Click here to Skip to main content
15,906,645 members

Comments by RanCohen (Top 6 by date)

RanCohen 16-Dec-10 14:18pm View    
That makes sense, thanks!
RanCohen 16-Dec-10 14:18pm View    
None of the above - I got a ClassCastException in my program and couldn't figure out why the compiler didn't block me in the first place...
And why is CodeProject the wrong place to ask this?
RanCohen 3-Jun-10 9:08am View    
Look at this for simplification as coordiantes in a 2D space. If I want to search for the closest match for (x1, y1), and assume that distance(x1, y1, 0, 0) is r1, I can look for the color with a distance to (0, 0) which is closest to r1. If the two colors distance is r2 between them, it means that the best possible match is between r1+r2, r1-r2 (its like a circle which has (x1, x2) in its circle and a radius of r2, and i am searching for the best match in that circle).
This is the same for a 3D space as well.
Is there an error in my logic?
RanCohen 2-Jun-10 8:08am View    
I think I thought of a somewhat simpler method - I look at the pixel as a coordinate in a space, and thus matching pixels are the ones that have the minimal distance between them. So instead of running on three lists, I can sort the array by the pixel distance to (0, 0, 0), and then do what you suggested - running up and down on the closest pixel, until the difference in the distances is greater then the minimal distance so far.
This is basicly the same algorithem, only now i run on a single list instead of three different lists that have to be interconnected.
RanCohen 29-May-10 7:37am View    
Let me see if I got it right - You are suggesting 3 lists: One sorted by R, G, B, One sortet by say G, B, R and a third sorted by B, R, G. I iterate the first until the difference in the R factor is greater then the total difference of the minimal RGB, Then I iterate the second list, then the third, with each time its faster then the last because the minimum is less in each iteration? It seems a good solution and its certainly less then iterating through the whole list, but is it realy much less then O(n)? Won't it be possible that in some cases i might run on most of the Red values before finding a match?