Packing convex polygons in minimum-perimeter convex hulls

Abstract

The problem of packing a given set of freely translated and rotated convex polygons in a minimum-perimeter convex polygon (in particular the minimum-perimeter convex hull) is introduced. A mathematical model of the problem using the phi-function technique is provided. Problem instances with up to 6 convex polygons are solved by the global NLP solver BARON to get a minimum-perimeter convex hull. Numerical experiments for larger instances are reported using the local NLP solver IPOPT.

Appendix: On an analytical solution for the minimum perimeter convex hull of two triangles

For the case of two triangles, we may expect geometrical characteristics of the optimal layout. Leaving the rigorous proof for our forthcoming paper, we present here these characteristics in the form of the following conjecture.

Conjecture

The minimal perimeter convex hull for two triangles corresponds to a layout with two tangent largest sides of the triangles (see Fig. 

Fig. 12

The minimum perimeter convex hull of two tangent triangles

12).

The motivation for this conjecture is as follows. If the triangles have no common points (totally separated), then we can move one triangle till they are tangent thus reducing the perimeter. Suppose that there is only one tangent point (a vertex of the triangle). Then we can rotate the corresponding triangle around its tangent vertex till the triangles have tangent sides and thus reducing the perimeter. Finally, among all layouts with tangent sides, the case of tangent largest sides corresponds to the minimal perimeter convex hull by the triangle inequality. The rigorous proof of this conjecture is on the way.

Suppose that we have two triangles with their largest tangent sides. Then to get a layout with minimum perimeter convex hull we must define the position of the vertex of one triangle on the tangent side of another (see Fig. 12). This can be done as follows.

Consider two triangles with vertices $A_{i}$, $B_{i}$, and $C_{i}$, sides $a_{i}$, $b_{i}$, and $c_{i}$, and internal angles $\alpha_{i}$, $\beta_{i}$, and $\gamma_{i}$, $i = 1,2$. The naming is such, that $c_{1} = A_{1} B_{1}$ is the largest side of the triangle 1 and that $c_{1}$ is greater than or equal to the largest side of the triangle 2. Therefore,$A_{1}$ and $B_{1}$ are vertices of the convex hull polygon ${\rm P}$.

Triangle 1 is placed such that its vertex $A_{1}$ is in the origin of the $x$$- y$-coordinate system, $B_{1}$ is at $(c,0)$, and $C_{1}$ is in the fourth quadrant with at $(x_{1} ,y_{1} )$ with $x_{1} > 0$ and $y_{1} < 0$. $C_{1}$ is another vertex of the convex hull polygon ${\rm P}$.

If we place triangle 2 such that $A_{2}$ and $B_{2}$ are on the positive axis with $0 \le x_{2A} < c$ and $0 < x_{2B} < c$, then $C_{2}$ at $(x_{2} ,y_{2} )$ completes the system of vertices of ${\rm P}$ (Fig. 12).

The $y_{2}$-coordinate follows from the triangle itself; it is just its height, i.e.,

$$ y_{2} = b_{2} \sin \alpha_{2} ,\alpha_{2} = arc\cos \frac{{b_{2}^{2} + c_{2}^{2} - a_{2}^{2} }}{{2b_{2} c_{2} }}, $$

where $(v_{{x_{2A} }} ,v_{{y_{2A} }} ) = (0,0)$, $(v_{{x_{2B} }} ,v_{{y_{2B} }} )$ and $(v_{{x_{2C} }} ,v_{{y_{2C} }} )$ are the intrinsic vertex coordinates of triangle 2.

As

$$ x_{2B} = x_{2A} + c, $$

and

$$ x_{2C} = x_{2} = x_{2A} + b_{2} \cos \alpha_{2} , $$

the convex hull perimeter ℓ = ℓ(P) depends only on one variable $x = x_{2A}$:

$$ l = a_{1} + b_{1} + \sqrt {(x + b_{2} \cos \alpha_{2} )^{2} + y_{2}^{2} } + \sqrt {(c_{1} - (x + b_{2} \cos \alpha_{2} ))^{2} + y_{2}^{2} } . $$

The contributions $a_{1}$ and $b_{1}$ are constant and can be computed a priori—for this arrangement of the two triangles. Other arrangements—in which the two triangles have one touching line—follow by permutations and can be evaluated similarly.

For simplification, let us introduce

$$ u = x + b_{2} \cos \alpha_{2} ,v = a_{1} + b_{1} $$

and follow up with

$$ l(u) = v + \sqrt {u^{2} + y_{2}^{2} } + \sqrt {(c_{1} - u)^{2} + y_{2}^{2} } . $$

Now we consider $l^{\prime}(u)$ and $l^{\prime\prime}(u)$ to find the local minimum $u_{*}$:

$$ l^{\prime}(u) = \frac{u}{{\sqrt {y_{2}^{2} + u^{2} } }} - \frac{{c_{1} - u}}{{\sqrt {y_{2}^{2} + (c_{1} - u)^{2} } }} $$

and

$$ l^{\prime\prime}(u) = \frac{1}{{\sqrt {u^{2} + y_{2}^{2} } }} + \frac{1}{{\sqrt {y_{2}^{2} + (c_{1} - u)^{2} } }} - \frac{{u^{2} }}{{(u^{2} + y_{2}^{2} )^{\frac{3}{2}} }} - \frac{{(c_{1} - u)^{2} }}{{(y_{2}^{2} + (c_{1} - u)^{2} )^{\frac{3}{2}} }}. $$

For $u_{*} = \frac{{c_{1} }}{2}$ the first derivative $l^{\prime}(u_{*} )$ vanishes. The second derivative takes the value

$$ l^{\prime\prime}(u) = 2\left[ {\frac{1}{{\sqrt {\left( {\frac{{c_{1} }}{2}} \right)^{2} + y_{2}^{2} } }}} \right. - \left. {\frac{{\left( {\frac{{c_{1} }}{2}} \right)^{2} }}{{\left( {\left( {\frac{{c_{1} }}{2}} \right)^{2} + y_{2}^{2} } \right)^{\frac{3}{2}} }}} \right] = \frac{{2y_{2}^{2} }}{{\left( {\left( {\frac{{c_{1} }}{2}} \right)^{2} + y_{2}^{2} } \right)^{\frac{3}{2}} }} \ge 0 $$

which is always positive, i.e., we have a local minimum at $u_{*} = \frac{{c_{1} }}{2}$, or

$$ x = \frac{{c_{1} }}{2} - b_{2} \cos \alpha_{2} = \frac{{c_{1} }}{2} - \frac{{b_{2}^{2} + c_{2}^{2} - a_{2}^{2} }}{{2c_{2} }}. $$