Is there a chance to make the bilinear interpolation faster? - c++

Question

First I want to provide you with some context.
I have two kind of images I need to merge. The first image is the background image with the format 8BppGrey and a resolution of 320x240. The second image is the forground image with the format 32BppRGBA and a resolution of 64x48.
Update
The github repo with an MVP is at the bottom of the question.
To do it I resize the second image with bilinear interpolation to the same size as the first one and then use blending to merge both to one image. Blending only happens when the alpha value of the second image is greater then 0.
I need to do it as fast as possible so my idea was to combine the resize and merge / blend process.
To achieve this I used the resize function from the writeablebitmapex repository and added merging / blending.
Everything works as expected but I want to decrease the execution time.
This are the current debug timings:
// CPU: Intel(R) Core(TM) i7-4810MQ CPU # 2.80GHz
MediaServer: Execution time in c++ 5 ms
MediaServer: Resizing took 4 ms.
MediaServer: Execution time in c++ 5 ms
MediaServer: Resizing took 5 ms.
MediaServer: Execution time in c++ 4 ms
MediaServer: Resizing took 4 ms.
MediaServer: Execution time in c++ 3 ms
MediaServer: Resizing took 3 ms.
MediaServer: Execution time in c++ 4 ms
MediaServer: Resizing took 4 ms.
MediaServer: Execution time in c++ 5 ms
MediaServer: Resizing took 4 ms.
MediaServer: Execution time in c++ 6 ms
MediaServer: Resizing took 6 ms.
MediaServer: Execution time in c++ 3 ms
MediaServer: Resizing took 3 ms.
Do I have any chance to increase the performance and lower the execution time of the resize / merge / blend process?
Are there some parts I maybe can parallelize?
Do I maybe have a chance to use some processor features?
A huge performance hit is the nested loop but I have no idea how I could write it better.
I would like to reach 1 or 2 ms for the whole process. Is this even possible?
Here's the modified visual c++ function I use.
pd is the backbuffer of the writeable bitmap I use to display the
result in wpf. The format I use is the default 32BppRGBA.
pixels is the int[] array of the 64x48 32BppRGBA image
widthSource and heightSource is the size of the pixels image
width and height is the target size of the output image
baseImage is the int[] array of the 320x240 8BppGray image
VC++ code:
unsigned int Resize(int* pd, int* pixels, int widthSource, int heightSource, int width, int height, byte* baseImage)
{
unsigned int start = clock();
float xs = (float)widthSource / width;
float ys = (float)heightSource / height;
float fracx, fracy, ifracx, ifracy, sx, sy, l0, l1, rf, gf, bf;
int c, x0, x1, y0, y1;
byte c1a, c1r, c1g, c1b, c2a, c2r, c2g, c2b, c3a, c3r, c3g, c3b, c4a, c4r, c4g, c4b;
byte a, r, g, b;
// Bilinear
int srcIdx = 0;
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
sx = x * xs;
sy = y * ys;
x0 = (int)sx;
y0 = (int)sy;
// Calculate coordinates of the 4 interpolation points
fracx = sx - x0;
fracy = sy - y0;
ifracx = 1.0f - fracx;
ifracy = 1.0f - fracy;
x1 = x0 + 1;
if (x1 >= widthSource)
{
x1 = x0;
}
y1 = y0 + 1;
if (y1 >= heightSource)
{
y1 = y0;
}
// Read source color
c = pixels[y0 * widthSource + x0];
c1a = (byte)(c >> 24);
c1r = (byte)(c >> 16);
c1g = (byte)(c >> 8);
c1b = (byte)(c);
c = pixels[y0 * widthSource + x1];
c2a = (byte)(c >> 24);
c2r = (byte)(c >> 16);
c2g = (byte)(c >> 8);
c2b = (byte)(c);
c = pixels[y1 * widthSource + x0];
c3a = (byte)(c >> 24);
c3r = (byte)(c >> 16);
c3g = (byte)(c >> 8);
c3b = (byte)(c);
c = pixels[y1 * widthSource + x1];
c4a = (byte)(c >> 24);
c4r = (byte)(c >> 16);
c4g = (byte)(c >> 8);
c4b = (byte)(c);
// Calculate colors
// Alpha
l0 = ifracx * c1a + fracx * c2a;
l1 = ifracx * c3a + fracx * c4a;
a = (byte)(ifracy * l0 + fracy * l1);
// Write destination
if (a > 0)
{
// Red
l0 = ifracx * c1r + fracx * c2r;
l1 = ifracx * c3r + fracx * c4r;
rf = ifracy * l0 + fracy * l1;
// Green
l0 = ifracx * c1g + fracx * c2g;
l1 = ifracx * c3g + fracx * c4g;
gf = ifracy * l0 + fracy * l1;
// Blue
l0 = ifracx * c1b + fracx * c2b;
l1 = ifracx * c3b + fracx * c4b;
bf = ifracy * l0 + fracy * l1;
// Cast to byte
float alpha = a / 255.0f;
r = (byte)((rf * alpha) + (baseImage[srcIdx] * (1.0f - alpha)));
g = (byte)((gf * alpha) + (baseImage[srcIdx] * (1.0f - alpha)));
b = (byte)((bf * alpha) + (baseImage[srcIdx] * (1.0f - alpha)));
pd[srcIdx++] = (255 << 24) | (r << 16) | (g << 8) | b;
}
else
{
// Alpha, Red, Green, Blue
pd[srcIdx++] = (255 << 24) | (baseImage[srcIdx] << 16) | (baseImage[srcIdx] << 8) | baseImage[srcIdx];
}
}
}
unsigned int end = clock() - start;
return end;
}
Github repo

Answer 1

One action that may speed up your code is to avoid type conversions from integer to float and vice versa. This can be achieved by having an int value in the suitable range instead of floats on range 0..1
Something like this:
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
int sx1 = x * widthSource ;
int x0 = sx1 / width;
int fracx = (sx1 % width) ; // range 0..width - 1
which turns into something like
l0 = (fracx * c2a + (width - fracx) * c1a) / width ;
And so on. A bit tricky but doable

Answer 2

Thank you for all the help but the problem was the managed c++ project. I transfered the function now to my native c++ library and used the managed c++ part only as a wrapper for the c# application.
After the compiler optimization the function is now finished in 1ms.
Edit:
I will mark my own answer for now as the solution because the optimization from #marom leads to a broken image.

Answer 3

The common way to speedup a resize operation with bilinear interpolation is to:
Exploit the fact that x0 and fracx are independent from the row and that y0and fracy are independent from the column. Even though you haven't pulled out the computation of y0 and fracy out of the x-loop, compiler optimization should take care of that. However, for x0 and fracx, one needs to pre-compute the values for all columns and store them in an array. Complexity for computing x0 and fracx becomes O(width) compared to O(width*height) without pre-computation.
Do the whole processing with integers by replacing floating point arithmetics by integer arithmetics, thereby using shift operations instead of integer divisions.
For better readability, I did not implement the pre-computation of x0 and fracx in the following code. Pre-computation is straight-forward anyways.
Note that FACTOR = 2048 is the max you can do with 32-bit signed integers here (2048 * 2048 * 255 is just fine). For higher precision, you should switch to int64_t and then increase FACTOR and SHIFT, respectively.
I placed the border check into the inner loop for better readability. For an optimized implementation one should remove it by iterating in both loops just before this case happens and add special handling for the border pixels.
In case someone is wondering what the + (FACTOR * FACTOR / 2) is for, it is for rounding in conjunction with the subsequent division.
Finally note that (FACTOR * FACTOR / 2) and 2 * SHIFT are evaluated at compile time.
#define FACTOR 2048
#define SHIFT 11
const int xs = (int) ((double) FACTOR * widthSource / width + 0.5);
const int ys = (int) ((double) FACTOR * heightSource / height + 0.5);
for (int y = 0; y < height; y++)
{
const int sy = y * ys;
const int y0 = sy >> SHIFT;
const int fracy = sy - (y0 << SHIFT);
for (int x = 0; x < width; x++)
{
const int sx = x * xs;
const int x0 = sx >> SHIFT;
const int fracx = sx - (x0 << SHIFT);
if (x0 >= widthSource - 1 || y0 >= heightSource - 1)
{
// insert special handling here
continue;
}
const int offset = y0 * widthSource + x0;
target[y * width + x] = (unsigned char)
((source[offset] * (FACTOR - fracx) * (FACTOR - fracy) +
source[offset + 1] * fracx * (FACTOR - fracy) +
source[offset + widthSource] * (FACTOR - fracx) * fracy +
source[offset + widthSource + 1] * fracx * fracy +
(FACTOR * FACTOR / 2)) >> (2 * SHIFT));
}
}
For clarification, to match the variables used by the OP, for instance, in the case of the alpha channel it is:
a = (unsigned char)
((c1a * (FACTOR - fracx) * (FACTOR - fracy) +
c2a * fracx * (FACTOR - fracy) +
c3a * (FACTOR - fracx) * fracy +
c4a * fracx * fracy +
(FACTOR * FACTOR / 2)) >> (2 * SHIFT));