Cross products Line segments The convex combination of two distinct points p

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1 CHAPTER Comutational Geometry Is the branch of comuter science that studies algorithms for solving geometric roblems. Has alications in many fields, including comuter grahics robotics, VLSI design comuter aided design statistics Deals with geometric objects such as oints, line segments, olygons, etc. Some tyical roblems considered: whether intersections occur in a set of lines. finding vertices of a conve hull for oints. whether a line can be drawn searating two sets of oints. whether one oint is visible from a second oint, given some olygons that may block visibility. otimal location of fire towers to view a region. closest or most distant air of oints. whether a oint is inside or outside a olygon. Advanced Algorithms, Feodor F. Dragan, Kent State University Cross roducts Line segments The conve combination of two distinct oints = (, y) and = (, y) is any oint = (, y) such that for some real number α with 0 α,, y ) = α(, y ) + ( α)(, ). ( y, the line segment joining and, is the set of all conve combinations of and. Intuition roblem: Show that if (,y) is a conve combination of (, y) and (, y) then y y y y α = = which is the standard equation of a line with sloe α. Cross roducts let and be oints on the lane The cross roduct corresonds to the signed area in the arallelogram. = det = y y =. y y y + Advanced Algorithms, Feodor F. Dragan, Kent State University (0,0)

2 If is negative, then o is counterclockwise from o. If is ositive, then o is clockwise from o. If 0, then o and o are collinear. = Cross roducts (cont.) y To determine if is clockwise from 0 0, we translate 0 to the origin and consider ' ' where ' = 0, ' = 0. ' ' = ( 0)( y y0) ( 0)( y y0). o Prohibited for If o is clockwise from (0,0) o Consider now whether two consecutive line segments and turn left or right at. 0 Check whether 0 is clockwise from 0 ( 0) ( 0) < 0 So, counterclockwise or left turn ( 0) ( 0) > 0 So, clockwise or right turn 0 0 Advanced Algorithms, Feodor F. Dragan, Kent State University Intersection of two line segments A two stage rocess is used. The first stage is called quick-rejection. Bounding Bo: The bounding bo of a geometric figure is the smallest rectangle that contains the figure and whose segments are arallel to and y ais. The bounding bo ( ˆ, ˆ ) of the line segment is the bo with lower lefthand oint ˆ ( ˆ, ˆ ), where = y ˆ = min(, ) yˆ = min( y, y ) and with uer right-hand oint ˆ = ( ˆ, yˆ ), where ˆ = ma(, ) yˆ = ma( y, y ) Intersection of bounding boes: Rectangles ( ˆ, ˆ ) and ( ˆ, ˆ 4) intersect if and only if the following is true: ( ˆ, ˆ 4) ˆ ˆ ) Λ( ˆ ˆ ) Λ( yˆ yˆ ) Λ( yˆ ˆ ) ( 4 4 y ŷ 4 this will not ass the quick rejection test ŷ ŷ ŷ ˆ, ˆ ) ( (0,0) ˆ ˆ ˆ 4 ˆ Advanced Algorithms, Feodor F. Dragan, Kent State University 4

3 Intersection of two line segments (cont.) Second stage: Decide whether each segment meets ( straddles ) the line containing the other. A segment straddles a line if lies on one side of the line and on the other side. (the segment straddles the line also if or lies on the line) Observation: Two segments intersect iff they ass the quick rejection test and each segment straddles the other. Testing straddle with cross roducts: we show how to check if straddles the line L determined by and If 4 does straddle the line containing and, then the following have different signs. ( ) ( ) ( ) ( ) Boundary cases where 4 straddles L At least one cross roduct is zero. Both cases ass the quick rejection test. 4 4 Advanced Algorithms, Feodor F. Dragan, Kent State University 5 Determining whether any air of segments intersect. Given n line segments, determine whether any air of them intersect. Sweeing: An imaginary vertical swee line moves across the -ais (which acts as a time ais) from left to right. All line segment endoints are considered in a left-to-right order. Whenever endoint is found, a check for an intersection is made. Algorithm runs in O(n logn) time, where n is the number of segments. Algorithm does not look for all intersections, but just whether or not there is at least one intersection. It takes Ω( n ) time in the worst case to find all intersections (see Eercise.-) Standard simlifying assumtions: no inut segment is vertical no three inut segments intersect at a single oint. Order definition: Let s and s be two segments s These segments are comarable at t if the vertical swee line with -coordinate t intersects both segments. s If s and s are comarable at t, then s is above s, written s > t s, if s intersects the swee line at t at a higher oint than s. t Advanced Algorithms, Feodor F. Dragan, Kent State University 6

4 > is a total order for each value of (refleive, antisymmetric, transitive) order > changes as changes (swee line status and event oint schedule). a segment obtains an order when its left endoint is encountered by swee line. a segment looses its order when swee line reaches its right endoint. swee line behavior near an intersection oint if the intersection oint is not on a swee line, then there eist lines v and w immediately to the right and left of the intersection. the segment that has the higher intersection at v reverses its order with other segment as swee line asses from v to w. if is the leftmost intersection oint and is at intersection of segments a and b, then a and b intersect v at adjacent oints a and b are called consecutive at v. Advanced Algorithms, Feodor F. Dragan, Kent State University 7 v a b w The Algorithm ANY-SEGMENTS-INTERSECTION(S) To maintain a total ordering T on the swee line as it moves, we need the following oerations: INSERT(T,s) insert segment s into T DELETE(T,s) delete segment s from T ABOVE(T,s) return the segment from T immediately above segment s BELOW(T,s) return the segment from T immediately below segment s Red-black trees can rovide a balanced search tree which can suort above oerations in O(log n) time (see Ch. in CLRS) () T emtyset () Sort segment endoints from left to right. Break ties by utting oint with lower y-coordinate first. () For each oint in the sorted list (in order) (4) if is left endoint of segment s then (5) INSERT(T,s) (6) if [ABOVE(T,s) eists and it intersects s] or [BELOW(T,s) eists and it intersects s] then return TRUE (7) if is the right endoint of segment then (8) if both [ABOVE(T,s) and BELOW(T,s) eist] and [ABOVE(T,s) intersects BELOW(T,s)] then return TRUE (9) DELETE(T,s) Animated demo (0) return FALSE htt:// Advanced Algorithms, Feodor F. Dragan, Kent State University 8 4

5 Comments and Correctness Sorting on line () will sort oints leicograhically and require O(n logn) time. A boundary case occurs if oint in algorithm also lies on another segment s. In this case s and s should be in consecutive order at. Correctness Theorem: A call to ANY-SEGMENTS-INTERSECTION(S) returns true iff there is an intersection among the segments of S. If a true is returned, an intersection has been found between the two segments. Suose a false is returned, but an intersection actually eists. Let be the leftmost intersection oint (break ties by taking lowest y-coordinate). Let a and b be segments intersecting at. Since no intersection occurs to left of, T gives the correct order rior to. that is, no reversals in order have occurred. Since no three segments intersect at the same oint, there eists a swee line z at which a and b become consecutive. z is to the left of or goes through. There is an endoint q on z that is the event at which a and b become consecutive. If is on swee line z, then =q. If not, then q is left of. Advanced Algorithms, Feodor F. Dragan, Kent State University 9 c z q a b Correctness and Runtime In either case, the order of T is correct before q is rocessed. Only two cases are ossible.. Either a or b is inserted into T at z and the other segment is above or below it. (In this case, intersection is detected by line (6).). Segments a and b are already in T and a segment c between them is deleted at z. (In this case, an intersection is detected by line (8).) In either case the intersection of a and b at is detected, so rocedure cannot return a FALSE, as suosed. c q a b Running Time: sort on line () takes O(n logn) time, using merge or hea sort. since there are n events (endoints) where swee line stos, the loo iterates at most n times. each iteration of loo takes O(log n) time, since red-black tree oerations take O(log n) time. Total time is O(n logn). READ Ch.. and Ch. in CLRS. Advanced Algorithms, Feodor F. Dragan, Kent State University 0 z 5

6 Conve Hull Algorithms Definitions and Proerties: Given n oints on the lane Q = { n,,..., }. Intersection of all conve sets containing Q Smallest conve set containing Q Intersection of all half-lanes containing Q Union of all triangles formed by oints of Q Smallest conve olygon containing Q All vertices of hull are some oints of Q rubber band etreme oint always unique not etreme oint NOTE: conve-hull(q) is the closed solid region, not just the boundary CH(Q) Advanced Algorithms, Feodor F. Dragan, Kent State University The Problem and Aroaches Problem: Given n oints on the lane Q = {,,..., Aroaches: Brute Force Gift Wraing QuickHull Graham Scan Incremental Divide and Conquer By Delaunay Triangulation & Voronoi Diagram n }, find CH(Q). Brute-Force Aroach Determine etreme edges: q for each air of oints,q Q do if all other oints lie on one side of line assing thru and q then kee edge (, q) Sort edges in counterclockwise order (we want outut in counterclockwise) Running time: O(n ) bad but not the worst yet Advanced Algorithms, Feodor F. Dragan, Kent State University 6

7 Gift Wraing or The Jarvis March Algorithm First suggested by Chand and Kaur (970). Then by Jarvis (97). Worst-case time: O(n ) Outut-sensitive time: O(nh), where h is the # of vertices of hull Can be etended to higher dimension was the rimary algorithm for higher dimensions for quite some time Method: Let 0 be the lowest oint (take leftmost oint in case of tie) The net conve hull verte has the least olar angle w.r.t. the ositive horizontal ray from 0. Measure in counterclockwise If tie, choose the furthest such oint Vertices,,, k are icked similarly where k is the highest oint (leftmost) The sequence 0,,,,, k is the right chain of CH(Q). To choose the left chain of CH(Q), start with k choose k+ as the oint with least angle w.r.t. the negative horizontal ray from k k again, measure in counterclockwise direction and if tie occurs, choose furthest such oint. continue in same fashion until you obtain k, k+,, t = 0 0 Advanced Algorithms, Feodor F. Dragan, Kent State University Comleity of Jarvis March For each verte from CH(Q), it takes O() time to comare the olar angle of q j and qi using cross-roduct O(n) time to find minimum olar angle O(n) total time If CH(Q) has h vertices, then the running time is O(nh). In worst case h=o(n), hence O( n ) is worst case time bound. Worst case occurs when O(n) of oints lie on the conve hull, e.g., all oints lie on a circle. If oints in Q are generated by random generator, then we eect h= O(logn) In ractice, Jarvis march is normally very fast. If h=o(log n), then this algorithm is asymtotically fastest. Theorem: Ω( n log n) is a lower bound for comuting the conve hull. Assume we want to sort n numbers,,..., } { n create n oints on lane i = ( i, yi) where yi = i ( i =,..., n). y construct CH = CH ({,,..., n}) get from descrition of CH original numbers in sorted order Conclusion: If we can construct CH in o(n logn) time then we can sort n numbers in o(n logn) time, which is imossible. Advanced Algorithms, Feodor F. Dragan, Kent State University 4 (0,0) 7

8 Graham Scan Algorithm First a base oint 0 is selected. Normally this is the oint with minimum y- coordinate (select leftmost in case of tie) Net all oints are sorted w.r.t. the olar angle they make with a half-ray with left endoint 0 and arallel to -ais. (!!! We do not need to comute those angles!!!) 4 remove Interior oints on ray cannot be conve hull oints and are removed during sort Remaining oints are stored in counterclockwise order w.r.t. 0 Let 0,,,,, m ( m n) be the sorted list of remaining oints. Clearly, 0 and are in CH(Q). Let S be a stack in which oints that are otentially conve hull oints will be stored. Initially S = 0. Remaining stes of the algorithm follow for i to m do while (the angle formed by oints NEXT_TO_TOP(S), TOP(S), and i make a non-left turn) POP(S) PUSH(S, i ) return S Advanced Algorithms, Feodor F. Dragan, Kent State University Eamle and Runtime Comutation Note straight line here. 4 is eliminated, as non-left turn. 8 CH(Q)={ 0,,, 5, 7, 8 } Requires O(n) to find 0 Sorting based on olar angle takes O(n logn) time Removal of n-m oints with dulicate angles takes O(n). For loo is eecuted m- times, hence O(n). Interior while statement is a roblem. It may iterate as many as O(n) time. Above observation can easily lead to an over-estimate of O( n ). Note that each ass through while statement, POP is eecuted. As in analysis of MULTIPOP, there is at most one POP oeration for each PUSH oeration (see amortized analysis) Since 0,, m are not oed, at most m- POP oerations occur. Note, both POP and test for while take O() time and m n. Hence amortized cost for each iteration of while loo is O(). Overall worst-case cost of for-loo is O(n) Worst-case running time of the algorithm is O(n logn) (= O(n logn) + O(n)). Advanced Algorithms, Feodor F. Dragan, Kent State University 6 8

9 Correctness of Graham Scan Claim : Each oint oed from stack S is not a verte of CH(Q) and lies inside new CH(S). Suose j is oed from S because k j i makes a non-left turn. Since oints are ordered in increasing olar angle w.r.t 0, the oints k j lie in counterclockwise order w.r.t. i i 0 and k j i contains j Then j cannot be a verte of CH(Q). Claim : The oints on stack S always form the vertices of a conve olygon. 0 The claim holds initially for oints 0,,. They form a triangle. The stack changes by oints being oed or ushed. If a oint is oed, then a verte is removed from conve olygon, leaving the resulting olygon conve. If a oint i is ushed, the resulting oints of S form a conve olygon. by choice of 0,, i lies above 0 (see diagram on net slide).. since a left turn occurs at j, i lies below j k. since i occurs later in the sequence than j, the olar angle of 0 i. is greater than the olar angle of 0 j. thus, i lies to the left of line 0 j,, i.e., i lies in the shaded region in diagram. so, after i is ushed on the stack, a conve region results. Advanced Algorithms, Feodor F. Dragan, Kent State University 7 j k Correctness of Graham Scan (cont.) j Diagram i k Claim : At the conclusion of the algorithm, the oints of S joined together consequently form a conve olygon that contains Q. By claim, all oints droed from S are not vertices of CH(Q). By claim, the oints of S are oints from Q that form a conve olygon By construction and claim, all oints of Q are either in S or lie interior to CH(S ) for some earlier S with CH(S ) CH(S). Then, at conclusion the oints of S joined together consequently give a conve olygon that contains all oints of Q. Animated demos htt:// htt:// htt:// htt:// Advanced Algorithms, Feodor F. Dragan, Kent State University 8 0 9

10 Closest Pair of Points Given n oints on the lane, find closest air of oints. The Euclidean distance between two oints =, ) and =, ) is ( y ( y d(, ) = ( ) + ( y y) An obvious but naïve aroach is to comute the distance between any two oints and take minimum. However, running time is n = O ( n ). A high-level descrition of a much better algorithm (at least for large sets) is given below. Let Q be a set of n lanar oints. If Q <4, then the distances between all airs of oints are comuted and the closest air is reorted. If Q >, then a Divide & Conquer & Combine rocedure is followed. Each recursive call receives as inut a set P Q arrays X and Y containing oints P sorted by and y coordinates, resectively Advanced Algorithms, Feodor F. Dragan, Kent State University 9 Divide and Conquer Stes Divide ste Find a vertical line L that bisects the oints in P into P and P with almost equal sizes. All oints in P are left of L or on L, all oints in P are right of L or on L. The array X is divided into arrays X and X which contain the oints of P and P, resectively, sorted by -coordinates. Likewise the array Y is slit into arrays Y and Y which contain the oints of P and P, resectively, sorted by y-coordinates. At the beginning, Q has to be resorted by -coordinates to obtain X and by y- coordinates to obtain Y. This we do before the first recursive call. Now, on each recursive call, the set P can easily be slit in linear time, giving sets P, P, X, and X. Both Y and Y can be formed by Y in linear time by walking through Y and utting each element in Y (or Y ) if it is in P (resectively, P ). Conquer Ste Make two recursive calls to find closest air of oints in P and P, resectively. First call gets inut (P, X, Y ) Second call gets inut (P,X,Y ) Set closest air returned for P and P have distance d and d, resectively. Set d = min{d, d } Advanced Algorithms, Feodor F. Dragan, Kent State University 0 0

11 Combine Combine Ste and Runtime Observe that if there is a air in P closer than d, then they must lie in a d d bo about L, as follows. There can be at most 8 oints in each stri. Find the array Y consisting of the oints of Y within d of L, sorted by their y- coordinate. d d If d<d, then the closest air with distance d is returned. Otherwise, the closest air formed by the two recursive calls is returned. Runtime The initial rerocessing ste costs O(n logn) The recurrence relation for the recursive ortion is T(n), T ( n / ) + O( n) if n > where T ( n) = O() if n Its solution is T(n)= O(n logn). This gives also the total running time. Advanced Algorithms, Feodor F. Dragan, Kent State University d d L d d L Possible coincident oints, one in P and one in P. Net, for each oint from Y, check to see if d(,q)<d for each of the 7 oints q that follow in Y. Find the closest-air distance d, comuted in receding ste over all oints in Y. READ Ch.. and.4 in CLRS. Homework 4: Problems: 0.5- (in CLR only),.- (. 99) (= roblem 5.- in CLR),.-4 (. 946) (= roblem 5.-4 in CLR),.-4 ( ) (= roblem 5.-4 in CLR),.4- (. 96) (= roblem 5.4- in CLR), Etra credit roblem:.-8 (. 940) (= roblem 5.-7 in CLR) Advanced Algorithms, Feodor F. Dragan, Kent State University