1 External Sorting Chapter 13. 2 Why Sort? A classic problem in computer science! Data requested...
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Transcript of 1 External Sorting Chapter 13. 2 Why Sort? A classic problem in computer science! Data requested...
![Page 1: 1 External Sorting Chapter 13. 2 Why Sort? A classic problem in computer science! Data requested in sorted order e.g., find students in increasing.](https://reader030.fdocuments.in/reader030/viewer/2022032723/56649d1c5503460f949f1ad8/html5/thumbnails/1.jpg)
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External Sorting
Chapter 13
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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 Sort-merge join algorithm involves sorting.
Problem: sort 1Gb of data with 1Mb of RAM. why not virtual memory?
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Using secondary storage effectively
General Wisdom : I/O costs dominate Design algorithms to reduce I/O
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2-Way Sort: Requires 3 Buffers
Phase 1: PREPARE. Read a page, sort it, write it. only one buffer page is used
Phase 2, 3, …, etc.: MERGE: three buffer pages used.
Main memory buffers
INPUT 1
INPUT 2
OUTPUT
DiskDisk
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Two-Way External Merge Sort
Idea: Divide and conquer: sort subfiles and merge into larger sorts
Input file
1-page runs
2-page runs
4-page runs
8-page runs
PASS 0
PASS 1
PASS 2
PASS 3
9
3,4 6,2 9,4 8,7 5,6 3,1 2
3,4 5,62,6 4,9 7,8 1,3 2
2,34,6
4,7
8,91,35,6 2
2,3
4,46,7
8,9
1,23,56
1,22,3
3,4
4,56,6
7,8
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Two-Way External Merge Sort
Costs for pass : all pages
# of passes : height of tree
Total cost : product of
above
Input file
1-page runs
2-page runs
4-page runs
8-page runs
PASS 0
PASS 1
PASS 2
PASS 3
9
3,4 6,2 9,4 8,7 5,6 3,1 2
3,4 5,62,6 4,9 7,8 1,3 2
2,34,6
4,7
8,91,35,6 2
2,3
4,46,7
8,9
1,23,56
1,22,3
3,4
4,56,6
7,8
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Two-Way External Merge Sort Each pass we read +
write each page in file. N pages in file => 2N
Number of passes
So total cost is:
log2 1N
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
9
3,4 6,2 9,4 8,7 5,6 3,1 2
3,4 5,62,6 4,9 7,8 1,3 2
2,34,6
4,7
8,91,35,6 2
2,3
4,46,7
8,9
1,23,56
1,22,3
3,4
4,56,6
7,8
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External Merge Sort
What if we had more buffer pages? How do we utilize them wisely ?
- Two main ideas !
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Phase 1 : Prepare
B Main memory buffers
INPUT 1
INPUT B
DiskDisk
INPUT 2
. . .. . .
• Construct as large as possible starter lists.
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Phase 2 : Merge
Compose as many sorted sublists into one long sorted list.
B Main memory buffers
INPUT 1
INPUT B-1
OUTPUT
DiskDisk
INPUT 2
. . .. . .
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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 1, 2, …, etc.: merge B-1 runs.
N B/
B Main memory buffers
INPUT 1
INPUT B-1
OUTPUT
DiskDisk
INPUT 2
. . . . . .
. . .
How can we utilize more than 3 buffer pages?
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Cost of External Merge Sort
Number of passes: Cost = 2N * (# of passes)
1 1 log /B N B
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Example Buffer : with 5 buffer pages File to sort : 108 pages
Pass 0: • Size of each run?• Number of runs?
Pass 1: • Size of each run?• Number of runs?
Pass 2: ???
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Example
Buffer : with 5 buffer pages File to sort : 108 pages
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
108 5/
22 4/
• Total I/O costs: ?
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Example Buffer : with 5 buffer pages File to sort : 108 pages
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
108 5/
22 4/
• Total I/O costs: 2*N * (4 passes)
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Number of Passes of External Sort
N B=3 B=5 B=9 B=17 B=129 B=257100 7 4 3 2 1 11,000 10 5 4 3 2 210,000 13 7 5 4 2 2100,000 17 9 6 5 3 31,000,000 20 10 7 5 3 310,000,000 23 12 8 6 4 3100,000,000 26 14 9 7 4 41,000,000,000 30 15 10 8 5 4
- gain of utilizing all available buffers- importance of a high fan-in during merging
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Optimizing External Sorting
Cost metric ?
I/O only (till now) CPU is nontrivial, worth reducing
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Internal Algorithm : Heap Sort
Quicksort is a fast way to sort in memory. An alternative is “tournament sort” (a.k.a.
“heapsort”) 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
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Internal Sort Algorithm
. . .124
2810
35
CURRENT SETINPUT
OUTPUT
1 input, 1 output, B-2 current setMain idea: repeatedly pick tuple in current set with smallest k value that is still greater than largest k value in output buffer and append it to output buffer
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Internal Sort Algorithm
. . .124
2810
35
CURRENT SETINPUT
OUTPUT
Input & Output? new input page is read in if it is consumed, output is written out when it is full
When terminate current run?When all tuples in current set are smaller than
largest tuple in output buffer.
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More on Heapsort
Fact: average length of a run in heapsort is 2B The “snowplow” analogy
Worst-Case: What is min length of a run? How does this arise?
Best-Case: What is max length of a run? How does this arise?
Quicksort is faster, but ...
B
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Optimizing External Sorting
Further optimization for external sorting. Blocked I/O Double buffering
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I/O for External Merge Sort
Thus far : do 1 I/O a page at a time But cost also includes real page read/write time.
Reading a block of pages sequentially is cheaper!
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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I/O for External Merge sort
Examplebuffer blocks = b pagesset one buffer block for input, one buffer block for outputmerge |B-b/b| runs in each pass
e.g., 10 buffer pages9 runs at a time with one-page input and output buffer
blocks4 runs at a time with two-page input and output buffer
block
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Double Buffering – Overlap CPU and I/O
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.
OUTPUT
OUTPUT'
Disk Disk
INPUT 1
INPUT k
INPUT 2
INPUT 1'
INPUT 2'
INPUT k'
block sizeb
B main memory buffers, k-way merge
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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: 15 minutes World record: 3.5 seconds
• 12-CPU SGI machine, 96 disks, 2GB of RAM
New benchmarks proposed: Minute Sort: How many can you sort in 1 minute? Dollar Sort: How many can you sort for $1.00?
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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 clusteredCould be a very bad
idea!
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Clustered B+ Tree Used for Sorting
Cost: root to left-most leaf,
then retrieve all leaf pages (Alternative 1)
For Alternative 2, 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")
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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!
(Directs search)
Data Records
Index
Data Entries("Sequence set")
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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.
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Summary
External sorting is important; DBMS may dedicate part of buffer pool for sorting!
External merge sort minimizes disk I/O costs: 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 runs rarely more than 2 or 3.
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Summary, cont.
Choice of internal sort algorithm may matter.
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.