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External Sorting
Why Sort?
§ A classic problem in computer science!
§ Data requested in sorted order
– e.g., find students in increasing gpa order
§ Sorting is first step in bulk loading B+ tree index.
§ Sorting useful for eliminating duplicate copies in a
collection of records (Why?)
§ Sort-merge join algorithm involves sorting.
§ Problem: sort 1Gb of data with 1Mb of RAM.
§ What is the minimum number of buffer pages needed to sort a file with arbitrary size?
§ Three.
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2-Way Sort: Requires 3 Buffers § Pass 1: Read a page, sort it, write it.
§ only one buffer page is used § Pass 2, 3, …, etc.:
§ three buffer pages used.
INPUT 1 INPUT 2
OUTPUT
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Disk
Main memory buffers
Disk
Two-Way External Merge Sort
• Each pass we read + write each page in file.
• N pages in the file => the number of passes
=élog2 Nù+1
• So toal cost is:
3,4 6,2 9,4 8,7 5,6 3,1 2
Inputfile
PASS 0
1-page runs
PASS 1
2-page runs
PASS 2
4-page runs
PASS 3
8-page runs
3,4
2,6
4,9
7,8
5,6
1,3
2
2,3
4,6
4,7
8,9
1,3
5,6
2
1,2
3,5
6
2,3
4,4
6,7
8,9
2N log N +1 ()
é2ù
1,2
2,3
3,4
4,5
6,6
7,8
9
• Idea: Divide and conquer: sort subfiles and merge
General External Merge Sort
☛ More than 3 buffer pages. How can we utilize them? • To sort a file with N pages using B buffer pages:
– Pass 0: use B buffer pages. Produce é N / Bù sorted runs of B pages each.
– Pass 2, …, etc.: merge B-1 runs.
INPUT 1
INPUT 2
.. .
INPUT B-1
OUTPUT
. . .
Disk
…
Disk
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B Main memory buffers
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9, 4
6, 2
4, 3
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9, 4
6
3, 2
4
Cost of External Merge Sort
• Number of passes:
• Cost=2N*(#ofpasses)1+élogB-1éN/Bùù
• E.g., with 5 buffer pages, to sort 108 page file: – Pass0: 108/5 =22sortedrunsof5pageseach
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éù
(last run is only 3 pages)
– Pass1: é22 / 4ù =6sortedrunsof20pageseach (last run is only 8 pages)
– Pass 2: 2 sorted runs, 80 pages and 28 pages
– Pass 3: Sorted file of 108 pages
Number of Passes of External Sort
N
B=3
B=5
B=9
B=17
B=129
B=257
100
1,000
10,000 100,000 1,000,000 10,000,000 100,000,000 1,000,000,000
7
10
13
17
20
23
26
30
4
5
7
9
10
12
14
15
3 4 5 6 7 8 9 10
2 3 4 5 5 6 7 8
1 2 2 3 3 4 4 5
1 2 2 3 3 3 4 4
Speed-up: Internal Sort Algorithm Quicksort is a fast way to sort in memory.
An alternative is “tournament sort” (a.k.a. “heapsort”): average run length is 2B.
Top: Read in B blocks
Output: move smallest record to output buffer
Read in a new record r
insert r into “heap”
if r not smallest,
then GOTO Output
else
remove r from “heap”
output “heap” in order; GOTO Top (next run)
This can be only effectively used in the
first pass. Virtually, make B be 2B.
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I/O for External Merge Sort
• … longer runs often means fewer passes!
• Actually, do I/O a page at a time
• In fact, read a block of pages sequentially!
• Suggests we should make each buffer (input/output) be a block of pages.
– But this will reduce fan-out during merge passes! – In practice, most files still sorted in 2-3 passes.
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Number of Passes of Optimized Sort
N
B=1,000
B=5,000
B=10,000
100
1,000
10,000 100,000 1,000,000 10,000,000 100,000,000 1,000,000,000
1 1 2 3 3 4 5 5
1 1 2 2 2 3 3 4
1 1 1 2 2 3 3 3
☛ Block size = 32, initial pass produces runs of size 2B.
Double Buffering
To reduce wait time for I/O request to complete, can
prefetch into `shadow block’.
– Potentially, more passes; in practice, most files
still sorted in 2-3 passes.
INPUT 1
INPUT 1′
INPUT 2
INPUT 2′
OUTPUT
OUTPUT’
INPUT k
INPUT k’
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Disk
b block size
Disk B main memory buffers, k-way merge
Sorting Records!
• Sorting has become a blood sport!
– Parallel sorting is the name of the game …
• Datamation: Sort 1M records of size 100 bytes
– Typical DBMS: several minutes minutes
– World record: several seconds
• 12-CPU SGI machine, 96 disks, 2GB of RAM
• New benchmarks proposed:
– Minute Sort: How many can you sort in 1 minute?
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Using B+ Trees for Sorting
• Scenario: Table to be sorted has B+ tree index on sorting column(s).
• Idea: Can retrieve records in order by traversing leaf pages.
• Is this a good idea?
• Cases to consider:
– B+ tree is clustered
– B+ tree is not clustered
Good idea!
Could be a very bad idea!
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Clustered B+ Tree Used for Sorting
• Cost: root to the left-most leaf, then retrieve all leaf pages (Alternative 1)
• If Alternative 2 is used? Additional cost of retrieving data records: each page fetched just once.
Index
(Directs search)
Data Entries
(“Sequence set”)
Data Records
☛ Always better than external sorting!
Unclustered B+ Tree Used for Sorting
Alternative (2) for data entries; each data entry contains rid of a data record. In general, one I/O per data record!
Index
(Directs search)
Data Entries
(“Sequence set”)
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Data Records
External Sorting vs. Unclustered Index
N
Sorting
p=1
p=10
p=100
100
1,000 10,000 100,000 1,000,000 10,000,000
200
2,000 40,000 600,000 8,000,000 80,000,000
100
1,000 10,000 100,000 1,000,000 10,000,000
1,000 10,000 100,000 1,000,000 10,000,000 100,000,000
10,000 100,000 1,000,000 10,000,000 100,000,000 1,000,000,000
☛ p: # of records per page
☛ B=1,000 and block size=32 for sorting ☛ p=100 is the more realistic value.
Summary
• External sorting is important; DBMS may dedicate part of buffer pool for sorting!
• External merge sort minimizes disk I/O cost:
– Pass 0: Produces sorted runs of size B (# buffer
pages). Later passes: merge runs.
– # of runs merged at a time depends on B, and block
size.
– Larger block size means less I/O cost per page.
– Larger block size means smaller # runs merged.
– In practice, # of passes rarely more than 2 or 3.
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Summary, cont.
• Choice of internal sort algorithm may matter: – Quicksort: Quick!
– Heap/tournament sort: slower (2x), longer runs
• The best sorts are wildly fast:
– Despite 40+ years of research, we’re still
improving!
• Clustered B+ tree is good for sorting; unclustered tree is usually very bad.
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