Reconstructing Generalized Staircase Polygons with Uniform Step Length

Transcription

1 Jornal of Graph Algorithms and Applications ol. 22, no. 3, pp (2018) DOI: /jgaa Reconstrcting Generalized Staircase Polygons with Uniform Step Length Nodari Sitchinaa 1 Darren Strash 2 1 Department of Information and Compter Sciences, Uniersity of Hawaii at Manoa, USA. 2 Department of Compter Science, Hamilton College, Clinton, NY, USA. Abstract Visibility graph reconstrction, which asks s to constrct a polygon that has a gien isibility graph, is a fndamental problem with nknown complexity (althogh isibility graph recognition is known to be in PSPACE). As far as we are aware, the only class of orthogonal polygons that are known to hae efficient reconstrction algorithms is the class of orthogonal conex fans (staircase polygons) with niform step lengths. We show that two classes of niform step length polygons can be reconstrcted efficiently by finding and remoing rectangles formed between consectie conex bondary ertices called tabs. In particlar, we gie an O(n 2 m)-time reconstrction algorithm for orthogonally conex polygons, where n and m are the nmber of ertices and edges in the isibility graph, respectiely. We frther show that reconstrcting a monotone chain of staircases (a histogram) is fixed-parameter tractable, when parameterized on the nmber of tabs, and polynomially solable in time O(n 2 m) nder alignment restrictions. As a conseqence of or reconstrction techniqes, we also get recognition algorithms for isibility graphs of these classes of polygons with the same rnning times. Sbmitted: Noember 2017 Reiewed: Janary 2018 Article type: Reglar paper Reised: March 2018 Pblished: September 2018 Accepted: April 2018 Commnicated by: F. Frati and K.-L. Ma Final: May 2018 This material is based pon work spported by the National Science Fondation nder Grant No addresses: nodari@hawaii.ed (Nodari Sitchinaa) dstrash@hamilton.ed (Darren Strash)

2 432 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons 1 Introdction Visibility graphs, sed to captre isibility in or between polygons, are simple bt powerfl tools in comptational geometry. They are integral to soling many fndamental problems, sch as roting in polygons, and art gallery and watchman problems, to name a few. Efficient, and een worst-case optimal, algorithms exist for compting a isibility graph from an inpt polygon [17]; howeer, comparatiely little is known abot the reerse direction: the so-called isibility graph recognition and reconstrction problems. In this paper, we stdy ertex-ertex isibility graphs, which are formed by isibility between pairs of ertices of a polygon. Gien a graph G = (V, E) on n = V ertices and m = E edges, the isibility graph recognition problem asks if G is the isibility graph of some polygon. Similarly, the isibility graph reconstrction problem asks s to constrct a polygon with G as a isibility graph. Srprisingly, recognition of simple polygons is only known to be in PSPACE [14], and it is still nknown if simple polygons can be reconstrcted in polynomial time. Therefore, crrent soltions are typically for restricted classes of polygons. Een characterizations are only known for special classes of isibility graphs [18, 21], and only for necessary conditions are known [16] for standard isibility graphs. 1.1 Special Classes of Polygons A well-known reslt de to ElGindy [12] is that eery maximal oterplanar graph is a isibility graph and a polygon can be reconstrcted from eery sch graph in polynomial time. Other special classes rely on a niqe configration of reflex and conex chains, which restrict isibility. For instance, spiral polygons [15] and tower polygons [8] (also called fnnel polygons), can be reconstrcted in linear time, and each consists of one and two reflex chains, respectiely. Frther, 2-spirals can be reconstrcted in polynomial time [2], as can a more general class of isibility graphs related to 3-matroids [3]. Finally conex polygons with a single conex hole [7] can be reconstrcted in polynomial time. For monotone polygons, Colley [9, 10] showed that if each face of a maximal oterplanar graph is replaced by a cliqe on the same nmber of ertices, then the reslting graph is a isibility graph of some ni-monotone polygon (monotone with respect to a single edge), and sch a polygon can be reconstrcted if the Hamiltonian cycle of the bondary edges is known. Howeer, not eery ni-monotone polygon (een those with niformly spaced ertices) has sch a isibility graph [13]. Finally, Eans and Saeedi [13] characterized terrain isibility graphs, which consist of a single monotone polygonal line. Orthogonal polygons. Srprisingly little is known abot isibility graphs of orthogonal polygons: we are only aware of reslts for orthogonal conex fans, which consist of a single staircase and an extra ertex. These are also known as staircase polygons. Abello and Eğecioğl [1] show that orthogonal conex fans

3 JGAA, 22(3) (2018) 433 with niform side-length can be reconstrcted in linear time; howeer, their constrction relies on a simpler definition of isibility, which allows for blocking ertices. Frthermore, Abello et al. [4] show that general orthogonal conex fans are recognizable in polynomial time, bt reconstrction is still open een for this simple class of orthogonal polygons. Other algorithms for orthogonal polygons se different isibility representations, which seem to be easier to work with. For example, researchers hae looked into edge-edge isibility and ertex-edge isibility. For orthogonal polygons, edge-edge isibility graphs [20, Section 7.3] hae been shown to consist of two disjoint trees and the polygon can be reconstrcted when these trees mesh well. For ertex-edge isibility, orthogonal polygons can be reconstrcted entirely from stabs [19] for each ertex, we are gien the edges hit by horizontal (or ertical) rays shot from the ertex in O(n log n) time. Howeer, we also mst be gien the Hamiltonian cycle of the bondary edges with the inpt. See Asano et al. [5] or Ghosh [16] for a thorogh reiew of reslts on isibility graphs. 1.2 Or Reslts In this work, we inestigate reconstrcting polygons consisting of mltiple niform step length staircases. To the best of or knowledge, the only pre isibility reconstrction reslt for orthogonal polygons is for a single niform step length staircase [1]. We first show that orthogonally conex polygons can be reconstrcted in O(n 2 m) time. We frther show that reconstrcting orthogonal ni-monotone polygons (also known as histograms) is fixed-parameter tractable, when parameterized on the nmber of horizontal bondary edges that are incident to two conex ertices in the polygon. We also proide an O(n 2 m) time algorithm nder alignment restrictions. As a conseqence of or reconstrction techniqe, we can also recognize the isibility graphs of these classes of polygons with the same rnning times. 2 Preliminaries Let P be a polygon on n ertices. We say that a point p sees a point q (or p and q are isible) in polygon P if the line segment pq does not intersect the exterior of P. Under this definition, isibility is allowed along edges and throgh ertices. If we restrict or attention to isibility between ertices of the polygon, we get a natral pairwise relationship between ertices, giing s a isibility graph. Definition 1 (isibility graph) A isibility graph G P = (V P, E P ) of polygon P has a ertex p V P for each ertex p of P, and an edge ( p, q ) E P if and only if ertices p and q are isible in P. De to the one-to-one relationship between polygon and graph ertices, we reference the ertex of a isibility graph as we wold a ertex of the polygon.

4 434 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons e 2 C 2 e 1 C 1 e 3 C 3 (a) (b) (c) Figre 1: Maximal conex regions on ertices of polygons (e.g., C 1, C 2, and C 3 ) are maximal cliqes in isibility graphs, and 1-simplicial edges (e.g., e 1, e 2, and e 3 ) are each in exactly one maximal cliqe. For or isibility graph discssion, we adopt standard notation for graphs and polygons. In particlar, for a graph G = (V, E), we denote the neighborhood of a ertex V by N() = { (, ) E}, the degree of by deg() = N(), and the nmber of ertices and edges by n = V and m = E, respectiely. For a isibility graph G P = (V P, E P ) of a polygon P, we call an edge in G P that is an edge of P a bondary edge. Other edges (diagonals in P ) are non-bondary edges. Finally, critical to or proofs is the fact that a maximal cliqe in G P corresponds to a maximal (in the nmber of ertices) conex region R P whose ertices are defined by ertices of P. We say that an edge (, ) is 1-simplicial if the common neighborhood N() N() is a cliqe, or eqialently (, ) is in exactly one maximal cliqe 1. The intition behind why we consider 1-simplicial edges is that, in orthogonal polygons with edges of niform length, bondary edges between conex ertices are 1-simplicial, with the ertices of the cliqe forming a rectangle. (See Figre 1.) Note, howeer, that non-bondary edges may also be 1-simplicial. For or rnning times, we rely on the following obseration. Obseration 1 We can test if (, ) is 1-simplicial and in a maximal k-cliqe in time O(kn) with an adjacency list data strctre, or time O(k 2 + n) with an adjacency matrix. Proof: Compte X = N() N() in O(n) time by marking ertices in N() and checking for marked ertices in N(), then check that X = k and X is a cliqe. This last step takes O(kn) time with an adjacency list, or O(k 2 ) with an adjacency matrix. 1 This is not to be confsed with simplicial edges, which are defined elsewhere to be edges (, ) sch that for eery w N() and x N(), w and x are adjacent. Instead, this definition parallels the sage for ertices: A ertex is simplicial if N() forms a cliqe, or eqialently is in exactly one maximal cliqe.

5 JGAA, 22(3) (2018) 435 We note that if k = O(n) this test takes time O(n 2 ) and, if k = O(1), the test takes time O(n). 3 Uniform-Length Orthogonally Conex Polygons We first trn or attention to a restricted class of orthogonal polygons that hae only niform-length (or eqialently, nit-length) edges. Let P be an orthogonal polygon with niform-length edges sch that no three consectie ertices on P s bondary are collinear, and frther let P be orthogonally conex 2. We call P a niform-length orthogonally conex polygon (UP). Note that eery ertex i on P s bondary is either conex or reflex. We call bondary edges between two conex ertices in a niform-length orthogonal polygon P tabs and the endertices of a tab tab ertices. We reconstrct the polygon by compting the clockwise ordering of ertices of the UP. Note that the bondary of a UP consists of for tabs connected ia staircases. For ease of exposition, we assme that the UP is embedded in R 2 with polygon edges axis-aligned. We call the tab with the largest y-coordinate the north tab, and we analogosly name the others the soth, east, and west tabs. Following this conention, we refer to the for bondary staircases as northwest, northeast, sotheast, and sothwest. Frthermore, note that it is possible for two (opposite) staircases to hae more ertices than the other two (also opposite) staircases. We call the staircases with more ertices long staircases and others short. We only consider polygons with more than 12 ertices, which eliminates many special cases. Smaller polygons can be soled in constant time ia brte force. We first introdce seeral strctral lemmas which help s to identify all conex ertices in a UP, which is key to or reconstrction. Lemma 1 For eery conex ertex in a UP, there is a conex ertex, sch that (, ) E P and (, ) is 1-simplicial. Proof: If is a tab ertex, then the tab ertex that is s bondary neighbor is conex and (, ) is 1-simplicial. Otherwise, withot loss of generality, sppose that is on the northwest staircase. Then there is a conex ertex on the sotheast staircase that is isible from. Edge (, ) is in exactly one maximal cliqe, consisting of,, the reflex ertices within the rectangle R defined by and as the opposite corners, and any other corners of R that are conex ertices of the polygon. Frthermore, we hae the following lemma, which allows s to differentiate between conex and reflex ertices. 2 That is, any two points in P can be connected by a staircase contained in P.

6 436 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons z w z w z w (a) w (b) w (c) w Figre 2: Illstration of the proofs of (a) Obseration 2 and (b)-(c) Obseration 3. The ertex on the northwest staircase can be reflex or conex. Lemma 2 For eery isible pair of ertices and in a UP, if or is a reflex ertex, then the edge (, ) E P is not 1-simplicial. Combining Lemmas 1 and 2 gies s a simple method to determine conex and reflex ertices: test if each edge is in more than one maximal cliqe. Mark all ertices incident to 1-simplicial edges as conex, and mark the rest as reflex. We proe Lemma 2 below, bt first we make some necessary obserations. Obseration 2 If a ertex on the northwest staircase of a UP sees a reflex ertex z on the sotheast staircase and the slope of the line spporting the line segment z is non-positie, then sees both (conex) bondary neighbors w and w of z. Proof: We will proe that w, the horizontal bondary neighbor of z, is isible from. The proof for isibility of w, the ertical bondary neighbor of z, is symmetric. See Figre 2(a) for an illstration. First, obsere that in a UP, no bondary edge on the northeast and the sothwest staircases blocks isibility between the ertices of the northwest staircase and the ertices of the sotheast staircase, and, conseqently, between and w. Next, by iewing as the origin of the Cartesian coordinate system, z lies in the sotheast qadrant (inclsie of the axis), since the slope of the line spporting z is non-positie. And since zw is a horizontal edge, w also lies in the sotheast qadrant. Therefore, no bondary edge of the northwest staircase blocks the isibility between and w (recall that or definition of isibility allows isibility along the polygon edges). Finally, by iewing w as the origin, we can similarly see that no edge of the sotheast staircase blocks the isibility between and w. Obseration 3 Let a reflex ertex be on the sothwest or northeast staircase of a UP and a reflex ertex z be on the sotheast staircase. If the slope of the line spporting the line segment z is non-positie, then sees both (conex) bondary neighbors w and w of z.

7 JGAA, 22(3) (2018) 437 t r w t b (a) (b) Figre 3: Illstration of the proof that an edge (, ) is not 1-simplicial when both and are reflex. The proof demonstrates the existence of a conex ertex w isible from both and. Proof: Consider the case when is on the sothwest staircase (the proof of the case when is on the northeast staircase is symmetric). See Figre 2(b)-(c) for an illstration. First, obsere that no bondary edge on the northwest and northeast staircase blocks the isibility between the ertices of the sothwest staircase and the ertices of the sotheast staircase, and, conseqently, between and w. Next, obsere that the slope of the line spporting the edges from any reflex ertex on the sothwest staircase and any ertex on the sotheast staircase is at least tan(π/4) and, therefore, no ertex on the sothwest staircase can block the isibility between them. Finally, by iewing w (resp., w ) as the origin of the Cartesian coordinate system, we can see that is in the northwest qadrant (inclsie of axis) and, therefore, no bondary edge on the sotheast staircase can block the isibility between w (resp., w ) and. We are ready to proe Lemma 2. Proof: Let (, ) E P and sppose that at least one of and is reflex. Case 1: Both and are reflex. Then they belong to a maximal cliqe consisting of all reflex ertices and no conex ertices. We will show that both and also see some conex ertex w, therefore,, and w are part of another maximal cliqe. For concreteness of exposition, we orient the polygon so is always on the northwest staircase. Frther, if all staircases hae the same nmber of ertices, we arbitrarily choose the northwest and sotheast staircases to be called short, and the others long. There are two cases to consider: (a) Both and are on short staircases. (See Figre 3(a)). Obsere, that eery reflex ertex on a short staircase sees all ertices on the other short staircase. Then, if and are on the same (northwest) staircase, there is a conex ertex w on the sotheast staircase, isible to both and. If is on the sotheast staircase, w is either of the two conex bondary neighbors of. Since w is a bondary neighbor of, it is isible from.

8 438 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons w w z w w z w w z w w Figre 4: Illstration of the proof that an edge (, ) is not 1-simplicial when is conex and is reflex. For each of the for cases of when is on the for different staircases, there is a pair of conex ertices w and w on the sotheast staircase, which are isible to both and, bt not to each other. (b) At least one of and is on a long staircase. (See Figre 3(b)). Vertex sees the right tab ertex t r of the north tab and the bottom tab ertex t b of the west tab. Since the northwest staircase is long, the nion of reflex ertices isible from t r and t b is the set of all reflex ertices in the polygon. Therefore, eery reflex ertex mst see at least one of t r or t b. Case 2: One of the ertices, or, is conex. Withot loss of generality, let it be (again, on the northwest staircase), and let be reflex. Let P be the polygon consisting of and all reflex ertices isible from. For eery reflex ertex in P, we will identify two conex ertices w and w on the sotheast staircase, which are isible to both and. Since w and w are conex ertices on the same staircase, they are inisible to each other and, therefore,,, w and,, w will be part of two distinct maximal cliqes. Let s consider the for possible locations for (see Figre 4): (a) is on the sotheast staircase. Let w and w be the two conex bondary neighbors of. Clearly, sees w and w. Moreoer, since sees and is on the opposite staircase from, the slope of the line spporting the segment is non-positie. By Obseration 2, mst also see w and w. (b) is on the sothwest staircase. Let z be the lowest reflex ertex on the sotheast staircase (adjacent to the soth tab s right ertex) and let w and w be the two conex bondary neighbors of z. Vertex z is isible from becase both and z are reflex. Since z is the lowest reflex ertex, the slope of the line spporting the segment z is non-positie (the slope is 0 if is the lowest reflex ertex on the sothwest staircase). Ths, by Obseration 3, w and w are isible from. Moreoer, since sees and is on the sothwest staircase, z mst be isible from with een smaller slope of the line spporting segment z, i.e., the slope is also non-positie. Therefore, by Obseration 2, w and w are also isible from. (c) is on the northeast staircase. Let z be the highest reflex ertex on the sotheast staircase (adjacent to the east tab s bottom ertex) and let w

9 JGAA, 22(3) (2018) 439 and w be the two conex bondary neighbors of z. Again, ertex z is isible from becase both and z are reflex. We can see that the slope of the line spporting the segment z is negatie by iewing as the origin of the Cartesian coordinate system and obsering that z is in the sotheast qadrant. Ths, by Obseration 3, w and w are isible from. At the same time, z is isible from becase is isible from and z is below and to the right of (or directly below). Moreoer, the slope of the line spporting segment z is also non-positie. Therefore, by Obseration 2, w and w are also isible from. (d) is on the northwest staircase. Since is isible from, mst be a reflex bondary neighbor of. Obsere, that becase P is bonded from all sides, it mst contains a non-empty sbset Z of reflex ertices from the sotheast staircase. Moreoer, there exists a ertex z Z, which is not axis-aligned with, i.e., the slope of the line spporting the segment z is strictly negatie. Let w and w be the two conex bondary neighbors of z. By definition of P, z is isible from, therefore, by Obseration 2, w and w are isible from. Again, z is isible from becase both and z are reflex. Since the slope of the line spporting segment z is strictly negatie and the polygon has niform length bondaries, the slope of the line spporting segment z mst be non-positie. Therefore, by Obseration 2, w and w are also isible from. Therefore, (, ) is not 1-simplicial. Lemma 2 states that only edges between conex ertices can be 1-simplicial. Combining this lemma with Lemma 1, we can identify all conex ertices by checking for each edge (, ) if N() N() is a cliqe in O(n 2 ) time, leading to the following lemma. Lemma 3 We can identify all conex and reflex ertices in a isibility graph of a UP in O(n 2 m) time. We now diide the class of niform-length orthogonal polygons into two classes of polygons, which we call reglar and irreglar. Definition 2 (reglarity) We call a UP reglar if each of its staircase bondaries hae the same nmber of ertices. Otherwise, we call it irreglar, consisting of two long and two short staircases. We restrict or attention to irreglar niform-length orthogonally conex polygons (IUPs); howeer, similar methods work for their reglar conterparts. 3.1 Irreglar Uniform-Length Orthogonally Conex Polygons Let G P be the isibility graph of IUP P. Or reconstrction algorithm first comptes the for tabs, then assigns the conex and reflex ertices to each

10 440 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons w (a) (b) (c) Figre 5: Steps to identify the tabs. (a) Each tab cliqe (highlighted) has three edges between conex ertices. (b) Vertical (also horizontal) non-bondary edges are not 1-simplicial. (c) After remoing these edges, the middle ertex on each remaining 2-path is a tab ertex, and in this instance the other tab ertex sees more reflex ertices than w. staircase. The following strctral lemma helps s to find the tabs. We assme that we hae already compted the conex and reflex ertices in O(n 2 m) time. Lemma 4 In eery IUP there are exactly for 7-ertex maximal cliqes that indiidally contain exactly three conex ertices. Each sch cliqe contains exactly one tab, and each tab is contained in exactly one of these cliqes. Proof: First note that each of the for tabs is in exactly one sch maximal 7-cliqe. Frther, any other cliqe that contains three conex ertices has at least nine ertices. Each conex ertex mst be on a different staircase with its two bondary neighbors, which are therefore distinct. We note that it is not necessary to identify the for tabs explicitly to contine with the reconstrction. There are only 7 4 = O(1) choices of tabs (one from each 7-cliqe of Lemma 4), ths we can try all possible tab assignments, contine with the reconstrction and erify that or reconstrction prodces a alid IUP P with the same isibility graph. Since there are only O(1) choices for tabs, this wold add no additional rnning time to or algorithm. Howeer, we show how to explicitly find the for tabs in the proof of the following lemma. Lemma 5 We can identify the for tabs of an IUP in O(nm) time. Proof: We compte the for 7-ertex maximal cliqes of Lemma 4 in O(nm) time sing Obseration 1. These cliqes hae exactly three conex ertices each, and tabs are incident to two conex ertices, narrowing or choice of tab down to 4 (3 2) = 12 edges. See Figre 5(a). For of these are ertical or horizontal

11 JGAA, 22(3) (2018) 441 (non-bondary) edges, which we can detect and eliminate, as they are not 1- simplicial: these edges are in both a cliqe with both tab ertices, and a cliqe containing a tab ertex, another conex ertex, and reflex ertices not in the tab cliqe. See Figre 5(b). We hae eight remaining edges to consider. These eight edges form for disjoint paths on two edges, and the middle ertex on each path is a tab ertex. Note that these middle tab ertices are on the long staircases. Let one of them be called. Now it remains to find s neighbor on its tab. Vertex has two candidate neighbors; let s call them and w. Jst for concreteness, let s say is the ertex of the north tab on the northeast staircase. See Figre 5(c). Sppose, withot loss of generality, that has more reflex neighbors than w, then is s neighbor on a tab, becase it sees reflex ertices on the whole northeast and sotheast staircases, while w sees only a sbset of those. Otherwise and w hae the same nmber of reflex neighbors, which only happens when w sees eery reflex ertex on the sotheast and northeast staircases. Then either or w has more conex neighbors. Sppose, withot loss of generality, that is a tab ertex, then has fewer conex neighbors than w. To see why, note that since is on the northeast (long) staircase, is on the northwest (short) staircase. Vertex has conex neighbors, w, and eery conex ertex on the sotheast (short) staircase. Likewise, w has conex neighbors,, eery conex ertex on the northeast (long) staircase (inclding ) and one ertex on the sotheast (short) staircase. We can perform these checks for all sch pairs and w, thereby compting all tabs. Note that, if we hae already distingished conex and reflex ertices, this step takes time O(nm): O(nm) time to compte the for cliqes containing tabs, and O(n) to cont the nmber of reflex and conex ertices isible from each tab ertex candidate. We pick one tab arbitrarily to be the north tab. We conceptally orient the polygon so that the northwest staircase is short and the northeast staircase is long. We do this by compting elementary cliqes, which identify the conex ertices on the short staircase. Definition 3 (elementary cliqe) An elementary cliqe in an IUP is a maximal cliqe that contains exactly three conex ertices: one from a short staircase, and one from each of the long staircases. (See Figre 6(a).) Lemma 6 We can identify the elementary cliqes containing ertices on the northwest staircase in O(nm) time. Proof: Each elementary cliqe C contains a 1-simplicial edge, as C is maximal and two conex ertices in C mst be on opposite staircases. Using Obseration 1, we compte all 1-simplicial edges in maximal cliqes on seen or nine ertices in O(nm) time, and keep the cliqes that are elementary cliqes those that contain three conex ertices (and either for or six reflex ertices). Let k NW be the nmber of conex ertices on the northwest staircase. We nmber these conex ertices from 0 to knw 1 in order along the northwest

12 442 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons b C 0 C 1 C 2 C 3 C 4 a (a) (b) (c) Figre 6: Elements of or reconstrction. (a) The elementary cliqes C 0,..., C 4 interlock along a short staircase. (b) The tab ertices a and b see niqe reflex ertices on long staircases. (c) Unassigned reflex ertices (sqares) are assigned by forming rectangles between the conex ertices and and their alreadyassigned reflex bondary neighbors (circles). staircase from the north tab to the west tab. Denote by C i the niqe elementary cliqe containing i. Then C 0 is the niqe maximal (elementary) cliqe containing the north tab. Frthermore, each cliqe C i contains a set of reflex ertices R i sch that R i C i+1 = 3, and for j {i 1, i, i + 1}, R i C j =. Therefore, gien the elementary cliqes C 0,..., C i, we can compte the elementary cliqe C i+1 by searching for the remaining elementary cliqe containing three reflex ertices in R i. (Note that the elementary cliqe C i 1 also contains ertices in R i, bt C i 1 has already been discoered.) Once we reach an elementary cliqe containing a tab, then we hae compted all elementary cliqes on the northwest staircase. This tab is the west tab and we are finished. Note that, if or sole prpose is to reconstrct the IUP P, we hae sfficient information. The nmber of elementary cliqes gies s the nmber of ertices on a short staircase of P, from which we can bild P. Howeer, in what follows, we can actally map all ertices to their positions in the IUP, a techniqe which we later se to bild a recognition algorithm for IUPs. First, we show how to assign all conex ertices from the elementary cliqes to each of the three staircases, sing isibility of the north and west tab ertices. Note, constrcting the elementary cliqes with Lemma 6 also gies s the west tab, since it is contained in the last elementary cliqe on the northwest staircase. Lemma 7 We can identify the conex ertices on the northwest staircase in O(n) time. Proof: The northwest staircase contains the conex ertices of the elementary cliqes from Lemma 6 that cannot be seen by any of the north or west tab ertices. The staircase frther contains the left ertex of the north tab and the

13 JGAA, 22(3) (2018) 443 top ertex of the west tab (which can be identified by the fact that they are tab ertices that do not see either ertex of the other tab). We can repeat the aboe process to identify the conex ertices of the sotheast staircase. Howeer, we might not yet be able to identify tabs as soth or east. Ths, we will obtain two possible orderings of the conex ertices on the sotheast staircase. Next, we show how to assign conex ertices to the long staircases. In the process we determine soth and east tabs, and conseqently, identify the correct ordering of conex ertices on the sotheast staircase. Lemma 8 We can assign the remaining conex ertices in O(n 2 ) time. Proof: Let W and E be the sets of all conex ertices on the sothwest and northeast staircases (which we are compting) and let W 0 and E 0 be the sets of conex ertices on the sothwest and northeast staircases that are already known from the elementary cliqes from Lemma 7. Note that the order of these ertices has not been determined. Let N c () denote the set of conex neighbors on the opposite staircase of a ertex. We now illstrate how to discoer ertices in W and E. Sppose for the moment that we can select the conex ertex w 0 W 0 that is the northernmost ertex in W 0. There is exactly one conex neighbor of w 0 that has not yet been discoered, and it is in N c (w 0 ) E. Call this conex ertex e. We note that a maximal cliqe (a rectangle) is formed between ertices w 0, e, and their reflex bondary neighbors, and therefore we hae discoered a new ertex in E by compting a rectangle spanning the polygon. See Figre 6(c). Howeer, note that we do not need to begin with the northernmost (conex) ertex in W 0 ; we can discoer ertices in E with any ertex w 0 W 0. All of w 0 s ndiscoered conex neighbors are in E, and at least one ertex in W 0 (e.g., the northernmost) has ndiscoered neighbors in E. Similarly, with ertices in E 0 we will discoer at least one new w W. We can then iteratiely bild sets W and E by compting the sets E i = ( w Wi 1 N c (w))\e i 1 and W i = ( e Ei 1 N c (e))\w i 1. We then identify all ertices of the sothwest and northeast staircases as W = i W i and E = i E i. To order the ertices along the sothwest staircase, note that the sets W i shold appear in order of increasing i from top to bottom. Also note that if one were to assign the ertices of W i to a staircase from top to bottom, each ertex w i in this order wold see fewer ertices of E i 1. Ths, we can order the ertices within each W i. The argment for ordering ertices of E i is symmetric. Compting the set E i takes time O( W i 1 n): we ealate all O(n) neighbors of each neighbor in W i 1. Likewise, compting W i takes time O( E i 1 n). Ths, oerall it takes O(( W + E ) n) = O(n 2 ) time to constrct (and order) sets W and E. We can now choose the soth and east tabs: a ertex on the east (soth) tab can see conex ertices on the sothwest (northeast) staircase. Now all conex ertices are assigned to staircases, and all that remains is to assign any remaining reflex ertices.

14 444 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons Lemma 9 We can assign the reflex ertices to each staircase in O(n 2 ) time. Proof: Now that the conex ertices are also ordered on the long staircases, we can assign the reflex ertices to each staircase. We first assign those reflex ertices that are seen from the tab ertices. Let a and b be ertices on different tabs sch that a sees b and ab contains all reflex ertices of a short staircase (see Figre 6(b)). For ease of discssion, we assme that a and b are on the west and north tabs, respectiely. Let R be the set of all reflex ertices of the IUP. The set R 0 = N(a) N(b) R contains all reflex ertices from the northwest staircase, pls two extra reflex ertices: the bondary neighbors of a and b, each on a different long staircase. The remaining ertices N(a) \ R 0 are on the northeast (long) staircase and N(b) \ R 0 are on the sothwest (long) staircase. Let k be the nmber of reflex ertices on the northwest staircase, then R 0 = k + 2 and N(b) \ R 0 = N(a) \ R 0 = k. Let l be the nmber of reflex ertices on the sothwest (or eqialently, northeast) staircase, and let w 1, w 2,..., w l, and e 1, e 2,..., e l be the reflex ertices on the sothwest and northeast staircases respectiely (which we are compting), where sbscripts indicate north to soth order along each staircase. In contrast to w 1 and e 1, we hae already discoered the ertices w 2,..., w k+1 and e 2,..., e k+1 thogh we do not yet know their order. Howeer, the reflex ertices on a single staircase can easily be ordered in time O(n 2 ) time by intersecting the neighborhoods of the (already-ordered) consectie conex ertices on the staircase. We first compte the reflex bondary neighbors of a and b, which will be w 1 and e 1, respectiely. Note that, w 1 is contained in a maximal 4-cliqe that is a degenerate rectangle (a horizontal line segment) containing a, w 1, e k+1, and e k+1 s conex bondary neighbor to the east, x, which is the (k + 1)-th conex ertex on the northeast staircase. The ertex w 1 can be fond in time O(n) by compting this 4-cliqe from the 1-simplicial edge (a, x). The ertex e 1 can be assigned analogosly. Note also that this gies s all reflex ertices on the northwest staircase, which consists of the set of ertices R 0 \ {w 1, e 1 }. To assign the remaining reflex ertices to their staircases, we will conceptally bild axis-aligned rectangles that span from one long staircase to the other, similar to the proof of Lemma 8. These rectangles are maximal cliqes consisting of two conex ertices and on different long staircases, two known reflex bondary neighbors of, and one known reflex bondary neighbor of (see Figre 6(c)). These rectangles contain one yet-nassigned reflex ertex, which is bondary neighbor of, which is then assigned to s staircase. Let be the (conex) bondary neighbor of both w 1 and w 2, and let be the bondary neighbor of e k+1 that is not a neighbor of e k. Then (, ) is a 1- simplicial edge in a maximal 6-cliqe (rectangle) with a single nassigned reflex ertex, which is e k+2. A symmetric argment can be sed to discoer w k+2 from e 1, e 2 and w k+1. We alternate discoering ertices from the sothwest and northeast staircases ntil we hae assigned all reflex ertices. Note that since conex ertices are known and in order along their staircases, the conex ertex can be be compted in O(1) time from : if is the i-th

15 JGAA, 22(3) (2018) 445 (a) (b) Figre 7: (a) A histogram with three tabs and (b) its decomposition into toching rectangles with a contact graph that is a tree. conex ertex on a long staircase, then is the (i + k)-th ertex on the opposite long staircase. Gien the two conex ertices and, it takes O(n) time to compte the maximal 6-cliqe of the rectangle containing them, and O(1) time to determine the new reflex ertex. We mst do this for O(n) rectangles. Hence, the total time to discoer these new reflex ertices is O(n 2 ). Obsere that within each staircase, bondary edges are formed only between conex ertices and their reflex neighbors. Ths, we can order reflex ertices on each staircase by iterating oer the staircase s conex ertices (order of which is determined in Lemmas 6-8) and we are done. The rnning time of or algorithm is dominated by the time to differentiate conex and reflex ertices, giing s the following reslt: Theorem 1 In O(n 2 m) time, we can reconstrct an IUP from its isibility graph. 4 Uniform-Length Histogram Polygons In this section we show how to reconstrct a more general class of polygons: those that consist of a chain of alternating p- and down-staircases with niform step length, which are monotone with respect to a single (longer) base edge. Sch polygons are niform-length histogram polygons [11], bt we simply call them histograms for breity (see Figre 7(a) for an example). We refer to the two conex ertices comprising the base edge as base ertices. Frthermore, we refer to top horizontal bondary edges incident to two conex ertices as tab edges or jst tabs and their incident ertices as tab ertices. We briefly note that, for histograms with two staircases, reconstrction can be done in linear time by a conting argment, similar to the staircase polygons [1]. (We gie sch an algorithm in Section ) Howeer, this constrction relies on the symmetry of the two staircases and that nearly all ertices hae niqe degrees. This symmetry breaks down when moing to general histograms, where it is not clear that any conting strategy will work.

16 446 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons 4.1 Oeriew of the Algorithm Eery histogram can be decomposed into axis-aligned rectangles whose contact graph is an ordered tree [11], as illstrated in Figre 7(b). In Section 4.2, we show that we can constrct the (nordered) contact tree T from the isibility graph G P in O(n 2 m) time by repeatedly peeling tabs from the histogram. We then show that each left-to-right ordering of T s k leaes (as well as a leftto-right orientation of the rectangles in the leaes) indces a histogram P. For each candidate polygon P (of k!2 k candidates), we then compte its isibility graph G P in O(n log n + m) time [17] and check if G P is isomorphic to G P. Instead of reqiring an expensie graph isomorphism check [6], we show how to se the ordering of T to qickly test if G P and G P are isomorphic. In Section 4.4 we show how to redce the nmber of candidate histograms from k!2 k to (k 2)!2 k 2, leading to the main reslt of or paper: Theorem 2 Gien a isibility graph G P of a histogram P with k 2 tabs, we can reconstrct P in O(n 2 m + (k 2)!2 k 2 (n log n + m)) time. Finally, we gie a faster reconstrction algorithm when the histogram has a binary contact tree, soling these instances in O(n 2 m) time (Section 4.4). 4.2 Rectanglar Decomposition and Contact Tree Constrction We constrct the contact tree T from G P by compting a set T of the k tab edges of G P (Lemma 11). Each tab (, ) is 1-simplicial and in a maximal 4-cliqe, since {, } (N() N()) is a 4-cliqe representing a nit sqare at the top of the histogram. Gien the set T of tab edges, or reconstrction algorithm picks an edge t from T and remoes the maximal 4-cliqe containing t. This is eqialent to remoing an axis-aligned rectangle in P, and, eqialently, remoing a leaf node from T. Moreoer, it associates that node of T with for ertices of P : two top ertices that are conex and two bottom ertices that are either both reflex or are both conex base ertices. This process might reslt in a new tab edge, which we identify and add to T Finding initial tabs We start by finding the k tabs. Recall that eery tab edge is 1-simplicial and in a maximal 4-cliqe. The conerse is not necessarily tre. Therefore, we begin by finding all 1-simplicial edges that are in maximal 4-cliqes as a set of candidate edges, and later exclde non-tabs from the candidates. Gien a isibility graph G P = (V P, E P ) of a histogram P and a maximal cliqe C V P, we call a ertex w C an isolated ertex with respect to P if there exists a tab edge (, ) E P, sch that (N() N()) C = {w}, i.e., of all ertices of C, only w is isible to some tab of P. Lemma 10 In a histogram, eery 1-simplicial edge in a maximal 4-cliqe contains either a tab ertex or an isolated ertex.

17 JGAA, 22(3) (2018) 447 t t t t (a) (b) (c) (d) Figre 8: Edges between and on a tab t s staircases, bt disjoint from t, are in a cliqe of size at least fie. Proof: Let l denote the leel of a ertex in P, which is its y-coordinate assming all niform edges hae nit length, where the base ertices are at leel 0. Consider an arbitrary 1-simplicial edge (, ) that is part of a maximal 4-cliqe C. We assme that neither nor is a tab ertex, otherwise we are done. Note that if l l 1 then and are in an axis-aligned rectangle in P defined by at least six ertices of G P (see Figre 8(a)), i.e., contradicting the assmption that (, ) is part of a maximal 4-cliqe. Withot loss of generality, sppose that lies aboe (i.e., l l 2) to the left of (the case of lying to the right of, or below, can be proen symmetrically). Since a top ertex does not see any ertices aboe it, mst be a bottom ertex. Ths, sees a ertex of some tab t. We will show that either t cannot see any other ertex of C, or C mst contain more than 4 ertices, contradicting the assmption that it is a 4-cliqe. Let R be a set of reflex ertices on s staircase on leels l + 1 p to and inclding l. Obsere, that cannot be part of the maximal 4-cliqe that contains both ertices of t (since it jst sees one of t s ertices), hence, we hae that 1 R l l ( R < l l when and are on the staircases of different tabs and the ertices of t are below ). Case 1: and belong to the staircases of a single tab t. (See Figre 8(b)- (d).) Consider the following cases: 1. is a top ertex: (See Figre 8(b).) Then and mst lie on different staircases and (, ) belongs to a cliqe consisting of at least six ertices: ertex, (l l ) 2 reflex ertices on s staircase,, and s two (reflex) bondary neighbors. 2. is a bottom ertex: (See Figres 8(c)-(d).) Then (, ) belongs to a cliqe consisting of at least fie ertices: (l l + 1) 3 bottom ertices on s staircase from leels l throgh l, and at least two ertices on the opposite staircase, e.g., the ertices of the ertical bondary edge spanning leels l and l + 1. Case 2: Vertices and belong to the staircases of different tabs t and t (see Figre 9). We call (, ) a crossing edge. Consider the following cases: 1. is reflex: Let (, ) and (, ) be the ertical and horizontal bondary edges, respectiely. Since and do not see each other, if sees both

18 448 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons w t w t w t w t t w t (a) (b) (c) Figre 9: Illstration for the proof of Lemma 10. (a)-(b) If is reflex, then the edge (, ) is in a cliqe of size greater than for. (c) If is conex and w blocks from seeing, then is an isolated ertex. and, then (, ) belongs to two cliqes and, therefore, cannot be 1- simplicial. Since is to the left of, sees. Therefore, there mst be some ertex w that is isible to both and and that blocks the iew from to. (a) is conex: (See Figre 9(a).) Then consider the ertical bondary edge (, w ). Clearly, sees w and (, ) belongs to a cliqe consisting of at least fie ertices: {, w,,, w }. (b) is reflex: (See Figre 9(b).) Then there mst be at least one ertex w C, bonding the conex region of C on the left (e.g., by line segments w and w ). Again, sees w and (, ) belongs to a cliqe consisting of at least fie ertices: {, w,,, w }. 2. is conex: (See Figre 9(c).) Let (, ) and (, ) be the ertical and horizontal bondary edges incident to, respectiely. Since sees and is below, mst also see. Moreoer, there mst be some ertex (isible from both and ), which bonds the conex region of C from the right. Let w be the closest sch ertex to the edge (, ). There are two cases to consider: (a) w is on s staircase: Then sees and (, ) belongs to a cliqe consisting of at least fie ertices: {,, w,, }. (b) w is not on s staircase: If C is a 4-cliqe, is not isible from, i.e., w is below, and C consists of {,, w, }. Since w bonds the conex region from the right, it cannot be isible from t. Moreoer, it also blocks and from t s iew. Ths, t sees only among the ertices of C, i.e., is an isolated ertex. Lemma 11 In a isibility graph of a histogram, tabs can be compted in time O(n 2 m). Proof: We begin by compting all 1-simplicial edges in maximal 4-cliqes, which takes time O(nm) by Obseration 1. Call this set of edges E sim, and the

19 JGAA, 22(3) (2018) 449 Figre 10: Illstrating all maximal 4-cliqes that contain 1-simplicial edges. These inclde tab cliqes (sqares) and non-tab cliqes (triangles). Note that tab cliqes can share ertices with zero, one, or two non-tab cliqes. set of their maximal cliqes C sim (see Figre 10). Then E sim contains the tabs, some edges that share a ertex with the tabs, and edges between staircases of different tabs (crossing edges) (which contain isolated ertices by Lemma 10). For all (non-incident) pairs of 1-simplicial edges e 1 and e 2 in maximal 4-cliqes C 1 and C 2, respectiely, we check if exactly one ertex of C 2 can be seen by an endertex of e 1. That is, we compte the set V 12 = { C 2 (, ) E and e 1 } and erify that V 12 = 1. If so, then C 2 is a non-tab cliqe and can be eliminated: If e 1 is a tab, then C 2 is a non-tab cliqe containing an isolated ertex (and a crossing edge); otherwise, e 1 is not a tab and its endertices see either zero or three ertices of any tab cliqe, and therefore C 2 cannot be a tab cliqe. Ths, if we compare all pairs of edges and cliqes, all 4-cliqes containing crossing edges will be eliminated, leaing only tab cliqes. We can do this check in O(nm) time by first storing, for each ertex, the edges {(, ) E sim } and cliqes {C C sim C}. Then for each edge (, ) in G P, we rn the isolated ertex check for each pair of edges and cliqes stored at the endertices and. Each check of all pairs takes O(n 2 ), and we do this for E sim = O(m) edges. If only disjoint cliqes remain after the preios step, then we hae compted exactly all k tabs. Otherwise, we need to eliminate non-tab edges that share a tab ertex. Note that non-tab edges form cliqes along the staircase incident to the tab. Since staircases in a histogram are disjoint, non-tab cliqes only intersect where they intersect a tab cliqe. Therefore, the remaining set of cliqes can be split into k mtally disjoint sets, where k is the nmber of tabs, and each set has at most three cliqes that intersect (see Figre 10), of which exactly one is a tab cliqe, and the other at most two cliqes contain a tab ertex, and its non-tab neighbors on the opposite staircase. Let S be one of these k sets. We can compte the tab in S as follows (three cases): 1. ( S = 1) Then the tab ertices see fewer ertices than non-tab (reflex) ertices. We can see this as follows: tab ertices see the ertices in their tab cliqe, pls the bottom ertices on the tab s staircases. The non-tab reflex ertices see the same ertices, pls at least two other ertices at their same leel, which tab ertices cannot see. See Figre 11(a). 2. ( S = 2) The two cliqes in S share three ertices, and two ertices and are in exactly one niqe cliqe each. One of these ertices (say ) is on the tab, and sees fewer ertices than the other non-tab ertex ()

20 450 N. Sitchinaa and D. Strash Reconstrcting Orthogonal Polygons w (a) (b) (c) Figre 11: Configrations of tab cliqes (sqares) oerlapping non-tab 4-cliqes (triangles) considered the case analysis in the proof of Lemma 11. Regions represent maximal 4-cliqes. Emboldened lines are 1-simplicial edges. Ble regions (and lines) represent tab cliqes (and tabs). (a) One (tab) cliqe. (b) Two oerlapping cliqes. (c) Three oerlapping cliqes. does. We can see this as follows: assme withot loss of generality that is on the left staircase. Then sees the three ertices of its tab cliqe, and all reflex ertices on the right staircase. Meanwhile is on the same side (left) as, and sees the same ertices as, pls a (conex) bondary neighbor w, which is on the leel between and, which cannot see. The remaining tab ertex is adjacent to along its 1-simplicial edge forming the cliqe in S. See Figre 11(b). 3. ( S = 3) The cliqes intersect symmetrically. The tab edge is formed between the two ertices that are in exactly two of these maximal cliqes. See Figre 11(c). After remoing cliqes with isolated ertices, there are at most 3k oerlapping cliqes in total, and they can be separated into k sets of cliqes incident to each tab in O(k) time by marking the ertices of each set, and collecting the intersecting sets. Then within each set, it takes O(1) time to find the tab cliqe (and the tab). Ths, the rnning time is dominated by the time to detect crossing edges in E sim : O(n 2 m). Note that top ertices cannot see the ertices aboe them. Therefore, only bottom ertices see tab ertices. Moreoer, eery bottom ertex sees at least one tab ertex. Ths, identifying all tabs immediately classifies ertices of G P into top ertices and bottom ertices Peeling tabs Let P be a polygon reslting from peeling tab cliqes (rectangles) from a histogram P. We call P a trncated histogram. See Figre 12(a) for an example. After peeling a tab cliqe, the reslting polygon does not hae niform step length and the isibility graph may no longer hae the properties on which Lemma 11 relied to detect initial tabs. Instead, we se the following lemma to detect newly created tabs dring tab peeling.

21 JGAA, 22(3) (2018) 451 t t a C t t b C t (a) (b) Figre 12: (a) A trncated histogram, created by iteratiely remoing six tabs (dashed) from a histogram. (b) When remoing C t : t a, t b form a tab if and only if t a, t b T \ C t, they see and, and C t = 4. Lemma 12 When remoing a tab cliqe from the isibility graph of a trncated histogram, any newly introdced tab can be compted in time O(n). Proof: Denote the remoed (tab) cliqe by C t and let t be its tab. Let, t be the non-tab ertices of C t. Since sees, (, ) is an edge in G P. Since top ertices can only see ertices at and below their own leel, besides the ertices of t, there are exactly two other top ertices in (remaining) G P that see and, namely, the top ertices t a and t b of P on the same leel as and (see Figre 12(b)). Since t a, t b are adjacent in G P, let t = (t a, t b ). When remoing C t from G P, we can compte t a and t b in time O(n) by selecting the only two top ertices adjacent to both and. Since t a and t b are the top ertices of a same rectangle R t, edge t is 1-simplicial and is in exactly one maximal cliqe C t = N(t a) N(t b ), which corresponds to the conex region R t. Finally, after C t is remoed, t is a newly created tab if and only if C t = 4, which can again be tested in time O(n) by compting the common neighborhood N(t a) N(t b ). With each tab cliqe (rectangle) remoal, we iteratiely bild the parentchild relationship between the rectangles in the contact tree T as follows. Using an array A, we maintain references to cliqes being remoed whose parents in T hae not been identified yet. When a tab cliqe C t is remoed from G P, the reference to C t is inserted into A[], where is one of the rectangle s bottom ertices. If the remoal of C t creates a new tab t = (t a, t b ), we identify C t in O(n) time sing Lemma 12. Recall that t sees all bottom ertices on the same leel. Ths, for eery bottom ertex N(t a) (in the original graph G P ), if A[] is non-empty, we set C t as the parent of the cliqe stored in A[] and clear A[]. This takes at most O(n) time for each peeling of a cliqe. We get the following lemma, where the time is dominated by the comptation of the initial tabs: Lemma 13 In O(n 2 m) time we can constrct the contact tree T of P, associate with each T the for ertices that define the rectanglar region of, and classify ertices of G P as top ertices and bottom ertices.