Lecture 24Query Execution

Monday, November 28, 2005

Outline

• Partitioned-base Hash Algorithms

• External Sorting• Sort-based Algorithms

Two Pass Algorithms Based on Hashing

• Idea: partition a relation R into buckets, on disk

• Each bucket has size approx. B(R)/M

M main memory buffers DiskDisk

Relation ROUTPUT

2INPUT

1

hashfunction

h M-1

Partitions

1

2

M-1

. . .

1

2

B(R)

• Does each bucket fit in main memory ?–Yes if B(R)/M <= M, i.e. B(R) <= M2

Hash Based Algorithms for

• Recall: (R) duplicate elimination

• Step 1. Partition R into buckets• Step 2. Apply to each bucket (may read in main memory)

• Cost: 3B(R)• Assumption:B(R) <= M2

Hash Based Algorithms for

• Recall: (R) grouping and aggregation

• Step 1. Partition R into buckets• Step 2. Apply to each bucket (may read in main memory)

• Cost: 3B(R)• Assumption:B(R) <= M2

Partitioned Hash Join

R |x| S• Step 1:

– Hash S into M buckets– send all buckets to disk

• Step 2– Hash R into M buckets– Send all buckets to disk

• Step 3– Join every pair of buckets

Hash-Join• Partition both

relations using hash fn h: R tuples in partition i will only match S tuples in partition i.

Read in a partition of R, hash it using h2 (<> h!). Scan matching partition of S, search for matches.

Partitionsof R & S

Input bufferfor Ri

Hash table for partitionSi ( < M-1 pages)

B main memory buffersDisk

Output buffer

Disk

Join Result

hashfnh2

h2

B main memory buffers DiskDisk

Original Relation OUTPUT

2INPUT

1

hashfunction

h M-1

Partitions

1

2

M-1

. . .

Partitioned Hash Join

• Cost: 3B(R) + 3B(S)• Assumption: min(B(R), B(S)) <= M2

Hybrid Hash Join Algorithm

• Partition S into k bucketst buckets S1 , …, St stay in memory

k-t buckets St+1, …, Sk to disk

• Partition R into k buckets– First t buckets join immediately with S

– Rest k-t buckets go to disk• Finally, join k-t pairs of buckets:(Rt+1,St+1), (Rt+2,St+2), …, (Rk,Sk)

Hybrid Join Algorithm

• How to choose k and t ?– Choose k large but s.t. k <= M

– Choose t/k large but s.t. t/k * B(S) <= M

– Moreover: t/k * B(S) + k-t <= M

• Assuming t/k * B(S) >> k-t: t/k = M/B(S)

Hybrid Join Algorithm

• How many I/Os ?• Cost of partitioned hash join: 3B(R) + 3B(S)

• Hybrid join saves 2 I/Os for a t/k fraction of buckets

• Hybrid join saves 2t/k(B(R) + B(S)) I/Os

• Cost: (3-2t/k)(B(R) + B(S)) = (3-2M/B(S))(B(R) + B(S))

Hybrid Join Algorithm

• Question in class: what is the real advantage of the hybrid algorithm ?

The I/O Model of Computation

• In main memory: CPU time– Big O notation !

• In databases time is dominated by I/O cost– Big O too, but for I/O’s– Often big O becomes a constant

• Consequence: need to redesign certain algorithms

• See sorting next

Sorting

• Problem: sort 1 GB of data with 1MB of RAM.

• Where we need this: – Data requested in sorted order (ORDER BY)

– Needed for grouping operations– First step in sort-merge join algorithm– Duplicate removal– Bulk loading of B+-tree indexes.

2-Way Merge-sort:Requires 3 Buffers in

RAM• Pass 1: Read a page, sort it, write it.

• Pass 2, 3, …, etc.: merge two runs, write them

Main memory buffers

INPUT 1

INPUT 2

OUTPUT

DiskDisk

Runs of length LRuns of length 2L

Two-Way External Merge Sort

• Assume block size is B = 4Kb

• Step 1 runs of length L = 4Kb• Step 2 runs of length L = 8Kb• Step 3 runs of length L = 16Kb• . . . . . .• Step 9 runs of length L = 1MB• . . .• Step 19 runs of length L = 1GB (why ?)

Need 19 iterations over the disk data to sort 1GB

Can We Do Better ?

• Hint:

We have 1MB of main memory, but only used 12KB

Cost Model for Our Analysis

• B(R): size of R in number of blocks

• M: Size of main memory (in # of blocks)

• E.g. block size = 4KB then:– B(R) = 250000– M = 250

External Merge-Sort

• Phase one: load M bytes in memory, sort– Result: runs of length M bytes ( 1MB )

M bytes of main memoryDiskDisk

.

.

.

. . .M

Phase Two

• Merge M – 1 runs into a new run (250 runs )

• Result: runs of length M (M – 1) bytes (250^2)

M bytes of main memoryDiskDisk

.

.

.

. . .Input M/B

Input 1

Input 2. . . .

Output

Phase Three

• Merge M – 1 runs into a new run• Result: runs of length M (M – 1)2

M bytes of main memoryDiskDisk

.

.

.

. . .Input M/B

Input 1

Input 2. . . .

Output

Cost of External Merge Sort

• Number of passes:

• How much data can we sort with 10MB RAM?– 1 pass 10MB data– 2 passes 25GB data

• Can sort everything in 2 or 3 passes !

⎡ ⎤⎡ ⎤B(R)/Mlog1 1M−+

External Merge Sort

• The xsort tool in the XML toolkit sorts using this algorithm

• Can sort 1GB of XML data in about 8 minutes

Two-Pass Algorithms Based on Sorting

• Assumption: multi-way merge sort needs only two passes

• Assumption: B(R) <= M2

• Cost for sorting: 3B(R)

Two-Pass Algorithms Based on Sorting

Duplicate elimination (R)• Trivial idea: sort first, then eliminate duplicates

• Step 1: sort chunks of size M, write– cost 2B(R)

• Step 2: merge M-1 runs, but include each tuple only once– cost B(R)

• Total cost: 3B(R), Assumption: B(R) <= M2

Two-Pass Algorithms Based on Sorting

Grouping: a, sum(b) (R)

• Same as before: sort, then compute the sum(b) for each group of a’s

• Total cost: 3B(R)• Assumption: B(R) <= M2

Two-Pass Algorithms Based on Sorting

R ∪ Sx = first(R)y = first(S)

While (_______________) do{ case x < y: output(x) x = next(R) case x=y:

case x > y;}

x = first(R)y = first(S)

While (_______________) do{ case x < y: output(x) x = next(R) case x=y:

case x > y;}

Completethe programin class:

Two-Pass Algorithms Based on Sorting

R ∩ Sx = first(R)y = first(S)

While (_______________) do{ case x < y:

case x=y:

case x > y;}

x = first(R)y = first(S)

While (_______________) do{ case x < y:

case x=y:

case x > y;}

Completethe programin class:

Two-Pass Algorithms Based on Sorting

R - S

Completethe programin class:

x = first(R)y = first(S)

While (_______________) do{ case x < y:

case x=y:

case x > y;

}

x = first(R)y = first(S)

While (_______________) do{ case x < y:

case x=y:

case x > y;

}

Two-Pass Algorithms Based on Sorting

Binary operations: R ∪ S, R ∩ S, R – S• Idea: sort R, sort S, then do the right thing

• A closer look:– Step 1: split R into runs of size M, then split S into runs of size M. Cost: 2B(R) + 2B(S)

– Step 2: merge M/2 runs from R; merge M/2 runs from S; ouput a tuple on a case by cases basis

• Total cost: 3B(R)+3B(S)• Assumption: B(R)+B(S)<= M2

Two-Pass Algorithms Based on Sorting

R |x|R.A =S.B S

x = first(R)y = first(S)

While (_______________) do{ case x.A < y.B:

case x.A=y.B:

case x.A > y.B;

}

x = first(R)y = first(S)

While (_______________) do{ case x.A < y.B:

case x.A=y.B:

case x.A > y.B;

}

Completethe programin class:

R(A,C) sorted on AS(B,D) sorted on B

Two-Pass Algorithms Based on Sorting

Join R |x| S• Start by sorting both R and S on the join attribute:– Cost: 4B(R)+4B(S) (because need to write to disk)

• Read both relations in sorted order, match tuples– Cost: B(R)+B(S)

• Difficulty: many tuples in R may match many in S– If at least one set of tuples fits in M, we are OK– Otherwise need nested loop, higher cost

• Total cost: 5B(R)+5B(S)• Assumption: B(R) <= M2, B(S) <= M2

Two-Pass Algorithms Based on Sorting

Join R |x| S• If the number of tuples in R matching those in S is small (or vice versa) we can compute the join during the merge phase

• Total cost: 3B(R)+3B(S) • Assumption: B(R) + B(S) <= M2