# Arithmetic-Analytic Representation of Peano Curve

Authors: Guangjun Yang; Xiaoling Yang; Ping Wang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

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 indexα=1/2, which implies that the curve is everywhere continuous and nowhere differentiable.


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## Body

## 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. [1] developed a matrix multiplication utilizing the Peano curves in designing a cache oblivious algorithm. Platos et al. [2] created a model of signal coverage based on optimized representation by space-filling curves to reduce memory consuming in computation. Sasidharan and Snir [3] 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 [4].Peano in 1890 [5] 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 1–3.Figure 1
Filling curve path 1.Figure 2
Filling curve path 2.Figure 3
Filling curve path 3.The most discussed is the one shown in Figure4.Figure 4
Filling curve path 4.For example, Moore in [6] and Milne in [7] conducted a very thorough study of it.More discussions of constructions of such curves can be found in Sagan [8] and Jaffard and Nicolay [9, 10]. Recently, Makarov and Podkorytov [11] 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 indexα=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. [8]) 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.Divide0,1 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=1∞tk/32k,tk∈0,1,2,…,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 segment0≤t≤12 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,13e−iπ/2z0t0+13+i,13e−iπz0t0+231+i,13e−iπ/2z0t0+1+i3,13z0t0+23,13eiπ/2z0t0+1+i3,13z0t0+231+i,which represent the nine diagonals in Figure 4 as functions of t0∈0,1. 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.Eliminatingt0 in (3) and (4), we get(5)z1t=13z032t,0≤t≤132,13eiπ/2z032t−1+1+i3,132≤t≤232,13z032t−1+23i,232≤t≤13,13e−iπ/2z032t−1+13+i,13≤t≤432,13e−iπz032t−4+231+i,432≤t≤532,13e−iπ/2z032t−5+1+i3,532≤t≤23,13z032t−6+23,23≤t≤732,13eiπ/2z032t−7+1+i3,732≤t≤832,13z032t−8+231+i,832≤t≤1.Denotez1=x1+iy1. Separating the real and imaginary parts in (5), we have(6)x1t=13x032t,0≤t≤132,−13x032t−1+13,132≤t≤232,13x032t−2,232≤t≤13,13x032t−3+13,13≤t≤432,−13x032t−4+23,432≤t≤532,13x032t−5+13,532≤t≤23,13x032t−6+23,23≤t≤732,−13x032t−7+1,732≤t≤832,13x032t−8+23,832≤t≤1,(7)y1t=13y032t,0≤t≤132,13y032t−1+13,132≤t≤232,13y032t−2+23,23≤t≤13,−13y032t−3+1,13≤t≤432,−13y032t−4+23,432≤t≤532,−13y032t−5+13,532≤t≤23,13x032t−6,23≤t≤732,13x032t−7+13,732≤t≤832,13x032t−8+23,832≤t≤1.Introducec1,c2∈0,1,2 in ternary form. Equations (6) and (7) can be simplified as, for τt∈0,1,(8)x=c1+1/21−−1c23+−1c23xτt,(9)y=−1c13c2−321−−1c1+−1c13yτt.Applying (8) and (9) again for xτt,yτt, and c3,c4∈0,1,2,(10)xt=c3+1/21−−1c43+−1c43xτ2t,yt=−1c33c4−321−−1c3+−1c33yτ2t.Substituting (10) back into (8) and (9), we have(11)xt=c1+1/21−−1c23+−1c232c3+121−−1c4+−1c2+c432xτ2t,yt=−1c13c2−321−−1c1+−1c1+c332c4−321−−1c3+−1c1+c332yτ2t.Continuing this process, with(12)φc2k−1,c2k=c2k−1−121−−1c2k,ψc2k−1,c2k=c2k−321−−1c2k−1,we obtain(13)xt=∑k=1n−1c2+⋯+c2k−23kφc2k−1,c2k+−1c2+⋯+c2n3nxτnt,yt=∑k=1n−1c2+⋯+c2k−13kψc2k−1,c2k+−1c1+⋯+c2n−13nyτnt.Since0≤xτnt≤1, 0≤yτnt≤1, sending n⟶∞, we arrive at the series expression of the Peano curve:(14)xt=∑k=1∞−1c2+⋯+c2k−23kφc2k−1,c2k,yt=∑k=1∞−1c2+⋯+c2k−13kψc2k−1,c2k,c2k−1,c2k∈0,1,2.Both series in (14) converge because φ≤3andψ≤3.


## 3. Properties of the Peano Curve

x
t
and
y
t in (14) are functions of c2k−1,c2k∈1,2,3 with(15)t=c13+c232+⋯+c2k−132k−1+c2k32k+⋯,ck∈0,1,2,3.Theorem 3.1.
Curve (14) fills a unit square up.Proof.
Let us first check the expression ofyt:(16)ψc2k−1,c2k3k=c2k−3/21−−1c2k−13k=c2k3k,ifc2k−1=0,2,2−c2k3k13k,ifc2k−1=1,c2k−23k−1+−1c2k−13kψc2k−1,c2k=c2k−23k−1+−1c2k−13kc2k−321−−1c2k−1=c2k−23k−1+c2k3k,ifc2k−1=0,2,c2k−23k−1+2−c2k3k+13k,ifc2k−1=1,2−c2k−23k−1+13k−1+−1c2k−13kψc2k−1,c2k=2−c2k−13k−1+c2k3k+13k,ifc2k−1=0,2,2−c2k−13k−1+c2k3k,ifc2k−1=1.
Denotingck∗=ck or 2−ck,(17)−1c13ψc1,c2+−1c1+c232ψc3,c4=c1∗3+c332,if−1c1+c3=1,c13+2−c332+132,if−1c1+c3=−1.
We have(18)yt=∑k=1∞−1c1+⋯+c2k−13kψc2k−1,c2k=∑k=1n−1c2k∗3k+c2n3n,if−1c1+⋯+c2k−1=1,2−c2n3n+13n,if−1c1+⋯+c2k−1=−1,that is,(19)yt=∑k=1∞−1c1+⋯+c2k−13kψc2k−1,c2k=∑k=1∞c2k∗3k,c2k∗∈0,1,2.
Similarly,(20)xt=∑k=1n−1c2+⋯+c2k−23kφc2k−1,c2k=c1+1/21−−1c23+−1c232c3+121−−1c4+⋯+−1c2+⋯+c2k−13k−1c2k−3+121−−1c2k−2+−1c2+⋯+c2k−23kc2k−1+121−−1c2k+⋯=c13+−1c232c3−321−−1c2+−1c2+c432c5−321−−1c4+⋯+−1c2+⋯+c2k−23kc2k−1−321−−1c2k−2+⋯=c13+∑k=2∞−1c2+⋯+c2k−23kψc2k−1,c2k.
Therefore,(21)xt=c13+∑k=2∞c2k−1∗3k=∑k=1∞c2k−1∗3k,c2k−1∗∈0,1,2.
Because the arbitrarity ofc′,c2k−1∗,c2k∗∈0,1,2,(22)xt=∑k=1∞c2k−1∗3k,yt=∑k=1∞c2k∗3k,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+1≤t′−t″<132l,(24)xt′−xt″≤232l+1<32t′−t″1/2,(25)yt′−yt″≤332l+1<32t′−t″1/2,which means x and y satisfy the Hölder inequality with index α=1/2.
In fact, from (15), m/32l≤t′,t″≤m+1/32l,m=0,1,…,32l, i.e.,(26)t′=∑k=12lck3k+∑k=l+1∞ck3k,t″=∑k=12lck3k+∑k=2l+1∞ck′3k,ck,ck′∈0,1,2.
Thus,(27)xt′=∑k=1l−1c2+⋯+c2k−23kφc2k−1,c2k+∑k=2l+1∞−1c2+⋯+c2k−23kφc2k−1,c2k,xt″=∑k=1l−1c2+⋯+c2k−23kφc2k−1,c2k+−1c2+⋯+c2k−23kφc2k−1,c2k,xt′−xt″≤∑k=l+1∞23k=23l∑k=1∞13k=13l−1=323l+1=32t′−t″1/2.
Similarly,(28)yt′−yt″≤32t′−t″1/2.
Inequalities (24) and (25) imply that xt and yt are continuous everywhere.
Next, we prove thatxt and yt are nowhere differentiable.
Take(29)t′=c13+c232+⋯+c2l−132l−1+c2l32l+m32l=m32l+∑k=l+1∞23k,t″=c13+c232+⋯+c2l−132l−1+c2l32l=m32l.
Then,(30)t′−t″=132l,xt′=∑k=1l−1c2+⋯+c2k−23kφc2k−1,c2k+−1c2+⋯+c2l∑k=2l+1∞φ2,23k=∑k=1l−1c2+⋯+c2k−23kφc2k−1,c2k+−1c2+⋯+c2l3k,xt″=∑k=1l−1c2+⋯+c2k−23kφc2k−1,c2k.
Therefore,(31)xt′−xt″=xm22l+132l−xm32l=13l=t′−t″1/2.
For anyt∈0,1 and any positive integer l, there exists m∈0,1,…,32l−1, such that(32)m32l≤t≤m+132l.
Ift=m/32l, it follows from (31) that(33)xt−xm+132l=xt−xt′=t−t′1/2.
Ifm/32l<t<m+1/32l, then either(34)xm32l−xt≥12xm32l−xm+132l=12⋅3l,orxt−xm+132l≥12xm32l−xm+132l=12⋅3l,must hold. Otherwise, we would have both(35)xm32l−xt<12xm32l−xm+132l,xt−xm+132l<12xm32l−xm+132l,which imply(36)xm32l−xt+xt−xm+132l<xm32l−xm+132l,contradicting the triangle inequality. Therefore, we can always take t′=m/32l or m+1/32l, such that t−t′≤1/32l and(37)xt−xt′≥12⋅22l≥12t′−t″1/2.
Therefore,(38)xt−xt′t−t′≥32l2⋅3l=3l2,which shows xt is not differentiable.
Similarly, we can also show thatyt is not differentiable.
Theorems 3.1 and 3.2 imply that the Peano curvex,y is a continuous curve that fills a unit square and it is nowhere differentiable.



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*Source: 6745202-2019-09-10.xml*