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(→‎Assignment 3)
Line 441: Line 441:
 
=== Assignment 3 ===
 
=== Assignment 3 ===
  
 +
__global__ void rotateKernel(int* oldImage, int* newImage, int rows, int cols, float rads) {
 +
 +
        //changed to coalesced access
 +
'''int c''' = blockIdx.x * blockDim.x + threadIdx.x;
 +
'''int r''' = blockIdx.y * blockDim.y + threadIdx.y;
 +
 +
int r0 = rows / 2;
 +
int c0 = cols / 2;
 +
float sinRads = sinf(rads);
 +
float cosRads = cosf(rads);
 +
 +
/*__shared__ int s[ntpb * ntpb];
 +
s[r * cols + c] = oldImage[r * cols + c];*/
 +
 +
if (r < rows && c < cols)
 +
{
 +
int r1 = (int)(r0 + ((r - r0) * cosRads) - ((c - c0) * sinRads));
 +
int c1 = (int)(c0 + ((r - r0) * sinRads) + ((c - c0) * cosRads));
 +
 +
if (inBounds(r1, c1, rows, cols))
 +
{
 +
newImage[r1 * cols + c1] = oldImage[r * cols + c];
 +
}
 +
}
 +
}
  
 
Using Coalesced Memory (changed matrix access from column to row)
 
Using Coalesced Memory (changed matrix access from column to row)

Revision as of 14:34, 8 April 2018


GPU610/DPS915 | Student List | Group and Project Index | Student Resources | Glossary

Three-Star

Team Members

  1. Derrick Leung
  2. Timothy Moy

Email All

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 program files (main/image.h/image.cpp)

pulled PGM sample files from here: https://userpages.umbc.edu/~rostamia/2003-09-math625/images.html

file sizes being 512x512, about 262 KB each file

Compiled to produce a flat profile and a call graph

>g++ -g -O2 -pg -o main main.cpp

>main a.pgm result.pgm

Note: Enlarged image by max permitted by program (5) to get more viewable results, since the profile without enlarging it produces non-significant results

Imageprofilesteps.png

The results of the flat profile:

Profileresults.png

The results of the call graph

Callgraphpt1.png

Callgraphpt2.png

Rotate image function is one of the longer running functions and looks like it has potential for parallelization.

Rotateimage.png

LZW Data Compression Algorithm

Timothy Moy profiled.

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

Raw Flat profile (50Mb Test file for compression):

Each sample counts as 0.01 seconds. 

 %   cumulative   self              self     total           
time   seconds   seconds    calls  us/call  us/call  name    
35.52      4.23     4.23                             compress(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >)
27.54      7.51     3.28 102062309     0.03     0.03  std::_Hashtable<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int>, std::allocator<std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int> >, std::__detail::_Select1st, std::equal_to<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::hash<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Prime_rehash_policy, std::__detail::_Hashtable_traits<true, false, true> >::_M_find_before_node(unsigned int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&, unsigned int) const
20.15      9.91     2.40 204116423     0.01     0.01  std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_replace_aux(unsigned int, unsigned int, unsigned int, char)
 8.23     10.89     0.98 49629412     0.02     0.05  std::__detail::_Map_base<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int>, std::allocator<std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int> >, std::__detail::_Select1st, std::equal_to<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::hash<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Prime_rehash_policy, std::__detail::_Hashtable_traits<true, false, true>, true>::operator[](std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&)
 4.28     11.40     0.51 52428800     0.01     0.01  std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_assign(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&)
 3.02     11.76     0.36 52436762     0.01     0.01  show_usage()
 1.26     11.91     0.15                             _Z22convert_char_to_stringB5cxx11PKci
 0.00     11.91     0.00     4097     0.00     0.00  std::_Hashtable<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int>, std::allocator<std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int> >, std::__detail::_Select1st, std::equal_to<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::hash<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Prime_rehash_policy, std::__detail::_Hashtable_traits<true, false, true> >::_M_insert_unique_node(unsigned int, unsigned int, std::__detail::_Hash_node<std::pair<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const, int>, true>*)
 0.00     11.91     0.00       22     0.00     0.01  std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_mutate(unsigned int, unsigned int, char const*, unsigned int)
 0.00     11.91     0.00        1     0.00     0.00  _GLOBAL__sub_I__Z18convert_int_to_binB5cxx11i
 0.00     11.91     0.00        1     0.00    28.13  std::_Hashtable<std::__cxx11::basic_

Summarized Flat Profile (50Mb Test file for compression): 

 %   cumulative   self              self     total           
time   seconds   seconds    calls  us/call  us/call  name    
35.52      4.23     4.23                             compress()
27.54      7.51     3.28 102062309    0.03     0.03  std::_Hashtable
20.15      9.91     2.40 204116423    0.01     0.01  std::__cxx11::basic_string
 8.23     10.89     0.98 49629412     0.02     0.05  std::__detail::_Map_base
 4.28     11.40     0.51 52428800     0.01     0.01  std::__cxx11::basic_string
 3.02     11.76     0.36 52436762     0.01     0.01  show_usage()
 1.26     11.91     0.15                             _Z22convert_char_to_stringB5cxx11PKci
 0.00     11.91     0.00     4097     0.00     0.00  std::_Hashtable
 0.00     11.91     0.00       22     0.00     0.01  std::__cxx11::basic_string
 0.00     11.91     0.00        1     0.00     0.00  _GLOBAL__sub_I__Z18convert_int_to_binB5cxx11i
 0.00     11.91     0.00        1     0.00    28.13  std::_Hashtable<std::__cxx11::basic_

Note how the compress() function takes up the largest amount of time (over one third), then the other functions which take up over 10% of the time are library functions. It is highly unlikely we could parallelize the library functions. The other functions that take up under 10% of the time will probably not give enough improvement in time to make a significant impact.

Thus, the function we should focus on is the compress function.

Summary of Compress() 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 source code:

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();

}

There are two loops which show possibility of parallelization: 

   for ( int unsigned i = 0 ; i < 256 ; i++ ){
       compress_dictionary[string(1,i)] = i;
   }

and 

   for(char& c: input){
       current_string = current_string + c; // Possible area for improvement via reduction
       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;
       }   
   }

The first for loop is constant and probably won't show much improvement if we parallelize it.


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];

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! The one thing we need to worry about is that order does seem to matter for the second for loop.

Conclusion

We decided to go with image profiling. It is a pretty simple parallelization since the transformation functions are matrix transformations which don't care about which element is processed first.

There are some possible issues with working with the simple-lzw-compression-algorithm and CUDA. You cannot use the C++ string type in a kernel because CUDA does not include a device version of the C++ String library that would be able run on the GPU. Even if it was possible to use string in a kernel, it's not something you would want to do because string handles memory dynamically, which would be likely to be slow.

https://stackoverflow.com/questions/26993351/is-there-a-penalty-to-using-char-variables-in-cuda-kernels?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa

Essentially, using chars on the gpu would require we use 8-bit arithmetic and need to convert from 32-bit arithmetic for operations. On top of that, the compress function refers to a map of strings/int pairs to shrink the size of the file. Even if we did manage the character operations, it would need to somehow use the string map to get the corresponding integer which could lead to being unable to use device memory for optimization.

Assignment 2

Original CPU Implementation: 

 void Image::rotateImage(int theta, Image& oldImage)
 /*based on users input and rotates it around the center of the image.*/
 {
   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;
 }

Parallelized Code (done by Timothy Moy, Derrick acted as consulting for how to use the program):

Kernels 

 __device__ bool inBounds(int row, int col, int maxRow, int maxCol) {
   if (row >= maxRow || row < 0 || col >= maxCol || col < 0)
     return false;
     //else
     return true;
 }
   
 __global__ void rotateKernel(int* oldImage, int* newImage, int rows, int cols, float rads) {
   int r = blockIdx.x * blockDim.x + threadIdx.x;
   int c = blockIdx.y * blockDim.y + threadIdx.y;
   
   int r0 = rows / 2;
   int c0 = cols / 2;
   float sinRads = sinf(rads);
   float cosRads = cosf(rads);
   
   /*__shared__ int s[ntpb * ntpb];
   s[r * cols + c] = oldImage[r * cols + c];*/
   
   if (r < rows && c < cols)
   {
     int r1 = (int)(r0 + ((r - r0) * cosRads) - ((c - c0) * sinRads));
     int c1 = (int)(c0 + ((r - r0) * sinRads) + ((c - c0) * cosRads));
   
     if (inBounds(r1, c1, rows, cols))
     {
       newImage[r1 * cols + c1] = oldImage[r * cols + c];
     }
   }
 }

Modified Function 

 void Image::rotateImage(int theta, Image& oldImage)
 /*based on users input and rotates it around the center of the image.*/
 {
   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;
   
   // workspace start
   // - calculate number of blocks for n rows assume square image
   int nb = (rows + ntpb - 1) / ntpb;
   
   // allocate memory for matrices d_a, d_b on the device
 
   // - add your allocation code here
   int* d_a;
   check("device a", cudaMalloc((void**)&d_a, rows* cols * sizeof(int)));
   int* d_b;
   check("device b", cudaMalloc((void**)&d_b, rows* cols * sizeof(int)));
   
   // copy h_a and h_b to d_a and d_b (host to device)
   // - add your copy code here
   check("copy to d_a", cudaMemcpy(d_a, oldImage.pixelVal, rows * cols * sizeof(int), cudaMemcpyHostToDevice));
   //check("copy to d_b", cudaMemcpy(d_b, tempImage.pixelVal, rows * cols * sizeof(int), cudaMemcpyHostToDevice));
   
   // launch execution configuration
   // - define your 2D grid of blocks
   dim3 dGrid(nb, nb);
   // - define your 2D block of threads
   dim3 dBlock(ntpb, ntpb);
   // - launch your execution configuration	
   rotateKernel<<<dGrid, dBlock >>>(d_a, d_b, rows, cols, rads);
   check("launch error: ", cudaPeekAtLastError());
   // - check for launch termination
   // synchronize the device and the host
   check("Synchronize ", cudaDeviceSynchronize());
   
   // copy d_b to tempImage (device to host)
   // - add your copy code here
   check("device copy to hc", cudaMemcpy(tempImage.pixelVal, d_b, rows * cols * sizeof(int), cudaMemcpyDeviceToHost));
   
   // deallocate device memory
   // - add your deallocation code here
   cudaFree(d_a);
   cudaFree(d_b);
   
   // reset the device
   cudaDeviceReset();
   // workspace end
   
   for(int i = 0; i < rows; i++)
   {
       for(int j = 0; j < cols; j++)
       {
           if(tempImage.pixelVal[i * M + j] == 0)
               tempImage.pixelVal[i * M + j] = tempImage.pixelVal[i * M + j+1];
       }
   }
   oldImage = tempImage;
 }

Profiling (Done by Derrick Leung)

Function CPU-Only GPU-CPU speedup(%)
Cuda Memory Allocation s 1164 ms %
Copy Image to Device memory s 6 ms %
Kernel s 0 ms %
Copy device image to host temp variable s 6 ms %
copy temp image to original image variable s 43 ms %
Total Rotation Time (no allocation, with memcpy) 1717ms 55ms %
Total Run Time 1775 ms 1294 ms %

Comparisons

Rotation Run Time (exclude memory allocation) Total Run Time
Size of Picture CPU-Only GPU-CPU speedup(%) CPU-Only GPU-CPU speedup ratio
512x512 67 ms 2ms 33.50 71 ms 372 ms .19
2x enlarged 265 ms 7 ms 37.85 277 ms 410 ms .67
3x enlarged 608 ms 23 ms 26.43 630 ms 427 ms 1.47
4x enlarged 1091 ms 37 ms 29.48 1129 ms 446 ms 2.53
5x enlarged 1717 ms 55 ms 31.22 1775 ms 476 ms 3.73

Assign2-rotate-comparison.png Assign2-total-comparison.png

Excel Sheet File:Assignment2 profile.xlsx.txt

Source Code: File:Image.cu.txt File:Image.h.txt File:Main.cpp.txt

Assignment 3

__global__ void rotateKernel(int* oldImage, int* newImage, int rows, int cols, float rads) { 

       //changed to coalesced access

int c = blockIdx.x * blockDim.x + threadIdx.x; int r = blockIdx.y * blockDim.y + threadIdx.y;

int r0 = rows / 2; int c0 = cols / 2; float sinRads = sinf(rads); float cosRads = cosf(rads);

/*__shared__ int s[ntpb * ntpb]; s[r * cols + c] = oldImage[r * cols + c];*/

if (r < rows && c < cols) { int r1 = (int)(r0 + ((r - r0) * cosRads) - ((c - c0) * sinRads)); int c1 = (int)(c0 + ((r - r0) * sinRads) + ((c - c0) * cosRads));

if (inBounds(r1, c1, rows, cols)) { newImage[r1 * cols + c1] = oldImage[r * cols + c]; } } }

Using Coalesced Memory (changed matrix access from column to row)

Uncoalesced Coalesced
Size of Picture memcpy rotate kernel total runtime memcpy rotate kernel total runtime
512x512 0.54ms 0.90ms 85.72ms 0.51ms 0.89ms 95.59ms
2x enlarged 1.80ms 3.55ms 99.66ms 1.76ms 3.54ms 103.11ms
3x enlarged 4.65ms 7.97ms 111.79ms 4.69ms 7.95ms 114.52ms
4x enlarged 8.22ms 14.15ms 134.32ms 7.90ms 14.13ms 114.33ms
5x enlarged 12.89ms 22.15ms 128.59ms 12.70ms 22.09ms 144.42ms

Changing the way memory is accessed doesn't seem to have any significant improvements/changes to time

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