On the Number of CrossingFree Matchings, (Cycles, and Partitions)


 David Morgan
 1 years ago
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1 O the Number of CrossigFree Matchigs, (Cycles, ad Partitios Micha Sharir Emo Welzl Abstract We show that a set of poits i the plae has at most O(1005 perfect matchigs with crossigfree straightlie embeddig The expected umber of perfect crossigfree matchigs of a set of poits draw iid from a arbitrary distributio i the plae is at most O(924 Seeral related bouds are deried: (a The umber of all (ot ecessarily perfect crossigfree matchigs is at most O(1043 (b The umber of leftright perfect crossigfree matchigs (where the poits are desigated as left or as right edpoits of the matchig edges is at most O(538 (c The umber of perfect crossigfree matchigs across a lie (where all the matchig edges must cross a fixed halig lie of the set is at most 4 These bouds are employed to ifer that a set of poits i the plae has at most O(8681 crossigfree spaig cycles (simple polygoizatios, ad at most O(1224 crossigfree partitios (partitios of the poit set, so that the coex hulls of the idiidual parts are pairwise disjoit 1 Itroductio Let P be a set of poits i the plae A geometric graph o P is a graph that has P as its ertex set ad its edges are draw as straight segmets coectig the correspodig pairs of poits The graph is crossigfree if o pair of its edges cross each other, ie, ay two edges are ot allowed to share ay poits other tha commo edpoits Therefore, these are plaar graphs with a plae embeddig gie by this specific drawig We are iterested i the umber of crossigfree geometric graphs o P of seeral special types Specifically, we cosider the umbers tr(p, of triagulatios (ie, maximal crossigfree graphs, pm(p, of crossigfree perfect matchigs, sc(p, of crossigfree spaig cycles, ad, cfp(p, of crossigfree partitios 1 (partitios Work by Micha Sharir has bee supported by the USIsrael Biatioal Sci Foudatio, by NSF Grat CCR , by a grat from the Israeli Acad of Sci for a Ceter of Excellece i Geom Comp at Tel Ai Ui, ad by the Herma Mikowski MINERVA Ceter for Geometry at Tel Ai Ui School of Computer Sciece, Tel Ai Uiersity, Tel Ai 69978, Israel, ad Courat Ist of Math Sci, 251 Mercer Street, NYC, NY 10012, USA Ist Theoretische Iformatik, ETH Zürich, CH8092 Zürich, Switzerlad 1 Our research was triggered by M a Kreeld askig about the umber of crossigfree partitios, ad, i the same week, by M Hoffma ad Y Okamoto askig about the umber of P, so that the coex hulls of the parts are pairwise disjoit We are cocered with upper bouds for the umbers listed aboe i terms of History This problem goes back to Newbor ad Moser [25] i 1980 who ask for the maximal possible umber of crossigfree spaig cycles i a set of poits 2 ; they gie a upper boud of ! but cojecture that the boud should be of the form c, c a costat This was established i 1982 by Ajtai, Chátal, Newbor, ad Szemerédi [4], who show 3 that there are at most crossigfree graphs 4 Further deelopmets were maily cocered with deriig progressiely better upper bouds for the umber of triagulatios 5 [29, 13, 28], so far culmiatig i a 59 upper boud by Satos ad Seidel [27] i 2003 It compares to Ω(848, the largest kow umber of triagulatios for a set of poits, recetly deried by Aichholzer et al [1]; this improes a earlier lower boud of 8 /poly( gie by García et al [17] (We let poly( deote a polyomial factor i Eery crossigfree graph is cotaied i some triagulatio (with at most 3 6 edges Hece, a c boud for the umber of triagulatios yields a boud of c < (8c for the umber of crossigfree graphs o a set of poits; with c 59, this is at most 472 To the best of our kowledge, all upper bouds so far o the umber of crossigfree graphs of arious types are deried ia a boud o the umber of triagulatios, of crossigfree spaig paths of a poit set (motiated by their quest for good fixed parameter algorithms for the plaar Euclidea Traelig Salesma Problem i the presece of a fixed umber of ier poits [10]; see also [19] 2 Akl s work [5] appeared earlier, but it refers to the mauscript by Newbor ad Moser, ad improes a lower boud (o the maximal umber of crossigfree spaig cycles of theirs 3 This paper is famous for its Crossig Lemma, proed i preparatio of the sigly expoetial boud The lemma gies a upper boud o the umber of edges a geometric graph with a gie umber of crossigs ca hae 4 For motiatio they metio also a questio of D Ais about the maximum umber of triagulatios a set of poits ca hae 5 Iterest was also motiated by the related questio (from geometric modelig [29] of how may bits it takes to ecode a triagulatio
2 Figure 1: 6 poits with 12 crossigfree perfect matchigs, the maximum possible umber; see [3] for the maximum umbers for up to te poits: 3 for 4 poits, 12 for 6, 56 for 8, ad 311 for 10 albeit i more refied ways Oe idea is to exploit the fact that graphs of certai types hae a fixed umber of edges; eg, sice a perfect matchig has 2 edges, we readily obtai pm(p ( 3 6 /2 tr(p < [14] A short historical accout of bouds o sc(p, with refereces icludig [5, 12, 17, 18, 20, 25, 26], ca be foud at the web site [11] (see also [8] The best boud published is 337 tr(p It relies o a 337 boud o the umber of cycles i a plaar graph [6] I the course of our iestigatios, we showed that a graph with m edges ad ertices has at most ( m cycles; hece, a plaar graph has at most 3 cycles The R Seidel proided us with a argumet, based o liear algebra, that a plaar graph has at most 6 < 245 spaig cycles Crossigfree partitios fit ito Figure 2: Graph of a crossigfree partitio the picture, sice eery such partitio ca be uiquely idetified with the graph of edges of the coex hulls of the idiidual parts these edges form a crossigfree geometric graph of at most edges; see Fig 2 The situatio is better uderstood for special cofiguratios, for example for P a set of poits i coex positio (the ertex set of a coex go, where the Catala umbers C m := m+1( 1 2m m = Θ(m 3/2 4 m, m N 0, play a promiet role I coex positio tr(p = C 2 (the EulerSeger problem, cf [30, pg 212] for its history, pm(p = C /2 for ee ([16], cf [30], sc(p = 1, ad cfp(p = C ([7] Crossigfree partitios for poit sets i coex positio costitute a wellestablished otio because of its may coectios to other problems, probably startig with plaar rhyme schemes i Becker s ote [7], cf [30, Solutio to 619pp] The geeral case was cosidered by [9] (uder the ame of pairwise liearly separable partitios for clusterig algorithms They show that that the umber of partitios ito k parts is O( 6k 12 for k costat Uder the assumptio of geeral positio (o three poits o a commo lie it is kow [17] that the umber of crossigfree perfect matchigs o a set of fixed size is miimized whe the set is i coex positio (Recetly, Aichholzer et al [1] showed that ay family of acyclic graphs has the miimal umber of crossigfree embeddigs o a poit set i coex positio With little surprise, the same holds for spaig cycles, but it does ot hold for triagulatios [21, 2, 23] For crossigfree partitios, this is ope Results We show the followig bouds, for a set P of poits i the plae: pm(p = O(1005, sc(p = O(8681, ad cfp(p = O(1224 Also, the expected umber of perfect crossigfree matchigs of a set of poits draw iid from ay distributio i the plae (where two radom poits coicide with probability 0 is O(924 The boud o the umber of crossigfree perfect matchigs is deried by a iductie techique that we hae adapted from the method that Satos ad Seidel [27] used for triagulatios (the adaptio howeer is far from obious We the go o to derie improed bouds o the umber of crossigfree matchigs of arious special types: (a The umber of all (ot ecessarily perfect crossigfree matchigs is at most O(1043 (b The umber of leftright perfect crossigfree matchigs (where the poits are desigated as left or as right edpoits of the matchig edges is at most O(538 (c The umber of perfect crossigfree matchigs across a lie (where all the matchig edges must cross a fixed halig lie of the set is at most 4 Fially, we derie upper bouds for the umbers of crossigfree spaig cycles ad crossigfree partitios of P i terms of the umber of certai types of matchigs of certai poit sets P that are costructed from P This yields the bouds as stated aboe We summarize the state of affairs i Table 1, (icludig lower bouds proofs are omitted here I work i progress, we are curretly refiig a tailored aalysis for spaig cycles ad trees, where the bouds ow stad at O(79 ad O(296, respectiely tr pm sc cfp P : 59 [27] P : 848 [1] 3 [17] 464 [17] 523 ma lrpm alpm rdpm P : P : Table 1: Etries c i the upper boud rows stad for O(c, ad etries c i the lower boud rows for Ω(c /poly(, where := P ma stads for all crossigfree matchigs, lrpm for perfect leftright crossigfree matchigs, alpm for perfect crossigfree matchigs across a lie, ad rdpm for the expected umber of perfect crossigfree matchigs of a set of iid poits
3 2 Matchigs: The Setup ad a Recurrece Let P be a set of poits i the plae i geeral positio, o three o a lie, o two o a ertical lie This is o costrait whe it comes to upper bouds o pm(p A crossigfree matchig M is a collectio of pairwise disjoit segmets whose edpoits belog to P Each poit of P is either matched, if it is a edpoit of a segmet of M, or isolated, otherwise The umber of matched poits is always ee If 2m poits are matched ad s poits are isolated, we call M a crossigfree m matchig or (m, smatchig We hae = 2m + s For m R we deote by ma m (P the umber of crossigfree matchigs of P with m segmets (this umber is 0 uless m {0, 1,, 2 }, ad by ma(p the umber of all crossigfree matchigs of P (ie ma(p = m ma m(p Recall pm(p = ma /2 (P Let M be a crossigfree (m, smatchig o a set P of = 2m + s poits, as aboe The degree d(p of a poit p P i M is defied as follows It is 0 if p is isolated i M Otherwise, if p is a left (resp, right edpoit of a segmet of M, d(p is equal to the umber of isible left (resp, right edpoits of other segmets of M, plus the umber of isible isolated poits; isible meas ertically isible from the relatie iterior of the segmet of M that has p as a edpoit Thus p ad the other edpoit of the segmet are ot couted i d(p See Fig 3 Each left (resp, right ed PSfrag replacemets poit u i M ca cotribute at u most 2 to the degrees of other poits: 1 to each of the left (resp, right edpoits of the segmets lyig ertically aboe ad below u, if there exist such z w Figure 3: Degrees i a matchig: d(u = 2, d( = 5, d(w = 1, d(z = 2 segmets Similarly, each isolated poit u ca cotribute at most 4 to the degrees of other poits: 1 to each of the edpoits of the segmets lyig ertically aboe ad below u It follows that p P d(p 4m + 4s There are may segmets ready for remoal The idea is to remoe segmets icidet to poits of low degree i a (m, smatchig (poits of degree at most 3 or at most 4, to be specific We will show that there are may such poits at our disposal The, i the ext step, we show that segmets with a edpoit of low degree ca be reiserted i ot too may ways These two facts will be combied to derie a recurrece for the matchig cout For i N 0, let i = i (M deote the umber of matched poits of P with degree i i M Hece, i 0 i = 2m Lemma 21 Let, m, s N 0, with = 2m + s eery (m, smatchig of ay set of poits, we hae ( s, ( s Proof Let P be the uderlyig poit set We hae i 0 i i = p P d(p 4s + 4m = 4s + i 0 2 i Therefore, 0 4s + i 0 (2 i i For κ R +, we add κ times = s + i 0 i to both sides to get κ (4 + κs + i 0 (2 + κ i i (4 + κs + 0 i<2+κ (2 + κ i i We set κ = 2 for (21 ad κ = 3 for (22 There are ot too may ways of isertig a segmet Fix some p P ad let M be a crossigfree matchig which leaes p isolated Now we match p with some other isolated poit such that the oerall matchig cotiues to be crossigfree For i N 0, let h i = h i (p, P, M be the umber of ways that ca be doe so that p has degree i after its isertio Lemma 22 (23 (24 4h 0 + 3h 1 + 2h 2 + h 3 24, 5h 0 + 4h 1 + 3h 2 + 2h 3 + h 4 48 Proof Let l i = l i (p, P, M be the umber of ways we ca match the poit p as a left edpoit of degree i First, we claim that l 0 {0, 1} To show this, form the ertical decompositio of M by drawig a ertical segmet up ad dow from each (matched or isolated poit of P \{p}, ad exted these segmets util they meet a edge of M, or else, all the way to ifiity; see Fig 4 We call these ertical segmets walls i order to distiguish them from the segmets i the matchig We obtai a decompositio of the plae itopsfrag ertical replacemets trapezoids Let τ be the trapezoid cotaiig p (assumig geeral u positio, p lies i the iterior of τ See Fig 4 We moe from τ to the right through ertical walls to adjacet trapezoids util we reach a ertical wall that is determied τ p I Figure 4: Isertig a segmet at p; d(p = 1 after isertio by a poit that is either a left edpoit or a isolated poit (if at all we may make our way to ifiity whe p caot be matched as a left edpoit to ay poit, i which case l i = 0 for all i Note that up to that poit there was always a uique choice for the ext trapezoid to eter Eery crossigfree segmet with p as its left edpoit will hae to go through all of these trapezoids It coects either to (which ca happe oly if is isolated,
4 or crosses the ertical wall up or dow from The former case yields a segmet that gies p degree 0 I the latter case, will cotribute 1 to the degree of p So p, if a optio, is the oly possible segmet that lets p hae degree 0 as a left edpoit (p will ot be a optio whe it crosses some segmet, or whe is a left edpoit We will retur to this setup whe we cosider degrees 1, i which case acts as a bifurcatio poit Before doig so, we first itroduce a fuctio f It maps eery oegatie real ector (λ 0, λ 1,, λ k of arbitrary legth k + 1 N to the maximum possible alue the expressio (25 λ 0 l 0 + λ 1 l λ k l k ca attai (for ay isolated poit i ay matchig of ay fiite poit set of ay size We hae already show that f(λ λ for λ R + 0 We claim that for all (λ 0, λ 1,, λ k (R + 0 k+1, with k 1, we hae { λ0 + f(λ (26 f(λ 0, λ 1,, λ k max 1,, λ k, 2f(λ 1,, λ k Assumig (26 has bee established, we ca coclude that f(1 1, f(2, 1 3, f(3, 2, 1 6, ad f(4, 3, 2, 1 12; that is, 4l 0 + 3l 1 + 2l 2 + l 3 12 ad the first iequality of the lemma follows, sice the same boud clearly holds for the case whe p is a right edpoit The secod iequality follows similarly from f(5, 4, 3, 2, 1 24 It remais to proe (26 Cosider a costellatio with a poit p that realizes the alue of f(λ 0, λ 1,, λ k We retur to the setup from aboe, where we hae traced a uique sequece of trapezoids from p to the right, till we ecoutered the first bifurcatio poit (if does ot exist the all l i aish Case 1: is isolated We kow that λ 0 l 0 λ 0 If we remoe from the poit set, the eery possible crossigfree segmet emaatig from p to its right has its degree decreased by 1 Therefore, λ 1 l λ k l k f(λ 1,, λ k, so the expressio (25 caot exceed λ 0 + f(λ 1,, λ k i this case Case 2: is a matched left edpoit The λ 0 l 0 = 0 (that is, we caot coect p to Possible crossigfree segmets with p as a left edpoit are discrimiated accordig to whether they pass aboe or below We first cocetrate o the segmets that pass aboe ; we call them releat segmets (emaatig from p Let l i be the umber of releat segmets that gie p degree i We carefully remoe isolated poits from P \ {p} ad segmets with their edpoits from the matchig M (eetually also the segmet of which is a left edpoit, so that i the ed all releat segmets are still aailable ad each oe, if iserted, makes the degree of p exactly 1 uit smaller tha its origial alue (this deletio process may create ew possibilities for segmets from p That will show λ 1 l λ k l k f(λ 1,, λ k The same will apply to segmets that pass below, usig a symmetric argumet, which gies the boud of 2f(λ 1,, λ k for (25 i this secod case The remoal process is performed as follows We defie a relatio o the set whose elemets are the edges of M ad the sigleto sets formed by the isolated poits of P \ {p}: a b if a poit a a is ertically isible from a poit b b, with a below b As is well kow (cf [15, Lemma 114], is acyclic Let + deote the trasitie closure of, ad let deote the trasitie reflexie closure of Let e be the segmet with as its left edpoit, ad cosider a miimal elemet a with a + e Such a elemet exists, uless e itself is a miimal elemet with respect to a is a sigleto: So it cosists of a isolated poit; with abuse of otatio we also deote by a the isolated poit itself a caot be a poit to which p ca coect with a releat edge Ideed, if this were the case, we add that edge e = pa ad modify to iclude e too; more precisely, ay pair i that ioles a is replaced by a correspodig pair iolig e, ad ew pairs iolig e are added (clearly, the relatio remais acyclic ad all pairs related uder + cotiue to be so related after e is icluded ad replaces a See Fig 5(a We hae e e (sice, by assumptio, the left edpoit of e is ertically isible below e, ad e + e (sice the right edpoit a of e satisfies a + e a cotradictio With a similar reasoig we ca rule out the possibility that a cotributes to the degree of p whe matched ia a releat edge pq Ideed, if this were the case, let e be the segmet directly aboe a, which is the first lik i the chai that gies a + e, ie, a e e (e must exist sice a + e After addig pq with a cotributig to its degree, we hae either a pq ad pq e (see Fig 5(b, or we hae pq a (see Fig 5(c I the former case, we hae a pq e e pq cotradictig the acyclicity of I the latter case, we hae pq a + e pq, agai a cotradictio So if we remoe a, the all releat edges from p remai i the game ad the degree of each of them (ie, the degree of p that the edge iduces whe iserted does ot chage p e e a p e e (a (b (c Figure 5: (a The poit a caot be coected to p ia a releat edge (b,c a caot cotribute from below (i (b or from aboe (i (c to the degree of p whe a releat edge pq is iserted a q p e e a q
5 a is a edge: It caot obstruct ay isolated poit or left edpoit below it from cotributig to the degree of a releat edge pq aboe (because a is miimal with respect to If a obstructs a cotributio to a releat edge pq from aboe, the we add pq, thus pq a which, together with PSfrag a replacemets + e ad e pq, cotradicts the acyclicity of (Fig 6 Agai, we ca remoe a without ay chages to releat possible edges from p We keep successiely re a moig elemets util e is miimal with respect to Note p q e that so far all the releat edges from p are still possible, ad the degree of p that ay of them iduces whe iserted has ot Figure 6: Edge a caot obstruct a poit from cotributig from aboe to the degree of p whe a releat edge pq is iserted chaged Now we remoe e with its edpoits This caot clear the way for ay ew cotributio to the degree of a releat edge I fact, ay such degree decreases by exactly 1 because disappears The claim is show, ad the proof of the lemma is completed Deriig a recurrece Lemma 23 Let, m N 0, such that m 2 ad s := 2m For eery set P of poits, we hae 12(s+2 3s ma m 1 (P if s < 3 ma m (P, ad 16(s+2 7s/3 ma m 1(P if s < 3 7 Let us ote right away that the first iequality supersedes the secod for s < 5 (ie m > 2 5, while the secod oe is superior for s > 5 Proof Fix the set P, ad let X ad Y be the sets of all crossigfree mmatchigs ad (m 1matchigs, respectiely, i P Let us cocetrate o the first iequality We defie a edgelabeled bipartite graph G o X Y as follows: Gie a mmatchig M, if p is a edpoit of a segmet e M ad d(p 3, the we coect M X to the (m 1matchig M \ {e} Y with a edge labeled (p, d(p; d(p is the degree label of the edge Note that M ad M \ {e} ca be coected by two (differetly labeled edges, if both edpoits of e hae degree at most 3 For 0 i 3, let x i deote the umber of edges i G with degree label i We hae (2 6s X 4x 0 + 3x 1 + 2x 2 + x 3 24(s + 2 Y }{{} }{{} ma m(p ma m 1(P The first iequality is a cosequece of iequality (21 of Lemma 21 The secod iequality is implied by iequality (23 i Lemma 22, as follows For a fixed (m 1matchig M i P, cosider a edge of G that is icidet to M ad is labeled by (p, i (if there is such a edge The p must be oe of the s+2 isolated poits of P (with respect to M, ad there is a way to coect p to aother isolated poit i a crossigfree maer, so that p has degree i i the ew matchig Hece, the cotributio by p ad M to the sum 4x 0 +3x 1 +2x 2 +x 3 is at most 24, by iequality (23 i Lemma 22, ad the right iequality follows The combiatio of both iequalities yields the first iequality the lemma By cosiderig edpoits up to degree 4 (istead of 3, we get the secod iequality For m, N 0, let ma m ( be the maximum umber of crossigfree mmatchigs a set of poits ca hae Lemma 24 For s, m, N 0, with = 2m + s, ma 0 (0 = 1, s ma m( 1, for s 1, 12(s+2 ma m ( 3s ma m 1 (, for s < 3, 16(s+2 7s/3 ma m 1(, for s < 3 7 Proof ma 0 (0 = 1 is triial The first of the three iequalities is implied by s ma m (P = p P ma m(p \ {p} ma m ( 1, for ay set P of poits The secod ad third iequalities follow from Lemma 23 3 Solig a Recurrece We derie a upper boud for a fuctio G G λ,µ : N 2 0 R +, for a pair of parameters λ, µ R +, µ 1, which satisfies (with s := 2m G(0, 0 = 1, { s G(m, 1, for s 1, (37 G(m, λ(s+2 µs G(m 1,, for s < µ The recurrece i Lemma 24 implies that a upper boud o G 12,3 (m, seres also as a upper boud for ma m (, ad the same holds for G 16,7/3 (m, We will see how to best combie the two parameter pairs, to obtai ee better bouds for ma m ( Later, we will ecouter other istaces of this recurrece, with other alues of λ ad µ We diide by λ m µ m The (37 becomes G(m, G(m, 1 λ m µ m µs λ m µ, for s 1, 1 m µ(s+2 µs G(m 1, λ m 1 µ m+1, for s < µ G(m, We set H(m, = H µ (m, := λ m µ Therefore, m still with the coetio s := 2m ad the assumptio µ 1, we hae (ote idepedece of λ H(0, 0 = 1, (38 H(m, { µs H(m, 1, for s 1, µ(s+2 µs H(m 1,, for s < µ
6 Lemma 31 Let m, N 0, with m 2 The H(m, ( m Proof H(0, 0 = 1 ( 0 0 forms the basis of a proof by iductio o ad m For all N 0, H(0, µ 1 = ( 0 follows, sice µ 1 Let 1 m m 2 If m µs the s µ < µ Hece, the secod iequality i (38 ca be applied, after which the first iequality ca be applied Hece, H(m, µ(s+2 µs µ(s+2 µs µ(s+2 m H(m 1, ( 1 m 1 = ( m H(m 1, 1 Otherwise, m > µs holds, which esures µs > m 0, ie, s 1 We ca therefore employ the first iequality of (38, ad obtai ( 1 m m ( H(m, µs H(m, 1 < = m By expadig alog the first iequality for a while before employig Lemma 31, we get H(m, µs k+1 µ(s k+1 H(m, k ( 1 k 1 ( k µ k i=0 i s i m = 1 ( k ( k (39 µ k ( k s m = 1 m ( (310 2m, for N0 k s ( 2m µ k ( m k m Whe we stop this uwidig of the recurrece, we could hae alteratiely proceeded oe ( more step, ad upper boud H(m, k by k k 1 µ(s k m, proided k < s As log as this expressio is smaller tha ( k m, we should ideed hae expaded further That is, we expad as log as ( < k k µ(s k ( k 1 m m k < µs+m µ 1 = m ( µ 1 = m ρ, for ρ := µ 1 That is, the best choice of k i (39 is (311 k = m m ρ = ρ I fact, if this suggested alue of k is egatie (or if ρ = 0, we should ot expad at all Istead, we try to expad alog the secod iequality of (38, to get (312 H(m, ( s 2 +k k ( 2µ s 2 k (, m k for N 0 k < 2µ s = m µ 1 2µ + 1; we employ here the usual geeralizatio of biomial coefficiets ( a to a R, amely, ( a k := a(a 1 (a k+1 k! k Rather tha optimizig the alue of k at which we stop the uwidig of the secod recurrece iequality of (38, we approximate it by (313 k = m µ 1 = m ρ, ad ote that it lies i the allowed rage, proided it is positie Whe m = ρ, both alues suggested for k i (311 ad (313 are 0, which idicates that we hae to cotet ourseles with the boud ( m from Lemma 31 Otherwise, it is clear which way to expad, sice m < ρ m ρ 0 ad m > ρ m ρ 0 We are ow ready for a improed boud For that we substitute k i (39 accordig to (311, ad i (312 accordig to (313 Lemma 32 Let m, N 0, where 2m, ad set ρ := µ 1 If m ρ, the 1 ( H µ (m, m/ρ ( m/ρ µ m/ρ ( m/ρ 2m m ad for m > ρ, we hae H µ (m, ( ρ 2 m ρ ( ( m 2 (1 µ 1 m ρ Thus, G λ,µ (m, G λ,µ (m, with ( λ m µ m/ρ m G λ,µ (m, := λ m µ m for m ρ m/ρ ( 2m m/ρ ρ, ad ( 2 ρ m ρ ( m 2 (1 1 µ m ρ for m > ρ ( m/ρ, m (, ρ Next we work out a umber of properties of the upper boud G λ,µ Estimates up to a polyomial factor I the followig deriatios, we sometimes use to deote equality up to a polyomial factor i We will frequetly use the followig estimate (implied by Stirlig s formula, cf [22, Chapter 10, Corollary ( α ( α α β β β (α β, for α, β R, α β α β 0 Big m We ote that for m 1 ρ 9] ( α β G λ,µ (m, = λ(s+2 µs (with s := 2m Sice λ(s+2 µs G λ,µ(m 1, < 1 s < 2λ λ+µ, the fuctio G λ,µ (m, maximizes m > (λ+µ 1+2λ 2(λ+µ for itegers m i the rage ρ m 2 at (314 m := (λ+µ 1+2λ 2(λ+µ = 2 2λ 2(λ+µ, uless this alue is ot i the proided rage Howeer, m 2 uless is ery small ( < 2λ Ad m ρ uless λ < µ 1
7 Small m With the idetity idicated i (310 we hae, for m ρ, that G ca also be writte as( (315 G λ,µ (m, = λ m µ m/ρ m ( 2m m ( m/ρ m m 2m m (4λ(µ 1 m( 2m This boud peaks (up to a additie costat at m := λ(µ 1 Note that m ρ for λ µ λ(µ 1 We summarize, that G λ,µ (m, attais its maximum up to a poly(factor oer m at { m if λ µ 1, ad (316 m = m otherwise I all applicatios i this paper we hae λ > µ 1, so the peak occurs at m 4 Matchig Bouds 41 Perfect Matchigs For perfect matchigs we cosider the case where is ee, m = 2, ad s = 0 We ote that i this case m/ = 1/2 > ρ, for ay alue of µ Hece, the secod boud of Lemma 32 applies We first calculate 2 2 (1 1 µ = 1 2µ, ad 2 ρ = 2 µ 1 1 = 2( Hece, G λ,µ ( 2, ( (λµ /2 ( = ( 1 = (λµ /2 2µ 1 2( 1 2( 1 2µ 1( µ 1 1 ( µ 1 2( µ 1 2µ( 2µ( 1 2µ ( µ 1 µ 1 ( µ (λ 1 2 (µ 1 µ 1 2µ (2µ 1 2µ µ Substitutig (λ, µ = (12, 3 ad (16, 7 3, as suggested by Lemma 24, we obtai the followig upper bouds for the umber of crossigfree perfect matchigs: ( ( G 12,3 2, = O(105129, ( G 16, 7 3 2, ( = O( While the secod boud is obiously superior, we remember that the recurrece with (λ, µ = (12, 3 is better for m > 2 5 (or s < 5 This obseratio leads to the followig better boud for P a set of poits ad for k = = 10, where we expad as i the first iequality of Lemma 23 ( k 1 pm(p ma /2 k (P i=0 12(2i+2 6i 4 k( /6 1 k G16,7/3 (/2 k, (2 20/21 3 2/7 5 1/ /14 = O( Perfect ersus all matchigs Recall from Lemma 23 that ma m (P 12(s+2 3s ma m 1(P Note that 12(s+2 3s < 1 for m > (ad i this rage is smaller tha the alteratie of the factor 12(s+2 3s fered i Lemma 23 That is, there are always fewer perfect matchigs tha there are matchigs More specifically, for sets P with := P ee, ad for k = = , we hae k 1 pm(p = ma /2 (P = ( k ( /6 k i=0 12(2i + 2 6i 1 ma/2 k (P ma /2 k (P ( ( 4 /30 1 1/5 ( 4 4/5 /6 5 5 ma (P 5 ( = 2 1/3 5 1/6 ma (P Therefore, pm(p ( 2 1/3 5 1/6 ma(p poly( = O(09635 ma(p, ie i eery poit set there are expoetially (i more crossigfree matchigs tha there are crossigfree perfect matchigs 42 All Matchigs Our cosideratios i the deriatio of the boud for perfect matchigs imply the followig upper boud for matchigs with m segmets G 16,7/3 (m,, m 2 5, ma m (P G 12,3 (m, G 16,7/3( 2 5,, otherwise G 12,3( 2 5 To determie where this expressio maximizes,, we ote that G 16,7/3 does ot peak i its small m rage (m 4 11 sice 16 > (recall (316 I the big m rage, it peaks at roughly (see (314, which exceeds 2 5 Therefore, the maximum occurs whe G 12,3 comes ito play, which peaks at roughly 7 15 For that alue the upper boud ealuates to (2 13/21 3 2/7 5 3/ /14 =O( Summig up Theorem 41 For P a set of poits i the plae (1 pm(p ( 2 20/21 3 2/7 5 1/ /14 poly( = O( ( (2 pm(p 2 1/3 5 1/6 ma(p poly( = O(09635 ma(p (3 ma(p ( 2 13/21 3 2/7 5 3/ /14 poly( = O( Radom Poit Sets Let P be ay set of N N poits i the plae, o three o a lie, ad let r N with r N If R is a subset of P chose uiformly at radom from ( P r m µ 1 N = 4 11, the, for λ = 16, µ = 7 3, ad proided N, ad r 2m, we hae, usig (315, E[ma m (R] = ( 1 N ma m (R = r R ( P r ( N 2m r 2m ( N r ma m (P
8 ( N 2m ( N (4λ(µ 1 m 2m r 2m ( N poly(m r m (4λ(µ 1 m( ( r 2m = m ( r 2m We see that if we sample r poits from a large eough set, the the expected umber of crossigfree matchigs obseres for all m the upper boud deried for the rage of small m Suppose ow that, for ee, we sample iid poits from a arbitrary distributio, for which we oly require that two sampled poits coicide with probability 0 The we ca first sample a set P of N > 11 8 poits, ad the choose a subset of size uiformly at radom from the family of all subsets of this size We obtai a set R of iid poits from the gie distributio If P is i geeral positio, by the argumet aboe the expected umber of perfect crossigfree matchigs is at most ( /2 If P exhibits colliearities, we perform a small perturbatio yieldig a set P ad the subset R Now the boud applies to R, ad also to R sice a sufficietly small perturbatio caot decrease the umber of crossigfree perfect matchigs Theorem 42 For ay distributio i the plae for which two sampled poits coicide with probability 0, the expected umber of crossigfree perfect matchigs of iid poits is at most ( /2 poly( = O( LeftRight Perfect Matchigs Here we assume that P is partitioed ito two disjoit subsets L, R ad cosider bipartite matchigs i L R such that, for each edge of the matchig, its left edpoit belogs to L ad its right edpoit to R We modify the defiitio of the degrees of the poits: If p L is a matched to a poit i R, the d(p is equal to the umber of left edpoits plus the umber of rightlabeled isolated poits that are ertically isible from (the relatie iterior of e A symmetric defiitio holds for right edpoits (Ituitiely, a rightlabeled isolated poit q has to cotribute oly to the degrees of leftlabeled poits, because, whe we isert a right edpoit, it caot coect to q, ad it does ot matter whether its icidet edge passes aboe or below q; that is, q does ot cause ay bifurcatio i the ways i which p ca be coected Sice isolated poits cotribute ow oly 2 to degrees of edpoits, we hae p P d(p 4m + 2s The aalysis further improes, because whe we reisert a poit p L, say, the correspodig umbers h i must be equal to l i, sice p ca oly be the left edpoit of a matchig edge A similar improemet holds for poits q R Hece, we ca boud the sum 4h 0 + 3h 1 + 2h 2 + h 3 by 12, rather tha 24; similarly, we hae 5h 0 + 4h 1 + 3h 2 + 2h 3 + h 4 24 That is, we hae for (λ, µ the pairs (6, 2 ad (8, 5 3 aailable ( k 1 i=0 6(2i+2 4i We ifer a boud of G 8,5/3 ( 2 k,, for k = 6, implyig Theorem 43 Let P be a set of poits i the plae ad assume that the poits are classified as left edpoits or right edpoits The umber of leftright ( perfect crossigfree matchigs i P is at most 2 7/10 3 3/20 7 7/10 poly( = O( Matchigs Across a Lie Cosider ext the special case of crossigfree bipartite perfect matchigs betwee two sets of 2 poits each that are separated by a lie Here we ca obtai a upper boud that is smaller tha the oe i Theorem 43 Theorem 44 Let be a ee iteger The umber of crossigfree perfect bipartite matchigs betwee two separated sets of 2 poits each i the plae is at most C 2 /2 < 4 ; (C m is the mth Catala umber Proof Let L ad R be the gie separated sets Without loss of geerality, take the separatig lie λ to be the yaxis, ad assume that the poits of L lie to the left of λ ad the poits of R lie to its right Let M be a crossigfree perfect bipartite matchig i L R For each edge e of M, let e L (resp, e R deote the portio of e to the left (resp, right of λ, ad refer to them as the left halfedge ad the right halfedge of e, respectiely We will obtai a upper boud for the umber of combiatorially differet ways to draw the left halfedges of a crossigfree perfect matchig i L R The same boud will apply symmetrically to the right halfedges, ad the fial boud will be the square of this boud I more detail, we igore R, ad cosider collectios S of 2 pairwise disjoit segmets, each coectig a poit of L to some poit o λ, so that each poit of L is icidet to exactly oe segmet For each segmet i S, we label its λedpoit by the poit of L to which it is coected The icreasig yorder of the λedpoits of the segmets thus defies a permutatio of L, ad our goal is to boud the umber of differet permutatios that ca be geerated i this way (I geeral, this is a strict upper boud o the quatity we seek We obtai this boud i the followig recursie maer Write m := L = 2 Sort the poits of L from left to right (we may assume that there are o ties they ca be elimiated by a slight rotatio of λ, ad let p 1, p 2,, p m deote the poits i this order Cosider the halfedge e 1 emaatig from the leftmost poit p 1 Ay other poit p j lies either aboe or below
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