Calculation of Lagrange multipliers for a simple vector support machine

Question

Firstly, I am new to vector machine support, so I'm sorry if I am going to solve this problem incorrectly. I am trying to implement a very simple SVM from scratch that uses the identifier core function to classify linearly shared data into one of two classes. As an example of the type of data that I will use, consider the chart below this document :

Using points (1,0), (3, 1) and (3, -1) as reference vectors, we know that the following is true to calculate the solution plane (Screenshot from the same document):

Which, when messing around and rebuilding a bit, gives us Lagrange multipliers of -3.5, 0.75 and 0.75, respectively.

I understand how this algebra works on paper, but I'm not sure about the best approach to its implementation. So my question is: how are SVM Lagrange multipliers calculated in practice ? Is there any algorithm that I am missing that can determine these values ​​for arbitrary linearly shared support vectors? Should I use a standard math library to solve linear equations (I implement SVM in java)? Would such a math library be slow for large-scale learning? Please note that this is a training exercise, so I'm not just looking for a ready-made SVM library.

Any other tips would be greatly appreciated!

EDIT 1 : LutzL made a good conclusion that half of the problem actually determines which points should be used as reference vectors, therefore, to simplify things for this question, have already been calculated.

Answer 1

Regardless of the kernel function, the determination of the coefficients leads to a quadratic optimization problem with linear positivity constraints. Which has terrible complexity if you implement naively testing all the boundary components, so you cannot avoid advanced optimization algorithms such as barrier methods or trust areas.

There are also heuristic approaches that try to keep the optimization problem in a small dimension by looking for points close to the dividing line and eliminate points that are probably far from it.