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General Overview

- Relational model - SQL
- Formal & commercial query languages
- Functional Dependencies
- Normalization
- Physical Design
- Indexing
- Query Processing and Optimization

Review: QP & O

SQL Query

Query

Processor

Parser

Query Optimizer

Algebraic

Expression

Execution plan

Evaluator

Data: result of the query

Review: QP & O

Query

Optimizer

Algebraic

Representation

Query Rewriter

Algebraic Representation

Data Stats

Plan Generator

Query Execution Plan

Selections Involving Comparisons

Query: Att K (r )

- A6 (primary index, comparison). (Relation is sorted on Att)
- For Att V(r) use index to find first tuple v and scan relation sequentially from there
- For AttV (r) just scan relation sequentially till first tuple > v; do not use index

Cost: EA5 =HTi + c / fr (where c is the cardinality of result)

HTi

k

...

k

Query: Att K (r )

Cardinality: More metadata on r are needed:

min (att, r) : minimum value of att in r

max(att, r): maximum value of att in r

Then the selectivity of Att = K (r ) is estimated as:

(or nr /2 if min, max unknown)

Intuition: assume uniform distribution of values between min and max

min(attr, r)

K

max(attr, r)

Plan generation: Range Queries

Att K (r )

A6: (secondary index, comparison).

Cost:

EA6 = HTi -1+ #of leaf nodes to read + # of file blocks to read

= HTi -1+ LBi * (c / nr) + c, if att is a candidate key

HTi

...

k+m

k, k+1

k+1

...

k+m

k

Plan generation: Range Queries

A6: (secondary index, range query). If att is NOT a candidate key

HTi

...

k+m

k, k+1

...

k

...

k

k+1

k+m

...

Cost: EA6 = HTi -1+ #of leaf nodes to read + #of file blocks to read +#buckets to read

= HTi -1+ LBi * (c / nr) + c + x

Join Operation

- Size and plans for join operation
- Running example: depositor customer

Metadata:

ncustomer = 10,000 ndepositor = 5000

fcustomer = 25 fdepositor = 50

bcustomer= 400 bdepositor= 100

V(cname, depositor) = 2500 (each customer has on average 2 accts)

cname in depositor a foreign key for customer

depositor(cname, acct_no)

customer(cname, cstreet, ccity)

Cardinality of Join Queries

- What is the cardinality (number of tuples) of the join?

E1: Cartesian product: ncustomer * ndepositor = 50,000,000

E2: Attribute cname common in both relations, 2500

different cnames in depositor

Size: ncustomer * (avg# of tuples in depositor with same cname)

= ncustomer * (ndepositor / V(cname, depositor))

= 10,000 * (5000 / 2500)

= 20,000

Cardinality of Join Queries

E3: cname is a foreign key for depositor on customer

Size: ndepositor * (avg # of tuples in customer with same cname)

= ndepositor * 1

= 5000

Note: If cname is a key for customer but it is NOT a foreign key for depositor,

(i.e., not all cnames of depositor are in customer) then 5000 an UPPER BOUND

Some customer names may not match w/ any customers in customer

Cardinality of Joins in general

Assume join: R S

- If R, S have no common attributes: nr * ns
- If R,S have attribute A in common:

(take min)

- If R, S have attribute A in common and:
- A is a candidate key for R: ≤ ns
- A is candidate key in R and candidate key in S : ≤ min(nr, ns)
- A is a key for R, foreign key for S: = ns

Nested-Loop Join

Query: R S

Algorithm 1: Nested Loop Join

Idea:

t1

u1

Blocks

of...

t2

u2

t3

u3

R

S

results

Compare: (t1, u1), (t1, u2), (t1, u3) .....

Then: GET NEXT BLOCK OF S

Repeat: for EVERY tuple of R

Nested-Loop Join

Query: R S

Algorithm 1: Nested Loop Join

for each tuple tr in R do for each tuple usin S dotest pair (tr,us) tosee if they satisfy the join condition if they do (a “match”), add tr• usto the result.

R is called the outerrelation and S the inner relation of the join.

Nested-Loop Join (Cont.)

Cost:

- Worst case, if buffer size is 3 blocks br + nrbsdisk accesses.
- Best case: buffer big enough for entire INNER relation + 2

br + bs DAs.

ncustomer = 10,000 ndepositor = 5000

fcustomer = 25 fdepositor = 50

bcustomer= 400 bdepositor= 100

- Assuming worst case memory availability cost estimate is
- 5000 400 + 100 = 2,000,100 disk accesses with depositor as outer relation, and
- 10000 100 + 400 = 1,000,400 disk accesses with customer as the outer relation.
- If smaller relation (depositor) fits entirely in memory, the cost estimate will be 500 disk accesses. (actually we need 2 more blocks)

Join Algorithms

Query: R S

Algorithm 2: Block Nested Loop Join

Idea:

t1

u1

Blocks

of...

t2

u2

t3

u3

R

S

results

Compare: (t1, u1), (t1, u2), (t1, u3)

(t2, u1), (t2, u2), (t2, u3)

(t3, u1), (t3, u2), (t3, u3)

Then: GET NEXT BLOCK OF S

Repeat: for EVERY BLOCK of R

Block Nested-Loop Join

- Block Nested Loop Join

for each block BRofR dofor each block BSof S do for each tuple trin BR do for each tuple usin Bsdo beginCheck if (tr,us) satisfy the join condition if they do (“match”), add tr• usto the result.

Block Nested-Loop Join (Cont.)

Cost:

- Worst case estimate: br bs + br block accesses.
- Best case: br+ bsblock accesses. Same as nested loop.
- Improvements to nested loop and block nested loop algorithms for a buffer with M blocks:
- 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
- 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)

Join Algorithms

Query: R S

Algorithm 3: Indexed Nested Loop Join

Idea:

t1

Blocks

of...

t2

t3

R

S

results

(fill w/

blocks of

S or index blocks)

For each tuple ti of R

if ti.A = K (A is the attribute R,S have in common)

then use the index to compute att = K (S )

Demands: index on A for S

Indexed Nested-Loop Join

Indexed Nested Loop Join

- For each tuple tRin the outer relation R, use the index to look up tuples in S that satisfy the join condition with tuple tR.
- Worst case: buffer has space for only one page of R, and, for each tuple in R, we perform an index lookup on s.
- Cost of the join: br + nr c
- Where c is the cost of traversing the index and fetching all matching s tuples for one tuple from 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.

Example of Nested-Loop Join Costs

Query: depositor customer

(cname, acct_no) (cname, ccity, cstreet)

Metadata:

customer: ncustomer = 10,000

fcustomer = 25 bcustomer = 400

depositor: ndepositor = 5000

fdepositor = 50 bdepositor = 100

V (cname, depositor) = 2500

i a primary index on cname (dense) for customer (fi = 20)

Minimal buffer

Plan generation for Joins

Algorithm 2: Block Nested Loop

1a: customer = OUTER relation

depositor = INNER relation

cost: bcustomer + bdepositor * bcustomer = 400 +(400 *100) = 40,400

1b: customer = INNER relation

depositor = OUTER relation

cost: bdepositor + bdepositor * bcustomer = 100 +(400 *100) = 40,100

Plan generation for Joins

Algorithm 3: Indexed Nested Loop

We have index on cname for customer. Depositor is the outer relation

Cost:

bdepositor + ndepositor * c = 100 +(5000 *c ) , c is the cost of evaluating a selection cname=K using index.

What is c? Primary index on cname, cname a key for customer

c = HTi +1

Plan generation for Joins

What is HTi ?

cname a key for customer. V(cname, customer) = 10,000

fi = 20, i is dense

LBi = 10,000/20 = 500

HTi ~ logfi(LBi) + 1 = log20 500 + 1 = 4

Cost of index nested loop is:

= 100 + (5000 * (4+1)) = 25,100 BA (cheaper than NLJ)

pS

Another Join StrategyQuery: R S

Algorithm: Merge Join

Idea: suppose R, S are both sorted on A (A is the common attribute)

A

A

2

2

3

5

1

2

3

4

...

...

Compare:

(1, 2) advance pR

(2, 2) match, advance pS add to result

(2, 2) match, advance pS add to result

(2, 3) advance pR

(3, 3) match, advance pS add to result

(3, 5) advance pR

(4, 5) read next block of R

Merge-Join

GIVEN R, S both sorted on A

- Initialization
- Reserve blocks of R, S into buffer reserving one block for result
- Pr= 1, Ps =1
- Join (assuming no duplicate values on A in R)

WHILE !EOF( R) && !EOF(S) DO

if BR[Pr].A == BS[Ps].A then

output to result; Ps++

else if BR[Pr].A < BS[Ps].A then

Pr++

else (same for Ps)

if Pr or Ps point past end of block,

read next block and set Pr(Ps) to 1

Merge-Join (Cont.)

- 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
- But....

What if one/both of R,S not sorted on A?

Ans: May be worth sorting first and then perform merge join (Sort-Merge Join)

Cost: bR + bS + sortR + sortS

External Sorting

Not the same as internal sorting

Internal sorting:

minimize CPU (count comparisons)

best: quicksort, mergesort, ....

External sorting:

minimize disk accesses (what we ‘re sorting doesn’t fit in memory!)

best: external merge sort

WHEN used?

1) SORT-MERGE join

2) ORDER BY queries

3) SELECT DISTINCT (duplicate elimination)

e

g

m

p

r

31

16

24

3

2

16

External SortingIdea:

1. Sort fragments of file in memory using internal sort (runs). Store runs on disk.

2. Merge runs. E.g.:

a

b

c

19

14

33

a

d

g

19

31

24

sort

a

a

b

c

d

d

d

e

g

m

p

r

g

a

d

c

b

e

r

d

m

p

d

a

14

19

14

33

7

21

31

16

24

3

2

16

24

19

31

33

14

16

16

21

3

2

7

14

merge

sort

b

c

e

14

33

16

merge

sort

a

d

d

14

7

21

d

m

r

21

3

16

sort

merge

a

d

p

14

7

2

External Sorting (cont.)

Algorithm

Let M = size of buffer (in blocks)

1. Sort runs of size M blocks each (except for last) and store. Use internal sort on each run.

2. Merge M-1 runs at a time into 1 and store. Merge for all runs.

3. if step 2 results in more than 1 run, goto step 2.

Run

m-1

Output

Run 1

Run 2

........

Run 3

External Sorting (cont.)

Cost: 2 bR * (logM-1(bR / M) + 1)

Intuition:

Step 1: create runs

every block read and written once

cost 2 bR I/Os

Step 2: Merge

every merge iteration requires reading and writing entire file (2 bR I/Os)

Total:

logM-1(bR / M)

Iteration #

---------------

1

2

3

.....

Runs Left to Merge

----------------------------

What if we need to sort?

Query: depositor customer

Merge-sort Join

Sorting depositor:

bdepositor = 100

Sort depositor = 2 * 100 * (log2(100 / 3) + 1)

= 1400

Same for customer.

Total: 100 + 400 + 1400 + 7200 = 9100 I/O’s!

Still beats BNLJ (40K), INLJ (25K)

Why not use SMJ always?

Ans: 1) Sometimes inner relation can fit in memory

2) Sometimes index is small

3) SMJ only work for natural joins, “equijoins”

Hash- joins

- Applicable only to natural joins, equijoins

Depends upon hash function h, used to partition both relations

must map values of joins attributes to { 0, ..., n-1} s.t. n = #partitions

Hash-Join Algorithm

Algorithm: Hash Join

- Partition the relation S using hashing function h so that each si fits in memory. Use 1 block of memory as the output buffer for each partition. (at least n blocks)

2. Partition R using h.

- For each partition #i (0,… n-1)
- Use BNLJ to compute the join between Ri and Si : Ri Si

(optimal since si fits in memory, inner relation)

S is called the build input and R is called the probe input.

Note: can reduce CPU costs by building in-memory hash index for each si

using a different hash function than h.

Hash Join

Partitioning:

must choose:

- # of partitions, n
- hashing function, h (each tuple {0, ..., n-1})

Goals (in order of importance)

1. Each partition of build relation should fit in memory

(=> h is uniform, n is large)

2. For partitioning step, can fit 1 output block of each partition in memory

(=> n is small (<= M-1))

Strategy:

Ensure #1.

Deal with violations of #2 when needed.

Hash Join

Goal #1: Partitions of build relations should fit in memory:

1

...

Memory

(M blocks)

n

n should be?

Ans: (reserving 2 blocks for R partition, output of BNLJ)

(In practice, a little large (fudge factor~1.2) as not all memory available

for partition joins)

Hash Join

Goal #2: keep n < M

what if not possible?

Recursive partitioning!

Idea:

Iteration #1: Partition S into M-1 partitions using h1

Iteration #2: Partition each partition of S into M-1 partitions using a different hash function h2

......

repeat until partition S into >=

Cost of Hash-Join

Cost:

case 1: No recursive partitioning

1. Partition S: bS reads and bS + n writes.

Why n?

2. Rartition R: bR reads and bR + n writes.

3. n partition joins: bR + bS + 2n Reads

- Total: 3(bR + bS) +4 n

Typically n is small enough (roughly ) so it can be ignored.

Cost of Hash-Join

case 2: Recursive Partitioning

Recall: partition build relation M-1 ways at each time.

So, total number of iterations:

logM–1(n) ~ logM–1(bS / M-2) ~ logM–1(bS / M-1) =

= logM–1bS - 1

- Cost:

1. partition S : 2 bS (logM–1bS - 1)

2. partition R: 2 bR (logM–1bS - 1)

3. n partition joins: bR + bS

Total cost estimate is:

2(bR + bS)(logM–1(bS)-1) + bR + bS

Example of Cost of Hash-Join

customer depositor

- Assume that memory size is M=3 blocks
- bdepositor= 100 and bcustomer= 400.
- depositor is to be used as build input.

NO

Recursive partitioning:

2(bcust + bdep) (log2(bdep) -1)+ bdep + bcust

= 1000 (6) + 500

= 6500 I/O’s !

Why ever use Sort-Merge-Join?

1) both input relations already sorted

2) skewless hash functions hard.

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