Various other suggestions were proposed for the problem of the thief location in a 2D region. Since we don't know the location of the thief, we need a watchman route that completely explores the region.
Joe proposed a variation of the problem, where we are given a set of regions on the plane and we wish to find a minimum tour that visits every one. Special cases are to visit every state in a map, or each line in a line arrangement.
Estie then pointed out the difference between the problem of visiting lines and line segments as there can be no constant factor approximation algorithm for line segments.
Joe explained that for halfspaces, the problem
can be done in polynomial time.
You can draw a circle around all vertices of the arrangement that
will meet every line. Then it can be transformed to a watchman
route. It was noted that Nielsen had found an error in an
watchman route algorithm, and had suggested an alternate 
algorithm
for it.
To visit all the lines in the arrangement, we can construct a polygon forcing the watchman to visit every line during the route, by making the polygon zig-zag around the endpoints of each line, thus creating as many essential regions as 2 times the number of lines.
Joe suggested another version for triangles in the unit square. Find a constant factor approximation tour to hit all the triangles in the unit square.
He also pointed out that the watchman route for a polygon with holes is a hard problem, as each hole may be a spiral that will force the watchman to pass over a specific point. This reduces TSP to the watchman route problem with holes.
Barry then suggested another version where we have triangulated rooms with one door per room. We have one thief and one guard. What is the strategy the guard should follow? Joe said this problem is known.
For general planar graphs, however, it is hard to do it, if the thief is mobile. It may engage the policeman into an infinite chase, and the problem now is to minimize the number of policemen needed to trap and catch the thief. This version was posed at the Stony Brook workshop last year and was solved at the workshop at Baltimore.
Steve then wandered if a constant number of guards suffices, but Estie pointed to the opposite.
Joe then asked about the situation with a rectangular partition, where the width and height of each rectangle is within certain bounds.
Estie pointed out that if there is a bound r on the ratio of the big / small rectangles, then there is an approximation algorithm with factor d r , where 5 < d < 20. For the triangle case, you can get similar approximation, except when there is no bound on the ratio of the angles of the triangle. Then it is not known.
Joe then pointed out that we can build the upper and lower envelope
of a line arrangement in 
time. Then an idea is to
construct a box that cuts all lines.
Another idea is to get the smallest sphere that hits all lines and consider the centerpoint as the origin. For each line, closest to the center point we can approximate the route with TSP. How bad can this be? Or get the minimum distance segment (lines are in 3D) for each pair of lines.
Estie then pointed out that the sphere may be very big. Joe then suggested we build a sphere that is tangent to 4 lines. We thus have ( n / 4 ) candidates. Let's fix 4 lines at a time. In constant time, we can find the smallest sphere that is touching all of them.
Among the (n /4 ) spheres choose those that stab all the lines and pick the smallest. Estie then pointed out that in 2D the optimal can be a constant factor times the perimeter of the circle.
Another idea is to take the intersection points of the lines with the sphere and do TSP approximation on them.