Let us have a graph. When we remove an edge, 2 ‘cars’ are created, one from each vertice of the edge. when these 2 cars meet they stop. The problem is to create a spanning tree so that the sum of the numbers of cars that pass through each vertice is minimal.
The added difficulty is that if a vertice has n cars passing from it, then the cost is K^n and not n*K.
some thoughts. We could find the shortest chordless cycles as a start but the position of those chordless cycles, ie whether they touch each other, changes the metric and thus what the shortest cycle is.
This is not a minimum spanning tree problem. I want to solve this because each car represents a varriable and the spanning tree is the most efficient way to compute an optimization problem. When 2 cars from the same edge meet and stop, I have a reduction of one varriable from the optimization.
edit:
The process is like this. We remove a number of edges to make the graph a spanning tree. Each removed edge creates 2 cars, one at each vertice of the removed edge, that need to meet each other. We fix a path for each of those twin cars. We then check how many cars (from all the edges that we removed) pass through each vertice. If the number of the cars that pass from a vertice is n, the cost is K^n where K is a constant. We then add all the costs and that is the global cost that needs to be minimized.
please tell me if something is unclear.
https://mathoverflow.net/questions/86301/spanning-tree-that-minimizes-a-dynamic-metric
Here’s one insight – cars will never pass through an articulation point, so you can break the graph up into its blocks and solve for each block separately (the minimum overall cost function is the sum of the minimum cost for each block).
To find the minimum cost for a block – you could enumerate all the spanning trees for that block and calculate the cost for each one – a brute force approach, but it should work.