Transcript External Sorting

External Sorting
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External Sorting
• Problem: Sort 1Gb of data with 1Mb of
RAM.
• When a file doesn’t fit in memory, there
are two stages in sorting:
1. File is divided into several segments, each of
which sorted separately
2. Sorted segments are merged
(Each stage involves reading and writing the file
at least once)
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Sorting Segments
Two possibilities depending on the number of disks:
1. Heapsort:
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optimal routine if only one disk drive is available.
It can be executed by overlapping the input/output
with processing
Each sorted segment will be the size of the available
memory.
2. Replacement selection:
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optimal for two or more disk drives.
Sorted segments are twice the size of memory.
Reading in and writing out can be overlapped
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Heapsort
What is a heap?
• A heap is a binary tree with the following
properties:
1. Each node has a single key and that key is greater than
or equal to the key at its parent node.
2. It is a complete binary tree. i.e. All leaves are on at
most 2 levels, leaves on the lowest level are at the
leftmost position.
3. Can be stored in an array; the root is at index 1, the
children of node i are at indexes 2*i, and 2*i+1.
Conversely, the parent of node j is stored at index j/2
(very compact: no need to store pointers)
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Example
Heap as a
binary tree:
10
35
45
60
40
50
Height = log n
20
25
30
55
Heap as an array:
10 35 20 45 40 25 30 60 50 55
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Heapsort Algorithm
• First Stage: Building the heap while reading the
file:
– While there is available space
• Get the next record from current input buffer
• Put the new record at the end of the heap
• Reestablish the heap by exchanging the new node with its
parent, if it is smaller than the parent: otherwise leave it, where
it should be. Repeat this step as long as heap property is
violated.
• Second stage: Sorting while writing the heap out
to the file:
– While there are records in heap
• Put the root record in the current output buffer.
• Replace the root by the last record in the heap.
• Restore the heap again, which has the complexity of O(log n)
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Example
• Trace the algorithm with:
48 70 30 19 50 45
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Heapsort
• How big is a heap?
– As big as the available memory.
• What is the time it takes to create the sorted
segments?
– Ignoring the seek time and assuming b blocks in the file,
where heap processing overlaps (approximately) with
I/O.
– The time for creating the initial sorted segments is
2b*btt (read in the segment and write out the runs)
– Note that the entire file has not been sorted yet. These
are just sorted segments, and the size of each segment is
limited to the size of the available memory used for this
purpose.
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Multiway Merging
• K-way merge: we want to merge K input lists to
create a single sequentially ordered output list. (K
is the order of a K-way merge)
• We will adapt the 2-way merge algorithm:
– Instead of two lists, keep an array of lists: list[0], list[1],
… list[k-1]
– Keep an array of the items that are being used from
each list: item[0], item[1], … item[k-1]
– The merge processing requires a call to a function (say
MinIndex) to find the index of the item with the
minimum value.
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Finding the minimum item
• When the number of lists is small (K 8)
sequential search among items works nicely.
(O(K))
• When the number of lists is large, we could place
the items in a priority queue (an array heap).
• The min value will be at the root (1st position in
array)
• Replace the root with the next value from the
associated list. This insert operation is O(log K)
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Merging as a way of Sorting Large Files
Let us consider the following example:
• File to be sorted:
– 8,000,000 records
– R = 100 bytes
– Size of the key = 10 bytes
• Memory available as a work area: 10MB (not
counting memory used to hold program, O.S., I/O
buffers etc.)
Total file size = 800MB
Total number of bytes for all keys = 80MB
So, we cannot do internal sorting nor keysorting.
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Basic idea
1. Forming runs (i.e. sorted subfiles):
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bring as many records as possible to main
memory, sort them using heapsort, save it into
a small file.
Repeat this until we have read all records
from the original file.
2. Do a multiway merge of the sorted
subfiles.
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Cost of Merge Sort
I/O operations are performed in the following times:
1. Reading each record into main memory for
sorting and forming the runs.
2. Writing sorted runs to disk.
These two steps are done as follows:
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Read a chunk of 10MB, write a chunk of 10Mb
(repeat this 80 times)
In terms of basic disk operations, we spend:
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For reading: 80 seeks + transfer time for 800 MB
Same for writing
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3. Reading runs into memory for merging. Read
one chunk of each run, so 80 chunks. Since
available memory is 10MB each chunk can have
(10,000,000/80)bytes = 125,000 bytes = 1250
records.
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How many chunks to be read for each run?
Size of run/size of chunk = 10,000,000/125,000= 80
Total number of basic seeks = Total number of
chunks (counting all runs) is 80 runs * 80 chunks/run
= 802 chunks = 6400 seeks.
Reading each chunk involves average seeking.
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4. Writing sorted file to disk: after the first
pass, the number of separate writes closely
approximate reads. We estimate two seeks
- one for reading and one for writing- for
each piece: 80* 80 pieces therefore
6400 seeks
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Sorting a File that is 10 times larger
• How is the time for merge phase affected if
the file is 80 million records?
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More runs: 800 runs
800-way merge in 10MB memory
i.e. divide the memory into 800 buffers.
Each buffer holds 1/800th of a run
So, 800 runs * 800 seeks/run = 640,000 seeks
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The cost of increasing the file size
• In general, for a K-way merge of K runs, the
buffer size for each run is
– (1/K) * size of memory space = (1/K) * size of each run
• So K seeks are required to read all of the records
in each run.
• Since there are K runs, merge requires K2 seeks.
• Because K is directly proportional to N it also
follows that the sort merge is an O(N2) operation.
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Improvements
There are several ways to reduce the time:
1. Allocate more hardware (e.g. Disk drives,
memory)
2. Perform merge in more than one step.
3. Algorithmically increase the lengths of the
initial sorted runs
4. Find ways to overlap I/O operations.
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Multiple-step merges
• Instead of merging all runs at once, we
break the original set of runs into small
groups and merge the runs in these groups
separately.
– more buffer space is available for each run;
hence fewer seeks are required per run.
• When all of the smaller merges are
completed, a second pass merges the new
set of merged runs.
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25 sets of 32 runs each
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Two-step merge of 800 runs
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Cost of multi-step merge
• 25 sets of 32 runs, followed by 25-way merge:
– Disadvantage: we read every record twice.
– Advantage: we can use larger buffers and avoid a large
number of disk seeks.
• Calculations:
First Merge Step:
– Buffer size = 1/32 run => 32*32 = 1024 seeks
– For 25 32-way merges=> 25 * 1024 = 25,600 seeks
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Second Merge Step:
– For each 25 final runs, 1/25 buffer space is allocated.
– So each input buffer can hold 4000 records (or 1/800
run)
– Hence, 800 seeks per run, so we end up making 25 *
800 = 20,000 seeks.
Total number of seeks for reading in two steps:
25600 + 20000 = 45,600
• What about the total time for merge?
– We now have to transmit all of the records 4 times
instead of two.
– We also write the records twice, requiring an extra
45,600 seeks.
• Still the trade is profitable (see sections 8.5.1-8.5.5
for actual times)
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Increasing Run Lengths
• Assume initial runs contain 200000 records.Then
instead of 800-way merge we need 400-way
merge.
• A longer initial run means
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fewer total runs,
a lower-order merge,
bigger buffers,
fewer seeks.
• How can we create initial runs that are twice as
large as the number of records that we can hold in
memory?
• => Replacement selection
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Replacement Selection
• Idea
– always select the key from memory that has the
lowest value
– output the key
– replacing it with a new key from the input list
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Input:
21,67,12, 5, 47, 16
Remaining input
21,67,12
21,67
21
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Front of input
Memory (P=3)
5
47 16
12 47 16
67 47 16
67 47 21
67 47 _
67 _
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Output run
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5
12,5
16,12,5
21,16,12,5
47, 21,16,12,5
67,47, 21,16,12,5
• What about a key arriving in memory too late to be
output into its proper position? => use of second heap
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Trace of replacement selection
Input:
( P = 3)
33, 18, 24,58,14,17,7,21,67,12,5,47,16
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Replacement Selection with two disks
Algorithm:
1. Construct a heap (primary heap) in the memory, while
reading records block by block from the first disk drive,
2. As we move records from the heap to output buffer, we
replace those records with records from the input buffer.
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If some new records have keys smaller than those already
written out, a secondary heap is created for them.
The other new records are inserted to the primary heap.
3. Repeat step 2 as long as there are records left in the
primary heap and there are records to be read.
4. When the primary heap is empty make the secondary
heap into primary heap and repeat steps 1-3.
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