A convex cover for closed unit curves has area at least 0.1
AA convex cover for closed unit curves hasarea at least 0.1
Bogdan Grechuk and Sittichoke Som-am
Abstract.
We improve a lower bound for the smallest area of convexcovers for closed unit curves from 0 . .
1, which makes it sub-stantially closer to the current best upper bound 0 . , and rectangle with side 0 . × . . . Mathematics Subject Classification (2010).
Primary 52C15; Secondary52A38.
Keywords.
Worm problem, Convex cover, Convex hull, Minimal-areacover, Closed curves.
1. Introduction
Moser’s worm problem is one of many unsolved problems in geometry posedby Leo Moser [11] in 1966 (see [12] for a new version). It asked about thesmallest area of region S in the plane which can be rotated and translatedto cover every unit arc. While this problem is still open, lower and upperbounds for this smallest area are available in the literature. The current bestupper bound 0 . S is strictly positive [10].This problem can be modified by restricting region S (e.g. triangle,rectangle, convex, non-convex, ect.) or restricting unit curves (e.g. closed,convex, ect.). Numerous problems in the worm family have been solved prior,see Wetzel [17] for a survey.If we require S to be convex, and cover arbitrary unit curves, the ex-istence of the solution is shown by Laidecker and Poole [9] using BlachkeSelection Theorem. In 2019, Panraksa and Wichiramala [14] showed that the a r X i v : . [ m a t h . M G ] A p r Bogdan Grechuk and Sittichoke Som-amWetzel’s sector is a cover for unit arcs which has an area of π/ ≈ . . S to cover all (not neces-sarily unit) curves of diameter one was posed by Henri Lebesgue in 1914 andis known as Lebesgue’s universal covering problem. In this case, the currentbest lower and upper bounds are 0 .
832 and 0 . closed unit curves bya convex region of the smallest area.In 1957, H.G. Eggleton [3] showed that the equilateral triangle of side √ π is the smallest triangle which can cover all unit closed curves and itsarea is √ π ≈ . π and (cid:113) − π is found by Shaer and Wetzel [15] and its area isabout 0 . . . . . . × . . . . . α to be the area of smallest cover of closed unit curves,then 0 . ≤ α ≤ . . In this paper, we use geometric methods combined with the Box searchalgorithm to prove that the area of convex cover for the line segment, circle,and certain rectangle of perimeter 1 is at least 0 .
1. convex cover for closed unit curves has area at least 0.1 3
Theorem 1.1.
The area of convex cover S for circle of perimeter , line oflength / , and rectangle of size . × . is at least . . By Theorem 1.1, we have 0 . ≤ α, which improves the previous bound 0 . S in Theorem 1.1 is about 0 . .
2. A systematic search for shape
Theorem 1.1 establishes a new lower bound for the area of a convex cover forclosed unit curves, by proving that a convex hull of three such curves, circle,line, and certain rectangle, must have area at least 0 .
1. In fact, the choice ofthese three curves was not an arbitrary or lucky one, but was the result of asystematic numerical search, which we will describe in this section.This section contains neither proofs nor any form of rigorous analysis,and the reader who is interested only in the proof of Theorem 1.1 can gostraight to the next sections. However, this section is important to understandhow we selected three curves considered in Theorem 1.1, and provides someguide were to not look when trying to substantially improve our lower bound.For each specific set of unit curves, we write a program which findsminimum area of a convex hull of these curves, where the minimum is takenwith respect to all translations and rotations of the curves. We then modifythe curves to make this minimal as large as possible. This result is a maximinproblem with highly non-smooth and non-convex objective function. We usedMatlab function, patternsearch, with random initial data to try to find theglobal optimal value in various cases.Let α be the area of smallest cover of closed unit curves. We will approx-imate curves by polygons, and it is clear that it suffices to consider convexpolygons only. Let us start with 2 polygons. Let F ( X , X ) be the area of con-vex hull of polygons X and X . Let T be the set of all orientation-preservingmotions T of the plane (that is, all possible compositions of rotations andtranslations). Then, given any two polygons X and X with unit perimeter,the solution of the minimization problemmin T ∈T F ( X , T ( X )) (2.1) Bogdan Grechuk and Sittichoke Som-amType ( n + m points) Optimal area Time (sec)2+3 0.072169 58893+3 0.072375 79953+4 0.085377 140804+4 0.085377 173703+5 0.087902 172965+5 0.087902 236849+19 0.095790 23982820+20 0.096120 2402589+50 0.096595 24117211+50 0.096605 251041 Table 1.
The numerical series for 2 objectsis the lower bound for α .Now, our aim to find polygons X and X for which this lower boundis as large as possible. Let N and N be a number of vertices in polygons X and X respectively. We assume that N and N are fixed but X and X can vary. Let X ( N ) be the set of all convex polygons with N vertices andunit perimeter. We consider optimization problem b ( N , N ) = max X ∈X ( N ) ,X ∈X ( N ) min T ∈T F ( X , T ( X )) . (2.2)It is clear that for any N , N , b ( N , N ) ≤ α, or, in words, b ( N , N ) is a lower bound for α . First, we construct a Matlab function NN2per to solve the minimizationproblem (2.1). The input of the function is 2 N + 2 N coordinates of verticesof X and X . The function first calculates the perimeters of the polygons,and scale them to make the perimeters to be equal to 1. Then it applies themotion T to X , which is described by three parameters: vector of transla-tion ( x , y ) and angle of rotation θ , and use Matlab function convhull toestimate F ( X , T ( X )). We then use function MultiStart in Matlab to solvethe minimization problem of the resulting convex hull area over parameters x , y , θ .Second, we solve the maximization problem (2.2) by finding maximalpossible output of function NN2per for fixed N and N . We start with someinitial configuration (for example, X is regular N -gon and X regular N -gon), and then use the patternsearch function.We repeat this procedure for various small values of N and N . Specifi-cally, we consider the cases of a line and a triangle (2+3 points), two triangles(3+3 points), a triangle and a rectangle (3+4 points), two rectangles (4+4points), and so on. The results are shown in Table 1. convex cover for closed unit curves has area at least 0.1 5 Figure 1.
The configuration for the 11 points and 50 pointswhich has area of 0 . . . For three objects, we use the same idea as for 2 objects. Let X be the thirdpolygon with N vertices which can be translated and rotated. We then solvethe maximin optimization problem b ( N , N , N ) = max X ∈X ( N ) ,X ∈X ( N ) ,X ∈X ( N ) , min T ∈T ,T ∈T F ( X , T ( X ) , T ( X )) , (2.3)where F now denotes the area of convex hull of three polygons. Again, b ( N , N , N ) is the lower bound for α .For the internal minimization problem, we now optimize over six param-eters which are x , y , θ and x , y , θ for translation and rotation of first andsecond polygons respectively. The maximization problem is now with respectto 2 N + 2 N + 2 N coordinates of polygon’s vertices.We start the cases of line and two triangles (2+3+3), three triangles(3+3+3), two triangles and a rectangle (3+3+4) and so on. We can see theresults in Table 2.Table 2 shows that the maximum lower bound for α we found is 0 . . n + m + k points) Optimal area Time (sec)2+3+3 0.072169 257413+3+3 0.073086 291782+3+4 0.087867 355183+3+4 0.087887 385493+3+5 0.088478 5130610+10+10 0.093546 2084392+4+50 0.1004 3270824+4+50 0.100403 5059717+7+50 0.100417 1022268 Table 2.
The numerical series for 3 objects
Figure 2.
The configuration for the 7 points, 7 points and50 points which has area of 0 . . . × . n -gon. We represent circle as regular 500-gon, line as a 2-gon,and then increased the number of vertices of the third polygon from 3 to 10.The results are presented in Table 3.Because n − n -gon with coinciding vertices,the lower bound improves by definition, and it is best for n = 10. However,Figure 3 shows that the resulting 10-gon is very similar to the same rectangleof size 0 . × . N = 7 , N = 7 , N = 50 convex cover for closed unit curves has area at least 0.1 7Type ( n points) Optimal area Time (sec)3 0.0970429 135084 0.1003 288655 0.100304 336526 0.100374 549797 0.100386 800778 0.100390 812859 0.100418 11910310 0.100473 123350 Table 3.
The numerical series for 3 objects when 2 objectsare fixed
Figure 3.
The configuration for the 500-gon, a strength lineand 10 points which has area of 0 . . . • Have found a line-rectangle-circle configuration, whose convex hull is,numerically, always grater than 0 .
1. This significantly improves the pre-vious best lower bound 0 . • Conclude that no other “simple” configuration of 3 objects gives a sub-stantially better bounds.In the following section, we give a rigorous prove for the first conclusion,and this is the main result of this work. The second conclusion is informaland has no proof. Bogdan Grechuk and Sittichoke Som-am
Figure 4.
The configuration X
3. Geometric analysis
Let C be a circle of perimeter 1, R be a rectangle of size u × v where u = 0 . v = 12 − u , and L be line of length 12 .Let us fix the center of circle to be C (0 , F be a regular 500-goninscribed in C , such that the sides of R are parallel to some longest diagonalsof F . Let X be a union F ∪ R ∪ L . X is called a configuration . Let H ( X ) bethe convex hull of X , and A ( X ) be the area of H ( X ). Our aim is to find aconfiguration X with the smallest A ( X ).Let R ( x , y ) be the center of R . By the symmetry of circle, we mayassume that x , y ≥
0. We have the vertices of R are R ( x − v/ , y + u/ R ( x + v/ , y + u/ R ( x + v/ , y − u/ R ( x − v/ , y − u/ L ( x , y ) be the center of L and θ be the angle between X axis and L L ,see Figure 4.The vertices of L are L ( x + cos( θ + π ) , y + sin( θ + π )) and L ( x + cos( θ ) , y + sin( θ )). Note that any configuration X is represented by x , y , x , y , and θ . Clearly, 0 ≤ θ ≤ π We define the function f : R → R by mapping vector ( x , y , x , y , θ )to A ( X ). Clearly, f is a continuous function. Since F is a subset of C , itsuffices to show that f ( x , y , x , y , θ ) ≥ . , ∀ x , y , x , y , θ. This would immediately imply Theorem 1.1.Next, we will apply result of Fary and Redei [4], and Lemma 1 andCorollary 2 in [7] to bound parameters x , y , x , y and θ . Lemma 3.1.
Let Z be a region of points z = ( x , y , x , y , θ ) in R satis-fying the inequalities ≤ x ≤ . , ≤ y ≤ . , − . ≤ x ≤ convex cover for closed unit curves has area at least 0.1 90 . , − . ≤ y ≤ . , ≤ θ ≤ π. If f ( z ) > . for all z ∈ Z , then infact f ( z ) > . for all z ∈ R .Proof. Let ψ ( x , y ) be the area of convex hull of F and R only. By [4], ψ isa convex function in both coordinates. By symmetry, ψ ( x , y ) = ψ ( x , − y ).This combined with convexity implies that ψ ( x , y ) ≥ ψ ( x , ≥ ψ (0 . , > . , whenever x ≥ . ψ ( x , y ) ≥ ψ (0 , y ) ≥ ψ (0 , . > . , whenever y ≥ . x , y ) be the area of convex hull of F and L only. By [4],if L moves in direction −−−→ L L , Φ( x , y ) attains minimum when C L isperpendicular to L L . Assume that | x | ≥ .
148 or | y | ≥ . (cid:112) x + y ≥ . Figure 5.
EKL L and l Let l be the line segment between C and L . Next, let points E, K ∈ F be such that line segments C L and EK are perpendicular, see Figure5. Then EKL L is trapezoid with bases lengths | EK | ≥ r cos (cid:0) π (cid:1) and | L L | = 0 .
5, where r = π . The area of F is S ( F ) = 500 r sin (cid:0) π (cid:1) . Thus, A ( X ) > (cid:18)
12 + 2 r cos (cid:16) π (cid:17)(cid:19) (cid:113) x + y + S ( F )2 > . (cid:3) Lemma 3.2.
Either f ( z ) > . , or F ∪ L ∪ R is a subset of a rectangle withside lengths . × . . Proof.
By Lemma 3.1, we can assume that z = ( x , y , x , y , θ ) ∈ Z . Let Y and Y be the points of configuration X = F ∪ L ∪ R with the lowest andhighest y -coordinates y ∗ and y ∗ , respectively. Because 0 ≤ y ≤ . , Y isbelow R . Let h , h be the height from Y , Y to R , respectively. h = 0 if Y is below or on R R . Let y ∗ − y ∗ > . A ( X ) > u ( − u ) + ( − u )( h + h ) = u ( − u ) + ( − u )( y ∗ − y ∗ − u ) > . Figure 6.
The configuration of y ∗ , y ∗ and R Let X and X be the points of configuration X = F ∪ T ∪ R withthe lowest and highest x -coordinates x ∗ and x ∗ , respectively. Let w = ( x, y )be any point in X . If w belongs to F , then | x | < . w is any point on rectangle, then | x | < . w is any pointon line, then | x | < . X , X ∈ L , then x ∗ − x ∗ ≤ .
5. Otherwise, x ∗ − x ∗ ≤ | x ∗ | + | x ∗ | < .
398 + 0 . < .
636 see Figure 7. (cid:3)
We prove Lipschitz continuity of f in Z in the following lemma Lemma 3.3.
For every ( x , y , x , y , θ ) ∈ Z , and any (cid:15) i ≥ , i = 1 , . . . , , | f ( x + (cid:15) , y + (cid:15) , x + (cid:15) , y + (cid:15) , θ + (cid:15) ) − f ( x , y , x , y , θ ) | ≤ (cid:88) i =1 (cid:15) i C i , with constants C = 0 . , C = 0 . , C = 0 . , C = 0 . , and C = 0 . . convex cover for closed unit curves has area at least 0.1 11 Figure 7.
The line shows the optimum possible position of
F, R, L
Proof.
Let g : R → R be a convex function on R . If C = max (cid:20) lim t →−∞ g ( t ) t , lim t → + ∞ g ( t ) t (cid:21) < ∞ , then | g ( t + (cid:15) ) − g ( t ) | ≤ C(cid:15), ∀ t, ∀ (cid:15) > . see Lemma 5 in [7].Let g ( x ) = f ( x , y , x , y , θ ), where y , x , y , θ are fixed. We havelim x →−∞ g ( x ) x = lim x → + ∞ g ( x ) x ≤ .
439 + u < C , where 0 .
439 comes from Lemma 3.2, while u is the height of R , see Figure 8. Figure 8.
The ratio between g ( x ) and x when x → + ∞ Hence, | f ( x + (cid:15) , y , x , y , θ ) − f ( x , y , x , y , θ ) | ≤ C (cid:15) . Similarly, with g ( y ) = f ( x , y , x , y , θ ) for fixed x , x , y , θ ,lim y →−∞ g ( y ) y = lim y → + ∞ g ( y ) y ≤ (0 . − u ) + ( x + 0 .
25 + r )2 < C , r = 1 / π and | x | ≤ . g ( x ) = f ( x , y , x , y , θ ),lim x →−∞ g ( x ) x = lim x → + ∞ g ( x ) x ≤ .
439 + ( y + u/ r )2 < C . where 0 .
439 come from Lemma 3.2 and 0 ≤ y ≤ . g ( y ) = f ( x , y , x , y , θ ),lim y →−∞ g ( y ) y = lim y → + ∞ g ( y ) y ≤ . x + (0 . − u ) / r )2 < C . where 0 ≤ x ≤ . Figure 9.
The ratio between g ( y ) and y when y → + ∞ Figure 10.
The ratio between g ( x ) and x when x → + ∞ convex cover for closed unit curves has area at least 0.1 13 Figure 11.
The ratio between g ( y ) and y when y → + ∞ see Figure 10. This implies that | f ( x , y + (cid:15) , x , y , θ ) − f ( x , y , x , y , θ ) | ≤ C (cid:15) , | f ( x , y , x + (cid:15) , y , θ ) − f ( x , y , x , y , θ ) | ≤ C (cid:15) , and | f ( x , y , x , y + (cid:15) , θ ) − f ( x , y , x , y , θ ) | ≤ C (cid:15) . Finally, we need to prove that | f ( x , y , x , y , θ + (cid:15) ) − f ( x , y , x , y , θ ) | ≤ C (cid:15) . (3.1)To prove the bound for C , we need the following claim. Claim 1.
The diameter d ( F ∪ R ) of
F ∪ R is less than 0 . d ( F ∪ R ) is maximal possible if R =(0 . , . R = (0 . , . F ∈ F be a pointwhich d ( x, R ) is maximum for all x ∈ F see Figure 12. By direct cal-culation, | R R | < . < | R F | . Hence, the diameter of F ∪ R is | R F | < . (cid:15) = (cid:15) , thenit also holds for (cid:15) = 2 (cid:15) . Therefore it is sufficient to prove (3.1) only forsufficiently small (cid:15) .Let L (cid:48) with endpoints L (cid:48) , L (cid:48) be the line L rotated around L by angle (cid:15) . Then | L L (cid:48) | = 2 | L L | sin( (cid:15) / < | L L | ( (cid:15) /
2) = | L L | (cid:15) = (cid:15) .Similarly, | L L (cid:48) | < (cid:15) .By selecting (cid:15) sufficiently small, we can ensure that all vertices ofpolygons H ( R, F, L ) and H ( R, F, L (cid:48) ) coincides, except possible the endpointsof L and L (cid:48) . Then area difference |A ( R, F, L (cid:48) ) − A ( R, F, L ) | is bounded bythe total area of four triangles X L X , X L (cid:48) X , X L X , X L (cid:48) X , where X i , i = 1 , , , H ( R, F, L ) adjacent to L , L , seeFigures 13 and 14.Let h , h be the height of triangle with respect to base X X . Let h , h be the height of triangle with respect to base X X see Figure 14.4 Bogdan Grechuk and Sittichoke Som-am Figure 12.
The longest between R and F Figure 13.
Polygon H ( R, F, L ) adjacent L , L By claim 1, We have |A ( R, F, L (cid:48) ) − A ( R, F, L ) | ≤ | h X X − h X X | + | h X X − h X X | = X X | h − h | + X X | h − h | < X X | L L (cid:48) | + X X | L L (cid:48) | ≤ × d ( F ∪ R ) × (cid:15) ≤ × . < . (cid:15) = C (cid:15) . (cid:3)
4. Computational results
Let Z be a region in Lemma 3.1. In this section, we prove that f ( z ) = f ( x , y , x , y , θ ) > . , ∀ z ∈ Z convex cover for closed unit curves has area at least 0.1 15 Figure 14.
Four triangles which L rotated by angle (cid:15) by using box-search method as described in [7]. The method works as follows.Let z ∗ be the center of a box B which has the form B = [ a , b ] × [ a , b ] × [ a , b ] × [ a , b ] × [ a , b ]. On every step, we check if f ( z ∗ ) − d C − d C − d C − d C − d C ≥ . , (4.1)where d i is a half of the length of [ a i , b i ]. If (4.1) holds, then inequality f ( z ) > . z ∈ B by Lemma 3.3.If (4.1) does not hold, we will select the largest length, say [ a , b ] andsplit B into two boxes B = [ a , ( a + b ) / × [ a , b ] × [ a , b ] × [ a , b ] × [ a , b ]and B = [( a + b ) / , b ] × [ a , b ] × [ a , b ] × [ a , b ] × [ a , b ]. Then, if (4.1)holds for both boxes B and B , then f ( z ) > . z ∈ B andfor every z ∈ B , hence it holds for every z ∈ B . If (4.1) does not hold foreither B or B (or both), we subdivide the corresponding boxes again andproceed iteratively.We start with B = Z , and, when the program halts, we are guaranteedthat f ( z ) > . , ∀ z ∈ Z .Let us illustrate the first step of this procedure. By Lemma 3.1, we startwith B = Z = [0 , . × [0 , . × [ − . , . × [ − . , . × [0 , π ].Then we check (4.1) for z ∗ = (0 . , . , , , π/ f ( z ∗ ) ≈ . f ( z ∗ ) − (cid:80) i =1 d i C i ≈ − . < .
1. Thus, we must select the maximum length to subdivide B into B and B . In this case, b − a = π ≈ .
14 is the maximum. Hence, we divide B into B = [ a , b ] × [ a , b ] × [ a , b ] × [ a , b ] × [ a , ( a + b ) /
2] and B =[ a , b ] × [ a , b ] × [ a , b ] × [ a , b ] × [( a + b ) / , b ]. Then we check (4.1) for B and for B and repeat this procedure iteratively.6 Bogdan Grechuk and Sittichoke Som-amWe run the Box-search algorithm [7] by using Matlab (cid:114) R2018a, see im-plementation details and Matlab code in Appendix. The programme haltsafter n = 527 , ,
566 iterations. This rigorously proves that the minimalarea A ( X ) is greater than 0 .
1. Numerically, the program returned value0 . x = 0 . , y =0 . , x = 0 . , y = − . , θ = 0 .
5. Main Theorems
Theorem 5.1. (Theorem 1.1 in the Introduction) The area of convex cover S for circle of perimeter , line of length / , and rectangle of size . × . is at least . .Proof. Let Z be a region in Lemma 3.1. The fact that Box-search algorithmhalted together with Lemma 3.3 implies that f ( z ) > . z ∈ Z . Then,by Lemma 3.1, f ( z ) > . z ∈ R . Thus, A ( F, R, L ) > .
1. Since F ⊂ C , A ( C, R, L ) ≥ A ( F, R, L ) > . (cid:3) Corollary 5.2.
Any convex cover for closed unit curves has area of at least . .Proof. Let S be a convex cover for closed unit curves. Then S can accom-modate C, R , and L , hence H ( C, R, L ) ⊂ S . Thus the area of S is at least A ( F, R, L ). However, A ( F, R, L ) > . (cid:3) Figure 15.
The convex hull of the configuration of the min-imum area with 0 . r n Figure 16.
The table of the percentage of r and n
6. Conclusion
In this work, we improve the lower bound for the area of convex covers forclosed unit arcs from 0 . .
1. First, we used the geometric methodto prove Lipschitz bounds for configuration function in 5 parameters whichrepresent the circle, 0 . × . . . Appendix A. Appendix
There are only two functions below which are used in Box-search method.First, function cvhN x , x , y , y , α ) is used to find the area of convexhull for F, R, L which are represented by x , x , y , y , α . Second, function checkminN B a , b , . . . , a , b . Finally, the Box-search results is shownin A.3. A.1. The box’s search results
Table 16 shown the percentage of the area of the initial box for which theinequality (4.1) is verified and the iteration number every 100 , ,
000 stepsby Box search program.Figure 17 illustrates the graphical process which ups to the number ofiterations.When the program finished, it displayed the result: [r,n,min,xx1,yy1,xx2,yy2,app]= checkminNB5(0,0.0741,0,0.0976,-0.148,0.148,-0.148,0.148,0,pi,0.1,0.0000001,0,0,0.11,0,0,0,0,0)r = 0.001990679394226 (this is the area of the initial box, 100% covered)8 Bogdan Grechuk and Sittichoke Som-am
Figure 17.
The graph of the percentage of r and n n = 527754566 (total number of iterations needed)min = 0.100438196959697 (the minimal area convex hull)xx1 = 0.004341796875000yy1 = 0.006481250000000xx2 = 0.004335937500000yy2 = -0.004335937500000app = 0.857111765231102 (the 5 coordinates for the optimal configura-tion) References [1] Brass, P. and Sharifi, M.,
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Miskolc MathematicalNotes. , 691–698 (2018).Bogdan GrechukSchool of Mathematics and Actuarial ScienceUniversity of Leicester, LeicesterLE1 7RHUnited Kingdome-mail: [email protected]