algorithms on secondary storage...

Worcester Polytechnic Institute

Algorithms on Secondary Storage“Sorting”

Chapter 13.

Professor E. Rundensteiner

CS542


Lesson: Using secondary storage effectively

─ Data too large to “live” in memory

─ “Regular” algorithms on small scale only

─ Must design algorithms for “large data”

─ I/O costs dominate – must reduce I/O


Why Sort?

• A classic problem in computer science!

• Data requested in sorted order via SQL:

─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?


2-Way Sort: Requires 3 Buffers

• PREPARE:

1 pass:

Read a page,

sort it, write it.

• MERGE:

Many passes:

Merge smaller

runs into

larger runs. Main memory buffer

INPUT 1

INPUT 2

OUTPUT

DiskDisk

Main memory buffer

INPUT /

OUTPUT

DiskDisk


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

2,3

4,6

4,7

8,9

1,3

5,6 2

2,3

4,4

6,7

8,9

1,2

3,5

6

1,2

2,3

3,4

4,5

6,6

7,8


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

2,3

4,6

4,7

8,9

1,3

5,6 2

2,3

4,4

6,7

8,9

1,2

3,5

6

1,2

2,3

3,4

4,5

6,6

7,8

Cost ?


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

2,3

4,6

4,7

8,9

1,3

5,6 2

2,3

4,4

6,7

8,9

1,2

3,5

6

1,2

2,3

3,4

4,5

6,6

7,8

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

2,3

4,6

4,7

8,9

1,3

5,6 2

2,3

4,4

6,7

8,9

1,2

3,5

6

1,2

2,3

3,4

4,5

6,6

7,8

• Each pass reads + writes each page in file.

• N pages in file => 2N

• Number of passes

• So total cost is:

log2 1N

2 12N Nlog


External Merge Sort

•What if we had more buffer pages?

•How do we utilize them wisely ?

- Two main ideas !


Phase 1 : Prepare

B Main memory buffers

INPUT 1

INPUT B

DiskDisk

INPUT 2

. . .. . .

• Construct as large as possible starter lists.


Phase 2 : Merge

Compose as many sorted sublistsinto one long sorted list.

B Main memory buffers

INPUT 1

INPUT B-1

OUTPUT

DiskDisk

INPUT 2

. . .. . .


Cost of External Merge Sort

•Cost = 2N * (# of passes)

•# of passes: 1 1 log /B N B


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?

─ etc ???


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

• Total I/O costs: ? 2*N * (4 passes)

108 5/

22 4/


Example

Buffer : with 5 buffer pages ; File to sort : 108 pages

Total IO costs

= 2 * N ( # of passes )

= 2N * (

= 2 * 108 * (1 + )

= 2 * 108 * (1 + log_4 ( 22 ))

= 2 * 108 * 4

5/1084log_

1 1 log /B N B


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

- gain of utilizing all available buffers

- importance of a high fan-in during merging


Optimizing External Sorting

•Cost metric ?

─I/O only (until now)

─CPU is nontrivial, worth reducing


Optimizing Internal Sorting: Phase 1

. . .12

4

28

103

5

CURRENT SETINPUT

OUTPUT

➢1 input, 1 output, B-2 current set buffer

➢Main 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


Internal Sort Algorithm

. . .124

2910

3

5

CURRENT SETINPUT

OUTPUT

8



. . .12

4

2

4

10

3

5

CURRENT SETINPUT

OUTPUT

9

8



. . .12

4

2

8

103

5

CURRENT SETINPUT

OUTPUT

➢Input & Output? new input page is read in, if it is fully 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.


Analysis of 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


Optimizing External Sorting: Phase 2

• Further optimization for external sorting:

─Blocked I/O

─Double buffering


Blocking of Buffer Pages: Ext. Sort

OUTPUT OUTPUT'

Disk

INPUT 1

INPUT k

INPUT 2

INPUT 1

INPUT 2

block size b

INPUT k

B main memory buffers

Disk

block size b

• Thus far : Do 1 I/O a page at a time

• But : Reading a block of pages sequentially is cheaper

• Make input/output a block of b pages


Example: Blocking for Merge Sort

• Idea : buffer blocks = b pages; 1 buffer block of b pages for input,

1 buffer block of b pages for outputthen we can merge |B-b/b| runs in that first pass

Example : 10 buffer pages

if one-page input and output buffer blocks,

then we produce 9 runs at a time

if two-page input and output buffer blocks,

then we produce 4 runs at a time


Blocking of Buffer Pages

OUTPUT OUTPUT'

Disk

INPUT 1

INPUT k

INPUT 2

INPUT 1

INPUT 2

block size b

INPUT k

B main memory buffers, k-way merge

Disk

block size b

Pro: Save total costs of “IOs”Cons: Reduce fan-out during merge passes

In practice, most files still sorted in 2-3 passes.


Double Buffering – Overlap CPU and I/O

• To reduce wait time for I/O request to complete, prefetch into `shadow block’.

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

Potentially, more passesIn practice, most files still sorted in 2-3 passes.


Using B+ Trees for Sorting

• Scenario: Table to be sorted has B+ tree index on sorting column(s).

• Idea: 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 Bad idea!


Clustered B+ Tree Used for Sorting

• Traversal of tree from root to left-most leaf,

• then if Alt. 1, retrieve all leaf pages in order

• If Alt. 2, additional cost of retrieving actual data records

Always better than external sorting!

(Directs search)

Data Records

Index

Data Entries("Sequence set")


Unclustered B+ Tree Used for Sorting

• Alternative (2) for data; 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")


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 (p=100 realistic)B=1,000 and block size=32 for sorting


Summary

• External sorting is important; DBMS may dedicate part of buffer pool for sorting!

• External merge sort minimizes disk I/O costs:

─In practice, # of phases rarely more than 3.

• Choice of internal sort algorithm may matter.

• Best sorts are wildly fast:

─Despite 40+ years of research, still improving!

• Clustered B+ tree is good for sorting; unclusteredtree is usually very bad.

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