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CS4402 – Parallel Computing

Lecture 9 – Sorting Algorithms (2)

Compare and Exchange Operation

Compare and Exchange Sorting

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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];

}

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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?

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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.

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Bubble Sort

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Bubble Sort

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Bubble Sort

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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.

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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

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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] ); }

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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?

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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

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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.

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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

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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

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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

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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”

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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

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Complexity Comparison for Parallel Sorting

Odd-Even Sort

Shell Sort

Merge Sort

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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.