Computational geometry

Transcription

1 Leture 23 Computtionl geometry Supplementl reding in CLRS: Chpter 33 exept 33.3 There re mny importnt prolems in whih the reltionships we wish to nlyze hve geometri struture. For exmple, omputtionl geometry plys n importnt role in Computer-ided design Computer vision Computer nimtion Moleulr modeling Geogrphi informtion systems, to nme just few Intersetion Prolem Given set of line segments S 1,..., S n in the plne, does there exist n intersetion? Humns, with their sophistited visul systems, re prtiulrly well-equipped to solve this prolem. The prolem is muh less nturl (ut still importnt, given the ove pplitions) for omputer. Input: Line segments S = {S 1,..., S n } in the plne, represented y the oordintes of their endpoints Output: Gol I: Detet. Determine whether there exists n intersetion. Gol II: Report. Report ll pirs of interseting segments. The ovious lgorithm is to hek eh pir of line segments for intersetion. This would tke Θ ( n 2) time. For Gol II, this n t e improved upon, s it ould tke Θ ( n 2) time just to report ll the intersetions (see Figure 23.1). We will fous on Gol I nd show tht there exists Θ(nlg n) lgorithm. For ese of exposition, we will ssume tht our line segments re in generl position, i.e., No two endpoints hve the sme x-oordinte. (In prtiulr, there re no perfetly vertil segments.) There re no instnes of three segments interseting t single point. Both of these ssumptions re unneessry nd n e dropped fter ouple of minor (ut reful) modifitions to the lgorithm.

2 Figure If the segments S 1,..., S n form squre grid, there ould e s mny s ( n 2 )( n2 ) = n 2 4 = Θ( n 2) intersetions. In other situtions, it ould hppen tht every pir of segments intersets Sweep line nd preorder Suppose you hve sheet of pper with the line segments S 1,..., S n drwn on it. Imgine drgging vertil ruler ross the pge from left to right (see Figure 23.2). This is extly wht we hve in mind when we sy, For eh x R, let L x e the vertil line t distne x from the y-xis. 1 We imgine x inresing over time, so tht L x represents sweep line whih moves ross the plne from left to right. For eh x R, there is nturl preorder 2 on the set of non-vertil line segments in the plne: for two non-vertil segments,, we sy x if the intersetion of (or its extension, if need e) with L x lies ove the intersetion of with L x. This reltion needn t e ntisymmetri euse it ould hppen tht nd interset L x t the sme point. In tht se, we sy tht oth x nd x ; ut of ourse, this does not imply tht =. Of ruil importne is the following ovious ft: Given x R nd line segments nd, there exists n O(1)-time lgorithm to hek whether x. 3 Also importnt is the following oservtion: Oservtion Suppose line segments nd interset t the point P = (x, y ). Then one of the following two sttements holds: (i) For x > x, we hve x. For x < x, we hve x. (ii) For x > x, we hve x. For x < x, we hve x. Conversely, if nd do not interset, then one of the following two sttements holds: (i) For ll x suh tht L x intersets nd, we hve x. (ii) For ll x suh tht L x intersets nd, we hve x. Another wy to view the seond prt of Oservtion 23.1 is s follows: 1 It took me while to figure out how to write down definition of L x tht didn t inlude the eqution x = x. 2 A preorder is inry reltion tht is reflexive nd trnsitive, ut not neessrily symmetri or ntisymmetri. 3 It is not so ovious, however, wht the est suh lgorithm is. Most nïve ttempts involve lulting slope, whih is numerilly unstle. To remedy this, the est lgorithms use ross produts to ompute orienttions without using division. Le 23 pg. 2 of 8

3 L 0.1 L 1.2 L 2.1 L 3.1 L 3.8 I d d d Figure A sweep line trversing from left to right. Notie tht x d x x for ll x I, ut d. Thought Experiment Consider finite set of line segments; for onreteness let s sy we re looking t the four segments {,,, d}. There is some intervl I (perhps empty) suh tht, for eh x I, L x intersets ll four segments (see Figure 23.2). For eh x I, the reltion x indues preorder struture on the set {,,, d}. Oservtion 23.1 implies tht, if there re no intersetions mong {,,, d}, then this preorder struture does not depend on the hoie of x. Thought Experiment 23.2 leds to the following strtegy, whih is quite ingenious. To introdue the strtegy, we first define the preorder dt struture. The preorder dt struture stores dynmi olletion T of ojets long with preorder on those ojets, whih we denote y. The supported opertions re 4 INSERT() inserts into the olletion DELETE() deletes from the olletion ABOVE() returns the element immeditely ove. If no suh element exists, returns NIL. BELOW() returns the element immeditely elow. If no suh element exists, returns NIL. The ovious issue with using this dt struture in onjuntion with our plne sweep is tht it only stores single preorder reltion, whih we hve denoted ove y (nd the usul ounterprts, >, <), wheres the reltion x hs the potentil to hnge s x vries. However, this needn t e prolem, s long s: 4 For ese of exposition, we hve osured one detil from view. It is possile for T to ontin two elements d, e suh tht d e nd e d. In order for the opertions ABOVE(d) nd BELOW(d) to work orretly, there must e some rule for reking ties. To this end, T will lso internlly store n order on its elements, suh tht lwys implies (ut not neessrily onversely). Then, the element immeditely ove is the element suh tht nd, whenever, we lwys hve. (Similrly for the element immeditely elow. ) The hoie of order does not mtter, s long s it is persistent through insertions nd deletions. Le 23 pg. 3 of 8

4 Condition During the lifetime 5 of ny prtiulr element, the rnking of with respet to the other elements of T does not ever need to e hnged. We re now prepred to give the lgorithm: Algorithm: DETECT-INTERSECTION(S ) 1 Sort the endpoints of segments in S y x-oordinte 2 T new preorder struture 3 Run the sweepline from left to right 4 for eh endpoint (x, y), y order of x-oordinte do 5 if (x, y) is the left endpoint of segment then 6 Insert into T, using the reltion x to deide where in the preorder should elong 7 Chek for intersetion with T.ABOVE() 8 Chek for intersetion with T.BELOW() 9 if there ws n intersetion then 10 return TRUE 11 else if (x, y) is the right endpoint of segment then 12 Chek T.ABOVE() for intersetion with T.BELOW() 13 if there ws n intersetion then 14 return TRUE 15 T.DELETE() 16 return FALSE This step needs justifition. See the proof of orretness. Proof of orretness. First, note the need to justify line 6. In implementtion form, T will proly store its entries in n ordered tle or tree; then, when T.INSERT() is lled, it will tke dvntge of T s internl order y using inry serh to insert in Θ ( lg T ) time. In line 6, we re sking T to use the reltion x to deide where to ple. This is fine, provided tht x grees with T s internl preorder when onsidered s reltion on the elements tht urrently elong to T. (Otherwise, inry serh would not return the orret result. 6 ) So, over the ourse of this proof, we will rgue tht Whenever line 6 is exeuted, the reltion x grees with T s internl preorder when onsidered s reltion on the elements tht urrently elong to T. From Thought Experiment 23.2 we see tht the ove sttement holds s long s the sweep line hs not yet pssed n intersetion point. Thus, it suffies to prove the following lim: Clim. DETECT-INTERSECTION termintes efore the sweep line psses n intersetion point. The lim oviously holds if there re no intersetions, so ssume there is n intersetion. Let s sy the leftmost intersetion point is P, where segments nd interset. Assume without loss of generlity tht s left endpoint lies to the right of s left endpoint (so tht gets inserted into T 5 By the lifetime of, I men the period of time in whih is n element of T. 6 Atully, in tht sitution, there would e no orret nswer; the very notion of inry serh on n unsorted list does not mke sense. Le 23 pg. 4 of 8

5 P Figure If the left endpoint of lies etween nd to the left of P nd the right endpoint of lies to the right of P, then must interset either or. efore does). If ever eomes djent to in T, then either lines 7 8 or line 12 will detet tht nd interset, nd the lgorithm will hlt. So we re free to ssume tht nd re never djent in T until fter the sweep line hs pssed P. This mens tht there exists line segment suh tht the left endpoint of lies etween nd 7 nd to the left of P, nd the right endpoint of lies to the right of P. Geometrilly, this implies tht must interset either or, nd tht intersetion point must lie to the left of P (see Figure 23.3). But this is impossie, sine we ssumed tht P ws the leftmost intersetion point. We onlude tht the lim holds. Atully, we hve done more thn prove the lim. We hve shown tht, if there exists n intersetion, then n intersetion will e reported. The onverse is ovious: if n intersetion is reported, then n intersetion exists, sine the only time we report n intersetion is fter diretly heking for one etween speifi pir of segments Running time The running time for this lgorithm depends on the implementtion of the preorder dt struture. CLRS hooses to use red lk tree 8, whih hs running time O (lg n) per opertion. Thus, DETECT-INTERSECTION hs totl running time Θ(nlg n) Finding the Closest Pir of Points Our next prolem is simple: Input: A set Q of points in the plne Output: The two points of Q whose (Euliden) distne from eh other is shortest. The nïve solution is to proeed y rute fore, proing ll ( n 2) pirs of points nd tking Θ ( n 2 ) time. In wht follows, we will exhiit sutle divide-nd-onquer lgorithm whih runs in Θ(nlg n) time. 7 Tht is, if the left endpoint of is (x, y), then either x x or x x. 8 For more informtion out red lk trees, see Chpter 13 of CLRS. Le 23 pg. 5 of 8

6 In 23.1, the pseudoode would not hve mde sense without few prgrphs of motivtion eforehnd. By ontrst, in this setion we will give the pseudoode first; the proof of orretness will eluidte some of the strnge-seeming hoies tht we mke in the lgorithm. This is more or less how the lgorithm is presented in 33.4 of CLRS. The lgorithm egins y pre-sorting the points in Q ording to their x- nd y-oordintes: Algorithm: CLOSEST-PAIR(Q) 1 X the points of Q, sorted y x-oordinte 2 Y the points of Q, sorted y y-oordinte 3 return CLOSEST-PAIR-HELPER(X, Y ) Most of the work is done y the helper funtion CLOSEST-PAIR-HELPER, whih mkes reursive lls to itself: Algorithm: CLOSEST-PAIR-HELPER(X, Y ) 1 if X 3 then 2 Solve the prolem y rute fore nd return 3 x the medin x-oordinte of X 4 Let X L X onsist of those points with x-oordinte x 5 Let X L X onsist of those points with x-oordinte > x 6 Let Y L Y onsist of those points whih re in X L 7 Let Y R Y onsist of those points whih re in X R 8 Find the losest two points in the left hlf 9 p L, q L CLOSEST-PAIR-HELPER(X L,Y L ) 10 Find the losest two points in the right hlf 11 p R, q R CLOSEST-PAIR-HELPER(X R,Y R ) 12 δ R distne from p L to q L 13 δ R distne from p R to q R 14 δ min{δ L,δ R } 15 Y those points in Y whose x-oordinte is within δ of x 16 Rell tht Y is lredy sorted y y-oordinte 17 ɛ 18 for i 1 to Y 1 do 19 p Y [i] 20 for q in Y [i + 1,...,min { i + 7, Y } ] do 21 if ɛ > distne from p to q then 22 ɛ distne from p to q 23 p p 24 q q 25 if ɛ < δ then 26 return p, q 27 else if δ R < δ L then 28 return p R, q R 29 else 30 return p L, q L Le 23 pg. 6 of 8

7 Left hlf Right hlf Figure CLOSEST-PAIR-HELPER divides the set X into left hlf nd right hlf, nd reurses on eh hlf Running time Within the proedure CLOSEST-PAIR-HELPER, everything exept the reursive lls runs in liner time. Thus the running time of CLOSEST-PAIR-HELPER stisfies the reurrene whih (y the Mster Theorem in 4.5 of CLRS) hs solution T(n) = 2 T ( n 2 ) + O(n), (23.1) T(n) = O (nlg n). (23.2) Note tht, if we hd deided to sort within eh reursive ll to CLOSEST-PAIR-HELPER, the O(n) term in (23.1) would hve insted een n O (nlg n) term nd the solution would hve een T(n) = O ( n(lg n) 2). This is the reson for reting helper proedure to hndle the reursive lls: it is importnt tht the lists X nd Y e pre-sorted so tht reursive lls need only liner-time opertions. Note lso tht, if insted of lines we hd simply heked the distne etween eh pir of points in Y, the O(n) term in (23.1) would hve insted een n O ( n 2) term, nd the solution would hve een T(n) = O ( n 2) Corretness CLOSEST-PAIR-HELPER egins y reursively lling itself to find the losest pirs of points on the left nd right hlves. Thus, lines re ostensily n ttempt to hek whether there exists pir of points p, q, with one point on the left hlf nd one point on the right hlf, whose distne is less thn tht of ny two points lying on the sme hlf. Wht remins to e proved is tht lines do tully hieve this ojetive. By the time we reh line 15, the vrile δ stores the shortest distne etween ny two points tht lie on the sme hlf. It is esy to see tht there will e no prolems if δ truly is the shortest possile distne. The se we need to worry out is tht in whih the losest pir of points ll it p, q hs distne less thn δ. In suh se, the x-oordintes of p nd q would oth hve to e within δ of x ; so it suffies to onsider only points within vertil strip V of width 2δ entered t x = x (see Figure 23.5). These re preisely the points stored in the rry Y on line 15. Sy p = Y [i] nd q = Y [ j], nd ssume without loss of generlity tht i < j. In light of lines we see tht, in order to omplete the proof, we need only show tht j i 7. Le 23 pg. 7 of 8

8 V δ δ Figure It suffies to onsider vertil strip V of width 2δ entered t x = x. δ S L S R Figure Eh of S L nd S R n hold t most 4 points. (Atully, in ompletely idel geometry, S R nnot ontin 4 points euse its left oundry is exluded. But sine the oordintes in omputtionl geometry prolem re typilly given s floting point numers, we re not lwys gurnteed orret hndling of edge ses.) Sy p = (p x, p y ). Let S L e the squre (inluding oundries) of side length δ whose right side lies long the vertil line x = x nd whose ottom side lies long the horizontl line y = p y. Let S R e the squre (exluding the left oundry) of side length δ whose left side lies long the vertil line x = x nd whose ottom side lies long the horizontl line y = p y. It is evident tht q must lie within either S L or S R (see Figure 23.6). Moreover, ny two points in the region S L re seprted y distne of t lest δ; the sme is true for ny two points in the region S R. Thus, y geometri rgument 9, S L nd S R eh ontin t most four points of Y. In totl, then, S L S R ontins t most eight points of Y. Two of these t-most-eight points re p nd q. Moreover, sine Y onsists of ll points in V sorted y y-oordinte, it follows tht the t-most-eight points of S L S R our onseutively in Y. Thus, j i 7. 9 One suh rgument is s follows. Divide S L into northest, northwest, southest nd southwest qudrnts. Eh qudrnt ontins t most one point of Y (why?), so S L ontins t most four points of Y. Le 23 pg. 8 of 8

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