1 external sorting. 2 why sort? a classic problem in computer science! data requested in sorted...
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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 Sort-merge join algorithm involves sorting.
Problem: sort 10GB of data with 1MB of RAM.
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Using secondary storage effectively
General Wisdom : I/O costs dominate Design algorithms to reduce I/O
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Overview on Merge-Sort Merge : Merge two sorted lists and repeatedly choose
the smaller of the two “heads” of the lists
Merge Sort: Divide records into two parts; merge-sort those recursively, and then merge the lists.
11 96 12 35 17 999481
11 969481 12 35 17 99
Can recursively divide again and again
Sort Sort
94 968111 12 17 35 99
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Overview on Merge-Sort (Cont’d)
Merge : Merge two sorted lists and repeatedly choose the smaller of the two “heads” of the lists
11 96 12 35 17 999481
11 969481 12 35 17 99
Can recursively divide again and again
Sort Sort
94 968111 12 17 35 99
Merge
11 12 17 35 81 94 96 99
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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
Main memory
1 buffer
1 buffer
1 buffer
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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
Pass 0 Only one memory block is needed
Pass I > 0 Only three memory blocks are needed
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Two-Way External Merge Sort
Costs for one 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
Notice: We ignored the CPU cost to sort a block in memory or merge two blocks
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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. Will reduce the number of needed passes
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Phase 2 : Merge
Merge as many sorted sublists into one long sorted list.
Keep 1 buffer for the output Use B-1 buffers to read from B-1 lists
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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Example: 2-Way Merge for 20 Runs
Number of passes = 5
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Example: 5-Way Merge for 20 Runs
Number of passes = 2
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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
#Pages in File
#Buffers available in main-memory
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Optimizing External Sorting
Cost metric ?
I/O only (till now) CPU is nontrivial, worth reducing
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Internal Main-Memory Sorting
Quicksort is a fast way to sort in memory.
An alternative is “heapsort”
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Optimizing External Sorting
Further optimization for external sorting. Blocked I/O Double buffering
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Blocked I/O for External Merge Sort
So far : 1 I/O one disk block to one memory buffer
Reading a several disk blocks sequentially is cheaper!
Suggests we should make each memory buffer (input/output) to hold several of disk blocks. But this will reduce fan-out during merge passes! In practice, most files still sorted in 2-3 passes.
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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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Summary
External sorting is important; DBMS may dedicate part of buffer pool for sorting!
External merge sort minimizes disk I/O costs: Two-Way External Sorting
• Only 3 memory buffers are needed Multi-Way External Sorting
• Depends on the number of memory buffers available
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Summary (Cont’d)
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’d)
Choice of internal sort algorithm may matter.
The best sorts are wildly fast: Quick & Heap sorts