New n Improve Spnning Rtios for Yo Grphs Luis Br Déprtement Informtique Université Lire e Bruxelles lrfl@ul..e Rolf Fgererg Deprtment of Computer Siene University of Southern Denmrk rolf@im.su.k Anré vn Renssen Shool of Computer Siene Crleton University nre@g.ss.rleton. Prosenjit Bose Shool of Computer Siene Crleton University jit@ss.rleton. Wh Loon Keng Deprtment of Computer Siene Lfyette College kengw@lfyette.eu Mirel Dmin Deprtment of Computing Sienes Villnov University mirel.min@villnov.eu Joseph O Rourke Deprtment of Computer Siene Smith College orourke@s.smith.eu Perouz Tslkin Sner Veronshot Shool of Siene n Shool of Computer Siene Engineering Crleton University Amerin University of sner@g.ss.rleton. Armeni perouz.tslkin@ul..e Ge Xi Deprtment of Computer Siene Lfyette College gexi@s.lfyette.eu ABSTRACT For set of points in the plne n fixe integer k > 0, the Yo grph Y k prtitions the spe roun eh point into k equingulr ones of ngle θ = 2π/k, n onnets eh point to nerest neighor in eh one. It is known for ll Yo grphs, with the sole exeption of Y 5, whether or not they re geometri spnners. In this pper we lose this gp y showing tht for o k 5, the spnning rtio of Y k is t most 1/(1 2 sin(3θ/8)), whih gives the first onstnt upper oun for Y 5, n is n improvement over the previous oun of 1/(1 2 sin(θ/2)) for o k 7. We further reue the upper oun on the spnning rtio for Y 5 from 10.9 to 2 + 3 3.74, whih flls slightly elow the lower oun of 3.79 estlishe for the spnning rtio of Θ 5 (Θ-grphs iffer from Yo grphs only in the wy they selet the losest neighor in eh one). This is the first suh seprtion etween Yo n Θ-grph with the sme Reserh supporte in prt y NSERC. Reserh supporte y NSF grnt CCF-1218814. Permission to mke igitl or hr opies of ll or prt of this work for personl or lssroom use is grnte without fee provie tht opies re not me or istriute for profit or ommeril vntge n tht opies er this notie n the full ittion on the first pge. Copyrights for omponents of this work owne y others thn the uthor(s) must e honore. Astrting with reit is permitte. To opy otherwise, or repulish, to post on servers or to reistriute to lists, requires prior speifi permission n/or fee. Request permissions from Permissions@m.org. SoCG 14, June 8 11, 2014, Kyoto, Jpn. Copyright is hel y the owner/uthor(s). Pulition rights liense to ACM. ACM 978-1-4503-2594-3/14/06...$15.00. numer of ones. We lso give lower oun of 2.87 on the spnning rtio of Y 5. Finlly, we revisit the Y 6 grph, whih plys prtiulrly importnt role s the trnsition etween the grphs (k > 6) for whih simple inutive proofs re known, n the grphs (k 6) whose est spnning rtios hve een estlishe y omplex rguments. Here we reue the known spnning rtio of Y 6 from 17.6 to 5.8, getting loser to the spnning rtio of 2 estlishe for Θ 6. 1. INTRODUCTION The omplete Eulien grph efine on point set S in the plne is the grph with vertex set S n eges onneting eh pir of points in S, where eh ege xy hs s weight the Eulien istne xy etween its enpoints x n y. Although this grph is useful in mny ifferent ontexts, its min isvntge is tht it hs qurti numer of eges. As suh, muh effort hs gone into the evelopment of vrious methos for onstruting grphs tht pproximte the omplete Eulien grph. Wht oes it men to pproximte this grph? One stnr pproh is to onstrut spnning sugrph with fewer eges (typilly liner) with the itionl property tht every ege e of the omplete Eulien grph is pproximte y pth in the sugrph whose weight is not muh more thn the weight of e. This gives rise to the notion of t-spnner. A t-spnner of the omplete Eulien grph is spnning sugrph with the property tht for eh pir of verties x n y, the weight of shortest pth in the sugrph etween x n y is t most t 1 times xy. The spnning rtio is the smllest t for whih the sugrph is t-spnner. Spn-
ners fin mny pplitions, suh s pproximting shortest pths or minimum spnning trees. For omprehensive overview of geometri spnners n their pplitions, we refer the reer to the ook y Nrsimhn n Smi [12]. One of the simplest wys of onstruting t-spnner is to first prtition the plne roun eh vertex x into fixe numer of ones 1 n then eges onneting x to losest vertex in eh one. Intuition suggests tht this woul yiel grph whose spnning rtio epens on the numer of ones. Inee, this is one of the first pproximtions of the omplete Eulien grph, referre to s Yo grphs in the literture, introue inepenently y Flinhugh n Jones [11] n Yo [13]. We enote the Yo grph y Y k where k is the numer of ones, eh hving ngle θ = 2π/k. Yo use these grphs to simplify omputtion of the Eulien minimum spnning tree. Flinhugh n Jones stuie their grph theoreti properties. Neither of them tully prove tht they re t-spnners. To the est of our knowlege, the first proof tht Yo grphs re spnners ws given y Althöfer et l. [1]. They showe tht for every t > 1, there exists k suh tht Y k is t-spnner. It ppers tht some form of this result ws known erlier, s Clrkson [8] lrey remrke in 1987 tht Y 12 is 1 + 3-spnner, leit without proviing proof or referene. Bose et l. [6] provie more speifi oun on the spnning rtio, y showing tht for k > 8, Y k is geometri spnner with spnning rtio t most 1/(os θ sin θ). This ws lter strengthene to show tht for k > 6, Y k is 1/(1 2 sin(θ/2))-spnner [4]. Dmin n Ruonis [9] showe tht Y 6 is 17.64-spnner, n Bose et l. [5] showe tht Y 4 is 663-spnner. For k < 4, El Moll [10] showe tht there is no onstnt t suh tht Y k is t-spnner. This leves open only the question of whether Y 5 is onstnt spnner. In this pper we lose this gp y showing tht for o k 5, the spnning rtio of Y k is t most 1/(1 2 sin(3θ/8)). This gives the first onstnt upper oun for Y 5 n implies tht Y k is onstnt spnner for ll k 4. For o k 7, our result lso improves on the previous oun of 1/(1 2 sin(θ/2)). A more reful nlysis llows us to reue the upper oun on the spnning rtio of Y 5 from 10.9 to 2 + 3 3.74. We lso give lower oun of 2.87 on the spnning rtio of Y 5. This omplements reent result on the spnning rtio of Θ 5, whih iffers from Y 5 only in the istne mesure it uses to selet the losest neighor in eh one: inste of Eulien istne, it projets eh vertex on the isetor of the one n selets the vertex with the losest projetion. Bose et l. [7] showe tht Θ 5 hs spnning rtio in the intervl [3.79, 9.96]. Beuse our upper oun of 3.74 on the spnning rtio of Y 5 is slightly lower thn the lower oun of 3.79 on the spnning rtio of Θ 5, this result estlishes the first seprtion etween the spnning rtio of Yo n Θ-grphs. For ll other k 4, it is unler whih of Θ k or Y k hs etter spnning rtio. Finlly, we revisit the Y 6 grph, whih plys prtiulrly importnt role s the trnsition etween the grphs (k > 6) for whih simple inutive proofs re known, n the grphs (k 6) whose est spnning rtios re estlishe y omplex rguments. Here we reue the known spnning rtio of Y 6 from 17.64 to 5.8, thus moving towr the spnning rtio of 2 estlishe for Θ 6 [3]. In ontrst to Y 5, we lso present 1 The orienttion of the ones is the sme for ll verties. lower oun of 2 on the spnning rtio of Y 6, showing tht it n never improve upon Θ 6 in this regr. 2. SPANNING RATIO OF Y k, FOR ODD k In this setion we stuy the spnning properties of the Yo grphs Y k efine on plne point set S y n o numer of ones k 5, eh of ngle θ = 2π/k. For k = 5 in prtiulr, this is the first result showing tht Y 5 is onstnt spnner. For o vlues k > 5, we improve the urrently known oun on the spnning rtio of Y k. We strt with few efinitions. For fixe k, let Q i() e the hlf-open one of ngle 2π/k with pex, inluing the ngle rnge [i, i + 1) 2π/k, for i = 0,..., k 1, where ngles re mesure ounterlokwise from the positive x-xis. The irete grph Y k inlues extly one irete ege from to losest point in Q i(), for eh i = 0,..., k 1. If there re severl eqully-losest points within Q i(), then ties re roken ritrrily. The grph Y k is the unirete version of Y k. For ny two points, S, let p(, ) enote the length of shortest pth in Y k from to. α Figure 1: If α is smll, there is lose reltion etween n. Lemm 1. Given three points,, n, suh tht n α < π, then (1 2 sin(α/2)). Proof. Let e the point on suh tht = (see Fig. 1). Sine forms n isoseles tringle, = 2 sin( /2) 2 sin(α/2). Now, y the tringle inequlity, + 2 sin(α/2) + = (1 2 sin(α/2)). Theorem 2. For ny o integer k 5, Y k hs spnning rtio t most t = 1/(1 2 sin(3θ/8)). Proof. Let, S e n ritrry pir of points. We show tht there is pth in Y k from to no longer thn t. For simpliity, let Q() enote the one with pex tht ontins, n let Q() enote the one with pex tht ontins. Rotte the point set S suh tht the isetor of Q() is in the iretion of the positive y-xis, s epite in Fig. 2. Assume without loss of generlity tht lies to the right of this isetor; the se when lies to the left of this isetor is symmetri. Let α e the ngle forme y the segment with the isetor of Q(), n let β e the ngle forme y with the isetor of Q(). Sine k is o, the isetor of Q() is
α Figure 2: Sine opposite ones re not symmetri, either α or β is smll. prllel to the right ounry of Q(). Hene, we hve tht α = θ/2 β. Assume without loss of generlity tht α is the smller of these two ngles (if not, we exhnge the roles of n ). It follows tht α θ/4. Our proof is y inution on the istne. In the se se is miniml mong ll istnes etween pirs of points, whih mens tht there is no point Q() tht is stritly loser to thn. Therefore either Y k, in whih se our proof for the se se is finishe, or there is point Q() suh tht = n Y k. In this ltter se, sine α θ/4 n k 5, the ngle etween n is t most θ/2 + α 3θ/4 3/4 (2π/5) = 3π/10. This is less thn π/3, whih implies tht <. This ontrits our ssumption tht is miniml. It follows tht Y k n the se se hols. For the inutive step, let Q() e suh tht Y k. If oinies with, then p(, ) = n the proof is finishe. So ssume tht. Beuse is the losest vertex to in this one, n euse θ/2+α 3θ/4, we n pply Lem. 1 to erive (1 2 sin(3θ/8)) = /t, whih is stritly less thn. Thus we n use the inutive hypothesis on to etermine pth etween n of length ( p(, ) + t + t ) = t. t Applying this result to Y 5 yiels spnning rtio of 1/(1 2 sin(3π/20)) 10.868. This is the first known upper oun on the spnning rtio of Y 5 n fully settles the question of whih Yo grphs re spnners. Corollry 3. The grph Y k is spnner if n only if k 4. Next we lower the upper oun on the spnning rtio of Y 5 y tking loser look t ll fesile onfigurtions. Theorem 4. The grph Y 5 hs spnning rtio t most 2 + 3 3.74. Here we lso use inution on the pirwise istnes etween pir of points in S. Consier the sme onfigurtion use in the proof of Thm. 2: Q() n Q() re points in S, n we seek short pth from n ; the isetor of Q() is ligne with the positive y-xis, n β lies to the right of this isetor; α n β re ngles s in Fig. 2, with α β. The ses where is miniml (se se) or Y 5 re s isusse in the proof of Thm. 2. So let Q() n Q() e in S suh tht Y 5 n Y5, n let φ =, n ψ = (see Fig. 3). Now, inste of pplying Lem. 1 for the mximum vlue of φ (s in the proof of Thm. 2), we pply Lem. 1 only for vlues φ θ or ψ θ, for some threshol ngle θ (to e etermine lter). These ses yiel spnning rtio of t 1/(1 2 sin(θ/2)). We hnle the remining ses ifferently, so for the reminer of the proof, we ssume tht φ > θ n ψ > θ. We ompute n ext vlue of θ shortly, ut for now we only nee tht θ/2 < θ < 3θ/4. This implies tht neither nor n lie to the right of, s this woul mke the orresponing ngle smller thn θ/2. First onsier the se where n interset. In this se, inste of iretly pplying n inutive rgument to either or, we oun the istne n use inution to show tht + t + t. To erive this oun, onsier the point suh tht = θ n = n the nloguously efine point (see Fig. 3). Let s e the intersetion point etween n. When n interset, the istne n e inrese y rotting towrs n towrs. Sine oth φ n ψ must e lrger thn θ, the worst se ours when φ = ψ = θ, leving n on the ounry of s. As is the longest sie of this tringle, it follows tht <. Using the ft tht the tringles s n s re similr n isoseles, we n ompute : = 2 s os θ = 2( s ) os θ ( = 2 2 os θ = (2 os θ 1) ) os θ Rell tht our im is to use inution on to otin short pth from to. We now ompute the spnning rtio t require for the inequlity +t + t to hol. By the inequlity ove, we hve tht + t + + t(2 os θ 1) +. This ltter term is oune ove y t for ny t 1/(1 os θ). So fr we erive two onstrints on t n θ: t 1/(1 2 sin(θ/2)) n t 1/(1 os θ). Beuse sin θ is inresing n os θ is eresing for ll vlues of θ uner onsiertion, we minimize t y hoosing θ suh tht 1/(1 2 sin(θ/2)) = 1/(1 os θ). This yiels θ = ros ( 3 1 ) 0.75 n t = 2 + 3 3.74. Now onsier wht hppens when one of or is short, uner some notion of short pture y the following lemm. Lemm 5. Let e tringle with ngle α = n longest sie. Let λ > 1 e rel onstnt. Then 2λ2 os α 2λ implies tht +λ λ. λ 2 1 Proof. The first inequlity ove implies λ > 1/ os α, otherwise woul e non-positive. By the lw of osines, = 2 + 2 2 os α. By sustituting this in the inequlity + λ λ, we see tht it only hols if 2λ2 os α 2λ, s stte y the lemm. λ 2 1
ψ s θ ψ φ θ φ () () () Figure 3: () Two verties of Y 5 with their losest verties. () The worst-se sitution when n ross. () The rottion to mximize when n o not ross. The only se left to onsier is when n re oth long, ut they o not interset. In this se, we gin seek to oun the istne. If we n show tht (2 os θ 1), we n pply the sme rgument s for the interseting se n we re one. Let e the point on the extension of with =, n let e the nloguous point on the extension of (see Fig. 3). If oes not interset, we n rotte wy from y inresing ψ. Similrly, if oes not interset, we n rotte wy from y inresing φ. Thus, the istne is mximize when φ + ψ is mximl, whih in our ontext hppens when φ + ψ = 3θ/2 = 3π/5. Note tht in most ses, rotting this fr moves the orresponing vertex pst the ounry of the one. But sine we re only trying to fin n upper oun, this is not prolem. Now let e the point on the line through with = 2t 2 os φ 2t, n let e the point on the line through with = 2t2 os ψ 2t. If lies on, Lem. 5 tells us tht + t t, whih is extly wht we nee. The only iffiulty is tht the lotion of hnge uring the rottion. But sine the rottion preserve n only inrese, the inequlity must hol for the onfigurtion efore the rottion s well. The sme rgument pplies for the se when lies on. The sitution where n lie on n, respetively, is hnle y the following lemm. s Proof. Assume without loss of generlity tht n = 1. Then 3π/10 3π/5 θ n θ 3π/10. Let e the point on the extension of with =, n let e the nloguous point on the extension of. Let s e the intersetion of n. Let e the point on the line through with = 2t2 os 2t, n let e the point on the line through with = 2t2 os 2t (see Fig. 4). Let 1 = 2t2 n 2 = 1. We erive sin(3π/5) (fter some lultions) Let s s = 1 sin, (1) = 1 sin(3π/5 ), (2) = 2 os(3π/5 ), (3) = 2 os. (4) x 1 = s sin = sin( + ) 2t2 os 2t, t 2 1 (5) x 2 = sin s = 1 sin( + ), (6) y 1 = s sin = sin( + ) 2t2 os 2t, t 2 1 (7) y 2 = s = 1 sin sin( + ). (8) Note tht the vlues of x 1 n y 1 oul e negtive if or lie pst s. Sustituting 1, 2, n (1) - (4) in the equlities ove yiels Figure 4: Illustrtion of Lem. 6 Lemm 6. Let,,, S. Let = n = suh tht > θ, > θ, n + = 3π/5. Let t = 2 + 3. If 2t 2 os 2t n 2t2 os 2t, then (2 os θ 1). x 1 = 2 os(3π/5 ) + 1 sin, (9) x 2 = 2 os(3π/5 ), (10) y 1 = 2 os 1 sin(3π/5 ), (11) y 2 = 2 os. (12) Rell tht 1 = 2t2, 2 = 1 sin(3π/5), n 3π/10 3π/5 θ. We verify the following:
2 x 1 = 2 2 sin(3π/5 ) + 1 os > 1.1 sin(3π/10) + 2.1 os(3π/5 θ) > 0, 2 x 2 = 2 sin(3π/5 ) > 0, 2 2 y 1 = 2 sin + 1 os(3π/5 ) 2 > 1.1 sin(3π/5 θ) + 2.1 os(3π/10) > 0, 2 y 2 = 2 sin > 0. 2 Therefore, y plugging in = 3π/10 or = 3π/5 θ s the lower- or upper-oun of into (9) - (12), we n verify the following rnges: 2 os(3π/10) + 1 sin(3π/10) x1, 2 os θ + 1 sin(3π/5 θ) x1, 2 os(3π/10) x2 2 os(3π/10) 1 sin(3π/10) y1, 2 os(3π/5 θ) 1 sin θ y1, 2 os(3π/10) y2, 2 os(3π/5 θ) y2. 2 os θ, (13) Speifilly, we n verify tht ( ) x 1 mx x2, y1, y 2, (14) whih implies (x 1 x 2 ) = x 1 x 2 > 0. By simply plugging in = 3π/10 into (5) n (6), we verify tht (x 1 x 2) > 0 when = 3π/10 n hene x 1 > x 2 for ll [3π/10, 3π/5 θ]. Similrly, we hve x 2 > 0 when = 3π/10, n hene y (13), x 2 > 0 for ll [3π/10, 3π/5 θ]. These together yiel x 1 > x 2 > 0. By the tringle inequlity, By (14), s + s = x 1 + y 1 = x 1 + y 1, s + s = x 1 + y 2 = x 1 + y 2, s + s = x 2 + y 1 x 1 + y 1, s + s = x 2 + y 2 x 1 + y 2. (x 1 + y 1 ) (x 1 + y 2 ) (x1) (x1) (y1) 0, (y2) 0. By plugging in = 3π/5 θ into (5), (7), n (8), one n esily verify tht x 1 + y 1 2 os θ 1 n x 1 + y 2 2 os θ 1 when is mximize. Therefore mx(x 1+ y 1, x 1+ y 2 ) 2 os θ 1 for ll [3π/10, 3π/5 θ], n hene mx(,,, ) 2 os θ 1 s require. This ompletes the proof for the upper oun. A proof estlishing the following lower oun n e foun in the full version of this pper [2]. Theorem 7. Y 5 hs spnning rtio t lest 2.87. 3. SPANNING RATIO OF Y 6 In this setion we fix k = 6 n show tht, for ny pir of points, S, p(, ) 5.8. We lso estlish lower oun of 2 for the spnning rtio of Y 6. Our proof is inutive n it relies on two simple lemms, whih we introue next. Let, S n let Y 6 e the ege from within the one tht inlues. The next two lemms will e relevnt in the ontext where we seek to oun p(, ) y pplying the inution hypothesis to p(, ). The si geometry is illustrte in Fig. 5. α x s h 1 Figure 5: Nottion for tringle. Here the imensions hve een normlize so tht = 1. Lemm 8. [Tringle] Let e lele s in Fig. 5, with, <, x = n s =. The rtio s/x is equl to some funtion t tht epens on α n β: r r s = t(α, β) = os(β/2) x os(α + β/2). (15) Proof. Normlize the tringle so tht = 1; this oes not lter the quntity we seek to ompute, s/x. Let = r to simplify nottion. Then x = 1 r n x 0 euse r = = 1. Note tht eh of the ngles n is stritly less thn π/2, euse n. Thus the projetion of onto is interior to the segment. Computing the ltitue h of in two wys yiels s sin α = r sin β. Also projetions onto yiel s os α + r os β = 1. Solving these two equtions simultneously yiels expressions for r n s s funtions of α n β: r = sin α sin α os β + os α sin β, s = sin β sin α os β + os α sin β Now we n ompute s/x = s/(1 r) s funtion of α n β. This simplifies to s lime. s x = os(β/2) os(α + β/2) β
The following lemm erives n upper oun on the funtion t(α, β) from Lem. 8, whih will e use in Thm. 10 to erive n optiml vlue for. Lemm 9. Let,, S stisfy the onitions of Lem. 8, n let t(α, β) e s efine in (15). Let (0, π/3) e fixe positive ngle. If α π/3, or β π/3, then t(α, β) t(π/3, π/3 ) = os(π/6 /2). sin(/2) Proof. The erivtive of t(α, β) with respet to α is t sin α + sin(α + β) = > 0. α 1 + os(2α + β) This mens tht, for fixe β vlue, t(α, β) rehes its mximum when α is mximum. Similrly, the erivtive of t(α, β) with respet to β is t β = sin α 2 os(α + β/2) > 0. 2 So for fixe vlue α vlue, t(α, β) rehes its mximum when β is mximum. Beuse, β. The sum of these two ngles is π α, therefore β π/2 α/2. This long with the erivtions ove implies tht, for fixe vlue α π/3, t(α, β) t(α, π/2 α/2) t(π/3, π/3 + /2) (we sustitute α = π/3 in this ltter inequlity). Next we evlute t(π/3, π/3 ) os(π/6 /2) sin(3/2) = t(π/3, π/3 + /2) os(π/6 + /2) sin(/2) > 1. It follows tht t(π/3, π/3 ) is mximl. We re now rey to prove the min result of this setion. Theorem 10. Y 6 hs spnning rtio t most 5.8. This result follows from the following lemm, with the vrile sustitute y the quntity 0 = 0.324 tht minimizes t(). (It n e esily verifie tht t() t(0.324) n t(0.324) < 5.8.) Lemm 11. Let (0, π/9) e stritly positive rel vlue. The grph Y 6 hs spnning rtio oune ove y { } os(π/6 /2) 2 t = t() = mx,. (16) sin(/2) 1 sin(2) sin(π/6+2) Proof. The proof is y inution on the pirwise istne etween pirs of points, S. Without loss of generlity let Q 0(). Bse se. We show tht, if is miniml, then Y 6 n so p(, ) =. If Y 6, then the lemm hols. So ssume tht Y 6; we will erive ontrition. Beuse Y6, there must e nother point Q 0() suh tht Y6 n =. Let α 1 n α 2 e the ngles tht n mke with the horizontl respetively. Beuse oth α 1, α 2 [0, π/3), neessrily α 1 α 2 < π/3. Thus < =, ontriting the ssumption tht is miniml. So in ft it must e tht Y 6, n the lemm is estlishe. Min ie of the inutive step. It hs lrey een estlishe tht Y 7 is spnner [4]; the setor ngles for Y 7 re 2π/7. The min ie of our inutive proof is to prtition the π/3-setors of Y 6 into peripherl ones of ngle, for some fixe (0, π/9), leving entrl setor of ngle π/3 2. (The -ones re the she regions in Fig. 6.) When n ege of Y 6 flls insie the entrl setor, inution will pply, euse n ege within the entrl setor mkes efinite progress towr the gol in tht setor (s it oes in Y 7), ensuring tht the remining istne to e overe is stritly smller thn the originl. This ie is pture y the flowing lemm. Lemm 12. [Inution Step] Let,, S suh tht n lie in the sme one with pex, n Y 6. Let α = n β =. If either α < π/3 or β < π/3, then we my use inution on p(, ) to onlue tht p(, ) t. Proof. This onfigurtion is epite in Fig. 5. Beuse Y6 n n lie in the sme one with pex, we hve tht. Beuse t lest one of α or β is stritly smller thn π/3, we hve tht <. Thus the onitions of Lem. 8 re stisfie, so we n use Lem. 8 to oun in terms of x = : sine /x < t, < tx. Beuse <, we my pply inution to oun p(, ): p(, ) t. Hene p(, ) + p(, ) tx + t = t(x + ) = t. We will heneforth use the symol Inut s shorthn for pplying Lem. 12 to tringle equivlent to tht in Fig. 5. Lem. 12 leves out Y 6 eges flling within the -ones, tht oul oneivly not mke progress towr the gol. For exmple, following one ege of n equilterl tringle leves one extly s fr wy from the other orner s t the strt. However, we will see tht when ll relevnt eges of Y 6 fll with the -ones ner π/3, the restrite geometri struture ensures tht progress towr the gol is inee me, n gin inution pplies. Inutive step. The inutive step proof first hnles the ses where eges of Y 6 irete from or from fll in the entrl portion of the relevnt setors, n so stisfy Lem. 9, n so Lem. 12 pplies. Rell tht Q 0() y our ssumption. If Y 6, then p(, ) = n we re finishe. Assuming otherwise, there must e point Q 0() suh tht Y 6 n. For the reminer of the proof, we re in this sitution, with Y 6 n. The proof now prtitions into three prts: (1) when only Q 0() is relevnt n les to Inut ; (2) when Q 2() les to Inut ; (3) when we fll into speil sitution, for whih inution lso pplies, ut for ifferent resons. (1) The Q 0 () setor. Consier s previously illustrte in Fig. 5. If either or is not in one of the -ones of Q 0(), then α = < π/3 : Inut.
β Figure 6: The -ones re she. () n in the sme -one. () n in ifferent -ones. () Q 0() n Q 2(). Here β is smll. Now ssume tht oth n lie in -ones of Q 0(). If they oth lie within the sme -one (Fig. 6), then gin α is smll: Inut. So without loss of generlity let lie in the lower -one, n in the upper -one of Q 0(); see Fig. 6. We nnot pply inution in this sitution euse the rtio s/x in Lem. 8 hs no upper oun. (2) The Q 2 () setor. Now we onsier Q 2(), the setor with pex t iming to the left of, n ssume tht Q 2(). Refer to Fig. 6. The se / Q 2() will e isusse lter (speil sitution). Beuse my suten n ngle s lrge s t with the horizontl, the upper 2-one of Q 2() eomes the relevnt region. If is not in the upper 2-one of Q 2() (s epite in Fig. 6), then stisfies Lem. 9 with β < π/3 : Inut. Note tht this onlusion follows even if is in the smll region outsie of n elow Q 2(): the ngle β t is then very smll. Assume now tht is in the upper 2-one of Q 2(). Let Q 2() e the point suh tht Y 6. We now onsier possile lotions for. If =, then p(, ) + 2, n we re finishe. So ssume heneforth tht is istint from. If is not in the upper -one of Q 0() (Fig. 7), then stisfies Lem. 9 with the roles of n reverse: tkes step towr, with the ngle t stisfying < π/3 : Inut. If is not in the upper 2-one of Q 2() (Fig. 7), then stisfies Lem. 9 gin with the roles of n reverse n this time the ngle t oune wy from π/3, < π/3 : Inut. Assume now tht is in the intersetion region etween the upper -one of Q 0() n the upper 2-one of Q 2(). Rell tht we re in the sitution where lies in the sme region, so it is lose to. See Fig. 7. This suggests the strtegy of following n, onnete y p(, ). We show tht in ft <, so the inutive hypothesis n e pplie to p(, ). More preisely, we show the following result. Lemm 13. Let,,, S e s in Fig. 7, with Y 6,, Q 0() n, Q 2(). If oth n lie ove the lower rys ouning the upper 2-ones of Q 0() n Q 2(), then for ny 0 π/9, sin(2). (17) sin(π/6 + 2) Note tht lies in the intersetion region etween the upper 2-ones of Q 0() n Q 2(), euse Q 0() Q 2() (y the sttement of the lemm). However, Lem. 13 oes not restrit the lotion of to the sme region. Inee, my lie either elow or ove the upper ry ouning Q 0(), s long s it stisfies the onition. (This onition must hol euse, re in the sme setor Q 2(), n Y6.) To keep the flow of our min proof uninterrupte, we efer proof of Lem. 13 to Setion 4.1. By Lem. 13 we hve <. Thus we n use the inution hypothesis to show tht p(, ) t. We know tht euse oth n re in Q 0() n Y 6. We lso know tht euse oth n re in Q 2() n Y 6. Let u n i e the upper n lower intersetion points etween the rys ouning Q 2() n the upper ry of Q 0(), s in Fig. 7. Note tht ui is equilterl, n euse lies in this tringle, we hve u = i. It follows tht. So in this sitution (illustrte in Fig. 7), we hve: p(, ) + p(, ) + 2 + p(, ) 2 + t sin(2) 2 + t sin(π/6 + 2) t. Here we hve pplie Lem. 13 to oun. Note tht the ltter inequlity ove is true for the vlue of t from (16). (3) Speil sitution. The only se left to isuss is the one in whih lies in the upper -one of Q 0() n to the right of the upper ry of Q 2(). This sitution is epite in Fig. 8. Next onsier Q 4(). Beuse Q 4(), there exists z Y 6, with z Q 4() n z. Clerly z Q 0() Q 5(). Note tht the isk setor D 0(, ) Q 0() with enter n rius must e empty, euse Y 6. Cse 3(). If z Q 0(), then z lies in the lower -one of Q 0() n to the right of D 0(, ), lose to. See Fig. 8. In this se we show tht the quntity on the right sie of inequlity (17) is loose upper oun on z, n tht similr inutive rguments hol here s well. Let the irumferene of D 0(, ) interset the right ry of Q 4() n the lower ry of Q 0() t points z n, respetively. Refer to
u 2 2 i Figure 7: () not in the upper -one of Q 0(): is smll. () not in the upper 2-one of Q 2(): is smll. () Lem. 13: <. D (,) 0 () z 60 o 60 o D (,) 0 Figure 8: Cse / Q 2(), z Q 0(): Lem. 13 pplies on. Fig. 8. Let e the ngle forme y with the upper ry of Q 0(). Then z = n z = /2. This implies tht oth n z lie in the intersetion region etween the lower -one of Q 0() n the right /2-one of Q 4(). Thus,,, z S stisfy the onitions of Lem. 13, with the roles of n reverse: z sin(2)/ sin(π/6 + 2). Arguments similr to the ones use for the se epite in Fig. 10 show tht z. This long with (euse Y 6) n the ove inequlity imply p(, ) + z + p(z, ) 2 + t z t for ny t stisfying the onitions stte y this lemm. 2 e u z v D (, ) 0 D (u, u ) 4 () i j D (u, u ) 4 () u z v () f 2 D (, ) 0 Figure 9: Cse / Q 2() n z Q 5() () z left of uv () z right of uv. z Cse 3(). Assume now tht z / Q 0(). Then z Q 5(), s epite in Fig. 9. In this se z lies in the isk setor D 4(, ) (euse z ) n elow the horizontl through (euse D 0(, ) is empty). This implies tht there exists e Y6, with e Q 5() n e z. Similrly, there exists f Y 6, with f Q 3() n f z. If e lies ove the lower 2-one of Q 5(), then e π/3, whih les to Inut n settles this se. Similrly, if f lies ove the lower -one of Q 3(), then f π/3, whih gin les to Inut. Otherwise, we show tht the following lemm hols. Lemm 14. Let,,, z S e in the onfigurtion epite in Fig. 9, with, z Y 6. Let e, f Y 6, with e in the lower 2-one of Q 5() n f in the lower -one of Q 3(). Then t lest one of the following is true: () e Q 3(), or () f Q 5(). We efer proof of Lem. 14 to Setion 4.2. Lem. 14 gurntees tht, if onition () hols, then e my not ross the lower ry ouning Q 3(). This se reues to one of the ses epite in Figs. 7 n 10, with e plying the role of n the pth pssing uner rther thn ove. Beuse e oes not ross the lower ry ouning Q 3(), the speil sitution epite in Fig. 8 (with e plying the role of ) my not our in this se. Similrly, onition () from Lem. 14 reues to one of the ses epite in Figs. 7 n 10, with the roles of n reverse n with f plying the role of ; the speil sitution epite in Fig. 8 (with f plying the role of ) my not our in this se. Hving exhuste ll ses, we onlue the proof. A proof estlishing the following lower oun n e foun in the full version of the pper [2]. Theorem 15. Y 6 hs spnning rtio t lest 2. 4. DEFERRED PROOFS 4.1 Proof of Lemm 13 Lemm 13. Let,,, S e s in Fig. 10, with Y 6,, Q 0() n, Q 2(). If oth n lie ove the lower rys ouning the upper 2-ones of Q 0() n Q 2(), then for ny 0 π/9, sin(2) sin(π/6 + 2)
Proof. Let u n v e the top n ottom points of the intersetion qurilterl R etween the upper 2-ones of Q 0() n Q 2(). See Fig. 10. Then R. For ny π/9, the ngles opposite to the igonl uv of R re oune elow y 5π/9, therefore uv is the imeter of R. u v () u 2 j i R Figure 10: Lem. 13: () uv is minimum when = 0. () uv. Assume first tht R s well. In this se, the quntity is oune ove y the length uv of the imeter of R. Let e the ngle forme y with the horizontl. We show tht uv is mximize when = 0. Set oorinte system with the origin t. Sle the point set S so tht = 1. Then the oorintes of re (os, sin ). The point u is t the intersetion of the two lines pssing through n with slopes tn π/3 n tn π/3 respetively, given y y = 3x n y = 3(x os ) + sin. Solving for x n y gives the oorintes of u 3 os + sin 3 os + sin x u = 2, y u =. 3 2 Similrly, the point v is t the intersetion of two lines given y y = tn(π/3 2)x n y = tn(π/3 2)(x os ) + sin. Solving for x n y gives the oorintes of v x v = y v = tn(π/3 2) os + sin, 2 tn(π/3 2) tn(π/3 2) os + sin. 2 We n now ompute uv = (x u x v) 2 + (y u y v) 2 s Figure 11: Lem. 13. The erivtive of uv with respet to, for, [0, π/9]. v () 2 funtion of n. The erivtive of this funtion with respet to is represente s grph in Fig. 11 for, [0, π/9]. Note tht this funtion is negtive on the given intervl, therefore uv inreses s ereses. Thus uv is mximum when = 0. We now set = 0 n ompute uv 3/2 ot(2 + π/6)/2 = sin(2)/ sin(2 + π/6) s lime. Assume now tht / R, so lies ove the upper ry ouning Q 0(). Let i e the intersetion point etween the upper ry ouning Q 0() n the lower ry ouning the upper 2-one of Q 2(). Then must lie outsie the isk D 2(, i ), euse lies outsie this isk (y ssumption) n (euse Y 6). Refer to Fig. 10. Let j e the intersetion point etween the lower ry ouning the upper 2-one of Q 2() n the irumferene of D 2(, u ). Then oth n lie in the strip elimite y D 2(, i ), D 2(, j ) n the two rys ouning the upper 2-one of Q 2(). Thus is no greter thn the imeter of this strip, whih we show to e no greter thn the imeter of R. For this, it suffies to show tht mx{ ui, uj, ij } uv. Beuse ui is n ege of R, ui is lerly no greter thn the imeter uv of R. Next we show tht uj uv. From the isoseles tringle uj we erive ujv = π/2. Angle uvj is exterior to uv, therefore uvj = vu + 2 π/6 + 2 (note tht vu = π/6 when is horizontl, otherwise vu < π/6). It follows tht ujv uvj for ny π/9. This long with the lw of sines pplie to ujv yiels uj uv. It remins to show tht ij < uv. We will in ft show tht ij < uj, whih long with the onlusion ove tht uj uv, yiels ij < uv. Angle uij is exterior to ui, therefore π/3 uij π/3 + 2. Erlier we showe tht ujv = π/2 7π/18, for ny π/9. It follows tht iuj π (7π/18+π/3) = 5π/18 is the smllest ngle of uij, therefore ij < uv. This ompletes the proof. 4.2 Proof of Lemm 14 Lemm 14. Let,,, z S e in the onfigurtion epite in Fig. 9, with, z Y 6. Let e, f Y 6, with e in the lower 2-one of Q 5() n f in the lower -one of Q 3(). Then t lest one of the following is true: () e Q 3() () f Q 5() Proof. We efine four intersetion points u, v, i n j s follows: u is t the intersetion etween the top rys of Q 0() n Q 2(); v is t the intersetion etween the isetor of u n the ounry of the isk setor D 4(u, u ); i is the foot of the perpeniulr from on the lower ry of Q 3(); n j is the foot of the perpeniulr from on the lower ry of Q 5(). Refer to Fig. 9. Note tht e i implies onition (), n f j implies onition (). We show tht the first hols if z lies to the left of or on uv, n the ltter hols if z lies to the right of or on uv (n so t lest one of the two onitions hols). We first show tht z D 4(u, u ). This follows immeitely from the inequlity uz + < z + u (whih n e erive using the tringle inequlity twie on the tringles inue y the igonls of uz), n the ft tht z (euse z Y 6). It follows tht uz < u, therefore z D 4(u, u ).
Conition (). Assume tht z lies to the left of uv (s in Fig. 9). Beuse z D 4(u, u ) is elow the horizontl through, zv is otuse n therefore z v (equlity hols when z oinies with v). Also e z, euse z n e re in the sme setor Q 5() n e Y 6. It follows tht e v. We now show tht v i, whih implies e i, thus settling this se. Let [0, ] e the ngle forme y with the horizontl through. Then i = π/3 n i = sin(π/3 ). The lw of sines pplie to uv tells us tht v sin π/6 = u sin uv = uv sin uv. Note tht uv = u u, euse v lies on the irumferene of D(u, u ) n lies outsie of this isk. This long with the ltter equlity ove yiels uv uv. The sum of these two ngles is 5π/6 (rell tht uv is the isetor of u), therefore uv 5π/12. Also note tht uv < π/2, euse v lies stritly elow the horizontl through (otherwise my not exist). It follows tht sin uv sin 5π/12. Sustituting this in the equlity ove yiels v u sin π/6/ sin 5π/12. The lw of sines pplie to tringle u yiels u = sin(π/3 + )/ sin π/3, whih sustitute in the previous equlity yiels sin(π/3 + ) sin π/6 v sin π/3 sin 5π/12. Thus the inequlity v i hols for ny stisfying sin(π/3 + ) sin π/6 sin π/3 sin 5π/12 sin(π/3 ). It n e esily verifie tht this inequlity hols for ny 23π/180, n in prtiulr for the vlues restrite y Lem. 11. Conition (). Assume now tht z lies to the right of uv (s in Fig. 9). In this se f z v. We now show tht v j, whih implies f j, thus settling this se. From the right tringle j with ngle j = π/3 +, we erive j = sin(π/3 + ). Next we erive n upper oun on v. From the isoseles tringle vu, hving ngle vu = π/6, we erive v = 2 u sin π/12. The lw of sines pplie to tringle u gives us u = sin(π/3 )/ sin π/3, whih sustitute in the previous equlity yiels v = 2 sin(π/3 ) sin π/12/ sin π/3. Thus the inequlity v j hols for ny vlue stisfying 2 sin(π/3 ) sin π/12 sin π/3 sin(π/3 + ). It n e verifie tht this inequlity hols for ny π/3, n in prtiulr for the vlues restrite y Lem. 11. 5. REFERENCES [1] I. Althöfer, G. Ds, D. Dokin, D. Joseph, n J. Sores. On sprse spnners of weighte grphs. Disrete & Computtionl Geometry, 9(1):81 100, 1993. [2] L. Br, P. Bose, M. Dmin, R. Fgererg, W. L. Keng, J. O Rourke, A. vn Renssen, P. Tslkin, S. Veronshot, n G. Xi. New n improve spnning rtios for Yo grphs. ArXiv e-prints, 2014. rxiv:1307.5829 [s.cg]. [3] N. Bonihon, C. Gvoille, N. Hnusse, n D. Ilinks. Connetions etween thet-grphs, Deluny tringultions, n orthogonl surfes. In Proeeings of the 36th Interntionl Workshop on Grph-Theoreti Conepts in Computer Siene (WG 2010), volume 6410 of Leture Notes in Computer Siene, pges 266 278, 2010. [4] P. Bose, M. Dmin, K. Douïe, J. O Rourke, B. Semone, M. Smi, n S. Wuhrer. π/2-ngle Yo grphs re spnners. ArXiv e-prints, 2010. rxiv:1001.2913 [s.cg]. [5] P. Bose, M. Dmin, K. Douïe, J. O Rourke, B. Semone, M. Smi, n S. Wuhrer. π/2-ngle Yo grphs re spnners. Interntionl Journl of Computtionl Geometry & Applitions, 22(1):61 82, 2012. [6] P. Bose, A. Mheshwri, G. Nrsimhn, M. Smi, n N. Zeh. Approximting geometri ottlenek shortest pths. Computtionl Geometry. Theory n Applitions, 29(3):233 249, 2004. [7] P. Bose, P. Morin, A. vn Renssen, n S. Veronshot. The θ 5-grph is spnner. In Proeeings of the 39th Interntionl Workshop on Grph-Theoreti Conepts in Computer Siene (WG 2013), pges 100 114, 2013. [8] K. L. Clrkson. Approximtion lgorithms for shortest pth motion plnning. In Proeeings of the 19th ACM Symposium on the Theory of Computing (STOC 1987), pges 56 65, 1987. [9] M. Dmin n K. Ruonis. Yo grphs spn thet grphs. Disrete Mthemtis, Algorithms n Applitions, 4(02):1250024, 2012. [10] N. M. El Moll. Yo spnners for wireless ho networks. PhD thesis, Villnov University, 2009. [11] B. E. Flinhugh n L. K. Jones. Strong onnetivity in iretionl nerest-neighor grphs. SIAM Journl on Algeri n Disrete Methos, 2(4):461 463, 1981. [12] G. Nrsimhn n M. Smi. Geometri Spnner Networks. Cmrige University Press, 2007. [13] A. C. C. Yo. On onstruting minimum spnning trees in k-imensionl spes n relte prolems. SIAM Journl on Computing, 11(4):721 736, 1982.