parallel sorting

1. Parallel Algorithm & Sorting in Parallel
Programming
Submitted By:-
Richa kumari,14MT-CS12
Submitted To:-
Dalpat songra

2. Contents:
1.1 Parallel algorithm
1.2 A Network for sorting
1.3 Sorting on a linear array
1.4 Sorting on the CRCW Model
1.5 Sorting on the CREW Model
1.6 Sorting on the EREW Model

3. 1.1 Parallel Algorithm:-
 A parallel algorithm or concurrent
algorithm, as opposed to a traditional
sequential algorithm, is an algorithm which
can be executed a piece at a time on many
different processing devices, and then
combined together again at the end to get
the correct result.

4. Parallel Sorting:-
 The fundamental operation of comparison-
based sorting is compare-exchange.
 The lower bound on any comparison-based
sort of n numbers is Θ(nlog n) .
 The sorted list is partitioned with the
property that each partitioned list is sorted
and each element in processor Pi's list is less
than that in Pj's list if i < j

5. Sorting: Parallel Compare Exchange Operation
A parallel compare-exchange operation. Processes Pi
and Pj send their elements to each other. Process Pi
keeps min{ai,aj}, and Pj keeps max{ai, aj}.

6. Quick Sort:-
 Quicksort is one of the most common sorting
algorithms for sequential computers because of
its simplicity, low overhead, and optimal
average complexity.
 Quicksort selects one of the entries in the
sequence to be the pivot and divides the
sequence into two - one with all elements less
than the pivot and other greater.
 The process is recursively applied to each of
the sublists.

7. Cont…
 Average optimal sequential complexity: O(n log n)
 Parallel efficiency limitations
 Partitions are unbalanced
 A single processor performs the initial
partitioning

8. Example of quicksort
 Let S = (6,5 ,9,2,4,3,5 , 1, 7,5,8 ).
T he first call to procedure Q U I C K S O R T
produces 5 as the median element of S, and hence
S1 = {2,4,3,1,5,5} and
S2 = {6,9,7,8,5}.
Note that S1 = 6 and S2= 5. A recursive call to Q U I
C K S O R T with S, as input produces the two
subsequences {2,1,3} and {4,5,5}. The second call
with S, as input produces {6,5,7}an d {9,8}. Further
recursive calls complete the sorting of these
sequences.

9. Quicksort algo….

10. COMPLEXITY OF QUICKSORT
For some constant c, we can express the running
time of procedure
QUICKSORT as
= O(n log n),

11. 1.2 A NETWORK FOR SORTING
 It is rather straightforward to use a collection of
merging networks
 to build a sorting network for the sequence S = {s1,
s2, . . . , sn), where n is a power of 2. The idea is the
following.
 In a first stage, a rank of n/2 comparators is used to
create n/2 sorted sequences each of length 2.
 In a second stage, pairs of these are now merged into
sorted sequences of length 4 using a rank of (2,2)-
merging networks. Again, in a

12. Conti….
 third stage, pairs of sequences of length 4 are
merged using (4,4)-merging networks into
sequences of length 8. The process continues until
two sequences of length n/2 each are merged by an
(n/2, n/2)-merging network to produce a single
sorted sequence of length n. The resulting
architecture is known as an odd-even sorting
network and is
 illustrated in Fig. for S = {8,4,7,2, 1,5,6,3). Note
that, as in the case of merging, the odd-even
sorting network is oblivious of its input.

13. FIG: ODD EVEN SORTING NETWORK FOR SEQUENCE OF EIGHT
ELEMENTS

14. The odd-even sorting network is impractical
for large input sequences :
(i) The network is extremely fast. It can sort a sequence of
length 2^20 within, on the order of, (20)2 time units.
This is to be contrasted with the time required by
procedure QUICKSORT, which would be in excess of
20 million time units.[(log n)^2]
(ii) The number of comparators is too high. Again for n =
2^20, the network would need on the order of 400 million
comparators.[n (log n)^2]
(iii) The architecture is highly irregular and the wires linking
the comparators have lengths that vary with n.

15. 1.3 SORTING ON A LINEAR ARRAY:
In this section we describe a parallel sorting algorithm
for an SIMD computer where the processors are
connected to form a linear array
FIG: LINEAR ARRAY CONNECTION

16. Odd-Even Transposition Sort
 Variation of bubble sort.
 Operates in two alternating phases, even phase
and odd phase.
 Even phase
Even-numbered processes exchange numbers
with their right neighbour.
 Odd phase
Odd-numbered processes exchange numbers with
their right neighbour.

17. 
Odd-Even Transposition Sort - example
Parallel time complexity: Tpar = O(n) (for P=n)

18. Algorithm

19. MERGE SPLIT:-
• Now consider the second approach. If N processors,
where N < n,
• Assume that each of the N processors in the linear array
holds a subsequence of S of length n/N.
•The comparison-exchange operations of procedure
ODD-EVEN TRANSPOSITION are now replaced with
merge-split operations on subsequences.
•Let Si denote the subsequence held by processor Pi.
Initially, the Si are random subsequences of S.

20. Sorting sequence of twelve elements using procedure
MERGE SPILIT:-

21. MERGE SPLIT ALGO:

22. Computational time complexity using n processors
 Parallel quicksort - O(n) but unbalanced processor
load, and communication can generate to O(nlogn)
 parallel sorting in network-O(n log^4 n)
Odd-even transposition sort- O(n^2)
 Parallel mergesplit - O(nlogn) but unbalanced
processor load and communication
Parallel sorting Conclusions:

23. 1.4 SORTING ON THE CRCW MODEL
 By this algorithm write conflicts problem can be
resolved.
 we shall assume that write conflicts are created
whenever several processors attempt to write
potentially different integers into the same address.
The conflict is resolved by storing the sum of these
integers in that address.

24. Cont......
 Assume that n^2 processors are available on such a
CRCW computer to sort the sequence
S = { s 1 , s2, . . . , sn).
 If two elements si and sj are equal, then si is taken to
be the larger of the two if i > j; otherwise sj is the
larger.

25. Cont....
procedure CRCW SORT (S)
Step 1: for i = 1 to n do in parallel
for j = 1 to n do in parallel
if (si > sj) or (si = sj and i > j )
then P(i, j) writes 1 in ci
else P(i, j ) writes 0 in ci
end if
end for ---
end for.
Step 2: for i = 1 to n do in parallel
P(i, 1 ) stores si in position 1 + ci of S
end for

26. Example: Let S = (5,2,4, 5) n=4 so n2 =16
Processor
0 1 1 0
0 0 0 0
0 1 0 0
1 1 1 0

27.  Update si array
 i: 1+ci position
 5: 1+2=3
 2: 1+0=1
 3:1+1=2
 4:1+3=4
Cont...

28. Cont......
Analysis:- Each of steps 1 and 2 consists of an
operation requiring constant time. Therefore
Running Time t(n) = O(1).
 Since p(n) = n2
 The cost of procedure CRCW SORT is:-
C(n)= O(n2) (which is not optimal)

29. 1.5 SORTING ON THE CREW MODEL
 Our purpose is to design an algorithm that is:
1. free of write conflicts.
2. uses a reasonable number of processors.
3. a running time that is small and adaptive.
4. a cost that is optimal.
 Assume that a CREW SM SIMD computer with N
processors PI, P2. . . , PN is to be used to sort the
sequence S = {s1 s2 . . . , sn), where N < n.

30. procedure CREW SORT (S)
Step 1: for i = 1 to N do in parallel
Processor Pi
(1.1) reads a distinct subsequence Si of S of size n/N
(1.2) QUICKSORT (Si)
(1.3) Si
1 <- Si
(1.4) Pi
1 <- Pi
end for.
O((n/N)log(n/N))
Algorithm:-

31. Cont…
Step 2 (2.1) u =1
(2.2) v = N
(2.3) while v > 1 do
(2.3.1) for m = 1 to |_v/2_| do in parallel
(i) Pu+1
m <- Pu
2m-1 U pu
2m
(ii) The processors in the set Pu+1
mperform
CREW MERGE (su
2m-1, su
2m, su+1
m)
end for
(2.3.2) if v is odd then
(1) pU+1
v/2 = pu
v
(ii) sU+1
v/2 = sU
V
end if
(2.3.3) u = u + 1
(2.3.4) V = v/2
end while.
O((n/N) + log n)
time

32. Example
 Let S = (2, 8, 5, 10, 15, 1, 12, 6, 14, 3, 11, 7, 9, 4, 13, 16) and N
= 4. Here N<n
Step1:- Subsequence Si created : n/N=>16/4= 4
And Quick sort apply for sorting elements
S1
1 ={2,5,8,10} S2
1 = {1,6,12,15}
S3
1= {3,7,11,14} S4
1 = {9,13,14,16}
Step2:- u=1 & v=N=4
for (m=1 to v/2)
P1
2=p1
1 U p2
1 =(p1,p2)=(1,2,5,6,8,10,12,15)
P2
2= p3
1 U p4
1 =(p3,p4)=(3,4,7,9,11,13,14,16)
4/2=2
CREW
MERGE ALGO
USED

33. Cont....
The processors {P1, P2,P3, P4} cooperate to merge S1
2 and
s2
2 into S1
3 = (1, 2,. . . , 16) by using CERW MERGE .
Analysis:- the total running time of procedure CREW
SOR'T is
t(n) = O((n/N)log(n/N)) + O((n/N)log N + log n log N)
= O((n/N)log n + log2n).
 Since p(n) = N, the cost is given by:-
c(n) = O(n log n + N log n^2).

34. 1.6 SORTING ON THE EREW MODEL:-
 Still, procedure CREW SORT tolerates multiple-
read operations. Our purpose in this section is to
deal with this third difficulty.
 We assume throughout this section that N
processors P1, P2 . . . , PN are available on an
EREW SM SIMD computer to sort the sequence S
= (s1, s2, . . . , sn)where N < n.

35. Cont….
 since N < n, N=n1-x where 0<x<1.
 Now mi =[ i(n/21/x)], for 1<=i<=21/x-1 .
 The mi can be used to divide S into 21/x subsequence of size
n/21/x .
 These subsequences, denoted by S1,S2,..., Sj, Sj+1,........S2j,
where j =2(1/x)-1
 Every subdivision process can now be applied recursively to
each of the subsequences Si until the entire sequence S is
sorted in nondecreasing order.
 K= 2(1/x)

36. Algorithm:-
procedure EREW SORT (S)
Step1 if |S| < k
then QUICKSORT (S)
else (1) for i = 1 to k - 1 do
PARALLEL SELECT (S, |i |s|/k|) [obtain mi]
end for
(2) Si = (s E S: s<=mi )
(3) for i = 2 to k - 1 do
Si ={s E S : mi-1<=s <=mi }
end for

37. Cont..
(4) Sk<= { s E S : s >=mk-1)
Step 2 for i = 1 to k/2 do in parallel
EREW SORT (Si)
end for
Step 3 for i = (k/2) + 1 to k do in parallel
EREW SORT (Si)
end for
end if.

38. Cont...
Let S = {5,9, 12, 16, 18,2, 10, 13, 17,4,7, 18, 18, 11, 3,
17,20,19, 14, 8, 5, 17, 1, 11, 15, 10, 6) (i.e., n = 27)
 Here N<n & N=n1-x => N=270.5 = 5 where 0<x<1
(x=0.5).
 K=21/x => k= 21/0.5
= 22
= 4
 During step 1 m1= 6 m2 = 11, and m3 = 17 are
computed.
 The four sub sequences S1 ,S2, S3 and S4 are created.
 In step 5 the procedure is applied recursively and
simultaneously to S1 and S2.
 Compute m1 = 2, m2= 4, and m3= 5, and the four
subsequence {1,2}, {3,4}, {5,5), and (6) are created
each of which is already in sorted order.

39. Cont....

40. Cont....
 Running Time t(n) = cnx + 2t(n/k)
= O(nx log n).
 Since p(n) = n1-x, the procedure's cost is given by
c(n) = p(n) x t(n) = O(n log n),
which is optimal.