IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi 10.1155/2019/6745202 6745202 Research Article Arithmetic-Analytic Representation of Peano Curve Yang Guangjun 1 Yang Xiaoling 2 https://orcid.org/0000-0003-1952-0076 Wang Ping 3 Lasiecka Irena 1 College of Mathematics and Statistics Yunnan University Kunming 650091 China ynu.edu.cn 2 College of Mathematics and Statistics Yunnan University of Finance and Economy Kunming 650221 China ynufe.edu.cn 3 Department of Mathematics Penn State University Schuylkill Haven PA 17972 USA psu.edu 2019 1092019 2019 01 06 2019 16 07 2019 06 08 2019 1092019 2019 Copyright © 2019 Guangjun Yang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this work, we obtained a nonmatrix analytic expression for the generator of the Peano curve. Applying the iteration method of fractal, we established a simple arithmetic-analytic representation of the Peano curve as a function of ternary numbers. We proved that the curve passes each point in a unit square and that the coordinate functions satisfy a Hölder inequality with index α=1/2, which implies that the curve is everywhere continuous and nowhere differentiable.

1. Introduction

Space-filing curves, such as the Peano curves, are geometrically interesting curves and have important applications, particularly in parallel computing. Bagga et al.  developed a matrix multiplication utilizing the Peano curves in designing a cache oblivious algorithm. Platos et al.  created a model of signal coverage based on optimized representation by space-filling curves to reduce memory consuming in computation. Sasidharan and Snir  showed how to reduce communication and improve the quality of partitions using a space-filling curve. Much more applications of space-filing curves can be found in Bader’s book titled Space-Filling Curves: An Introduction with Applications in Scientific Computing .

Peano in 1890  geometrically constructed a continuous curve, later called the Peano curve, that fills the unit square 0,1×0,1. The idea of the construction is to divide each square into 9 smaller equal squares continuously and to determine a path, or curve, so it goes through each square. The limit of this path is the Peano curve. Many different such paths can be designed as shown in Figures 13.

Filling curve path 1.

Filling curve path 2.

Filling curve path 3.

The most discussed is the one shown in Figure 4.

Filling curve path 4.

For example, Moore in  and Milne in  conducted a very thorough study of it.

More discussions of constructions of such curves can be found in Sagan  and Jaffard and Nicolay [9, 10]. Recently, Makarov and Podkorytov  constructed a nonsymmetric plane Peano curves whose coordinate functions satisfy the Lipschitz conditions. However, there has not been any analytic expression of the Peano curve with which we can study them analytically and more thoroughly.

The goal of this work is to use fractal iteration method to establish an analytic expression of the iterated function system (IFS), then a series representation of the Peano curve; thus, we are able to discuss various properties of the curve. Specifically, we will prove the coordinate functions of the curve satisfy a Hölder inequality with index α=1/2, which shows that the Peano curve is everywhere continuous and nowhere differentiable. We will also show the analytically constructed curve passes each point in a unit square.

2. Peano Curve Generated by Iteration

Sagan (cf. ) studied the Peano curve with an IFS expressed in matrix form. Matrix multiplication makes it very complicated to work with the IFS. In this section, we will obtain the IFS for the Peano curve in an analytic (series) form, which allows us not only to obtain more convenient coordinate functions xt,yt for the curve but also to prove the nondifferentiability of the functions.

Divide 0,1 into 9 equal subintervals, each of which is divided into 9 equal subintervals. We have, k-steps later, 32k subintervals Iik, each with length 1/32k. Similarly keep dividing 0,1×0,1 into 9 smaller equal squares, k-steps later, we have 32k squares Sik each with side length 1/3k. The problem is to find for t=k=1tk/32k,tk0,1,2,,9(1)x=φt,y=ψt,mapping Iik to Sik.

The initiator of the Peano curve is a diagonal of a unit square and can be obtained by stretching line segment 0t12 times and rotating 45 counterclockwise, represented in complex coordinate system by(2)z0=2eiπ/4=1+it.

The generator of the Peano curve can be expressed as(3)z1t=13z0t0,13eiπ/2z0t0+1+i3,13z0t0+23i,13eiπ/2z0t0+13+i,13eiπz0t0+231+i,13eiπ/2z0t0+1+i3,13z0t0+23,13eiπ/2z0t0+1+i3,13z0t0+231+i,which represent the nine diagonals in Figure 4 as functions of t00,1. Divide 0,1 into nine equal sections:(4)t=t032,t032+132,t032+232,t032+13,t032+13+132,t032+13+232,t032+23,t032+23+132,t032+23+232.

Eliminating t0 in (3) and (4), we get(5)z1t=13z032t,0t132,13eiπ/2z032t1+1+i3,132t232,13z032t1+23i,232t13,13eiπ/2z032t1+13+i,13t432,13eiπz032t4+231+i,432t532,13eiπ/2z032t5+1+i3,532t23,13z032t6+23,23t732,13eiπ/2z032t7+1+i3,732t832,13z032t8+231+i,832t1.

Denote z1=x1+iy1. Separating the real and imaginary parts in (5), we have(6)x1t=13x032t,0t132,13x032t1+13,132t232,13x032t2,232t13,13x032t3+13,13t432,13x032t4+23,432t532,13x032t5+13,532t23,13x032t6+23,23t732,13x032t7+1,732t832,13x032t8+23,832t1,(7)y1t=13y032t,0t132,13y032t1+13,132t232,13y032t2+23,23t13,13y032t3+1,13t432,13y032t4+23,432t532,13y032t5+13,532t23,13x032t6,23t732,13x032t7+13,732t832,13x032t8+23,832t1.

Introduce c1,c20,1,2 in ternary form. Equations (6) and (7) can be simplified as, for τt0,1,(8)x=c1+1/211c23+1c23xτt,(9)y=1c13c23211c1+1c13yτt.

Applying (8) and (9) again for xτt,yτt, and c3,c40,1,2,(10)xt=c3+1/211c43+1c43xτ2t,yt=1c33c43211c3+1c33yτ2t.

Substituting (10) back into (8) and (9), we have(11)xt=c1+1/211c23+1c232c3+1211c4+1c2+c432xτ2t,yt=1c13c23211c1+1c1+c332c43211c3+1c1+c332yτ2t.

Continuing this process, with(12)φc2k1,c2k=c2k11211c2k,ψc2k1,c2k=c2k3211c2k1,we obtain(13)xt=k=1n1c2++c2k23kφc2k1,c2k+1c2++c2n3nxτnt,yt=k=1n1c2++c2k13kψc2k1,c2k+1c1++c2n13nyτnt.

Since 0xτnt1, 0yτnt1, sending n, we arrive at the series expression of the Peano curve:(14)xt=k=11c2++c2k23kφc2k1,c2k,yt=k=11c2++c2k13kψc2k1,c2k,c2k1,c2k0,1,2.

Both series in (14) converge because φ3 and ψ3.

3. Properties of the Peano Curve

x t  and  y t in (14) are functions of c2k1,c2k1,2,3 with(15)t=c13+c232++c2k132k1+c2k32k+,ck0,1,2,3.

Theorem 3.1.

Curve (14) fills a unit square up.

Proof.

Let us first check the expression of yt:(16)ψc2k1,c2k3k=c2k3/211c2k13k=c2k3k,if c2k1=0,2,2c2k3k13k,if c2k1=1,c2k23k1+1c2k13kψc2k1,c2k=c2k23k1+1c2k13kc2k3211c2k1=c2k23k1+c2k3k,if c2k1=0,2,c2k23k1+2c2k3k+13k,if c2k1=1,2c2k23k1+13k1+1c2k13kψc2k1,c2k=2c2k13k1+c2k3k+13k,if c2k1=0,2,2c2k13k1+c2k3k,if c2k1=1.

Denoting ck=ck or 2ck,(17)1c13ψc1,c2+1c1+c232ψc3,c4=c13+c332,if 1c1+c3=1,c13+2c332+132,if 1c1+c3=1.

We have(18)yt=k=11c1++c2k13kψc2k1,c2k=k=1n1c2k3k+c2n3n,if 1c1++c2k1=1,2c2n3n+13n,if 1c1++c2k1=1,that is,(19)yt=k=11c1++c2k13kψc2k1,c2k=k=1c2k3k,c2k0,1,2.

Similarly,(20)xt=k=1n1c2++c2k23kφc2k1,c2k=c1+1/211c23+1c232c3+1211c4++1c2++c2k13k1c2k3+1211c2k2+1c2++c2k23kc2k1+1211c2k+=c13+1c232c33211c2+1c2+c432c53211c4++1c2++c2k23kc2k13211c2k2+=c13+k=21c2++c2k23kψc2k1,c2k.

Therefore,(21)xt=c13+k=2c2k13k=k=1c2k13k,c2k10,1,2.

Because the arbitrarity of c,c2k1,c2k0,1,2,(22)xt=k=1c2k13k,yt=k=1c2k3k,fills the square S=0,1×0,1 up.

Theorem 3.2.

Curve (14) is everywhere continuous and nowhere differentiable.

Proof.

We will first show for(23)132l+1tt<132l,(24)xtxt232l+1<32tt1/2,(25)ytyt332l+1<32tt1/2,which means x and y satisfy the Hölder inequality with index α=1/2.

In fact, from (15), m/32lt,tm+1/32l,m=0,1,,32l, i.e.,(26)t=k=12lck3k+k=l+1ck3k,t=k=12lck3k+k=2l+1ck3k,ck,ck0,1,2.

Thus,(27)xt=k=1l1c2++c2k23kφc2k1,c2k+k=2l+11c2++c2k23kφc2k1,c2k,xt=k=1l1c2++c2k23kφc2k1,c2k+1c2++c2k23kφc2k1,c2k,xtxtk=l+123k=23lk=113k=13l1=323l+1=32tt1/2.

Similarly,(28)ytyt32tt1/2.

Inequalities (24) and (25) imply that xt and yt are continuous everywhere.

Next, we prove that xt and yt are nowhere differentiable.

Take(29)t=c13+c232++c2l132l1+c2l32l+m32l=m32l+k=l+123k,t=c13+c232++c2l132l1+c2l32l=m32l.

Then,(30)tt=132l,xt=k=1l1c2++c2k23kφc2k1,c2k+1c2++c2lk=2l+1φ2,23k=k=1l1c2++c2k23kφc2k1,c2k+1c2++c2l3k,xt=k=1l1c2++c2k23kφc2k1,c2k.

Therefore,(31)xtxt=xm22l+132lxm32l=13l=tt1/2.

For any t0,1 and any positive integer l, there exists m0,1,,32l1, such that(32)m32ltm+132l.

If t=m/32l, it follows from (31) that(33)xtxm+132l=xtxt=tt1/2.

If m/32l<t<m+1/32l, then either(34)xm32lxt12xm32lxm+132l=123l,or xtxm+132l12xm32lxm+132l=123l,must hold. Otherwise, we would have both(35)xm32lxt<12xm32lxm+132l,xtxm+132l<12xm32lxm+132l,which imply(36)xm32lxt+xtxm+132l<xm32lxm+132l,contradicting the triangle inequality. Therefore, we can always take t=m/32l or m+1/32l, such that tt1/32l and(37)xtxt1222l12tt1/2.

Therefore,(38)xtxttt32l23l=3l2,which shows xt is not differentiable.

Similarly, we can also show that yt is not differentiable.

Theorems 3.1 and 3.2 imply that the Peano curve x,y is a continuous curve that fills a unit square and it is nowhere differentiable.

Data Availability

The iteration data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Bagga S. Girdhar A. Trivedi M. C. Yang Y. Z. RMI approach to cluster based cache oblivious peano curves Proccedings of the Second International Conference on Computational Intelligence & Communication Technology (CICT) February 2016 Ghaziabad, India IEEE Xplore Platos J. Kromer P. Snasel V. Efficient area association using space filling curves Proceedings of the International Conference on Intelligent Networking and Collaborative Systems September 2015 Taipei, Taiwan IEEE Xplore Sasidharan A. Snir M. Space-filling curves for partitioning adaptively refined meshes Mathematics and Computer Science 2015 Bader M. Space-Filling Curves: An Introduction with Applications in Scientific Computing 2013 Berlin, Germany Springer-Verlag Peano G. Sur une courbe, qui remplit toute une aire plane Mathematische Annalen 1890 36 1 157 160 10.1007/bf011994382-s2.0-0005083863 Moore E. H. On certain crinkly curves Transactions of the American Mathematical Society 1900 1 1 72 90 10.1090/s0002-9947-1900-1500526-42-s2.0-0001955089 Milne S. C. Peano curves and smoothness of functions Advances in Mathematics 1980 35 2 129 157 10.1016/0001-8708(80)90045-62-s2.0-0000822667 Sagan H. Space-Filling Curve 1994 Berlin, Germany Springer-Verlag Jaffard S. Nicolay S. Pointwise smoothness of space-filling functions Applied and Computational Harmonic Analysis 2009 26 2 181 199 10.1016/j.acha.2008.04.0022-s2.0-59649084186 Jaffard S. Nicolay S. Recent developments in fractals and related fields Space-Filling Function and Daveport Series 2010 Berlin, Germany Springer-Verlay Makarov B. M. Podkorytov A. N. On the coordinate functions of peano curves St. Petersburg Mathematical Journal 2017 28 1 115 125 10.1090/spmj/14412-s2.0-85010471371