[SOLVED] CS计算机代考程序代写 algorithm SQL data structure database DBMS Overview

DBMS Overview

>>
DBMS Overview

DBMSs

Query Evaluation

Mapping SQL to RA

Mapping Example

Query Cost Estimation

Implementations of RA Ops

DBMS Architecture

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❖ DBMSs

DBMS = DataBase Management System

Our view of the DBMS so far …

A machine to process SQL queries.

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❖ DBMSs (cont)

One view of DB engine: “relational algebra virtual machine”

Machine code for such a machine:

selection (σ)projection (π)join (⨝, ×)
union (∪)intersection (∩)difference (-)
sortinsertdelete

For each of these operations:

various data structures and algorithms are available

DBMSs may provide only one, or may provide a choice

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❖ Query Evaluation

The path of a query through its evaluation:

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❖ Mapping SQL to RA

Mapping SQL to relational algebra, e.g.

— schema: R(a,b) S(c,d)
select a as x
from R join S on (b=c)
whered = 100
— could be mapped to
Tmp1(a,b,c,d) = R Join[b=c] S
Tmp2(a,b,c,d) = Sel[d=100](Tmp1)
Tmp3(a) = Proj[a](Tmp2)
Res(x)= Rename[Res(x)](Tmp3)

In general:

SELECT clause becomes projection

WHERE condition becomes selection or join

FROM clause becomes join

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❖ Mapping Example

Consider the database schema:

Person(pid, name, address, …)
Subject(sid, code, title, uoc, …)
Terms(tid, code, start, end, …)
Courses(cid, sid, tid, …)
Enrolments(cid, pid, mark, ..)

and the query: Courses with more than 100 students in them?

which can be expressed in SQL as

select s.sid, s.code
from Course c join Subject s on (c.sid=s.sid)
join Enrolment e on (c.cid=e.cid)
groupby s.sid, s.code
having count(*) > 100;

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❖ Mapping Example (cont)

The SQL might be compiled to

Tmp1(cid,sid,pid) = Course Join[c.cid = e.cid] Enrolment
Tmp2(cid,code,pid)= Tmp1 Join[t1.sid = s.sid] Subject
Tmp3(cid,code,nstu) = GroupCount[cid,code](Tmp2)
Res(cid,code) = Sel[nstu > 100](Tmp3)

or, equivalently

Tmp1(cid,code)= Course Join[c.sid = s.sid] Subject
Tmp2(cid,code,pid)= Tmp1 Join[t1.cid = e.cid] Enrolment
Tmp3(cid,code,nstu) = GroupCount[cid,code](Tmp2)
Res(cid,code) = Sel[nstu > 100](Tmp3)

Which is better?

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❖ Query Cost Estimation

The cost of evaluating a query is determined by

the operations specified in the query execution plan

size of relations   (database relations and temporary relations)

access mechanisms   (indexing, hashing, sorting, join algorithms)

size/number of main memory buffers   (and replacement strategy)

Analysis of costs involves estimating:
the size of intermediate results

then, based on this, cost of secondary storage accesses

Accessing data from disk is the dominant cost in query evaluation

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❖ Query Cost Estimation (cont)

Consider execution plans for:
σc (R  ⨝d  S  ⨝e  T)
where R(c,d), S(d,e), T(e)

Tmp1(c,d,e):=hash_join[d](R,S)
Tmp2(c,d,e):=sort_merge_join[e](tmp1,T)
Res(c,d,e) :=binary_search[c](Tmp2)

or

Tmp1(d,e):=sort_merge_join[e](S,T)
Tmp2(c,d,e):=hash_join[d](R,Tmp1)
Res(c,d,e) :=linear_search[c](Tmp2)

or

Tmp1(c,d):=btree_search[c](R)
Tmp2(c,d,e):=hash_join[d](Tmp1,S)
Res(c,d,e) :=sort_merge_join[e](Tmp2,T)

All produce same result, but have different costs.

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❖ Implementations of RA Ops

Sorting   (quicksort, etc. are not applicable)

external merge sort   (cost O(NlogBN) with B memory buffers)

Selection   (different techniques developed for different query types)
sequential scan   (worst case, cost O(N))

index-based   (e.g. B-trees, cost O(logN), tree nodes are pages)

hash-based   (O(1) best case, only works for equality tests)

Join   (fast joins are critical to success of relational DBMSs)
nested-loop join   (cost O(N.M), buffering can reduce to O(N+M))

sort-merge join   (cost O(NlogN+MlogM))

hash-join   (best case cost O(N+M.N/B), with B memory buffers)

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❖ DBMS Architecture

Most RDBMSs are client-server systems:

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<< ∧❖ DBMS Architecture (cont)Layers within the DBMS server:COMP9315 21T1 ♢ DBMS Overview ♢ [11/11]Produced: 15 Feb 2021