Upper bound

If ${\displaystyle K}$ is a bounded convex set to be covered, then its boundary ${\displaystyle \partial K}$ forms an opaque set whose length is the perimeter ${\displaystyle |\partial K|}$. Therefore, the shortest possible length of an opaque set is at most the perimeter. For sets ${\displaystyle K}$ 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.^[5] 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.^[4] 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.^[3][4]

Lower bound

There are several proofs that an opaque set for any convex set ${\displaystyle K}$ must have total length at least ${\displaystyle |\partial K|/2}$, 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 ${\displaystyle K}$ 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 ${\displaystyle K}$ (which form a vanishing fraction of all lines), a line that intersects ${\displaystyle K}$ crosses its boundary twice. Therefore, if a random line intersects ${\displaystyle K}$ with probability ${\displaystyle p}$, the expected number of boundary crossings is ${\displaystyle 2p}$. But each line that intersects ${\displaystyle K}$ intersects its opaque set, so the expected number of intersections with the opaque set is at least ${\displaystyle p}$, which is at least half that for ${\displaystyle K}$. By the Crofton formula, the lengths of the boundary and barrier have the same proportion as these expected numbers.^[6]

This lower bound of ${\displaystyle |\partial K|/2}$ 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 ${\displaystyle |\partial K|/2}$ is best possible.^[6] However, getting closer to ${\displaystyle |\partial K|/2}$ in this way involves considering a sequence of shapes rather than just a single shape, because for any convex set ${\displaystyle K}$ that is not a triangle, there exists a ${\displaystyle \delta }$ such that all opaque sets have length at least ${\displaystyle |\partial K|/2+\delta }$.^[7]

Specific shapes

For a triangle, as for any convex polygon, the shortest connected opaque set is its minimum Steiner tree.^[8] In the case of a triangle, this tree can be described explicitly: if the widest angle of the triangle is ${\displaystyle 2\pi /3}$ (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.^[9] 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.^[10]

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 ${\displaystyle 1+{\sqrt {3}}\approx 2.732}$. However, a shorter, disconnected opaque forest is known, with length ${\displaystyle {\sqrt {2}}+{\tfrac {1}{2}}{\sqrt {6}}\approx 2.639}$. 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,^[11] but it was already described in 1962 and 1964 by Jones.^[12][13] It is known to be optimal among forests with only two components,^[5][14] and has been conjectured to be the best possible more generally, but this remains unproven.^[7] The perimeter-halving lower bound of 2 for the square, already proven by Jones,^[12][13] can be improved slightly, to ${\displaystyle 2.00002}$, for any barrier that consists of at most countably many rectifiable curves,^[7] improving similar previous bounds that constrained the barrier to be placed only near to the given square.^[6]

The case of the unit circle was described in a 1995 Scientific American column by Ian Stewart, with a solution of length ${\displaystyle 2+\pi }$,^[15] optimal for a single curve or connected barrier^[8][16][17] but not for an opaque forest with multiple curves. Vance Faber and Jan Mycielski credit this single-curve solution to Menachem Magidor in 1974.^[8] By 1980, E. Makai had already provided a better three-component solution, with length approximately ${\displaystyle 4.7998}$,^[18] rediscovered by John Day in a followup to Stewart's column.^[19] The unknown length of the optimal solution has been called the beam detection constant.^[20]

Approximation

By the general bounds for opaque forest length in terms of perimeter, the perimeter of a convex set approximates its shortest opaque forest to within a factor of two in length. In two papers, Dumitrescu, Jiang, Pach, and Tóth provide several linear-time approximation algorithms for the shortest opaque set for convex polygons, with better approximation ratios than two:

• For general opaque sets, they provide an algorithm whose approximation ratio is at most
  $${\displaystyle {\frac {1}{2}}+{\frac {2+{\sqrt {2}}}{\pi }}\approx 1.5868.}$$

  
  The general idea of the algorithm is to construct a "bow and arrow" like barrier from the minimum-perimeter bounding box of the input, consisting of a polygonal chain stretched around the polygon from one corner of the bounding box to the opposite corner, together with a line segment connecting a third corner of the bounding box to the diagonal of the box.^[4]
• For opaque sets consisting of a single arc, they provide an algorithm whose approximation ratio is at most
  $${\displaystyle {\frac {\pi +5}{\pi +2}}\approx 1.5835.}$$

  
  The resulting barrier is defined by a supporting line of the input shape. The input projects perpendicularly onto an interval of this line, and the barrier connects the two endpoints of this interval by a U-shaped curve stretched tight around the input, like the optimal connected barrier for a circle. The algorithm uses rotating calipers to find the supporting line for which the length of the resulting barrier is minimized.^[4]
• For connected opaque sets, they provide an algorithm whose approximation ratio is at most ${\displaystyle 1.5716}$. This method combines the single-arc barrier with special treatment for shapes that are close to an equilateral triangle, for which the Steiner tree of the triangle is a shorter connected barrier.^[4]
• For interior barriers, they provide an algorithm whose approximation ratio is at most ${\displaystyle 1.7168}$.^[24] The idea is to use a generalization suggested by Shermer of the structure of the incorrect earlier algorithms (a Steiner tree on a subset of the points, together with height segments for a triangulation of the remaining input),^[23] with a fast approximation for the Steiner tree part of the approximation.^[24]

Additionally, because the shortest connected interior barrier of a convex polygon is given by the minimum Steiner tree, it has a polynomial-time approximation scheme.^[4]

Coverage

The region covered by a given forest can be determined as follows:

• Find the convex hull of each connected component of the forest.
• For each vertex ${\displaystyle p}$ of the hull, sweep a line circularly around ${\displaystyle p}$, subdividing the plane into wedges within which the sweep line crosses one of the hulls and wedges within which the sweep line crosses the plane without obstruction. The union of the covered wedges forms a set ${\displaystyle C_{p}}$.
• Find the intersection of all of the sets ${\displaystyle C_{p}}$. This intersection is the coverage of the forest.

If the input consists of ${\displaystyle n}$ line segments forming ${\displaystyle m}$ connected components, then each of the ${\displaystyle n}$ sets ${\displaystyle C_{p}}$ consists of at most ${\displaystyle 2m}$ wedges. It follows that the combinatorial complexity of the coverage region, and the time to construct it, is ${\displaystyle O(m^{2}n^{2})}$ as expressed in big O notation.^[25]

Although optimal in the worst case for inputs whose coverage region has combinatorial complexity matching this bound, this algorithm can be improved heuristically in practice by a preprocessing phase that merges overlapping pairs of hulls until all remaining hulls are disjoint, in time ${\displaystyle O(n\log ^{2}n)}$. If this reduces the input to a single hull, the more expensive sweeping and intersecting algorithm need not be run: in this case the hull is the coverage region.^[26]