Fast marching method applyed in segmentation : how to deal with a nabla operator ?

Hello everyone ! Happy new year ! I'm working on the fast marching that i have implemented in order to deal with the search of minimal paths . Now i want to apply my algorithm on image segmentation. My teacher adviced me to use a special function for the speed cij = 1/(1+\nabla I(x,y)^2) . My question is that i don't really understand this formula, i don't know how to deal with the nabla operator. He told me that the speed would be equal to 0 on the edge of the surface i want to segment. Can someone help me, explain me how to do this ? I hope that i was clear enough. Thanks for your help !

Answer 1

[Star Strider]

‘Nabla’ is the inverted-Delta symbol. I don’t understand what you’re doing and it how it applies to your problem, so I refer you to the Wikipedia article on the mathematical interpretation of the Nabla symbol for guidance.

See the documentation for the gradient and del2 functions.