Marc Marc - 1 year ago 87
C++ Question

Calculate the gradient for an histogram in c++

I calculated the histogram(a simple 1d array) for an 3D grayscale Image.
Now I would like to calculate the gradient for the this histogram at each point. So this would actually mean I have to calculate the gradient for a 1D function at certain points. However I do not have a function. So how can I calculate it with concrete x and y values?

For the sake of simplicity could you probably explain this to me on an example histogram - for example with the following values (x is the intensity, and y the frequency of this intensity):

x1 = 1; y1 = 3

x2 = 2; y2 = 6

x3 = 3; y3 = 8

x4 = 4; y4 = 5

x5 = 5; y5 = 9

x6 = 6; y6 = 12

x7 = 7; y7 = 5

x8 = 8; y8 = 3

x9 = 9; y9 = 5

x10 = 10; y10 = 2

I know that this is also a math problem, but since I need to solve it in c++ I though you could help me here.

Thank you for your advice

Answer Source

I think you can calculate your gradient using the same approach used in image border detection (which is a gradient calculus). If your histogram is in a vector you can calculate an approximation of the gradient as*:

for each point in the histogram compute 
     gradient[x] = (hist[x+1] - hist[x])

This is a very simple way to do it, but I'm not sure if is the most accurate.

  • approximation because you are working with discrete data instead of continuous


Other operators will may emphasize small differences (small gradients will became more emphasized). Roberts algorithm derives from the derivative calculus:

lim delta -> 0 = f(x + delta) - f(x) / delta

delta tends infinitely to 0 (in order to avoid 0 division) but is never zero. As in computer's memory this is impossible, the smallest we can get of delta is 1 (because 1 is the smallest distance from to points in an image (or histogram)).


lim delta -> 0 to lim delta -> 1

we get

f(x + 1) - f(x) / 1 = f(x + 1) - f(x) => vet[x+1] - vet[x]
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download