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Rotation on the digital plane

Carolin Hannusch Faculty of Informatics, University of Debrecen, Hungary hannusch.carolin@inf.unideb.hu  and  Attila Pethő Faculty of Informatics, University of Debrecen, Hungary petho.attila@unideb.hu
Abstract.

Let AφA_{\varphi} denote the matrix of rotation with angle φ\varphi of the Euclidean plane, FLOOR the function, which rounds a real point to the nearest lattice point down on the left and ROUND the function for rounding off a vector to the nearest node of the lattice. We prove under the natural assumption φkπ2\varphi\not=k\frac{\pi}{2} that the functions FLOORAφFLOOR\circ A_{\varphi} and ROUNDAφROUND\circ A_{\varphi} are neither surjective nor injective. More precisely we prove lower and upper estimates for the size of the sets of lattice points, which are the image of two lattice points as well as of lattice points, which have no preimages. It turns out that the density of that sets are positive except when sinφ±cosφ+r,r\sin\varphi\not=\pm\cos\varphi+r,r\in{\mathbb{Q}}.

1. Introduction

The digital plane is a lattice whose elements are points with integer coordinates, the so called lattice points. The values of continuous functions can be represented only approximatively, rounding its computed value to a lattice point. Rounding is a mapping q:22q\;:\;{\mathbb{R}}^{2}\mapsto{\mathbb{Z}}^{2}. The discrete variant of the function f:22f\;:\;{\mathbb{R}}^{2}\mapsto{\mathbb{R}}^{2} is qfq\circ f. Of course there are plenty discrete variants of ff. One of the most studied function of the plane is the rotation, which is a 2×22\times 2 real matrix with eigenvalues cosφ+isinφ\cos\varphi+i\sin\varphi and cosφisinφ\cos\varphi-i\sin\varphi. Let us denote it by QφQ_{\varphi}. There exists an invertible real matrix QQ such that Qφ=QAφQ1Q_{\varphi}=QA_{\varphi}Q^{-1} with

Aφ=(cosφsinφsinφcosφ).A_{\varphi}=\left(\begin{array}[]{rr}\cos\varphi&-\sin\varphi\\ \sin\varphi&\cos\varphi\end{array}\right).

For (x,y)2(x,y)\in{\mathbb{R}}^{2} the sequence of points {fn(x,y)}n=0\{f^{n}(x,y)\}_{n=0}^{\infty} is called an orbit111Some author calls this sequence trajectory. of ff generated by (x,y)(x,y). The orbits of QφQ_{\varphi} generated by any non-zero points lie on an ellipse, those of AφA_{\varphi} on the unit circle. Plainly QφQ_{\varphi} is a bijective mapping on 2{\mathbb{R}}^{2}, but usually not bijective on 2{\mathbb{Z}}^{2}. Combining it with a rounding, which results the function qQφq\circ Q_{\varphi} we obtain a mapping of 22{\mathbb{Z}}^{2}\mapsto{\mathbb{Z}}^{2}, which is a discrete rotation in wide sense. Many interesting and hard questions appear: is a discrete rotation injective, surjective or bijective? How are its orbits? All orbits of QφQ_{\varphi} are bounded, but this is not at all clear for its discrete variants. The investigation of such questions have long tradition, see the early example [6].

FLOOR222Rounds a real point to the nearest lattice point down on the left., ROUND333Stands for rounding off a vector to the nearest node of the lattice. and TRUNC 444Denotes the coordinate-wise truncation of the fractional part of a vector towards the zero point. are eminent examples of rounding functions. Discretizing the rotation with them the resulted mappings 22{\mathbb{Z}}^{2}\mapsto{\mathbb{Z}}^{2} are more or less different. Kozyakin et al [9] gave a good overview on the results concerning discretized rotations, especially on ROUNDAφROUND\circ A_{\varphi} and TRUNCAφTRUNC\circ A_{\varphi}. Diamond et al [7] proved that if φkπ2,k\varphi\not=k\frac{\pi}{2},k\in{\mathbb{Z}} then all orbits of TRUNCAφTRUNC\circ A_{\varphi} eventually gets into the zero point. The situation is very different with ROUNDAφROUND\circ A_{\varphi}. Kozyakin et al [9] proved among others that if the rotation angle φ\varphi is such that the rows of all the nonnegative powers of the matrix AφA_{\varphi} are rationally independent then the density of lattice points with empty full preimages is positive. They used measure theoretic approach, which allow to prove much more general results too. For other probabilistic results on discrete rotations we refer to [6, 13, 14, 8].

In former investigations of the second author with different coauthors [1, 2, 3, 4] a kind of discrete rotation appears as a natural generalization of positional number systems. It was FLOORBφFLOOR\circ B_{\varphi} with

Bφ=(011λ),λ=2cosφ.B_{\varphi}=\left(\begin{array}[]{rr}0&1\\ -1&-\lambda\end{array}\right),\quad\lambda=-2\cos\varphi.

We come back to this function later, but before we discuss some properties of the FLOOR function. It commutes with the additive group of translations of 2{\mathbb{Z}}^{2} and the full preimage of zero is [0,1[×[0,1[[0,1[\times[0,1[, which is Jordan measurable, thus FLOOR, like ROUND, is a quantizer in the sense of [9]. Hence the discretized rotation FLOORAφ\circ A_{\varphi} has similar properties as ROUNDAφ\circ A_{\varphi}. This holds, among others, for the above mentioned property of preimages.

In this note we prove under the natural assumption φkπ2\varphi\not=k\frac{\pi}{2} that the function FLOORAφFLOOR\circ A_{\varphi} is neither surjective nor injective. More precisely we prove lower density estimates for the sets of lattice points, which are the image of two lattice points as well as of lattice points, which have no preimages. It turns out, see Theorems 3.1 and 4.2, that these densities are positive except when sinφ±cosφ+r,r\sin\varphi\not=\pm\cos\varphi+r,r\in{\mathbb{Q}}. This means that the number of such lattice points lying in a box symmetric to the origin and of side length 2M+12M+1 is O(M2)O(M^{2}). In contrast in the exceptional case this number is only O(M)O(M). In Section 6 we indicate that the same results hold to ROUNDAφROUND\circ A_{\varphi} too. We use in the proof elementary results of uniform distribution theory and properties of primitive Pythagorean triplets. Our results are more precise than those in [9].

There are discrete rotations which are bijective. Trivial examples are FLOORAφFLOOR\circ A_{\varphi} with φ=kπ2,k\varphi=k\frac{\pi}{2},k\in{\mathbb{Z}}. More interesting are the functions FLOORBφ,0φ<2πFLOOR\circ B_{\varphi},0\leq\varphi<2\pi. Reeve-Black and Vivaldi [12] claim that a generic discrete rotation is neither injective nor surjective. Our results justify this claim for the function FLOORAφFLOOR\circ A_{\varphi}. To prove similar characterization for FLOORQφFLOOR\circ Q_{\varphi} is a challenging problem. We expect that apart the previous examples only the transpose of BφB_{\varphi} lie in the exceptional set.

Despite many efforts and interesting results, we have deterministic knowledge only on the orbits of FLOORBφFLOOR\circ B_{\varphi}. For the eleven values 2cosφ=λ=0,±1,(±1±5)/2,±2,±32\cos\varphi=\lambda=0,\pm 1,(\pm 1\pm\sqrt{5})/2,\pm\sqrt{2},\pm\sqrt{3} all orbits are periodic, see [10, 2, 3]. Generally it was proved by Akiyama and Pethő [4] that for any φ\varphi there are infinitely many periodic orbits. This is still far from the conjecture that all orbits are periodic see [2].

The matrix AφA_{\varphi} through Aφ(a,b),(a,b)2A_{\varphi}\cdot(a,b),(a,b)\in{\mathbb{R}}^{2} 555To be precise we had to write Aφ(a,b)TA_{\varphi}\cdot(a,b)^{T} instead of Aφ(a,b)A_{\varphi}\cdot(a,b), where (a,b)T(a,b)^{T} denote the transpose of the vector (a,b)(a,b), i.e., a column vector. As in the article we should do this often, and from the context it will be clear whether the actual vector is a row or a column vector, we avoid this extra notation. induces a linear mapping on 2{\mathbb{R}}^{2}, which we will denote by AφA_{\varphi} too.

FLOOR and integer part .\lfloor.\rfloor are the same functions, in the sequel we will use the later. To simplify our notation we define rφ:22r_{\varphi}\;:\;{\mathbb{Z}}^{2}\mapsto{\mathbb{Z}}^{2} by

rφ(a,b)\displaystyle r_{\varphi}(a,b) =\displaystyle= Aφ(a,b)\displaystyle\lfloor A_{\varphi}(a,b)\rfloor
=\displaystyle= (acosφbsinφ,asinφ+bcosφ).\displaystyle(\lfloor a\cos\varphi-b\sin\varphi\rfloor,\lfloor a\sin\varphi+b\cos\varphi\rfloor).

We computed the orbits of rφr_{\varphi} for many choice of the angle and the starting point and found always periodicity. For the angle φ=π4\varphi=\frac{\pi}{4} we found infinitely many starting points which generate short periodic orbit, see Theorem 5.2. Based on our numerical and theoretical results we propose the following conjecture

Conjecture 1.1.

Every orbit of rφr_{\varphi} is periodic.

We also use the fractional part function, i.e, {x}=xx\{x\}=x-\lfloor x\rfloor. Both functions will be applied coordinate wise to the points of the real vector spaces. We use the same notation to these extended functions. Let U=[0,1[×[0,1[U=[0,1[\times[0,1[ and U¯=[0,1]×[0,1]\bar{U}=[0,1]\times[0,1], then obviously rφ(a,b)=(x,y)2r_{\varphi}(a,b)=(x,y)\in{\mathbb{Z}}^{2} if and only if Aφ(a,b)=(x,y)+uA_{\varphi}\cdot(a,b)=(x,y)+u for some uUu\in U. The third equivalent expression is {Aφ(a,b)}=Aφ(a,b)rφ(a,b)\{A_{\varphi}\cdot(a,b)\}=A_{\varphi}\cdot(a,b)-r_{\varphi}(a,b).

2. Preliminary results

In order to prove our main results we need some tools from uniform distribution theory. Let 𝐚=(a1,,an),𝐛=(b1,,bn)n{\bf a}=(a_{1},\ldots,a_{n}),{\bf b}=(b_{1},\ldots,b_{n})\in{\mathbb{R}}^{n} be such that 0aj<bj1,j=1,,n0\leq a_{j}<b_{j}\leq 1,\;j=1,\ldots,n then we set B𝐚,𝐛=[a1,b1[××[an,bn[B_{{\bf a,b}}=[a_{1},b_{1}[\times\ldots\times[a_{n},b_{n}[. This is a box with side lengths b1a1,,bnanb_{1}-a_{1},\ldots,b_{n}-a_{n}, whose volume is plainly j=1n(bjaj)\prod_{j=1}^{n}(b_{j}-a_{j}). For a sequence of nn-dimensional real vectors X=(𝐱m)X=({\bf x}_{m}) set

A(X,B𝐚,𝐛,N)=|{m: 0mN,{𝐱m}B𝐚,𝐛}|,A(X,B_{{\bf a,b}},N)=|\{m\;:\;0\leq m\leq N,\{{\bf x}_{m}\}\in B_{{\bf a,b}}\}|,

where |S||S| denotes the cardinality of the set SS.

The sequence XX is called uniformly distributed modulo 11, shortly uniformly distributed if

limNA(X,B𝐚,𝐛,N)N=j=1n(bjaj)\lim_{N\to\infty}\frac{A(X,B_{{\bf a,b}},N)}{N}=\prod_{j=1}^{n}(b_{j}-a_{j})

holds for all 𝐚,𝐛n{\bf a,b}\in{\mathbb{R}}^{n} with the above property. Notice that if XX is uniformly distributed then there exist for any B𝐚,𝐛B_{{\bf a,b}} a constant c=c𝐚,𝐛>0c=c_{\bf a,b}>0 such that

(2.1) A(X,B𝐚,𝐛,N)>cNA(X,B_{{\bf a,b}},N)>cN

holds for all large enough NN. Indeed, one may set c=j=1n(bjaj)εc=\prod_{j=1}^{n}(b_{j}-a_{j})-\varepsilon for some ε>0\varepsilon>0.

The following theorem is a bit modified version of Theorem I, p.64. of Cassels [5] and it plays a crucial role in this paper.

Theorem 2.1.

Let Lj(𝐱)L_{j}({\bf x}) for 1jm1\leq j\leq m be homogeneous linear forms in the nn variables 𝐱=(x1,,xn){\bf x}=(x_{1},\ldots,x_{n}). Suppose that the only set of integers u1,,umu_{1},\ldots,u_{m} such that

u1L1(𝐱)++umLm(𝐱)u_{1}L_{1}({\bf x})+\dots+u_{m}L_{m}({\bf x})

has integer coefficients in x1,,xnx_{1},\ldots,x_{n} is u1==um=0u_{1}=\dots=u_{m}=0. Then the set of vectors 𝐳(𝐱)=(L1(𝐱),,Lm(𝐱)){\bf z}^{({\bf x})}=(L_{1}({\bf x}),\dots,L_{m}({\bf x})) for integral 𝐱{\bf x} is uniformly distributed modulo 11.

Now we formulate the main lemma of this paper.

Lemma 2.2.

Let 0<t1,t210<t_{1},t_{2}\leq 1 and set L1(x1,x2)=x1cosφx2sinφL_{1}(x_{1},x_{2})=x_{1}\cos\varphi-x_{2}\sin\varphi, L2(x1,x2)=x1sinφ+x2cosφL_{2}(x_{1},x_{2})=x_{1}\sin\varphi+x_{2}\cos\varphi. If φkπ2,k\varphi\not=k\frac{\pi}{2},k\in{\mathbb{Z}} then there exist constants c1,c2>0c_{1},c_{2}>0 depending only on φ,t1,t2\varphi,t_{1},t_{2} such that the number of solutions (x1,x2)2,|x1|,|x2|M(x_{1},x_{2})\in{\mathbb{Z}}^{2},|x_{1}|,|x_{2}|\leq M of the system of inequalities

(2.2) 0{L1(x1,x2)}\displaystyle 0\leq\{L_{1}(x_{1},x_{2})\} <\displaystyle< t1\displaystyle t_{1}
(2.3) 0{L2(x1,x2)}\displaystyle 0\leq\{L_{2}(x_{1},x_{2})\} <\displaystyle< t2\displaystyle t_{2}

lie between c1M2c_{1}M^{2} and c2M2c_{2}M^{2} except when cosφ=±sinφ+r,r\cos\varphi=\pm\sin\varphi+r,\;r\in{\mathbb{Q}} in which case it lies between c1Mc_{1}M and c2Mc_{2}M.

The same statement holds for the number of solutions in pairs of odd integers.

Proof.

There are only (2M+1)2(2M+1)^{2} integer pairs with |x1|,|x2|M|x_{1}|,|x_{2}|\leq M, thus the upper estimate c2M2c_{2}M^{2} is obvious. Hence in the sequel we have to deal with the upper bound in the exceptional case and with the lower bound.

In the proof we have to distinguish three cases according the arithmetic nature of cosφ\cos\varphi and sinφ\sin\varphi.

Case 1. 1,sinφ,cosφ1,\sin\varphi,\cos\varphi are {\mathbb{Q}}-linearly independent. Then the linear forms L1(x1,x2),L2(x1,x2)L_{1}(x_{1},x_{2}),L_{2}(x_{1},x_{2}) satisfy the assumptions of Theorem 2.1, thus the points ({L1(a,b)},{L2(a,b)})(\{L_{1}(a,b)\},\{L_{2}(a,b)\}) for a,ba,b\in{\mathbb{Z}} is uniformly distributed in [0,1[2.\left[0,1\right[^{2}.

There are (2M+1)2(2M+1)^{2} points (a,b)2(a,b)\in{\mathbb{Z}}^{2} with |a|,|b|M|a|,|b|\leq M, thus setting B=[0,t1[×[0,t2[B=[0,t_{1}[\times[0,t_{2}[ and N=(2M+1)2N=(2M+1)^{2} in (2.1) we obtain the statement immediately.

If ({L1(a,b)},{L2(a,b)})(\{L_{1}(a,b)\},\{L_{2}(a,b)\}) for a,ba,b\in{\mathbb{Z}} is uniformly distributed in [0,1[2\left[0,1\right[^{2} then the same holds if (a,b)(a,b) runs through a sublattice of 2{\mathbb{Z}}^{2}, which proves the second assertion. In Case 2 we use the uniformly distributed property of some sequence, hence for them the second assertion holds by the above remark.

Case 2. cosφ=r1sinφ+r2,\cos\varphi=r_{1}\sin\varphi+r_{2}, where r1,r2,r_{1},r_{2}\in{\mathbb{Q}}, r10r_{1}\neq 0, and sinφ.\sin\varphi\notin{\mathbb{Q}}. There exist integers p1,p2,qp_{1},p_{2},q with q>0,q>0, such that r1=p1qr_{1}=\frac{p_{1}}{q} and r2=p2q.r_{2}=\frac{p_{2}}{q}. For any (a,b)2(a,b)\in{\mathbb{Z}}^{2} there exist u,v,s,t, 0s,t<qu,v,s,t\in{\mathbb{Z}},\;0\leq s,t<q such that a=uq+sa=uq+s and b=vq+tb=vq+t. With these notations we obtain

L1(a,b)\displaystyle L_{1}(a,b) =\displaystyle= (ap1qb)sinφ+ap2q\displaystyle\left(a\frac{p_{1}}{q}-b\right)\sin\varphi+\frac{ap_{2}}{q}
=\displaystyle= (up1vq)sinφ+(sp1qt)sinφ+ap2q\displaystyle(up_{1}-vq)\sin\varphi+\left(\frac{sp_{1}}{q}-t\right)\sin\varphi+\frac{ap_{2}}{q}

and similarly

L2(a,b)=(uq+vp1)sinφ+(s+tp1q)sinφ+bp2q.L_{2}(a,b)=(uq+vp_{1})\sin\varphi+\left(s+\frac{tp_{1}}{q}\right)\sin\varphi+b\frac{p_{2}}{q}.

Fix s,ts,t and set f1={(sp1qt)sinφ+ap2q}f_{1}=\left\{\left(\frac{sp_{1}}{q}-t\right)\sin\varphi+\frac{ap_{2}}{q}\right\} and f2={(s+tp1q)sinφ+bp2q}.f_{2}=\left\{\left(s+\frac{tp_{1}}{q}\right)\sin\varphi+b\frac{p_{2}}{q}\right\}. We have 0f1,f2<1.0\leq f_{1},f_{2}<1. With these notations we have to count the number of solutions of the system of inequalities

0\displaystyle 0 \displaystyle\leq {(up1vq)sinφ+f1}<t1\displaystyle\left\{(up_{1}-vq)\sin\varphi+f_{1}\right\}<t_{1}
0\displaystyle 0 \displaystyle\leq {(uq+vp1)sinφ+f2}<t2\displaystyle\left\{(uq+vp_{1})\sin\varphi+f_{2}\right\}<t_{2}

in the integers u,vu,v with |u|,|v|Mq1|u|,|v|\leq\frac{M}{q}-1. We have to distinguish two subcases

Case 2.1. p1=±qp_{1}=\pm q. We have up1vq=p1(uv)=±q(uv)=±(uq+vp1)up_{1}-vq=p_{1}(u\mp v)=\pm q(u\mp v)=\pm(uq+vp_{1}), thus the coefficients of sinφ\sin\varphi in the last inequalities are up to sign equal and may assume at most O(M)O(M) different values. As sinφ\sin\varphi\notin{\mathbb{Q}} the sequence ({nsinφ})(\{n\sin\varphi\}) is by Theorem 2.1 uniformly distributed in [0,1[[0,1[, thus each inequality, hence the system too, has at most O(M)O(M) solutions.

Choosing s=t=0s=t=0 the integers aa and bb are divisible by qq, hence f1=f2=0f_{1}=f_{2}=0. If, for example, p1=qp_{1}=q then we may assume without loss of generality 0<t1t20<t_{1}\leq t_{2}. Our first inequality has at least (t1/2)(M/q)(t_{1}/2)(M/q) solutions in u,v,|u|,|v|Mq1<Mqu,v\in{\mathbb{Z}},\;|u|,|v|\leq\frac{M}{q}-1<\frac{M}{q} provided MM is large enough.

Case 2.2. p1±qp_{1}\not=\pm q. Setting D=q2p12D=q^{2}-p_{1}^{2} this is equivalent to D0D\not=0. As sinφ\sin\varphi\notin{\mathbb{Q}} and DD\in{\mathbb{Z}} we have DsinφD\sin\varphi\notin{\mathbb{Q}}, hence the sequence ({nDsinφ})(\{nD\sin\varphi\}) is uniformly distributed in [0,1[[0,1[. Thus the inequality

0uDsinφ+ft0\leq uD\sin\varphi+f\leq t

has for any fixed 0f<1, 0<t<10\leq f<1,\;0<t<1 at least O(M)O(M^{\prime}) solutions in u,|u|Mu\in{\mathbb{Z}},\;|u|\leq M^{\prime}, where the positive constant indicated by the OO notation depends only on sinφ\sin\varphi and on tt. Hence the system of the (independent) inequalities

0\displaystyle 0 \displaystyle\leq {u1Dsinφ+f1}<t1\displaystyle\left\{u_{1}D\sin\varphi+f_{1}\right\}<t_{1}
0\displaystyle 0 \displaystyle\leq {u2Dsinφ+f2}<t2\displaystyle\left\{u_{2}D\sin\varphi+f_{2}\right\}<t_{2}

has at least O(M2)O(M^{\prime 2}) solutions in (u1,u2)2,|u1|,|u2|M(u_{1},u_{2})\in{\mathbb{Z}}^{2},\;|u_{1}|,|u_{2}|\leq M^{\prime}.

For all such pairs the system of linear equations

up1vq\displaystyle up_{1}-vq =\displaystyle= u1(q2p12)\displaystyle u_{1}(q^{2}-p_{1}^{2})
uq+vp1\displaystyle uq+vp_{1} =\displaystyle= u2(q2p12)\displaystyle u_{2}(q^{2}-p_{1}^{2})

has unique solution in u,vu,v\in{\mathbb{Z}}, i.e. (u,v)(u,v) solves our original system of equations. Using Cramer’s rule we obtain

u=(u1p1+u2q)(q2p12)p12+q2andv=(u2p1u1q)(q2p12)p12+q2.u=\frac{(u_{1}p_{1}+u_{2}q)(q^{2}-p_{1}^{2})}{p_{1}^{2}+q^{2}}\quad\mbox{and}\quad v=\frac{(u_{2}p_{1}-u_{1}q)(q^{2}-p_{1}^{2})}{p_{1}^{2}+q^{2}}.

Thus |u|,|v|2qM|u|,|v|\leq 2qM^{\prime}. Hence choosing M=M2q2M^{\prime}=\frac{M}{2q^{2}} we can produce at least O(M2)O(M^{2}) integer solutions of (2.2) and (2.3).

Case 3. sinφ,cosφ.\sin\varphi,\cos\varphi\in{\mathbb{Q}}. Although the statement is the same as in the other cases, we have to use different tools in the proof, because Theorem 2.1 does not hold. We have a kind of discrete uniform distribution treated systematically in Narkiewicz [11]. First we set sinφ=p1q,cosφ=p2q,\sin\varphi=\frac{p_{1}}{q},\;\cos\varphi=\frac{p_{2}}{q}, where p1,p2,q,p1,p2,q0.p_{1},p_{2},q\in{\mathbb{Z}},\;p_{1},p_{2},q\neq 0. Then

1=sin2φ+cos2φ=p12q2+p22q2,1=\sin^{2}\varphi+\cos^{2}\varphi=\frac{p_{1}^{2}}{q^{2}}+\frac{p_{2}^{2}}{q^{2}},

which implies p12+p22=q2,p_{1}^{2}+p_{2}^{2}=q^{2}, i.e. p1,p2,qp_{1},p_{2},q is a Pithagorean triple, and we may assume that it is primitive, i.e. gcd(p1,q)=gcd(p2,q)=1.\gcd(p_{1},q)=\gcd(p_{2},q)=1. Then there are u,v,u,v\in{\mathbb{Z}}, uv(mod2),u\not\equiv v\pmod{2}, gcd(u,v)=1\gcd(u,v)=1 such that p1=u2v2,p2=2uvp_{1}=u^{2}-v^{2},p_{2}=2uv or p1=2uv,p2=u2v2p_{1}=2uv,p_{2}=u^{2}-v^{2} and q=u2+v2.q=u^{2}+v^{2}. We work out in the sequel only the first possibility, because the alternative case can be handled analogously.

With these notations and (a,b)2(a,b)\in{\mathbb{Z}}^{2} we have

L1(a,b)\displaystyle L_{1}(a,b) =\displaystyle= a2uvu2+v2bu2v2u2+v2\displaystyle a\frac{2uv}{u^{2}+v^{2}}-b\frac{u^{2}-v^{2}}{u^{2}+v^{2}}
L1(a,b)\displaystyle L_{1}(a,b) =\displaystyle= au2v2u2+v2+b2uvu2+v2.\displaystyle a\frac{u^{2}-v^{2}}{u^{2}+v^{2}}+b\frac{2uv}{u^{2}+v^{2}}.

Because gcd(2uv,u2+v2)=1\gcd(2uv,u^{2}+v^{2})=1 there exists an integer 0<h<u2+v2=q0<h<u^{2}+v^{2}=q such that

(2.4) 2uvhu2v2(modu2+v2).2uvh\equiv u^{2}-v^{2}\pmod{u^{2}+v^{2}}.

For fixed a,ba,b\in{\mathbb{Z}} there exist d,d1d,d_{1}\in{\mathbb{Z}} such that ahb=d(u2+v2)+d1,a-hb=d(u^{2}+v^{2})+d_{1}, where 0d1<u2+v2.0\leq d_{1}<u^{2}+v^{2}. Then

a(u2v2)+2buv=bh(u2v2)+d(u4v4)+d1(u2v2)+2buv.a(u^{2}-v^{2})+2buv=bh(u^{2}-v^{2})+d(u^{4}-v^{4})+d_{1}(u^{2}-v^{2})+2buv.

Multiplying this by 2uv2uv and taking (2.4) into account we obtain

2uv(a(u2v2)+2buv)\displaystyle 2uv(a(u^{2}-v^{2})+2buv) \displaystyle\equiv 2uvhb(u2v2)+2uvd1(u2v2)+b(2uv)2\displaystyle 2uvhb(u^{2}-v^{2})+2uvd_{1}(u^{2}-v^{2})+b(2uv)^{2}
\displaystyle\equiv b((u2v2)2+4u2v2)+2uvd1(u2v2)\displaystyle b((u^{2}-v^{2})^{2}+4u^{2}v^{2})+2uvd_{1}(u^{2}-v^{2})
\displaystyle\equiv b(u2+v2)2+2uvd1(u2v2)\displaystyle b(u^{2}+v^{2})^{2}+2uvd_{1}(u^{2}-v^{2})
\displaystyle\equiv 2uvd1(u2v2)(modu2+v2).\displaystyle 2uvd_{1}(u^{2}-v^{2})\pmod{u^{2}+v^{2}}.

Since gcd(2uv,u2+v2)=1\gcd(2uv,u^{2}+v^{2})=1, we obtain

(2.5) a(u2v2)+2buvd1(u2v2)(modu2+v2).a(u^{2}-v^{2})+2buv\equiv d_{1}(u^{2}-v^{2})\pmod{u^{2}+v^{2}}.

On the other hand by (2.4) we get

2uva(u2v2)b\displaystyle 2uva-(u^{2}-v^{2})b \displaystyle\equiv b(2uvh(u2v2))+2uvd1\displaystyle b(2uvh-(u^{2}-v^{2}))+2uvd_{1}
\displaystyle\equiv 2uvd1(modu2+v2).\displaystyle 2uvd_{1}\pmod{u^{2}+v^{2}}.

Hence

{L1(a,b)}={2uvd1u2+v2}\{L_{1}(a,b)\}=\left\{\frac{2uvd_{1}}{u^{2}+v^{2}}\right\}

and

{L2(a,b)}={(u2v2)d1u2+v2}.\{L_{2}(a,b)\}=\left\{\frac{(u^{2}-v^{2})d_{1}}{u^{2}+v^{2}}\right\}.

Choosing d1=0d_{1}=0, which implies a=hb+d(u2v2)a=hb+d(u^{2}-v^{2}) we obtain that the pair (a,b)2(a,b)\in{\mathbb{Z}}^{2} solves the system of inequalities (2.2), (2.3).

Finally choosing b,db,d\in{\mathbb{Z}} such that |b|M2q<M2h|b|\leq\frac{M}{2q}<\frac{M}{2h} and |d|M2|p1|=M2|u2v2||d|\leq\frac{M}{2|p_{1}|}=\frac{M}{2|u^{2}-v^{2}|} we obtain |a|M|a|\leq M, hence such (a,b)2(a,b)\in{\mathbb{Z}}^{2} pairs not only solves the system of inequalities (2.2),(2.3), but satisfies |a|,|b|M|a|,|b|\leq M too. Plainly the number of such pairs is at least O(M2)O(M^{2}) and the lemma is completely proved.

Choosing bb odd aa is odd too for all odd or even dd depending on the parities of hh and u2v2u^{2}-v^{2}. The argument of the last paragraph proves the second assertion in this case.

3. Injectivity of digital rotation

Before stating the main result of this section we introduce a notation: Tφ(M)T_{\varphi}(M) denotes the number of (x,y)2(x,y)\in{\mathbb{Z}}^{2} such that |x|,|y|M|x|,|y|\leq M and (x,y)(x,y) is the image by rφr_{\varphi} of two different grid points. If φ=kπ2,k\varphi=k\frac{\pi}{2},k\in{\mathbb{Z}} then rφ=Aφr_{\varphi}=A_{\varphi}, thus it is bijective, thus Tφ(M)=0T_{\varphi}(M)=0. Otherwise, we prove that Tφ(M)T_{\varphi}(M) tends to infinity.

Theorem 3.1.

If φkπ2,k\varphi\not=k\frac{\pi}{2},k\in{\mathbb{Z}} then there exist constants c3,c4>0c_{3},c_{4}>0 depending only on φ\varphi such that c3M2Tφ(M)c4M2c_{3}M^{2}\leq T_{\varphi}(M)\leq c_{4}M^{2} except when cosφ=±sinφ+r,r\cos\varphi=\pm\sin\varphi+r,\;r\in{\mathbb{Q}}, in which case c3MTφ(M)c4Mc_{3}M\leq T_{\varphi}(M)\leq c_{4}M hold.

Proof.

Let (a1,b1),(a2,b2)2(a_{1},b_{1}),(a_{2},b_{2})\in{\mathbb{Z}}^{2} such that rφ(a1,b1)=rφ(a2,b2)r_{\varphi}(a_{1},b_{1})=r_{\varphi}(a_{2},b_{2}). Then

a1cosφb1sinφ\displaystyle\lfloor a_{1}\cos\varphi-b_{1}\sin\varphi\rfloor =\displaystyle= a2cosφb2sinφ\displaystyle\lfloor a_{2}\cos\varphi-b_{2}\sin\varphi\rfloor
a1sinφ+b1cosφ\displaystyle\lfloor a_{1}\sin\varphi+b_{1}\cos\varphi\rfloor =\displaystyle= a2sinφ+b2cosφ,\displaystyle\lfloor a_{2}\sin\varphi+b_{2}\cos\varphi\rfloor,

which implies

(a1a2)cosφ(b1b2)sinφ\displaystyle(a_{1}-a_{2})\cos\varphi-(b_{1}-b_{2})\sin\varphi =\displaystyle= u1\displaystyle u_{1}
(a1a2)sinφ+(b1b2)cosφ\displaystyle(a_{1}-a_{2})\sin\varphi+(b_{1}-b_{2})\cos\varphi =\displaystyle= u2\displaystyle u_{2}

for some 1<u1,u2<1-1<u_{1},u_{2}<1. After squaring and adding these equations we obtain

(a1a2)2+(b1b2)2=u12+u22<2.(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}=u_{1}^{2}+u_{2}^{2}<2.

As a1a2a_{1}-a_{2} and b1b2b_{1}-b_{2} are integers the right hand side can be either zero ore one. The first case is excluded because either a1a2a_{1}-a_{2} or b1b2b_{1}-b_{2} is nonzero. Thus we have the following four possibilities:

a1a21001b1b20110\begin{array}[]{c|c|c|c|c}a_{1}-a_{2}&-1&0&0&1\\ \hline\cr b_{1}-b_{2}&0&-1&1&0\end{array}

This implies that if two grid points have the same image by rφ,r_{\varphi}, then they are neighbors.

Now, we show that for each φ\varphi there exist infinitely many neighbors such that rφ(a1,b1)=rφ(a2,b2).r_{\varphi}(a_{1},b_{1})=r_{\varphi}(a_{2},b_{2}).

First, we assume a1=a2a_{1}=a_{2} and b2=b1+1.b_{2}=b_{1}+1. Inserting them into the starting equations we obtain

(3.1) a1cosφb1sinφ=a1cosφb1sinφsinφ\lfloor a_{1}\cos\varphi-b_{1}\sin\varphi\rfloor=\lfloor a_{1}\cos\varphi-b_{1}\sin\varphi-\sin\varphi\rfloor

and

(3.2) a1sinφ+b1cosφ=a1sinφ+b1cosφ+cosφ\lfloor a_{1}\sin\varphi+b_{1}\cos\varphi\rfloor=\lfloor a_{1}\sin\varphi+b_{1}\cos\varphi+\cos\varphi\rfloor

If φ]3π2,2π[,\varphi\in\left]\frac{3\pi}{2},2\pi\right[, then sinφ<0\sin\varphi<0 and cosφ>0\cos\varphi>0 and the system of equations (3.1) and (3.2) holds if and only if

(3.3) 0{L1(a1,b1)}={a1cosφb1sinφ}<1+sinφ0\leq\{L_{1}(a_{1},b_{1})\}=\{a_{1}\cos\varphi-b_{1}\sin\varphi\}<1+\sin\varphi

and

(3.4) 0{L2(a1,b1)}={a1sinφ+b1cosφ}<1cosφ.0\leq\{L_{2}(a_{1},b_{1})\}=\{a_{1}\sin\varphi+b_{1}\cos\varphi\}<1-\cos\varphi.

Observe that L1,L2L_{1},L_{2} are the linear forms introduced in Lemma 2.2, further setting t1=1+sinφ,t2=1cosφt_{1}=1+\sin\varphi,\;t_{2}=1-\cos\varphi the lemma implies our statement in this case.

The other cases can be handled similarly we give only the important data for repeating the argument.

If a2=a1+1a_{2}=a_{1}+1 and b2=b1,b_{2}=b_{1}, then (3.1), (3.2) reads [L1(a1,b1)]=[L1(a1,b1)+cosφ][L_{1}(a_{1},b_{1})]=[L_{1}(a_{1},b_{1})+\cos\varphi] and [L2(a1,b1)]=[L2(a1,b1)+sinφ][L_{2}(a_{1},b_{1})]=[L_{2}(a_{1},b_{1})+\sin\varphi]. If φ]0,π2[\varphi\in]0,\frac{\pi}{2}[ then sinφ,cosφ>0\sin\varphi,\cos\varphi>0 and setting t1=1cosφ,t2=1sinφt_{1}=1-\cos\varphi,\;t_{2}=1-\sin\varphi we can apply Lemma 2.2.

If a2=a1a_{2}=a_{1} and b2=b11,b_{2}=b_{1}-1, then (3.1), (3.2) reads [L1(a1,b1)]=[L1(a1,b1)+sinφ][L_{1}(a_{1},b_{1})]=[L_{1}(a_{1},b_{1})+\sin\varphi] and [L2(a1,b1)]=[L2(a1,b1)cosφ][L_{2}(a_{1},b_{1})]=[L_{2}(a_{1},b_{1})-\cos\varphi]. If φ]π2,π[\varphi\in]\frac{\pi}{2},\pi[ then sinφ>0\sin\varphi>0 and cosφ<0\cos\varphi<0 and setting t1=1sinφ,t2=1+cosφt_{1}=1-\sin\varphi,\;t_{2}=1+\cos\varphi we are done by Lemma 2.2.

Finally if a2=a11a_{2}=a_{1}-1 and b2=b1,b_{2}=b_{1}, then (3.1), (3.2) reads [L1(a1,b1)]=[L1(a1,b1)cosφ][L_{1}(a_{1},b_{1})]=[L_{1}(a_{1},b_{1})-\cos\varphi] and [L2(a1,b1)]=[L2(a1,b1)sinφ][L_{2}(a_{1},b_{1})]=[L_{2}(a_{1},b_{1})-\sin\varphi]. If φ]π,3π2[\varphi\in]\pi,3\frac{\pi}{2}[ then sinφ,cosφ<0\sin\varphi,\cos\varphi<0 and setting t1=1+cosφ,t2=1+sinφt_{1}=1+\cos\varphi,\;t_{2}=1+\sin\varphi finishes the proof by Lemma 2.2. ∎

4. Surjectivity of digital rotation

In the last section we proved that the digital rotation is usually not injective. Now we prove that usually it is not surjective either. To achieve our goal we need an elementary geometric lemma. To state and prove it we introduce some notation. For a point (a,b)2(a,b)\in\mathbb{R}^{2} put T(a,b)=(a,b)+UT_{(a,b)}=(a,b)+U and T¯(a,b)=(a,b)+U¯\bar{T}_{(a,b)}=(a,b)+\bar{U}, where ++ means here translation. The squares T(a,b),(a,b)2T_{(a,b)},\;(a,b)\in{\mathbb{Z}}^{2} are disjoint and their union cover 2\mathbb{R}^{2}. As AφA_{\varphi} is a rotation the sets Aφ(T(a,b))A_{\varphi}(T_{(a,b)}) are squares too with the same properties. Thus there exists for any (n,m)2(n,m)\in\mathbb{Z}^{2} unique (a,b)2(a,b)\in{\mathbb{Z}}^{2} such that (n,m)Aφ(T(a,b))(n,m)\in A_{\varphi}(T_{(a,b)}).

Lemma 4.1.

Let (n,m)2.(n,m)\in{\mathbb{Z}}^{2}. Then (n,m)rφ(a,b)(n,m)\neq r_{\varphi}(a,b) for each (a,b)2(a,b)\in{\mathbb{Z}}^{2} if and only if (n,m)Aφ(T(a,b))(n,m)\in A_{\varphi}(T_{(a,b)}) and Aφ(a+εa,b+εb)T(n,m)A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})\notin T_{(n,m)} for each εa,εb{0,1}.\varepsilon_{a},\varepsilon_{b}\in\{0,1\}. Moreover in this case the points Aφ(a+εa,b+εb),εa,εb{0,1}A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b}),\varepsilon_{a},\varepsilon_{b}\in\{0,1\} belong in some order to the horizontal and vertical neighbour squres to T(n,m)T_{(n,m)}.

Proof.

Necessity: If Aφ(a+εa,b+εb)T(n,m)A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})\in T_{(n,m)} for some εa,εb{0,1},\varepsilon_{a},\varepsilon_{b}\in\{0,1\}, then rφ(a+εa,b+εb)=(n,m),r_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})=(n,m), thus (n,m)(n,m) is the image of some point.


Sufficiency: We have T(a,b)=(a,b)+T(0,0),T_{(a,b)}=(a,b)+T_{(0,0)}, therefore Aφ(T(a,b))=Aφ(a,b)+Aφ(T(0,0)),A_{\varphi}(T_{(a,b)})=A_{\varphi}(a,b)+A_{\varphi}(T_{(0,0)}), since AφA_{\varphi} is linear. The same holds for the closure of T(a,b),T_{(a,b)}, denoted by T(a,b)¯.\overline{T_{(a,b)}}. In Figure 1 we show the four main situations of the rotated unitgrid.

xxyyAφ(0,1)A_{\varphi}(0,1)Aφ(1,1)A_{\varphi}(1,1)Aφ(1,0)A_{\varphi}(1,0)0<φ<π20<\varphi<\frac{\pi}{2}xxyyAφ(0,1)A_{\varphi}(0,1)Aφ(1,1)A_{\varphi}(1,1)Aφ(1,0)A_{\varphi}(1,0)π2<φ<π\frac{\pi}{2}<\varphi<\pi
xxyyAφ(0,1)A_{\varphi}(0,1)Aφ(1,1)A_{\varphi}(1,1)Aφ(1,0)A_{\varphi}(1,0)π<φ<3π2\pi<\varphi<\frac{3\pi}{2}xxyyAφ(0,1)A_{\varphi}(0,1)Aφ(1,1)A_{\varphi}(1,1)Aφ(1,0)A_{\varphi}(1,0)3π2<φ<2π\frac{3\pi}{2}<\varphi<2\pi
Figure 1. The situation of Aφ(T(0,0))A_{\varphi}(T_{(0,0)})

We give the proof of the lemma in detail only for the case 3π2<φ<2π\frac{3\pi}{2}<\varphi<2\pi, the other cases can be handled similarly. Assume that (n,m)Aφ(T(a,b))(n,m)\in A_{\varphi}(T_{(a,b)}), but Aφ(a+εa,b+εb)T(n,m)A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})\notin T_{(n,m)} for each εa,εb{0,1}.\varepsilon_{a},\varepsilon_{b}\in\{0,1\}. Then we have

Aφ(a,b+1)y<m, Aφ(a+1,b)ym+1,A_{\varphi}(a,b+1)_{y}<m,\mbox{ }A_{\varphi}(a+1,b)_{y}\geq m+1,
Aφ(a,b)x<n, Aφ(a+1,b+1)xn+1.A_{\varphi}(a,b)_{x}<n,\mbox{ }A_{\varphi}(a+1,b+1)_{x}\geq n+1.

Here and in the sequel, Aφ(.,.)x,Aφ(.,.)yA_{\varphi}(.,.)_{x},A_{\varphi}(.,.)_{y} denote the xx (respectively yy) coordinate of the corresponding point. Notice that the two strong inequalities are due to the assumption Aφ(a+εa,b+εb)T(n,m)A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})\notin T_{(n,m)} for each εa,εb{0,1}.\varepsilon_{a},\varepsilon_{b}\in\{0,1\}.

We show that T(n,m)Aφ(T(a,b)¯)T_{(n,m)}\setminus A_{\varphi}(\overline{T_{(a,b)}}) is the union of three disjoint triangles H1,H2,H3H_{1},H_{2},H_{3}. The triangle H1H_{1} is bordered by the lines x=n,y=m+1x=n,y=m+1 and by the line sequent between the points Aφ(a,b)A_{\varphi}(a,b) and Aφ(a+1,b)A_{\varphi}(a+1,b). Similarly H2H_{2} is bordered by the lines x=n+1,y=m+1x=n+1,y=m+1 and by the line sequent between the points Aφ(a+1,b)A_{\varphi}(a+1,b) and Aφ(a+1,b+1)A_{\varphi}(a+1,b+1). Finally the borders of H3H_{3} are the lines x=n,y=m+1x=n,y=m+1 and the line segment between the points Aφ(a,b+1)A_{\varphi}(a,b+1) and Aφ(a+1,b+1)A_{\varphi}(a+1,b+1).

That are proper triangles. For example, look at H1.H_{1}. The triangle with vertices Aφ(a+1,b),Aφ(a+1,b+1)A_{\varphi}(a+1,b),\;A_{\varphi}(a+1,b+1) and the intersection of the lines x=Aφ(a+1,b)xx=A_{\varphi}(a+1,b)_{x} and y=Aφ(a+1,b+1)yy=A_{\varphi}(a+1,b+1)_{y} is rectangular and the legth of its hypotenuse is 11, thus Aφ(a+1,b+1)xAφ(a+1,b)x<1A_{\varphi}(a+1,b+1)_{x}-A_{\varphi}(a+1,b)_{x}<1, which together with the above inequalities implies Aφ(a+1,b)x>nA_{\varphi}(a+1,b)_{x}>n. As Aφ(a,b)x<nA_{\varphi}(a,b)_{x}<n the line x=nx=n has an intersection with the line segment between Aφ(a,b)A_{\varphi}(a,b) and Aφ(a+1,b),A_{\varphi}(a+1,b), which is different from the end points. Similarly we have Aφ(a,b)yAφ(a+1,b)y<1,A_{\varphi}(a,b)_{y}-A_{\varphi}(a+1,b)_{y}<1, i.e. Aφ(a,b)y<m+1,A_{\varphi}(a,b)_{y}<m+1, hence the line y=m+1y=m+1 intersects the line segment between Aφ(a,b)A_{\varphi}(a,b) and Aφ(a+1,b).A_{\varphi}(a+1,b). In this case it may happen that the intersection point is Aφ(a+1,b).A_{\varphi}(a+1,b).

As a byproduct we proved Aφ(a+1,b)T(n,m+1)A_{\varphi}(a+1,b)\in T_{(n,m+1)}, i.e rφ(a+1,b)=(n,m+1)r_{\varphi}(a+1,b)=(n,m+1). Performing similar arguments for H2H_{2} and H3H_{3} we obtain that Aφ(a+1,b+1)T(n+1,m)A_{\varphi}(a+1,b+1)\in T_{(n+1,m)}, i.e rφ(a+1,b+1)=(n+1,m)r_{\varphi}(a+1,b+1)=(n+1,m) and Aφ(a,b+1)T(n,m1)A_{\varphi}(a,b+1)\in T_{(n,m-1)}, i.e rφ(a,b+1)=(n,m1)r_{\varphi}(a,b+1)=(n,m-1) respectively. Finally, one can prove Aφ(a,b)T(n1,m)A_{\varphi}(a,b)\in T_{(n-1,m)}, i.e rφ(a,b)=(n1,m)r_{\varphi}(a,b)=(n-1,m) too, hence the second assertion is proved.

No we finalize the proof of the first assertion. We assume that there exists (c,d)2,(c,d)\in{\mathbb{Z}}^{2}, such that rφ(c,d)=(n,m).r_{\varphi}(c,d)=(n,m). Then Aφ(c,d)T(n,m)A_{\varphi}(c,d)\in T_{(n,m)} and (c,d)T(a,b)¯,(c,d)\notin\overline{T_{(a,b)}}, i.e. Aφ(c,d)T(n,m)Aφ(T(a,b)¯),A_{\varphi}(c,d)\in T_{(n,m)}\setminus A_{\varphi}(\overline{T_{(a,b)}}), hence Aφ(c,d)A_{\varphi}(c,d) is contained in one of the triangles H1,H2,H3.H_{1},H_{2},H_{3}. We have H1Aφ(T(a,b1)),H_{1}\subseteq A_{\varphi}(T_{(a,b-1)}), which contains only the grid point Aφ(a,b1),A_{\varphi}(a,b-1), therefore (c,d)=(a,b1).(c,d)=(a,b-1). In contrast, Aφ(a,b1)x<Aφ(a,b)<n,A_{\varphi}(a,b-1)_{x}<A_{\varphi}(a,b)<n, i.e. Aφ(a,b1)x<n=Aφ(c,d)x,\lfloor A_{\varphi}(a,b-1)_{x}\rfloor<n=\lfloor A_{\varphi}(c,d)_{x}\rfloor, which is a contradiction. We have H2Aφ(T(a+1,b))H_{2}\subseteq A_{\varphi}(T_{(a+1,b)}) and H3Aφ(T(a,b+1)),H_{3}\subseteq A_{\varphi}(T_{(a,b+1)}), which implies (c,d)=(a+1,b)(c,d)=(a+1,b) and (c,d)=(a,b+1)(c,d)=(a,b+1) respectively. The proof of the case 3π2<φ<2π\frac{3\pi}{2}<\varphi<2\pi is finished. As we mentioned above the three other cases can be handled similarly. ∎

Similarly to Section 3 we introduce the function Nφ(M)N_{\varphi}(M), which is the number of grid points (n,m)(n,m), such that |n|,|m|M|n|,|m|\leq M and which are images of no grid points under the mapping rφr_{\varphi}. If φ=kπ2,k\varphi=k\frac{\pi}{2},k\in{\mathbb{Z}} then rφr_{\varphi} is bijective, hence Nφ(M)=0N_{\varphi}(M)=0. By the next theorem this cannot happen otherwise.

Theorem 4.2.

If φkπ2,k\varphi\not=k\frac{\pi}{2},k\in{\mathbb{Z}} then there exist constants c5,c6>0c_{5},c_{6}>0 depending only on φ\varphi such that c5M2Tφ(M)c6M2c_{5}M^{2}\leq T_{\varphi}(M)\leq c_{6}M^{2} except when cosφ=±sinφ+r,r\cos\varphi=\pm\sin\varphi+r,\;r\in{\mathbb{Q}}, in which case c5MTφ(M)c6Mc_{5}M\leq T_{\varphi}(M)\leq c_{6}M hold.

An immediate consequence is

Corollary 4.3.

If φkπ2\varphi\not=k\frac{\pi}{2} then rφr_{\varphi} has infinitely many different orbits.

Proof of Theorem 4.2.

Let (n,m)2(n,m)\in{\mathbb{Z}}^{2}, which is the image of no grid points under rφr_{\varphi}. Then, by Lemma 4.1, there exists (a,b)2(a,b)\in{\mathbb{Z}}^{2} such that (n,m)Aφ(T(a,b))(n,m)\in A_{\varphi}(T_{(a,b)}) and Aφ(a+εa,b+εb)T(n,m)A_{\varphi}(a+\varepsilon_{a},b+\varepsilon_{b})\notin T_{(n,m)} for all εa,εb{0,1}\varepsilon_{a},\varepsilon_{b}\in\{0,1\}. In the same lemma we proved that Aφ(a,b),Aφ(a,b+1),Aφ(a+1,b),Aφ(a+1,b+1)A_{\varphi}(a,b),A_{\varphi}(a,b+1),A_{\varphi}(a+1,b),A_{\varphi}(a+1,b+1) belong in some order to the four unit squares left, right, top and down to the unit square T(n,m)T_{(n,m)}. Depending on the size of φ\varphi we distinguish four cases.

Case 1. 0φ<π20\leq\varphi<\frac{\pi}{2}. Then by Lemma 4.1 Aφ(a,b)T(n,m1),Aφ(a,b+1)T(n1,m),Aφ(a+1,b)T(n+1,m),Aφ(a+1,b+1)T(n,m+1)A_{\varphi}(a,b)\in T_{(n,m-1)},A_{\varphi}(a,b+1)\in T_{(n-1,m)},A_{\varphi}(a+1,b)\in T_{(n+1,m)},A_{\varphi}(a+1,b+1)\in T_{(n,m+1)}, which means

0\displaystyle 0 \displaystyle\leq acosφbsinφn<1\displaystyle a\cos\varphi-b\sin\varphi-n<1
0\displaystyle 0 \displaystyle\leq acosφ(b+1)sinφ(n1)<1\displaystyle a\cos\varphi-(b+1)\sin\varphi-(n-1)<1
0\displaystyle 0 \displaystyle\leq (a+1)cosφbsinφ(n+1)<1\displaystyle(a+1)\cos\varphi-b\sin\varphi-(n+1)<1
0\displaystyle 0 \displaystyle\leq (a+1)cosφ(b+1)sinφn<1.\displaystyle(a+1)\cos\varphi-(b+1)\sin\varphi-n<1.

Rearranging we obtain the system of inequalities

(4.1) 0\displaystyle 0 \displaystyle\leq acosφbsinφn<1\displaystyle a\cos\varphi-b\sin\varphi-n<1
(4.2) sinφ1\displaystyle\sin\varphi-1 \displaystyle\leq acosφbsinφn<sinφ\displaystyle a\cos\varphi-b\sin\varphi-n<\sin\varphi
(4.3) 1cosφ\displaystyle 1-\cos\varphi \displaystyle\leq acosφbsinφn<2cosφ\displaystyle a\cos\varphi-b\sin\varphi-n<2-\cos\varphi
(4.4) sinφcosφ\displaystyle\sin\varphi-\cos\varphi \displaystyle\leq acosφbsinφn<1+sinφcosφ.\displaystyle a\cos\varphi-b\sin\varphi-n<1+\sin\varphi-\cos\varphi.

As 0φ<π20\leq\varphi<\frac{\pi}{2} we have sinφ,cosφ>0\sin\varphi,\cos\varphi>0. Under this assumption we have sinφ1<sinφcosφ<1cosφ\sin\varphi-1<\sin\varphi-\cos\varphi<1-\cos\varphi and 0<1cosφ0<1-\cos\varphi holds too. Hence max{sinφ1,sinφcosφ,1cosφ,0}=1cosφ\max\{\sin\varphi-1,\sin\varphi-\cos\varphi,1-\cos\varphi,0\}=1-\cos\varphi.

Similarly, sinφ<1<2cosφ\sin\varphi<1<2-\cos\varphi and sinφ<1+sinφcosφ\sin\varphi<1+\sin\varphi-\cos\varphi, thus min{sinφ,1,2cosφ,1+sinφcosφ}=sinφ\min\{\sin\varphi,1,2-\cos\varphi,1+\sin\varphi-\cos\varphi\}=\sin\varphi. Hence the inequalities (4.1)-(4.4) hold if and only if

1cosφacosφbsinφn<sinφ.1-\cos\varphi\leq a\cos\varphi-b\sin\varphi-n<\sin\varphi.

After multiplying by 22 and adding cosφsinφ1\cos\varphi-\sin\varphi-1 we obtain

(4.5) 1cosφsinφ(2a+1)cosφ(2b+1)sinφ2n3<sinφ+cosφ1.1-\cos\varphi-\sin\varphi\leq(2a+1)\cos\varphi-(2b+1)\sin\varphi-2n-3<\sin\varphi+\cos\varphi-1.

Performing the analogous computation for mm we get that a,b,ma,b,m\in{\mathbb{Z}} satisfy the requirements if and only if

(4.6) 1cosφsinφ(2a+1)sinφ+(2b+1)cosφ2m1<sinφ+cosφ1.1-\cos\varphi-\sin\varphi\leq(2a+1)\sin\varphi+(2b+1)\cos\varphi-2m-1<\sin\varphi+\cos\varphi-1.

As sinφ+cosφ1>0\sin\varphi+\cos\varphi-1>0, hence 1sinφcosφ<01-\sin\varphi-\cos\varphi<0 we can apply Lemma 2.2 tothe system of inequalities (4.5) and (4.5) with x1=2a+1x_{1}=2a+1 and x2=2b+1x_{2}=2b+1, which proves the theorem in this case.


Case 2. π2φ<π\frac{\pi}{2}\leq\varphi<\pi. Then by Lemma 4.1 Aφ(a,b)T(n+1,m),Aφ(a,b+1)T(n,m1),Aφ(a+1,b)T(n,m+1),Aφ(a+1,b+1)T(n1,m)A_{\varphi}(a,b)\in T_{(n+1,m)},A_{\varphi}(a,b+1)\in T_{(n,m-1)},A_{\varphi}(a+1,b)\in T_{(n,m+1)},A_{\varphi}(a+1,b+1)\in T_{(n-1,m)}. Then

0\displaystyle 0 \displaystyle\leq acosφbsinφ(n+1)<1\displaystyle a\cos\varphi-b\sin\varphi-(n+1)<1
0\displaystyle 0 \displaystyle\leq acosφ(b+1)sinφn<1\displaystyle a\cos\varphi-(b+1)\sin\varphi-n<1
0\displaystyle 0 \displaystyle\leq (a+1)cosφbsinφn<1\displaystyle(a+1)\cos\varphi-b\sin\varphi-n<1
0\displaystyle 0 \displaystyle\leq (a+1)cosφ(b+1)sinφ(n1)<1.\displaystyle(a+1)\cos\varphi-(b+1)\sin\varphi-(n-1)<1.

Rearranging the inequalities we obtain

(4.7) 1\displaystyle 1 \displaystyle\leq acosφbsinφn<2\displaystyle a\cos\varphi-b\sin\varphi-n<2
(4.8) sinφ\displaystyle\sin\varphi \displaystyle\leq acosφbsinφn<1+sinφ\displaystyle a\cos\varphi-b\sin\varphi-n<1+\sin\varphi
(4.9) cosφ\displaystyle-\cos\varphi \displaystyle\leq acosφbsinφn<1cosφ\displaystyle a\cos\varphi-b\sin\varphi-n<1-\cos\varphi
(4.10) sinφcosφ1\displaystyle\sin\varphi-\cos\varphi-1 \displaystyle\leq acosφbsinφn<sinφcosφ.\displaystyle a\cos\varphi-b\sin\varphi-n<\sin\varphi-\cos\varphi.

We have sinφ>0,cosφ<0\sin\varphi>0,\cos\varphi<0, because π2<φ<π\frac{\pi}{2}<\varphi<\pi.

Thus max{sinφ,1,cosφ,sinφcosφ1}=1\max\{\sin\varphi,1,-\cos\varphi,\sin\varphi-\cos\varphi-1\}=1. Similarly, min{2,1+sinφ,1cosφ,sinφcosφ}=sinφcosφ\min\{2,1+\sin\varphi,1-\cos\varphi,\sin\varphi-\cos\varphi\}=\sin\varphi-\cos\varphi. Hence the inequalities (4.7)-(4.10) hold if and only if

1acosφbsinφn<sinφcosφ.1\leq a\cos\varphi-b\sin\varphi-n<\sin\varphi-\cos\varphi.

Multiplying by 22 and adding sinφ+cosφ1-\sin\varphi+\cos\varphi-1 we obtain

1+cosφsinφ(2a+1)cosφ(2b+1)sinφ2n1<sinφcosφ1.1+\cos\varphi-\sin\varphi\leq(2a+1)\cos\varphi-(2b+1)\sin\varphi-2n-1<\sin\varphi-\cos\varphi-1.

Performing the analogous computation for mm we get that a,b,ma,b,m\in{\mathbb{Z}} satisfy the requirements if and only if

1+cosφsinφ(2a+1)sinφ(2b+1)cosφ2m1<sinφcosφ1.1+\cos\varphi-\sin\varphi\leq(2a+1)\sin\varphi-(2b+1)\cos\varphi-2m-1<\sin\varphi-\cos\varphi-1.

As sinφ>0\sin\varphi>0 and cosφ<0\cos\varphi<0 we have sinφcosφ1>0\sin\varphi-\cos\varphi-1>0, hence 1sinφ+cosφ<01-\sin\varphi+\cos\varphi<0, thus we may apply Lemma 2.2 to the last system of inequalities, which completes the proof in the second case,


Case 3. π<φ<3π2\pi<\varphi<\frac{3\pi}{2}. Then by Lemma 4.1 Aφ(a,b)T(n,m+1),Aφ(a,b+1)T(n+1,m),Aφ(a+1,b)T(n1,m),Aφ(a+1,b+1)T(n,m1)A_{\varphi}(a,b)\in T_{(n,m+1)},A_{\varphi}(a,b+1)\in T_{(n+1,m)},A_{\varphi}(a+1,b)\in T_{(n-1,m)},A_{\varphi}(a+1,b+1)\in T_{(n,m-1)}. The same computation as in Cases 1. and 2. lead to the system of inequalities

1+sinφ+cosφ\displaystyle 1+\sin\varphi+\cos\varphi \displaystyle\leq (2a+1)cosφ(2b+1)sinφ2n1<sinφcosφ1\displaystyle(2a+1)\cos\varphi-(2b+1)\sin\varphi-2n-1<-\sin\varphi-\cos\varphi-1
1+sinφ+cosφ\displaystyle 1+\sin\varphi+\cos\varphi \displaystyle\leq (2a+1)cosφ+(2b+1)sinφ2m1<sinφcosφ1\displaystyle(2a+1)\cos\varphi+(2b+1)\sin\varphi-2m-1<-\sin\varphi-\cos\varphi-1

As sinφ,cosφ,1+sinφ+cosφ<0\sin\varphi,\cos\varphi,1+\sin\varphi+\cos\varphi<0 we have sinφcosφ1>0-\sin\varphi-\cos\varphi-1>0 and can apply Lemma 2.2 again.


Case 4. 3π2<φ<2π\frac{3\pi}{2}<\varphi<2\pi. Then by Lemma 4.1 Aφ(a,b)T(n1,m),Aφ(a,b+1)T(n,m+1),Aφ(a+1,b)T(n,m1),Aφ(a+1,b+1)T(n+1,m)A_{\varphi}(a,b)\in T_{(n-1,m)},A_{\varphi}(a,b+1)\in T_{(n,m+1)},A_{\varphi}(a+1,b)\in T_{(n,m-1)},A_{\varphi}(a+1,b+1)\in T_{(n+1,m)}. The same computation as in Cases 1. and 2. lead to the system of inequalities

1+sinφcosφ\displaystyle 1+\sin\varphi-\cos\varphi \displaystyle\leq (2a+1)cosφ(2b+1)sinφ2n1<sinφ+cosφ1\displaystyle(2a+1)\cos\varphi-(2b+1)\sin\varphi-2n-1<-\sin\varphi+\cos\varphi-1
1+sinφcosφ\displaystyle 1+\sin\varphi-\cos\varphi \displaystyle\leq (2a+1)cosφ+(2b+1)sinφ2m1<sinφcosφ1\displaystyle(2a+1)\cos\varphi+(2b+1)\sin\varphi-2m-1<\sin\varphi-\cos\varphi-1

As sinφ,cosφ,1+sinφcosφ<0\sin\varphi,-\cos\varphi,1+\sin\varphi-\cos\varphi<0 we have sinφ+cosφ1>0-\sin\varphi+\cos\varphi-1>0 and can apply Lemma 2.2 again, which completes the proof of the theorem. \Box

5. Orbits with short periodicity

Our first goal was to study the periodicity of the orbits of rφr_{\varphi}, which seems to be very difficult. As a first step we examined some other properties of rφr_{\varphi}, but did not forget the ultimate goal. In this section we present a small finding, which corresponds to φ=π4.\varphi=\frac{\pi}{4}. We show that there are infinitely many a,a\in{\mathbb{Z}}, such that the orbit of rφr_{\varphi} generated by (a,0)(a,0) is periodic of length 8,8, i.e.

rφ8(a,0)=(a,0).r_{\varphi}^{8}(a,0)=(a,0).

Nevertheless we could present other examples, but this already shows that we do not have yet the necessary technique to prove much more general results.

In the next lemma we collected those identities, which are necessary to prove our periodicity result. Their proofs are one step direct computation. We denote, as usual in these notes, the fractional part of xx by {x}.\{x\}.

Lemma 5.1.

Let aa\in{\mathbb{Z}} and set ω=12a.\omega=\lfloor\frac{1}{\sqrt{2}}a\rfloor. Suppose 2ω=a1.\lfloor\sqrt{2}\omega\rfloor=a-1. Then

(5.1) 12a+12=ω\left\lfloor-\frac{1}{\sqrt{2}}a+\frac{1}{\sqrt{2}}\right\rfloor=-\omega
(5.2) 12a12={ω, if a=1ω1, otherwise \left\lfloor\frac{1}{\sqrt{2}}a-\frac{1}{\sqrt{2}}\right\rfloor=\begin{cases}\omega,\mbox{ if }a=1\\ \omega-1,\mbox{ otherwise }\end{cases}
(5.3) 12a+12={ω, if {12a}<112ω+1, if {12a}112\left\lfloor\frac{1}{\sqrt{2}}a+\frac{1}{\sqrt{2}}\right\rfloor=\begin{cases}\omega,\mbox{ if }\{\frac{1}{\sqrt{2}}a\}<1-\frac{1}{\sqrt{2}}\\ \omega+1,\mbox{ if }\{\frac{1}{\sqrt{2}}a\}\geq 1-\frac{1}{\sqrt{2}}\end{cases}
(5.4) 12a+2={ω, if {12a}>21ω+1, if {12a}21\left\lfloor-\frac{1}{\sqrt{2}}a+\sqrt{2}\right\rfloor=\begin{cases}-\omega,\mbox{ if }\{\frac{1}{\sqrt{2}}a\}>\sqrt{2}-1\\ -\omega+1,\mbox{ if }\{\frac{1}{\sqrt{2}}a\}\leq\sqrt{2}-1\end{cases}
(5.5) 2ω+2={a, if {2ω}+{2}<1a+1, if {2ω}+{2}1\left\lfloor\sqrt{2}\omega+\sqrt{2}\right\rfloor=\begin{cases}a,\mbox{ if }\{\sqrt{2}\omega\}+\{\sqrt{2}\}<1\\ a+1,\mbox{ if }\{\sqrt{2}\omega\}+\{\sqrt{2}\}\geq 1\end{cases}
(5.6) 2ω+12={a, if {2ω}>12a+1, if {2ω}12\left\lfloor-\sqrt{2}\omega+\frac{1}{\sqrt{2}}\right\rfloor=\begin{cases}-a,\mbox{ if }\{\sqrt{2}\omega\}>\frac{1}{\sqrt{2}}\\ -a+1,\mbox{ if }\{\sqrt{2}\omega\}\leq\frac{1}{\sqrt{2}}\end{cases}
Theorem 5.2.

Let a,a\in{\mathbb{Z}}, ω=12a\omega=\lfloor\frac{1}{\sqrt{2}}a\rfloor and suppose 2ω=a1.\lfloor\sqrt{2}\omega\rfloor=a-1. If {12a}[112,12],\{\frac{1}{\sqrt{2}}a\}\in\left[1-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right], then rφ8(a,0)=(a,0).r_{\varphi}^{8}(a,0)=(a,0). There exist infinitely many aa\in{\mathbb{Z}} satisfying the assumptions.

Proof.

First we prove that the assumptions imply {2ω}[112,12]\{\sqrt{2}\omega\}\in\left[1-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]. Indeed, [2ω]=a1[\sqrt{2}\omega]=a-1 means a12ω<aa-1\leq\sqrt{2}\omega<a (equality is only possible if a=1a=1), hence 0<a2ω<120<\frac{a}{\sqrt{2}}-\omega<\frac{1}{\sqrt{2}}. Further {12a}[112,12]\{\frac{1}{\sqrt{2}}a\}\in\left[1-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right] is equivalent to the sequence of inequalities

112\displaystyle 1-\frac{1}{\sqrt{2}} <\displaystyle< a2ω<12\displaystyle\frac{a}{\sqrt{2}}-\omega<\frac{1}{\sqrt{2}}
12\displaystyle-\frac{1}{\sqrt{2}} <\displaystyle< 2ωa<1+12\displaystyle\sqrt{2}\omega-a<-1+\frac{1}{\sqrt{2}}
112\displaystyle 1-\frac{1}{\sqrt{2}} <\displaystyle< 2ω(a1)<12,\displaystyle\sqrt{2}\omega-(a-1)<\frac{1}{\sqrt{2}},

which proves the claim.

Now we prove the second assertion. As 2\sqrt{2} is irrational, the sequence {a2}\left\{\frac{a}{\sqrt{2}}\right\} is by Theorem 2.1 uniformly distributed, thus there are infinitely many aa\in{\mathbb{Z}} satisfying {12a}[112,12]\{\frac{1}{\sqrt{2}}a\}\in\left[1-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right].

Now we turn to prove the first assertion. It is obvious that rφ(a,0)=(ω,ω).r_{\varphi}(a,0)=(\omega,\omega). Further by the assumption 2ω=a1\lfloor\sqrt{2}\omega\rfloor=a-1 we have rφ2(a,0)=(0,a1).r_{\varphi}^{2}(a,0)=(0,a-1). Further we have rφ3(a,0)=rφ(0,a1)=(12a+12,12a12).r_{\varphi}^{3}(a,0)=r_{\varphi}(0,a-1)=(\lfloor-\frac{1}{\sqrt{2}}a+\frac{1}{\sqrt{2}}\rfloor,\lfloor\frac{1}{\sqrt{2}}a-\frac{1}{\sqrt{2}}\rfloor). Thus by equations (5.1) and (5.2) we get rφ3=(ω,ω1).r_{\varphi}^{3}=(-\omega,\omega-1). Further, by equation 5.6 we have rφ4(a,0)=(a+1,1).r_{\varphi}^{4}(a,0)=(-a+1,-1). After that we get rφ5(a,0)=(ω+1,ω1)r_{\varphi}^{5}(a,0)=(-\omega+1,-\omega-1) by equation (5.4). Then rφ6(a,0)=(1,a).r_{\varphi}^{6}(a,0)=(1,-a). Further we get by equations (5.1) and (5.3) rφ7(a,0)=(ω+1,ω1).r_{\varphi}^{7}(a,0)=(\omega+1,-\omega-1). Finally, using equation (5.5) we have rφ8(a,0)=rφ(ω+1,ω1)=(2ω+2,0)=(a,0).r_{\varphi}^{8}(a,0)=r_{\varphi}(\omega+1,-\omega-1)=(\lfloor\sqrt{2}\omega+\sqrt{2}\rfloor,0)=(a,0).

Remark 5.3.

There exist infinitely many natural numbers a,a, which fulfill the conditions in Theorem 5.2, therefore there are infinitely many orbits with short periodicity.

We only give one class of starting points for the rotation by 4545^{\circ} in order to achieve periodicity of 8.8. There may exist many other starting points with the same periodicity.

6. Remarks on ROUNDAφROUND\circ A_{\varphi}

Following the proofs of Theorems 3.1 and 4.2 one can prove similar statements for the function ROUNDAφROUND\circ A_{\varphi}. To perform such a project one has to adjust Lemma 2.2 according the new rounding function. This is straight forward if one of cosφ\cos\varphi and sinφ\sin\varphi is irrational, but needs some computation otherwise. To save space we leave this to the interested reader.

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