O Radom Lie Segmets i the Uit Square Thomas A. Courtade Departmet of Electrical Egieerig Uiversity of Califoria Los Ageles, Califoria 90095 Email: tacourta@ee.ucla.edu I. INTRODUCTION Let Q = [0, 1] [0, 1] deote the uit square ad let L be a set of lie segmets i Q. Two lie segmets are said to be crossig if they itersect at ay poit. A subset of lie segmets is called o-crossig if o two segmets i the subset are crossig. Cosider the sceario where the edpoits of the lie segmets are radomly distributed, idepedetly ad uiformly, i Q. Defie N(L ) to be the size of the largest o-crossig subset of segmets. Formally: N(L ) = max { U : l 1, l 2 do ot cross for all l 1, l 2 U}, where U deotes the umber of lie segmets i the subset U. Further, defie L(L ) to be the maximum sum-legth over all o-crossig subsets of L. Formally: { } L(L ) = max l : l 1, l 2 do ot cross for all l 1, l 2 U, where l deotes the legth of lie segmet l. I this short ote, we prove that N(L ) ad L(L ) are both Θ( ) asymptotically almost surely. It is possible to exted these results to the case where lie segmets are required to be separated by some miimum distace. Partial characterizatios of this more geeral case are give i Sectio II-B. A. No-Crossig Lie Segmets II. MAIN RESULTS AND PROOFS Our first two results are that N(L ) ad L(L ) both behave roughly like : Theorem 1: Asymptotically almost surely, /2 N(L ) 15. Theorem 2: Asymptotically almost surely, /7 L(L ) 22. While we have made o sigificat efforts to optimize the coefficiets of the bouds i Theorems 1 ad 2, mior modificatios to our argumets ca yield somewhat tighter results. Additioal techiques are likely required i order to close the gap completely. The case where lie segmets are required to be separated by some miimum distace is discussed i Sectio II-B. Proof of Theorem 1: Throughout the proof, we make the distictio betwee left ad right edpoits of lie segmets. This is somewhat arbitrary, but simplifies the argumet sigificatly. Thus, a lie segmet is geerated accordig to the followig process: (Step 1) the left edpoit is chose uiformly from Q, ad (Step 2) the right edpoit is chose idepedetly ad uiformly from Q.
Claim 1: With probability tedig to 1 as, N(L ) /2. Partitio Q ito disjoit horizotal strips, each havig height 1/ ad width 1. Note that if a lie segmet l is cotaied i a sigle strip, the it will ot itersect lie segmets cotaied i ay other strip. The N(L ) Y, where Y is the umber of strips that cotai lie segmets. Observe that Pr [Lie l i strip j] = Pr [{left edpoit of l i strip j} {right edpoit of l i strip j}] = Pr [left edpoit of l i strip j] Pr [right edpoit of l i strip j] = 1. Where we used the fact that the probability a give poit falls i a particular strip is 1/ ad poits are chose idepedetly. The, the probability that a give strip does ot cotai ay lie segmets is: Further, ote that: (1 1/) 1/e. Pr [{Lie l ot i strip i} {Lie l ot i strip j}] = 1 Pr [{Lie l i strip i} {Lie l i strip j}] = 1 (Pr [Lie l i strip i] + Pr [Lie l i strip j]) = 1 2/. Where we used the fact that the evets {Lie l i strip i} ad {Lie l i strip j} are disjoit. The for ay pair of strips (i, j), the probability that either strip i or strip j cotais ay lie segmets is (1 2/) 1/e 2. Let X i be the idicator radom variable takig value 1 if strip i cotais o lie segmets ad takig the value 0 otherwise. Note that EX i e 1, Var(X i ) e 1 (1 e 1 ), ad Cov(X i, X j ) = E [X i X j ] E [X i ] E [X j ] = (1 2/) (1 1/) 2 e 2 e 2 + o(1) = o(1). The, lettig X = i=1 X i be the umber of strips that do t cotai ay lie segmets, ad otig that Chebyshev s iequality yields: Var(X) = Var(X i ) + Cov(X i, X j ) i i j ( ( 1 1 1 ) ) + o(1) + ( 1)o(1) e e + o(1), Pr [ X EX 1 ] 100 Var(X) 10 + o(1) 100 0.
Therefore, with probability tedig to 1, ( X 1 + 1 ) 10 + o(1) e Hece, Y = X /2 with probability tedig to 1. This proves the claim. We briefly remark that it is possible, with probability tedig to 1, to fid Ω ɛ ( ) o-crossig segmets 1 of legth greater tha 1 ɛ. The basic idea is to modify the above proof so that the left (right) edpoit of each lie segmet is withi distace ɛ/2 of the left (right) edge of Q. We omit the details i the iterest of brevity. Claim 2: With probability tedig to 1 as, N(L ) 15. From [1], there exists a absolute costat c such that for ay 2k poits i the plae, the umber of o-crossig left-right 2 perfect matchigs is upper-bouded by c 29 k. Cosider ay realizatio of lie segmets i the plae ad further cosider the 2k (k left ad k right) edpoits correspodig to ay subset S cosistig of k lie segmets. Coditioed o the locatios of the left ad right edpoits, every left-right perfect matchig of these 2k poits is equally likely, ad thus the probability that these k segmets are o-crossig is upper bouded by: Pr [S is o-crossig] c 29k sice there are left-right perfect matchigs o the 2k edpoits. Stirlig s formula states lim k 2πk ( k e ) k = 1, 2. ad thus c 29 k ( ) k 29 e o(1). k Recallig the crude upper boud ( ) ( k e ) k, k a uio boud gives: ( ) c 29 k Pr[ k o-crossig segmets] k ( 29 e 2 o(1) Lettig k = 15, we have Pr[ 15 ( 29 e 2 o-crossig segmets] o(1) 15 2 k 2 o(1) (.96) 15 0. This proves the claim ad completes the proof of Theorem 1. ) k ) 15 1 A fuctio f() is said to be Ω ɛ(g()), if there exists a iteger 0 ad a costat c ɛ depedig oly o ɛ so that f() c ɛg() for all > 0. 2 A left-right perfect matchig distiguishes betwee left edpoits ad right edpoits i edges. I other words, a edge is oly allowed to match a left edpoit to a right edpoit.
Proof of Theorem 2: We claim that, asymptotically almost surely, there exists a subset of o-crossig lie segmets U L satisfyig l > 1, (1) 7 where l deotes the legth of lie segmet l. Ideed, cosider the o-crossig set of lie segmets costructed i the proof of the lower boud of Theorem 1. Deote this set U. Let the edpoits of a lie segmet l U be give i Cartesia form: (x 1, y 1 ) ad (x 2, y 2 ). Note that l x 1 x 2 := L l, where the radom variable L l is idepedet from the evet {l U}, sice this evet oly depeds o y 1 ad y 2. Moreover, {L l } is a set of idepedet idetically distributed radom variables, each with expectatio 1/3 ad fiite variace. Therefore, by the weak law of large umbers ( ) 1 Pr L l 1 U 3 > 1 0. 21 Some basic algebra combied with the fact that U /2 a.a.s. proves the claim. The estimate give i (1) is essetially the best possible. Ideed, the upper boud of Theorem 1 combied with the fact that l 2 yields l < 22 for ay o-crossig subset U a.a.s. This completes the proof of the theorem. B. Extesio to d-disjoit Lie Segmets Two apparetly stroger results ca be deduced via a slight modificatio of the previous proofs. We state the results ad sketch the proofs i this sectio. First, defie the distace betwee two lie segmets l 1, l 2 L as the miimum distace betwee ay poit i l 1 ad ay poit i l 2. Formally, d(l 1, l 2 ) := if x y, x l 1,y l 2 where x y is the Euclidea distace betwee poits x, y Q. A set U of lie segmets is said to be d-disjoit if all pairs of lie segmets i U are at least distace d apart. The N d (L ) = max { U : d(l 1, l 2 ) > d for all l 1, l 2 U} is the size of the largest d-disjoit subset of lie segmets i L. Similarly, defie { } L d (L ) = max l : d(l 1, l 2 ) > d for all l 1, l 2 U. Note that N(L ) = N 0 (L ) ad L(L ) = L 0 (L ). With these defiitios i had, we state our last two results: Theorem 3: Asymptotically almost surely, N d() (L ) = Θ( ) if d() = O( 1/4 ). Theorem 4: Asymptotically { almost surely: Θ(1/d()) if d() = Ω(1/ ) L d() (L ) = Ω( ) if d() = o(1/ ) L d (L ) < 9/d.
Proof of Theorem 3: Repeat the costructive part i the proof of Theorem 1, except divide Q ito squares of size 1/4 1/4. Asymptotically almost surely, this yields at least /2 disjoit squares cotaiig lies. Sice d() = O( 1/4 ), we ca fid a costat fractio of the boxes cotaiig lies that are pairwise distace d() apart. This shows that we ca fid Ω( ) lie segmets which are d()-disjoit. Sice N d() (L ) is o-icreasig i d(), the upper boud of Theorem 1 proves the coverse part. Proof of Theorem 4: To prove the lower bouds, repeat the costructive part i the proof of Theorem 1. Assume first that d() = Ω(1/ ). Now, we ca partitio the strips (takig every (d() ) th strip) ito classes of roughly equal size so that all strips withi a class are separated by a distace greater tha d(). By the pigeo-hole priciple, at least oe of these classes cotais at least /(2 M ) d()-disjoit lie segmets a.a.s., where M is the umber of classes i the partitio. Some simple arithmetic reveals that M is roughly d(). Thus, there exists Ω(1/d()) d()-disjoit lie segmets a.a.s. If d() = o(1/ ), takig every other strip is sufficiet. Hece there exist Ω( ) d()-disjoit lie segmets i this case. I both cases, we ca apply the weak law of large umbers (as i the proof of Theorem 2) to give the desired lower estimate for L d() (L ). For the coverse, let U be a subset of d-disjoit lie segmets, ad for l U, geerate a rectagle of dimesio d l by wideig the segmet equally o both sides. Sice U is a subset of d-disjoit lie segmets, these rectagles must be pairwise disjoit. Sice the sum of the areas of the rectagles is at most slightly larger tha the area of Q (rectagles ca exted beyod the uit square), a simple volumetric argumet the shows that L d (L ) (1 + d) 2 /d. Without loss of geerality, d 2, thus L d (L ) < 9/d. III. CONCLUDING REMARKS We have give several cocetratio results for radom lie segmets i the uit square. First, we show that the maximum umber of o-crossig segmets behaves roughly like. Secod, we show that there is a o-crossig subset of lie segmets whose legths sum to roughly, which is the maximum possible. Fially, if lie segmets are required to be separated by some miimum distace d, we obtai a partial characterizatio of the previous two quatities. These results ca be applied to the study of user capacity ad trasport capacity i radom geometric graphs. A iterestig directio for further work would be to study N d() (L ) whe d() = ω( 1/4 ), ad L d() (L ) whe d() = o(1/ ) more thoroughly. Aother importat directio for future work would be to examie the behavior of a maximal subset of o-crossig (or d-disjoit) lie segmets that is chose greedily. We cojecture that a greedily selected o-crossig (or d-disjoit) maximal subset of segmets exhibits behavior which is order-idetical to a optimally chose subset. Experimetal evidece appears to support this cojecture. ACKNOWLEDGEMENTS This research was performed while the author was with Trellisware Techologies. The author gratefully ackowledges that, i the proof of Claim 1, the idea to divide Q ito strips is due to Michael Neely of USC. The author would also like to thak Tom Halford of Trellisware Techologies for the may coversatios from which these results traspired. REFERENCES [1] M. Sharir ad E. Welzl, O the umber of crossig-free matchigs, cycles, ad partitios, SIAM J. Comput. 36(3) (2006), 695-720. [2] N. Alo ad J. Specer. The probabilistic method (2ed). New York: Wiley-Itersciece (2000). ISBN 0-471-37046-0.