1. Query Processing and Optimization

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

3. Review: QP & O SQL Query Query Processor Parser Query Optimizer
Algebraic
Expression
Execution plan
Evaluator
Data: result of the query

4. Review: QP & O Query Optimizer Algebraic Query Rewriter Representation
Algebraic Representation
Plan Generator
Data Stats
Query Execution Plan

5. 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 AttV (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

6. Query: Att  K (r ) How big is c?
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)

7. 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, k+1
k+m
...
k+1
k+m k

8. Plan generation: Range Queries
A6: (secondary index, range query). If att is NOT a candidate key
HTi
...
k, k+1
k+m
... k
k+1
k+m k
...
...

9. Cost: EA6 = HTi -1+ #of leaf nodes to read + #of file blocks to read +#buckets to read
= HTi -1+ LBi * (c / nr) + c + x

10. Join Operation Metadata: ncustomer = 10,000 ndepositor = 5000
Size and plans for join operation
Running example: depositor customer
Metadata:
ncustomer = 10, ndepositor = 5000
fcustomer = fdepositor = 50
bcustomer= 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)

11. 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

12. 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

13. 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

14. Nested-Loop Join Algorithm 1: Nested Loop Join Idea: Query: R S 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

15. Nested-Loop Join Algorithm 1: Nested Loop Join
Query: R S
Algorithm 1: Nested Loop Join
for each tuple tr in R do for each tuple us in S do test pair (tr,us) to see if they satisfy the join condition if they do (a “match”), add tr • us to the result.
R is called the outer relation and S the inner relation of the join.

16. Nested-Loop Join (Cont.)
Cost:
Worst case, if buffer size is 3 blocks br + nr  bs disk accesses.
Best case: buffer big enough for entire INNER relation + 2
br + bs DAs.
ncustomer = 10, ndepositor = 5000
fcustomer = fdepositor = 50
bcustomer= bdepositor= 100
Assuming worst case memory availability cost estimate is
5000  = 2,000,100 disk accesses with depositor as outer relation, and
10000  = 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)

17. Join Algorithms Algorithm 2: Block Nested Loop Join Idea: Query: R S
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

18. Block Nested-Loop Join
for each block BR of R do for each block BS of S do for each tuple tr in BR do for each tuple us in Bs do begin Check if (tr,us) satisfy the join condition if they do (“match”), add tr • us to the result.

19. Block Nested-Loop Join (Cont.)
Cost:
Worst case estimate: br  bs + br block accesses.
Best case: br + bs block 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)

20. Join Algorithms Algorithm 3: Indexed Nested Loop Join Idea: Query: R S
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

21. Indexed Nested-Loop 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.
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.

22. Example of Nested-Loop Join Costs
Query: depositor customer
(cname, acct_no) (cname, ccity, cstreet)
Metadata:
customer: ncustomer = 10,000
fcustomer = bcustomer = 400
depositor: ndepositor = 5000
fdepositor = bdepositor = 100
V (cname, depositor) = 2500
i a primary index on cname (dense) for customer (fi = 20)
Minimal buffer

23. 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

24. 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

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

26. Another Join Strategy Algorithm: Merge Join
Query: R S
Algorithm: Merge Join
Idea: suppose R, S are both sorted on A (A is the common attribute) A A 1 2 3 4 pR pS 2 3 5
...
...
Compare:
(1, 2) advance pR
(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

27. 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

28. 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

29. 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)

30. External Sorting Idea:
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
sort a d g 19 31 24 g a d c b e r m p 24 19 31 33 14 16 21 3 2 7 a b c d e g m p r 14 19 33 7 21 31 16 24 3 2
merge d e g 31 16 24
sort b c e 14 33 16
merge
sort a d 14 7 21 d m r 21 3 16
sort
merge m p r 3 2 16 a d p 14 7 2

31. 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 1
Run 2
Run 3
Run
m-1
Output

32. 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

33. 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: = 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”

34. 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

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

36. Hash Join Partitioning: must choose: Goals (in order of importance)
# 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.

37. Hash Join Goal #1: Partitions of build relations should fit in memory:
...
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)

38. 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 >=

39. 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: (bR + bS ) +4 n
Typically n is small enough (roughly ) so it can be ignored.

40. 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–1 bS - 1
Cost:
1. partition S : bS (logM–1 bS - 1)
2. partition R: bR (logM–1 bS - 1)
3. n partition joins: bR + bS
Total cost estimate is:
2(bR + bS )( logM–1(bS)-1) + bR + bS

41. 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.