aa r X i v : . [ m a t h . M G ] M a r Abstract
The Serpinsky-Knopp curve is characterized as the only curve (up to isometry) that maps aunit segment onto a triangle of a unit area, so for any pair of points in the segment, the squareof the distance between their images does not exceed four times the distance between them. keywords plane Peano curves, square-to-linear ratio, locality Sierpinski- Knopp curve, space-fillingcurves 1 bout the Serpinsky-Knopp curve.
E. V. Shchepin
The known Serpinsky-Knopp curve [1] maps the segment to an isosceles rectangular triangle insuch a way that one half of the segment is mapped to half of the triangle, and the other half — theother half. In this case, any of the” halves ” of the curve is similar to the entire curve. Therefore,each of the halves, in turn, is divided in half, so that the first quarter of the segment is displayed onthe first quarter of the triangle, the second — on the second, etc.The halving process continues indefinitely, resulting in a continuous map-ping of the unit segment s : [0 , → ∆ onto a triangle, such that the image ofany interval of type [ k − n , k n ], where k ≤ n are natural numbers, is a trian-gle, similar to the original with a similarity coefficient of 2 n/ . The segment[ k n , k +12 n ], following [2], is called fractal period of the curve s ( t ), and the re-striction of the curve to a fractal period — by the fraction of the order n of this curve. Different fractions of the same order are isometric with eachother. In figure 1, the broken line shows the order of passage of the fourth-order fractions of theSerpinsky-Knopp curve.In the future, we will treat the variable t as time. The square-linear ratio (SLR) of a pair of points p ( t ) , p ( t ) of a plane curve is called, according to [2], the ratio of the square of the Euclidean distancebetween their images to the time interval between them: | p ( t ) − p ( t ) | t − t (1)The similarity of fractions preserves the square-linear ratio. The exact upper bound of a SLR pairof curve points is called the locality of the curve. Among all currently known curves that map a unitsegment to a flat set of unit area, the Serpinsky-Knopp curve has the smallest locality(see[3]). Thefact that this curve has locality 4 is proved in the article [4]. We will give a simpler proof here. Lemma 1.
The maximum SLR of the s ( t ) curve is reached.Proof. Similar to the proof of theorem 1 of [5].
Lemma 2.
If the angle ∠ p ( t ) p ( t ) p ( t ) is not obtuse and t < t < t , then SLR pair p ( t ) p ( t ) isnot less than the maximum of the SLR pairs p ( t ) p ( t ) and p ( t ) p ( t ) . And if the angle is sharp, it isless.Proof. Indeed, the condition for an angle is equivalent to an inequality | p ( t ) − p ( t ) | + | p ( t ) − p ( t ) | ≤ | p ( t ) − p ( t ) | , ∗ Steklov Math. Institute [email protected] † This work was supported from a grant to the Steklov International Mathematical Center in the framework of thenational project ”Science” of the Russian Federation.. roof that the locality of s ( t ) is equal to four. Consider a pair of points s ( t ) s ( t ′ ) of the Sierpinski-Knopp curve with a maximum SLR. Let the points A and C be the ends of the fraction the largestorder containing this pair. Then s ( t ) and s ( t ′ ) lie in different the halves of this faction. The middleof the fraction is B . According to the lemme 2 angle ∠ s ( t ) Bs ( t ′ ) must be straight, and The SLR ofboth pairs s ( t ) B and Bs ( t ′ ) is also maximum. Thus s ( t ) lies on the side of the fraction, and B is itsvertex. Consider the largest-order fraction containing the pair s ( t ) , B . In that case s ( t ) lies on thefirst half of the hypotenuse of this fraction. Since the rectangular vertex B ′ of this fraction passesbetween s ( t ), then the angle ∠ s ( t ) , B ′ b must be straight. And this is only possible if s ( t ) is the vertexof the fraction. But in this case, the SLR of the s ( t ) s ( t ′ ) pair is equal to four.The main result of the article is as follows: Theorem 1.
There is a unique, up to isometry, mapping of a unit segment onto a triangle of a unitarea whose locality is four.
The proof of the theorem is preceded by several lemmas.
Lemma 3.
If a triangle of unit area has sides a, b, c that satisfy the inequalities c ≤ ≥ a + b , then a = b = √ , c = 2 .Proof. Consider a triangle with the sides a ≤ b ≤ c , which satisfies the inequalities of the Lemma andhas the maximum area. Then, for him, the Lemma inequalities will obviously reverse the equality c = 2, a + b = 4. Moreover, this triangle will have equal sides a = b . Indeed, if a = b , then thesame height isosceles triangle with the base c and the sides a , b has the same area and satisfies thestrict inequality a + b <
4, and therefore cannot have a maximum area. Where do we get that themaximum area of a triangle satisfying the Lemma’s inequalities is a rectangular isosceles, and othertriangles that satisfy the Lemma’s inequalities have a smaller area.
Lemma 4.
If the curve p ( t ) : [ a, b ] → R of locality 4 passes through all three vertices of the rectangularisosceles triangle ABC , whose area is equal to the length of the segment, then the beginning and endof the segment [ a, b ] pass to the sharp-angled vertices A , B a triangle, and the middle of the segment— into its rectangular vertex C .Proof. Since the area of the triangle
ABC is equal to b − a , its cathets are equal to p b − a ), andthe hypotenuse is equal to 2 √ b − a . Let t A , t B , and t C denote the points of passing the vertices of thetriangle. Then by virtue of the condition on the locality of the curve | t B − t A | ≥ | B − A | = b − a ,where | t A − t B | = b − a , which is possible only if the moments t A and t B coincide with the ends of thesegment. Further, the inequalities | t A − t C | ≥ | C − A | = ( b − a ) = | B − A | ≤ | t B − t C | entailequality distances from t C to both ends of the segment. A ′ C ′ A C B Proof of the theorem.
Let a triangle with the sides a ≤ b ≤ c of the unit area isan image of the p ( t ) curve with locality 4. Let’s show that in this in this case, thetriangle is rectangular and isosceles. The condition for locality 4 gives inequalities a + b ≤ ≥ c . So on based on the 3 Lemma, we conclude that the the triangleis isosceles and rectangular with a hypotenuse equal to 2.It follows from the Lemma 4 that the ends of the curve p ( t ) coincide withsharp-angled vertices triangle image. Consider the Serpinsky-Knopp curve s ( t )with the same image as p ( t ), the beginning of which s (0) coincides with p (0). In this case, the sameLemma implies that p (1) = s (1). We will prove by induction on n that p (cid:18) k n (cid:19) = s (cid:18) k n (cid:19) (2)3or all binary-rational points of a unit segment. For n = 0, our statement is proved above. Assumethat the equality (2) is true for all k ≤ n . To prove the induction step, it is sufficient to make surethat p (cid:18) k + 12 n +1 (cid:19) = s (cid:18) k + 12 n +1 (cid:19) (3)when k < n . Consider the ABC fraction of the s ( t ) curve of order n , So that A = p ( k n ) and C = p ( k +12 n are its sharp-angled vertices, and B = s ( k +12 n +1 ) — rectangular. It follows from the Lemma4 that if B ∈ p (cid:20) k n , k + 12 n (cid:21) , (4)then B = p ( k +12 n +1 ). Therefore, for a complete proof of (2), it is sufficient to provethe inclusion (4). Consider all fractions of the n -th order of the curve s ( t ), containing B . Thereare no more than four of them. All of them are contained in a square with the center B and the side AC . Let’s denote the vertices of this square A ′ and C ′ (see figure 2). Others are located at a distancenot less than the length of AC . Therefore, p - images of their fractal periods, coinciding at the endswith s ( t ). it is too far from B (at least the length of AC ) to enable it. Therefore, B can belong toimages of only the following four curve segments of type p [ i n , i +12 n ]: these are segments of the curvewith ends in AC , AA ′ , A ′ C ′ , CC ′ . According to the Lemma 2 these curve segments are containedin circles with diameters AC , AA ′ , A ′ C ′ , CC ′ respectively. But the point B is not internal to theunion of three circles with diameters AA ′ , A ′ C ′ , CC ′ . Therefore, there are points arbitrarily close to B belonging to p [ k n , k +12 n ]. Due to the closedness of the latter we conclude that B itself belongs toit. Thus, it is established that the considered curve coincides with the Serpinsky-Knopp curve in allbinary-rational points. From the continuity of the curves, their complete coincidence follows. References [1]
Hans Sagan , ”Space-Filling Curves.”, Universitext. Springer, 1994.[2]
Shchepin E. , “On fractal Peano Curves”, Proc. Steklov Inst. Math., (2004), 272–280[3]
Haverkort H., F. van Walderveen , Locality and bounding-box quality of two dimensionalspace-filling curves. Computational Geometry, Theory and Applications 43(2) (2010) 131–147.[4]
Niedermeier R., Reinhardt K., Sanders P. , Towards optimal locality in mesh-indexings,Discrete Applied Mathematics 117 (2002) 211–237.[5]