September 30, :24 WSPC/Guidelines Y4Spanner

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1 Septemer 30, :24 WSPC/Guielines Y4Spnner Interntionl Journl of Computtionl Geometry & Applitions Worl Sientifi Pulishing Compny π/2-angle YAO GAPHS AE SPANNES POSENJIT BOSE Shool of Computer Siene, Crleton University Ottw, Cn MIELA DAMIAN Deprtment of Computer Siene, Villnov University Villnov, USA KAIM DOUÏEB Shool of Computer Siene, Crleton University Ottw, Cn JOSEPH O OUKE Deprtment of Computer Siene, Smith College Northmpton, USA orourke@s.smith.eu BEN SEAMONE Shool of Mthemtis n Sttistis, Crleton University Ottw, Cn semone@onnet.rleton. MICHIEL SMID Shool of Computer Siene, Crleton University Ottw, Cn mihiel@ss.rleton. STEFANIE WUHE Institute for Informtion Tehnology, Ntionl eserh Counil Ottw, Cn stefnie.wuhrer@nr-nr.g. We show tht the Yo grph Y 4 in the L 2 metri is spnner with streth ftor 8 2( ). Enroute to this, we lso show tht the Yo grph Y4 in the L metri is plne spnner with streth ftor 8. Keywors: Yo grph; Y4; spnner. 1

2 Septemer 30, :24 WSPC/Guielines Y4Spnner 2 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer 1. Introution Let V e finite set of points in the plne n let G = (V, E) e the omplete Eulien grph on V. We will refer to the points in V s noes, to istinguish them from other points in the plne. The Yo grph 8 with n integer prmeter k > 0, enote Y k, is efine s follows. Any k eqully-seprte rys strting t the origin efine k ones. Pik set of ritrry, ut fixe ones. Trnslte the ones to eh noe u V. In eh one with pex u, pik shortest ege uv, if there is one, n to Y k the irete ege uv. Ties re roken ritrrily. Note tht the Yo grph iffers from the Θ-grph in how the shortest ege is hosen. While the Yo grph hooses the shortest ege in terms of the Eulien istne, the Θ-grph hooses the ege whose projetion on the isetor of the one is shortest. Most of the time we ignore the iretion of n ege uv; we refer to the irete version uv of uv only when its origin (u) is importnt n unler from the ontext. We will istinguish etween Y k, the Yo grph in the Eulien L 2 metri, n Yk, the Yo grph in the L metri. Unlike Y k however, in onstruting Yk ties re roken y lwys seleting the most ounterlokwise ege; the reson for this hoie will eome ler in Setion 2. The length of pth is the sum of the lengths of its onstituent eges. For given sugrph H G n fixe t 1, H is lle t-spnner for G if, for ny two noes u, v V, the shortest pth in H from u to v is no longer thn t times the length uv of uv. The vlue t is lle the iltion or the streth ftor of H. If t is onstnt, then H is lle length spnner, or simply spnner. The lss of grphs Y k hs een muh stuie. Bose et l. 2 showe tht, for k 1 9, Y k is spnner with streth ftor. In ef. 1 we improve the streth os 2π k sin 2π k ftor n showe tht, in ft, Y k is spnner for ny k 7. eently, Dmin n uonis 4 showe tht Y 6 is 17.7-spnner. Moll 6 showe tht Y 2 n Y 3 re not spnners, n tht Y 4 is spnner with streth ftor 4(2 + 2), for the speil se when the noes in V re in onvex position (see lso ef. 3 ). The uthors onjeture tht Y 4 is spnner for ritrry point sets. In this pper, we settle their onjeture n prove tht Y 4 is spnner with streth ftor 8 2( ). The pper is orgnize s follows. In Setion 2, we prove tht the grph Y4 is spnner with streth ftor 8. In Setion 3 we estlish severl properties for the grph Y 4. Finlly, in Setion 4, we use the properties of Setion 3 to prove tht, for every ege in Y4, there exists pth etween n in Y 4 not muh longer thn the Eulien istne etween n. By omining this with the result of Setion 2, we onlue tht Y 4 is spnner. 2. Y 4 in the L Metri In this setion we fous on Y4, whih hs nier struture ompre to Y 4. First we prove tht Y4 is plne grph. Then we use this property to show tht Y4 is n 8-spnner. To e more preise, we prove tht for ny two noes n, the grph Y4 ontins pth etween n whose length (in the L -metri) is t

3 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 3 most 8. We nee few efinitions. We sy tht two eges n properly ross (or ross, for short) if they shre point other thn n enpoint (,, or ); we sy tht n interset if they shre point (either n interior point or n enpoint). Q () 1 P () 1 Q () 2 S(,) Q () Q () 3 4 () () j Fig. 1. () Definitions: Q i (), P i () n S(, ). () Lemm 1: n nnot ross. Throughout the pper, we will use the following nottion: for eh noe V, x() is the x-oorinte of n y() is the y-oorinte of ; Q 1 (), Q 2 (), Q 3 () n Q 4 () re the four qurnts t, epite in Fig. 1; eh qurnt is hlf-open n hlf-lose, inluing ll points on the lokwise ounry xis (with respet to the qurnt isetor through ), n exluing ll points on the ounterlokwise ounry xis; P i () is the pth tht strts t n follows the irete Yo eges in qurnt Q i ; P i (, ) is the supth of P i () tht strts t noe n ens t noe ; is the L istne etween n, efine s mx{ x() x(), y() y() }; sp(, ) is shortest pth in Y4 etween n ; S(, ) is the open squre with orner whose ounry ontins ; n S(, ) is the ounry of S(, ). These efinitions re epite in Fig. 1. Lemm 1. Y 4 is plne grph. Proof. The proof is y ontrition. Assume the opposite. Then there re two eges, Y4 tht ross eh other. Sine Y4, S(, ) must e empty of noes in V, n similrly for S(, ). Let j e the intersetion point etween n. Then j S(, ) S(, ), mening tht S(, ) n S(, ) must overlp. However, neither squre my ontin,, or. It follows tht S(, ) n S(, ) oinie, mening tht n lie on S(, ) (see Fig. 1). Sine intersets, n must lie on opposite sies of. Thus either or lies ounterlokwise from. Assume without loss of generlity tht lies ounterlokwise from ; the other se is ientil. Beuse S(, ) oinies with S(, ), we hve tht =. In this se however, Y4 woul rek the tie etween n y seleting the most ounterlokwise ege, whih is. This ontrits tht Y 4.

4 Septemer 30, :24 WSPC/Guielines Y4Spnner 4 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer Theorem 1. Y 4 is n 8-spnner in the L metri. Proof. We show tht, for ny pir of points, V, sp(, ) < 8. The proof is y inution on the pirwise L -istne etween the points in V. Assume without loss of generlity tht Q 1 (), n = x() x() (i.e., lies elow the igonl of S(, ) inient to ). Consier the se in whih is losest (in the L metri) pir of points in V. This is the se se for our inution. If Y4, then sp(, ) =. Otherwise, there must e Y4, with =. ell tht Y4 reks ties y lwys seleting the most ounterlokwise ege, so must e ounterlokwise of. Also rell tht Q 1 () oes not inlue the vertil oorinte xis through, therefore lies stritly to the right of. It follows tht < (see Fig. 2), ontrition. S i r j i A e j r i A A () () () () r i e A (e) j Fig. 2. () Bse se. () empty () non-empty, P r P 2 () = {j} () non-empty, P r P 2 () =, e ove r (e) non-empty, P r P 2 () =, e elow r. Assume now tht the inutive hypothesis hols for ll pirs of points loser (in the L metri) thn. If Y4, then sp(, ) = n the proof is finishe. If / Y4, then the squre S(, ) must e nonempty. Let A e the retngle s in Fig. 2, where n re prllel to the igonls of S(, ). If A is nonempty, then we n use inution to prove tht sp(, ) 8 s follows. Pik A ritrry. Then + = x() x() + x() x() =, n y the inutive hypothesis sp(, ) sp(, ) is pth in Y4 no longer thn = 8 ; here represents the ontention opertor. Assume now tht A is empty. Let e t the intersetion etween the line supporting n the vertil line through (see Fig. 2). We isuss two ses, epening on whether is empty of points or not. Cse 1: is empty of points. Let P 1 (). We show tht P 4 () nnot ontin n ege rossing. Assume the opposite, n let st P 4 () ross. Note tht st P 4 () lso implies st P 4 (s), whih long with the ft tht st rosses, implies tht s is either vertilly ligne, or to the left of.. Sine is empty, s must lie ove n t elow. It follows tht n t re in the sme qurnt

5 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 5 Q 4 (s) (rell tht this qurnt inlues the ownwr ry from s). Furthermore, st y(s) y(t) > y(s) y() = s, ontriting the ft tht st Y4. We hve estlishe tht P 4 () oes not ross, whih implies tht P 4 () must exit S(, ) through its right ege. Also note tht P 2 () nnot ross, euse is empty of points, n ny point left of is L -frther from thn. It follows tht P 2 () exits S(, ) through its top ege. This together with the ft tht P 4 () exits S(, ) through its right ege, implies tht P 4 () n P 2 () must meet in point i P 4 () P 2 () (see Fig. 2). Now note tht P 4 (, i) P 2 (, i) x() x() + y() y() < 2. Thus we hve tht sp(, ) P 4 (, i) P 2 (, i) < + 2 = 3. Cse 2: is nonempty. In this se, we seek short pth from to tht oes not ross to the unersie of, to voi osillting pths tht ross ritrrily mny times. Let r e the rightmost point tht lies insie. Arguments similr to the ones use in Cse 1 show tht P 3 (r) nnot ross n therefore it must meet P 1 () in point i. Then P r = P 1 (, i) P 3 (r, i) is pth in Y4 of length P r < x() x(r) + y() y(r) < + 2 = 3. (1) The term 2 in the inequlity ove results from the ft tht y() y(r) y() y() 2. Consier first the simpler sitution in whih P 2 () meets P r in point j P 2 () P r (see Fig. 2). Let P r (, j) e the supth of P r extening etween n j. Then P r (, j) P 2 (, j) is pth in Y4 from to, therefore sp(, ) P r (, j) P 2 (, j) < 2 y(j) y() + 5. Consier now the se when P 2 () oes not interset P r. We rgue tht, in this se, Q 1 (r) my not e empty. Assume the opposite. Then no ege st P 2 () my ross Q 1 (r). This is euse, for ny suh ege, sr < st, ontriting st Y4. This implies tht P 2 () intersets P r, gin ontrition to our ssumption. This estlishes tht Q 1 (r) is nonempty. Let r P 1 (r). The ft tht P 2 () oes not interset P r implies tht lies to the left of. The ft tht r is the rightmost point in implies tht lies outsie (see Fig. 2). It lso implies tht P 4 () shres no points with. This long with rguments similr to the ones use in se 1 show tht P 4 () n P 2 () meet in point j P 4 () P 2 (). Thus we hve foun pth P = P 1 (, i) P 3 (r, i) r P 4 (, j) P 2 (, j). (2) extening from to in Y 4. If r = x() x(r), then r < x() x() =, n the pth P hs length P 2 y() y() + < 7. (3) In the ove, we use the ft tht y() y() = y() y(r) + y(r) y() < + 2. Suppose now tht r = y() y(r). (4)

6 Septemer 30, :24 WSPC/Guielines Y4Spnner 6 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer In this se, it is unler whether the pth P efine y (2) is short, sine r n e ritrrily long ompre to. Let e e the lokwise neighor of long the pth P (e n my oinie). Then e lies elow, n either e P 4 (), or e P 2 (e) (or oth). If e lies ove r, or t the sme level s r (i.e., e Q 1 (r), s in Fig. 2), then y(e) y(r) < y() y(r). (5) Sine r P 1 (r) n e is in the sme qurnt of r s, we hve r re. This long with inequlities (4) n (5) implies re > y(e) y(r), whih in turn implies re = x(e) x(r), n so r. Then inequlity (3) pplies here s well, showing tht P < 7. If e lies elow r (s in Fig. 2e), then e y() y(e) y() y(r) = r. (6) Assume first tht e P 2 (e), or e = x(e) x(). In either se, e er < 2. This long with inequlity (6) shows tht r < 2. Sustituting this upper oun in (2), we get P 2 y() y() + 2 < 8. Assume now tht e P 2 (e), n e = y(e) y(). Then ee P 2 (e) nnot go ove (otherwise e < ee, ontriting ee P 2 (e)). This long with the ft e P 4 () implies tht P 2 (e) intersets P r in point k. eefine P = P r (, k) P 2 (e, k) P 4 (e, j) P 2 (, j). Then P is pth in Y4 from to of length P 2 y(r) y() + 5. This theorem will e employe in Setion Y 4 in the L 2 Metri In this setion we estlish si properties of Y 4. The ultimte gol of this setion is to show tht, if two eges in Y 4 ross, there is short pth etween their enpoints (Lemm 8). We egin with few efinitions. Let Q(, ) enote the infinite qurnt with origin t tht ontins. For pir of noes, V, efine reursively irete pth P( ) from to in Y 4 s follows. If =, then P( ) = null. If, there must exist Y 4 tht lies in Q(, ). In this se, efine P( ) = P( ). ell tht represents the ontention opertor. This efinition is illustrte in Fig. 3. Fisher et l. 5 show tht P( ) is well efine n lies entirely insie the squre entere t whose ounry ontins. For ny pth P n ny pir of noes, P, let P [, ] e the supth of P from to. Let (, ) e the lose xis-ligne retngle with igonl (we permit (, ) to e egenerte retngle, when is either horizontl or vertil). For fixe pir of noes, V, efine pth P ( ) s follows. Let e V e the first noe long P( ) tht is not stritly interior to (, ). Then

7 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 7 Q () 2 y Q () 1 (,) x Q(, ) Q () Q () 3 4 () h () P ( ) e Fig. 3. Definitions. () Q(, ) n P( ). () P ( ). P ( ) is the supth of P( ) tht extens etween n e. In other wors, P ( ) is the pth tht follows the Y 4 eges pointing towrs, trunte s soon s it rehes or leves (, ). Formlly, P ( ) = P( )[, e]. This efinition is illustrte in Fig. 3. Our proofs will mke use of the following two propositions. Proposition 1. The sum of the lengths of rossing igonls of non-egenerte (neessrily onvex) qurilterl is stritly greter thn the sum of the lengths of either pir of opposite sies: + > + + > +. This n e prove y prtitioning the igonls into two piees eh t their intersetion point, n then pplying the tringle inequlity twie. Proposition 2. For ny tringle, the following inequlities hol: < 2 + 2, if < π/2 2 = 2 + 2, if = π/2 > 2 + 2, if > π/2 This proposition follows immeitely from the Lw of Cosines pplie to tringle. Lemm 2. For eh pir of noes, V, P ( ) 2. (7) Furthermore, eh ege of P ( ) is no longer thn. Proof. Let e one of the two orners of (, ), other thn n. Let e P ( ) e the lst ege on P ( ), whih neessrily intersets (, )

8 Septemer 30, :24 WSPC/Guielines Y4Spnner 8 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer (note tht it is possile tht e = ). efer to Fig. 3. Then e, otherwise e oul not e in Y4. Sine lies in the retngle with igonl, we hve tht, n similrly for eh ege on P ( ). This estlishes the ltter lim of the lemm. For the first lim of the lemm, let p = P ( )[, ]. Sine e, we hve tht P ( ) p. Sine p lies entirely insie (, ) n onsists of eges pointing towrs, we hve tht p is n xy-monotone pth (i.e., ny line prllel to oorinte xis intersets p in t most one point). It follows tht p +, whih is oune ove y 2. j Fig. 4. Lemm 3: if n ross, they nnot oth e in Y 4. Lemm 3. Let,,, V e four isjoint noes suh tht, Y 4, Q i () n Q i (), for some i {1, 2, 3, 4}. Then n nnot ross. Proof. We my ssume without loss of generlity tht i = 1 n is to the left of. The proof is y ontrition. Assume tht n ross eh other. Let j e the intersetion point etween n (see Fig. 4). Sine j Q 1 () Q 1 (), it follows tht Q 1 () n Q 1 (). Thus, euse otherwise, nnot e in Y 4. By Proposition 1 pplie to the qurilterl, + < +. This long with implies tht <, ontriting tht Y 4. The next four lemms (4 7) eh onern pir of rossing Y 4 eges, ulminting (in Lemm 8) in the onlusion tht there is short pth in Y 4 etween pir of enpoints of those eges. We hoose to efer the proofs of lemms 4 6 to the ppenix, for etter unerstning of the logil flow of our nlysis. Lemm 4. Let,, n e four isjoint noes in V suh tht, Y 4, n rosses. Then (i) the rtio etween the shortest sie n the longer igonl of the qurilterl is no greter thn 1/ 2, n (ii) the shortest sie of the qurilterl is stritly shorter thn either igonl. Lemm 5. Let,,, e four istint noes in V, with Q 1 (), suh tht (i) Q1 () n Q 2 () re in Y 4 n ross eh other, n (ii) is shortest

9 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 9 sie of qurilterl. Then P ( ) n P ( ) hve nonempty intersetion. Lemm 6. Let,,, e four istint noes in V, with Q 1 (), suh tht (i) Q1 () n Q 3 () re in Y 4 n ross eh other, n (ii) is shortest sie of qurilterl. Then P ( ) oes not ross. The next lemm relies on ll of Lemms 2 6. Lemm 7. Let,,, V e four istint noes suh tht Y 4 rosses Y4, n let xy e shortest sie of the qurilterl. Then there exist two pths P x n P y in Y 4, where P x hs x s n enpoint n P y hs y s n enpoint, with the following properties: (i) P x n P y hve nonempty intersetion. (ii) P x + P y 3 2 xy. (iii) Eh ege on P x P y is no longer thn xy. Proof. Assume without loss of generlity tht Q 1 (). We isuss the following exhustive ses: (1) Q 1 (), n Q 1 (). In this se, n nnot ross eh other (y Lemm 3), so this se is finishe. x x () y () () () (e) P (y ) y P ( y) Fig. 5. Lemm 7: (, ) Q 1 () () Q 2 () () Q 4 (). (2) Q 1 (), n Q 2 (), s in Fig. 5. Sine Y 4,. Sine rosses, n, Q 2 (). Sine Y 4,. These long

10 Septemer 30, :24 WSPC/Guielines Y4Spnner 10 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer with Lemm 4 imply tht n re the only nites for shortest ege of. Assume first tht is shortest ege of. By Lemm 3, P = P ( ) oes not ross, euse P Q 2 () n Q 2 () re in the qurnts of ientil inies. It follows from Lemm 5 tht P n P = P ( ) hve nonempty intersetion. Furthermore, y Lemm 2, P 2 n P 2, n no ege on these pths is longer thn, proving the lemm true for this se. Consier now the se when is shortest ege of (see Fig. 5). Note tht is elow (otherwise, Q 2 () n > ) n, therefore, Q 1 (). By Lemm 3, P = P ( ) oes not ross, euse P Q 1 () n Q 1 (). If P = P ( ) oes not ross, then P n P hve nonempty intersetion, proving the lemm true for this se. Otherwise, there exists xy P ( ) tht rosses (see Fig. 5). Define P = P ( ) P (y ) P = P ( y). By Lemm 3, P (y ) oes not ross, euse they re oth in qurnt Q 2. Then P n P must hve nonempty intersetion. We now show tht P n P stisfy onitions (i) n (iii) of the lemm. Proposition 1 pplie on the qurilterl xy tells us tht x + y < xy +. We lso hve tht x, sine Y 4 n x is in the sme qurnt of s. This long with the inequlity ove implies y < xy. Beuse xy P ( ), y Lemm 2 we hve tht xy, whih long with the previous inequlity shows tht y <. This long with Lemm 2 shows tht onition (iii) of the lemm is stisfie. Furthermore, P (y ) y 2 n P ( y) y 2. It follows tht P + P 3 2. (3) Q 1 (), n Q 3 (), s in Fig. 5. Then mx{, }, n y Lemm 4 is not shortest ege of. The se when is shortest ege of is settle y Lemms 3 n 2: Lemm 3 tells us tht P = P ( ) oes not ross, (euse they re oth in Q 1,) n P = P ( ) oes not ross (euse they re oth in Q 3 ). It follows tht P n P hve nonempty intersetion. Furthermore, Lemm 2 gurntees tht P n P stisfy onitions (ii) n (iii) of the lemm. Consier now the se when is shortest ege of ; the se when is shortest is symmetri. By Lemm 6, P ( ) oes not ross. If P ( ) oes not ross, then this se is settle: P = P ( ) n P = P ( ) stisfy the three onitions of the lemm. Otherwise, let xy P ( ) e the ege rossing. Arguments similr to the ones use in se 1 ove show tht P = P ( ) P (y ) n P = P ( y) re two pths tht stisfy the onitions of the lemm. (4) Q 1 (), n Q 4 (), s in Fig. 5. Note tht horizontl refletion of Fig. 5, followe y rottion of π/2, epits se ientil to se (2), Fig. 5, whih hs lrey een settle.

11 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 11 (5) Q 2 (), s in Fig. 5. Note tht Fig. 5 rotte y π/2 epits se ientil to se (2), Fig. 5 (with the roles of n swithe), whih hs lrey een settle. (6) Q 3 (). Then it must e tht Q 1 (), otherwise nnot ross. By Lemm 3 however, n my not ross, unless one of them is not in Y 4. (7) Q 4 (). By Lemm 3, my not lie in Q 1 (), therefore must e in Q 2 (), s in Fig. 5e. Note tht vertil refletion of Fig. 5e epits se ientil to se (2), Fig. 5 (with the roles of n swithe), so this se is settle s well. We re now rey to estlish the min lemm of this setion, showing tht there is short pth etween the enpoints of two interseting eges in Y 4. Lemm 8. Let,,, V e four istint noes suh tht Y 4 rosses Y4, n let xy e shortest sie of the qurilterl. Then Y 4 ontins pth p(x, y) onneting x n y, of length p(x, y) xy. Furthermore, no ege on p(x, y) is longer thn xy. Proof. Let P x n P y e the two pths whose existene in Y 4 is gurntee y Lemm 7. By onition (iii) of Lemm 7, no ege on P x n P y is longer thn xy. By onition (i) of Lemm 7, P x n P y hve nonempty intersetion. If P x n P y shre noe u V, then the pth p(x, y) = P x [x, u] P y [y, u] is pth from x to y in Y 4 no longer thn 3 2 xy ; the length restrition follows from gurntee (ii) of Lemm 7. Otherwise, let P x n P y e two eges rossing eh other. Let x y e shortest sie of the qurilterl, with x P x n y P y. Lemm 7 tells us tht xy n xy. These long with Lemm 4 imply tht x y xy / 2. (8) This enles us to erive reursive formul for omputing pth p(x, y) Y 4 s follows: { x, if x = y p(x, y) = (9) P x [x, x ] P y [y, y ] p(x, y ), if x y. Next we use inution on the length of xy to prove the lim of the lemm. The se se orrespons to x = y. In this se p(x, y) egenertes to point n p(x, y) = 0. To prove the inutive step, pik shortest sie xy of qurilterl, with, Y 4 rossing eh other, n ssume tht the lemm hols for ll suh sies shorter thn xy. Let p(x, y) e the pth etermine reursively s in (9). By the inutive hypothesis, we hve tht p(x, y ) ontins no eges longer thn x y xy, n p(x, y ) x y 6 2 xy. (10) 2

12 Septemer 30, :24 WSPC/Guielines Y4Spnner 12 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer This ltter inequlity follows from (8). Also rell tht no ege on P x n P y is longer thn xy, whih together with formul (9) n the rguments ove, implies tht no ege on p(x, y) is longer thn xy. Sustituting inequlities 10 n (ii) from Lemm 7 in formul (9) yiels This ompletes the proof. p(x, y) ( ) xy = xy. 4. Y 4 n Y 4 The finl step of our nlysis is to prove tht every iniviul ege of Y 4 is spnne y short pth in Y 4. This, long with the result of Theorem 1, estlishes tht Y 4 is spnner. Fix n ege Y 4. Cll n ege or pth t-short (with respet to ) if its length is within onstnt ftor t of. In our proof tht is spnne y t-short pth in Y 4, we will mke use of the following three sttements (prove in the Appenix). S1 If xy is t-short, then P (x y), n therefore its reverse, P 1 (x y) re t 2-short y Lemm 2. S2 If xy Y 4 is t 1 -short n zw Y 4 is t 2 -short, n if xy intersets zw, Lemm 4(ii) n Lemm 8 show tht there is t 3 -short pth etween ny two of the enpoints of these eges, with t 3 = t 1 + t 2 + 3(2 + 2) mx(t 1, t 2 ). S3 If p(x, y) is t 1 -short pth n p(z, w) is t 2 -short pth n these two pths interset, then y S2 there is t 3 -short pth P etween ny two of the enpoints of these pths, with t 3 = t 1 + t 2 + 3(2 + 2) mx(t 1, t 2 ). Lemm 9. Fix n ege Y 4. There is pth p(, ) Y 4 etween n, of length p(, ) t, for t = Proof. For the ske of lrity, we only prove here tht there is short pth p(, ) etween n, n n efer the lultion of the tul streth ftor t to the Appenix. We refer to n ege or pth s short if its length is within onstnt ftor of. Assume without loss of generlity tht Q 1 (). If Y 4, then p(, ) = n the proof is finishe. So ssume the opposite, n let e the ege in Y 4 tht lies in Q 1 (); sine Q 1 () is nonempty, exists. Beuse Y 4 n is in the sme qurnt of s, we hve tht (i) < 2 (ii). (11) Inequlity (ii) ove follows immeitely from the Lw of Cosines, whih implies tht 2 < (euse the ngle forme y n is stritly less thn π/2), n the ft tht. Thus oth n re short. An this

13 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 13 in turn implies tht P ( ) is short y S1. We next fous on P ( ). For simpliity, we ssume tht is ounterlokwise of ; the sitution when lies lokwise of is symmetril. Let / (, ) e the other enpoint of P ( ). We istinguish three ses. P ( ) e P ( ) s P ( ) P ( ) () () P (e ) P ( ) x r e () Fig. 6. Lemm 9: () Cse 1: P ( ) n hve nonempty intersetion. () Cse 2: P ( ) n hve n empty intersetion. () Cse 3: P ( ) n hve non-empty intersetion. Cse 1: P ( ) n interset (see Fig. 6). Then y S3 there is short pth p(, ) etween n. Cse 2: P ( ) n o not interset, n P ( ) n o not interset (see Fig. 6). Note tht euse is the enpoint of the short pth P ( ), the tringle inequlity on implies tht is short, n therefore P ( ) is short, y S1. We onsier two ses: (i) P ( ) intersets. Then y S3 there is short pth p(, ). So p(, ) = p(, ) P 1 ( ) is short. (ii) P ( ) oes not interset. Then P ( ) must interset P ( ) P ( ). Next we estlish tht is short. Let e e the lst ege of P ( ), n so inient to (note tht e n my oinie). Beuse P ( ) oes not interset, n re in the sme qurnt for e. It follows tht e e n e < π/2. These oservtions long with Proposition 2 for e imply tht 2 < e 2 + e 2 2 e 2 < 2 2 (this ltter inequlity uses the ft tht e > π/2, whih implies tht e < ). It follows tht 2 2 (y (11)ii). (12) Thus is short, n y S1 we hve tht P ( ) is short. Sine P ( ) intersets the short pth P ( ) P ( ), there is y S3 short pth

14 Septemer 30, :24 WSPC/Guielines Y4Spnner 14 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer p(, ), n so is short. p(, ) = p(, ) Cse 3: P ( ) n o not interset, n P ( ) intersets (see Fig. 6). If P ( ) intersets t, then p(, ) = P ( ) P ( ) is short. So ssume otherwise, in whih se there is n ege e P ( ) tht rosses. Then Q 1 (), e Q 3 () Q 4 (), n e n re in the sme qurnt for. Note however tht e nnot lie in Q 3 (), sine in tht se e > π/2, whih woul imply e >, whih in turn woul imply e / Y 4. So it must e tht e Q 4 (). Next we show tht P (e ) oes not ross. Assume the opposite, n let rs P (e ) ross. Then r Q 4 (), s Q 1 () Q 2 (), n s n re in the sme qurnt for r. Arguments similr to the ones ove show tht s / Q 2 (), so s must lie in Q 1 (). Let δ e the L istne from to. Let x e the projetion of r on the horizontl line through. Then rs rx + δ rx + x > r (y the tringle inequlity) Beuse n s re in the sme qurnt for r, the inequlity ove ontrits rs Y4. We hve estlishe tht P (e ) oes not ross. Then P ( e) must interset P = e P (e ). Note tht e is short euse it is in the short pth P ( ). Thus e is short (euse e < i + ei < + e, where i is the intersetion point etween n e), n so P ( e) n P (e ) re short, y S1. Then the short pth P ( e) intersets either e or P (e ), eh of whih is short, n y S3 there is short pth p(, e). Then p(, ) = p(, e) P 1 ( ) P 1 ( ) is short. Strightforwr lultions etile in the ppenix show tht, in eh of these ses, the streth ftor for p(, ) oes not exee Our min result follows immeitely from Theorem 1 n Lemm 9: Theorem 2. Y 4 is t-spnner, for t 8 2( ). 5. Conlusion Our results settle long-stning open prolem, sking whether Y 4 is spnner or not. We nswer this question positively, n estlish loose streth ftor of 8 2( ). Fining tighter streth ftors for oth Y4 n Y 4 remin interesting open prolems. Estlishing whether or not Y 5 is spnner is lso open.

15 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 15 Aknowlegements. We thnk the reviewers for their reful reing n useful omments. The first uthor ws supporte y NSEC. The seon uthor ws supporte in prt y NSF grnt CCF n y Villnov s Center of Exellene in Enterprise Tehnology. eferenes 1. P. Bose, M. Dmin, K. Douïe, J. O ourke, B. Semone, M. Smi n S. Wuhrer, π/2-angle Yo Grphs re Spnners, Tehnil eport (2010) rxiv: v1. 2. P. Bose, A. Mheshwri, G. Nrsimhn, M. Smi n N. Zeh, Approximting geometri ottlenek shortest pths, Computtionl Geometry: Theory n Applitions (2004) 29: M. Dmin, N. Moll n V. Piniu, Spnner properties of π/2-ngle Yo grphs, in Pro. of the 25th Europen Workshop on Computtionl Geometry (Mrh 2009) pp M. Dmin n K. uonis, Yo grphs spn Thet grphs, in Pro. of the 4th Annul Interntionl Conferene on Comintoril Optimiztion n Applitions (Deemer 2010) pp M. Fisher, T. Lukovszki n M. Ziegler, Geometri serhing in wlkthrough nimtions with wek spnners in rel time, in Pro. of the 6th Annul Europen Symposium on Algorithms (1998) pp N. Moll, Yo spnners for wireless ho networks, M.S. Thesis, Deprtment of Computer Siene, Villnov University (Deemer 2009). 7. J.W. Green, A note on the hors of onvex urve, Portuglie Mthemti (1951) 10(3): A.C.-C. Yo, On onstruting minimum spnning trees in k-imensionl spes n relte prolems, SIAM Journl on Computing (1982) 11(4):

16 Septemer 30, :24 WSPC/Guielines Y4Spnner 16 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer 6. Appenix 6.1. Proof of Lemm 4 For ny noe V, let D(, r) enote the open isk entere t of rius r, n let D(, r) enote the ounry of D(, r). Proof. The first prt of the lemm is well-known ft tht hols for ny qurilterl (see ef. 7, for instne). For the seon prt of the lemm, let e the shorter of the igonls of, n ssume without loss of generlity tht Q 1 (). Imgine two isks D = D(, ) n D = D(, ), s in Fig. 7. If either or elongs to D D, then the lemm follows: shortest qurilterl ege is shorter thn. 1 1 k 2 4 i l 3 () j 3 () Fig. 7. Lemm 4 () / () 1. So suppose tht neither nor lies in D D. In this se, we use the ft tht rosses to show tht nnot e n ege in Y 4. Define the following regions (see Fig. 7): 1 = (Q 1 () Q 2 ())\(D D ) 2 = (Q 2 () Q 3 ())\(D D ) 3 = (Q 4 () Q 3 ())\(D D ) 4 = (Q 1 () Q 4 ())\(D D ). If the noe is not insie ny of the regions i, for i = {1, 2, 3, 4}, then the noes n re in the sme qurnt of s. In this se, note tht either > π/2 or > π/2, whih implies tht either or is stritly smller thn. These together show tht / Y 4. So ssume tht is in i for some i {1, 2, 3, 4}. In this sitution, the noe must lie in the region j, with j = (i + 2) mo 4 (with the unerstning tht

17 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 17 0 = 4 ), euse otherwise, either (i) n re in the sme qurnt of n < or (ii) n re in the sme qurnt of n <. Either se ontrits the ft Y 4. Consier now the se 1 n 3 ; the other ses re trete similrly. Let i n j e the intersetion points etween D n the vertil line through. Similrly, let k n l e the intersetion points etween D n the vertil line through (see Fig. 7). Sine ij is imeter of D, we hve tht ij = π/2 n similrly kl = π/2. Also note tht ij = π/2, mening tht >. Similrly, kl = π/2, mening tht >. These long with the ft tht t lest one of n is in the sme qurnt for s, imply tht / Y 4. This ompletes the proof Proof of Lemm 5 Proof. The proof onsists of two prts showing tht the following lims hol: (I) Q 2 () n (II) P ( ) oes not ross. Before we prove these two lims, let us rgue tht they re suffiient to prove the lemm. Lemm 3 n lim (I) imply tht P ( ) nnot ross, euse P ( ) Q 2 () n Q 2 () re in qurnts of ientil inies. As result, P ( ) intersets the left sie of the retngle (, ). Consier the lst ege xy of the pth P ( ). If this ege rosses the right sie of (, ), then lim (II) implies tht y is in the wege oune y n the upwrs vertil ry strting t ; this further implies tht y <, ontriting the ft tht is n ege in Y 4. Therefore, xy intersets the ottom sie of (, ), n the lemm follows (see Fig. 8). To prove the first lim (I), we oserve tht the lemm ssumptions imply tht Q 1 () Q 2 (). Therefore, it suffies to prove tht is not in Q 1 (). Assume to the ontrry tht Q 1 (). Sine Q 1 (), it must e tht Q 2 (); otherwise, π/2, whih implies >, ontriting the ft tht Y 4. Let i n j e the intersetion points etween n D(, ), where i is to the left of j. Sine ij > π/2, we hve <. This, together with the ft tht n re in the sme qurnt Q 2 (), ontrits the ssumption tht is n ege in Y 4. This ompletes the proof of lim (I). Next we prove lim (II) y ontrition. Thus, we ssume tht there is n ege xy on the pth P ( ) tht rosses. Then neessrily x (, ) n y Q 1 () Q 4 (). If y Q 4 (), then xy > π/2, mening tht xy > x, ontrition to the ft tht xy Y 4. Thus, it must e tht y Q 1 (), s in Fig. 8. This implies tht y, euse Y 4. The ontrition to our ssumption tht xy rosses will e otine y proving tht xy > x. Inee, this inequlity ontrits the ft tht xy Y 4, euse oth n y re in Q 4 (x), n Y 4 woul hve pike x in ple of xy. Let δ e the istne from x to the horizontl line through. Our intermeite gol is to show tht δ / 2. (13)

18 Septemer 30, :24 WSPC/Guielines Y4Spnner 18 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer δ x i j () y P ( ) () P ( ) Fig. 8. () Lemm 5: xy P ( ) nnot ross. We lim tht < π/2. Inee, if this is not the se, then <, ontriting the ft tht is n ege in Y 4. By similr rgument, n using the ft tht is n ege in Y 4, we otin the inequlity < π/2. We now onsier two ses, epening on the reltive lengths of n. (1) Assume first tht >. If π/2, then >, ontriting the ft tht is n ege in Y 4 (rell tht n re in the sme qurnt of ). Therefore, we hve < π/2. So fr we hve estlishe tht three ngles of the onvex qurilterl re ute. It follows tht the fourth one ( ) is otuse. Proposition 2 pplie to tells us tht 2 > , where the ltter inequlity follows from the ssumption tht is shortest sie of (n, therefore, ). Thus, we hve tht / 2. This long with the ft tht x (, ) implies inequlity (13). (2) Assume now tht. Let i e the intersetion point etween n the horizontl line through (refer to Fig. 8). Note tht i π/2 n i π/2 (these two ngles sum to π). This long with Proposition 2 pplie to tringle i shows tht 2 i 2 + i 2. Similrly, Proposition 2 pplie to tringle i shows tht 2 i 2 + i 2. The two inequlities ove long with our ssumption tht imply tht i i, whih in turn implies tht i /2, euse i + i =. Sine x is elow i (otherwise, x <, ontriting the ft tht is n ege in Y 4 ), we hve δ i. It follows tht δ /2. Finlly we erive ontrition using the now estlishe inequlity (13). Let j e the orthogonl projetion of x onto the vertil line through (thus j = δ).

19 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 19 Note tht jy < π/2, euse y Q 4 (x). By Proposition 2 pplie to jy, we hve y 2 < j 2 + jy 2 = δ 2 + jy 2. Sine y n re in the sme qurnt of, n sine Y 4, we hve tht y. This long with the inequlity ove n (13) implies tht jy / 2 δ. By Proposition 2 pplie to xjy, we hve xy 2 > xj 2 + jy 2 xj 2 + δ 2 = xj 2 + j 2 = x 2. It follows tht xy > x, ontriting our ssumption tht xy Y Proof of Lemm 6 Proof. We first show tht / Q 3 (). Assume the opposite. Sine Q 1 () n x x y > y x y () xy > x () Fig. 9. Lemm 6: () P ( ) oes not ross. () If is not the shortest sie of, the lemm onlusion might not hol. Q 3 (), we hve tht > π/2. This implies tht <, whih long with the ft tht, Q 3 () ontrit the ft tht Y 4. Also note tht / Q 1 (), sine in tht se n oul not interset. In the following we isuss the se Q 2 (); the se Q 4 () is symmetri. A first oservtion is tht must lie elow ; otherwise < (sine > π/2), whih woul ontrit the ft tht Y 4. We now prove y ontrition tht there is no ege in P ( ) rossing. Assume the ontrry, n let xy P ( ) e suh n ege. Then neessrily x (, ) n xy Q 4 (x). Note tht y nnot lie elow ; otherwise x < xy (sine xy > π/2), whih woul ontrit the ft tht xy Y 4. Also y must lie outsie D(, ) Q(, ), otherwise oul not e in Y 4. These together show tht y sits to the right of. See Fig. 9. Then the following inequlities regring the qurilterl xy must hol:

20 Septemer 30, :24 WSPC/Guielines Y4Spnner 20 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer (i) y >, ue to the ft tht y > π/2. (ii) x ( x = if x n oinie). If x n re istint, the inequlity x > follows from the ft tht x (sine x is outsie D(, )), n Proposition 1 pplie to the qurilterl x: + x < x + Inequlities (i) n (ii) show tht y n x re longer thn sies of the qurilterl, n so they must e longer thn the shortest sie of, whih y ssumption (ii) of the lemm is : min{ x, y } x (this ltter inequlity follows from the ft tht x (, )). Also note tht y, sine Y 4 n y lies in the sme qurnt of s. The ft tht oth igonls of xy re in Y 4 enles us to pply Lemm 4(ii) to onlue tht y is not shortest sie of the qurilterl xy. Thus x is shortest sie of the qurilterl xy, n we n use Lemm 4(ii) to lim tht x < min{ xy, } xy. This ontrits our ssumption tht xy Y 4. Fig. 9() shows tht the lim of the lemm might e flse without ssumption (ii) Clultions for the streth ftor of p(, ) in Lemm 9 We strt y omputing the streth ftor of the short pths lime y sttements S2 n S3. S2 If xy Y 4 n zw Y 4 re short, n if xy intersets zw, then there is short pth P etween ny two of the enpoints of these eges, of length P xy + zw + 3(2 + 2) mx{ xy, zw }. (14) This upper oun n e erive s follows. Let ij e shortest sie of the qurilterl xzyw. By Lemm 8, Y 4 ontins pth p(i, j) no longer thn 6( 2+1) ij. By Lemm 4, ij mx{ xy, zw }/ 2. These together with the ft tht P xy + zw + p(i, j) yiel inequlity (14). S3 Here we prove tighter version of this sttement: If p(x, y) n p(z, w) re short pths tht interset, then there is short pth P etween ny two of the enpoints of these pths, of length P p(x, y) + p(z, w) + 3(2 + 2) mx{ xy, zw }. (15) This follows immeitely from S2 n the ft tht no ege of p(x, y) p(z, w) is longer thn mx{ xy, zw } (y Lemm 8). Cse 1: P ( ) n interset. Then y S3 we hve

21 Septemer 30, :24 WSPC/Guielines Y4Spnner π/2-angle Yo Grphs re Spnners 21 p(, ) P (, ) + + 3(2 + 2) mx{, } (2 + 2) 2 (y (7), (11)ii) = 3( ) 3( ) (y (11)i). Cse 2(i): P ( ) n o not interset; P ( ) n o not interset; n P ( ) intersets. By S3, there is short pth p(, ) of length p(, ) P (, ) + + 3(2 + 2) mx{, } (2 + 2) mx{, } (y (7)). (16) Next we estlish n upper oun on. By the tringle inequlity, < + 3 (y (12)). (17) Sustituting this inequlity in (16) yiels p(, ) ( ). (18) Thus p(, ) = p(, ) P 1 ( ) is pth in Y 4 of length p(, ) p(, ) + 2 (y (7)) p(, ) + 2 (y (11)ii) ( ) (y (18)) ( ) (y (11)i). Cse 2(ii): P ( ) n o not interset; P ( ) n o not interset; n P ( ) oes not interset. Then P ( ) must interset P ( ) P ( ). By S3 there is short pth p(, ) of length p(, ) P ( ) + P ( ) + P ( ) + 3(2 + 2) mx{,, } ( + + ) 2 + 3(2 + 2) mx{,, } (y (7)). Inequlities (11)ii, (12) n (17) imply tht mx{,, } 3. Sustituting in the ove, we get p(, ) ( ) 2 + 9(2 + 2) ( ) (y (11)i). Thus p(, ) = p(, ) is pth in Y 4 from to of length p(, ) ( ) ( ) (y (11)i).

22 Septemer 30, :24 WSPC/Guielines Y4Spnner 22 Bose, Dmin, Douïe, O ourke, Semone, Smi n Wuhrer Cse 3: P ( ) n o not interset, n P ( ) intersets. If P ( ) intersets t, then p(, ) = P ( ) P ( ) is lerly short n oes not exee the spnning rtio of the lemm. Otherwise, there is n ege e P ( ) tht rosses, n P ( e) intersets e P (e ) (s estlishe in the proof of Lemm 9). If P ( e) intersets e, then y S3 there is short pth p(, e) of length p(, e) P ( e) + e + 3(2 + 2) mx{ e, e } (19) Otherwise, if P ( e) intersets P (e ), then y S3 there is short pth p(, e) of length p(, e) P ( e) + P (e ) + 3(2 + 2) e (20) A loose upper oun on e n e otine y employing Proposition 1 to the qurilterl e: e + < + e < +. Sustituting the upper oun for from (17) yiels e < (21) By Lemm 2, e (sine e P ( )), whih long with (17) implies Sustituting inequlities (7), (21) n (22) in (19) yiels e 3. (22) p(, e) ( ). Sustituting inequlities (7) n (21) in (20) gives p(, e) ( ), whih is looser upper oun tht pplies to oth ses. Then p(, ) = p(, e) P 1 is pth from to of length ( ) P 1 ( ) p(, ) ( ) (y (23), (17), (11)) = ( ).