Covering Polygons with Rectangles

Abstract

A well-known and well-investigated family of hard optimization problems deals with nesting, i.e., the non-overlapping placing of polygons to be cut from a rectangle or the plane whilst minimizing the waste. Here we consider the in some sense inverse problem of a subsequent step in production technology: given a set of polygons in the plane and an axis-aligned rectangle (modeling a gripping device), we seek the minimum number of copies of the rectangle such that every polygon is completely covered by at least one copy of the rectangle. As motions of the given rectangle for obtaining the copies we investigate the cases of translation in x-direction, of arbitrary translation and of arbitrary translation combined with rotation. We give a generic algorithm for all three cases which leads to a polynomial-time algorithm for the first case. The other two cases are NP-hard so we introduce a rather straightforward algorithm for the second case and two different approaches to the third one. Finally, we give experimental results and compare them to the theoretical analysis done before.

A Deferred Proofs

1.1 A.1 Proof of Lemma 1

Proof

Let p be an arbitrary point of $\mathbf {P}$ with minimum x-coordinate. Then there is a rectangle $R_j\in \mathbf {C}$ containing p. Translating this rectangle in positive x-direction till p lies on its left side we obtain a rectangle $R_j'$ with the required properties.

1.2 A.2 Proof of Lemma 2

Proof

Let $p_1$ be a point of $\mathbf {P}$ with minimum x-coordinate and $P_1$ a polygon of $\mathbf {P}$ which has $p_1$ as a vertex. Then there is a rectangle $R_j\in \mathbf {C}$ containing $P_1$. Now we translate $R_j$ in positive x-direction till $p_1$ lies on the left side of the translated rectangle $\hat{R}_j$. Clearly, we have $\mathsf{cov}(\hat{R}_j,\mathbf {P})\supseteq \mathsf{cov}(R_j,\mathbf {P})$. Subsequently, we translate $\hat{R}_j$ in negative y-direction till a point $p_2$ with the following properties lies on the upper side of the translated rectangle $R_j'$:

• 1. All polygons of $\mathbf {P}$ in $\hat{R}_j$ with $p_2$ as a vertex are contained in $R_j'$, and

• 2. $p_2$ is a point with maximum y-coordinate fulfilling the above requirements.

Then we have $\mathsf{cov}(R_j',\mathbf {P})\supseteq \mathsf{cov}(\hat{R}_j,\mathbf {P})\supseteq \mathsf{cov}(R_j,\mathbf {P})$, so $\mathbf {C}\backslash \{R_j\}\cup \{R_j'\}$ is indeed an optimal translational cover of $\mathbf {P}$. Moreover, $p_1$, $p_2$ and $P_1$ meet their requirements by construction, and for $P_2$ we can chose an arbitrary polygon that has $p_2$ as a vertex and is contained in $R_j'$.

In Fig. 1, $R_j$ corresponds to the dotted rectangle, $\hat{R}_j$ to the dashed one, and the final rectangle $R_j'$ is drawn with a full line.

1.3 A.3 Proof of Lemma 3

Proof

Let $R_j\in \mathbf {C}$ be a rectangle containing $P_{pi}$. We apply to $R_j$ similar translations as in Lemma 2 but do not translate in x- and y-direction but in directions parallel to adjacent sides of $R_j$. Doing so, we end up with a rectangle $\hat{R}_j$ and two (not necessarily distinct!) points $p_1$ and $p_2$ with the following properties:

• 1. $\hat{R}_j$ contains $P_{pi}$,

• 2. $p_1$ and $p_2$ lie on adjacent sides of $\hat{R}_j$,

• 3. $\mathsf{cov}(\hat{R}_j,\mathbf {P})\supseteq \mathsf{cov}(R_j,\mathbf {P})$, and

• 4. there are polygons $P_1,P_2\in \mathbf {P}$ such that for $i\in \{1,2\}$, $p_i$ is a vertex of $P_i$ and $P_i$ is contained in $\hat{R}_j$.

Now we perform a general motion of $\hat{R}_j$ combined of a clockwise rotation and suitable translation which keeps $p_1$ and $p_2$ on their respective sides. There are two cases:

• 1. $p_1$ and $p_2$ coincide. Then the described general motion is a simple rotation of $\hat{R}_j$ around $p_1$. This rotation is continued until a point $p_3$ lies on a side of the resulting rectangle $R_j'$ such that the following properties hold:

  • (a) $R_j'$ contains $P_{pi}$,

  • (b) there are polygons $P_1,P_3\in \mathbf {P}$ such that for $i\in \{1,3\}$ $p_i$ is a vertex of $P_i$ and $P_i$ is contained in $R_j'$, and

  • (c) $\mathsf{cov}(R_j',\mathbf {P})\supseteq \mathsf{cov}(\hat{R}_j,\mathbf {P})$.

• 2. $p_1$ and $p_2$ are distinct. Here we continue the general motion till one of the following two cases concerning the resulting rectangle $R_j'$ occurs:

  • (a) $p_1$ or $p_2$ coincide with a vertex $p_3$ of $R_j'$, or

  • (b) there is a point $p_3$ on a side of $R_j'$ such that

    • i $R_j'$ contains $P_{pi}$,

    • ii there are polygons $P_1,P_2,P_3\in \mathbf {P}$ such that for $i\in \{1,2,3\}$ $p_i$ is a vertex of $P_i$ and $P_i$ is contained in $R_j'$, and

    • iii $\mathsf{cov}(R_j',\mathbf {P})\supseteq \mathsf{cov}(\hat{R}_j,\mathbf {P})$.

Now, after possibly necessary renamings, $p_1$, $p_2$ and $p_3$ together with $R_j'$ meet the requirements of the lemma.

An example situation of the proof above is shown in the left part of Fig. 2: the dotted rectangle corresponds to $R_j$, the dashed one to $\hat{R}_j$ and the fully lined to $R_j'$. Moreover, $P_1$ is the left triangle while $P_2$ and $P_3$ coincide here in the right triangle. $p_1$ equals the point of the left triangle lying on the rectangle and $p_2$ and $p_3$ are the points of the right triangle lying on the rectangle. An extreme situation where $p_1$ and $p_2$ coincide with two opposite vertices of $R_j$ is illustrated by the right part of the same figure (note that two opposite vertices of a rectangle lie on adjacent sides).

1.4 A.4 Upper Bound of the Branching Degree for the General Case

Let the candidate rectangles for the general case be computed as described in Subsect. 7.1, and let us replace $n_{pi}$ by n for better readability. According to Sperner’s theorem, there are at most ${\left( {\begin{array}{c}n\\ \lfloor (n/2\rfloor \end{array}}\right) }$ inclusion-maximal coverable sets containing $P_{pi}$. Ignoring the asymptotically irrelevant Gaussian brackets and expanding the binomial coefficient yields $\frac{n!}{(n/2)!^2}$ which amounts in the asymptotically view to $\frac{\sqrt{2\pi n}\,\left( \frac{n}{e}\right) ^{n}}{\left( \sqrt{2\pi n/2}\,\left( \frac{n/2}{e}\right) ^{n/2}\right) ^2}$ by Stirling’s formula. Elementary calculus leads now to the result from Subsect. 7.1.