Let G0 be the union of and its bitangents. Notice that a real-world robot capable of moving along points of G0 needs to be able to follow walls (boundary of R), and move down bitangent lines. Although the former is not that hard to implement, the latter poses a problem. However, it would not be that hard to buld a robot that could follow bitangent lines closely on the side of the bitangent on which the corresponding gaps are split. If such a robot would start approaching the bitangent, the two gaps about to merge would get closer and closer - a condition detectable by the robot. The robot could then start moving directly toward or away from the gaps about to merge, which would cause it to follow the bitangent. We therefore define a set of points G as the set of points through which our limited robot can manouver. For simplicity, assume that G is equivalent to the union of the boundary of R shrunk by some small , and the bitangents of shifted by the same to the side navigable by our robot (the side where the gaps associated with the bitangent are split). Also note that a robot navigating through G would be able to localize itself sufficiently by noticing it has come up to a bitangent or . Both of these capabilities must already be in the robot, since it must perform wall and bitangent line following.
We will now show that a robot capable of moving only along points in G is capable of clearing R if it is possible to clear R by a single pursuer.
If R can be cleared by a single pursuer, then there exists a solution that starts and ends on points belonging to G.
This is trivial since before executing any general solution ,the robot can go from a point in G to , execute the solution until R is cleared, and then go to some point in G.
For any path , such that if and only if , there exists a path such that g(t0) = p(to), g(t1) = p(t1), and the information state at g(t1) is equisuperior to the information state at p(t1) (we define this to mean that executing g will clear all gaps cleared by executing p, while not contaminating any gaps executing p does not contaminate).
Proof: Due to the nature of p, we can see that there exists a simle closed loop , such that both p(t0) and p(t1) sit on L, and p(t) is in the interior of L for t0<t<t1. Then we know that p(t1) can be reached through L, but we still need to show that it can be done in a way that produces an equisuperior information state. Let's first suppose that L is composed of line segments surrounding n bitangent lines enumerated clockwise, and with p(t0) directly outside b1. Remember that due to our definition of G, 'surrounding' means that if a pursuer moves from any point in L towards the inside of L, a pair of gaps will merge (also remember that a cleared gap can only be contaminated if it merges with a contaminated gap). If the pursuer starts moving from p(t0) along L clockwise, the first pursuit event (ignoring inflection line crossings) will be a split caused by crossing b2 (let's name this event S2), followed by a merge caused by crossing b1 (M1). Using the same notation, going all the way around L would cause events in precisely that order. If it were the case that the part of L we supposed was outside bi was actuallty following a part of or the bitangent corresponding to bi was outside L, the pair of events Si,Mi would simply never occur. Note that throughout this sequence of events, the only time information can be lost is at M1 (if M1 is triggered). This is because prior to every other merge, there was already a split, and between the two there can be no events that will contaminate any of the now split gaps, which means that the merged gap will not be contaminated unless it was contaminated before the split. Now we have two cases to deal with.
First, if p triggers M1 then the pursuer can start from p(t0), safely go all the way around L, and then continue around L to reach p(t1). Going all the way around L will cross all inflection lines going through L and therefore clear all gaps executing p would have cleared. Also, if doing so caused contamination upon crossing M1 then executing p would have also caused the same contamination.
Second, if p does not trigger M1, then the pursuer can start from p(t0), visit all points on L that can be reached without triggering M1, and then continue around L to p(t1). This path will cross all inflection points that p crosses, and therefore clear all gaps executing p would have cleared.
In either case, upon reaching p(t1) we have ensured all requirements for an equisuperior information state.
A robot capable of navigating only through G can clear any clearable region R.
Proof: Take any path that clears R. By lemma 1, there exists that clears R and starts and ends on G. By lemma 2 we can replace all parts of that do not lie on G with ones that do, and still ensure that all gaps are cleared at the end of . Since the mentioned robot can navigate through all points in , it can clear R.