1. Chap 8. Cosequential Processing and the Sorting of Large Files

2. Chapter Objectives(1)
Describe a class of frequently used processing activities known as cosequential process
Provide a general object-oriented model for implementing varieties of cosequential processes
Illustrate the use of the model to solve a number of different kinds of cosequential processing problems, including problems other than simple merges and matches
Introduce heapsort as an approach to overlapping I/O with sorting in RAM

3. Chapter Objectives(2)
Show how merging provides the basis for sorting very large files
Examine the costs of K-way merges on disk and find ways to reduce those costs
Introduce the notion of replacement selection
Examine some of the fundamental concerns associated with sorting large files using tapes rather than disks
Introduce UNIX utilities for sorting, merging, and cosequential processing

4. Contents 8.1 Cosequential operations
8.2 Application of the OO Model to a General Ledger Program
8.3 Extension of the OO Model to Include Multiway Merging
8.4 A Second Look at Sorting in Memory
8.5 Merging as a Way of Sorting Large Files on Disk
8.6 Sorting Files on Tape
8.7 Sort-Merge Packages
8.8 Sorting and Cosequential Processing in Unix

5. Cosequential operations
Coordinated processing of two or more sequential lists to produce a single list
Kinds of operations
merging, or union
matching, or intersection
combination of above

6. Matching Names in Two Lists(1)
So called “intersection operation”
Output the names common to two lists
Things that must be dealt with to make match procedure work reasonably
initializing that is to arrange things
methods that are getting and accessing the next list item
synchronizing between two lists
handling EOF conditions
recognizing errors
e.g. duplicate names or names out of sequence

7. Matching Names in Two Lists(2)
In comparing two names
if Item(1) is less than Item(2), read the next from List 1
if Item(1) is greater than Item(2), read the next name from List 2
if the names are the same, output the name and read the next names from the two lists

8. Cosequential match procedure(1)
PROGRAM: match
Item(1)
Item(1) < Item(2)
List 1
same
name
use input() & initialize() procedure
List 2
Item(1) > Item(2)
Item(2)

9. Cosequential match procedure(2)
int Match(char * List1, char List2, char *OutputList) {
int MoreItems; // true if items remain in both of the lists
// initialize input and output lists
InitializeList(1, List1); InitializeList(2, List2);
InitializeOutput(OutputList);
// get first item from both lists
MoreItems = NextItemInLIst(1) && NextItemInList(2);
while (MoreItems) { // loop until no items in one of the lists
if(Item(1) < Item(2) ) MoreItems = NextItemInList(1);
else if (Item(1) == Item (2) ) {
ProcessItem(1); // match found
MoreItems = NextItemInList(1) && NextItemInList(2); }
else MoreItems = NextItemInList(2); // Item(1) > Item(2)
FinishUp(); return 1;

10. General Class for Cosequential Processing(1)
template <class ItemType> class CosequentialProcess
// base class for cosequential processing
{ public:
// the following methods provide basic list processing
// these must be defined in subclasses
virtual int InitializeList (int ListNumber, char *LintName) = 0;
virtual int InitializeOutput (char * OutputListName) = 0;
virtual int NextItemInList (int ListNumber) = 0;
// advance to next item in this list
virtual ItemType Item(int ListNumber) = 0;
// return current item from this list
virtual int ProcessItem(int ListNumber) = 0;
// process the item in this list
virtual int FinishUp() = 0; // complete the processing
// 2-way cosequential match method
virtual int Match2Lists (char *List1, char * List2, char *OutputList); };

11. General Class for Cosequential Processing(2)
A Subclass to support lists that are files of strings, one per line
class StringListProcess : public CosequentialProcess<String &>
{ public:
StringListProcess (int NumberOfLists); // constructor
// Basic list processing methods
int InitializeList (int ListNumber, char * List1);
int InitializeOutput(char * OutputList);
int NextItemInList (int ListNumber); // get next
String & Item (int ListNumber); // return current
int ProcessItem (int ListNumber); // process the item
int FinishUp(); // complete the processing
protected:
ifstream * List; // array of list files
String * Items; // array of current Item from each list
ofstream OutputLsit;
static const char * LowValue; //used so that NextItemInList() doesn’t // have to get the first item in an special way
static const char * HighValue; };

12. General Class for Cosequential Processing(3)
Appendix H: full implementation
An example of main
#include “coseq.h”
int main() {
StringListProcess ListProcess(2); // process with 2 lists
ListProces.Match2Lists (“list1.txt”, “list2.txt”, “match.txt”); }

13. Merging Two Lists(1)
Based on matching operation
Difference
must read each of the lists completely
must change MoreItems behavior
keep this flag set to true as long as there are records in either list
HighValue
the special value (we use “\xFF”)
come after all legal input values in the files to ensure both input files are read to completion

14. Merging Two Lists(2)
Cosequential merge procedure based on a single loop
This method has been added to class CosequentialProcess
No modifications are required to class StringListProcess
template <class ItemType>
int CosequentialProcess<ItemType> :: Merge2Lists
(char * List1Name, char * List2Name, char * OutputList) {
int MoreItems1, MoreItems2; // true if more items in list InitializeList (1 List1Name);
InitializeList (2, List2Name);
InitializeOutput (OutputListName);
MoreItems1 = NextItemInLIst(1);
MoreItems2 = NextItemInList(2);
while(MoreItems1 || MoreItems2) {
(continued … )

15. Merging Two Lists(3)
while (MoreItems(1) || MoreItems(2) ) { // if either file has more
if (Item(1) < Item(2)) { // list 1 has next item to be processed
ProcessItem(1);
MoreItem1 = NextItemInList(1); }
else if (Item(1) == Item(2) ) {
MoreItems1 = NextItemInList(1);
MoreItems2 = NextItemInList(2);
else // Item(1) > Item(2) {
ProcessItem(2);
MoreItem2 = NextItemInList(2);
FinishUp(); return 1;

16. Cosequential merge procedure(1)
PROGRAM: merge
NAME_1
NAME_2
List 1
List 2
OutputList
(Item(1) < Item(2) )or match
Item(1) > Item(2)

17. Summary of the Cosequential Processing Model(1)
Assumptions
two or more input files are processed in a parallel fashion
each file is sorted
in some cases, there must exist a high key value or a low key
records are processed in a logical sorted order
for each file, there is only one current record
records should be manipulated only in internal memory

18. Summary of the Cosequential Processing Model(2)
Essential Components
initialization - reads from first logical records
one main synchronization loop
- continues as long as relevant records remain
selection in main synchronization loop
Input files & Output files are sequence checked by comparing the previous item value with new one
if (Item(1) > Item(2) then
else if ( Item(1) < Item(2)) then
else /* current keys equal */
endif

19. Summary of the Cosequential Processing Model(3)
Essential components (cont’d)
substitute high values for actual key when EOF
main loop terminates when high values have occurred for
all relevant input files
no special code to deal with EOF
I/O or error detection are to be relegated to supporting method so the details of these activities do not obscure the principal processing logic

20. 8.2 The General Ledger Program (1)
Figure 8.7 Sample journal entries
Acct.No. Check No. Date Description Debit/credit
/02/97 Auto expense
/02/97 Tune-up and minor repair 78.70
/02/97 Rent
/02/97 Rent for April
/04/97 Advertising
/04/97 Newspaper ad re:new product 87.50
/02/97 Office expense
/02/97 Printer cartridge
/02/97 Auto expense
/09/97 Oil change

21. 8.2 The General Ledger Program (2)
Figure 8.8 Sample ledger printout showing the effect of posting from the journal
Checking account #1
/02/97 Auto expense
/02/97 Rent
/04/97 Advertising
/02/97 Auto expense Prev.bal: New bal:
Checking account # /02/97 Office expense Prev.bal: New bal:
Advertising expense /04/97 Newspaper ad re:new product Prev.bal:25.00 New bal:
Auto expenses
04/02/97 Tune-up and minor repair 78.70
/09/97 Oil change Prev.bal: New bal:
이전 잔액
현재 잔액

22. 8.2 The General Ledger Program (3)
Figure 8.9 List of journal transactions sorted by account number
Acct.No. Check No. Date Description Debit/credit
/02/97 Auto expense
/02/97 Rent
/04/97 Advertising
/02/97 Auto expense
/02/97 Office expense
/04/97 Newspaper ad re:new product 87.50
/02/97 Tune-up and minor repair 78.70
/09/97 Oil change
/02/97 Printer cartridge
/02/97 Rent for April

23. 8.2 The General Ledger Program (4)
Figure 8.10 Conceptual view of cosequential matching of the ledger and journal files
Ledger List Journal List
Checking account # Auto expense Rent Advertising Auto expense
Checking account # Office expense
Advertising expense Newspaper ad re: new product
510 Auto expenses Tune-up and minor repair Oil change

24. 8.2 The General Ledger Program (5)
Figure 8.11 Sample ledger printout for the first six accounts
Checking account #1
/02/97 Auto expense
04/02/97 Rent
/02/97 Auto expense
/04/97 Advertising
Prev.bal: New bal:
Checking account # /02/97 Office expense Prev.bal: New bal:
Advertising expense /04/97 Newspaper ad re:new product Prev.bal:25.00 New bal:
Auto expenses
04/02/97 Tune-up and minor repair 78.70
/09/97 Oil change Prev.bal: New bal:
Bank charges Prev.bal: 0.00 New bal: 0.00
520 Books and publications Prev.bal: New bal: 87.40

25. 8.2 The General Ledger Program(6)
The ledger (master) account number
The journal (transaction) account number
Class MasterTransactionProcess (Fig 8.12)
Subclass LedgeProcess (Fig 8.14)

26. 8.2 The General Ledger Program (7)
Template <class ItemType>
class MasterTransactionProcess: Public CosequentialProcess<ItemType>
// a cosequential process that supports master/transaction processing
{public:
MasterTransactionProcess(); // constructor
Virtual int ProcessNewMaster() = 0; //processing when new master read
Virtual int ProcessCurrentMaster() = 0;
Virtual int ProcessEndMaster() = 0;
Virtual int ProcessTransactionError()= 0;
//cosequential processing of master and transaction records
int PostTransactions (char * MasterFileName, char * TransactionFileName, char * OutputListName); };

27. 8.2 The General Ledger Program (8)
while (MoreMasters || MoreTransactions)
if (Item(1) < Item(2)) { // 이 마스터 레코드를 끝낸다
ProcessEndMaster();
MoreMasters = NextItemInList(1);
if (MoreMasters)
ProcessNewMaster();
} else if (Item(1) == Item(2)) { // 마스터와 일치하는 트랜잭션
ProcessCurrentMaster(); // 마스터를 위한 또다른 트랜잭션
ProcessItem(2); // 트랜잭션 레코드 출력
MoreTransactions = NextItemInList(2);
} else { // Item(1) > Item(2) 마스터를 갖지 않는 트랜잭션
ProcessTransactionError(); }

28. 8.2 The General Ledger Program (9)
int LedgerProcess::ProcessNewMaster()
{ // 헤더를 프린트하고 마지막 달의 잔액을 설정
ledger.PrintHeader(OutputList);
ledger.Balances[MonthNumber] = ledger.Balances[MonthNumber-1]; }
int LedgerProcess::ProcessCurrentMaster()
{ // 이 달의 잔액에 트랜잭션의 양을 더한다
ledger.Balances[MonthNumber] += journal.Amount;
int LedgerProcess::ProcessEndMaster()
{ // 잔액을 출력한다
PrintBalances(OutputList,
ledger.Balances[MonthNumber-1],
ledger.Balances[MonthNumber]);

29. 8.3 A K-way Merge Algorithm
A very general form of cosequential file processing
Merge K input lists to create a single, sequentially ordered output list
Algorithm
begin loop
determine which list has the key with the lowest value
output that key
move ahead one key in that list
in duplicate input entries, move ahead in each list
loop again

30. 8.3 Selection Tree for Merging Large Number of Lists
K-way merge
nice if K is no larger than 8 or so
if K > 8, the set of comparisons for minimum key is expensive
loop of comparison (computing)
Selection Tree (if K > 8)
time vs. space trade off
a kind of “tournament” tree
the minimum value is at root node
the depth of tree is log2 K

31. 8.3 Selection Tree 7, 10, 17....List 0 7 9, 19, 23....List 1 711
18, 22, List 3
input 5
12, 14, List 4 5
5, 6, List 5 5
15, 20, List 6 8
8, 16, List 7

32. 8.4 A Second Look at Sorting in Memory
Read the whole file from into memory, perform sorting, write the whole file into disk
Can we improve on the time that it takes for this RAM sort?
perform some of parts in parallel
selection sort is good but cannot be used to sort entire file
Using Heap technique!
processing and I/O can occur in parallel
keep all the keys in heap
Heap building while reading a block
Heap rebuilding while writing a block

33. 8.4 Overlapping processing and I/O : Heapsort
a kind of binary tree, complete binary tree
each node has a single key, that key is less than or equal to the key at its parent node
storage for tree can be allocated sequentially
so there is no need for pointers or other dynamic overhead for maintaining the heap

34. 8.4 A heap in both its tree form and as it would be stored in an array
(1)
* n, 2n, 2n+1 positions B
(2) c
(3) E
(4) H
(5) I
(6) D
(7) F
(9) G
(8) A B C E H I D G F

35. 8.4 Class Heap and Method Insert(1)
{ public:
Heap(int maxElements);
int Insert (char * newKey);
char * Remove();
protected:
int MaxElements; int NumElements;
char ** HeapArray;
void Exchange (int i, int j); // exchange element i and j
int Compare (int i, int j) // compare element i and j
{ return strcmp(Heaparray[i], HeapArray[j]); } };

36. 8.4 Class Heap and Method Insert(2)
int Heap::Insert(char * newKey) {
if (NumElements == MaxElements) return FALSE;
NumElements++; // add the new key at the last position
HeapAray[NumElements] = newKey;
// re-order the heap
int k = NumElements; int parent;
while(k > 1) { // k has a parent
parent = k/2;
if (Compare(k, parent) >= 0) break;
// HeapArray[k] is in the right place
// else exchange k and parent
Exchange(k, parent);
k = parent; }
return;

37. Heap Building Algorithm(1)
input key order : F D C G H I B E A
New key to
be inserted
Heap, after insertion
of the new key
Selected heaps
in tree form F F D
D F C C
C F D D F G
C F D G H
C F D G H
(continued....)

38. Heap Building Algorithm(2)
input key order : F D C G H B E A
New key to
be inserted
Heap, after insertion
of the new key
Selected heaps
in tree form I
C F D G H I C F D B
B F C G H I D G H I E
B E C F H I D G B A
A B C E H I D G F F C H G I D
(continued....)

39. Heap Building Algorithm(3)
input key order : F D C G H B E A
New key to
be inserted
Heap, after insertion
of the new key
Selected heaps
in tree form A
A B C E H I D G F A C B D H I E G F

40. Illustration for overlapping input with heap building(1)
(Free ride of main memory processing: heap building is faster than IO!)
Total RAM area allocated for heap
First input buffer. First part of heap is built here. The
first record is added to the heap, then the second record
is added, and so forth
Second input buffer. This buffer is being filled
while heap is being built in first buffer.

41. Illustration for overlapping input with heap building(2)
(One Heap is growing during IO time!)
Second part of heap is built here. The first record is
added to the heap, then the second record, etc
Third input buffer. This buffer is filled while heap is being
built in second buffer
Third part of heap is built here
Fourth input buffer is filled while heap is being
built in third buffer

42. 8.4 Sorting while Writing to the File
Heap rebuilding while writing a block
(Free ride of main memory processing)
Retrieving the keys in order (Fig 8.20)
while( there is no elements)
get the smallest value
put largest value into root
decrease the # of elements
reorder the heap
Overlapping retrieve-in-order with I/O
retrieve-in-order a block of records
while writing this block,
retrieve-in-order the next block

43. 8.4 Method Remove char * Heap::Remove()
{ // 가장 작은 원소를 제거하고, 힙을 재정렬시키고, 가장 작은 원소를 리턴한다
// 리턴하기 위해 가장 작은 값을 val에 넣는다
char * val = HeapArray[1];
HeapArray[1] = HapArray[NumElements]; // 루트에 가장 큰 값을 넣는다
NumElements--; // 원소의 수를 감소시킨다
// 교환과 하향이동에 의해 힙을 재정렬한다
int k = 1; // 가장 큰 값을 포함하는 힙의 노드
int newK; // 가장 큰 값으로 교환되는 힙의 노드
while (2*k <= NumElements) { // k는 최소 하나의 자식을 가짐
// k의 가장 작은 자식 인덱스로 newK를 설정
if (Comapre(2*k, 2*k+1) > 0) newK = 2*k;
else newK = 2*k+1;
// k와 newK가 순서대로 되어 있는지를 비교
if (Compare(k, newK) < 0) break; // 순서대로 되어 있는 경우
Exchange(k, newK); // k와 newK가 순서대로 되지 않은 경우
k = newK; //트리의 아래 방향으로 진행}
return val; }
Figure 8.20

44. 8.5 Merging as a Way of Sorting Large Files on Disk
Keysort: holding keys in memory
Two Shortcomings of Keysort
substantial cost of seeking may happen after keysort
cannot sort really large files
e.g. a file with 800,000 records, size of each record: 100 bytes,
size of key part: 10 bytes, then 800,000 X 10 => 8G bytes!
cannot even sort all the keys in RAM
Multiway merge algorithm
small overhead for maintaining pointers, temporary variables
run: sorted subfile
using heap sort for each run
split, read-in, heap sort, write-back

45. ............. Sorting through the creation of runs
and subsequential merging of runs
800,000 unsorted records
80 internal sorts
80runs, each containing 10,000 sorted records
Merge
800,000 records in sorted order

46. Multiway merging (K-way merge-sort)
Can be extended to files of any size
Reading during run creation is sequential
no seeking due to sequential reading
Reading & writing is sequential
Sort each run: Overlapping I/O using heapsort
K-way merges with k runs
Since I/O is largely sequential, tapes can be used

47. How Much Time Does a Merge Sort Take?
Assumptions
only one seek is required for any sequential access
only one rotational delay is required per access
Four I/Os
during the sort phase
reading all records into RAM for sorting, forming runs
writing sorted runs out to disk
during the merge phase
reading sorted runs into RAM for merging
writing sorted file out to disk

48. Four Steps(1)
Step1: Reading records into RAM for sorting and forming runs
assume: 10MB input buffer, 800MB file size
seek time --> 8msec, rotational delay --> 3msec
transmission rate --> MB/msec
Time for step1:
access 80 blocks (80 X 11)msec + transfer 80 blocks (800/0.0145)msec
Step2: Writing sorted runs out to disk
writing is reverse of reading
time that it takes for step2 equals to time of step1

49. Four Steps(2) Step3: Reading sorted runs into RAM for merging
10 MB of RAM is for storing runs. 80 runs
reallocate each of 80 buffers 10MB RAM as 80 input buffers
access each run 80 buffers to read all of it
Each buffer holds 1/80 of a run (0.125MB)
total seek & rotational time --> 80 runs X 80 seeks
--> 6400 seeks X 11 msec = 70 seconds
transfer time --> 60 seconds (800MB  MB/msec)
total time = total seek & rotation time + transfer time

50. Four Steps(3) Step4: Writing sorted file out to disk
need to know how big output buffers are
with 20,000-byte output buffers,
total seek & rotation time = 4,000 x 11 msec
transfer time is still 60 seconds
Consider Table 8.1 (359pp)
What if we use keysort for 800M file? --> 24hrs 26mins 40secs
80,000,000 bytes
4,000 seeks
20,000 bytes per seek

51. : Effect of buffering on the number of seeks required 800MB file
1st run = 80 buffers’ worth(80 accesses)
2nd run = 80 buffers’ worth(80 accesses)
800MB file
800,000
sorted records :
80 buffers(10MB)
80th run = 80 buffers’ worth(80 accesses)

52. Sorting a Very Large File
Two kinds of I/O
Sort phase
I/O is sequential if using heapsort
Since sequential access is minimal seeking, we cannot algorithmically speed up I/O
Merge phase
RAM buffers for each run get loaded, reloaded at predictable times -> random access
For performance, look for ways to cut down on the number of random accesses that occur while reading runs
you can have some chance here!

53. The Cost of Increasing the File Size
K-way merge of K runs
Merge sort = O(K2) ( merge op. -> K2 seeks )
If K is a big number, you are in trouble!
Some ways to reduce time!! (8.5.4, , 8.5.6)
more hardware (disk drives, RAM, I/O channel)
reducing the order of merge (k), increasing buffer size of each run
increase the lengths of the initial sorted runs
find the ways to overlap I/O operations

54. Hardware-base Improvements
Increasing the amount of RAM
longer & fewer initial runs
fewer seeks
Increasing the number of disk drives
no delay due to seek time after generation of runs
assign input and output to separate drives
Increasing the number of I/O channels
separate I/O channels, I/O can overlap
Improve transmission time

55. Decreasing the Num of Seeks Using Multiple-step Merges
K-way merge characteristics
a selection tree is used
the number of comparisons is N*log K
(K-way merge with N records)
K is proportional to N
O(N*log N) : reasonably efficient
Reducing seeks is to reduce the number of runs
give each run a bigger buffer space
multiple-step merge provides the way without more RAM

56. Multiple-step merge(1)
Do not merge all runs at one time
Break the original set of runs into small groups and Merge runs in these group separately
Leads fewer seeks, but extra transmission time in second pass
Reads every record twice
to form the intermediate runs & the final sorted file
Similar to have selection tree in merging n lists!!

57. Two-step merge of 800 runs ...... 25 sets of 32 runs each 32 runs
(25 sets X 32 runs) = 800 runs
......
32 runs
25 sets of 32 runs each

58. Multiple-step merge(2)
Essence of multiple-step merging
increase the available buffer space for each run
extra pass vs. random access decrease
Can we do even better with more than two steps?
trade-offs between the seek&rotation time and the transmission time
major cost in merge sort
seek, rotation time, transmission time, buffer size, number of runs

59. Increasing Run Lengths Using Replacement Selection(1)
Facts of Life
Want to use the heap sort in memory
Want to allocate longer output runs
Can we pack the longer output runs using the heap sort in memory?
Replacement Selection
Idea
always select the key from memory that has the lowest value
output the key
replace it with a new key from the input list
use 2 heaps in the memory buffer
(continued...)

60. Increasing Run Lengths Using Replacement Selection(2)
Implementation
step1: read records and sort using heap sort
this heap is the primary heap
step2: write out only the record with the lowest value
step3: bring in new record and compare its key with that of record just output
step3-a: if the new key is higher, insert new record into its proper in the primary heap along with the other records selected for output
step3-b: if the new key is lower, place the record in a secondary heap with key values lower than already written out
step4: repeat step 3 while there are records in the primary heap and there are records to be read in. When the primary heap is empty, make the secondary heap into the primary heap and repeat step2 & step3

61. Example of the principle underlying replacement selection
Input:
21, 67, 12, 5, 47, 16
Front of input string
(Heap sort!)
Remaining input
Memory(p=3)
Output run - 5
12, 5
16, 12, 5
21, 16, 12, 5
47, 21, 16, 12, 5
67, 47, 21, 16, 12, 5
21, 67, 12
21, 67 21 -

62. Replacement Selection(1)
What happens if a key arrives in memory too late to be output into ins proper position relative to the other keys? (if 4th key is 2 rather than 12)
use of second heap, to be included in next run
refer to page 335 Figure 8.25
Two questions
Given P locations in memory, how long a run can we expect replacement selection to produce, on the average?
On the average, we can expect a run length of 2P
Knuth provides an excellent description (page )
(continued...)

63. Comparisons of access times required to sort 8 million records
both RAM sort and replacement selection
Total Seek &
Rotation Delay
Time
# of Records
per Seek to
Form Runs
Size of
Runs
Formed
# of Seeks
Required to
Form Runs
Merge
Order
Used
Total
Number
of Seeks
Approach
(hr) (min)
800 RAM
sorts followed
by an 800-way
merge
10,000
10,000
800
1,600
681,600 4 58
Replacement
selection followed
by 534-way merge
(records in random
order)
2,500
15,000
534
6,400
521,134 3 48
Replacement
selection followed
by 200-way merge
(records partially
ordered)
2,500
40,000
200
200
206,400 1 30

64. Step-by-step op. of replacement selection with 2 heaps
working to form two sorted runs(1)
Input
33, 18, 24, 58, 14, 17, 7, 21, 67, 12, 5, 47, 16
Front of input string
(Heap sort!)
Remaining input
Memory(P=3)
Output run(A)
33, 18, 24, 58, 14, 17, 7, 21, 67, 12
33, 18, 24, 58, 14, 17, 7, 21, 67
33, 18, 24, 58, 14, 17, 7, 21
33, 18, 24, 58, 14, 17, 7
33, 18, 24, 58, 14, 17
33, 18, 24, 58, 14
33, 18, 24, 58
( 7)
67 (17) ( 7)
(14) (17) ( 7) - 5
12, 5
16, 12, 5
21, 16, 12, 5
47, 21, 16, 12, 5
67, 47, 21, 16, 12, 5

65. Step-by-step op. of replacement selection
working to form two sorted runs(2)
Remaining input
Memory(P=3)
Output run(B)
First run complete; start building the second
33, 18, 24, 58
33, 18, 24
33, 18 - - - 7
14, 7
17, 14, 7
18, 17, 14, 7
24, 18, 17, 14, 7
33, 24, 18, 17, 14, 7
58, 33, 24, 18, 17, 14, 7

66. Replacement Selection Plus Multiple Merging
8.5 Merging as a Way of Sorting Large Files on Disk
Replacement Selection Plus Multiple Merging
Total number of seeks is less than for the one-step merges
The two-step merge requires transferring the data two more times than do the one-step merge
the two-step merges & replacement selection are still better, but the results are less dramatic
refer to table of the next slide

67. Comparison of merges, considering transmission times(1) :1-step merge
8.5 Merging as a Way of Sorting Large Files on Disk
Comparison of merges, considering transmission times(1)
:1-step merge
Approach
Number of
Records per
Seek to
Form Runs
Merge
Pattern
Used
Number
of Seeks
for Sorts
and Merges
Seek +
Rotational
Delay
Time(min)
Total
Passes
over the
File
Total
Trans-
mission
Time(min)
Total of Seek,
Rotation, and
Transmission
Times(min)
RAM sorts
800-
way
10,000
681,700
298 4 43
341
replacement
selection
(records in
random order)
534-
way
2,500
228 4 43
341
521,134
replacement
selection
(records part
-ially ordered)
2,500
200-
way 90 4 43
206,400
341
(continued...)

68. Comparison of merges, considering transmission times(2) :2-step merge
8.5 Merging as a Way of Sorting Large Files on Disk
Comparison of merges, considering transmission times(2)
:2-step merge
Approach
Number of
Records per
Seek to
Form Runs
Merge
Pattern
Used
Number
of Seeks
for Sorts
and Merges
Seek +
Rotational
Delay
Time(min)
Total
Passes
over the
File
Total
Trans-
mission
Time(min)
Total of Seek,
Rotation, and
Transmission
Times(min)
25 x 32
-way
(one 25-way)
RAM sorts
10,000
127,200 56 6 65
121
replacement
selection
(records in
random order)
19 x 28
-way
(one 19-way)
2,500 55 6 65
120
124,438
20 x 10
-way
(one 20-way)
replacement
selection
(records part
-ially ordered)
2,500
110,400 48 6 65
113

69. Using Two Disks with Replacement Selection
8.5 Merging as a Way of Sorting Large Files on Disk
Using Two Disks with Replacement Selection
Two disk drives
input & output can overlap
reduce transmission by 50%
seeking is virtually eliminated
Sort phase
the run selection & output can overlap
Merge phase
output disk becomes input disk, and vice versa
seeking will occur on input disk, output is sequential
substantially reducing merge & transmission time

70. heap Memory organization for replacement selection disk1 input buffers
8.5 Merging as a Way of Sorting Large Files on Disk
Memory organization for replacement selection
disk1
input
buffers
heap
disk2
output
buffers

71. More Drives? More Processors?
8.5 Merging as a Way of Sorting Large Files on Disk
More Drives? More Processors?
More drives?
Until I/O becomes so fast that processing cannot keep up with it
More processors?
mainframes
vector and array processors
massively parallel machines
very fast local area networks

72. Effects of Multiprogramming
8.5 Merging as a Way of Sorting Large Files on Disk
Effects of Multiprogramming
Increase the efficiency of overall system by overlapping processing and I/O
Effects are very hard to predict

73. A Concept Toolkit for External Sorting
8.5 Merging as a Way of Sorting Large Files on Disk
A Concept Toolkit for External Sorting
For in-RAM sorting, use heapsort
Use as much RAM as possible
Use a multiple-step merge when
the number of initial runs is so long that seek and rotation time is much greater than transmission time
Use replacement selection when
possibility of partially ordered
Use more than one disk drive and I/O channel
read/write can overlap
Look for ways to take advantage of new architecture and systems
parallel processing or high-speed networks

74. Sorting Files on Tape Tape contains runs
Balanced Merge with several tape drivers
Tape contains runs
T R1 R3 R5 R7 R9
Step T R2 R4 R6 R8 R10 T T
Figure 8.28 (2 way-balanced 4 tape merge)
P is the number of passes, N is the number of runs, k is the number of input drivers ==> then, P = ceiling of (logkN)
4 tape drivers (2 for input, 2 for output), 10 runs ==> 4 passes
20 tape drivers (10 for input, 10 for output), 200 runs ==> 3 passes

75. Sorting Files on Tape Other ways of Balanced Merge
(Fig 8.30) T T T T4
Step
Step
Step
Step
(Fig 8.31) T T T T4
Step
Step …
Step …
Step …
Step

76. K-way Balanced Merge on Tapes
Some difficult questions
How does one choose an initial distribution that leads readily to an efficient merge pattern?
Are there algorithmic descriptions of the merge patterns, given an initial distribution?
Given N runs and J tape drives, is there some way to compute the optimal merging performance so we have a yardstick against which to compare the performance of any specific algorithm?

77. Unix: Sorting and Cosequential Processing
Sorting in Unix
The Unix sort command
The qsort library routine
Cosequential processing utilities in Unix
Compares: cmp
Difference: diff
Common: comm

78. Let’s Review !! 8.1 Cosequential operations
8.2 Application of the Model to a General Ledger Program
8.3 Extension of the Model to Include Multiway Merging
8.4 A Second Look at Sorting in Memory
8.5 Merging as a Way of Sorting Large Files on Disk
8.6 Sorting Files on Tape
8.7 Sort-Merge Packages
8.8 Sorting and Cosequential Processing in Unix