3 ms·

Can SVM fit a circle with variable origin and variable radius to a set?

• ajj 17y ago
  It can in principle, and this is exactly where the kernel trick is useful. The standard SVM formulation can only give linear classifiers. But, if you project your data into feature space (a higher, possibly infinite dimensional space), a linear separator in that space can be a circle in your original space. Since you can not do explicit computations in an infinite dimensional space, the kernel trick lets you get away without doing them at all. You can thus get an inner product value of two infinite dimensional vectors using a kernel function. So classifiers that only require inner product values and never the explicit vectors can exploit the kernel trick. i.e., SVM, logistic regression, etc. That being said, choosing the appropriate kernel function is not always straightforward for your data.
  • olefoo 17y ago
    Can you recommend a good book that would get mathematically literate laymen a functional understanding of how to pick classifiers for data sets?
  • jules 17y ago
    Can you give a kernel function that would let you fit a circle with variable position and radius? The examples always use a fixed position.
    • ajj 17y ago
      The Radial Basis Function kernel K(x,y) = exp(-(x-y)^2/gamma) is a very general kernel function that can get you this. Finally, whether you actually get such a classifier depends on your data, and as I had mentioned setting the parameters (like gamma) is not very straightforward. Typically you may have to try with various values and see what works for your data. Train your SVM (or any kernel method) and see what classifier it gives (run the classifier on data points near the boundary you want to detect to see where the exact boundary lies). This would only be to see how the classification function looks like.