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Three-Star

Revision as of 08:21, 21 February 2018 by Dleung25 (talk | contribs) (Image Profiling)

Three-Star

Team Members

  1. Derrick Leung
  2. Timothy Moy

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Progress

Assignment 1

Image Profiling

Chosen to profile image profiling as shown here: http://www.dreamincode.net/forums/topic/76816-image-processing-tutorial/ , using the sample programs (main/image.h/image.cpp) Slightly modified main.cpp to accomodate larger images. Had to expand a PGM image (to about 83~MB size) to return any meaningful result (Using a regular sized PGM image of 11KB yielded absolutely no meaningful results to the human eye - all 0's on the flat profile/call graph)

The results of the flat profile:

Flat profile:

Each sample counts as 0.01 seconds.

 %   cumulative   self              self     total
time   seconds   seconds    calls  ms/call  ms/call  name
55.49      1.01     1.01        3   336.67   336.67  Image::Image(Image const&)
28.57      1.53     0.52        6    86.67    86.67  Image::Image(int, int, int)                                                                                      
 9.34      1.70     0.17        1   170.00   290.00  readImage(char*, Image&)
 6.59      1.82     0.12 16000000     0.00     0.00  Image::setPixelVal(int, int, int)                                                                                      
 0.00      1.82     0.00       30     0.00     0.00  Image::getPixelVal(int, int)                                                                                      
 0.00      1.82     0.00        9     0.00     0.00  Image::~Image()
 0.00      1.82     0.00        8     0.00     0.00  std::operator|(std::_Ios_Openmode, std::_Ios_Openmode)                                                                                        
 0.00      1.82     0.00        6     0.00     0.00  writeImage(char*, Image&)
 0.00      1.82     0.00        6     0.00     0.00  Image::getImageInfo(int&,int&, int&)                                                                                         
 0.00      1.82     0.00        6     0.00     0.00  Image::operator=(Image con                                                                                        
 0.00      1.82     0.00        2     0.00     0.00  std::cos(float)
 0.00      1.82     0.00        2     0.00     0.00  std::sin(float)
 0.00      1.82     0.00        1     0.00     0.00  _GLOBAL__sub_I__ZN5ImageC2Ev                                                                                        
 0.00      1.82     0.00        1     0.00     0.00  readImageHeader(char*, int&, int&, int&, bool&)                                                                                        
 0.00      1.82     0.00        1     0.00     0.00  __static_initialization_and_destruction_0(int, int)                                                                                        
 0.00      1.82     0.00        1     0.00    86.67  Image::getSubImage(int, int, int, int, Image&)                                                                                        
 0.00      1.82     0.00        1     0.00    86.67  Image::negateImage(Image&)
 0.00      1.82     0.00        1     0.00    86.67  Image::rotateImage(int, Image&)                                                                                        
 0.00      1.82     0.00        1     0.00    86.67  Image::shrinkImage(int, Image&)                                                                                        
 0.00      1.82     0.00        1     0.00    86.67  Image::enlargeImage(int, Image&)                                                                                        
 0.00      1.82     0.00        1     0.00   336.67  Image::reflectImage(bool,Image&)                                                                                         
 0.00      1.82     0.00        1     0.00     0.00  Image::inBounds(int, int)


Out of all the functions tested, reflectImage has the largest ms/call. Below is the code for reflectImage:

void Image::reflectImage(bool flag, Image& oldImage) /*Reflects the Image based on users input*/ {

   int rows = oldImage.N;
   int cols = oldImage.M;
   Image tempImage(oldImage);
   if(flag == true) //horizontal reflection
   {
       for(int i = 0; i < rows; i++)
       {
           for(int j = 0; j < cols; j++)
               tempImage.pixelVal[rows - (i + 1)][j] = oldImage.pixelVal[i][j];
       }
   }
   else //vertical reflection
   {
       for(int i = 0; i < rows; i++)
       {
           for(int j = 0; j < cols; j++)
               tempImage.pixelVal[i][cols - (j + 1)] = oldImage.pixelVal[i][j];
       }
   } 
   
   oldImage = tempImage;

}

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