Partition points in a plane with a straigth line
There is a brute $O(n^3)$ algorithm.

There are $\binom n 2 = \frac {n(n-1)} 2$ lines over pairs of points. If a line goes over $k$ points, there are $2(k+1)$ ways to rotate the line by a negligible angle, so that no points go over it and each rotation partitions the $k$ points differently.

For each pair of vertices (v1, v2),
  rotate the space so that v1.y = v2.y = 0
  let UP be the sum of values of vertices with positive y
  let DOWN be the same for negative y
  let L be a list of vertices with y = 0, sorted by x

  For each of L.length+1 ways to cut L in two,
    let LEFT be the sum of values of vertices in L to the left of the cut
    let RIGHT be the same for vertices to the right

    check whether one of UP + LEFT, DOWN + RIGHT or UP + RIGHT, DOWN + LEFT
     is a better partition than your previous best.