Post on 08-Jan-2018
1
CS4402 – Parallel Computing
Lecture 9 – Sorting Algorithms (2)
Compare and Exchange Operation
Compare and Exchange Sorting
2
3
4
5
Compare and Exchange Operation
Take place between processors rank1, rank2.
Each processor keeps the sub-array a=(a[i],i=0,1,…,n).
if(rank is rank1){MPI_Send(&a,n,MPI_INT,rank2, tag1,MPI_COMM_WORLD);MPI_Recv(&b,n,MPI_INT,rank2, tag2,MPI_COMM_WORLD,&status);c = merge(n,a,n,b);for(i=0;i<n;i++)a[i]=c[i];
} if(rank is rank2){MPI_Send(&a,n,MPI_INT,rank2, tag2,MPI_COMM_WORLD);MPI_Recv(&b,n,MPI_INT,rank2, tag1,MPI_COMM_WORLD,&status);c = merge(n,a,n,b);for(i=0;i<n;i++)a[i]=c[i+n];
}
6
Compare and Exchange Operation
Complexity?
What amount of computation is being used?
What amount of communication takes place?
CAN YOU FIND ARGUMENTS TO PROVE
THAT THIS IS OPTIMAL OR EFFICIENT?
7
Compare and Exchange Algorithms
Step 1. The array is scattered onto p sub-arrays.
Step 2. Processor rank sorts a sub-array.At any time the processors keep the sub-arrays sorted.
Step 3. While is not sorted / is needed compare and exchange between some processors
Step 4. Gather of arrays to restore a sorted array.
8
Bubble Sort
9
Bubble Sort
10
Bubble Sort
11
12
13
14
Odd-Even Sort1. Scatter the array onto processors.2. Sort each sub-array aa.3. Repeat for step=0,1,2,…, p-1
if (step is odd){if(rank is odd)exchange(aa,n/size,rank, rank+1); if(rank is even) exchange(aa,n/size,rank-1, rank);
} if (step is even){
if(rank is even)exchange(aa,n/size,rank, rank+1); if(rank is odd) exchange(aa,n/size,rank-1, rank);
}4. Gather the sub-arrays back to root.
15
Odd-Even SortSimple Remarks:
- Odd-Even Sort uses size rounds of exchange.
- Odd-Even Sort keeps all processors busy … or almost all.
- The complexity is given by
- Scatter and Gather the array n/size elements
- Sorting the array n/size elements
- Compare and Exchange process size rounds involving n/size
elements
16
if( rank == 0 ){
array = (double *) calloc( n, sizeof(double) ); srand( ((unsigned)time(NULL)+rank) );
for( x = 0; x < n; x++ ) array[x]=((double)rand()/RAND_MAX)*m;}
MPI_Scatter( array, n/size, MPI_DOUBLE, a, n/size, MPI_DOUBLE, 0, MPI_COMM_WORLD );
merge_sort(n/size,a);
for(i=0;i<size;i++){
if( (i+rank)%2 ==0 ){ if( rank < size-1 ) exchange(n/size,a,rank,rank+1,MPI_COMM_WORLD); } else { if( rank > 0 ) exchange(n/size,a,rank-1,rank,MPI_COMM_WORLD);
} MPI_Barrier(MPI_COMM_WORLD)
}
MPI_Gather( a, n/size, MPI_DOUBLE, array, n/size, MPI_DOUBLE, 0, MPI_COMM_WORLD );
if( rank == 0 ){ for( x = 0; x < n; x++ ) printf( "Output : %f\n", array[x] ); }
17
Comments on Odd-EvenFeatures of the algorithm:
- Simple and quite efficient.
- In p steps of compare and exchange the array is sorted out
- Why???
- The number of steps can be reduced if test “array sorted” but still in O(p).
- C&E operations only between neighbors.
Can we do C&E operations between other processors?
18
Odd-Even Sort Complexity
Stage 0. To sort out the scattered array
Stage 1. Odd-Even for p levels
Scatter and Gather
Total computation complexity
comTpn
pn log
commcomcomcomm TnTnTpnT
pnp
2222
commcom Tnpn
Tnpn
pn
222log
commTpn
2
19
isSorted(n, a, comm)The parallel routine int isSorted(int n, double *a, MPI_Comm comm)
1. Test if the processors have all the local arrays in order.
2. rank1 < rank2 elements of rank1 < rank2.
3. If the answer if yes then no exchange is needed.
How to do it?1. The test is done at the root.
2. The test is done collectively by all processors.
20
isSorted(n,a,comm) – Strategy 1The test is done collectively by all processors
1. Send last to the right processor
2. Receive last from the left processor
3. Test if last > a[0] then answer = 0
4. All_Reduce answer by using MIN
21
isSorted(n,a,comm) – Strategy 2The test is done at the root.
1. Gather the first elements to the root.
2. Gather the last elements to the root.
3. If rank == 0 then1. For size-1 times do
- test if last[i] > first[i+1]
• Broadcast the answer
22
Shell Sort It is based on the notion of “shell/group” of consecutive processors.
- C&E take place between equally extreme procs. - The shell is then divided into 2.
(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15) #(shell)=p
(0 1 2 3 4 5 6 7) (8 9 10 11 12 13 14 15) #(shell)=p/2 (0 1 2 3) (4 5 6 7) (8 9 10 11) (12 13 14 15) #(shell)=p/4 (0 1) (2 3) (4 5) (6 7) (8 9) (10 11) (12 13) (14 15) #(shell)=p/8
- There are log(p) levels of division.
For the level l we have- there are pow(2,l) shells each of size p/pow(2,l).
- The shell k contains the processors
1
2)1(,...,1
2,
2 lll
pkpkpk
23
Shell SortShell Sort is based on two stages:
Stage 1. Divide the shells
for l=0,1,2, log(p)
- exchange in parallel between extreme processors in each shell.
Stage 2. Odd-Even
for l=0,1,2, …,p
- if rank and l are both even then exchange in parallel betw rank and rank+1
- if rank and l are both odd then exchange in parallel betw rank and rank+1
- test “array sorted”
24
Shell Sort Complexity
Stage 0. To sort out the scattered array
Stage 1. Odd-Even for l levels
Catch the average complexity of l is in this case O(log^2(p)) so that in average the shell can be
Scatter and Gather
Total computation complexity
comTpn
pn log
comcommcomcomm TppnTp
pnTl
pnTl
pn
22 log2log222
commcom Tppn
pnTp
pn
pn
pn
22 log22log2log
commTpn
2
25
Complexity Comparison for Parallel Sorting
Odd-Even Sort
Shell Sort
Merge Sort
commcom Tppn
pnTp
pn
pn
pn
22 log22log2log
commcom Tnpn
Tnpn
pn
222log
commcom TpnnT
pnn
pn
pn
2222log
26
AssignmentDescription: Write a MPI program to sort out an array:
1. Use a MPI method to compare and exchange
2. Use a MPI method to test isSorted()
3. Use the odd-even sort.
4. Evaluate the performances of the program in a readme.doc
General Points:1. It is for 10% of the marks.
2. Deadline on Monday 2/12/2013 at 5 pm.
3. The following elements must be submitted by email to j.horan@4c.ucc.ie:1. The c program name with your name and student number e.g. SabinTabirca_111111111.c.
2. The Makefile file
3. Readme.doc in which you have 1) to give your student details and 2) to state the performances.