title: Recursive Relation Help for dynamic programming 2d Plane algorithm
body: <p>So I've been working on an algorithm. The task I'm trying to accomplish is: consider a 2D plane there are targets that are randomly distributed between a y upper bound and lower bound. This set is T. T1 is marked with coordinates(X,Y). There is a set S of sensors that are guaranteed to cover every target each sensor has a radius 1 and an (X,Y) coordinate. Each target has a cost (c) that is a weight. So my task is to find a minimum weight or cost for a set S' that covers each sensor.</p>
  
  <p>So I know that there is a recursive relation between the disks that "dominate" or "control" other disks but I'm having trouble seeing a property that shows how to utilize the dominance of the disks. I got this far:   </p>
  
  <p>Let D+ be the set of disks whose centre lies above the strip (upper disks).</p>
  
  <p>Let D- be the set of disks whose centre lies below the strip (lower disks).</p>
  
  <p>Consider an upper disk d, and d intersects a vertical line L. Another upper disk d' is said to be controlled or dominated by d, if one of the following holds:
  (1)d' does not intersect L
  (2)the lower intersection endpoint of d' and L is higher than the lower intersection endpoint of d and L
  (3)the lower intersection endpoint of d' and L is identical to the lower intersection endpoint of d and L, but the centre of d' is on the right of the centre of d.</p>
  
  <p>Similarly, for a lower disk d, and d intersects a vertical line L. Another lower disk d' is said to be controlled or dominated by d, if one of the following holds:
  (1)d' does not intersect L
  (2)the upper intersection endpoint of d' and L is lower than the upper intersection endpoint of d and L
  (3)the upper intersection endpoint of d' and L is identical to the upper intersection endpoint of d and L, but the centre of d' is on the right of the centre of d.
  Example:<img src="https://i.stack.imgur.com/yeQc1.png" alt="three sensors with radius 1"></p>
  
  <p>But I’m having trouble finalizing the algorithm. any help? Is that clear?</p>
  
  <p>Thanks,
  Christopher</p>