13. External Sorting
Transcript of 13. External Sorting
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13. External Sorting Motivation 2-way External Sort: Memory, passes,cost General External Sort: Memory, passes, cost Optimizations
Snowplow Double Buffering Forecasting Using a B+ tree index
Bucket Sort Intergalactic Standard Reference
Graefe, Implementing Sorting in Database Systems http://portal.acm.org/citation.cfm?id=1132964
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Learning Objectives Derive formula for cost of external merge sort Derive amount of memory needed to sort a
file in 2 passes, using merge or bucket sort Describe algorithm for generating longer
initial runs and identify its best and worst cases
Describe forecasting and why it is useful Identify when indexes should be used for
sorting Identify the pros and cons of external bucket
vs merge sort.
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Why sort?
A classic problem in computer science! Exercises many software and hardware features
Data is often 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 some query processing algoritms (Chapter 14)
Problem: sort 1Gb of data with 1Mb of RAM. why not virtual memory?
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Sort algorithms? If the data can fit in memory, which sort
algorithm is best? But most DBMS files will not fit in available memory
If the data is larger than memory, try the same alg. Suppose
• for this data, your sort alg. requires 220 random memory accesses
• Memory access takes 1 microsec, disk takes 10 millisecs. How much time is required to do your sort
algorithm’s memory accesses?• If there is enough memory to hold the data?• If the data is four times the size of memory?
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External Sorts
Definition: When data is larger than memory. An aside: What is “Memory”?
Physical memory? The DBMS is not the only player We concluded that most in-memory sort algs
won’t be effective for external sorting. What sort algorithms are best for external sort?
Sort-based Hash-based
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2-Way External Merge Sort: Memory?
Pass 0: Read a page, sort it, write it. How many buffer pages needed?
Pass 1, 2, 3, …, etc.: How many buffer pages needed?
Main memory buffers
INPUT 1
INPUT 2OUTPUT
Result ofPass k+1
Result ofPass k
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2-Way External Merge Sort: Passes?
Assume file is N pages Run = sorted subfile What happens in pass Zero?
How many runs are produced? What is the cost in I/Os?
What happens in pass 1? What happens in pass i? How many passes are required? What is the total cost?
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Two-Way External Merge Sort: Cost
Each pass we read + write each page in file.
N pages in the file => the number of passes
So total cost is:
Idea: Divide and
conquer: sort subfiles and merge
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
4,78,9
1,35,6 2
2,34,46,78,9
1,23,56
1,22,33,44,56,67,8
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I/Os
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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. How many sorted
runs of B pages each are produced? Cost?
Pass 1,2, …, etc.: merge B-1 runs. • How many runs are created after pass i? Cost
of pass i?• How many passes? Total Cost?
B Main memory buffers
INPUT 1
INPUT B-1
OUTPUT
Disk
INPUT 2. . .Disk
. .
.. . .
Suppose we have more than 3 buffer pages.
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Cost of External Merge Sort Number of passes: Cost = 2N * (# of passes) E.g., with 5 buffer pages, to sort 108 page file Number of passes is (1 + log4 108/5 ) = 4 Cost is 2(108)*4
Pass 0: Output is = 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
1 1 log /B N B
108 5/
22 4/
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How much memory is needed to sort a file in one or two passes?
N = number of data pages, B = memory pages available
To sort in One pass: N B To Sort in Two passes: 1+logB-1N/B 2
N/B (B-1)1
Approximating B-1 by B, this yields N B For example, if pages are 4KBytes, a 4GByte
file can be sorted in Two Passes with ? buffers.
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Sorting in 2 passes: graphical proof
File is B pages wide Each run is B pages
File is x pages high Merge x runs in pass
1 xB since we must
merge x runs in B pages of memory
So N=xB BB or NB
The File, N pages
Each run, 1xB pages
B
x
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Memory required for 2-pass sort
N = # Pages in File[File size in Bytes]
B = # Buffer Pages [ #Bytes]to sort in Two passes
2**10 [ 4Meg ] 2**5 [128K]
2**20 [4 Gig] 2**10 [ 4 Meg]
2**26 [256 Gig ] 2**13 [32 Meg]
Assuming page size of 4K
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Can we always sort in 1 or 2 passes?
Assume only one query is active, and there is at least 1 gig of physical RAM.
Yes: DB2,Oracle, SQLServer, MySQL They allocate all available memory to queries Tricky to manage memory allocation as
queries need more memory during execution No: Postgres
Memory allocated to each query, for sort and other purposes, is fixed by a config parameter.
Sort memory is typically a few meg, in case there may be many queries executing.
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Extremes of SortingOne Pass: N =B, 1-pass sort, cost = N
N BOriginal Data
Sorted Data
Quick-sort
Two Pass: N = B2, 2-pass sort, cost = 3N
N
Original Data
B
Quicksort into B runs, length B
12..B
B
B
B
Runs
B
Merge
Sorted Data
N
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Extreme Problems
Most sorting of large data falls between these two extremes
If we apply the intergalactic standard sort-merge algorithm, in every textbook, the cost for any dataset with B<N<B2 will be 3M.
Must we always pay that large price? Might there be an algorithm that is a
hybrid of the two extremes?
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Hybrid Sort when N 3B The key idea of hybrid sort is don’t waste memory. Here is an example of the hybrid sort algorithm
when N is approximately 3B.
One+ Pass: M 3B, 2-pass sort, cost = 3M – 2B
Original Data
B
Quicksort into runs, length B, leaving the last run in memory
B B
RunsB
Merge the runs on disk with the run in memory
Sorted Data
N
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Hybrid Sort in general Let k = N/B Arrange R so the last run is B-k pages Cost is N + 2(N – (B-k)) = 3N -2B + 2(N/B) = 3N – 2( B-N/B) When N=B, cost is N+1. When N=B2, cost is 3N
Original Data
B
Quicksort into runs, length B, leaving the last run in memory
B B
Runs
B
Merge the runs on disk with the run in memory
Sorted Data
N12..K-1
12..K
B-k pages
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13.3.1 Maximizing initial runs
Defn: making initial runs as long as possible.
Why is it helpful to maximize initial runs?
If initial run size is doubled, what is the time savings?
How can you maximize initial runs? What algorithm is best?
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Replacement Selection Initialize empty priority queues CUR, NEXT Read B buffer pages of data into CUR Do
Pop record s with smallest key from CUR to current run
// key of s is now highest key in current run If key of next input r >= key of s
• //Can put in current run• insert r into CUR
else• insert r into NEXT
If CUR is empty• interchange NEXT and CUR and start a new run
Until (input is empty)
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More on Replacement Selection
Cf. Knuth, vol 3 [442], page 255. Theorem: average length of a run in replacement
sort is 2B 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 ...2B
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How can we prove the Theorem?
Begin with some modeling assumptions Data to be sorted are real numbers between 0 and
1 Data appear at a uniform rate and distribution A snowplow picks up one datum as one falls Picking up a datum == pop( ) off the queue Each datum is infinitesimal Each run begins when the plow passes zero
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B
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Snowplow: Conclusion
The figures on the previous page show that At any time after run 0, the amount of snow
= size of memory = B. After the first run, the volume of snow
removed in one circuit is 2B. Cf. Larson and Graefe [471]
In spite of memory management problems, the snowplow optimization is very effective.
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13.4 I/O for External Merge Sort
What else can we do to improve performance? We have assumed I/O is done a page at a time Text suggests reading a block of pages
sequentially. Pass 0: No problem Pass 1,2,…: lowers fanin Sometimes a win
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External Sort’s jerky behavior
Recall that each input is one page What happens after the last record on a
page is output?
B Main memory buffers
INPUT 1
INPUT B-1
OUTPUT
Disk
INPUT 2. . .Disk
. .
.. . .
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Double Buffering To reduce wait time for I/O request to
complete, can prefetch into `shadow block’. This could increase the number of passes In practice, most files still sorted in 1-2
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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Forecasting Cf. Knuth, vol 3, pg 324-7
Double Buffering requires Doubling memory
What a huge waste! Most shadow buffers lie
idle, unused, wasted. How can we forecast which
shadow buffers will be needed first?
Forecasting can achieve performance of double buffering with little memory
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A B C
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Sorting Records!
Sorting has become a blood sport! Parallel sorting is the name of the game ...
www.research.microsoft.com/barc/SortBenchmark
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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 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")
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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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Bucket Sort Suppose search key values are 0-K B pages in memory, N pages in the file Pass 0: Partition the file into B-1 intervals
Inervals are not runs! If the interval fits in one page, sort it
B Main memory buffers
OUTPUT 1
OUTPUT B-1Disk
OUTPUT 2. . .
Disk
. .
.
[0,K/(B-1))
[K/(B-1),2K/(B-1))
[(B-2)K/(B-1),K)
INPUT
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Bucket sort cost What happens after pass 0
? intervals, each ? long After pass 1?
? intervals, each ? long How much memory is required to sort in
two passes? Each interval is at most one page, or ? Same as for external merge sort
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External Merge Sort vs External Bucket Sort
Approximately the same I/O cost Same memory requirement for two passes Same number of passes required to sort
Bucket sort has less CPU cost Bucketizing is much cheaper than
sorting/merging But bucket sort is subject to skew Thus merge sort is used in practice