A fully general, exact algorithm for nesting irregular shapes

Abstract

This paper introduces a fully general, exact algorithm for nesting irregular shapes. Both the shapes and material resource can be arbitrary nonconvex polygons. Moreover, the shapes can have holes and the material can have defective areas. Finally, the shapes can be arranged using both translations and arbitrary rotations (as opposed to a finite set of rotation angles, such as 0$^\circ$and 180$^\circ$). The insight that has made all this possible is a novel way to relax the constraint that the shapes not overlap. The key idea is to inscribe a few circles in each irregular shape and then relax the non-overlap constraints for the shapes by replacing them with non-overlap constraints for the inscribed circles. These relaxed problems have the form of quadratic programming problems (QPs) and can be solved to optimality to provide valid lower bounds. Valid upper bounds can be found via local search with strict non-overlap constraints. If the shapes overlap in the solution to the relaxed problem, new circles are inscribed in the shapes to prevent this overlapping configuration from recurring, and the QP is then resolved to obtain improved lower bounds. Convergence to any fixed tolerance is guaranteed in a finite number of iterations. A specialized branch-and-bound algorithm, together with some heuristics, are introduced to find the initial inscribed circles that approximate the shapes. The new approach, called “QP-Nest,” is applied to three problems as proof of concept. The most complicated of these is a problem due to Milenkovic that has four nonconvex polygons with 94, 72, 84, and 74 vertices, respectively. QP-Nest is able prove global optimality when nesting the first two or the first three of these shapes. When all four shapes are considered, QP-Nest finds the best known solution, but cannot prove optimality.

Appendix

1.1 Algorithm for inscribing circles and computing maximum penetrations

Figure 29 illustrates the steps in the branch-and-bound procedure for inscribing a circle in a polygon in such a way that: (i) the circle lies entirely inside the polygon and (ii) the circle does not intersect any previously inscribed circles. The figure shows six steps, labeled (a)–(f), that we now explain.

Fig. 29

The steps in the basic circle filling procedure

• (a) The starting point for the algorithm is a polygon with one or more previously inscribed circles (here just one circle).

• (b) We cover the original polygon with triangles. The triangular cover is initialized by finding the bounding box of the polygon and dividing it into two triangles. We have grayed out the original polygon and circle to focus attention on the triangles.

• (c) In each iteration, one triangle is selected and subdivided by connecting the midpoint of the longest side to the opposite vertex. Here we show how this is done for the lower right triangle.

• (d) As the iterations proceed, it is possible that some triangle will have an empty intersection with the polygon; an example is the shaded triangle. Any such triangle is deleted, as it cannot contain the center of any inscribed circle.

• (e) For each triangle in the partition, we compute a lower and upper bound on the radius of a circle that has a center in the triangle, lies within the polygon, and does not intersect any previously inscribed circles. Here we illustrate the computation of the lower bound for the thick-lined triangle. We begin by finding the centroid of the triangle, here shown as the black dot. If the centroid lies inside the original polygon (which may not always be the case), we compute the minimum distance from the centroid to each side and to each existing inscribed circle; these distances are shown with dotted lines. The minimum of all these distances is the radius of the largest circle that can be inscribed having the centroid as its center. This is therefore a lower bound on the radius of an inscribed circle with a center in the triangle. We keep track of the largest circle so found, as this is our best feasible solution. If the centroid does not lie in the polygon, we skip the computation of the lower bound.

• (f) Here we illustrate the computation of the upper bound for the thick-lined triangle. For each side of the polygon, we determine which vertex of the triangle is furthest from the side (i.e, maximizes the minimum distance between the side and the vertex). In the figure, the dotted lines connect each side to the furthest vertex of the triangle. The furthest distance will be the distance between the triangle vertex and either (i) the closest endpoint of the side or (ii) the projection of the triangle vertex on the side of the polygon. Similarly, for each circle, we compute the distance from the perimeter of the circle to the furthest vertex. Clearly, these are the largest possible distances from a point in the triangle to a polygon side or circle perimeter. The minimum of all these distances is thus a valid upper bound on the radius of an inscribed circle with a center in the triangle that does not intersect any previously inscribed circles.

The entire procedure starts with the initial two triangles in (b), computing the lower and upper bound as discussed. From this point we proceed iteratively, selecting the triangle with the largest upper bound, bisecting it, and computing the bounds for the two child rectangles. Of course, if a child triangle lies outside the polygon, as in (d), we skip the bound calculation and simply delete it. The stopping criterion is the gap between radius of the largest circle successfully drawn from a triangle centroid (our best solution) and the overall upper bound (the largest upper bound of any undeleted triangle). When this gap is sufficiently small, we stop. I normally use a very small tolerance; for example, in the Milenkovic problems the tolerance was $10^{-4}$. In general, the tolerance used should be several (say three) orders of magnitude smaller than the accuracy $\epsilon$ that one desires for the nest length.

We can compute the maximum penetration of polygon $A$ into polygon $B$ using the following modification of the branch-and-bound algorithm:

• The initial bounding box is the bounding box of both polygons $A$ and $B$.

• We delete a triangle if it either has no intersection with polygon $A$ or has no intersection with polygon $B$.

• We only compute a lower bound for a triangle if the centroid lies in both polygons $A$ and $B$. If this happens, we find the shortest distance from each side of polygon $B$ to the centroid. The minimum of these “shortest distances to the centroid” is the radius of a circle that can be grown from the centroid and still lie within polygon $B$. It is a valid lower bound on the maximum penetration.

• To compute an upper bound for a triangle, for each side of polygon $B$, we find the furthest vertex of the triangle. The minimium of these “furthest distances” is the upper bound for the triangle.

Finally, we can compute the largest circle that can be inscribed in the overlap of two polygons with the following modifications of the branch-and-bound algorithm:

• The initial bounding box is the bounding box of both polygons $A$ and $B$.

• We delete a triangle if it either has no intersection with polygon $A$ or has no intersection with polygon $B$.

• We only compute a lower bound for a triangle if the centroid lies in both polygons $A$ and $B$. If this happens, we find shortest distance from each side of polygons $A$ and $B$ to the centroid. The minimum of these “shortest distances to the centroid” is the radius of a circle that can be grown from the centroid and still lie within both polygons $A$ and $B$. It is a valid lower bound on the radius of a circle inscribed in the overlap of $A$ and $B$.

• To compute an upper bound for a triangle, for each side of polygons $A$ and $B$, we find the furthest vertex of the triangle. The minimium of these “furthest distances” is the upper bound for the triangle.

1.2 Proof of theorem 2

To set up the proof, for each polygon, assume we have computed the distance between any two vertices and between any vertex and a non-adjacent side. Let $s_{\min }$ denote the very smallest of these distances. Because there are only a finite number of vertices and sides, and because the sides do not intersect, the minimum distance $s_{\min }$ must exist and be strictly greater than zero. Also compute the smallest (most acute) angle $\theta _{\min }$ of any vertex. Figure 30 illustrates the computation of $s_{\min }$ and $\theta _{\min }$.

Fig. 30

The value of $s_{\min }$ in the proof of Theorem 2 is the smallest of a the minimum distance between any two vertices and b the minimum distance between any vertex and a non-adjacent side

Now suppose Theorem 2 were false. Then there would exist some number $\delta >0$ with the property that, in every iteration, one shape would penetrate another by at least $\delta$. This means that in every iteration there would exist a point $x$ in some shape $A$ that penetrates some other shape $B$ by $\delta$. By definition, then, the circle with center $x\in A$ and radius $\delta$ would lie entirely in $B$ (see Fig. 31a).

Fig. 31

If Theorem 2 were false, there would always exist some point $x$ in some shape $A$ that penetrates some other shape $B$ by $\delta$

Now let $\lambda =\min (\delta ,s_{\min }/2)$ and let $C$ denote the circle with center $x$ and radius $\lambda$. This circle obviously lies entirely inside $B$ (see Fig. 31b). Moreover, because the diameter of $C$ is less than or equal to $s_{\min }$, at most one vertex of $A$ can lie in $C$ (if there were two vertices inside $C$, then these vertices would be closer together than $s_{\min }$, which is impossible).

We may now distinguish three cases, as illustrated in Fig. 32. In case 1 there are no vertices of shape $A$ in $C$. In case 2 there is exactly one vertex and this vertex is not $x$. Finally, in case 3 there is exactly one vertex and this vertex is $x$. In all three cases, we can find a circle that lies in the intersection of shapes $A$ and $B$, as shown in Fig. 32.

Fig. 32

Three cases considered in the proof of Theorem 2

Clearly, the smallest overlap of shape $A$ with circle $C$ (and hence with $B$) occurs in case 3 when the angle at the vertex is the smallest (most acute) angle $\theta _{\min }$. Thus, the smallest possible overlap is a wedge with angle $\theta _{\min }$ from a circle of radius $\lambda$. Beside the two sides forming the wedge, no other side from polygon $A$ can be inside circle $C$, because we have assumed that the radius of circle $C$ is smaller than the smallest distance from a vertex to any non-adjacent side; this is why the entire wedge must be inside shape $A$ and, hence, the entire wedge is in the overlap of polygons $A$ and $B$. A little trigonometry (see Fig. 33) shows that one can inscribe a circle in this wedge with radius:

$$\begin{aligned} r=\frac{\lambda \sin (\theta _{\min }/2)}{1+\sin (\theta _{\min }/2)}>0. \end{aligned}$$

(20)

Moreover, this is true no matter what iteration we are on, because we have assumed that in each iteration some polygon penetrates another by at least $\delta$. But if we can always inscribe a circle of radius $r$ inside the overlap region, then the radius of the largest possible inscribed circle would not go to zero, which contradicts Theorem 1! $\square$

Fig. 33

Computing the radius of the largest circle that can be inscribed in a wedge

As an aside, note that “asymptotically,” as the iterations proceed and the maximum penetration $\delta$ becomes so small that it is less than $s_{\min }/2$, then we will have $\lambda = \min (\delta , s_{\min }/2)=\delta$. Equation (20) then implies that the ratio $\delta /r_{max}$—the ratio of the maximum penetration to the radius of the largest inscribed circle—is indeed bounded, as stated in Eq. (15) in the text. In particular, we have:

$$\begin{aligned} \frac{\delta }{r_{max}} \le \frac{\delta }{r} = \frac{\lambda }{r} = \frac{1+\sin (\theta _{\min }/2)}{\sin (\theta _{\min }/2)} \end{aligned}$$

(21)