In 1926, Jarník introduced the problem of drawing a convex n-gon with vertices having integer coordinates. He constructed such a drawing in the grid [1, c ·n 3/2]2 for some constant c > 0, and showed that this grid
size is optimal up to a constant factor. We consider the analogous problem of drawing the double circle, and prove that it can be done within the same grid size. Moreover, we give an O(n log n)-time algorithm to construct such a point set.Consejo Nacional de Ciencia y Tecnologia (México)Programa de Apoyo a Proyectos de Investigación e Innovación Tecnológica (Universidad Nacional Autónoma de México)Comisión Nacional de Investigación Científica y Tecnológica (Chile)Fondo Nacional de Desarrollo Científico y Tecnológico (Chile

Bereg, Sergey

Díaz Báñez, José Miguel (Coordinador)

Fabila Monroy, Ruy

Flores Peñaloza, David

Garijo Royo, Delia (Coordinador)

Lopez, Mario A.

Márquez Pérez, Alberto (Coordinador)

Pérez Lantero, Pablo

Urrutia Galicia, Jorge (Coordinador)

English

idUS. Depósito de Investigación Universidad de Sevilla

XV Spanish Meeting on Computational Geometry, June 26-28, 2013Drawing the double circle on a grid of minimum sizeS. Bereg∗1, R. Fabila-Monroy†2, D. Flores-Peñaloza‡3, M.A. Lopez§4, and P. Pérez-Lantero¶51Department of Computer Science, University of Texas at Dallas, USA.2CINVESTAV, Instituto Politécnico Nacional, Mexico.3Departamento de Matemáticas, Facultad de Ciencias, UNAM, Mexico.4Department of Computer Science, University of Denver, USA.5Escuela de Ingeniería Civil en Informática, Universidad de Valparaíso, Chile.AbstractIn 1926, Jarník introduced the problem of drawing aconvex n-gon with vertices having integer coordinates.He constructed such a drawing in the grid [1, c ·n3/2]2for some constant c > 0, and showed that this gridsize is optimal up to a constant factor. We considerthe analogous problem of drawing the double circle,and prove that it can be done within the same gridsize. Moreover, we give an O(n log n)-time algorithmto construct such a point set.1 IntroductionGiven n ≥ 3, a double circle is a set P ={p0, p1, . . . , pn−1, p′0, p′1, . . . , p′n−1} of 2n planar pointsin general position such that: (1) p0, p1, . . . , pn−1 areprecisely the vertices of the convex hull of P labelledin counterclokwise order around the boundary; (2)point p′i is close to the segment joining pi with pi+1;(3) the line passing through pi and p′i separates pi+1from P ; and (4) the line passing through p′i and pi+1separates pi from P (see Figure 1). Subindices aretaken modulo n. The double circle has been consid-ered in combinatorial geometry and it is conjecturedto have the least number of triangulations [1, 2].Drawing an n-vertex convex polygon with integervertices can be easily done by considering the n points(1, 1), (2, 4), (3, 9), . . . , (n, n2) as the vertices of thepolygon. In this case the size of the integer pointset is equal to n2 − 1 = Θ(n2), where size refers tothe smallest N such that the point set can be trans-lated to lie in the grid [0, N ]2. In 1926, Jarník [5]showed how to draw an n-vertex convex polygon with∗Email: besp@utdallas.edu.†Email: ruyfabila@math.cinvestav.edu.mx. Partially sup-ported by grant 153984 (CONACyT, Mexico).‡Email: dflorespenaloza@gmail.com. Partially supported bygrants 168277 (CONACyT, Mexico) and IA102513 (PAPIIT,UNAM, Mexico).§Email: mlopez@du.edu.¶Email: pablo.perez@uv.cl. Partially supported by grantCONICYT, FONDECYT/Iniciación 11110069 (Chile).p0p1p2p3 p4p5p′0p′1p′2p′3p′4p′5Figure 1: A double circle of twelve points.size N = O(n3/2) and proved that this bound is op-timal. In recent years the so-called Jarník polygonsand extensions of them have been studied [3, 6].Given any integer point (i, j), we say that (i, j)is visible (from the origin) if the interior of the linesegment joining the origin and (i, j) contains no lat-tice points. Observe that (i, j) is visible if and onlyif gcd(i, j) = 1, where gcd(i, j) denotes the great-est common divisor of i and j. We consider pointsas vectors as well, and vice versa. A Jarník poly-gon is formed by choosing a natural number Q, andtaking the set VQ of visible vectors (i, j) such thatmax{|i|, |j|} ≤ Q [4, 5, 6]. The polygon is thenthe unique (up to translation) convex polygon whoseedges, viewed as vectors, are precisely the elementsof VQ, that is, the vertices can be obtained by start-ing from an arbitrary point and adding the vectorsof VQ, one by one, in counterclockwise order, to thepreviously computed vertex (see Figure 2).V2Figure 2: A Jarník polygon (right) and its generatingvectors V2 (left).We study how to draw a 2n-point double circle withinteger points using the smallest size N . We present65Drawing the double circle on a grid of minimum sizeanO(n log n)-time algorithm that correctly constructsthe double circle with size within O(n3/2), where thatbound is also optimal. In Section 2 we show our al-gorithm, and in Section 3 its correctness is proved.Finally, in Section 4, we state future work.2 Double circle constructionObserve that a simple construction with quadraticsize is as follows: Consider the function f(x) = x2+x.For i = 1, . . . , 2n − 1, add the point (i, f(i)) if i isodd, and the point (i, f(i) + 2) otherwise. The finalpoint is (n, f(2n−1)+f(1)2 − 1) = (n, 2n2 − n), i.e., thepoint just below the midpoint of the segment con-necting (1, f(1)) and (2n− 1, f(2n− 1)). The size ofthe resulting point set is N = f(2n − 1) − f(1) =(2n− 1)2 + (2n− 1)− 2 = 4n2 − 2n− 2 = Θ(n2).We say that a sequence V of vectors is symmetricif V contains an even number of vectors sorted coun-terclockwise around the origin, and for every vector ain V its opposite vector −a is also in V . Observe thatany sequence of vectors defining a Jarník polygon issymmetric. For any sequence V = [v1, v2, . . . , v2t] of2t vectors let the point set P(V ) := {p1, p2, . . . , p2t},where p1 = v1 and pi = pi−1 + vi for i = 2, . . . , 2t.Note that if we sort the elements of VQ around theorigin then the elements of P(VQ) are the vertices ofthe Jarník polygon. Furthermore, if V is symmetricthen the elements of P(V ) are in convex position. Letsequence alt(V ) := [v2, v1, v4, v3, . . . , v2t, v2t−1] (seeFigure 3 for an example with t = 8). For any scalarλ let the sequence λV := [λv1, λv2, . . . , λv2t].Figure 3: P(alt(V4)).The idea is to generate a suitable symmet-ric sequence V of 2n vectors and then buildthe point set P(alt(V )) as the double cir-cle point set, up to some transformation ofthe elements of alt(V ). A (not optimal) ex-ample is V = [(1, 1), (1, 2), . . . , (1, n), (−1,−1),(−2,−2), . . . , (−1, n)] for even n ≥ 4. The point setP(alt(V )) is in fact a double circle but its size is equalto 1 + 2 + . . .+ n = Θ(n2) (see Figure 4).The construction in which the resulting point set isa double circle of size O(n3/2) is based on the nextFigure 4: A naive construction for n = 4 showing bothvectors (left) and the resulting point set (right).two algorithms:VisibleVectors(n): With input n ≥ 3, the sym-metric sequence V of 2n visible vectors, sorted coun-terclockwise around the origin, is generated so as tosatisfy the next two invariants. Let Bt := {p ∈ Z2 :‖p‖1 ≤ t}, k := maxv∈V ‖v‖1, and (even) s be thenumber of visible vectors of Bk−1: (i) all visible vec-tors of Bk−1 are in V , and (ii) the other elements of Vare generated as follows, until 2n− s elements are ob-tained: for i = 1, . . . , k− 1 generate vectors (i, k− i),(−i,−(k− i)), (−i, k− i), (i,−(k− i)) in this order, ifand only if gcd(i, k − i) = 1. Refer to Algorithm 2.1for a pseudo-code.BuildDoubleCircle(n): With input n ≥ 3,build a 2n-point double circle. First, set sequencesV :=VisibleVectors(n) and [v′1, v′2, . . . , v′2n] :=alt(V ). Then, the sequence W = [w1, w2, . . . , w2n] of2n vectors is created as follows: for i = 1, 3, . . . , 2n−1set wi = (1−λ)v′i+λv′i+1 and wi+1 = λv′i+(1−λ)v′i+1,where λ = 1/3. Finally, build the 2n-point setP((1/λ)W ) as the double circle.Algorithm 2.1: VisibleVectors(n)k ← 1, V ← [(1, 0), (−1, 0), (0, 1), (0,−1)]repeatk ← k + 1for i← 1 to k − 1doj ← k − iif gcd(i, j) = 1thenif length(V ) < 2nthen V ← V + [(i, j), (−i,−j)]if length(V ) < 2nthen V ← V + [(−i, j), (i,−j)]until length(V ) = 2nSort V counterclockwise around originreturn (V )3 Construction correctnessLet V = [v1, v2, . . . , v2n] be the (circular) sequence ofvectors obtained by executing VisibleVectors(n),66XV Spanish Meeting on Computational Geometry, June 26-28, 2013for n ≥ 3. For every i = 1, 3, 5, . . . , 2n−1 we say thatthe pair of vectors vi, vi+1 is a pair of alt(V ).Lemma 1 (Chapter 2 of [4]) Given a naturalnumber Q, the number |VQ| of vertices of the Jarníkpolygon is equal to4 + 4Q∑i=1Q∑j=11gcd(i,j)=1=24Q2π2+O(Q logQ).The size S(Q) of the Jarník polygon is equal to1 + 2Q∑i=1Q∑j=1igcd(i,j)=1= 1 + 2Q∑i=1Q∑j=1jgcd(i,j)=1=6Q3π2+O(Q2 logQ).Lemma 2 V is symmetric and point set P(V ) hassize O(n3/2).Proof. Observe that for every vector a in V , −a isalso in V since in algorithm VisibleVectors the vec-tors are added to sequence V in pairs, and each pairconsists of two opposite vectors. Then V becomessymmetric once the elements of V are sorted coun-terclockwise around the origin. On the other handVb k−12 c⊂ V ⊂ Vk, where k = maxv∈V ‖v‖1. Then wehave |Vb k−12 c| ≤ 2n ≤ |Vk|, which implies k = Θ(√n)by Lemma 1. By the same lemma we obtain:n∑i=1x(vi),n∑i=1y(vi) < 1 + 2k∑i=1k∑j=1igcd(i,j)=1= S(k)= Θ(k3) = Θ(n3/2)Hence, the size of P(V ) is O(n3/2). Let o denote the origin of coordinates. Given twopoints p, q let `(p, q) denote the line passing through pand q and directed from p to q, and pq denote the seg-ment joining p and q. Given three points p = (xp, yp),q = (xq, yq), and r = (xr, yr), let ∆(p, q, r) denotethe triangle with vertices at p, q, and r; A(p, q, r) de-note de area of ∆(p, q, r); and turn(p, q, r) denote theso-called geometric turn (going from p to r passingthrough q) whereturn(p, q, r) =∣∣∣∣∣∣xp yp 1xq yq 1xr yr 1∣∣∣∣∣∣and A(p, q, r) = 12 |turn(p, q, r)|. Extending this no-tation, let ∆(p, q) := ∆(o, p, q), A(p, q) := A(o, p, q),and turn(p, q) := turn(o, p, q). We use the so-calledPick’s theorem:Theorem 3 (Pick’s theorem [7]) The area of anysimple polygon H with lattice vertices is equal to i +b/2−1, where i and b are the numbers of lattice pointsin the interior and the boundary of H, respectively.Lemma 4 For every two consecutive vectors a1, a2 ofV we have A(a1, a2) = 1/2.Proof. Suppose ∆(a1, a2) contains a lattice point pdifferent from o, a1, and a2. Then p cannot belongto segments oa1 and oa2, and segment op containsa visible point q (possibly equal to p). If ‖q‖1 <max{‖a1‖1, ‖a2‖1} then q must belong to V by in-variant (i) of algorithm VisibleVectors. Other-wise, we have ‖q‖1 = max{‖a1‖1, ‖a2‖1}. Supposew.l.o.g. that ‖a1‖1 < ‖a2‖1, and let the point q′ de-note the intersection of `(o, q) with the segment s con-necting a1 to a2. Observe that q′ = δa1 + (1 − δ)a2for some δ ∈ (0, 1), and further that ‖q‖1 ≤ ‖q′‖1 =‖δa1 + (1− δ)a2‖1 ≤ δ‖a1‖1 + (1− δ)‖a2‖2 < ‖a2‖1,which is a contradiction. Then we must have that‖q‖1 = ‖a1‖1 = ‖a2‖1, which implies that q, a1, a2 be-long to a same quadrant since in this case q is at theinterior of the segment s. Therefore, q must belong toV by invariant (ii) of algorithm VisibleVectors. Inboth cases, the fact that q belongs to V contradictsthe fact that a1 and a2 are consecutive vectors of V .Hence A(a1, a2) = 1/2 by Pick’s theorem. Let λ ∈ (0, 1/2). Given a pair a, b of vectors leth(λ, a, b) := (1− λ)a+ λb (see Figure 5).abh(λ, a, b)h(λ, b, a)Figure 5: Two vectors a and b, and the vectors h(λ, a, b)and h(λ, b, a).Lemma 5 Let a1, a2, a3, a4 be four consecutive vec-tors of V such that a1, a2 and a3, a4 are pairs ofalt(V ). Let λ ∈ (0, 1/2), q1 = h(λ, a2, a1), q2 =q1 + h(λ, a1, a2), q3 = q2 + h(λ, a4, a3), and q4 =q3 + h(λ, a3, a4). Then q2 is to the right of `(o, q1)and both q3 and q4 are to the left of `(o, q1).Proof. (Refer to Figure 6.) We have turn(q1, q2) =turn(q1, q1 +h(λ, a1, a2)) = turn((1−λ)a2 +λa1, a1 +a2) = (1 − λ) turn(a2, a1) + λ turn(a1, a2) = 2(2λ −1)A(a1, a2) < 0, which implies that q2 is to the rightof the line `(o, q1). On the other hand:turn(q1, q3)= turn(h(λ, a2, a1), h(λ, a2, a1) + h(λ, a1, a2) +h(λ, a4, a3))= turn(h(λ, a2, a1), h(λ, a1, a2)) +turn(h(λ, a2, a1), h(λ, a4, a3))67Drawing the double circle on a grid of minimum size= turn((1− λ)a2 + λa1, (1− λ)a1 + λa2)) +turn((1− λ)a2 + λa1, (1− λ)a4 + λa3))= (1− λ)2 turn(a2, a1) + λ2 turn(a1, a2) +(1− λ)2 turn(a2, a4) + λ(1− λ) turn(a2, a3) +λ(1− λ) turn(a1, a4) + λ2 turn(a1, a3)= 2((2λ− 1)A(a1, a2) + (1− λ)2A(a2, a4) +λ(1− λ)A(a2, a3) + λ(1− λ)A(a1, a4) +λ2A(a1, a3))= 2(12(2λ− 1) + (1− λ)2A(a2, a4) + (1)λ(1− λ)A(a2, a3) + λ(1− λ)A(a1, a4) +λ2A(a1, a3))≥ (2λ− 1) + (1− λ)2 + λ(1− λ) + (2)λ(1− λ) + λ2= 2λ > 0where equation (1) follows from Lemma 4 and equa-tion (2) follows from the fact that by Pick’s the-orem the area of any non-empty triangle with lat-tice vertices is at least 1/2. Therefore, q3 isto the left of `(o, q1). Similarly, since we havethat turn(ai, aj) > 0 (i = 1, 2; j = 3, 4) thenturn(h(λ, a2, a1), h(λ, a3, a4)) > 0, which implies thatq4 is to the left of `(o, q1) given that q3 is to theleft of `(o, q1). By symmetry, it can be proved thatturn(q4, q3, q1) < 0 and turn(q4, q3, o) < 0, implyingthat both q1 and o are to the right of `(q4, q3). oq1q2q3q4`(o, q1)h(λ, a2, a1)h(λ, a1, a2)h(λ, a4, a3)h(λ, a3, a4)Figure 6: Proof of Lemma 5.Theorem 6 There is an O(n log n)-time algorithmthat for all n ≥ 3 builds a double circle of 2n pointsin the grid [0, N ]2 where N = O(n3/2).Proof. Execute the algorithm BuildDoubleCir-cle with input n, being V the result of callingVisiblePoints(n), building the point set P of 2npoints. Observe that λ = 1/3 implies that pointwi/λ = 3wi is integer for i = 1 . . . 2n, and thenall elements of P are integer points. By Lemma 5point set P is a double circle. The size of P(V ) isO(n3/2) by Lemma 2, and since all elements of P be-long to the polygon with vertices P(3V ) the size Nof P is also O(n3/2). Finally, translate P to lie in thegrid [0, N ]2. In algorithm VisiblePoints the timecomplexity is dominated by: (1) computing gcd(i, j)for O((√n)2) = O(n) pairs i, j; and (2) sorting vec-tors V counterclockwise around the origin. In Case(1) the time complexity is O(n log n) since gcd(i, j)consumes O(log(min{i, j})) = O(log√n) = O(log n)time. Case (2) consumes O(n log n) time as well.Since the time complexity of VisiblePoints dom-inates the time complexity of the main algorithmBuildDoubleCircle, the result follows. 4 Future workWe are working on extending these results to buildother known point sets in integer points of small size,such as the double convex chain, the Horton set, andothers. We plan to eventually release a software li-brary supporting many of these constructions.AcknowledgementsThe problem studied here were introduced and partiallysolved during a visit to University of Valparaiso funded byproject CONICYT Fondecyt/Iniciación 11110069 (Chile).The authors would like to thank anonymous referees fortheir valuable comments.References[1] O. Aichholzer, F. Hurtado, and M. Noy. A lower boundon the number of triangulations of planar point sets.Computational Geometry, 29(2):135–145, 2004.[2] O. Aichholzer, D. Orden, F. Santos, and B. Speck-mann. On the number of pseudo-triangulations of cer-tain point sets. Journal of Combinatorial Theory, Se-ries A, 115(2):254–278, 2008.[3] I. Bárány and N. Enriquez. Jarník’s convex lat-tice n-gon for non-symmetric norms. MathematischeZeitschrift, 270:627–643, 2012.[4] M. N. Huxley. Area, lattice points, and exponentialsums. Oxford Science Publications. The ClarendonPress Oxford University Press, New York, 1996.[5] V. Jarník. Über die Gitterpunkte auf konvexen Kur-ven. Mathematische Zeitschrift, 24:500–518, 1926.[6] G. Martin. The Limiting Curve of Jarník’s Polygons.Transactions of the American Mathematical Society,355(12):4865–4880, 2003.[7] G. Pick. Geometrisches zur Zahlenlehre. Sitzungs-berichte des Deutschen Naturwissenschaftlich-Medicinischen Vereines für Böhmen "Lotos" in Prag.,19:311–319, 1899.68

Drawing the double circle on a grid of minimum size

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Drawing the double circle on a grid of minimum size

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