When the region to be covered is a
convex set, the length of its shortest opaque set must be at least half its perimeter and at most its perimeter. For some regions, additional improvements to these bounds can be made.
Upper bound If K is a bounded convex set to be covered, then its
boundary \partial K forms an opaque set whose length is the perimeter |\partial K|. Therefore, the shortest possible length of an opaque set is at most the perimeter. For sets K that are strictly convex, meaning that there are no line segments on the boundary, and for interior barriers, this bound is tight. Every point on the boundary must be contained in the opaque set, because every boundary point has a
tangent line through it that cannot be blocked by any other points. The same reasoning shows that for interior barriers of
convex polygons, all
vertices must be included. Therefore, the
minimum Steiner tree of the vertices is the shortest
connected opaque set, and the
traveling salesperson path of the vertices is the shortest
single-curve opaque set. However, for interior barriers of non-polygonal convex sets that are not strictly convex, or for barriers that are not required to be connected, other opaque sets may be shorter; for instance, it is always possible to omit the longest line segment of the boundary. In these cases, the perimeter or Steiner tree length provide an
upper bound on the length of an opaque set.
Lower bound There are several proofs that an opaque set for any convex set K must have total length at least |\partial K|/2, half the perimeter. One of the simplest involves the
Crofton formula, according to which the length of any curve is proportional to its expected number of intersection points with a random line from an appropriate
probability distribution on lines. It is convenient to simplify the problem by approximating K by a strictly convex superset, which can be chosen to have perimeter arbitrarily close to the original set. Then, except for the tangent lines to K (which form a vanishing fraction of all lines), a line that intersects K crosses its boundary twice. Therefore, if a random line intersects K with probability p, the expected number of boundary crossings is 2p. But each line that intersects K intersects its opaque set, so the expected number of intersections with the opaque set is at least p, which is at least half that for K. By the Crofton formula, the lengths of the boundary and barrier have the same proportion as these expected numbers. This lower bound of |\partial K|/2 on the length of an opaque set cannot be improved to have a larger constant factor than 1/2, because there exist examples of convex sets that have opaque sets whose length is close to this lower bound. In particular, for very long thin rectangles, one long side and two short sides form a barrier, with total length that can be made arbitrarily close to half the perimeter. Therefore, among lower bounds that consider only the perimeter of the coverage region, the bound of |\partial K|/2 is best possible. However, getting closer to |\partial K|/2 in this way involves considering a sequence of shapes rather than just a single shape, because for any convex set K that is not a triangle, there exists a \delta such that all opaque sets have length at least |\partial K|/2+\delta.
Specific shapes For a
triangle, as for any convex polygon, the shortest connected opaque set is its minimum Steiner tree. In the case of a triangle, this tree can be described explicitly: if the widest angle of the triangle is 2\pi/3 (120°) or more, it uses the two shortest edges of the triangle, and otherwise it consists of three line segments from the vertices to the
Fermat point of the triangle. However, without assuming connectivity, the optimality of the Steiner tree has not been demonstrated. Izumi has proven a small improvement to the perimeter-halving lower bound for the
equilateral triangle. For a
unit square, the perimeter is 4, the perimeter minus the longest edge is 3, and the length of the minimum Steiner tree is 1+\sqrt3\approx 2.732. However, a shorter, disconnected opaque forest is known, with length \sqrt2+\tfrac12\sqrt6\approx 2.639. It consists of the minimum Steiner tree of three of the square's vertices, together with a line segment connecting the fourth vertex to the center.
Ross Honsberger credits its discovery to Maurice Poirier, a Canadian schoolteacher, but it was already described in 1962 and 1964 by Jones. It is known to be optimal among forests with only two components, and has been conjectured to be the best possible more generally, but this remains unproven. The perimeter-halving lower bound of 2 for the square, already proven by Jones, can be improved slightly, to 2.00002, for any barrier that consists of at most countably many
rectifiable curves, improving similar previous bounds that constrained the barrier to be placed only near to the given square. The case of the
unit circle was described in a 1995
Scientific American column by
Ian Stewart, with a solution of length 2+\pi, optimal for a single curve or connected barrier but not for an opaque forest with multiple curves.
Vance Faber and
Jan Mycielski credit this single-curve solution to
Menachem Magidor in 1974. By 1980, E. Makai had already provided a better three-component solution, with length approximately 4.7998, rediscovered by John Day in a followup to Stewart's column. The unknown length of the optimal solution has been called the
beam detection constant. ==Algorithms==