Tolga Birdal

Hey Joao ,

Well, you are right at some point but in Sobel-like edges you are considering only 3x3 neighborhoods, while more advanced derivatives such as Simoncelli or 5 point stencil use a larger neighborhood and they are proved to be more robust against noise.

Actually my questions was whether you evaluated any other derivatives or not. Thanks for the response. I would appreciate if you could try others and give a feedback. I happened to realize that the algorithm only works with Sobel-like derivatives.

/tolga

Thank you for this code.

Please note that "close all" is not expected by a user from within a function. Plotting in a specific figure/axes object can be accomplished instead by explicitly opening one and passing the handle to "plot":
fig = figure;
ax = axes('Parent', fig);
plot(ax, ...)

Also "hold on" is interfering with the current settings by the user, if this function is called to add a curve to an existing figure. It would be better to first get the current hold state (if an axes object handle is passed as argument), then change it to "on" in order to add the curve and finally restore it to its original state:
http://www.mathworks.com/matlabcentral/fileexchange/36641-take-restore-hold

The function would benefit if it had arguments, so that the user can plot whichever Cassini curve they prefer.
If arguments are included, then the help text can be more explicit about how the function is used.
Referring to a link for the help text is not the usual practice.

Note also that the provided link is broken.

As a demonstration alternative, the function could plot several curves as it does now, if no arguments are provided, otherwise plot the curve prescribed by the arguments.

I have found many cases where this function fails. There is a bug, somewhere in the code.
Could the author fix it? Otherwise unreliable.

Here is a counterexample:

X = 49936;
Y =-78883;
XV = [49370 54355 54355 49370];
YV = [-8406 -8406 -77126 -77126];
inpolygon(X,Y,XV',YV')

ans =

0

>> inpolygon_fast(X,Y,XV',YV')

ans =

1

The correct answer is that of inpolygon,
since the point is located outside the rectangle.
Use the following code to check it out yourself:

plot(XV,YV,'-b');
hold
plot(X,Y,'.r')