CS计算机代考程序代写 algorithm External Sorting

External Sorting

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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.

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Main memory buffers

INPUT 1

INPUT 2

OUTPUT

DiskDisk

Two-Way External Merge Sort

• Each pass we read + write
each page in file.

• N pages in the file => the
number of passes

• So toal cost is:

• Idea: Divide and conquer:
sort subfiles and merge

é ù= +log2 1N

é ù( )2 12N Nlog +

Input file

1-page runs

2-page runs

4-page runs

8-page runs

PASS 0

PASS 1

PASS 2

PASS 3

3,4 6,2

3,4 2,6

2,3
4,6

4,7
8,9

1,3
5,6 2

1,2
3,5
6

9

1,2
2,3
3,4
4,5
6,6
7,8

2,3
4,4
6,7
8,9

5,64,9 7,8

9,4 8,7 5,6 3,1 2

1,3 2

General External Merge Sort

• To sort a file with N pages using B buffer pages:
– Pass 0: use B buffer pages. Produce sorted runs of B pages

each.
– Pass 2, …, etc.: merge B-1 runs.

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é ùN B/

B Main memory buffers

INPUT 1

INPUT B-1

OUTPUT

DiskDisk

INPUT 2
. . . . . .. . .

☛More than 3 buffer pages. How can we utilize them?

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4, 3

6, 2

9, 4

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4 6 9, 4

3, 2

Cost of External Merge Sort
• Number of passes:
• Cost = 2N * (# of passes)
• E.g., with 5 buffer pages, to sort 108 page file:

– Pass 0: = 22 sorted runs of 5 pages each
(last run is only 3 pages)

– Pass 1: = 6 sorted runs of 20 pages each
(last run is only 8 pages)

– Pass 2: 2 sorted runs, 80 pages and 28 pages
– Pass 3: Sorted file of 108 pages

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é ùé ù1 1+ -log /B N B

é ù108 5/

é ù22 4/

Number of Passes of External Sort

N B=3 B=5 B=9 B=17 B=129 B=257
100 7 4 3 2 1 1
1,000 10 5 4 3 2 2
10,000 13 7 5 4 2 2
100,000 17 9 6 5 3 3
1,000,000 20 10 7 5 3 3
10,000,000 23 12 8 6 4 3
100,000,000 26 14 9 7 4 4
1,000,000,000 30 15 10 8 5 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 1 1
1,000 1 1 1
10,000 2 2 1
100,000 3 2 2
1,000,000 3 2 2
10,000,000 4 3 3
100,000,000 5 3 3
1,000,000,000 5 4 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.

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OUTPUT

OUTPUT’

Disk Disk

INPUT 1

INPUT k

INPUT 2

INPUT 1′

INPUT 2′

INPUT k’

block size
b

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 Good idea!
– B+ tree is not clustered 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.

☛ Always better than external sorting!

(Directs search)

Data Records

Index

Data Entries
(“Sequence set”)

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!

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(Directs search)

Data Records

Index

Data Entries
(“Sequence set”)

External Sorting vs. Unclustered Index

N Sorting p=1 p=10 p=100
100 200 100 1,000 10,000
1,000 2,000 1,000 10,000 100,000
10,000 40,000 10,000 100,000 1,000,000
100,000 600,000 100,000 1,000,000 10,000,000
1,000,000 8,000,000 1,000,000 10,000,000 100,000,000
10,000,000 80,000,000 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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