Does the shortest path from an initial point to a goal point, in an environment with obstacles, also gives the minimum sum of turning angles?
It doesn't seem to be obvious, but I couldn't come up with a counter example! If it is indeed true, is there a proof which states that?
Note: the shortest path means minimum sum of Euclidean distance. Additional context: I was using A* on a visibility graph to find the optimal path. Two different cost functions were used. 1. Just the Euclidean distance and 2. Euclidean distance + theta, turning angle, multiplied by some constant.
The following image, constructed by the original poster, demonstrates that the claim is false. The shortest path (green) has a larger total turning angle than the longer path (red).