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Three-Star

Revision as of 00:20, 21 February 2018 by Tmoy (talk | contribs) (Assignment 1)

Three-Star

Team Members

  1. Derrick Leung
  2. Timothy Moy

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Progress

Assignment 1

Calculation of Pi

Chosen to profile calculation of pi as shown here: https://computing.llnl.gov/tutorials/parallel_comp/#ExamplesPI .

Based on the example program MPI Program in C: mpi_pi_reduce.c https://computing.llnl.gov/tutorials/mpi/samples/C/mpi_pi_reduce.c .

LZW Data Compression Algorithm

Timothy Moy profiled.

Original algorithm: https://codereview.stackexchange.com/questions/86543/simple-lzw-compression-algorithm

Summary of Profiles

Size (MB) Compress() time in seconds
10 0.96
15 1.35
20 1.8
25 2.14
30 2.64
35 3.16
40 3.45
45 4.24
50 4.23

The compress function seems to have some room for improvement as can be seen in the source code below

void compress(string input, int size, string filename) {

   unordered_map<string, int> compress_dictionary(MAX_DEF);
   //Dictionary initializing with ASCII
   for ( int unsigned i = 0 ; i < 256 ; i++ ){
   compress_dictionary[string(1,i)] = i;
   }
   string current_string;
   unsigned int code;
   unsigned int next_code = 256;
   //Output file for compressed data
   ofstream outputFile;
   outputFile.open(filename + ".lzw");
   // Possible area for improvement via reduction
   for(char& c: input){
   current_string = current_string + c;
   if ( compress_dictionary.find(current_string) ==compress_dictionary.end() ){
           if (next_code <= MAX_DEF)
               compress_dictionary.insert(make_pair(current_string, next_code++));
           current_string.erase(current_string.size()-1);
           outputFile << convert_int_to_bin(compress_dictionary[current_string]);
           current_string = c;
       }   
   }   
   if (current_string.size())
           outputFile << convert_int_to_bin(compress_dictionary[current_string]);
   outputFile.close();

}

Note the comment above the second for loop notes we can do something like this:

for (int i = 1; i < n; i+=) a[0] += a[i];

changed to

for (int s = 1; s <= n/2; s*=2) for(int j = 0; j < n; j +=2 * s) a[j] += a[j + s];

The first for loop is constant and probably won't show much improvement if we parallelize it. As such, the major hotspot in this function is the second for loop. This is especially true since the file might be very large and we may be dealing with millions of characters!

Assignment 2

Assignment 3