New Results in Sona Drawing: Hardness and TSP Separation
Man-Kwun Chiu, Erik D. Demaine, Yevhenii Diomidov, David Eppstein, Robert A. Hearn, Adam Hesterberg, Matias Korman, Irene Parada, Mikhail Rudoy
CCCCG 2020, Saskatoon, Canada, August 5–7, 2020
New Results in Sona Drawing: Hardness and TSP Separation
Man-Kwun Chiu ∗ Erik D. Demaine † Yevhenii Diomidov † David Eppstein ‡ Robert A. Hearn § Adam Hesterberg † Matias Korman ¶ Irene Parada (cid:107)
Mikhail Rudoy ∗∗ In memoriam Godfried Toussaint (1944–2019)
Abstract
Given a set of point sites, a sona drawing is a singleclosed curve, disjoint from the sites and intersectingitself only in simple crossings, so that each boundedregion of its complement contains exactly one of thesites. We prove that it is NP-hard to find a minimum-length sona drawing for n given points, and that such acurve can be longer than the TSP tour of the same pointsby a factor > . In April 2005, Godfried Toussaint visited the secondauthor at MIT, where he proposed a computationalgeometric analysis of the “sona” sand drawings of theTshokwe people in the West Central Bantu area of Africa.Godfried encountered sona drawings, in particular theethnomathematical work of Ascher [1] and Gerdes [7],during his research into African rhythms. Together withhis then-student Perouz Taslakian, we came up with aformal model of sona drawing of a set P of point sites — a closed curve drawn in the plane such that1. wherever the curve touches itself, it crosses itself;2. each crossing involves only two arcs of the curve;3. exactly one site is in each bounded face formed bythe curve; and4. no sites lie on the curve or within its outside face. ∗ Institut f¨ur Informatik, Freie Universit¨at Berlin, Germany,[email protected]. This work was supported in part byERC StG 757609. † CSAIL, Massachusetts Institute of Technology, USA, { edemaine,diomidov,achester } @mit.edu ‡ Computer Science Department, University of California, Irvine,[email protected]. This work was supported in part by the USNational Science Foundation under grant CCF-1616248. § [email protected] ¶ Tufts University, USA, [email protected] (cid:107)
TU Eindhoven, The Netherlands, [email protected] ∗∗ CSAIL, MIT, USA. Now at Google Inc.
Our first paper on sona drawings appeared atBRIDGES 2006 [5], detailing the related cultural prac-tices, proving and computing combinatorial results anddrawings, and posing several open problems. In early2006, we brought these open problems to Godfried’sBellairs Winter Workshop on Computational Geometry,where a much larger group tackled sona drawings, re-sulting in a CCCG 2006 paper later the same year [4].Next we highlight some of the key prior results and openproblems as they relate to the results of this paper.
Sona vs. TSP.
Every TSP tour can be easily convertedinto a sona drawing of roughly the same length: insteadof visiting a site, loop around it, except for one sitethat we place slightly interior to the tour [5, Lemma 11].Conversely, every sona drawing can be converted into aTSP tour of length at most a factor π +2 π ≈ . + √ ≈ . √ π ( √ √ π ( √ ≈ . L and L ∞ metrics, wherewe prove that the worst-case TSP/sona separation factoris exactly 1 . Length minimization.
The relation to TSP implies aconstant-factor approximation algorithm for finding theminimum-length sona drawing on a given set of sites.But is this problem NP-hard? In Section 3, we prove NP-hardness for L , L , and L ∞ metrics, settling [5, OpenProblem 5] and [4, Open Problem 4]. Grid drawings.
The last variant we consider is whenthe sona drawing is restricted to lie along the edges of aunit-square grid, while sites are at the centers of cells ofthe grid. Not all point sets admit a grid sona drawing;however, if we scale the sites’ coordinates by a factor of3, then they always do [4, Proposition 10]. A naturalremaining question [4, Open Problem 3] is which pointsets admit grid sona drawings. In Section 4, we provethat this question is in fact NP-hard. a r X i v : . [ c s . C G ] J u l nd Canadian Conference on Computational Geometry, 2020
We first show an example which gives a large TSP/sonaseparation factor under the L metric in the plane. Theorem 1
There exists a set of sites for whichthe length of the minimum-length TSP tour is √ π ( √ √ π ( √ ≈ . times the length of theminimum-length sona drawing. The full proof can be found in Appendix A.
Sketch of Proof.
We construct a problem instancewhose minimum-length TSP tour is longer than itsminimum-length sona drawing by a factor within ε of √ π ( √ √ π ( √ . Our construction is illustrated in Fig-ure 1 and follows a fractal approach with ε − levels (forsimplicity, we assume that ε − is an integer).1. We start by defining a few auxiliary points:(a) The initial set of auxiliary points A isthe intersection between a slightly shiftedinteger lattice and the L ball B (0 , ε − )of radius ε − centered at the origin.That is, A = (cid:8) ( i +12 , j +12 ) : i, j ∈ Z (cid:9) ∩ (cid:8) ( x, y ) : | x | + | y | ≤ ε − (cid:9) . In Figure 1a, theauxiliary points are exactly the intersectionsof solid red lines.(b) Then, in Step i (starting with i = 1), for eachauxiliary point p ∈ A i − , we add five points to A i : p itself, and four new points at distance (cid:16) √ (cid:17) i from p in each of the four cardinaldirections. Let A = A ε − . By construction, wehave | A i | = 5 i | A | and | A | = 2 ε − + O ( ε − ).We note that set A contains auxiliary points(not sites). These points will not be part ofthe instance.2. We now use the auxiliary points to create some sites(the isolated black points of Figure 1):(a) For any i ≥ P i of sites as follows:for each auxiliary point q ∈ A i − we add thefour sites whose x and y coordinates each differfrom q by (cid:16) √ (cid:17) i . We note that all theadded sites are distinct sites.(b) We define P as the set of integer lattice pointsin B (0 , ε − ). Equivalently, for each auxiliarypoint q ∈ A we add the sites whose x and y coordinates each differ from q by , but we donot add sites that lie outside B (0 , ε − ), whichaffects O ( ε − ) sites (out of Ω( ε − ) sites of P ).In this case, the sites created by differentauxiliary points may lie in the same spot. In total, P contains only one site per aux-iliary point of A (except for O ( ε − ) auxil-iary points near the boundary). Thus, | P i | =4 · | A i − | = 4 · i − · | A | (for i ≥
1) and | P | = | A | + O ( ε − ) = (2 ε − + O ( ε − )).Let P (1) = P ∪ P ∪ · · · ∪ P ε − .3. Next, we place additional sona sites packing linesegments and/or curves. Whenever we pack anycurve, we place sites spaced at a distance δ smallenough that the length of the shortest path thatpasses within δ of all of them is within a factor(1 − ε ) of the length of the curve.(a) Solid lines as drawn in red in Figure 1a: foreach x ∈ { i + : − ( ε − + 1) ≤ i ≤ ε − and i ∈ Z } , we pack the vertical line segment withendpoints ( x, ε − +1 −| x | ) and ( x, | x |− ε − − y for the horizontal linesegments.(b) In Step i of the above recursive definition (start-ing with i = 1), when we create four new aux-iliary points of A i from a point p ∈ A i − , wealso pack the boundary of the region withinEuclidean distance (cid:16) √ (cid:17) i of the squarewhose vertices are the four auxiliary pointsof A i . Note that this boundary region forms asquare with rounded corners as in Figure 1b.With these extra points we preserve the invari-ant that the auxiliary points are exactly theintersections of packed curves.Let P (2) be the set of sites created in Step 3 in ourconstruction, and P = P (1) ∪ P (2) . This is a completedescription of the construction.In the full proof we show that the length of the packedcurves is a (1 + ε )-approximation of the total lengthof the minimum-length sona drawing of P . Carefulcalculations then yield that the length of the minimum-length sona drawing is (2 ε − + O ( ε − )) (cid:16) π √ − (cid:17) . We then argue that the minimum-length TSP hasan additional length of (2 ε − + O ( ε − ))(2 √ π √ − +2 √ π √ − = π ( √ √ √ π ( √ ≈ . L metric inthe plane showing that the ratio between the lengths ofthe minimum-length TSP and the minimum-length sonadrawing can be strictly greater than 1.5. Our next resultshows that this cannot be the case for the L and L ∞ metrics in the plane. CCG 2020, Saskatoon, Canada, August 5–7, 2020 (a) First step of our construction for ε − = 2: points of A lie in the intersection of solid lines (packed segments). Inthe construction, P contains thirteen sites (shown as blackdots). (b) Final construction for ε − = 2. Sets P , P , and P areshown as black dots of varying sizes. Additional sites of P (2) pack lines and rounded squares nearby the auxiliary pointsof A , A , and A . Figure 1: Recursive construction of sites requiring ≈ . Theorem 2
For the Manhattan ( L ) and the Chebyshev( L ∞ ) metrics, the minimum-length TSP tour for a set ofsites P has length at most . times that of the minimum-length sona drawing for P . Moreover, this bound is tightfor both metrics. Proof.
The proof of the upper bound on the lengthof the minimum-length TSP tour follows the lines ofthe (unpublished) proof of [4, Theorem 12]. Let P = { p , . . . , p n } be a set of n sites, S ( P ) the minimum-lengthsona drawing for P , and TSP( P ) the minimum-lengthTSP tour for P . The sona drawing S ( P ) must have n bounded faces, each containing a site of P . Let f i bethe face of S ( P ) containing the site p i . In this proof,for an edge-weighted graph H , | H | denotes the sum ofthe weights/lengths of all the edges of H . In particular, | S ( P ) | denotes the length of the sona drawing S ( P ).For each site p i , let c ( p i ) be the closest point in S ( P )to p i and r i the distance between p i and c ( p i ). By thedefinition of c ( p i ), the open disk centered at p i andwith radius r i does not intersect S ( P ). This impliesthat the length of the boundary of f i is at least theperimeter of a disk with radius r i , that for both the L and the L ∞ metrics is 8 r i . That is, | f i | ≤ r i . Moreover,the sum of the lengths of all the faces is 2 | S ( P ) | , so | f | + · · · + | f n | < | S ( P ) | since we do not sum thelength of the unbounded face. We define a multigraph G whose vertex set is theunion of the set of sites P , the set of vertices of S ( P ),and { c ( p i ) ∈ S : p i ∈ P } . The edge set of G is theunion of the set of edges of S and two parallel edges { p i , c ( p i ) } for each p i ∈ P . The weight of each edge isits length in the drawing. By the observations above, | G | = | S ( P ) | + 2 r + · · · + 2 r n ≤ | S ( P ) | + | f | / · · · + | f n | / < | S ( P ) | + | S ( P ) | / . | S ( P ) | .To obtain the desired upper bound on | TSP( P ) | itremains to show that | TSP( P ) | ≤ | G | . By construction,since S ( P ) is Eulerian, so is G . An Euler tour of G defines a TSP tour for the vertices of G by skippingvertices that were already visited (as in the Christofides1.5-approximation algorithm for TSP on instances wherethe distances form a metric space [3]). This TSP tourhas length at most | G | and can be shortcut so that itonly visits the sites of P . By the triangle inequality, thelength of the tour does not increase with these shortcuts.Thus, we have that | TSP( P ) | ≤ | G | < . | S ( P ) | .The construction for the matching this bound is similarto the one in the proof of Theorem 1, but simpler. Anillustration can be found in Figure 1a. For every ε > P ε (the set of TSP vertices/sonasites) such that | TSP( P ε ) | ≥ (1 . − ε ) | S ( P ε ) | .We fix k = (cid:100) / (2 ε ) (cid:101) . The set of sites P ε includesevery integer lattice point ( x, y ) such that | x | + | y | ≤ k .Consider drawing Q resulting from the union of the axis-2 nd Canadian Conference on Computational Geometry, 2020 aligned unit squares centered at these sites. It is easyto see, for example by rotating the construction, thatso far we have added ( k + 1) + k sites to P ε and thatthe length of Q is 4( k + 1) . Straightforward compu-tations show that k +1) +( k +1) + k k +1) = 5 / k k +1) ≥ / ε +1) = 1 . − ε + ε (4 ε +3)(2 ε +1) > . − ε . Thus, adense-enough packing of sites along Q yields the desiredresult. (cid:3) We next consider sona drawings on the sphere. By thedefinition of sona drawings in the plane, the unboundedface contains no sites. For the sphere we consider thefollowing analogue: if there is a face that contains inits interior a half-sphere then this face contains no sites.Note that there is at most one such face. The followingtheorem shows a tight upper bound on the TSP/sona sep-aration factor for drawings on the sphere. (We considerthe usual metric inherited from the Euclidean metricin R .) Theorem 3
For drawings on the sphere, the length ofthe minimum-length TSP tour for a set of sites P isat most times the length of the minimum-length sonadrawing for P . Moreover, this bound is tight. Proof.
The proof of the upper bound on the length ofthe minimum-length TSP tour again follows the lines ofthe (unpublished) proof of [4, Theorem 12]. It only differsslightly from the first part of the the proof of Theorem 2.Using the same notation, in this case, the distance r i between a site p i ∈ P and its closest point c ( p i ) in S ( P )corresponds to the length of the shortest arc on the greatcircle through p i and c ( p i ). The open disk centered at p i and with radius r i is an open spherical cap that doesnot intersect S ( P ). Assuming that the sphere has radius ρ , the boundary of this cap has length 2 πρ sin( r i /ρ ).Since the face containing a site cannot contain a half-sphere in its interior we have that 0 ≤ r i /ρ ≤ π/ x ) /x in the interval 0 ≤ x ≤ π/ ρ/r i sin( r i /ρ ) ≥ /π sin( π/
2) = 2 /π .This implies that 2 πρ sin( r i /ρ ) ≥ r i . Thus, the face f i of S ( P ) containing the site p i has length | f i | ≥ r i .Moreover, | f | + · · · + | f n | ≤ | S ( P ) | . With the samearguments and defining the same multigraph as in theproof of Theorem 2 we obtain that | TSP( P ) | ≤ | S ( P ) | .The construction showing that this bound is tightplaces two sites on the north and south poles of thesphere and packs the equator densely with sites. Thenthe minimum-length sona drawing goes along the equatorwhile the minimum-length TSP must reach both poles,yielding a 2 − ε TSP/sona separation factor. (cid:3)
In this section and Appendix B, we prove that findinga sona drawing of minimum length for given sites is NP-hard, even when the sites lie on a polynomially sizedgrid. The complexity of minimum-length sona drawingwas posed as an open problem in 2006 by Damian etal. [4, Open Problem 4]. We use a reduction from theproblem of finding a Hamiltonian cycle in a grid graph(a graph whose n vertices are a subset of the points inan integer grid, and whose edges are the unit-length linesegments between pairs of vertices), proven NP-completeby Itai, Papadimitriou, and Szwarcfiter [8].Let V be the set of n vertices in a hard instance forHamiltonian cycle in grid graphs. If V is a yes instance,its Hamiltonian cycle forms a Euclidean traveling sales-man tour with length exactly n . If it is a no instance,the shortest Euclidean traveling salesman tour throughits vertices has length at least 1 for every grid edge, andlength at least √ n + √ − ≈ n + 0 . L distance, theincrease in length is larger, at least 1. Our reductionreplaces each point of V by two points, close enoughtogether to make the increase in length from convertinga TSP to a sona drawing negligible with respect to thisgap in tour length. Theorem 4
It is NP-hard to find a sona drawing fora given set of sites whose length is less than a giventhreshold L , for any of the L , L , and L ∞ metrics. While minimizing the length of sona drawings in generalis hard, if we restrict the drawing to lie on a grid, theneven determining the existence of a sona drawing is hard.Given n sites at the centers of some cells in the unit-square grid, a grid sona drawing is a sona drawingwhose edges are drawn as polygonal lines along theorthogonal grid lines (like orthogonal graph drawing).We show that finding a grid sona drawing for a givenset of sites is NP-hard by a reduction from Planar CNFSAT [9]. In this section we view the grid as a graph, thus by edge we mean a unit segment of a grid line, and by vertex we mean a grid vertex — these terms are distinctfrom “sona edge” etc. We say that an edge is either on or off according to as it belongs in the sona drawing.The subgraph of the grid that is on is the path graph .Observe that two grid-adjacent sites always require theedge between them to be on; otherwise both would bein the same sona face ( connected ). Also, every vertexmust have even degree in the path graph.Here we are concerned with internal properties of thegadgets. Their exteriors are lined with unconnected CCG 2020, Saskatoon, Canada, August 5–7, 2020
AB C D FE
Figure 2: Wiregadget Figure 3:Constant gadget Figure 4: Turn /Split / Invertgadgetedges; we will show later how to connect them.
Wire.
The wire gadget is shown in Figure 2. One ofedges A and B must be on, otherwise two sites wouldbe connected. Assume without loss of generality that A is on. Now, suppose C is off. Then D must be too,to preserve even vertex degree. Then, B and E mustboth be on to prevent sites from being connected, butthis is impossible with C off. Therefore C and D areon. The same reasoning shows that the entire line A , D ,etc. is on, as well as edges C , F , etc. Then E must beoff to prevent an empty face, and thus the entire line B , E , etc. is off. The wire thus has two states: the upperline can be on and the lower off, or vice-versa. We canextend a wire as long as necessary. An unconnected wireend serves as a variable.In Figures 2 through 5, all marked edge states (red foron, gray for off) are forced by the indicated wire states.These marks were generated by computer search, butare easy to verify by local analysis. Constant.
The gadget shown in Figure 3 forces theattached wire to be in the up state: the edge betweenthe two left sites must be on, forcing the rest.
Turn / Split / Invert.
The gadget shown in Figure 4is multi-purpose. If we view the left wire as the input,then the upper and lower outputs represent turned sig-nals, and the right output represents an inverted signal.(Unused outputs can be left unattached, thus uncon-strained.)Any wire state forces all the others. Given that theleft wire is in the up state, suppose the right wire is alsoup. Then the top wire and bottom wire must be in thesame left/right state, otherwise we will have degree-threevertices in the middle. But this would leave the centralsite connected to another site, so the right wire is forceddown. Then, if the (top, bottom) wires are not in the(left, right) states, again the central site will be connectedto another one. (This figure contains multiple loops, butthese will be eliminated in the final configuration byadding more edges.)
OR.
Figure 5 shows the OR gadget. The upper wire isinterpreted as an output, with the left state representingtrue; the other wires are inputs, with true representedas down on the left wire and up on the right wire. (Wecan easily adjust truth representations between gadgetswith inverters.) If either input is true, then the outputmay be set true, as shown. If both inputs are false, theoutput may be set false. Figure 5e shows that settingthe output to true when both inputs are false is notpossible: all marked edge states are forced by the wireproperties, but two sites are left connected.
Theorem 5
It is NP-hard to find a grid sona drawingfor a given set of sites at the centers of grid cells.
Proof.
Given a CNF Boolean formula with a planarincidence graph, we connect the above gadgets to rep-resent this graph: unconstrained wire ends representvariables, and are connected to splitters and inverters toreach clause constructions. A clause is implemented withchained OR gadgets, with the final output constrained tobe true with a constant gadget. By the gadget propertiesdescribed above, we will be able to consistently choosewire states if and only if the formula is satisfiable.We must still show that all edges can be joined to-gether into a single closed loop, while retaining the sonaproperties. Our basic strategy for connecting loose endsis to border each gadget with “crenellations”, as shownin Figure 6. This figure also shows how to pass pairs ofpath segments across a wire without affecting its internalproperties, which we will use to help form a single loop.Adding crenellations to the other gadgets is straight-forward, and we defer explicit figures to Appendix C,with one exception. (The crenellations do add a parityconstraint when wiring gadgets together; we show in theappendix how to shift parity.) When we use the gadgetin Figure 4 to turn a wire, it will be useful to use thecrenellated version in Figure 7. With the connectionsto other gadgets on the left and top, the right and bot-tom portions are unconstrained. We can place edges asshown, so that they leave the gadget identically regard-less of which state it is in. Then, all paths entering fromthe left or the top leave on the bottom or the right asloose edges, except that in Figure 7a, one path connectsthe left to the top. If we connect a right turn to the topport, this path will also terminate in an unconnectededge. If every wire contains a left turn and matchingright turn, then every path in the sona graph must endin two unconnected edges in turn gadgets, because thereare no internal loops in any of the gadgets.The space occupied by the loose ends of a turn lieseither in an internal face of the wiring graph, or on itsexterior. We route the interior ends to pass-throughpairs as shown in Figure 6, so all unconnected edges2 nd Canadian Conference on Computational Geometry, 2020 (a) false + true → true (b) true + true → true (c) true + false → true (d) false + false → false (e) Bad state Figure 5: OR gadgetFigure 6: Wire gadget with crenellations and pass-through (a) State 1 (b) State 2
Figure 7: Crenellated turnwind up on the outer border of the graph. Becausethe terminal edges are placed identically in Figures 7aand 7b, we can plan their routing without knowing thewire states. As a result, we can place additional sites asrequired for the property that a single site lies in eachinternal sona face. (We can lengthen the wires as neededto create additional routing space in the internal faces.)Now we are in a state where all paths end on theexterior of the construction. If we join these pathstogether without crossing, the number of extra sitesneeded in the outer face is just the number of paths. Weplace that many sites in a widely spaced grid (spacingproportional to number of paths) surrounding the innerconstruction. Then, we can complete the path greedilyby repeatedly connecting one outer path end to one ofits neighboring path ends, surrounding one of the addedsites. Only one of its two neighboring path ends cancome from the same path, so there’s always another oneto connect to. The wide grid spacing of the outer sites means there is always room to route the connection. (cid:3)
Acknowledgments
Thanks to Godfried Toussaint for introducing us (andcomputational geometry) to sona drawings. This re-search was initiated during the Virtual Workshop onComputational Geometry held March 20–27, 2020, whichwould have been the 35th Bellairs Winter Workshop onComputational Geometry co-organized by E. Demaineand G. Toussaint if not for other circumstances. Wethank the other participants of that workshop for helpfuldiscussions and providing an inspiring atmosphere.
References [1] Marcia Ascher.
Mathematics Elsewhere: AnExploration of Ideas Across Cultures . PrincetonUniversity Press, 2002.[2] Molly Baird, Sara C. Billey, Erik D. Demaine,Martin L. Demaine, David Eppstein, S´andorFekete, Graham Gordon, Sean Griffin, Joseph S. B.Mitchell, and Joshua P. Swanson. Existence andhardness of conveyor belts. Electronic preprintarXiv:1908.07668, 2019. URL:https://arXiv.org/abs/1908.07668.[3] Nicos Christofides. Worst-case analysis of a newheuristic for the travelling salesman problem.Technical report, Graduate School of IndustrialAdministration, Carnegie Mellon University, 1976.[4] Mirela Damian, Erik D. Demaine, Martin L.Demaine, Vida Dujmovi´c, Dania El-Khechen,Robin Flatland, John Iacono, Stefan Langerman,Henk Meijer, Suneeta Ramaswami, Diane L.Souvaine, Perouz Taslakian, and Godfried T.Toussaint. Curves in the sand: Algorithmicdrawing. In
Proceedings of the 18th AnnualCanadian Conference on Computational Geometry(CCCG 2006) , pages 11–14, Kingston, Ontario,August 2006. URL:https://cccg.ca/proceedings/2006/cccg4.pdf.
CCG 2020, Saskatoon, Canada, August 5–7, 2020 [5] Erik D. Demaine, Martin L. Demaine, PerouzTaslakian, and Godfried T. Toussaint. Sanddrawings and Gaussian graphs.
Journal ofMathematics and The Arts , 1(2):125–132, June2007. Originally at BRIDGES 2006.doi:10.1080/17513470701413451.[6] Erik D. Demaine, Joseph S. B. Mitchell, andJoseph O’Rourke. Problem 33: Sum of SquareRoots. In
The Open Problems Project . SmithCollege, September 19 2017. URL:https://cs.smith.edu/ ∼ jorourke/TOPP/P33.html.[7] Paulus Gerdes. The ‘sona’ sand drawing traditionand possibilities for its educational use. In Geometry From Africa: Mathematical andEducational Explorations , pages 156–205. TheMathematical Association of America, 1999.[8] Alon Itai, Christos H. Papadimitriou, andJayme Luiz Szwarcfiter. Hamilton paths in gridgraphs.
SIAM Journal on Computing ,11(4):676–686, 1982. doi:10.1137/0211056.[9] David Lichtenstein. Planar formulae and their uses.
SIAM Journal on Computing , 11(2):329–343, 1982.URL: https://doi.org/10.1137/0211025,doi:10.1137/0211025.[10] Joseph O’Rourke. Advanced problem 6369.
American Mathematical Monthly , 88(10):769, 1981.doi:10.2307/2321488.
A Separation from TSP Tour under the L Metric
Theorem 1
There exists a set of sites for whichthe length of the minimum-length TSP tour is √ π ( √ √ π ( √ ≈ . times the length of theminimum-length sona drawing. Proof.
We construct a problem instance whoseminimum-length TSP tour is longer than its minimum-length sona drawing by a factor within ε of √ π ( √ √ π ( √ . Our construction is illustrated in Fig-ure 1 and follows a fractal approach with ε − levels (forsimplicity, we assume that ε − is an integer).1. We start by defining a few auxiliary points:(a) The initial set of auxiliary points A isthe intersection between a slightly shiftedinteger lattice and the L ball B (0 , ε − )of radius ε − centered at the origin.That is, A = (cid:8) ( i +12 , j +12 ) : i, j ∈ Z (cid:9) ∩ (cid:8) ( x, y ) : | x | + | y | ≤ ε − (cid:9) . In Figure 1a, theauxiliary points are exactly the intersectionsof solid red lines. (b) Then, in Step i (starting with i = 1), for eachauxiliary point p ∈ A i − , we add five points to A i : p itself, and four new points at distance (cid:16) √ (cid:17) i from p in each of the four cardinaldirections. Let A = A ε − . By construction, wehave | A i | = 5 i | A | and | A | = 2 ε − + O ( ε − ).We note that set A contains auxiliary points(not sites). These points will not be part ofthe instance.2. We now use the auxiliary points to create some sites(the isolated black points of Figure 1):(a) For any i ≥ P i of sites as follows:for each auxiliary point q ∈ A i − we add thefour sites whose x and y coordinates each differfrom q by (cid:16) √ (cid:17) i . We note that all theadded sites are distinct sites.(b) We define P as the set of integer lattice pointsin B (0 , ε − ). Equivalently, for each auxiliarypoint q ∈ A we add the sites whose x and y coordinates each differ from q by , but we donot add sites that lie outside B (0 , ε − ), whichaffects O ( ε − ) sites (out of Ω( ε − ) sites of P ).In this case, the sites created by differentauxiliary points may lie in the same spot.In total, P contains only one site per aux-iliary point of A (except for O ( ε − ) auxil-iary points near the boundary). Thus, | P i | =4 · | A i − | = 4 · i − · | A | (for i ≥
1) and | P | = | A | + O ( ε − ) = (2 ε − + O ( ε − )).Let P (1) = P ∪ P ∪ · · · ∪ P ε − .3. Next, we place additional sona sites packing linesegments and/or curves. Whenever we pack anycurve, we place sites spaced at a distance δ smallenough that the length of the shortest path thatpasses within δ of all of them is within a factor(1 − ε ) of the length of the curve.(a) Solid lines as drawn in red in Figure 1a: foreach x ∈ { i + : − ( ε − + 1) ≤ i ≤ ε − and i ∈ Z } , we pack the vertical line segment withendpoints ( x, ε − +1 −| x | ) and ( x, | x |− ε − − y for the horizontal linesegments.(b) In Step i of the above recursive definition (start-ing with i = 1), when we create four new aux-iliary points of A i from a point p ∈ A i − , wealso pack the boundary of the region withinEuclidean distance (cid:16) √ (cid:17) i of the squarewhose vertices are the four auxiliary pointsof A i . Note that this boundary region forms asquare with rounded corners as in Figure 1b.2 nd Canadian Conference on Computational Geometry, 2020
With these extra points we preserve the invari-ant that the auxiliary points are exactly theintersections of packed curves.Let P (2) be the set of sites created in Step 3 in ourconstruction, and P = P (1) ∪ P (2) . This is a completedescription of the construction.We now find the minimum-length sona drawing of P .Each pair of consecutive points in a packed curve mustbe in a separate sona region, so any sona drawing mustpass between them; in particular, any sona drawing mustpass within δ of each of them, and so the length of anyvalid sona drawing is at least 1 − ε times the length of thepacked curves. Also, there’s a valid sona drawing that’sat most 1+ ε times the length of the packed curves: followall the packed curves exactly, adding small loops aroundthe sona sites of the packed curves as necessary (loopssmall enough to lengthen the curve by a factor of at most1 + ε ). The graph of packed curves is Eulerian (becauseit’s defined as a union of boundaries of regions, which arecycles), so the TSP tour can follow an Eulerian circuitthrough it. At an intersection of packed curves, we havetwo options for the sona drawing (as shown in the insetimages of Figure 1). We can have one sona path crossover the other in the Eulerian circuit (by including everysite of the packed curve in a small loop). Alternatively,we can have one sona path cross over the other in twoplaces p and q at the intersection, and leaving one sonasite of the packed curve out of a small loop to be thesona site of the extra region between p and q . In eithercase, we conclude that there is a valid sona path thatfollows an Eulerian circuit of the packed curves within(1 + ε ). Note that, although our description focused inthe sites of P (2) , this is a valid sona tour for P since thesites of P (1) lie in different faces.The total length of the packed curves is hence a (1+ ε )-approximation of the total length of the minimum-lengthsona drawing.The total length of the packed segments of P (2) (thesquare lattice) is 4 ε − + O ( ε − ), since the area of theregion | x | + | y | < ε − and the number of lattice pointsin it are each 2 ε − + O ( ε − ).Now we bound the length of the packed curves(rounded squares). In Step i of the construction (startingwith i = 1), we added a packed curve that is the bound-ary of the region within Euclidean distance (cid:16) √ (cid:17) i of a square of side length (cid:16) √ (cid:17) i (with a total lengthof (4 + π ) (cid:16) √ (cid:17) i ).Recall that we added one such curve for each of thepoints of A i − and that | A i | = 5 i − (2 ε − + O ( ε − )).Thus, the total length of the sona drawings introducedat Step i is (4 + π ) (cid:16) √ (cid:17) i i − (2 ε − + O ( ε − )).For ε small, this series is well-approximated by an infinite geometric series with sum(2 ε − + O ( ε − )) (cid:32) π (1+ √ · (cid:16) − √ (cid:17) (cid:33) =(2 ε − + O ( ε − )) (cid:16) π √ − (cid:17) , and adding in the length ofthe packed segments of P (2) (the square lattice) gives(2 ε − + O ( ε − )) (cid:16) π √ − (cid:17) . We have approximated the minimum length of a validsona drawing; now we approximate the minimum lengthof a TSP tour.Any TSP tour must also come within δ of every pointon every packed curve, which requires a length at least(2 ε − + O ( ε − )) (cid:16) π √ − (cid:17) as above. Also, the TSPtour must visit each site of P (1) . We observe someproperties of this set: • Set P (1) is defined so that sites are far from eachother. Specifically, the Euclidean ball centered atany site p ∈ P i of radius r i = (cid:16) √ (cid:17) i does notcontain other sona sites. This means that we mustinclude at least 2 r i in the length of the TSP tour foreach point in P i , for the part of the tour that passesfrom the boundary of this ball to P i and then backto the boundary. • There are 4 · i − (2 ε − + O ( ε − )) sites in P i (for i ≥
1) and (2 ε − + O ( ε − )) sites in P .When ε tends to zero, the additional length neededin the TSP tour is(2 ε − + O ( ε − ))(1 + (cid:88) i ≥ r i · · i − )= (2 ε − + O ( ε − ))(1 + 45 (cid:88) i ≥ (cid:18)
53 + 2 √ (cid:19) i )= (2 ε − + O ( ε − ))(1 + 43 + 2 √ · − √ )= (2 ε − + O ( ε − ))(1 + 42 √ − ε − + O ( ε − ))(2 √ . So, the total length of the TSP tour is at least(2 ε − + O ( ε − )) (cid:16) π √ − + 2 √ (cid:17) . Hence theratio of the length of the TSP tour to the lengthof the sona drawing is, ignoring lower-order terms, π √ − +2 √ π √ − = π ( √ √ √ π ( √ ≈ . (cid:3) B Complexity of Length Minimization
Theorem 4
It is NP-hard to find a sona drawing fora given set of sites whose length is less than a giventhreshold L , for any of the L , L , and L ∞ metrics. CCG 2020, Saskatoon, Canada, August 5–7, 2020
Figure 8: Local modifications to convert a grid Hamilto-nian cycle into a short sona drawing for a set of doubledsites
Proof.
Let V be the set of n vertices in a hard instancefor finding a Hamiltonian cycle in grid graphs. Wemay form a hard instance of the minimum-length sonadrawing problem for L or L distances by replacingeach vertex in V by a pair of sites, one at the originalvertex position and the other at distance less than ε fromit, where ε = Θ(1 /n ) is chosen to be small enough that4 nε < √ −
1. We set L = n + 2 nε . For L ∞ distance,we use a hard instance for L distance, rotated by 45 ◦ .If V is a yes-instance for Hamiltonian cycle, let C be aHamiltonian cycle of length n for V . We may form a sonadrawing of length less than L by modifying C within aneighborhood of each pair of sites so that, for all butone of these pairs, it makes two loops, one surroundingeach site (Figure 8), and so that for the remaining pair itmakes one loop around one of the two sites and surroundsthe other point by the face formed by C itself. In thisway, each face of the modified curve surrounds a singlesite of our instance. Each of these local modifications to C may be performed using additional length less than2 ε , so the total length of the resulting sona drawing isless than L .If V is a no instance for Hamiltonian cycle, let C be any sona drawing for the resulting instance of theminimum-length sona drawing problem. Then C mustpass between each pair of sites in the instance, and bymaking a local modification of length at most 2 ε neareach pair, we can cause it to touch the point in thepair that belongs to V itself. Thus, we have a curve oflength | C | + 2 nε touching all points of V . Because V isa no instance, the length of this curve must be at least n + √ −
1, from which it follows that the length of C is at least n + √ − − nε ≥ L . (cid:3) By scaling the sites by a factor of Θ(1 /ε ) = O ( n ) wemay obtain a hard instance of the minimum-length sonadrawing problem in which all sites lie in an integer gridwhose bounding box has side length O ( n ).It is possible to represent a minimum-length sonadrawing combinatorially, as a conveyor belt [2] formedby bitangents and arcs of infinitesimally small disks centered at each site, and to verify in polynomial timethat a representation of this form is a valid sona drawing.However, this does not suffice to prove that the decisionversion of the minimum-length sona drawing problembelongs to NP. The reason is that, when the sites haveinteger coordinates, the limiting length of a sona drawing,represented combinatorially in this way, is a sum ofsquare roots (distances between pairs of given points)and we do not know the computational complexity oftesting inequalities involving sums of square roots [6, 10].(Euclidean TSP has the same issue.) C Crenellations for Grid Drawing
Figures 9, 10, and 11 show how to add crenellations to theConstant gadget, an unconstrained wire end (variable),and the OR gadget, respectively. The crenellated Split/ Invert is the same as in Figure 7, extended in theobvious way for ports that are used. In no case dothe crenellations affect the internal properties describedin the main text; these figures simply show that it ispossible to add the crenellations appropriately.As mentioned in the main text, the crenellations doadd a parity constraint when connecting gadgets withwires; we can no longer make wires of arbitrary length,but must match the crenellations to the gadgets at eachend. In order to do that we need one additional gadget,an inverting turn, shown in Figure 12. Unlike in Figure 7,the wire state is switched during the turn. Observe thatin Figure 7, turning does not change crenellation parity,but the straight-through path, which would invert ifnot terminated, does change crenellation parity. Theinverting turn also does not change crenellation parity.Therefore, to change the crenellation parity of a wire,we can invert it (straight through), changing the parity,and add a sequence left inverting turn, right turn, rightturn, left turn to restore the original line of the wire.2 nd Canadian Conference on Computational Geometry, 2020
Figure 9: Crenellated Constant gadget (a) State 1 (b) State 2
Figure 10: Crenellated unconstrained wire end (a) false + true → true (b) true + true → true(c) true + false → true (d) false + false → falsefalse