Using openMP to get the index of minimum element parallelly

Question

I tried to write this code

float* theArray; // the array to find the minimum value
int   index, i;
float thisValue, min;

index    = 0;
min = theArray[0];
#pragm         


        

Answer 1

Basic Idea

This can be accomplished without any parellelization-breaking critical or atomic sections by creating a custom reduction. Basically, define an object that stores both the index and value, and then create a function that sorts two of these objects by only the value, not the index.

Details

An object to store an index and value together:

typedef std::pair IndexValuePair;

You can access the index by accessing the first property and the value by accessing the second property, i.e.,

IndexValuePair obj(0, 2.345);
unsigned int ix = obj.first;  // 0
float val = obj.second; // 2.345

Define a function to sort two IndexValuePair objects:

IndexValuePair myMin(IndexValuePair a, IndexValuePair b){
    return a.second < b.second ? a : b;
}

Then, construct a custom reduction following the guidelines in the OpenMP documentation:

#pragma omp declare reduction \
(minPair:IndexValuePair:omp_out=myMin(omp_out, omp_in)) \
initializer(omp_priv = IndexValuePair(0, 1000))

In this case, I've chosen to initialize the index to 0 and the value to 1000. The value should be initialized to some number larger than the largest value you expect to sort.

Functional Example

Finally, combine all these pieces with the parallel for loop!

// Compile with g++ -std=c++11 -fopenmp demo.cpp
#include 
#include 
#include 

typedef std::pair IndexValuePair;

IndexValuePair myMin(IndexValuePair a, IndexValuePair b){
    return a.second < b.second ? a : b;
}

int main(){

    std::vector vals {10, 4, 6, 2, 8, 0, -1, 2, 3, 4, 4, 8};
    unsigned int i;

    IndexValuePair minValueIndex(0, 1000);

    #pragma omp declare reduction \
        (minPair:IndexValuePair:omp_out=myMin(omp_out, omp_in)) \
        initializer(omp_priv = IndexValuePair(0, 1000))

    #pragma omp parallel for reduction(minPair:minValueIndex)
    for(i = 0; i < vals.size(); i++){

        if(vals[i] < minValueIndex.second){
            minValueIndex.first = i;
            minValueIndex.second = vals[i];
        }
    }

    std::cout << "minimum value = " << minValueIndex.second << std::endl;   // Should be -1
    std::cout << "index = " << minValueIndex.first << std::endl;    // Should be 6


    return EXIT_SUCCESS;

}