joinspjm/co572/lectures/joins.pdf · revision youshouldknowaboutjoins...

Joins

Holger Pirk

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Purpose of this lecture

You shouldUnderstand and know the basic join algorithms: nested loop (withvariants), hash, sort-mergeBe able to select an appropriate join implementation for a givensituation

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Why Join?

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Why care?

Joins are everywhereIn part due to to whole normalization business

I These are mostly Foreign-Key joins (we’ll talk about those in thecontext of indexing)

In part because combining (joining) data produces valueI These are more complicated (and interesting)

ExamplesFind users that have bought the same productsFind the shortest route visting 5 of London’s best sightsFind advertisements that worked

I (lead to users searching for a specific term within a timeframe)

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Revision

You should know about joinsJoins are basically cross products with selections on top

I select R.r, S.s from R,S where R.id = S.id

Quick quiz: join or cross productselect R.r, S.s from R,S where R.r = R.id

Quick quiz: join or cross productselect R.r from R,S where R.r = "something"

Quick quiz: join or cross productselect R.r from R,S where R.r = S.s

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Left and Right Outer JoinsLeft Join

A left join RLon S returns every row in R , even if no rows in S match. In

such cases where no row in S matches a row from R , the columns of S arefilled with NULL values.

Right Join

A right join RRon S returns every row in S , even if no rows in R match. In

such cases where no row in R matches a row from S , the columns of R arefilled with NULL values.

Outer Join

An outer join ROon S returns every row in R , even if no rows in S match,

and also returns every row in S even if no row in R matches.

ROon S ≡ (R

Lon S) ∪ (R

Ron S)

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Revision Quiz

What are the bounds of for the size of the output (O) of this queryselect * from R full outer join S on (R.r = S.s)

AnswersA |R| ≤ |O| ≤ |R| × |S |B max(|R|, |S |) ≤ |O| ≤ max(|R|, |S |, |R| × |S |)C |R|+ |S | ≤ |O| ≤ max(|R|+ |S |, |R| × |S |)D |S | ≤ |O| ≤ |R| × |S |

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On matching criteria

The matching functionselect * from R full outer join S on (R.r = S.s)

The matching function need not be equalityIf it is, we call the join an equi-join (these are the most importantjoins)If it is an inequality constraint (< or >), we call them inequality joins

select count(*) from event, marker where event.timebetween marker.time and marker.time+60

If it is an <> (!= in C syntax), we call it an anti-joinAll other joins are called Theta joins

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Nested Subqueries in where-clauses: the In operator

Definitiontrue if the attribute value is contained in the setsubquery must only return a single column

The simple case: statically defined setsselect * from Customer where name IN ("holger", "sam")

The correlated caseselect Name from Customer where CustomerID IN (select CustomerID from Order)

equivalent toselect distinct Name from Customer, Orderwhere Customer.CustomerID = Order.CustomerID

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Join algorithms

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Nested Loop - the naive one

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Nested Loop - the naive one

Implementationfor (size_t i = 0; i < leftRelationSize; i++) {

auto leftInput = left[ i];for (size_t j = 0; j < rightRelationSize; j++) {

auto rightInput = right[ j];if(leftInput[leftAttribute] == rightInput[rightAttribute])

writeToOutput({leftInput, rightInput});}

}

Example dataR: 10, 17, 7, 16, 12, 8, 13

S: 8, 16, 12, 1, 17, 2, 7

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Worksheet

ParametersAssume a nested loop join of two disk-resident tables in N-ary formThe first table has 500,000 tuples of three integer attributesThe second has 50,000 tuples of two integer attributesPages are plain (unslotted) 4KBThe page buffer can hold 10 pagesThe replacement strategy of the buffer manager is LRU

How many page replacements need to be performed?

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Calculation

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Sort-Merge Joins - a join for special cases

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Implementation (assuming values are unique and sorted)auto leftI = 0;auto rightI = 0;while (leftI < leftInputSize && rightI < rightInputSize) {

auto leftInput = left[ leftI];auto rightInput = right[ rightI];if(leftInput[leftAttribue] < rightInput[rightAttribue])

leftI++;else if(rightInput[rightAttribue] < leftInput[leftAttribue])

rightI++;else {

writeToOutput({leftInput, rightInput});rightI++;leftI++;

}}

Example dataR = {7, 8, 10, 12, 13, 16, 17};

S = {1, 2, 7, 8, 12, 16, 17};

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Sort-Merge Joins - a join for special casesExample data

R = {7, 8, 10, 12, 13, 16, 17};, S = {1, 2, 7, 8, 12, 16, 17};

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Why Sort-Merge Joins works

InvariantsAssume, w.l.o.g., that the value on the rightis less than the value on the leftAll values succeeding the value on the rightare greater than the value on right⇒ No value beyond the value on the rightcan be a join partner⇒ The value on the left has no joinpartners succeeding the value on the right⇒ The cursor on the left can be advanced

Visualisation

>7>5

57

Merge Direction

Comparison

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Worksheet: Sort-Merge Equality Joins

ParametersYou have two sorted relations with a single attribute

I R: 2, 5, 6, 7, 9, 11, 12I S: 1, 4, 5, 6, 10, 11, 13

Evaluate the query select r,s from R,S where r = sI Use a sort-merge joinI You can assume the values in each relation are unique

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Calculation

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Hash joins - the special case that is so important

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Hash joins - the special case that is so importantImplementationextern int* hashTable;

for (size_t i = 0; i < buildSideSize; i++) {auto buildInput = build[ i];auto hashValue = hash(buildInput[buildAttribute]);while (hashTable[hashValue])

hashValue = nextSlot(hashValue);hashTable[hashValue] = buildInput;

}

for (size_t i = 0; i < probeSideSize; i++) {auto probeInput = probe[ i];auto hashValue = hash(probeInput[probeAttribute]);while (hashTable[hashValue] &&

hashTable[hashValue][buildAttribute] != probeInput[probeAttribute])hashValue = nextSlot(hashValue);

if(hashTable[hashValue][buildAttribute] == probeInput[probeAttribute])writeToOutput({hashTable[hashValue], probeInput});

}

NomenclatureWe distinguish build-side (the side that is buffered in the hashtable)and probe-side (the one used to look up tuples in the hashtabl)

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Hash join details. . . the hash functionHash-function requirements

Pure no stateKnown output domain we need to know the range of generated valuesContiguous output domain we do not want holes in the output domain

Nice to haveUniform all values should be equally likely

Typical examplesCRC32MD5MurmurHash

I This is one of the fastest "decent" hash-functionsModulo-Division

I Arguably the simplest function you could have

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Conflict Handling

When a slot is already filled but there is space in the table. . .We need to put the value somewhere. . .The conflict handling strategy prescribes where

RequirementsLocality (but not too much :-))No holes (probe all slots)

Many exist - let’s talk about threeLinear probingQuadratic probingRehashing: specifically, cyclic group probing

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Linear Probing

DescriptionWhen a slot is filled, try the next one (distance 1). . .. . . and the next one (distance 2). . .. . . continue until you find one that is free (3,4,5,6, etc.)

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Hash-join with modulo hashing and linear probing

Implementationextern int* hashTable;

for(size_t i = 0; i < buildSideSize; i++) {auto buildInput = build[i];auto hashValue = buildInput[buildAttribute] % 10; // hash-functionwhile(hashTable[hashValue])

hashValue = (hashValue++ % 10); // probe functionhashTable[hashValue] = buildInput;

}

for(size_t i = 0; i < probeSideSize; i++) {auto probeInput = probe[i];auto hashValue = probeInput[probeAttribute] % 10;while(hashTable[hashValue] && //

hashTable[hashValue][buildAttribute] != probeInput[probeAttribute])hashValue = (hashValue++ % 10);

if(hashTable[hashValue][buildAttribute] == probeInput[probeAttribute])writeToOutput({hashTable[hashValue], probeInput});

}

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Hash-join with modulo hashing and linear probingExample data (, linear probing)

auto hash = [](auto v){return v % 10; };

auto probe = [](auto v) { return (v + 1) % 10; }

Probe = {7, 8, 10, 12, 13, 16, 17}; Build = {1, 2, 7, 8, 12, 16, 17};

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Linear Probing

DescriptionWhen a slot is filled, try the next one (distance 1). . .. . . and the next one (distance 2). . .. . . continue until you find one that is free (3,4,5,6, etc.)

DisadvantagesLeads to long probe-chains for high-locality dataFor example, 9,8,7,6,5,4,3,2,2

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Quadratic Probing

DescriptionWhen a slot is filled, try the next one (distance 1). . .. . . double the distance (distance 2). . .. . . continue until you find one that is free (4, 8, 16, etc.)

AdvantagesSimpleGood locality in the first three probes

I Bad after that (that is the point)

DisadvantagesThe first few probes are still likely to incur conflictsFor example, 9,8,7,6,5,4,3,2,2

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Rehashing, specifically, cyclic group probing

The challenge:Randomize the probesMake them deterministicMake sure all slots are probed

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Cyclic group probing

Hardcore math to the rescueThere is a way to generate a (pseudorandom) sequence of numbers

I while making sure every number in a range is generatedA multiplicative group of integers modulo n

I The equation is f (x) = (x × g) mod n for certaing g and n

How do we determine g and n (n is the size of our hash table)I n = pk with p an odd prime and k > 0I There is no easy way to calculate the primitive roots (g) for a given nI Using Mathematica: PrimitiveRootList@11 = {2,6,7,8}I group = 6, 3, 7, 9, 10, 5, 8, 4, 2, 1, 6, ...

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Worksheet: Hash-building

Parameters (Hashtable size = 11)You have two relations with a single attribute

I Probe: 6, 7, 9, 12, 11, 15, 2I Build: 1, 4, 5, 6, 15, 11, 14

Evaluate the query select r,s from R,S where s = rI Use a hash-joinI You can assume the values in each relation are unique

The hash function is f (x) = x mod 10 + 1

Which is the best probing strategy in this case(with respect to the comparisons in the build phase)?

I Linear ProbingI Quadratic ProbingI Rehashing with the cyclic group f (x) = (x × 6) mod 11

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Calculation

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Hash Joins practicalities

Cyclic groups are the theory. . .The practice is a lot less elegantPeople simply rehash using some arbitrary hash-function

I (often the same they used for initial hashing)I does this guarantee hole-freeness?

Hashing is expensiveEspecially good hashing

I Lots of CPU cycles (often more expensive than multiple data accesses)

Slots are often allocated in bucketsBuckets are slots with space for more than one tupleYou will sometimes see people implementing buckets as linked lists

I A horrible idea if you care about lookup performance (inserts are okay)

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Hash Joins practicalities

Hashtables are arrays tooThey occupy spaceThey are usually overallocated by at least a factor two

I i.e., you allocate twice as many slots as (estimated) tuple inputs(obviously adapting the hash-function)

They are probed randomly in the probe phase (a lot)I You really want to make sure they stay in the buffer pool/cache

For this class, assume that, if the hashtable does not fit, every accessis a page fault (consistent with what we established previously)Rule of thumb: use Hash Joins when one relation is much smaller thanthe other

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Partitioning. . . for buffer pool friendlyness

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Hashjoins

RecallRule of thumb: use Hash Joins when one relation is much smaller thanthe other

How do you make sure of that?You partition!

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Partitioning - when you don’t have enough memory

Fundamental premise:Sequential access is much cheaper than random access

I Difference grows with the page sizeI Assume: Random value access cost cI Sequential value access cost c

pagesizeI Why?

Assume your hashtable does not fit in the buffer poolI.e., if the relation is larger than half the buffer poolIt can pay off to invest in an extra pass for partitioning

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Partitioning - an example

AssumeYour smaller relation is exactly the size of the buffer pool minus onepage

I Hashtable is overallocated by a factor 2I Probability of a page fault upon hash is roughly 50 percent

Partitioning it into 4 smaller partitions costs you one scanI Plus writing output (which is also one scan)

Now, you join the large relation with each of the 4 smaller, buffer poolresident relations

I Costs: 4 scans over the larger relationI Remember: we assume |buidl | << |probe|, i .e., costsroughlyquadrupleBut: each of the hash lookups is now hitting the buffer pool

I Disk seek latency: 3ms, RAM access latency: 30ns – a factor of 100K

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Partitioning - an example

Visualization

1412175

9…

inputBuffer Pool

(size 5)2011

partitioning function (mod 4)

Pages on disk

when page is filled(semi-random but

without bandwidth waste)

for every tuple

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Partitioning

BonusYou can parallelize the processing of each of the smaller joins

I because they are disjointYou can partition the larger relation as well. . .

I . . . and only join the overlapping partitionsI this is the state of the art in join processing

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Blocked Nested Loops - the slightly smarter NLJ

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Blocked Nested Loops - the slightly smarter

Implementation (assuming all pages are filled)auto tuplesOnLeftInputPages = 512;auto tuplesOnRightInputPages = 1024;for (size_t i = 0; i < leftRelationSize; i+= tuplesOnLeftInputPages)

for (size_t j = 0; j < rightRelationSize; j+= tuplesOnRightInputPages)for (size_t i_page = 0; i_page < tuplesOnLeftInputPages; i_page++) {

auto leftInput = left[ i+i_page];for (size_t j_page = 0; j_page < tuplesOnRightInputPages; j_page++) {

auto rightInput = right[ j+j_page];if(leftInput[leftAttribute] == rightInput[rightAttribute])

writeToOutput({leftInput, rightInput});}

}

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Visualization

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Worksheet

ParametersAssume a nested loop join of two disk-resident tables in N-ary formThe first table has 500,000 tuples of three integer attributesThe second has 50,000 tuples of two integer attributesPages are plain (unslotted) 4KBThe page buffer can hold 10 pagesThe replacement strategy of the buffer manager is LRU

How many page replacements need to be performed?

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Calculation

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Indexed Nested Loops - when you know what you’re joining

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AssumptionsOne side has an indexWhat is an index, you ask?

I Well, let’s, for now, assume it is sortedI We’re going to talk about this more later

The other is not

IdeaScan the unindexed side (like you would in the nested loops case)Look up values on the inner side using the appropriate lookup function

I In the case of a sorted relation, that means binary searching

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PropertiesSequential I/O on the unindexed sideQuasi-random on the indexed-sideParallelizable over the values on the unindexed side

EffortΘ(|unindexed | × | log indexed |)O(|left| × |right|)Note: I/O behaviour on the indexed side is not-quite random

DisadvantagesWe are basically cheating: we pushed part of the computation intoanother componentMaintaining sortedness is expensive

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Recall

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Ensuring sortedness

In C++class Table {

vector<Tuple> storage;bool isSorted = true;bool isDense = true;

public:void insert(Tuple t) {

storage.push_back(t);sort(storage.begin(), storage.end(), [](auto l, auto r) { return l.id <

r.id; });↪→isDense = true;for(size_t i = 1; i < storage.size(); i++)

isDense &= storage[i].id == storage[i - 1].id + 1;};vector<Tuple> findByKey(int id);

};

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Conclusion

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So, which join algorithm should I use?

My DBMS Professor:The answer is always sorting or hashing

The truth is more complicatedSort-Merge join if

I Relations are sortedI Relations are unsorted but have similar sizeI If you are evaluating an inequality-join

Indexed nested loop join if one relation has an index (e.g., is sorted)Hashjoin if one relation is much smaller than the other (less than 10%)Nested loop join if one relation is tiny (less than 20 values)Blocked nested loops join for theta-joins

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Joins vs. Cross Products

Function calls

Cross Product Hash Join0

5.0×1010

1.0×1011

1.5×1011

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Joins vs. Cross Products

What factor is Hashjoin bounding performance?

CPU I/O0

5.0×106

1.0×107

1.5×107

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The end

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joinspjm/co572/lectures/joins.pdf · revision youshouldknowaboutjoins...

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