ISSN 1063-4541, Vestnik St. Petersburg University. Mathematics, 2016, Vol. 49, No. 4, pp. 334–339. © Allerton Press, Inc., 2016. Original Russian Text © V.S. Kalnitsky, 2016, published in Vestnik Sankt-Peterburgskogo Universiteta. Seriya 1. Matematika, Mekhanika, Astronomiya, 2016, No. 4, pp. 595–602.
MATHEMATICS
Orthogonal Triangulation of Polygons V. S. Kalnitsky St. Petersburg State University, Universitetskaya nab. 7–9, St. Petersburg, 199034 Russia *e-mail:
[email protected] Received April 23, 2016
Abstract—In this study, the question whether any convex polygon can be divided using an orthogonal grid into right-angled triangles is answered in the negative. Moreover, it is demonstrated that there exists a convex pentagon which cannot be even approximated by divisible pentagons. Keywords: division of polygonal domains, rectangular grid, triangulation of plane domains. DOI: 10.3103/S1063454116040075
1. INTRODUCTION The operation of dividing polygons occurs in various applied problems, for example, computer graphics and automated design. One of the objectives of division is to replace some, as a rule convex, polygonal domain by fragments for which test and numerical problems are executed in a maximally simple and unified way. Most known algorithms operate with triangles. The question whether it is always possible to assume that a polygon belongs to some rectangular grid and is comprised of rectangles and right-angled triangles of this grid was probably put forward for the first time by N.Yu. Dodonov in a series of papers [1–3] on the reconstruction of parametric surfaces defined in a tabulated form. In spite of the fact that the answer to this question is positive for triangles (obviously) and convex quadrangles, we prove here that in a seven-dimensional space of convex pentagons the set of indivisible pentagons has a nonzero interior. Thus, the possibility of approximating an indivisible polygon by divisible ones is denied. The space of pentagons can be described as follows: the set of ordered five-point subsets of the Euclidean plane defining convex pentagons with the given counter-clockwise side tracing is parameterized to the plane motion by the set of 10 coordinates in which the first point has the coordinates (0, 0), and the second has the form (x, 0); i.e., it has seven free parameters defining the local coordinate system. In spite of the simplicity of formulation, we did not find the proof of this fact or even the formulation of the problem under discussion in earlier publications than those indicated above. For practical purposes we formulate the calculable geometrical conditions which guarantee that a convex pentagonal domain is divisible and can be approximated by divisible domains. 2. SUFFICIENT DIVISIBILITY CONDITION Let us consider a finite rectangular grid on a plane, i.e., two finite sets of straight lines perpendicular to each other. Definition 1. We call a polygon in the plane right-angle triangulated or divided if all its vertices lie in the grid nodes, all sides lie on the straight lines of the sets or diagonals of the grid cells. The polygon is divisible if there exists its rectangular triangulation by some finite grid. Successive construction of the grid can be described as tracing the trajectory emerging from the vertex which rotates by the right angle when it meets the polygon side. The rotation stops when it meets either a vertex or a side perpendicular to the direction of motion. Two mutually perpendicular directions of motion are preset. Let us consider the convex obtuse-angled pentagon ABCDE. Let us construct five perpendiculars from its vertices (Fig. 1). We require that D' ∈ AB. Let us denote the lengths of the perpendiculars as follows: a = AA’, b = BB’, c = CC’, d = DD’, e = EE’. 334
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D A'
B'
E C
A
E'
D'
B
C'
Fig. 1. Divisibility condition.
Theorem 1. If the inequality (c – e)(b – a) > 0 is satisfied, then the pentagon can be divided along the side AB. Proof. Let us assume for definiteness that b > a, c > e. It follows from this assumption that a trajectory emerging from D meets AB at D', and a trajectory emerging from B meets AB after meeting ED above A'. Let us study in turn the behavior of the trajectories emerging from the vertices C, A, E. (1) The trajectory emerging from C and E. If c > a, then the trajectory emerging from C meets ED above A' and ends after meeting AB. It is necessary to study the only variant a > c > e. Let us consider an arbitrary point C1 ∈ [CB’] on DC, and C1' , its projection onto AB. The position of C1 on the side is determined by the difference of the heights x = C1C1' – CC’. Of interest are the points for which x < a – c. Such a point after successively meeting the segments EA’, EA, and BC meets CB’ at the point C2 with the height difference ϕ(x) = C2C 2' – CC’. Let us calculate the function
ϕ( x) = μ x + λ, where
(b − c)e (b − c)(c − e)a > 0, λ = > 0. (a − e)c (a − e)c If ϕ(x) < a – c, we repeat the tracing operation. If the inequality is transformed into the equality or changes the meaning, this trajectory after meeting ED reaches AB (end of construction). After n steps we have μ=
ϕ ( x) = μ x + n
n
1−μ λ. 1−μ n
Here, the index n in ϕn(x) denotes composition. If μ ≥ 1, then the sequence {ϕn(x)} increases unrestrictedly; therefore, at the final step it is larger than (a – c). If μ < 1, we have
ϕ n( x) ⎯⎯ → λ . n→∞ 1−μ For proving the finiteness of the trajectory it is sufficient to demonstrate that λ/(1 – μ) > a – c. Let us multiply both parts of the inequality by the denominator of the left-hand part and take into account the definition of the numbers λ, μ,
(b − c)(c − e)a (a − c)(ac − be) . > (a − e)c (a − e)c After canceling the positive denominator and collecting terms, we obtain the correct inequality (b − a)(a − e) > 0. VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
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Table 1 Side AB BC CD DE EA
a
b
c
e
(c − e)(b − a)
≈29.78 ≈29.21 ≈32.04 ≈27.52 ≈32.01
≈28.73 ≈30.30 ≈31.30 ≈27.97 ≈32.19
26.2 ≈21.23 ≈28.30 ≈21.82 ≈24.27
20.5 ≈26.75 ≈23.49 ≈23.45 ≈27.42
a, and then a small shift of C upward, keeping b > a, yields the inequality c > e. 2. EXISTENCE OF AN INDIVISIBLE PENTAGON In this section we prove the sufficient condition of indivisibility along a side. Theorem 2. If the inequality (c – e)(b – a) < 0 is satisfied, then the pentagon is indivisible along the side AB. Proof. Let us trace the dynamics of the segment BC (Fig. 2) based on the assumption that c > e and b < a. Otherwise, the side AE would be considered. After meeting the sides of the sample successively thrice, the image of the segment BC lies strictly inside it. Thus, we constructed a contracting mapping of the segment into itself. It is clear that neither the trajectory from the vertex B nor the trajectory from the vertex C can be finite. Q.E.D. Theorem 3. The convex obtuse-angled pentagon with the vertex coordinates {A(0; 0), B(27; 0), C(30, 6; 26, 2), D(11; 40), E(−10; 20, 5)} cannot be divided along its sides. Proof. Due to Theorem 2, it is sufficient to prove that each vertex is projected onto the opposite side, calculate the quantities a, b, c, and e for each side, and apply the criterion. Table 1 gives the corresponding values. According to the results of calculations (the calculations were performed using the GeoGebra code), this pentagon cannot be divided along its sides. Q.E.D. Let us prove a stronger proposition: there does not exist a direction along which this pentagon can be divided. For proving this fact we need an additional construction. 3. STRUCTURAL PORTRAIT Let us consider a polygon divisible into right-angled triangles, i.e., a polygon situated at the nodes of some finite grid. Each band between the neighboring vertical or horizontal straight lines of the grid can be extended (compressed) horizontally (or vertically), and the other part of the plane just shifted. Thus, a piecewise linear transformation of the plane changing the width of the grid band is formed. Using such transformations in a sequence, we can obtain the same width of the horizontal and vertical bands of the grid (square grid). Any triangulated polygon can be reduced to a polygon in the square grid, and vice versa. This means that a polygon can be triangulated if and only if it is an image of some polygon in a square grid under the action of some finite family of piecewise linear transformations of the above type. Definition 2. An image of a triangulated polygon upon transition to a square grid is called the structural portrait of the polygon. Let us list some elementary properties of these transformations: (1) the transformations preserve the vertical and horizontal directions: (2) the diagonals of the grid cells remain rectilinear; VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
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E A B D
C
Fig. 3. Structural portrait.
D
C E
A
B
Fig. 4. Behavior of trajectories.
(3) the segments’ growth and reduction are preserved. These properties, in particular, imply that upon transition to the structural portrait the neighboring sides with the same monotonicity sign merge into one segment on the diagonals of the grid squares. What is the structural portrait of the pentagon described in the previous case based on the assumption that it is divisible? Since the direction of division is not parallel to its sides, all sides of the pentagon are inclined with respect to this direction, and only two of the sides are inclined in the same direction. These two sides merge into one and the vertex between them lies on this side. This portrait in a square grid represents a rectangular tetragon at an angle of 45° to the grid direction (Fig. 3). Note an important property of the obtained portrait. If the trajectories emerging from A meet each other on CD, the trajectories from B and E first meet the vertices C and D. If the trajectories from A meet before meeting the side CD, then the trajectories from B and E first meet the sides ED and BC, respectively. Otherwise, both trajectories from B and E first meet the side DC. Thus, the behavior of these trajectories, as regards their intersection with the sides, is symmetric. Theorem 4. For a pentagon from Theorem 3 there does not exist a direction along which it can be divided. Proof. Let us assume that the pentagon is divisible and consider the vertex A as the unbending vertex in the structural portrait (Fig. 4). Since the angle between the chords EC and DB toward the side DC is obtuse, the trajectories orthogonal to each other emerging from the vertices E and B first meet DC; i.e., the direction of the division not always go between these chords. The property of the structural portrait implies that the trajectory rectangle resting on the angle A cannot have the vertex inside the pentagon or on the side DC. Let us prove that the trajectories emerging from each of the vertices in the admissible directions always meet inside the pentagon, which is impossible for the structural portrait. The proof will be based on the following fact of analytical geometry which can be easily established. Lemma 1. For any acute angle and point A inside this angle, any pair of orthogonal trajectories emerging from the point A in such a way that they meet different sides of the angle, and after meeting them intersect at the point lying on some hyperbola (Fig. 5). VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
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A
Fig. 5.
The proof is based on direct calculation. It follows from this lemma that if a straight line is drawn through the ends of any two rectangles, the vertices of all rectangles between them lie at one side of this line because of the fact that the half-hyperbola is concave. Let us prove the theorem. Since the limiting possible trajectory directions emerging from a vertex are the diagonals of the pentagon emerging from the other vertices or the directions of adjacent sides, it is sufficient to prove that for these directions the trajectories meet inside the pentagon. By drawing the line through these points, we obtain some intermediate direction for which the corresponding vertex is separated from the pentagon side by the constructed segment (to show the direction in which the hyperbola is concave). Let us consider the constructions from the vertex A. The limiting directions are BD and EC. The vertex of the rectangle directed along BD has the following approximate coordinates: (19.8827; 33.1469) and is situated inside the pentagon. The vertex of the rectangle directed along EC has the following coordinates: (23.8316; 30.2157) and is also situated inside the pentagon. The intermediate direction determined by the point with the coordinates (–5.8728; 24.3324) on the side ED yields the point (22.0533; 31.0725) separated from the side DC by the constructed segment. Let us present the results of the calculations for the other vertices in the form of a table where the columns contain the following data: the coordinates of the vertex and the result (inside/outside) for the first limiting direction, the coordinates of the vertex and the result (inside/outside) for the second direction, and the coordinates of the vertex and the result (separated/not separated from the side) for the intermediate direction. The limiting directions are: for the vertex B, EB, and BC; for the vertex C, AD, and EB; for the vertex D, ED, and DC; and for the vertex E, AC, and EA. The obtained results show that in all directions the possible structural portrait is not realized. Theorem 4 is proved. Q.E.D. Corollary 2. The pentagon from Theorem 3 cannot be approximated by divisible pentagons. This proposition follows from the fact that a small wiggling of the parameters preserves the inequalities in Theorem 3 and the positions of the vertices of the inscribed rectangles with respect to the sides. Thus, all pentagons sufficiently close to the defined one are indivisible. REFERENCES 1. N. Yu. Dodonov and A. V. Masalskikh, “Reconstruction of parametric surfaces specified with tabular data by means of shifts and dilations of one function,” Izv. Tul. Gos. Univ. Estestv. Nauki, No. 1-1, 232–248 (2014). 2. N. Yu. Dodonov and A. V. Masalskikh, “Applying a class of summatory type approximation aggregates for parametric surfaces reconstruction,” Izv. Tul. Gos. Univ. Estestv. Nauki, No. 2, 59–75 (2014). 3. N. Yu. Dodonov, “On uniform approximation in R2 of continuous Banach functions,” Vestn. S.-Peterb. Univ., Ser. 1: Mat., Mekh., Astron., No. 2, 11–24 (2005).
Translated by E. Baldina VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
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