Reputation: 8135
I read about it on Wikipedia, theory sounds good, but I don't know to apply to practice.
I have an small example like this one:
Original Image Matrix
1 2
3 4
If I want to double size the image, then the new matrix is
x x x x
x x x x
x x x x
x x x x
Now, the fun part is how to transfer old values in original matrix to the new matrix, I intend to do like this
1 x 2 x
x x x x
3 x 4 x
x x x x
Then applying the Bi cubic Interpolation on it (at this moment just forget about using 16 neighbor pixel, I don't have enough space to demonstrate such a large matrix here).
Now my questions are:
1. Do I do the data transferring (from old to new matrix) right? If not, what should it look like?
2. What should be the value of x variables in the new matrix? to me , this seems correct because at least we have some values to do the calculation instead of x notations.
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
3. Will all of the pixels in new matrix be interpolated? Because the pixels at the boundary do not have enough neighbor pixels to perform the calculation.
Thank you very much.
Upvotes: 0
Views: 1109
Reputation: 308500
Interpolation means estimating a value for points that don't physically exist. You need to start with a coordinate system, so let's just use two incrementing integers for X position and Y position.
0, 0 1, 0
0, 1 1, 1
Your output requires 4x4 pixels which should be spaced at 0.5 intervals instead of the 1.0 intervals of the input:
-0.25,-0.25 0.25,-0.25 0.75,-0.25 1.25,-0.25
-0.25, 0.25 0.25, 0.25 0.75, 0.25 1.25, 0.25
-0.25, 0.75 0.25, 0.75 0.75, 0.75 1.25, 0.75
-0.25, 1.25 0.25, 1.25 0.75, 1.25 1.25, 1.25
None of the coordinates in the output exist in the input, so they'll all need to be interpolated.
The offset of the first coordinate of -0.25 is chosen so that the first and last coordinate will be equal distances from the edges of the input, and is calculated by the difference between the output and input intervals divided by 2. For example if you wanted a 10x zoom the interval is 0.1, the initial offset is (0.1-1)/2 and the points would be (-0.45, -0.35, -0.25, -0.15, ... 1.35, 1.45).
The Bicubic algorithm will require data for points outside of the original image. The easiest solution comes when you use a premultiplied alpha representation for your pixels, then you can just use (0,0,0,0) as the value for anything outside the image boundaries.
Upvotes: 1