Query Processing

Query Processing Overview

Main weapons are:

clever implementations techniques for operators
exploiting ‘equivalencies’ of relational operators
using cost models to choose among alternatives

Workflow

Selections

Schema for Examples

Sailors(sid: integer, sname: string, rating: integer, age: real)
Reserves(sid: integer, bid: integer, day: dates, rname: string)

Sailors(S) Each tuple is 50 bytes long, 80 tuples per page, 500 pages

Reserves(R) Each tuple is 40 bytes long, 100 tuples per page, 1000 pages

Simple Selections

The best way to preform a selection depends on:

available indexes/access paths
expected size of the result (number of tuples and/or number of pages)

Estimate result size (reduction factor)

Size of result approximated as:

$size\_of\_relation \times \prod(reduction\_factors)$

Reduction factor is usually called selectivity. It estimates what portion of the relation will qualify for the given predicate, i.e. satisfy the given condition.

This is estimated by the optimizer.

Alternatives for Simple Selections

With no index, unsorted: We must scan the whole relation, i.e. perform Heap Scan.
Cost = $[R]$
With no index, but file is sorted:
Cost = cost of binary search + $[R_{condition}]$ = $\log_2([R]) + (RF\times [R])$
With an index on selection attribute:
Use index to find qualifying data entries, then retrieve corresponding data records
Cost depends on the number of qualifying tuples
Clustering is important when calculating the total cost
1. Clustered index Cost = $([I]+[R])\times RF$
2. Unclustered index Cost = $([I]+|R|))\times RF$

General Selection Conditions

Typically queries have multiple predicates (conditions)

Example: day<8/9/94 AND rname='Paul' AND bid=5 AND sid=3

A B+ tree index matches (a combination of) predicates that involve only attributes in a prefix of the search key (upon which we build the index).

Index on <a, b, c> matches predicates on (a, b, c) (a, b) (a)

This implies that only reduction factors of the matching predicates(or primary conjuncts) that are part of the prefix will be used to determine the index cost.

Selection Approach

Find the cheapest access path
An index or file scan with the least estimated page I/O
Retrieve tuples using it
Predicates that match this index reduce the number of tuples retrieved (and impact the cost)
Apply the predicates that don't match the index (if any) later on
These predicates are used to discard some retrieved tuples, but do not affect number of tuples/pages fetched (nor the total cost)
In this case selection over other predicates is said to be done “on-the-fly”

Example: day < 8/9/24 AND bid = 5 AND sid = 3

A B+ tree index on day can be used: $RF = RF(day)$ . Then bid=5 and sid=3 must be checked for each retrieved tuple on the fly

Similarly, a hash index on <bid, sid> could be used $\prod RF=RF(bid)\times RF(sid)$ . Then, day<8/9/94 must be checked on the fly.

Projections

Issue with projection is removing duplicates

SELECT DISTINCT R.sid, R.bid
FROM Reserves R;

Projection can be done based on sorting or hashing. By sorting, the duplicates are adjacent to each other. By hashing, the duplicates fall on the same bucket.

Sorting

Basic approach is to use sorting

Scan R, extract only the needed attributes
Sort the result set(typically using external merge sort)
Remove adjacent duplicates

Usually, we bring data from disk to memory to sort it. But what if the data is too large to fit in memory?

A Simple Two-way Merge Sort

When sorting a file, several sorted subfiles are typically generated in intermediate steps. Each sorted file is called a run.

In the first pass, the pages in the file are read in one at a time. After a page is read in, the records on it are sorted and the sorted page is written out.

In subsequent passes, pairs of runs from the output of the previous pass are read in and merged to produce runs that are twice as long.

If the number of pages in the input file is $2^k$ , for some $k$ , then:

Pass $0$ produces $2^k$ sorted runs of one page each.

Pass $1$ produces $2^{k-1}$ sorted runs of two pages each.

Pass $2$ produces $2^{k-2}$ sorted runs of four pages each.

…

Pass $k$ produces $1$ sorted run of $2^k$ pages.

In each pass, we read every page in the file, process it, and write it out. We have 2 I/Os per page, per pass. The overall cost is $2N(\lceil \log_{2}^{N}\rceil+1)$ I/Os.

Only three buffer pages in the memory is needed (Two inputs and one(most cases, but not necessarily) output). In order to merge 2 sorted runs, we only need two buffer pages. We compare the records respectively. Once one page of one sorted run is run out, we can load a new page. Therefore, to merge 2 sorted runs, only 2 buffer pages are needed. Beyond 2 buffer pages, 1 output page is needed.

External Merge Sort

Two optimizations as opposed to Two-way merge sort:

In the initial “conquer pass”, we load B pages and sort them at a time.
Merge more than 2 pages at a time. Because we need a output buffer page, we can merge $B-1$ pages at a time.

The number of I/Os needed is $2N\times(1+\lceil \log _{B-1}^{\frac{N}{B}}\rceil)$ .

Sort runs: Make each $B$ pages sorted (called runs)

Merge runs: Make multiple passes to merge runs

Pass 2: Produce runs of length $B(B-1)$ pages
Pass 3: Produce runs of length $B(B-1)^2$ pages
…
Pass P: Produce runs of length $B(B-1)^P$ pages

Projection with external sort:

Scan R, extract only the needed attributes
Sort the result set using EXTERNAL SORT
Remove adjacent duplicates

WriteProjectedPages = $[R] \times PF$

PF: Projection Factor says how much are we projecting, ratio with respect to all attributes ( e.g. keeping $\frac{1}{4}$ of attributes, or $10\%$ of all attributes )

Sorting Cost = $2\times$ NumPasses $\times$ ReadProjectedPages

Hashing

Scan $R$ , extract only the needed attributes
Hash data into buckets
Apply hash function $h_1$ to choose one of $B-1$ output buffers
Remove adjacent duplicates from a bucket
2 tuples from different partitions guaranteed to be distinct

Projection based on External Hashing

Partition data into $B-1$ partitions with $h_1$ hash function

Load each partition, hash it with another hash function( $h_2$ ) and eliminate duplicates

Partitioning phase
- Read $R$ using one input buffer
- For each tuple:
  Discard unwanted fields
  Apply hash function $h_1$ to choose one of $B-1$ output buffers
- Result is $B-1$ partitions (of tuples with no unwanted fields)
  2 tuples from different partitions guaranteed to be distinct
Duplicate elimination phase
- For each partition
  Read it and build an in-memory hash table
  Using hash function $h_2$ on all fields
  While discarding duplicates
- If partition does not fit in memory
  Apply hash-based projection algorithm recursively to this partition

Joins

To find matches between two lists of records, it is important to sort them first. Otherwise, the process can be fairly inefficient.

Table A       Table B
ID            ID
1             7
2             3
7             8
3             2
9             9

Join techniques we will cover:

Nested-loops join
Sorted-merge join
Hash join

Cross product is very expensive. So if you want to perform a condition join, do $R \times S$ followed by a selection is inefficient.

Consider two inputs of tables:

The left input is called the outer input and right input is called the inner input.

Have nothing to do with inner/ outer joins!

Join is associative and communicative

Simple Nested Loops Join

For each tuple in the outer relation R, we scan the entire inner relation S.

Pseudo Code:

foreach tuple in R do
	foreach tuple s in S do
		if ri == sj then add<r, s> to result

Cost = $[Outer]+|Outer|\times [Inner]$

Page-Oriented Nested Loops Join

For each page of R, get each page of S. Write out matching pairs of tuples $<r,s>$ , where $r$ is in R page and S is in S-page.

Pseudo code:

foreach page bR in R do
	foreach page bS in S do
		foreach tuple r in bR do
			foreach tuple in bS do
				if ri == sj then add<r, s> to result

Cost = $[Outer]+[Outer]\times [Inner]$

Block Nested Loops Join

Page-oriented NL doesn’t exploit extra memory buffers

Use one page as an input buffer for scanning the inner S, one page as the output buffer, and use the remaining pages to hold ‘block’ of outer R.

For each matching tuple $r$ in R-block, $s$ in S-page, add $<r,s>$ to result. Then read next R-block, scan S, etc.

Cost = $[Outer]+\# Blocks(Outer)\times [Inner]$

$\#Blocks(Outer)=\lceil\frac{[Outer]}{B-2}\rceil$

Index Nested Loop Join

foreach record ri in R:
	foreach record sj in S where condi(ri,sj) == true:
		yield <ri, sj>

Cost = $[R]+|R|\times(cost\,to\,look\,up\,matching\,records\,in\,S)$

Sort-Merge Join

It's cumbersom to handle the identical keys in $R$ and $S$ .

function sort_merge_join():
  sort(R)
  sort(S)
  p := first tuple of R
  q := first tuple of S
  result := []
  while !(either p or q reaches the end of its relation):
    if R[p].join_attribute < S[q].join_attribute:
      increment(p)
    else if R[p].join_attribute > S[q].join_attribute:
      increment(q)
    else (R[p].join_attribute == S[q].join_attribute):
      mark(q)
    while R[p].join_attribute == S[q].join_attribute:
      result.add(join(R[p], S[q]))
      increment(q)
    restore(q)
    increment(p)
  return result

Average cost = $Sort(R)+Sort(S)+[R]+[S]$

Worst case cost = $Sort(R)+Sort(S)+|R|\times[S]$

$Sort(R)=2\times \#Passes\times [R]$

Hash join

Requires equality predicate.

Naive Hash Join

construct a hash table with a size of $B-2$ pages for $R$ . Look up records of $S$ in the hash table. The cost is $[R]+[S]$ . The problem is that $R$ needs to be fit in the memory.

Grace Hash Join

Two phases:

Partition (divide)
Build and Probe (conquer)

Partition both relations using hash function $h_1$ . $R$ tuples in partition $I$ will only match $S$ tuples in partition $I$ . If the partition doesn't fit in the memory, we recursively partiton it. Write the partition into the disk.

Read in a partition of tuples of $R$ and build a hash table using hash function $h_2$ . Scan matching partition of S, probe hash table for matches.

In Partition phase, we read+write both relations.

In Build and Probe phase, we read both relations.

Cost = $3\times [R]+3\times [S]$

Code Example for Grace Hash Join

#include <functional>
#include <unordered_map>
#include <vector>

using Record = std::pair<uint32_t, std::string>;
using Table = std::vector<Record>;

// Hash function for partitioning input tables
struct Hash1 {
    size_t operator()(const Record& r) const {
        // TODO: Implement your own hash function here
        // Use the join key value of the record to compute its hash value
        // Return a size_t value that represents the hash value
    }
};

// Hash function for constructing the hash table
struct Hash2 {
    size_t operator()(const Record& r) const {
        // TODO: Implement your own hash function here
        // Use the join key value of the record to compute its hash value
        // Return a size_t value that represents the hash value
    }
};

// Recursive partitioning function
void partition(const Table& table, std::vector<Table>& partitions, size_t depth) {
    // TODO: Implement the recursive partitioning function here
    // Use the Hash1 hash function to determine which partition each record belongs to
    // Continue partitioning until the desired depth is reached
}

// Grace hash join function
Table grace_hash_join(const Table& table1, const Table& table2) {
    // Partition the two input tables using the Hash1 hash function
    std::vector<Table> partitions1(depth);
    partition(table1, partitions1, depth);

    std::vector<Table> partitions2(depth);
    partition(table2, partitions2, depth);

    // Construct the hash table using the Hash2 hash function
    std::unordered_map<uint32_t, std::vector<std::string>, Hash2> hash_table;
    for (const auto& record : table2) {
        auto& bucket = hash_table[record.first];
        bucket.push_back(record.second);
    }

    // Perform the join operation using the partitioned tables and hash table
    Table result;
    for (const auto& partition : partitions1) {
        for (const auto& record1 : partition) {
            auto it = hash_table.find(record1.first);
            if (it != hash_table.end()) {
                for (const auto& record2 : it->second) {
                    result.emplace_back(record1.first, record1.second + record2);
                }
            }
        }
    }

    return result;
}

General Join Conditions

Equalities over serveral attributes

For inequality conditions, hash join and sort merge join is not applicable. Block NL quite likely to be the best join method here.

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Last updated 11 months ago