointed to
requires an I/O for each record
Linear file scan may be cheaper if many records are
to be fetched!
50
Join Operation
Several different algorithms to implement joins
Block nested-loop join
Indexed nested-loop join
Merge-join
Hash-join
51
Block Nested-Loop Join
Variant of nested-loop join in which every block of inner
relation is paired with every block of outer relation.
for each block Br of r do begin
for each block Bs of s do begin
for each tuple tr in Br do begin
for each tuple ts in Bs do begin
Check if (tr,ts) satisfy the join condition
if they do, add tr • ts to the result.
end
end
end
end
52
Block Nested-Loop Join (Cont.)
Worst case estimate: br  bs + br block accesses.
Each block in the inner relation s is read once for each block in the
outer relation (instead of once for each tuple in the outer relation
Best case: br + bs block accesses.
Improvements to nested loop and block nested loop algorithms:
In block nested-loop, use M — 2 disk blocks as blocking unit for
outer relations, where M = memory size in blocks; use remaining
two blocks to buffer inner relation and output
Cost = br / (M-2)  bs + br
Optimizations:
If equi-join attribute forms a key or inner relation, stop inner loop on
first match
Scan inner loop forward and backward alternately, to make use of
the blocks remaining in buffer (with LRU replacement)
Use main-memory hash table for the outer relation (to decrease CPU
cost)
53
Indexed Nested-Loop Join
Index lookups can replace file scans if
join is an equi-join or natural join and
an index is available on the inner relation’s join attribute
Can construct an index just to compute a join.
For each tuple tr in the outer relation r, use the index to look up tuples
in s that satisfy the join condition with tuple tr.
Cost of the join: br + nr  c
Where c is the cost of traversing index and fetching all matching s tuples
for one tuple or r
c can be estimated as cost of a single selection on s using the join
condition.
If indices are available on join attributes of both r and s,
use the relation with fewer tuples as the outer relation.
54
Merge-Join
Sort both relations on their join attribute (if not already sorted on the
join attributes).
Merge the sorted relations to join them
Join step is similar to the merge stage of the sort-merge algorithm.
Main difference is handling of duplicate values in join attribute — every
pair with same value on join attribute must be matched
55
Merge-Join (Cont.)
Can be used only for equi-joins and natural joins
Each block needs to be read only once (assuming all tuples for any
given value of the join attributes fit in memory)
Thus number of block accesses for merge-join is
br + bs + the cost of sorting if relations are unsorted.
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Hash-Join
Applicable for equi-joins and natural joins.
A hash function h is used to partition tuples of both relations
h maps JoinAttrs values to {0, 1, ..., n}, where JoinAttrs denotes the
common attributes of r and s used in the natural join.
r0, r1, . . ., rn denote partitions of r tuples
Each tuple tr  r is put in partition ri where i = h(tr [JoinAttrs]).
s0,, s1. . ., sn denotes partitions of s tuples
Each tuple ts s is put in partition si, where i = h(ts [JoinAttrs]).
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Hash-Join (Cont.)
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Hash-Join (Cont.)
r tuples in ri need only to be compared with s tuples in si
Need not be compared with s tuples in any other partition,
since:
an r tuple and an s tuple that satisfy the join condition will have the
same value for the join attributes.
If that value is hashed to some value i, the r tuple has to be in ri
and the s tuple in si.
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Hash-Join Algorithm
The hash-join of r and s is computed as follows.
1. Partition the relation s using hashing function h. When
partitioning a relation, one block of memory is reserved as
the output buffer for each partition.
2. Partition r similarly.
3. For each i:
(a) Load si into memory and build an in-memory hash index on it
using the join attribute. This hash index uses a different hash
function than the earlier one h.
(b) Read the tuples in ri from the disk one by one. For each tuple
tr locate each matching tuple ts in si using the in-memory hash
index. Output the concatenation of their attributes.
Relation s is called the build input and
r is called the probe input.
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Hash-Join algorithm (Cont.)
The number of partitions n and the hash function h is chosen
such that each si should fit in memory.
In order for each partition of the build input to fit in memory: n ≥
bs/M
Also M ≥ n+1 because for each partition we should have one buffer
page (plus one page for input buffer)
In order to satisfy these conditions: M > sqrt(bs)
The probe relation partitions need not fit in memory
Recursive partitioning required if number of partitions n is
greater than number of pages M of memory.
Rarely required: e.g., recursive partitioning not needed for relations
of 1GB or less with memory size of 2MB, with block size of 4KB.
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