Three-Star
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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 1~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)
Rotated and negated the image.
>g++ -g -O2 -pg -omain main.cpp >main baboonsizetwo >gprof -p -b main>main.flt
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 30.00 0.03 0.03 writeImage(char*, Image&) 30.00 0.06 0.03 Image::rotateImage(int, Image&) 10.00 0.07 0.01 2 5.00 5.00 Image::Image(int, int, int) 10.00 0.08 0.01 2 5.00 5.00 Image::operator=(Image const&) 10.00 0.09 0.01 Image::negateImage(Image&) 10.00 0.10 0.01 Image::Image(Image const&) 0.00 0.10 0.00 2 0.00 0.00 Image::~Image() 0.00 0.10 0.00 1 0.00 0.00 _GLOBAL__sub_I__ZN5ImageC2Ev
Out of all the functions tested, rotateImage takes up the largest amount of time. Below is the code for rotateImage:
void Image::rotateImage(int theta, Image& oldImage) {
int r0, c0;
int r1, c1;
int rows, cols;
rows = oldImage.N;
cols = oldImage.M;
Image tempImage(rows, cols, oldImage.Q);
float rads = (theta * 3.14159265)/180.0;
r0 = rows / 2;
c0 = cols / 2;
for(int r = 0; r < rows; r++)
{
for(int c = 0; c < cols; c++)
{
r1 = (int) (r0 + ((r - r0) * cos(rads)) - ((c - c0) * sin(rads)));
c1 = (int) (c0 + ((r - r0) * sin(rads)) + ((c - c0) * cos(rads)));
if(inBounds(r1,c1))
{
tempImage.pixelVal[r1][c1] = oldImage.pixelVal[r][c];
}
}
}
for(int i = 0; i < rows; i++)
{
for(int j = 0; j < cols; j++)
{
if(tempImage.pixelVal[i][j] == 0)
tempImage.pixelVal[i][j] = tempImage.pixelVal[i][j+1];
}
}
oldImage = tempImage;}
Main section to be parallelized.
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!