1. Chapter 13: Query Processing
Overview
Measures of Query Cost
Selection Operation
Sorting
Join Operation
Other Operations
Evaluation of Expressions

2. Basic Steps in Query Processing
1. Parsing and translation
2. Optimization
3. Evaluation

3. Parsing, Translation and Evaluation
Parsing and translation:
Checks syntax, verifies relations, attributes, etc.
Translates the query into its internal form, which is then translated into relational algebra.
Evaluation:
Executes the (chosen) query-evaluation plan.
Returns results.

4. Optimization
Observation #1: A given relational algebra expression has many equivalent expressions:
balance2500(balance(account)) balance(balance2500(account))
Observation #2: Each relational algebraic operation can be evaluated using one of several different algorithms:
Could use an index on balance to find accounts with balance < 2500, or perform complete relation scan and discard accounts with balance  2500
Thus a relational-algebra expression can be evaluated in many ways.

5. Optimization (Cont.)
Annotated expression specifying detailed evaluation strategy is called an evaluation-plan (also called a query execution plan, query plan, etc).
The process of Query Optimization involves choosing the evaluation plan that has lowest “cost” from amongst all equivalent evaluation plans.
A cost estimate is calculated using statistical information from the database catalog.
Number of tuples in each relation, size of tuples, etc.

6. Optimization (Cont.) Chapter 13 covers: Chapter 14 covers:
How to measure query cost.
Algorithms for individual relational algebra operations.
How to combine algorithms for individual operations in order to evaluate a complete expression.
Chapter 14 covers:
We study how to optimize queries, that is, how to find an evaluation plan with lowest estimated cost.

7. Measures of Query Cost
Cost is generally defined as total elapsed time for query execution.
Many factors contribute to cost:
disk accesses
CPU time
network communication
Sequential vs. random I/O
In database systems the predominant cost is typically disk access, which are also relatively easy to estimate.

8. Measures of Query Cost, Cont.
Disk access time is comprised of:
Number of disk seeks * average-seek-time
Number of blocks read * average-block-read-time
Number of blocks written * average-block-write-time
Cost to write a block is greater than cost to read a block because data is read back after being written to ensure that the write was successful.
For simplicity our cost measure will be a function of the number of block transfers to/from disk, seek time tS and block transfer time tT.
Real DBMSs take other factors into account.

9. Measures of Query Cost (Cont.)
Cost depends on the size of the buffer in main memory:
Having more memory reduces disk accesses.
Amount of available buffer space depends on several factors, i.e., other concurrent processes, machine constraints, etc) and is therefore hard to determine in the abstract:
We often use worst case estimates, assuming only the minimum amount of memory is available for query execution.
Other times the authors will do average case, or a somewhat peculiar mix of average and worst case.
We do not include the cost to writing the final output to disk in our cost formulae.

10. Statistical Information for Cost Estimation
nr: number of tuples in a relation r.
br: number of blocks containing tuples of r.
sr: size of a tuple of r.
fr: blocking factor of r — i.e., the number of tuples that fit into one block.
If tuples of r are stored together physically in a file, then:

11. Statistical Information for Cost Estimation
More generally:
tT : average transfer time
tS : average block access time (seek plus rotational latency)

12. Selection Operation
Search algorithms that perform selections without using an index are referred to as file scans.
Algorithm A1 (linear search): Scan each file block and test all records to see whether they satisfy the selection condition.
Cost estimate = tS + br * tT
note that henceforth the term “estimate” will be assumed
If the selection is on a candidate key attribute, cost = tS + (br /2) * tT
Linear search can be applied regardless of
selection condition or
ordering of records in the file, or
availability of indices
Note that linear search is sometimes referred to as a table scan or a file scan, although the latter term includes other algorithms.
For an arbitrary
predicate, this is
best, worst and
average case.
This is average case.

13. 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 (number of disk blocks to be scanned):
log2(br) * (tS + tT)
The cost of locating the first tuple using a binary search.
Plus the number of blocks containing records that satisfy selection condition .
Will see how to estimate this cost in Chapter 14

14. Selections Using Indices
Search algorithms that use an index to perform a selection are referred to as index scans.
Selection condition must involve (at least part of) the search-key of index.
A3 (primary index on candidate key, equality): Retrieve a single record that satisfies the corresponding equality condition.
Cost = (HTi + 1) * (tS + tT), where HTi represents the “height” of index i
A4 (primary index on nonkey, equality) Retrieve multiple records.
Records will be on consecutive blocks.
Cost = HTi * (tS + tT) + tS + b * tT, where b is the number of blocks containing retrieved records.

15. Selections Using Indices, Cont.
A5 (equality on search-key of secondary index).
Retrieve a single record if the search-key is a candidate key.
Cost = (HTi + 1) * (tS + tT),
Retrieve multiple records if search-key is not a candidate key.
Cost = (HTi + n) * (tS + tT), where n is the number of records retrieved
The book does not include block reads for the bucket.
Worst case - assumes each record is on a different block.

16. Selections Involving Comparisons
Can implement selections of the form AV (r) or A  V(r) by using
File Scan - linear or binary search (if sorted).
Index Scan - using indices as specified in the following.
A6 (primary index, comparison).
A  V(r) use index to find first tuple  v and scan sequentially from there.
AV (r) scan relation sequentially until first tuple > v; do not use index.
A7 (secondary index, comparison).
A  V(r) use index to find first index entry  v and scan index sequentially from there, to find pointers to records.
AV (r) scan leaf pages of index using pointers to records, until reaching first entry > v.
Either case requires an I/O for each record, worst case.

17. Implementation of Complex Selections
Conjunction: 1 2. . . n(r)
A8 (conjunctive selection using one index).
Select one of the conditions i and algorithms A1 through A7 that results in the least cost for i (r).
Test other conditions on tuples after fetching them into the buffer.
A9 (conjunctive selection using multiple-key index).
Use appropriate composite (multiple-key) index if available.
A10 (conjunctive selection by intersection of identifiers).
Requires indices with record pointers.
Use corresponding index for each condition, and take intersection of all the obtained sets of record pointers, and then fetch records from data file.
If some conditions do not have appropriate indices, apply test in memory.

18. Algorithms for Complex Selections
Disjunction:1 2 . . . n (r).
A11 (disjunctive selection by union of identifiers).
Use corresponding index for each condition, and take union of all the obtained sets of record pointers, and then fetch records from data file.
Applicable only if all conditions have available indices.
Otherwise use linear scan.
Could sort pointers before retrieving records.
Negation: (r)
Use linear search.
If very few records satisfy , and an index is applicable to , find satisfying records using leaf-level of index and fetch from the data file.

19. Sorting
Option #1: Use an existing applicable ordered index (e.g., B+ tree) to read the relation in sorted order.
Option #2: Build an index on the relation, and then use the index to read the relation in sorted order.
Option #3: For relations that fit in memory, techniques like quick-sort can be used.
Option #4: For relations that don’t fit in memory, external sort-merge is a good choice.

20. External Sort-Merge Let M denote memory size (in blocks).
Create sorted runs:
Let i be 0 initially. Repeat until the end of the relation:
(a) Read M blocks of relation into memory
(b) Sort the in-memory blocks
(c) Write sorted data to run Ri;
(d) i = i + 1;
Let the final value of i be denoted by N;
The end result is N runs, numbered 0 through N-1, where each run, except the last, contains M blocks.

21. External Sort-Merge, Cont.
Merge the runs (N-way merge):
// We assume (for now) that N < M.
// Use N blocks of memory to buffer input runs, and 1 block to buffer output.
Read the first block of each run into its buffer block;
repeat
Select the first record (in sort order) among all buffer blocks;
Write the record to the output buffer block;
If the output buffer block is full then write it to disk;
Delete the record from its input buffer block;
If the buffer block becomes empty then
read the next block (if any) of the run into the buffer;
until all input buffer blocks are empty;

22. External Sort-Merge, Cont.
Original Table
on disk
Sorted Runs
each has M blocks
Final Sorted Table
Buffer
Runs are merged
block at a time
Sort R0
Sort R1
Sort R2
Sort
RN-1
Output block

23. External Sort-Merge, Cont.
If N  M, several merge passes are required.
In each pass, contiguous groups of M - 1 runs are merged.
A pass reduces the number of runs by a factor of M -1, and creates runs longer by the same factor.
E.g. If M=11, and there are 90 runs, one pass reduces the number of runs to 9, each 10 times the size of the initial runs.
Repeated passes are performed till all runs have been merged into one.

24. Example: External Sorting Using Sort-Merge

25. External Merge Sort (Cont.)
Cost analysis:
Total number of merge passes required: logM–1(br/M).
Disk accesses for initial run creation as well as in each pass is 2br
For final pass, we ignore final write cost for all operations since the output of an operation may be sent to the parent operation without being written to disk
Thus total number of disk accesses for external sorting:
br ( 2 logM–1(br / M) + 1)

26. Join Operation Algorithms for implementing joins:
Nested-loop join
Block nested-loop join
Indexed nested-loop join
Merge-join
Hash-join
Various versions of the above
Choice of algorithm is based on cost estimate.
Examples use the following information:
customer: 10,000 rows, 400 blocks
depositor: 5000 rows, 100 blocks

27. Nested-Loop Join To compute the theta join r  s
for each tuple tr in r do begin for each tuple ts in s do begin test pair (tr,ts) to see if they satisfy the join condition  if they do, add tr • ts to the result. end; end;
r is the outer relation and s the inner relation.
Does not use or require indices.
Can be used with any kind of join condition.
Expensive - examines every pair of tuples in the two relations.

28. Nested-Loop Join (Cont.)
In the worst case, if there is enough memory only to hold one block of each relation, the number of disk accesses is: nr  bs + br
If both relations fit entirely in memory.
Best case: br + bs
If only the smaller of the two relations fits entirely in memory then use that as the inner relation and the bound still holds.
Assuming worst case:
5000  = 2,000,100 with depositor as outer relation.
10000  = 1,000,400 with customer as the outer relation.
Assuming best case:
= 500
or even if one relation fits entirely in memory

29. Block Nested-Loop Join
Embellished version of nested-loop join:
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; 29

30. Block Nested-Loop Join (Cont.)
Worst case occurs when there are just two blocks for input, in which case the number of block accesses is: br  bs + br
Best case: br + bs
Assuming worst case:
100 * = 40,100 (depositor as the outer relation)
400 * = 40,400 (customer as the outer relation)
Assuming best case:
= 500 (same as with nested-loop join)
When would a nested-loop join be preferable?
When a sorted result is required (for an outer operation)

31. Improvements to Nested-Loop and Block Nested-Loop Algorithms
Use a block nested-loop, but with modifications:
Use M - 2 disk blocks as the blocking unit for the outer relation, where M = memory size in blocks
Use one buffer block to buffer the inner relation
Use one buffer block to buffer the output
Worst case: br / (M-2)  bs + br
Best case is still the same: bs + br
If the join is an equi-join, and the join attribute is a candidate key on the inner relation, then 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).

32. Indexed Nested-Loop Join
Index scans can replace file scans if:
the join is an equi-join or natural join, and
an index is available on the inner relation’s join attribute
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.
In the worst case the buffer has space for only one block of r and one block of the index for s.
Worst case: br + nr  c
Where c is the cost to search the index and retrieve all matching tuples for each tuple or r
Worst case will assume index blocks
are not in buffer, best case will assume
that some are (perhaps close to br+bs
if index is pinned in buffer?)

33. Indexed Nested-Loop Join, Cont.
Other Options:
If a supporting index does not exist than it can be constructed “on-the-fly.”
If indices are available on the join attributes of both r and s, then use the relation with fewer tuples as the outer relation.
Or perhaps use the relation which has a primary index on it as the inner relation…

34. Example of Index Nested-Loop Join Costs
Compute depositor customer, with depositor as the outer relation.
Suppose customer has a primary B+-tree index on customer-name, which contains 20 entries in each index node.
Since customer has 10,000 tuples, the height of the tree is 4, and one more access is needed to find the actual data.
Recall that depositor has 5000 tuples
Cost of indexed nested loops join:
* 5 = 25,100 disk accesses.
Worst case

35. Merge-Join Applicable for equi-joins and natural joins.
Sort both relations on their join attribute (if not already sorted).
Merge the sorted relations to join them
Join step is similar to the merge stage of the sort-merge algorithm.
Main difference is in how duplicate values in join attribute are treated — every pair with same value on join attribute must be matched
Detailed algorithm is in the book (see page 546)

36. Merge-Join (Cont.) Can be used only for equi-joins and natural joins
Worst case number of block accesses:
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.

37. Hybrid Merge-Join Applicable if: Algorithm Outline:
Join is an equi-join or a natural join
One relation is sorted
The other has a secondary B+-tree index on the join attribute
Algorithm Outline:
Merge the sorted relation with the leaf entries of the B+-tree.
Sort the result on the addresses of the unsorted relation’s tuples.
Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples.
Sequential scan more efficient than random lookup.
Not really a “scan” of the whole relation.

38. Hash-Join Applicable for equi-joins and natural joins.
Let h be a hash function mapping JoinAttrs to {0, 1, ..., n-1}.
h is used to partition tuples of both relations:
The tuples from r are partitioned into r0, r1, . . ., rn-1
Each tuple tr  r is put in partition ri where i = h(tr [JoinAttrs]).
The tuples from s are partitioned into s0,, s1. . ., sn-1
Each tuple ts s is put in partition si, where i = h(ts [JoinAttrs]).
*Note that the book uses slightly different notation.

39. Hash-Join (Cont.)
Tuples in ri need only to be compared with tuples in si:
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.

40. The hash-join of r and s is computed as follows.
Hash-Join Algorithm
The hash-join of r and s is computed as follows.
Partition the relation s using hashing function h.
// When partitioning a relation, one block of memory is used as output for
// each partition, and one block is used for input.
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 (block at a time).
- For each tuple tr locate each matching tuple ts in si using the in-memory hash index.
- Output the concatenation of their attributes.
A simplified version of the algorithm would
simply read si into memory (the buffer), and
then read in tuples from ri, one at a time.
Relation s is called the build relation and r is called the probe relation.

41. Hash-Join algorithm (Cont.)
The value n (the number of partitions) and the hash function h is chosen such that each si should fit in memory.
Typically n is chosen as bs/M * f where f is a “fudge factor”, typically around 1.2
The probe relation partitions ri need not fit in memory
Average size of a partition si will be just less than M blocks using the above formula for n, thereby allowing room for the index.

42. Hash-Join algorithm (Cont.)
If the build relation s is very large, then the value of n given by the above formula may be greater than M-1.
The number of buckets is > the number of buffer pages.
In such a case, the relation s can be recursively partitioned:
Instead of partitioning n ways, use M – 1 partitions for s
Further partition the M – 1 partitions using a different hash function
Use same partitioning method on r
Rarely required: e.g., recursive partitioning not needed for relations of 1GB or less with memory size of 2MB, with block size of 4KB.
In order to avoid recursive partitioning we must have:
M > bs/M * f
which roughly simplifies to:
M > square root of bs

43. Cost of Hash-Join
If recursive partitioning is not required: cost of hash join is (br + bs) + 4  n
If recursive partitioning is required, the number of passes required for partitioning s is:
logM–1(bs) – 1
The number of partitions of r is the same as for s.
The number of passes for partitioning of r is the same as for s.
Worst case: (ignoring partially filled blocks):
2(br + bs)logM–1(bs) – 1 + br + bs

44. Cost of Hash-Join, Cont.
Because of the inner term in this expression, it is best to choose the smaller relation as the build relation.
If the smaller relation can fit in main memory, it can be used as the build relation and n can be set to 1 and the algorithm does not partition the relations into temporary files, but may still build an in-memory index.
Cost estimate goes down to br + bs.

45. Handling of Overflows
Even if s is recursively partitioned hash-table overflow can occur, i.e., some partition si may not fit in memory.
Many tuples in s with same value for join attributes
Bad hash function
Partitioning is said to be skewed if some partitions have significantly more tuples than some others.

46. Handling of Overflows, Cont.
Overflows can be handed in a variety of ways:
Resolution (during the build phase):
Partition si is further partitioned using different hash function.
Partition ri must be similarly partitioned.
Avoidance (during build phase):
Partition build relation into many partitions, then combine them
Most such approaches fail with large numbers of duplicates:
Another option is to use block nested-loop join on overflowed partitions.

47. Example of Cost of Hash-Join
customer depositor
Assume that memory size is 20 blocks
bdepositor= 100 and bcustomer = 400.
depositor is used as build input:
Partitioned into five (bdepositor/M) partitions, each of size 20 blocks.
This partitioning can be done in one pass.
customer is used as the probe input:
Partitioned into five partitions, each of size 80.
This is also done in one pass.
Therefore total cost: 3( ) = 1500 block transfers
Ignores cost of writing partially filled blocks

48. Hybrid Hash–Join
Useful when memory sized are relatively large, and the build input is bigger than memory.
Hybrid hash join keeps the first partition of the build relation in memory.
Can be generalized beyond what is described here.
Keep the first two partitions in memory, if possible.

49. Hybrid Hash–Join, Cont.
With memory size of 25 blocks, depositor can be partitioned into five partitions, each consisting of 20 blocks.
Division of memory (during partitioning):
1 block is used for input, and 1 block each for buffering 4 of the partitions.
The 5th partition is maintained in the remaining 20 blocks of the buffer.
Customer is similarly partitioned into five partitions each of size 80; the first is used right away for probing, instead of being written out and read back.
Cost of 3( ) = 1300 block transfers for hybrid hash join, instead of 1500 with plain hash-join.
Hybrid hash-join most useful if M >>

50. Complex Joins Join with a conjunctive condition: r 1 2...  n s
Use either nested loop or block nested loop join.
Compute the result of one of the simpler joins r i s
final result comprises those tuples in the intermediate result that satisfy the remaining conditions
1   i –1  i +1   n
Join with a disjunctive condition:
r 1  2 ...  n s
Compute as the union of the records in individual joins r  i s:
(r 1 s)  (r 2 s)   (r n s)

51. Other Operations Duplicate elimination: Projection:
Sorting - duplicates will come adjacent to each other, and all but one set of duplicates can be deleted.
Optimization: duplicates can be deleted during run generation as well as at intermediate merge steps in external sort-merge.
Hashing - similarly, duplicates will end up in the same bucket.
Projection:
Perform projection on each tuple followed by duplicate elimination.
Use the leaf-level of a B+ tree to extract required attributes.

52. Other Operations, Cont.
Aggregation: (implemented similar to duplicate elimination)
Sorting or hashing can be used to bring tuples in the same group together, and then aggregate functions can be applied on each group.
Optimization: combine tuples in the same group during run generation and intermediate merges, by computing partial aggregate values.

53. Other Operations, Cont.
Set operations (,  and ): use a variant of merge-join after sorting, or a variant of hash-join.
Hashing:
1. Partition both relations using the same hash function, thereby creating r0, .., rn-1 and s0,.., sn-1
2. For each partition i build an in-memory hash index on ri (using a different hash function) after it is brought into memory.
3. – r  s: Add tuples in si to the hash index if they are not already in it. Tuples in the hash index comprise the result.
– r  s: Output tuples in si to the result if they are already there in the hash index.
– r – s: For each tuple in si, if it appears in the hash index, delete it. At end of si add remaining tuples in the hash index to the result.

54. Other Operations, Cont. Outer join can be computed either as
A join followed by addition of null-padded non-participating tuples.
By modifying the join algorithms.
Modifying merge join to compute r s
Modify merge-join to compute r s: During merging, for every tuple tr from r that does not match any tuple in s, output tr padded with nulls.
Right outer-join and full outer-join can be computed similarly.
Modifying hash join to compute r s
If r is probe relation, output non-matching r tuples padded with nulls
If r is build relation, when probing keep track of which r tuples matched s tuples. At end of si output non-matched r tuples padded with nulls

55. Evaluation of Expressions
So far we have seen algorithms for individual operations.
Alternatives for evaluating an entire expression:
Materialization: Evaluate a relational algebraic expression from the bottom-up, explicitly generating and storing the results of each operation.
Pipelining: Evaluate operations in a multi-threaded manner, i.e., pass tuples resulting from one operation to the next (parent), as input, while the first operation is still being executed.

56. Materialization Example:
In a (completely) materialized evaluation, the expression
would be computed and stored explicitly. The join with customer would then be computed and store explicitly. Finally the projection onto customer-name would be computed. 56

57. Materialization Materialized evaluation is always possible.
Cost of writing/reading results to/from disk can be quite high:
Until now, cost formulas for operations ignored cost of output.

58. Pipelining
Evaluate several operations simultaneously in a multi-threaded manner, passing the results of one operation to the next.
In the previous example, don’t store (materialize) result of:
Instead, pass tuples directly to the join.
Similarly, don’t store result of join, pass tuples directly to projection.
Pipelining may not always be possible or easy:
sort, hash-join.
Pipelines can be demand driven or producer driven.

59. Pipelining (Cont.) In demand driven or lazy evaluation:
System repeatedly requests a tuple from the top-level operation.
Each operation requests tuples from child operations as required, in order to output its next tuple.
In between calls, each operation has to maintain “state” so it knows what to return next.
In produce-driven or eager evaluation:
Operators produce tuples automatically and pass them to their parent:
A buffer is maintained between operators; the child puts tuples in the buffer, and the parent removes tuples from the buffer.
If a buffer is full, the child waits until there is space, and then generates more tuples.
The system schedules operations that have space in their output buffer and can process more input tuples.

60. Complex Joins Join involving three relations: loan borrower customer
Strategy 1. Compute borrower customer; use result to compute loan (borrower customer)
Strategy 2. Computer loan borrower first, and then join the result with customer.
Strategy 3. Perform the two joins at once:
Build an index on loan for loan-number, and on customer for customer-name.
For each tuple t in borrower, look up the corresponding tuples in customer and the corresponding tuples in loan.
Each tuple of borrower is examined exactly once.