Using sentinels to detect intersections of convex and nonconvex polygons

Using sentinels to detect intersections of convex and nonconvex polygons

W.F. Mascarenhas; E.G. Birgin

Computer Science Department, Institute of Mathematics and Statistics University of São Paulo, Rua do Matão, 1010, Cidade Universitária 05508-090 São Paulo, SP, Brazil E-mails: [e-mail]/ [e-mail]

ABSTRACT

We describe finite sets of points, called sentinels, which allow us to decide if isometric copies of polygons, convex or not, intersect. As an example of the applicability of the concept of sentinel, we explain how they can be used to formulate an algorithm based on the optimization of differentiable models to pack polygons in convex sets.

Mathematical subject classification: 90C53, 65K05.

Key words: sentinels, polygons, intersection, packing, nonlinear programming.

1 Introduction

In [1] we propose a nonlinear programming approach to pack arbitrary polygons in convex sets (not necessarily polygons). This approach is based on the observation that if the interior of translated and rotated copies P' and Q' of the polygons P and Q in Figure 1 intersect then either the interior of P' contains one of the points or the interior of Q' contains some . Motivated by this fact, we say that the p_i and q_i are sentinels for {P,Q} with respect to translationsand rotations.

The observation above leads to the following algorithm to pack translated and rotated copies of P₁, P₂, ..., P_N on a convex set C:

1. We parameterize P_i by the coordinates (x_i,y_i) of its barycenter and its rotation angle Φ_i .

2. We define differentiable functions ψ_ij(x_i,y_i,θ_i,x_j,y_j,θ_j) based on the distance of the sentinels of P_i to P_j so that ψ_ij(x_i,y_i,θ_i,x_j,y_j,θ) is zero if all the sentinels of P_i are outside the interior of P_j and ψ_ij(x_i,y_i,θ_i,x_j,y_j,θ) is positive otherwise. (See [1] for an example of ψ_ij-functions for identical rectangles.)

3. We find approximations to the solution of the feasibility problem

For each solution of problem (1) we obtain a packing of P₁, P₂, ..., P_Nin C.

In [1] we describe the nonlinear programming aspects of the approach above in detail, from the theoretical and practical perspectives. In [5, 7, 8], Stoyan's Φ-functions are introduced. A Φ-function for a pair of polygons is defined as a function whose value is positive if the polygons overlap and zero otherwise. Our functions ψ are analogous to Stoyan's Φ-functions in the sense that they are Φ-functions defined through the usage of sentinels. Sentinels can also be used to detect the intersection of rotated and translated copies of polygons [6].Finally, the concept of sentinel leads to packings which are not necessarily lattice-like [3].

In the present work we focus on the geometric aspects of the novel concept of sentinel. We formalize this concept for arbitrary families of polygons and discuss their existence and complexity. Sentinel is a neat concept, but unfortunately some polygons require an infinite number of them. For instance, ifinstead of the polygons P and Q in Figure 1 we consider translated and rotated copies of a single triangle T then no finite set of points {t₁,...,t_n}, no matterhow large, would be enough to detect all the intersections of T and a translated copy T' of it: see Figure 2; we can always translate T' (maintaining the overlapping with T) in order to have outside T and t_a outside of T' no matter how near we define t_a from t_b and from .

Small internal angles are the main reason why we may need an infinite number of sentinels. More precisely, in the next section we show that if P is a finite family of convex and non-convex polygons P such that all the internal angles of P are bigger than or equal to π/2 then we can assign a finite set S(P) ⊂ ² to each P ∈ P in such way that if T,U: ²→ ² are isometries, P,Q ∈ P and the interior of T(P) and U(Q) intersect then either T(S(P)) intersects the interior of U(Q) or U(S(Q)) intersects the interior of T(P). In Section 2 we also define the terms we use throughout the paper and present basic results about the existence of sentinels. In Section 3 we discuss sentinel assignments for rectangles, which are the main motivation behind [1]. We present minimal sets of sentinels for the families of rectangles that motivated [1] and lead us to define the concept of sentinel. The last section contains concluding remarks.

2 Sentinels

In this section we formalize the concept of sentinel and prove the existence of finite sets of sentinels for relevant families of polygons. We start by defining the basic terms we use:

Notation 1. If u,v ∈ ² then we call the segment with extremes u and v by uv.

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Convention 1. A polygon P is defined in terms of an integer n > 3 and vertices p_i ∈ ², for i ∈ , such that

(a) p_{i + n} = p_i for all i and p_{i + j} ≠ p_i if j is not a multiple of n.

(b) p_i ∉ p_i-1p_i+1 for all i.

(c) If the segments p_i p_i+1 and p_j p_j+1 intersect then either

(i) i ≡ j mod n, or

(ii) i ≡ j + 1 mod n and p_i p_i+1∩ p_j p_j+1 = {p_i}, or

(iii) i + 1 ≡ j mod n and p_i p_i+1∩ p_j p_j+1 = {p_j}.

These conditions imply that P is a Jordan curve. We denote the interior of this curve by int(P) and its border by border(P). Moreover, we assume that the points p_i are in counterclockwise order and use n(P) to denote n.

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Notation 2. Sub(²) denotes the set whose elements are the subsets of ².

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Using this terminology we can formalize the concept of sentinel:

Definition 1. Let P be a family of polygons and let T be a family of transformations T: ²→ ² such that T(P) is a polygon for all P ∈ P. We say that a function S: P → Sub(²) is a sentinel assignment for P with respect to T if for all T, U ∈ T and P, Q ∈ P such that int(T(P)) ∩ int(U(Q)) ≠ Ø we have that either int(T(P)) ∩U(S(Q)) ≠ Ø or int(U(Q)) ∩T(S(P)) ≠ Ø. In this context, we say that the elements of S(P) are the sentinels of P.

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In this paper we care only about two families T of transformations:

Notation 3. The isometries of ² are transformations of the form T(x) = Qx + d, where Q is a 2 ×2 orthogonal matrix and d ∈ ². We call the set of such isometries by I₂. We define as the set of transformations T above with det Q = 1.

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In this section we provide sufficient conditions on the family P of polygons to guarantee the existence of a sentinel assignment S for P with respect to I₂ such that S(P) is finite for all P ∈ P. In the next section we discuss the analogous question when P is a family of rectangles and the family of transformations is . These conditions are expressed in terms of the parameter α (P):

Notation 4. We call the smallest internal angle of the polygon P by α (P).

The relevance of α (P) is illustrated in Figure 3. According to this figure, we can decide whether b is inside, over or outside the circle with diameter ac by looking at the angle α. This figure is the motivation for the condition α (P) > π/2 used throughout this paper.

Besides α (P) our results are formulated using the sets Δ(P):

Definition 2. Let P be a polygon with n > 4 vertices (remember that we are dealing with convex and nonconvex polygons). We define Δ(P) as the set formed by the positive δ's such that every segment uv connecting disjoint sides of P has length bigger than δ .

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Unfortunately, Δ(P) is not as simple as α(P), starting with the fact that it is a set and not a number. Please, pay much attention to this fact, because it does not follow the computational geometry tradition of using numbers to characterize properties of polygons. For instance, the supremum of our sets Δ(P) could be related to the several attempts to quantify the fatness of a polygon (see [4]) but our approach is different: we look at the whole set Δ(P) , not only at its supremum.

If P is convex and α(P) > π/2 then Δ(P) = (0,µ(P)], where µ(P) is the length of P's shortest side. In this particular case the arguments below could be rephrased in terms of µ(P). However, in [1] we care mainly about rectangles and if R is a rectangle then Δ(R) = (0,µ(R)) does not contain µ(R). Therefore, in order to unify the treatment of rectangles and polygons with bigger internal angles we chose to use Δ(P).

Using α(P) and Δ(P) we can state the key results behind most arguments in this paper. Given a family P of polygons and a function S: P → Sub(²), Theorems 1 and 2 give sufficient conditions for S to be a sentinel assignment for P with respect to I₂. Lemmas 1, 2 and 3 present technical results used to prove the theorems.

Lemma 1. Let P and Q be polygons with α(P) > π/2 and a(Q) > π/2. Suppose u, v, w ∈ q_k q_k+1 are such that ║u- q_k║ < ║v - q_k║ < ║w - q_k║ and

(a) u,w ∉ (P), (b) v ∈ int(P) and (c) ║u - w║∈ Δ(P) ∩ Δ(Q).

If no vertex of P is in the interior of Q then there exists a vertex p_j of P with p_j-1 p_j ∩ uv = {x} and p_jp_j+1 ∩ vw = {y} (see Fig. 4). Moreover, x and y are such that xy - {x,y} ⊂ (P),

Proof of Lemma 1. Let xy be the biggest segment contained in q_kq_k+1 such that v ∈ xy and xy - {x,y} ⊂ int(P) and consider the sides p_i p_i+1 and p_j p_j+1 such that x ∈ p_i p_i+1 and y ∈ p_j p_j+1. Items (a) and (b) imply that xy ⊂ uw. Item (c) shows that ║x -y║ < ║u - w ║ ∈ Δ(P). Since x ∈ p_ip_i+1 and y ∈ p_j p_j+1, the definition of Δ implies that p_i p_i+1∩ p_j p_j+1≠ Ø.

Let p ∈ {p_j,p_j+1} be the common vertex of p_i p_i+1 and p_j p_j+1. According to Figure 3, p belongs to the disk D = { z ∈ ² with ║2 z - (x + y) ║ < ║x - y║ }. If L and R are the open half planes on the left and right sides of the line q_k q_k+1 then, in principle, we have the two possibilities regarding p described in Figure 4. However, the situation on the left side of this figure does not occur, because D ∩L ⊂ (Q) and there is no vertex of P in int(Q) by hypothesis.In fact, notice that D ∩ L cannot intersect sides of Q which are nonconsecutive to q_k q_k+1 because (c) implies that the distances from such sides to q_k q_k+1is bigger than the radius of D. Moreover, D ∩ L does not intersect the sides q_k-1 q_k and q_k+1 q_k+2 either, because the internal angles of Q are at least π/2. This implies that D ∩ L is contained in the interior of Q and, thus, p ∉ D ∩ L as we claimed. Therefore, we must have p = p_j = p_i+1∈ D ∩ R as described in the right side of Figure 4 and the bounds in (2) holds because the diameter of D is at most ║u - w║. (For further reference, as p_i+1≡ p_j, we will referto p_ip_i+1 as p_j-1p_j.)

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Lemma 2. Let P, Q and {x} = q_kq_k+1∩ p_j-1 p_j as in Lemma 1. Assume that x ≠ q_k and let c ∈ border(P) ∩ border(Q) be the first point different from x encountered when walking x p_j-1 from x to p_j-1 (see Fig. 5)¹. Then, either

or

Proof of Lemma 2. Let q_m q_m+1 be the side of Q containing c. We will first show that q_m q_m+1≠ q_k q_k+1 and q_mq_m+1≠ q_k+1 q_k+2. Since x ∈ q_k q_k+1 -{ q_k}, c ∈ q_m q_m+1 and Ø ≠ xc -{x,c} ⊂ (Q) we have that x and c are in different sides of Q and, in consequence, q_m q_m+1≠ q_k q_k+1. Considering the triangle with vertices xp_jy (right hand side of Fig. 4) and knowing that xp_jy = p_j-1p_jp_j+1 > π /2 we have that q_k x c ≅ p_j x y < p/2. Then q_k x c < π/2 implies that q_k+1 xc > π /2. This and q_k q_k+1q_k+2> π/2 imply that the half lines x + γ₁ (c - x) and q_k+1+ γ₂ (q_k+2- q_k+1), γ₁,γ₂> 0, never intersect. Thus, c ∉ q_k+1q_k+2 and q_mq_m+1≠ q_k+1 q_k+2.

Therefore, we have only these two possibilities regarding q_mq_m+1:

• q_m q_m+1 = q_k-1q_k. In this case ║x - c║ > max{║c - q_k║, ║q_k - x║} because xc is the biggest side in the (nonempty) triangle with vertices x q_k c, since the angle x q_k c is equal to q_k+1q_k q_k-1, which is at least π/2.

• q_m q_m+1≠ q_k-1q_k. In this case, by definition of Δ, ║x - c║ > δ for all δ ∈ D(Q). This implies that ║x - c║ > sup Δ(Q).

Clearly, a symmetric result can be obtained for {y} = q_k-1q_k ∩ p_jp_j+1 defined in Lemma 1.

Part of the hypothesis of the main theorem in this section requires that each polygon in P has an internal sentinel outside the forbidden region F(P,δ) that we now describe. Suppose P is a polygon and δ > 0 satisfies δ < ║p_i+1 - p_i║ for all i. For each vertex p_i of P we define the forbidden corner FC(P,i,δ ) as FC(P,i,δ) = { x = λu + (1 -λ)v for λ ∈ (0,1), u ∈ p_i-1p_i, v ∈ p_ip_i+1, uv - {u,v}⊂ int(P) and ║u - v║ < δ}

(see Fig. 6) and we define the forbidden region

The next lemma shows that we can always find internal sentinels outside the forbidden regions for any polygon with n > 3 vertices.

Lemma 3. If P is a polygon with n > 3 vertices and δ ∈ Δ(P) then the set int(P) - F(P,δ) is not empty.

Proof of Lemma 3. We analyze convex and non-convex polygons separately.

Let us start with P convex. For each i = 0,...,n-1 let P_i be the convex polygon with vertices {p₀,...,p_n-1} - {p_i}. Note that

P_i∩ P_j∩ P_k⊃ {p₀,...,p_n-1} - {p_i, p_j,p_k} ≠ Ø

for every triplet i, j, k. Therefore, Radon's theorem implies that

The set A does not contain vertices or sides of P. Thus, A ⊂ int(P). Moreover, the interior of the ears of P do not intersect A. It is clear then that int(P) - F(P,δ) ⊃ A ≠ Ø and we are done with the convex case.

If P is not convex then there exists at least one i such that the anglep_i-1p_i p_{i + 1} is bigger than π. Let ε ∈ (0, δ/4) be such that

belongs to int(P). Geometrically (see Fig. 7), s is a point in the bisectrix of the angle p_i-1p_i p_{i + 1} very close to p_i:

To complete this proof we will show that s ∈ int(P) -F(P, δ ), i.e., it does not belong to any forbidden corner FC(P,j,δ) with p_j-1p_j p_j+1 < π. Consider a forbidden corner FC(P,j,δ) as above. Let u ∈ p_j-1 p_j and v ∈ p_j p_j+1 be such that s ∈ uv, uv - {u,v} ∈ int(P). To prove that s ∉ FC(P,j,δ) is enough to show that ║u - v║ > δ. Noticing that p_j-1 p_jp_j+1 < π < p_i-1p_i p_i+1 implies i ≠ j, we are left with the three cases:

(i) j = i + 1 (described in Fig. 7a),

(ii) j = i - 1 (described, after reflection, in Fig. 7a),

(iii)j ∉ {i - 1, i, i + 1} (described in Fig. 7b).

Cases (i) and (ii) differ only by a reflection and we will threat both as case (i).

Let us start with case (i). Since P has more than three vertices, we have that p_i-1p_i and p_i+1p_i+2 are not consecutive. Thus, by definition of Δ(P) we have that ║p_i - v║ > δ. Now,

v p_i u > s p_i u = p_i-1p_i p_i+1 / 2 > π/ 2.

This implies that ║u - v║ > ║p_i - v║ > δ and we are done with cases (i) and (ii).

Finally, let us handle case (iii). In this case the sides p_ip_i+1 and p_j p_{j +1} are not consecutive. Thus, by definition of δ, ║p_i - v║ > δ. As a consequence,

Analogously, since the sides p_i-1p_i and p_j-1p_j are not consecutive we get that ║p_i - u ║ > δ and then

Combining (6) and (7) we get

║ u - v║ = ║u - s║ + ║ s - v ║ > δ/2 + δ/ 2 = δ

and the proof is complete.

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Now we can address the main question of this section: Given a family P of polygons and a function S: P → Sub(²), give sufficient conditions for S to be a sentinel assignment for P with respect to I₂.

Theorem 1. Let P be a family of polygons such that, for each P ∈ P, α(P) > π/2. Let S be a function from P to Sub(²). Assume that for all P ∈ P

(a) S(P) contains the vertices of P;

and that there exists δ ∈ ∩_P∈PΔ(P) for which

(b) for each side p_i p_i+1 of P there exist n_i∈ and s_j≡ p_i + (γ_j / Γ) (p_i+1-p_i) ∈ S(P), where Γ = ║p_i+1-p_i║, such that γ_j∈ [0,Γ] for j = 1,...,n_i, if n_i ≠ 0 then γ₁ = δ and γn_i = Γ- δ, γ_j <γ_j+1 and | γ_j - γ_j+1 | < δ for j = 1,...,n_i-1;

(c) he set S(P) contains a point in int(P) -F(P, δ).

Then S is a sentinel assignment for P with respect to I₂.

Proof of Theorem 1. Consider members A and B of P and T, U ∈ I₂. Define P = T(A), Q = U(B), (P) = T(S(A)) and (Q) = U(S(B)). Since T and U are isometries, hypotheses (a), (b) and (c) also apply to P and (P) and Q and (Q). To prove Theorem 1, we assume that there exists a point z ∈ (P) ∩ int(Q) and show that either

Hypothesis (c) implies that there exists s ∈ (P) ∩ (int(P) - F(P,δ)). If s ∈ int(Q) then the first condition in (8) is satisfied and we are done. Thus, we assume that s∉ int(Q). Since s and z are in the interior of the Jordan curve P, there exists a continuous path contained in the interior of P connecting s to z. This path intersects border(Q) at some point v because s ∉ int(Q) and z ∈ int(Q). If v ∈ (Q) then the second condition in (8) is satisfied and we are done again. Thus, we only need to care about the case in which v ∉ (Q) and v is the first intersection of the path above and border(Q), that is, there exists a continuous function Φ:[0,1] → ² such that

If q_k q_k+1 is the side of Q that contains v, then by hypotheses (a) and (b) thereexist u,w ∈ (Q) ∩q_k q_k+1 such that v ∈ uw, ║u- w║ < δ and ║ u - q_k ║ < ║ w - q_k ║. If u or w belong to int(P) then the second condition in (8) is satisfied and we are done. To complete this proof, let us now derive a contradiction from the assumption that hypothesis (c) holds and neither u nor w belong to int(P). In this case Lemma 1 implies that there exist p_j-1, p_j and p_j+1, vertices of P, and {x} = uv ∩p_j-1p_j, {y} = vw ∩p_j p_j+1 and xy - {x,y} ⊂ int(P) like the ones in Figure 8.

This figure is also accurate regarding the fact that s is in the interior of the triangle τ = x p_j y. In fact, (9) implies that the path Φ([0,1]) does not cross x p_j nor p_j y and Φ([0,1)) does not touch yx. Moreover, if ε is small Φ(1 - ε) is close to v and outside Q. This implies that Φ(1 - ε) ∈ int(τ) for e small. Since Φ([0,1)] does not touch border(τ) we have that Φ([0,1)) ⊂ int(τ). In particular, s = Φ(0) ∈ int(τ). However, equation (2) shows that ║x -p_j║ < δ and ║p_j - y║ < δ. This implies that int(τ) ⊂ FC(P,j,δ) and we deduce that s ∈ F(P,δ). This conclusion contradicts our choice of s at the beginningof this proof.

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Theorem 1 implies that if P is a finite family of polygons P with α(P) > π/2 then there exists a sentinel assignment S for P with respect to I₂ such that S(P) is finite for all P ∈ P: take an arbitrary choice of δ ∈ ∩ _P∈P Δ(P), for example,

and for each P ∈ P choose a finite set S(P) of sentinels that satisfy the conditions (a)-(c) in Theorem 1. Lemma 3 guarantees that it is possible to find a point satisfying condition (c) of Theorem 1.

Using Theorem 1 you can prove the following corollary to justify our claim regarding the intersections of rotated and translated copies of the pentagon and the hexagon in the first page of this paper:

Corollary 1. Let δ > 0 and let P be a family of convex polygons such that, for all P ∈ P we have that α(P) > π/2 and all sides of P have length δ. For P ∈ P take s ∈ int(P) - F(P,δ) and define S(P) = {s,p₁,p₂,...,p_n(P)}. The function S is a sentinel assignment for P regarding I₂. (S satisfies hypotheses (a) and (c) by definition and hypothesis (b) with n_i = 0.)

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We have seen that Theorem 1 can be used to construct sentinel assignments for certain families of polygons or to verify that a given assignment S that satisfies the hypotheses of Theorem 1 is a sentinel assignment. However, there is a simple and important case that is not covered by Theorem 1. Consider a family P of identical squares P of side d. In this case we have that Δ(P) = (0,d) for all P ∈ P. Let δ = Consider a sentinel assignment S for P such that, for each P ∈ P, S(P) is given by a sentinel in each vertex of P, a sentinel in the middle of each side of P and a sentinel in int(P)- F(P,δ). This S satisfies hypotheses (a) and (b) but does not satisfy hypothesis (c) of Theorem 1. Therefore, Theorem 1 can not be used to certify that S is, in fact, a sentinel assignment for P.The theorem below is similar to Theorem 1 except for the fact that hypothesis (c) is replaced by a couple of other hypotheses satisfied by the sentinel assignment described above.

Theorem 2.

Let P be a family of polygons such that, for all P ∈ P, α(P) > π/2. Let S be a function from P to Sub(²). Assume that for all P ∈ P

(a) >S(P) contains the vertices of P;

and that there exists δ ∈ ∩_P∈PΔ(P) that satisfies

(b) for each side p_i p_i+1 of P there exist n_i∈ and s_j≡ p_i + (γ_j / Γ) (p_i+1-p_i) ∈ S(P), where Γ = ║p_i+1-p_i║, such that γ_j∈ [0,Γ] for j = 1,...,n_i, if n_i ≠ 0 then γ₁ = δ and γn_i = Γ- δ, γ_j <γ_j+1 and | γ_j - γ_j+1 | < δ for j = 1,...,n_i-1;

(c) S(P) ∩ int(P) ≠ Ø; and

(d) ║p_i+1 - p_i║ > 2δ for all i ∈ .

Then S is a sentinel assignment for P with respect to I₂.

Proof of Theorem 2. This proof follows very closely the proof of Theorem 1. Thus, we will assume that there exists a point z ∈ int(P) ∩ int(Q) and show that either

By hypothesis (c), there exists s ∈ (P) ∩ int(P) and let v ∈ int(P) be the first point that belongs to border(Q) encountered when walking the continuous path from s to z trought the interior of P. If s ∈ int(Q) or if v ∈ (Q) then (10) holds and we are done. Therefore, we only need to care about the case in which there exists a continuous function Φ:[0,1] → ² such that

Let q_k q_k+1 be the side of Q that contains v. Then, hypotheses (a) and (b) imply that there exist u,w ∈ (Q) ∩ q_k q_k+1 such that v ∈ uw, ║u- w║ < δ and ║ u - q_k ║ < ║ w - q_k║. If (Q) ∩ int(P) ≠ Ø we are done. So, we assume that u,w ∉ int(P). In this case Lemma 1 implies that there exist p_j-1, p_j and p_j+1, vertices of P, and {x} = uv ∩p_j-1p_j and {y} = vw ∩p_j p_j+1 like the ones in right hand side of Figure 4. Hypothesis (d) and the fact that ║u - w║ < δ imply that either u ≠ q_k or w ≠ q_k+1. By symmetry, assume that u ≠ q_k which implies that x ≠ q_k (in fact, it implies that ║x - q_k║ > δ). In this case, by Lemma 2, there exists c ∈ border(P) ∩border(Q), the first point different from x encountered when walking x p_j-1 from x to p_j-1 (see Fig. 5), such that either (3) or (4) holds. By (3) and the fact that ║x - q_k║ > δ or by (4) and the fact that δ ∈ Δ(Q), we have that ║x - c║ > δ.So, we have xc ⊂ int(Q) ∩p_j-1 p_j and ║x - c║ > δ. By hypothesis (b) there exists t ∈ xc ∩ (P) and, thus, we have verified (10).

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3 Sentinels for rectangles

In this section we present optimal sentinel assignments for rectangles with respect to . We assume that a = a(R) = ║r₀ - r₁║ is the longest side of the rectangle R and b = b(R) = ║r₀ - r₃║ is the smallest side. We define sentinel assignments that depend on two parameters δ and ρ: given a rectangle R we define S_δρ(R) as the set of points indicated by circles in Figure 9.

Formally, given a rectangle R and δ and ρ with 0 < δ, ρ < ║r₃ - r₀║ we define S_δρ(R) as the set formed by:

1. The vertices r₀, r₁, r₂, r₃ of R.

2. The points s_i = r_i + δ(r_{i + 1} - r_i)/║r_i+1 -r_i║ for i = 0,1,2 and 3.

3. If b > 2δ, the points x_i= s₃+ i (r₀ - s₃) /n_γ and y_i = s₁ + i (r₂ - s₁) / n_γfor 1 < i < n_γ ,with n_γ = ⌈(b - δ) / δ⌉ .

4. If a > 2δ, the points t_i = ρ(r₃ - r₀)/b + (s₀ -r₀) + i Δ/ n_τ, for 1 < i < n_τ, with Δ = r₁ - r₀- 2 (s₀ - r₀) and n_τ= é⌈║Δ║ /δ⌉.

5. If a < 2 δ, the point s = (r₀ + r₁ + r₂ + r₃)/4.

The following theorem is the main result in this section:

Theorem 3. Let R be a family of rectangles and suppose δ ∈ ∩_R∈RΔ(R). If for all _R∈R we have ρ(R) ∈ (0,b(R)) such that

then the function S: R → Sub(²) given by S(R) = S_δρ(R)(R) is a sentinel assignment for R with respect to .

If all rectangles in R∈R have δ < b(R) < 2δ then the sets S_δρ(R)(R) contain only one element in each smaller side of R and the sentinel assignment S above in Theorem 3 is optimal, i.e., if S' is another sentinel assignment for R with respect to then S'(R) has at least as many elements as S(R). In fact, S'(R) must contain the vertices of R and at least one element in each side of R. Moreover, if S'(R) does not contain as many elements as S(R) then it is possible to superimpose a copy of R rotated by π/2 and R in order to contradict the definition of sentinel assignment.

We end this paper with the proof of Theorem 3:

Proof of Theorem 3. The sets S_δρ(R) and the condition (12) are invariant under , in the sense that if T ∈ and R ∈ R then S_δρ(T(R)) = T( S_dr(R)) and b(R) = b(T(R)). Therefore, to prove Theorem 3 it is enough to show that if H and R are rectangles with ║h₀ - h₃║ > ║h₁ - h₂║, ║r₀ - r₃║ > ║r₁ - r₂║ and

and (H) ∩ int(R) ≠ Ø then

If a(H) < 2δ and a(R) < 2δ then (14) is a consequence of Theorem 1 applied to P = {H,R}. Thus, we can assume that a(H) = ║h₁ - h₀║ > 2δ. Moreover, if s_i are the sentinels described in Figure 9 for H and {h₀,h₁,h₂,h₃, s₀,s₁,s₂,s₃}∩ int(R) ≠ Ø then (14) is satisfied. Therefore, we only need to analyze the case

We leave to the reader, the verification of the fact that if h₀ h₁ and r₀ r₁ are parallel then (14) is satisfied². From now on we analyze the case in which H is horizontal and R is a rotated rectangle as in Figure 10 or a rotated rectangle with a(R) < 2 δ.

To avoid conflicting names, we rename R's sentinels as in Figure 10: the s sentinels for R are called z's and the t sentinels are called w. Notice that in all polygons R in Figure 9 if r_i is a vertex of R then there exists a sentinel z_i at a distance δ from r_i in the counterclockwise direction along R. Repeating the argument in the second paragraph of the proof of Theorem 1 with u = h₀, w = s₀and q_k+1 = h₁, we deduce that if int(R) ∩h₀s₀ ≠ Ø then {z₀,z₁,z₂,z₃} ∩int(H) ≠ Ø and (14) is satisfied. Therefore, from now on we assume that int(R) ∩ h₀ s₀= Ø. If R ∩ (h₀ s₀- {h₀,s₀}) ≠ Ø then we have only two possibilities:

• h₀ s₀ and r₁ r₂ are parallel. In this case h₀ s₀ must be contained in the line r₁ r₂ and you can check that either (a) s₁ ∈ int(R), (b) z₂∈ int(H) or (c) s₁ = z₂ and t₁∈ int(R).

• h₀ s₀and r₁ r₂ are not parallel. In this case, since we are assuming that r₀r₁and h₀ h₁ are not parallel and that int(R) ∩ h₀ s₀ = δ, there is a vertex r_i ∈ h₀s₀ - {h₀,s₀} and (15) implies that z_i ∈ int(H).

Thus, if R ∩ (h₀ s₀ - {h₀,s₀}) ≠ Ø then (14) is satisfied and we can assume that

Now let us analyze the intersections of R with the side h₃ h₀ of H. Since this side is populated with the sentinels h₃, s₃, x_i and h₀, which are at most δ apart, Lemma 1 yields that if int(R) ∩ h₃ h₀ ≠ Ø then we have a situation like the one in the right side of Figure 4, with q_k = h₃, q_k+1 = h₀ and q_k+2 = h₁ and p_j as one vertex r_i of R. We claim that in this case z_i ∈ int(H). In fact, since ║z_i - p_i║ = δ is at least as big as the diameter of D we have that z_i ∉ r_iy. Let δ be the point defined in items (iv)-(vi) of Lemma 1. We have this three possibilities:

• d satisfies (iv): This case case contradicts (16) and need not to be considered.

• d satisfies (v). In this case we have that

  and ║r_i - d ║ >δ. This inequality combined with z_i ∉ z_i y and ║z_i - r_i║ = δ implies that z_i ∈ yd -{y,d} ⊂ int(H). Thus, z_i ∈ int(H) in case (v).

• d satisfies (vi): In this case d ∈ h₀ h₁ ∩ R and since R ∩ (h₀ s₀ - {h₀,s₀}) = Ø we must have d ∈ s₀ h₁. This implies that ║d - h₀║ > δ and, since q_k+1 = h₀ in our context, (iv) implies that ║y - d║ > δ. We can then use the same argument following (17) and conclude that z_i ∈ int(H).

In summary, we have shown that if int(R) ∩ h₃ h₀ ≠ Ø then (14) holds. Let us then assume that int(R) ∩ h₃ h₀ = Ø. Using this assumption and (16) it is easy to show that if R ∩ (h₃ h₀ -{h₃,h₀}) ≠ Ø then there exists r_i ∈ (h₃ h₀ -{h₃,h₀}) and z_i ∈ int(H). Thus, we can assume that R ∩ (h₃ h₀ - {h₃,h₀}) ≠ Ø. This assumption, the fact that int(H) ∩ int(R) ≠ Ø and (16) imply that R ∩ (h₃ h₀ ∩ h₀ s₀ - {h₃,s₀}) = Ø. Using symmetry we can resume our conclusions up to this point as the statement that either

or (14) is satisfied and we assume (18) from now on.

Equation (18) shows that if {r₀,r₁,r₂,r₃} ∩ int(H) = Ø and int(H) ∩ int(R) ≠ Ø then one of the sides r₀ r₁ or r₂r₃ crosses both segments s₀h₁ and s₂h₃,at points e and f, say. Since ║f - e║ > δ we conclude that if R is "short", i.e., if b(R) < 2 δ, then {z₀,s₂} ∩ int(H) ≠ Ø. Therefore, we can assume that

b(R) > 2 δ and {r₀,r₁,r₂,r₃,z₀,z₂} ∩ int(H) = Ø

and that z₀ is on or above the line h₂h₃ and z₂ is on or below the line h₀h₁ (These details are illustrated in Fig. 10.) We then have our three final possibilities:

(i) If r₀ r₁ ∩(s'₀ h₃ - {s'₀}) ≠ Ø then R must be inclined to the left as illustrated in Figure 10, because r₀ r₁ ∩ h₀ s₀ - {s₀} = Ø. As a result, the point 0 is above h₂h₃ and 0 is below h₀ h₁ and if ef = 01∩ H then ║e - f║ > δ. Since the segment ef - {e,f} is populated with sentinels w_i which are less δ apart and ║₀- w₁║ < δ and ║₂ - wn_w ║ < δ at least one w_i belongs to int(H) and (14) is satisfied.

(ii) The case r₂ r₃ ∩(s'₂ h₁ - {s'₂}) ≠ Ø is symmetric to item (i).

(iii) If r₀ r₁ ∩ (s'₀ h₃ - {s'₀}) = Ø and r₂ r₃ ∩ (s'₂ h₁ - {s'₂}) = Ø then we are in the situation described on the right of Figure 10. In this case we can exchange H and R and repeat the argument in item (i).

□

4 Concluding remarks

We defined the concept of sentinels and found finite sentinel assignments for finite families of polygons with internal angles bigger than or equal to π/2. We presented optimal assignments for some families of rectangles. It would be interesting to characterize optimal sentinel assignments for more general families of polygons. For instance, the sentinel assignments for the rectangles suggest that by locating sentinels along well chosen lines in the interior of the polygons it is possible to produce smaller sentinel assignments.

The concept of sentinel provides a new approach, based on nonlinear programming, for solving a large variety of packing problems to optimality. Defining sentinels for a given set of polygons may be a hard task. Modelling the problem of finding the sentinels set as a mathematical programming problem will be the subject of future research. It may allow the straightforward application of sentinels-based packing techniques on real applications.

5 Acknowledgments

We would like to thank Professor I. Bárány for his suggestions and for encouraging us to improve our initial work on sentinels and obtain the proofs we present in this work.

Received:02/XI/09

Accepted: 19/I/10

#CAM-144/09.

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1

Note that

p_j-1p_j ∩

uv =

p_j-1p_j ∩

q_kq_k+1 = {

x} ≠

q_k implies that the segments

q_kq_k+1 and

p_j-1p_j are not colinear and the point

c is well defined.

2

when doing that, do not forget hypothesis (12).

Publication Dates

• Publication in this collection
  22 July 2010
• Date of issue
  June 2010

History

• Received
  02 June 2009
• Accepted
  19 Jan 2010