An Optimal Algorithm for the Separating Common Tangents of two Polygons

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1 An Optimal Algorithm for the Separating Common Tangents of two Polygons arxiv:5.436v [cs.cg] 2 Nov 25 Miel Abrahamsen Department of Computer Science, University of Copenhagen Universitetsparen 5 DK-2 Copanhagen Ø Denmar miab@di.u.d September, 28 Abstract We describe an algorithm for computing the separating common tangents of two simple polygons using linear time and only constant worspace. A tangent of a polygon is a line touching the polygon such that all of the polygon lies to the same side of the line. A separating common tangent of two polygons is a tangent of both polygons where the polygons are lying on different sides of the tangent. Each polygon is given as a read-only array of its corners. If a separating common tangent does not exist, the algorithm reports that. Otherwise, two corners defining a separating common tangent are returned. The algorithm is simple and implies an optimal algorithm for deciding if the convex hulls of two polygons are disjoint or not. This was not nown to be possible in linear time and constant worspace prior to this paper. An outer common tangent is a tangent of both polygons where the polygons are on the same side of the tangent. In the case where the convex hulls of the polygons are disjoint, we give an algorithm for computing the outer common tangents in linear time using constant worspace. Introduction The problem of computing common tangents of two given polygons has received some attention in the case where the polygons are convex. For instance, it is A preliminary version of this paper appeared at SoCG 25 []. In the case where the convex hulls of the polygons are not disjoint, it is not clear that the algorithm for separating common tangents terminates within the given bound on the running time. Here, we give a correct algorithm and simplify the proof of correctness slightly. Research partly supported by Miel Thorup s Advanced Grant from the Danish Council for Independent Research under the Sapere Aude research career programme.

2 necessary to compute outer common tangents of disjoint convex polygons in the classic divide-and-conquer algorithm for the convex hull of a set of n points in the planebypreparataandhong[3]. Theygiveanavelineartimealgorithmforouter common tangents since that suffices for an O(n log n) time convex hull algorithm. The problem is also considered in various dynamic convex hull algorithms[6, 9, 2]. Overmars and van Leeuwen [2] give an O(log n) time algorithm for computing an outer common tangent of two disjoint convex polygons when a separating line is nown, where each polygon has at most n corners. Kirpatric and Snoeyin [] give an O(logn) time algorithm for the same problem, but without using a separatingline. Guibasetal.[8]giveanΩ(log 2 n)lowerboundonthetimerequired to compute an outer common tangent of two intersecting convex polygons, even if it is nown that they intersect in at most two points. They also describe an algorithm achieving that bound. Touissaint [4] considers the problem of computing separating common tangents of convex polygons and notes that the problem occurs in problems related to visibility, collision avoidance, range fitting, etc. He gives a linear time algorithm. Guibas et al. [8] give an O(logn) time algorithm for the same problem. All the here mentioned wors mae use of the convexity of the polygons. If the polygons are not convex, one can use a linear time algorithm to compute the convex hulls before computing the tangents [7, ]. However, if the polygons are given in read-only memory, it requires Ω(n) extra bits to store the convex hulls. In this paper, we also obtain linear time while using only constant worspace, i.e. O(logn) bits. For the outer common tangents, we require the convex hulls of the polygons to be disjoint. There has been some recent interest in constant worspace algorithms for geometric problems, see for instance [2, 3, 4, 5]. The problem of computing separating common tangents is of special interest because these only exist when the convex hulls of the polygons are disjoint, and our algorithm detects if they are not. Thus, we also provide an optimal algorithm for deciding if the convex hulls of two polygons are disjoint or not. This was to the best of our nowledge not nown to be possible in linear time and constant worspace prior to our wor.. Notation and some basic definitions Given two points a and b in the plane, the closed line segment with endpoints a and b is written ab. When a b, the line containing a and b which is infinite in both directions is written L(a, b). Define the dot product of two points x = (x,x ) and y = (y,y ) as x y = x y +x y, and let x = ( x,x ) be the counterclocwise rotation of x by the angle π/2. Now, for three points a, b, and c, we define T (a,b,c) = sgn((b a) (c b)), where sgn is the sign function. T(a,b,c) is if c is to the left of the directed line from a to b, if a, b, and c are collinear, and if c is to the right of the directed line from a to b. We see that T(a,b,c) = T(b,c,a) = T (c,a,b) = T (c,b,a) = T (b,a,c) = T (a,c,b). We also note that if a and b are on the line L(a,b) and appear in the same order as a and b, i.e., (b a) (b a ) >, then T (a,b,c) = T(a,b,c) for every point c. 2

3 P P Figure : Two polygons P and P and their four common tangents as thic lines. The edges of the convex hulls which are not edges of P or P are dashed. The left half-plane LHP(a, b) is the closed half plane with boundary L(a, b) lying to the left of directed line from a to b, i.e., all the points c such that T (a,b,c). The right half-plane RHP(a,b) is just LHP(b,a). Assume for the rest of this paper that P and P are two simple polygons in the plane with n and n corners, respectively, where P is defined by its corners p [],p [],...,p [n ] in clocwise or counterclocwise order, =,. Indices of the corners are considered modulo n, so that p [i] andp [j] arethe same corner when i j (mod n ). We assume that the corners are in general position in the sense that P and P have no common corners and the union of corners =, {p [],...,p [n ]} contains no three collinear corners. A tangent of P is a line l such that l and P are not disjoint and such that P is contained in one of the closed half-planes defined by l. The line l is a common tangent of P and P if it is a tangent of both P and P. A common tangent is an outer common tangent if P and P are on the same side of the tangent, and otherwise the tangent is separating. See Figure. For a simple polygon P, we let H(P) be the convex hull of P. The following lemma is a well-nown fact about H(P). Lemma. For a simple polygon P, H(P) is a convex polygon and the corners of H(P) appear in the same cyclic order as they do on P. The following lemma states follore properties of tangents of polygons. Lemma 2. A line is a tangent of a polygon P if and only if it is a tangent of H(P). Under our general position assumptions, the following holds: If one of H(P ) and H(P ) is completely contained in the other, there are no outer common tangents of P and P. Otherwise, there are two or more. There are exactly two if P and P are disjoint. If H(P ) and H(P ) are not disjoint, there are no separating common tangents of P and P. Otherwise, there are exactly two. 3

4 P s (2) s () s (7) P s () Figure 2: Algorithm running on two polygons P and P. The corners p [s (i) ] are mared and labeled as s (i) for the initial values s () and after each iteration i where an update of s happens. The segments p [s (i) ]p [s (i) ] on the temporary line are dashed. 2 Computing separating common tangents In this section, we assume that the corners of P and P are both given in counterclocwise order. We prove that Algorithm returns a pair of indices (s,s ) such that the line L(p [s ],p [s ]) is a separating common tangent with P contained in RHP(p [s ],p [s ]) for =,. If the tangent does not exist, the algorithm returns NULL. The other separating common tangent can be found by a similar algorithm if the corners of the polygons are given in clocwise order and = is changed to = in line 3. Algorithm : SeparatingCommonTangent(P,P ) s ; t ; s ; t ; u 2 while t < 3n or t < 3n 3 if T (p u [s u ],p u [s u ],p u [t u ]) = 4 if t u 2n u 5 return NULL 6 s u t u 7 t u s u + 8 t u t u + 9 u u return (s,s ) The algorithm traverses the polygons in parallel one corner at a time using the indices t and t. We say that the indices (s,s ) define a temporary line, which is the line L(p [s ],p [s ]). We update the indices s and s until the temporary line is the separating common tangent. At the beginning of an iteration of the loop at line 2, we traverse one corner p u [t u ] of P u, u =,. If the corner happens to be on the wrong side of the intermediate line, we mae the temporary line pass through that corner by updating s u to t u and we reset t u to s u +. The reason for resetting t u is that a corner of P u which was on the correct side of 4

5 the old temporary line can be on the wrong side of the new line and thus needs be traversed again. We show that if the temporary line is not a separating common tangent after each polygon has been traversed twice, then the convex hulls of the polygons are not disjoint. Therefore, if a corner is found to be on the wrong side of the temporary line when a polygon is traversed for the third time, no separating common tangent can exist and NULL is returned. Let s (i) be the value of s after i =,,... iterations, =,. We always have s () = due to the initialization of s. See Figure 2. Assume that s is updated in line 6 in iteration i. The point p [s (i) ] is in ]). the half-plane LHP(p [s (i ) ],p [s (i ) ]), but not on the line L(p [s (i ) ],p [s (i ) Therefore, we have the following observation. Observation 3. When s is updated, the temporary line is rotated counterclocwise around s by an angle less than π. Assume in the following that the convex hulls of P and P are disjoint so that separating common tangents exist. Let (r,r ) be the indices that define the separating common tangent such that P is contained in RHP(p [r ],p [r ]), i.e., (r,r ) is the result we are going to prove that the algorithm returns. SinceH(P )isconvex, thetemporarylinealwaysdividesh(p )intotwoconvex parts. If we follow the temporary line from p [s ] in the direction towards p [s ], we enter H(P ) at some point x and thereafter leave H(P ) again at some point y. We clearly have x = y if and only if the temporary line is a tangent to H(P ), since if x = y and the line was no tangent, H(P ) would only be a line segment. The part of the boundary of H(P ) counterclocwise from x to y is in RHP(p [s ],p [s ]) whereas the part from y to x is on LHP(p [s ],p [s ]). We therefore have the following observation. Observation 4. Let d be the index of the corner of H(P ) strictly after y in counterclocwiseorder. Thereexists acorner p [t] ofp suchthat T(p [s ],p [s ],p [t]) = if and only if T (p [s ],p [s ],p [d]) =. Let c be the index of the first corner of H(P ) when following H(P ) in counterclocwise orderfromy,c =,...,n. Ify isitselfacornerofh(p ), we have p [c ] = y. ByObservation4we see thatt(p [s ],p [s ],p [c ]) with equality if and only if p [c ] = p [s ] = y. Let c () be c when only line has been executed. Consider now the value of c after i =,2,... iterations. Let c (i) = c andaddn toc (i) untilc(i) c (i ). This gives a non-decreasing sequence of indices c (),c(),... of the first corner of H(P ) in LHP(p [s ],p [s ]). Actually, we proveinthefollowingthatweneedtoaddn toc (i) atmostoncebeforec(i) c (i ). If r < c () we add n to r. Thus we have = s () c () r < 2n. The following lemma intuitively says that the algorithm does not jump over the correct solution and it expresses the main idea in our proof of correctness. Lemma 5. After each iteration i =,,... and for each =, we have s (i) c (i) r < 2n. Furthermore, the test in line 4 is never positive. 5

6 s (i ) P w y s (i ) x r c (i ) v s (i) r P c (i) Figure 3: An update of s happens in iterationifrom s (i ) to s (i) and p [c ] moves forward on H(P ) from p [c (i ) ] to p [c (i) ]. The relevant corners are mared and labeled with their indices. The polygon C from the proof of Lemma 5 is drawn with thic lines. Proof. We prove the lemma for =. From the definition of r, we get that = s () c () r < 2n. Since the sequence s (),s(),... is non-decreasing, the inequality s (i) is true for every i. Now, assume inductively that s (i ) c (i ) r and consider what happens during iteration i. If neither s nor s is updated, the statement is trivially true from the induction hypothesis, so assume that an update happens. By the old temporary line we mean the temporary line defined by (s (i ),s (i ) ) and the new temporary line is the one defined by (s (i),s (i) ). The old temporary line enters H(P ) at some point x and exits at some point y when followed from p [s (i ) ]. Liewise, let v be the point where the new temporary line exits H(P ) when followed from p [s (i) ]. The point x exists since the convex hulls are disjoint. Assume first that the variable u in the algorithm is, i.e., a corner of the polygon P is traversed. In this case s (i ) = s (i). We now prove s (i) c (i). Assume that p [s (i ) ] p [c (i ) ]. The situation is depicted in Figure 3. In this case T (p [s (i ) ],p [s (i ) ],p [c (i ) ]) =. Hence, the update happens when p [c (i ) ] is traversed or earlier, so s (i) c (i ) c (i). Assume now that p [s (i ) ] = p [c (i ) ]. We cannot have c (i) = c (i ) since T (p [s (i) ],p [s (i) ],p [c (i ) ]) = T (p [s (i ) ],p [s (i ) ],p [s (i) ]) =, therefore c (i) > c (i ). Consider the corner p [c ] on H(P ) following p [c (i ) ] in counterclocwise order, c > c (i ). Due to the minimality of c, we have c c (i). By Observation 4, T (p [s (i ) ],p [s (i ) ],p [c ]) =. Therefore, s must be updated when p [c ] is traversed or earlier, so s (i) c c (i). For the inequality c (i) r, consider the new temporary line in the direction from p [s (i ) ] to p [s (i) ]. We prove that v is in the part of H(P ) from y counterclocwise to r. The point p [s (i) ] is in the polygon Q defined by the segment xy together with the part of H(P ) from y counterclocwise to x. Therefore, the new temporary line enters and exits Q. It cannot exit through the segment xy, 6

7 since the old and new temporary lines intersect at p [s (i ) ], which is in H(P ). Therefore, v must be on the part of H(P ) from y to x. If r is on the part of H(P ) from x counterclocwise to y, then v is on the part from y to r as we wanted. Otherwise, assume for contradiction that the points appear in the order y, p [r ], v, x counterclocwise along H(P ), where p [r ] v x. The endpoints of the segment p [s (i ) ]x are on different sides of the tangent defined by (r,r ), so the segment intersects the tangent at a point w. The part of H(P ) from p [r ] to x and the segments xw and wp [r ] form a simple polygon C, see Figure 3 for an example. The new temporary line enters C at the point v, so it must leave C after v. The line cannot cross H(P ) after v since H(P ) is convex. It also cannot cross the segment xw at a point after v since the old and the new temporary line cross before v, namely at p [s (i ) ]. The tangent defined by (r,r ) and the new temporary line intersect before v since the endpoints of the segment p [s (i ) ]v are on different sides of the tangent. Therefore, the line cannot cross the segment wp [r ] at a point after v. Hence, the line cannot exit C. That is a contradiction. Therefore, v is on the part of H(P ) from y to p [r ] and hence the first corner p [c (i) ] of H(P ) after v must be before or coincident with p [r ], so that c (i) r. Assume now that u = in the beginning of iteration i, i.e., a corner of the other polygon P is traversed. In that case, we have s (i) = s (i ) c (i ) c (i), and we need only prove c (i) r. Observation 3 gives that v is in the part of H(P ) from y to x, since the new temporary line is obtained by rotating the old temporary line counterclocwise around p [s (i ) ] by an angle less than π. That v appears before p [r ] on H(P ) counterclocwise from y follows from exactly the same arguments as in the case u =. We have nowhere used the test at line 4 to conclude that s < 2n. Hence, the test is never positive. This completes the proof. We are now ready to prove that Algorithm has the desired properties. Theorem 6. If the polygons P and P have separating common tangents, Algorithm returns a pair of indices (s,s ) defining a separating common tangent such that P is contained in RHP(p [s ],p [s ]) for =,. If no separating common tangents exist, the algorithm returns NULL. The algorithm runs in linear time and uses constant worspace. Proof. Assume first that thealgorithmreturns (s,s ). We nowthat s < 2n for each =,,sinceweneverupdates tovaluesaslargeas2n. Therefore, wehave that p [t] RHP(p [s ],p [s ]) for each =, and each t {2n,...,3n }. Hence the pair (s,s ) indeed defines the separating common tangent. Assume now that there exists a separating common tangent. By Lemma 5, a pair (s,s ) is returned. As we already saw, this means that (s,s ) defines the separating common tangent. If an update happens in iteration i, the sum s +s is increased by at least i j, 2 where j was the previous iteration where an update happened. Inductively, we see that when the final update of s and s happens, there has been at most 2(s + s ) iterations. After the final update, at most 3n s +3n s iterations follow. In total, the algorithm performs 3n +s +3n +s 5(n +n ) iterations. 7

8 s (9) s (6) P s () s (5) s () s (9) s (3) s (6) s () s (4) s (2) s (2) P s () Figure 4: Algorithm 2 running on two polygons P and P. The corners p [s (i) ] are mared and labeled as s (i) for the initial values s () and after each iteration i where an update of s happens. The segments p [s (i) ]p [s (i) ] on the temporary line are dashed. 3 Computing outer common tangents In this section, we assume that two polygons P and P are given such that their convex hulls are disjoint. We assume that the corners p [],...,p [n ] of P are given in counterclocwise order and the corners p [],...,p [n ] of P are given in clocwise order. We say that the orientation of P and P is counterclocwise and clocwise, respectively. We prove that Algorithm 2 returns two indices (s,s ) that define an outer common tangent such that P and P are both contained in RHP(p [s ],p [s ]). Algorithm 2: OuterCommonTangent(P,P ) s ; t ; s ; t ; u 2 while t < 2n or t < 2n 3 if T (p [s ],p [s ],p u [t u ]) = 4 s u t u 5 t u s u + 6 t u t u + 7 u u 8 return (s,s ) As in the case of separating common tangents, we define s (i) as the value of s after i =,,... iterations of the loop at line 2 of Algorithm 2. See Figure 4. For this algorithm, we get a slightly different analogue to Observation 3: Observation 7. When s is updated, the temporary line is rotated around s in the orientation of P by an angle less than π. Let y be the point where the temporary line enters H(P ) when followed from p [s ] and x the point where it exits H(P ). We have the following analogue of Observation 4. 8

9 l l x P r r y s (i ) r s (i ) P Figure 5: The area A from the proof of Lemma 9 in grey. The relevant corners are mared and labeled with their indices. Observation 8. Let d be the index of the corner of H(P ) strictly after y following the orientationof P. Thereexists acorner p [t] ofp suchthat T(p [s ],p [s ],p [t]) = if and only if T (p [s ],p [s ],p [d]) =. Let c be the index of the first corner of H(P ) after y following the orientation of P, where p [c ] = y if y is itself a corner of H(P ). By Observation 8, we have T (p [s ],p [s ],p [c ]) with equality if and only if p [c ] = p [s ] = y. Define a non-decreasing sequence c (),c(),... of the value of c after i =,,... iterations as we did for separating tangents. Also, let the indices (r,r ) define the outer common tangent that we want the algorithm to return such that c () r < 2n. We can now state the analogue to Lemma 5 for outer common tangents. Lemma 9. After each iteration i =,,... and for each =, we have s (i) c (i) r < 2n. Proof. Assume = and the induction hypothesis s (i ) c (i ) r. The inequality s (i) c (i) can be proven exactly as in the proof of Lemma 5. Therefore, consider the inequality c (i) r and assume that an update happens in iteration i. Let the old temporary line and the new temporary line be the lines defined by the indices (s (i ),s (i ) ) and (s (i),s (i) ), respectively. Let y and x be the points where the old temporary line enters and exits H(P ) followed from p [s (i ) ], respectively, and let v be the point where the new temporary line enters H(P ). The points y and v exist since the convex hulls of P and P are disjoint. Assume first that the variable u in the algorithm equals when the update happens. We prove that v is in the part of H(P ) from y to p [r ] following the orientation of P, which is counterclocwise. The point p [s (i) ] is in the simple polygon Q bounded the part of H(P ) from y counterclocwise to x and the segment xy. Therefore, the new temporary line must enter Q to get to p [s (i) ]. It cannot enter through xy, since the old and new temporary line cross at p [s (i ) ] which is not in H(P ) by assumption. Therefore, it must enter through the part 9

10 of H(P ) from y to x, so v is in this part. If r is not in the part of H(P ) from y to x, it is clearly true that v is in the part from y to p [r ]. Otherwise, assume for contradiction that the points appear on H(P ) in the order y,p [r ],v,x and p [r ] v x. Let l be the half-line starting at p [r ] following the tangent away from p [r ], and let l be the half-line starting at x following the old temporary line away from p [s (i ) ]. The part of H(P ) from p [r ] to x and the half-lines l and l define a possibly unbounded area A outside H(P ), see Figure 5. We follow the new temporary line from p [s (i ) ] towards v. The point p [s (i ) ] is not in A and the new temporary line exits A at v since it enters H(P ) at v, so it must enter A somewhere at a point on the segment p [s (i ) ]v. It cannot enter through H(P ) since H(P ) is convex. It cannot enter through l since v and p [s (i ) ] are on the same side of the outer common tangent. It cannot enter through l since the old and new temporary line intersect in p [s (i ) ], which is not in A. That is a contradiction, so v is on the part of H(P ) from y to p [r ]. Hence, the first corner after y is coincident with or before p [r ], i.e., c (i) r. Assume now that u = in the beginning of iteration i so that a corner of the polygon P is traversed. Observation 7 gives that v is on the part of H(P ) from y counterclocwise to x. It follows that v appears before p [r ] on H(P ) counterclocwise fromy fromexactly thesame arguments asinthecase u =. Lemma. If p [s ] p [r ] or p [s ] p [r ], then T(p [s ],p [s ],p [t]) = for some =, and some index t {s +,...,r }. Proof. Assume that T(p [s ],p [s ],p [r ]) for =,, since otherwise, we are done. Liewise, assume that all of the part P [s,r ] of P from p [s ] to p [r ] is in RHP(p [s ],p [s ]). The part P [s,r ] separates p [s ] from p [r ] in the set W = RHP(p [s ],p [s ]) RHP(p [r ],p [r ]). Since the part P [s,r ] of P from p [s ] to p [r ] cannot cross P [s,r ] or L(p [r ],p [r ]), it must exit and enter W through points on L(p [s ],p [s ]) when followed from p [s ], and hence the claim is true. We can now prove the stated properties of Algorithm 2. Theorem. If the polygons P and P have disjoint convex hulls, Algorithm 2 returns a pair of indices (s,s ) defining an outer common tangent such that P and P are contained in RHP(s,s ). The algorithm runs in linear time and uses constant worspace. Proof. Assume that the pair (s,s ) does not define the outer common tangent. By Lemma, an update of s or s happens when p [r ] or p [r ] is traversed or before. By Lemma 9, the algorithm does not terminate before p [r ] and p [r ] has been traversed. Hence, when the algorithm terminates, (s,s ) defines the outer common tangent. Lie in the proof of Theorem 6, we see inductively that when the final update of s and s happens, there has been at most 2(s +s ) iterations. After that, at most 2n s +2n s iterations follow. Hence, the algorithm terminates after at most 4n +4n iterations.

11 s () s () P P Figure 6: Two polygons P and P where Algorithm 2 does not wor for the initial values of s and s as shown. The correct tangent is drawn as a dashed line. 4 Concluding Remars We have described an algorithm for computing the separating common tangents of two simple polygons in linear time using constant worspace. We have also described an algorithm for computing outer common tangents using linear time and constant worspace when the convex hulls of the polygons are disjoint. Figure 6 shows an example where Algorithm 2 does not wor when applied to two disjoint polygons with overlapping convex hulls. In fact, if there was no bound on the values t and t in the loop at line 2, the algorithm would update s and s infinitely often and never find the correct tangent. An obvious improvement is to find an equally fast and space efficient algorithm which does not require the convex hulls to be disjoint. An algorithm for computing an outer common tangent of two polygons, when such one exists, also decides if one convex hull is completely contained in the other. Together with the algorithm for separating common tangents presented in Section 2, we would have an optimal algorithm for deciding the complete relationship between the convex hulls: if one is contained in the other, and if not, whether they are disjoint or not. However, eeping in mind that it is harder to compute an outer common tangent of intersecting convex polygons than of disjoint ones [8], it would not be surprising if it was also harder to compute an outer common tangent of general simple polygons than simple polygons with disjoint convex hulls when only constant worspace is available. Acnowledgments We would lie to than Mathias Tejs Bæ Knudsen for pointing out the error in the algorithm for separating common tangents in the preliminary version of the paper []. References [] M. Abrahamsen. An optimal algorithm for the separating common tangents of two polygons. In 3st International Symposium on Computational Geometry (SoCG 25), pages

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