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Following the recent trends in robotics, we will try to find some
small set of capabilities needed by a robot to solve the
pursuit-evasion problem. That is, we will try to limit our robot to
make robot construction and solution execution as simple and cheap as
possible. We start from the observation that
the robot does not need to be able to visit all the cells in the cell
decomposition of *R*.
Let *G*_{0} be the union of and its bitangents.
Notice that a real-world robot capable of moving along
points of *G*_{0} 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.

**Lemma 415**

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*.

**Lemma 431**

For any path , such that if and only if , there
exists a path such that *g*(*t*_{0}) = *p*(*t*_{o}), *g*(*t*_{1}) = *p*(*t*_{1}), and
the information state at *g*(*t*_{1}) is equisuperior to the information state
at *p*(*t*_{1}) (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*(*t*_{0})
and *p*(*t*_{1}) sit on *L*, and
*p*(*t*) is in the interior of *L* for *t*_{0}<*t*<*t*_{1}. Then we know that *p*(*t*_{1})
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*(*t*_{0}) directly outside *b*_{1}.
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*(*t*_{0}) along *L* clockwise, the first pursuit event
(ignoring inflection line crossings) will be
a split caused by crossing *b*_{2} (let's name this event *S*_{2}),
followed by a merge caused by crossing *b*_{1} (*M*_{1}). 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 *b*_{i} was
actuallty following a part of or the
bitangent corresponding to *b*_{i} was outside *L*, the pair of events
*S*_{i},*M*_{i} would simply never occur. Note that throughout this
sequence of events, the only time information can be lost is at
*M*_{1} (if *M*_{1} 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 *M*_{1} then the pursuer can start from *p*(*t*_{0}), safely
go all the way around *L*, and then continue around *L*
to reach *p*(*t*_{1}). 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 *M*_{1} then executing
*p* would have also caused the same contamination.

Second, if *p* does not trigger *M*_{1}, then the pursuer can start from
*p*(*t*_{0}), visit all points on *L* that can be reached without
triggering *M*_{1}, and then continue around *L* to *p*(*t*_{1}). 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*(*t*_{1}) we have ensured all requirements for an
equisuperior information state.

**Theorem 489**

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*.

** Next:** Computing a solution
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*Stjepan Rajko*

*5/7/2000*