Computing the Convex Hull of. W. Hochstattler, S. Kromberg, C. Moll

Transcription

1 Reort No A new Linear Time Algorithm for Comuting the Convex Hull of a Simle Polygon in the Plane by W. Hochstattler, S. Kromberg, C. Moll 994 W. Hochstattler, S. Kromberg, C. Moll Universitat zu Koln Weyertal D{5093 Koln Germany Telefon (0) 470{60 (603) e{mail WH@MI.UNI-KOELN.DE SK@MI.UNI-KOELN.DE CM@MI.UNI-KOELN.DE

2 99 Mathematics Subject Classication: 68U05, 5N0 Keywords: convex hull, linear time algorithm, simle olygon, lanar geometry

3 A new Linear Time Algorithm for Comuting the Convex Hull of a Simle Polygon in the Plane W. Hochstattler, S. Kromberg, C. Moll Abstract The roblem of determining the convex hull of a simle olygon has received a lot of attention in the early eighties. The rst linear time algorithm for this task roosed by Sklansky in [S7] was based on the simle idea of removing all left turns while moving around the olygon in clockwise orientation. This algorithm was shown to fail in some cases. Since then several correct, yet more comlicated linear algorithms have been ublished and classes of olygons have been determined for which Sklansky's original algorithm can be used. In our note we show how to mend Sklansky's Algorithm in a simle way and rove the correctness of the resulting algorithm. As an alication we show how to comute a rectangle of smallest area containing a given simle olygon in linear time. Introduction The roblem of determining the convex hull of a simle olygon has received a lot of attention in the early eighties. The rst linear time algorithm for this task was roosed by Sklansky in [S7]. Sklansky's Algorithm can be described as follows: Start at an extreme oint of the olygon. Delete all left turns while moving around the olygon in clockwise direction. After each ste backtrack until the ath from the starting oint to the oint currently considered given by the not (yet) deleted oints has right turns only. The asset of this algorithm is the simlicity of its idea and that it requires only a minimum amount of global information about the given simle olygon in advance { its orientation and an extreme oint. Its idea is based on the observation that at each left turn q of a simle clockwise oriented olygon there exists a line segment containing q whose extremities are oints belonging to the interior of the olygon so that q cannot be an extreme oint of the convex hull of the olygon (see Figure ). The roblem with Sklansky's Algorithm is that it may return a non{simle olygon and that it hence fails in some cases. An examle it fails on is given in Figure. Note that there also is a local reason for the failure of Sklansky's Algorithm on the examle given in Figure. Although the original olygon had a left turn at and hence cannot be an extreme oint of the convex hull of the olygon, the olygon stored at the time the

4 q Figure : A counterexamle to Sklansky's Algorithm algorithm rocesses has a right turn at and hence is not deleted. On this examle Sklansky's Algorithm thus in a sense violates its very idea. The algorithm resented in this note sulements Sklansky's algorithm in a simle way assuring that the ath u to the currently considered oint reresented by the oints which have been tentatively acceted as extreme oints is always simle and consists of right turns only. Like Sklansky's and dierent from other (correct) O(n)-algorithms for comuting the convex hull of a simle olygon our algorithm only requires that the simle olygon be given in clockwise orientation and that the starting oint is known to be an extreme oint of the convex hull. From then on the decision whether or not to delete a oint is based solely on local information. Our algorithm thus can not only be regarded as being simler than the algorithms in [MA79], [GS83], [BE84], it also rovides new insight into the structure of the roblem. Denitions and Notation In this note the term oint refers to a oint in the Euclidean lane reresented by its coordinates. For two oints = (x; y) and q = (x 0 ; y 0 ) in the lane [; q] denotes the line segment f + (? )q j [0; ]g and det(; q) denotes xy 0? x 0 y. If and q are dierent, L(; q) denotes the directed line through and q directed from to q. We say a oint r is to the right of (resectively on, left of) L(; q) if det(? q; r? q) is ositive (resectively zero, negative) and denote by H(; q) the halfsace of oints on or to the right of L(; q). The term olygon refers to a iecewise linear closed curve in the lane. A olygon is reresented by a sequence P = : : : n of oints in the order they are encountered as the curve is traversed, it being understood that for i n the curve traverses [ i ; i+ ]

5 i j tail (a) tail K P ( i, j ) (b) head head run Figure : Some K P ( i ; j ) from i to i+ (indices read modulo n). The oints i ( i n) are referred to as the turn oints of the olygon, i is a right turn (resectively left turn) or a convex turn oint (resectively reex turn oint) of P if i+ is to the right (resectively left) of L( i? ; i ). A olygon is convex if all its turn oints are convex. Given a olygon P = : : : n and i; j f; : : :; ng, we let P ath P ( i ; j ) denote the ath in P from i to j which follows the orientation of P. P ath P ( i ; j ) is convex if all turn oints dierent from i and j that are visited by the ath are convex turn oints of P. If P ath P ( i ; j ) is simle and all turn oints visited by the ath are contained in H( i ; j ), we let K P ( i ; j ) denote the union of the bounded regions enclosed by the olygon i i+ : : : j and its boundary. A few examles are shown in Figure. The convex hull of a set S of oints is denoted by conv(s). It is well known that conv(s) is a (convex) olyhedron, if S is nite. The vertices of the olyhedron conv(s) are also called extreme oints of conv(p ). The set of extreme oints is uniquely characterized as the minimal subset E of S with conv(e) = conv(s). A simle olygon is clockwise oriented if its interior lies to the right as the olygon is traversed. It follows from the Jordan Curve theorem that for a simle clockwise oriented olygon P = : : : n the boundary of its convex hull is reresented by the subsequence of extreme oints of : : : n. 3

6 inut: A sequence of oints : : : n stored in a doubly linked list reresenting a clockwise oriented simle olygon with the roerty that is an extreme oint of conv(p ) outut:a subsequence i (= ) : : : ik reresenting the simle convex olygon that is the boundary of conv(p ) () run := succ() () if run = sto (3) if succ( run ) is to the right of L(red( run ); run ) then (4) run := succ( run ) and go to () (5) else (6) head := succ( run ) tail := red( run ) (7) if head is on or to the left of L(red( tail ); tail ) then (8) tail := red( tail ) and go to (7) (9) L 0 := L( tail ; head ) (0) if succ( head ) is on or to the right of L 0 and to the left of L(red( head ); head ) then () head := succ( head ) () if head is on or to the right of L 0 then go to () (3) else go to (7) (4) succ( tail ) := head red( head ) := tail (5) run := head and go to () 3 The Algorithm Figure 3: The Algorithm We now give an informal descrition of our algorithm ointing out where it sulements Sklansky's algorithm. The algorithm requires that the inut olygon is given in clockwise orientation and that the starting oint is an extreme oint of the convex hull. As, however, it is easy to nd such a oint and determine the orientation of a simle olygon in linear time, this assumtion does not restrict its generality. The inut olygon is stored in a doubly linked list S. During oeration the algorithm deletes oints from S. When it stos the contents of S reresents the simle convex olygon that is the boundary of the convex hull of the inut olygon. At any time during oeration our algorithm kees a variable run set to some turn oint of the inut olygon P. At run we do not have that the currently stored olygon has a right turn at run while the original olygon had a left turn at run. We leave S unchanged and move to the next oint, if the currently stored olygon has a right turn at run. If not, we initialize head to the successor of the currently considered oint run and tail to its redecessor. Now we alternate two stes as follows (see Figure ): (a) If the ath given by the (current) redecessors of tail followed by head has a left turn at tail, we backtrack tail until the resulting ath has right turns only. (b) If the successor of head belongs to K P ( tail ; head ), i.e. lies inside the olygon formed by the original ath from tail to head and [ tail ; head ], we advance head to the rst 4

7 (ii) q (iii) s S r q Figure 4: The situations considered in Lemma ((ii) & (iii)) oint outside of this olygon. If none of the two conditions holds, we remove the oints between tail and head and set run to head. The algorithm stos once it reaches the starting oint. 4 Correctness of the Algorithm The roof of the correctness of our algorithm is based on the following two lemmas. In the rst lemma we collect the facts from lanar geometry we need. Most imortant for our uroses is the contents of Lemma (ii) a icture of which is given in Figure 4. The second lemma contains crucial observations about certain subsequences of the inut sequence of our algorithm occurring during oeration which lead to a roof of the correctness of our algorithm. In what follows we abbreviate K P ( i ; j ) by K( i ; j ). Lemma Let s and P = : : : s be a sequence of airwise distinct oints. (i) If is an extreme oint of conv(p ) and P ath P ( ; s ) is convex then P ath P ( ; s ) is simle i 8 3 i s 8 j i? : i is to the right of L( j ; j+ ). (ii) If P ath P ( ; s ) is simle, i is on or to the right of L( ; s ) for all i s and and q are oints such that is on or to the right of L( ; s ), = K( ; s ) and [; q] \ K( ; s ) 6= ; then also [; q] \ [ i ; i+ ] 6= ; for some i s?. (iii) Let S denote a directed line, H the halfsace of oints on or to the right of S and let ; q; r be three oints such that is on S, q and r are to the right of S and r is on or to the left of L(; q). Then H \ H(; q) H \ H(; r). 5

8 Lemma Let P = : : : n be the inut olygon of the algorithm. (a) If P 0 = i : : : ik of either ( i < i < : : : < i k n) denotes the subsequence of P consisting or the currently stored redecessors of run (= ik ) in line () the currently stored redecessors of tail (= ik? ) followed by head (= ik ) in line (9), then the following holds: (i) i =, k and i is to the right of L( n ; ) (ii) P ath P 0( i ; ik ) is simle and convex, (iii) 8 k? : P ath P ( i ; i+ ) T = H( i ; i+ ), (iv) 8 k 8? : i = K( i ; i+ ), (v) all oints j that were discarded (i.e. j i k ; j = P 0 ) are not extreme oints of conv(p ). If ik = run, we furthermore have that (vi) succ( run ) = K( ik? ; ik ): (b) The algorithm stos. If P 0 = i : : : ik the list when the algorithm stos, then ( i < i < : : : < i k n) is the contents of (i) i = and k 3, (ii) P 0 reresents a simle convex olygon, (iii) all oints that were discarded are not extreme oints of conv(p ). Proof. The roof is by induction. Initially run =, is the current sequence of redecessors of run and (i){(vi) hold trivially. Let P 0 = i : : : ik denote a subsequence of P of the kind considered in (a) and suose (i){(vi), resectively (i){(v) hold. From the assumtion that is an extreme oint of conv(p ) it follows that there exists a directed line containing such that all turn oints of P dierent from are to the right of that line. Let L denote some such line and let H be the half lane of oints on or to the right of L. We have to consider every situation that might ossibly occur when the algorithm asses line (9) or (). The case requiring the most involved roof occurs when the algorithm iterates stes () and (). As in all other cases the roof of (i)-(vi) (res. (i)-(v)) is either trivial or uses the same arguments we only give a roof for this case. Thus, suose we have that ik = head ; ik? = tail, ik + = succ( head ) K( ik? ; ik ) and (i){(v) hold. Observe that from the simlicity of P it follows that we have ik + K( ik? ; ik ) if and only if ik + is on or to the right of L( ik? ; ik ) and to the left of L( ik?; ik ). In the situation considered the algorithm starts iterating lines () and () 6

9 (a) (b) s (c) n i + i i - k i k j L H() L n Figure 5: Situations considered in the roof of Lemma until the new head is to the left of L( ik? ; ik ). Thus, we rst have to show that there always exists s fi k ; : : :; n; g such that s is to the left of L( ik? ; ik ). As the boundary of K( ik? ; ik ) consists { with the excetion of [ ik? ; ik ] { of edges of the original olygon P, it suces to show that we cannot have f r+ ; r+ ; : : :; n ; g K( ik? ; ik ): If this were the case, it would follow that we must have = i k? since is assumed to be an extreme oint of conv(p ) and hence [ n ; ] K( ; i ) (See Figure 5 (a)). This, however, would imly that n is to the left of L( ; ) contradicting the assumtion that P is clockwise oriented. Thus, either there exists n s > r + such that s = K( ik? ; ik ) or f r+ ; : : :; n g K( ik? ; ik ) and = K( ik? ; ik ). We consider the two cases searately. Case (): 9 i k < s n : s = K( ik? ; ik ) In this situation the algorithm udates head to s where s is smallest between i k and n such that s = K( ik? ; ik ), and backtracks tail. First we claim that tail is not backtracked beyond. If s = n the claim follows from inductive assumtion (i). Thus, suose s 6= n and s is on or to the left of L( i ; i ). Then by inductive assumtion we are in the situation of Lemma (iii), where L is as above (see Figure 5 (b)). Hence s is to the right of L( n ; ). Thus, the sequence to consider is P 00 = i : : : ik 0 s where k 0 k? and the new tail = ik 0. By what already has been shown this sequence satises (i). In order to show (iv) we rst claim that j = K( i ; i+ ) for all i k j < s and all k 0?. This holds for i k by inductive assumtion (iv) and follows inductively for 7

10 the other j using Lemma (ii) and j K( ik? ; ik ) conv(f j i k? i k g) k\ 0? = H( i ; i+ ); by inductive assumtion (iii). (See Figure 5 (c)). Another alication of Lemma (ii) then also roves s = K( i ; i+ ) for all k 0? and hence (iv). Now, suose P ath P 00 ( i ; s ) is not simle, i.e. there exists < k 0? such that [ ik 0 ; s ] \ [ i ; i+ ] 6= ;: We then by (iv) and inductive assumtion (ii) { using Lemma (i) { have that s is not contained in the olyhedron H \ k\ 0? = H( i ; i+ ): The one{dimensional faces of this olyhedron are line segments contained in either L, or [ i ; i+ ] ( k 0? ), or L( ik 0? ; ik 0). As it follows from inductive assumtion (iii) that s? is contained in the interior of this olyhedron we therefore must have that [ s? ; s ] meets some one{dimensional face of this olyhedron in a oint dierent from s? and s. As both of these oints are contained in H and H( ik 0? ; ik 0) we deduce that we must have [ i ; i+ ] \ [ s? ; s ] 6= ; for some < k 0? and hence K( i ; i+ ) \ [ s? ; s ] 6= ; for some < k 0? : Using Lemma (ii) this contradicts the simlicity of P. As P ath P 00( i ; s ) is also convex by construction we thus get (ii). An alication of Lemma (iii) with = ik 0 ; q = ik 0 + ; r = s and S = L( ik 0? ; ik 0 ) in case k 0, resectively S = L in case k 0 = then also shows (iii) for P 00 (See Figure 6 (d)). In order to rove (v) we only have to to show that i is not an extreme oint of conv(p ) for all k 0 < k the claim being clear from (iii) for the other oints. Suose to the contrary that i is an extreme oint. Let j denote its redecessor in the subsequence of P reresenting the boundary of the convex hull. By inductive assumtion (v) we have j = i for some <. Since for? it follows from inductive assumtion (ii) that i+ is to the left of L( i ; i ) we must have that j = i?. But as s is on or to the left of L( i? ; i ) and s = [ i? ; i ] by (iv) which already has been roved we have reached a contradiction to the fact that P is all on or to the right of L( i? ; i ), which suorts a face of conv(p ). Case (): f ik : : : n g K( ik? ; ik ) and = K( ik? ; ik ) In this situation the algorithm udates head to and backtracks tail. Dierent from case () we have to argue that at least three oints stay in the list. Since [ n ; ]\[ ik? ; ik ] 6= ; and n is on or to the right of L( ik? ; ik ) and is to the left of L( ik? ; ik ) we must have that ik? is to the left of L( n ; ). Hence, since all oints are to the right of L the oint ik 0 that tail is backtracked to is to the left of L( n ; ), too, and therefore by inductive assumtion (i) we get that ik 0 is to the right of L( ; i ) and k 0 3. We also get that for 8

11 (d) r (e) q S i k n ik- tail Figure 6: Situations considered in the roof of Lemma the new head = and tail = ik 0 that succ( head ) = i is to the right of L( tail ; head ) and to the right of L(red( head ; head )) since red( head ) = n (See Figure 6 (e)). Hence run is udated to and the algorithm stos. That the sequence i i : : : ik 0 the algorithm returns reresents a simle olygon follows by the same arguments given above to rove simlicity of the ath considered in case (). Again this olygon is convex by construction. That all oints discarded cannot be extreme oints of conv(p ) also follows similarly as above. Theorem The algorithm nds the convex hull of a simle olygon in linear time. Proof. The oerativeness of the algorithm was shown in Lemma. It has linear running time since for every constant time oeration we either move one oint further or discard a oint and oints once discarded are never considered again. 5 An Alication: Comuting the Smallest Box Containing a Simle Polygon in Linear Time Our original interest in linear time algorithms for comuting the convex hull of a simle olygon arose from the following cutting stock roblem. We have to cut work ieces of a olygonal shae from some sheet. In a rst ste the sheet is rocessed by a machine able to cut vertically or horizontally only. This yields the roblem of determining a rectangle of least ossible area containing a given simle olygon. In order to show that this is ossible in linear time it suces { by what has already been shown { to show that one can nd a 9

12 rectangle of smallest area containing a given simle convex olygon in linear time, as any rectangle containing a set of oints in the lane must also contain their convex hull. Given a oint and a line (-segment) M in the lane we say that covers M, if M. Crucial for the solution of the task at hand is the following roosition. Proosition Let P be a simle convex olygon in the lane. If R is a rectangle of minimum area circumscribing P, then at least one edge of P is contained in the boundary of R. Proof. Since R is assumed to be of least area containing P we obviously must have that each of the four edges of R is covered by at least one of the turn oints of P. The claim follows, if we show that some edge of R is covered by two turn oints of P. Thus, we have to show that none of the following situations occurs for a rectangle of smallest ossible area circumscribing P : () Exactly four turn oints ; ; 3 ; 4 of P are contained in the boundary of R and every edge of R is covered by exactly one of the i. () There are exactly three turn oints of P contained in the boundary of R one of which equals a vertex of R and the other two each cover exactly one of the two edges of R not containing that vertex. (3) Two oosite vertices of R are turn oints of P and no other turn oint of P is contained in the boundary of R. Suose () holds and consider the convex quadrangle Q whose vertices are the i. The situation is shown in gure 7 and we use the notation given there. The volume of Q is x. Hence, the area vol(r) of R is minimal if and only if the sum of the areas of the four triangles ( i ; r i ; i+ ); i = ; ; 3; 4 (indices read modulo 4) is minimal. Each of these triangles has area given by If we choose = as free arameter then Hence vol(r) = f() = vol(q) + 4 l i sin( i ) cos( i ) = 4 l i sin( i ): =? + 3 =?? + 4 = 3??? 3 + : 4X i= 0 Xi? l i + (i? )? Considering the second derivative of this we nd 0 4X Xi? f 00 () =? l i + (i? )? i= 0 j= j A : j= j A :

13 r ax? a?? 4??? l l 4 4 aaaaaaaaaaaa x 3 x??? XXXX l 3 X l XXXXXXXXX x 3 r r 3 Figure 7: The situation in the roof of roosition 3 r 4 4 But this exression is strictly negative since each of the i ]0; [. As, by assumtion, can vary in both directions R cannot be a rectangle of minimum area circumscribing Q and a fortiori not a rectangle of least ossible area circumscribing P contradicting the assumtion. The cases () and (3) are done along the same line. We now briey describe how roosition yields a linear time algorithm for determining a rectangle of least ossible area containing a given simle convex olygon. Similar O(n)- algorithms are described in [FS83] and [MS] Let P = : : : n reresent a simle clockwise oriented convex olygon and let e i = [ i ; i+ ] denote the edges of P. For the rectangle R i of least ossible area containing P that contains e i on its boundary let f ij ( j 4) denote the four edges of R i listed in clockwise order where e i f i. From Proosition we know that among the R i there is a rectangle of smallest ossible area containing P. Traversing P in clockwise orientation starting at i let l i denote the rst turn oint of P encountered that lies on f i, u i the rst turn oint of P on f i3 and r i the rst turn oint of P on f i4. Note that R i is uniquely determined by (e i ; l i ; u i ; r i ) and that its reresentation as the sequence of its vertices listed in clockwise order and its area may be comuted in constant time from these data. We have that j = l i () ( i+? i ) T ( j+? j ) 0 and ( i+? i ) T ( j? j? ) > 0; j = u i () det( i+? i ; j+? j ) 0 and det( i+? i ; j? j? ) < 0; j = r i () ( i+? i ) T ( j+? j ) 0 and ( i+? i ) T ( j? j? ) < 0: If l = j, we furthermore have that there exists some < k n and some subsequence j j : : : jk of j j + : : : n : : : j? j such that the sequence l : : : l n equals j : : : j j : : : j : : : jk : : : jk. It follows that, given l, it is ossible to comute the sequence l : : :l n in O(n). In fact, this sequence can be determined by comuting at most n of the n dot roducts considered in the characterisation of the l i above. Similarly, given u, resectively r, it is ossible to determine the sequences u : : :u n and r : : : r n in linear time. As it obviously is also ossible to comute (e ; l ; u ; r ) in O(n) it follows from the above

14 discussion that one may nd all (e i ; l i ; u i ; r i ) in linear time and hence the R i of least ossible area may be determined in linear time, too. Note: After nishing the aer we learned about the linear time algorithm for determining the convex hull of a simle olygon by Graham and Yao [GY83] that we missed while doing the research. That algorithm avoids the roblems of Sklansky's algorithm in exactly the same way our does, i.e. when running into a K P ( i ; j ) { in [GY83] suggestively called a \ocket" { the currently considered turn oint is advanced to the rst turn oint outside of the ocket. However, although the main idea is the same that algorithm diers from ours in that it avoids \curling in" by using a air of antiodal extreme oints one of which is the starting oint for the rocessing and the other is used in each iteration of the outer loo to determine a second scanline. The consequence of this dierence is that our algorithm should be faster than Graham and Yao's on inut olygons that are \close" to being convex already. In articular, if the inut olygon is convex, our algorithm detects this by considering each turn oint exactly once while the algorithm from [GY83] considers each turn oint twice { once for each of the two scan lines. On the other hand, our algorithm does sueruous work on each instance of \curling in" as in these situations it rst tentatively accets turn oints and discards them later while the algorithm from [GY83] discards these oints right away. References [BE84] B. Bhattacharya, H. Elgindy: A new Linear Convex Hull Algorithm for Simle Polygons. IEEE Trans. Inform. Theory, vol. IT-30, no. 85 { 88, 984 [FS83] H. Freeman, R. Shaira: Determining the Minimum-Area Encasing Rectangle for an Arbitrary Closed Curve. Comm. ACM, 8 (7), 409 { 43, 975 [GS83] S. Ghosh, R. Shyamasundar: A Linear Time Algorithm for Obtaining the Convex Hull of a Simle Polygon. Patt. Recogn., vol. 6, no. 6, 587 { 59, 983 [GY83] R. Graham, F. Yao: Finding the Convex Hull of a Simle Polygon. J. Algo. 4, 34 { 33, 983 [MS] R. Martin, P. Stehenson: Containment Algorithms for Objects in Rectangular Boxes. rerint [MA79] D. McCallum, D. Avis: A Linear Algorithm for Finding the Convex Hull of a Simle Polygon. Inform. Proc. Lett., vol. 9, no. 5, 0 { 06, 979 [S7] J. Sklansky: Measuring concavity on a rectangular mosaic., 355 { 364, 97 IEEE Trans. Comut.