Parity-constrained Triangulations with Steiner points

Transcription

1 Parity-contrained Triangulation with Steiner point Victor Alarez October 31, 2012 Abtract Let P R 2 be a et of n point, of which k lie in the interior of the conex hull CH(P) of P. Let u call a triangulation T of P een (odd) if and only if all it ertice hae een (odd) degree, and peudo-een (peudo-odd) if at leat the k interior ertice hae een (odd) degree. On the one hand, triangulation haing all it interior ertice of een degree hae one nice property; their ertice can be 3-colored, ee [1, 2, 3]. On the other hand, odd triangulation hae recently found an application in the colored erion of the claic Happy Ending Problem of Erdő and Szekere, ee [4]. In thi paper we how that there are et of point that admit neither peudoeen nor peudo-odd triangulation. Neerthele, we how how to contruct a et of SteinerpointS = S(P)ofizeatmot k 3 +c,whereciapoitiecontant,uchthat a peudo-een (peudo-odd) triangulation can be contructed on P S. Moreoer, we alo how that een (odd) triangulation can alway be contructed uing at mot n 3 +c Steiner point, where again c i a poitie contant. Our contruction hae the property that eery Steiner point lie in the interior of CH(P). 1 Introduction Let P R 2 be a et of n point in general poition and let Γ : P {0,1} be an aignment of paritie to the element of P, where 0 mean een and 1 mean odd. Let G be a traight-edge plane graph with ertex et P. We ay that a ertex P of G i happy, with repect to G, if and only if the degree of in G i of the parity aigned to by Γ. If a ertex i not happy w.r.t. G we will ay that i unhappy. If the graph G i clear from context we will jut ay that ertice are happy or unhappy without referring to G. Gien P and Γ, the problem of finding plane graph on P that maximize the number of happy ertice ha recently receied ome attention. In [5] it wa hown that one can alway contruct a tree, a 2-connected outerplanar graph, and a pointed peudotriangulation that make all but at mot three ertice happy. For the cla of triangulation it wa hown that one can alway contruct one that make eentially 2n 3 of it Fachrichtung Informatik, Unierität de Saarlande, alarez@c.uni-aarland.de. 1

2 ertice happy, but a configuration of point with paritie wa alo hown where at leat n 108 ertice remain unhappy regardle of the choen triangulation. The contruction of thi lower bound require the ue of both paritie, but the author pointed out that there are two particular cae that might accept triangulation with a many a n o(n) happy ertice. Thee two particular cae are the one where the paritie aigned to the element of P by Γ are either all een or all odd repectiely. Howeer, howing that in thoe particular cae n o(n) ertice can be made happy turned out to be ery challenging. In the ame paper the author howed that in the all-een cae, a triangulation that make at leat 2n 3 ertice happy can alway be contructed. They alo howed that in the all-odd cae a fraction of happy ertice can alway be enured. Thee two pecial cae, all een or all odd, are of ignificant interet ince they hae intereting propertie and/or application. For example, it i well known that a connected graph G haing all it ertice of een degree i Eulerian. If on top of it G happen to be a triangulation a well, then G i alo 3-colorable, ee [1, 2] for a general reference on 3-colorable planar graph. Thoe two propertie are uually conidered nice in a graph, and they are characterized only by the parity of the degree of it ertice. For 3-colorability of triangulation a lightly weaker characterization i known: T i a triangulation haing all it interior ertice of een degree if and only if T i 3-colorable, ee [3] for example. The application related to the all-odd cae i a little bit more intricate and we refer the reader to [4] where thi application i hown. Let P bea before. We will ay that atriangulation T of P i een, or odd, if andonly if the degree of eery ertex of T i een, or odd repectiely. If only the interior ertice of T are een, or odd, we will ay that T i peudo-een, or peudo-odd repectiely. Thi define four kind of triangulation of P: Een, peudo-een, odd, peudo-odd. In thi paper we attack the following problem: Gien P and one kind T of triangulation of the four mentioned aboe, contruct a et S = S(P,T ) uch that a triangulation of kind T can alway be contructed on P S. Thu, the problem attacked in thi paper can be een a the Steiner-point erion of the one regarding triangulation preented in [5]. With thi idea in mind, the following lemma i worth noting: Lemma 1. There are arbitrarily large et of point that, without the ue of Steiner point, admit neither peudo-een nor peudo-odd triangulation. Proof. Let P be a et of point like the one hown to the left in Figure 1 where the ize of the conex polygon Q hown in gray i 1 mod 3. Let P be the only point not in Q. It i clear that in any triangulation of P, point mut be adjacent to eery ertex of Q, that i, without a triangulation of Q, eery ertex of Q ha degree three. Now, it i well known that eery triangulation of a polygon ha at leat two ear, i.e., a triangle formed by three conecutie ertice of the polygon. Thi mean that, regardle of what triangulation of Q we chooe, there will alway be a ertex of Q whoe adjacencie are only it two neighbor in Q and. Thu no peudo-een triangulation of P exit. See to the right in Figure 1. 2

3 Q u w u w Figure 1: To the left we hae a configuration in which all hown adjacencie are forced, and it accept neither peudo-een nor peudo-odd triangulation. To the right we how in red one of the ear of the hown triangulation of Q. To how that P doe not admit a peudo-odd triangulation either it uffice to how that regardle of what triangulation of Q we chooe, there will alway be at leat one interior point of P haing een degree. It i not hard to erify that in an een triangulation, the ize of the outer face mut be 0 mod 3, a proof can be found in [3]. Since Q 1 mod 3, then Q doe not admit an een triangulation, o in eery triangulation of Q there will be at leat two ertice of odd degree, there mut be an een number of them. So aume that there i a triangulation of Q in which the only two ertice of odd degree are the two neighbor u,w of in CH(P). Thu we could add a point p outide CH(P), below the edge uw, and add the adjacencie pu,pw. Thi implie that the et of point Q = Q {p} ha an een triangulation, where upw i an ear. But Q 2 mod 3, which clearly contradict the neceary condition on the ize of the outer face of an een triangulation. Therefore, in any triangulation of Q there mut be at leat one interior point q P of odd degree. The force adjacency q implie that q get een degree in a triangulation of P, which i what we wanted to proe. 2 Our contribution By Lemma 1 the ueof Steiner point i ometime neceary if we init in contructing any of the four kind of triangulation mentioned before(een, peudo-een, odd, peudoodd). The releant iue now i not whether we can contruct the triangulation we are intereted in, but rather with how many Steiner point can we achiee uch contruction, the le, the better. The reult we are going to how are the following: Theorem 1. Let P R 2 be a et of n point in general poition where k of them are interior point. Then a et S = S(P) of interior Steiner point of ize at mot k can alway be contructed uch that P S accept a peudo-een triangulation. Theorem 2. Let P be a before. Then a et S = S(P) of interior Steiner point of ize at mot n can alway be contructed uch that P S admit an een triangulation. 3

4 Theorem 3. Let P R 2 be a et of n point in general poition where k of them are interior point. Then a et S = S(P) of interior Steiner point of ize at mot k 3 +c, with c a poitie contant, can alway be contructed uch that P S accept a peudo-odd triangulation. Theorem 4. Let P be a before. Then a et S = S(P) of interior Steiner point of ize at mot n 1 3 +c, with c a poitie contant, can alway be contructed uch that P S admit an odd triangulation. The proof of all theorem will be contructie. The ret of the paper i diided a follow: In Section 3 we tart with ome preliminarie. In Section 4 and 5 we how algorithm that imply Theorem 1, 2, and Theorem 3, 4 repectiely. We cloe the paper in Section 6 we ome oberation and concluion. 3 Pre-proceing of P Let u quickly recall that gien a polygon P, a ertex of P i called reflex if the internal angle at i larger than π. We will call it conex otherwie. Alo, by a uitable rotation of P we can aume that the ertex of CH(P) with the lowet y-coordinate i unique. The following pre-proceing of P will be done: Uing a a piot we will ort each interior point of P by it angle with repect to. Let p 1,...,p k, be a labeling, from left to right with repect to thi angular order, of the interior point of P. Let p 0,p k+1 be the left and right neighbor of on CH(P) repectiely. We contruct a imple polygon P from P \{} a follow: We add each edge p i p i+1, for 0 i k. We call thi the lower part of P and we will denote it by L(P). Alo, we conider the edge of CH(P)\{p 0,p k+1 } and we call thi the upper part of P, which will be denoted by U(P), ee Figure 2. Next we will can L(P) from left to right and we will conider the larget polygonal chain that can be formed uing conecutie conex ertice of L(P). Note that for each uch polygonal chain, the left and right ertice mut be reflex ertice of P, ee Figure 3. Now, for each chain, we will make adjacent it two endertice, thu forming conex polygon Q j, with j 0, from thoe conex polygonal chain. Thee conex polygon can be thought a big ear hanging from L(P). We will triangulate the ret of P outide thee Q j arbitrarily, ee Figure 2. If there i no conex ertex of P in L(P), then thi arbitrary triangulation of P i a triangulation of all P. Thi i all the pre-proceing that will be done. From here eery algorithm will complete a triangulation of P in it own way. 4 Een and peudo-een triangulation The following oberation i ery ueful when working with 3-colorable triangulation: 4

5 p 0 p k+1 p 0 p k+1 Figure 2: To the left we hae the polygon P on n 1 ertice in gray. The conex polygon formed by canning L(P) from left to right are hown dahed. Note that each pair of conecutie conex polygon hare at mot one ertex. To the right we ee a triangulation T(P) of P. The dahed edge are the only one that are not arbitrary. Oberation 1. Let T be a 3-colored triangulation with outer face C, not necearily conex, and let u,,w be three conecutie ertice of C. If the degree of in T i een, then the color of u i different than the color of w. If the degree of in T i odd, then both u,w hae the ame color. We can now continue with the algorithm. Let u triangulate each Q j, if any exit, a follow: Take it ertex with the lowet y-coordinate, breaking tie arbitrarily, and join it withtraight-line egmenttoallotherertice ofq j, incaethatq j hamorethanthree ertice. Thi i uually called a fan triangulation. Thee fan triangulation along the arbitrary triangulation outide the Q j complete a triangulation T(P) of polygon P. It i well-known that triangulation T(P) can be 3-colored, ee [6], thu we will do it, and obere that the only colorle ertex i, the one we ued in the beginning to ort P angularly. Clearly, if a triangulation of P i 3-colorable, then it mut be at leat peudo-een, thu our idea now i to complete a 3-colorable triangulation of P uing T(P) and it 3-coloring. So at all time we will ue a 3-coloring a a meaure of the correctne of our algorithm. From thi point on, our contruction i done by cae analyi. We will aume without lo of generality that the color at our dipoal are {0,1,2}. Note that a T(P) i already 3-colored, if all the interior ertice of P are colored by only two color, ay 0,1, we could ue color 2 for without iolating the 3-coloring of T(P), and hence, uing the traight-line egment that connect with each ertex of L(P), we obtain a 3-colorable triangulation of P. Neerthele, in general it i not going to happen that the interior ertice can be colored uing only two color, hence we need to do omething ele in uch cae. We will proceed in a line-weep fahion from p 0 to p k+1 with repect to the angular order around. 5

6 Let u fix the color of a the color of the mallet chromatic cla in L(P) uing the 3-coloring of T(P), and ay that color i i 1 without lo of generality, 0 i 2. Note that the point in L(P) with color i are the one cauing trouble to complete the deired triangulation, hence we will handle thoe point depending on their kind in P, namely if they are reflex or conex ertice of P. We will keep the inariant that, by the time we are proceing an interior point, edge i preent, and all interior point p r, with r < j, hae een degree already, except poibly for p 0. Alo note that by thi time, if the degree of i odd it i becaue +1 ha color i, due to Oberation 1. Let u tart now with our cae analyi, we will aume that we are currently proceing the interior point, 1 j < k, o again, we will aume that the edge i already preent in the partial triangulation of P contructed o far. We hae the following cae: (1) Point +1 i of color i, jut a, and i a reflex ertex of P. If point and +2 are of different color then we can jut add the edge +2, ince +1 ha already een degree in T(P), ee to the left in Figure 3. Thu we can add the edge +2 and moe to +2. If on the other hand, and +2 hae the ame color, we introduce one Steiner point, below L(P), of the third color different than the one of and, which will be adjacent to,+1,+2 and, a hown in the middle in Figure 3. Thu we can again moe to +2 and continue p l Q Figure 3: The point i currently being proceed. Point +1 i of the ame color i of. If and +2 hae the ame color, then one Steiner point uffice to be able to moe to +2. To the right the conex polygon Q i hown in gray. Point +1 i the piot of the fan triangulation of Q. (2) Point +1 i of color i again but a conex ertex of P. If and +2 hae the ame color we proceed jut a before, introducing one Steiner point, below L(P) and of the third aailable color, which will be again adjacent to,+1,+2 and. We moe to +2 and continue, ee in the middle in Figure 3. 1 In the figure, unle otherwie tated, we will ue color {i,i + 1,i + 2} = {black,red,blue}, with arithmetic modulo 3. 6

7 So let u aume that and +2 hae different color, ay w.l.o.g. i + 1 and i + 2 repectiely. Note that, a +1 i a conex ertex of P, it mut be part of one of the conex polygon Q 1,...Q r, the big ear, we contructed in the pre-proceing phae. Let u denote thi one conex polygon imply by Q, and it rightmot ertex by p l, l j +2, according to the angular order around. We hae now the following ub-cae: (2.1) Vertex +1 wa ued a a piot in the fan triangulation of Q, ee to the right in Figure 3. Thi in particular mean that +1 i the only ertex of color i in Q. If l > j + 2 then we can re-triangulate Q by contructing the fan triangulation of Q with piot at p l 1, thi change the 3-coloring of Q only between +1 and p l 1, the former receie color i + 2 while the latter i the new unique ertex of Q of color i. Thu the only thing we did wa to moe the ertex of color i to the right, and hence we can join to all ertice +1,...,p l 2 uing traight-line egment. If l = j +2 then Q i actually a triangle, and eerything i till alid, i connected to p l 2 =, ee to the left in Figure 4. We now further ditinguih between the following cae: (2.1.1) Point p l i of color i + 1, p l+1 i of color i and p l+2 i of color i + 2, ee in the middle in Figure 4. We introduce two Steiner point 1, 2 of color i+2,i+1 repectiely and we will make the following adjacencie: (1) 1 get adjacent to p l 1,p l,p l+1 and 2, and (2) 2 get adjacent to p l 2,p l 1, 1,p l+1,p l+2,. Obere that thee adjacencie can alway be done without introducing any croing. Moreoer, note that with two Steiner point we complete the een degree of each point in the region,...,p l+2 in which there were originally two point of color i. Thu we can moe to p l+2 and continue. (2.1.2) Point p l and p l+1 a before, but p l+2 i of color i+1. We will proceed a before except that thi time the adjacencie of 1, 2 are a follow: (1) 1 get adjacent to 2,p l 1,p l,p l+1,p l+2 and, and (2) 2 get adjacent to p l 2,p l 1, 1 and. A before, we alo introduce the adjacencie +1,...,p l 2 and p l+2. We can again moe to p l+2. See to the right in Figure 4 for the final configuration. (2.1.3) Point p l a before but p l+1 i of color i + 2 intead. Note that in thi cae, from to p l+1 there i only one ertex of color i, namely p l 1, thu we will introduce only one Steiner point. Alo obere that ince p l i a reflex ertex of P we can add the adjacency p l 1 p l+1. Finally we make adjacent to p l 2,p l 1,p l+1,, and we make, a before, adjacent to +1,...,p l 2 and p l+1. See to the left in Figure 5 for the final configuration. The following three cae are complementary: (2.1.4) Point p l i of color i+2, p l+1 i of color i and p l+2 i of color i+1. 7

8 p l 1 p l p l p p l p l+2 p l+2 l 1 p l 1 Q p 2 j Figure 4: If +1 wa ued a a piot to triangulate a conex polygon that can be cut from P, then we can ue p l 1 a the new piot without changing the color of or anything to it left. Note that p l mut be necearily a reflex ertex of P. In the middle we ee the final configuration in the cae that p l+1 i of color i and p l+2 i of color i+2. To the right we ee the final configuration when p l+1 i of color i and p l+2 i of color i+1. p l p p l 1 l q = q +2 2 Figure 5: To the left we ee the final configuration in the cae that +1 wa a piot of color i and p l+1 i of color i + 2. In the middle and to the right we hae that, if +1 of color i wa not a piot and it neighbor hae different color from each other, then one of them mut necearily be a piot, +2 in thi cae. So we hae to go back and remoe ome adjacencie that will allow u to introduce the Steiner point appropriately. Quadrilateral i hown in gray. (2.1.5) Point p l and p l+1 a before and p l+2 i of color i+2. (2.1.6) Point p l a before and p l+1 i of color i+1. Howeer, thee cae are the ymmetric cae of (2.1.1), (2.1.2) and (2.1.3) repectiely, where p l get the other poible color, and thu the contruction get hifted color. The olution, a can eaily be erified, i alo of hifted color. The two lat figure of Figure 4 and the firt one of Figure 5 can be ued a reference if we hift their color. (2.2) In thi cae +1 of color i wa not ued a a piot in the fan triangulation of Q. Thi mean that and +2 are adjacent, ince they are part of Q. So 8

9 we hae two cae depending on whether or +2 i the piot in the fan triangulation of Q. (2.2.1) Point i the piot in Q. Then conider the point two poition ahead, thati+3,+4. Letuaumew.l.o.g. thattheyexit, otherwie+2 = p k+1, which i the right neighbor of on CH(P), and,+1,+2 are the lat three point that the algorithm will proce. It i then eay to erify that two interior Steiner point uffice to finih. Now, conider the pattern of color of,...,+4. The firt three color are fixed i+1,i,i +2. If we ee i,i +1 next, then the pattern matche what we ee in cae (2.1.4) 2, which we know how to ole uing two Steiner point, but we would alo get rid of two point of color i. If we ee next i,i + 2, then we ee the ame a in cae (2.1.5). If +3 i of a color different than i then it mut necearily be of color i + 1, ince +2, of color i + 2, and +3 are adjacent. But ince, of color i +1, i the piot of Q, then +3 cannot be a ertex of Q, o +2 i actually a reflex ertex of P, and we would find ourele eeing what we ee in cae (2.1.6). (2.2.2) Point +2 i the piot in Q. Thi cae i trickier ince for the point,...,+4 we could ee the pattern of color i+1,i,i+2,i+1,i, which we do not know how to ole locally uing only two Steiner point. What we will do i to not look ahead but to ee behind. Since the edge i currently in the triangulation being built, there i exactly one triangle uing thi edge. Let q {,} be the third ertex of uch triangle. Note that q lie to the left of the edge and hence it already ha een degree, moreoer, the color of q i i+2, ince, hae color i+1,i repectiely. Now we hae the following two cae: Vertex q i a Steiner point, or the quadrilateral = q,,+1, i conex. Let u conider only the cae that i conex, if that i not the cae then q i a Steiner point and it can be moed a pleaed to make conex without affecting anything. We will flip the edge for the edge q+1 and introduce one Steiner point of color i + 1 that will be adjacent to q,+1,+2,, ee in the middle of Figure 5. If q i not a Steiner point and i non-conex, then it i not hard to ee that the only poibility i q = 2, and 1 mut then be a reflex ertex of P of color i. Note then that the edge e = 2 mut be preent in the triangulation, by cae (1), and thu 1 i adjacent to no Steiner point. Hence we will remoe e and we will introduce one Steiner point 1 of color i+2 that i adjacent to 1,,+1, 2, where 2 i another Steiner point of color i + 1 that i adjacent to 2 Which in turn i the ymmetric cae of (2.1.1). 9

10 2, 1, 1,+1,+2,. We can now moe to +2 and continue. See to the right in Figure 5. Note that the color i of wa choen a the color of the mallet chromatic cla in L(P), o it cardinality can be at mot k+2 3. Now, it i important to obere that we aumed that the point that we proce i neither p 0 nor p k+1 of CH(P), ince the algorithm tarted with j 1. So it could happen that at leat one of thoe extreme point i of the ame color i of, which would gie a conflict in the 3-coloring of the triangulation we are contructing. Let u ee how can we deal with thi kind of ituation. Aume without lo of generality that p 0 i of color i. Then, before tart proceing the firt interior point p 1, introduce one Steiner point inide CH(P), o cloe to that the angular order p 1,...,p k w.r.t. i alo kept by. Thi Steiner point will replace in the algorithm, o it will get color i a well. Now ymbolically delete and run the algorithm. When the algorithm end we will till hae the conflict of the monochromatic edge p 0, we could imultaneouly hae the ame conflict with edge p k+1. p 1 p 1 p k p 0 p k+1 p 0 p k+1 Figure 6: Polygon P hown in light gray. In the figure color i = black, and the color white mean that thoe point are omehow 3-colored without conflicting with the black point. The iibility region of i hown in dark gray. Put back and remoe the conflicting edge from the contruction. We will proceed depending on what ee, haing the edge of the contruction a obtacle. That i, if edge p 0 i the only one with conflict, then ee the polygonal chain p 0,p 1,,p k+1, ee to the left in Figure 6. If the edge p k+1 alo ha conflict then ee the polygonal chain p 0,p 1,,p k,p k+1, ee to theright in Figure 6. Theolution will dependon whether p 0 i the only conflict or not, and whether ha een degree, in the triangulation contructed by the algorithm, or not. The four cae, hown in olid line, and their olution, hown with dahed line, can be een in Figure 7. Obere that at mot one more Steiner point i ued for the olution, and that, are both interior Steiner point. So can be charged to p 0, which i one of the k+2 3 point of L(P) of color i, and would be imply one interior Steiner point that cannot be charged to anything. Finally, in the analyi of cae (2.1.1) to (2.1.6) we alway aumed that point p l+1 alway exited. Thi might not alway be the cae ince we could hae p l = p k+1, but in thi cae we could afely aume that p l+1 = or p l+1 =, depending on conflict on CH(P), o the econd Steiner point we introduce in thoe cae i alo an interior 10

11 p 1 p 1 p 0 p k+1 p 0 p k+1 p 1 p k p 1 p k p 0 p k+1 p 0 p k+1 Figure 7: Polygon P hown in gray. On the top part we hae the olution for the cae where p 0 i the only conflict and the degree of i odd, left, or een, right. Below we hae the olution for the cae when both edge p 0, p k+1 are in conflict and the degree of i odd, left, or een, right. In the figure color i = black. Steiner point that cannot be charged to anything. Hence the contruction ued at mot +2 interior Steiner point, and Theorem 1 follow. k Extenion to een triangulation Theorem 1 guarantee a 3-colorable triangulation, which will be at leat peudo-een, but it might not necearily be completely een, and thi i becaue when we chooe an arbitrary triangulation for a part of P, ome ertice of CH(P) might get odd degree. Thu in order to contruct an een triangulation we hae to do ome more work. A we mentioned before, in an een triangulation the ize of the outer face mut be multiple of three. Thu the firt thing we will do, if neceary, i to complete CH(P) uing at mot two Steiner point o that we fulfill CH(P) 0 mod 3. Let again the unique ertex of CH(P) with the mallet y-coordinate, and ort all P \{} angularly, from left to right, around. Let p 1,...,p n 1 be the point of P \{} in thi orted order. The main idea behind the contruction i to encloe P in a bigger polygon Q o that all p 0,...,p n 1 are interior point, and then run the algorithm of Theorem 1, which will guarantee that all p 0,...,p n 1 will hae een degree. The contruction i done in uch a way that CH(P) appear in the contruction, at the end we can complete the degree of to an een degree, if neceary, and uch that by remoing Q we do not detroy 11

12 any parity, ince Q, upon deletion, will take with it an een number of adjacencie per affected point of CH(P), o the degree of thoe affected point will remain een at the end. So let u jump to the actual contruction. Encloe CH(P) in a bigger polygon Q with the following propertie: (1) The ize of Q i CH(P) + 1, (2) The ertex of Q with mallet y-coordinate i alo and i unique, (3) The polygon C formed by Q CH(P) can be triangulated uing a zigzag tarting at. That i, if we denote the ertice of CH(P) in clockwie order tarting with by q 1 =,q 2,...,q k, with k 3 and k 0 mod 3, and we denote the ertice of Q the ame way by q 1 =,q 2,...,q k,q k+1, then the zig-zag i q 1 = q 1 =,q 2,q 2,q 3,...,q k,q k,q k+1, = q 1 = q 1, which along the edge of C complete a triangulation of C itelf. See to the left in Figure 8. Q C Q C q 3 q k 1 q 3 q k 1 q 3 q 2 = p 1 q k = p n 1 q k q 3 q 2 = p 1 q k = p n 1 q k q 2 q k+1 q 2 q k+1 = q 1 = q 1 = q 1 = q 1 Figure 8: To the left: The polygon Q i the outer face of the contruction hown. Obere that it doe not hae to necearily be conex. The conex hull of P, CH(P), i hown in dark gray, and C i hown in light gray, along with it zig-zag triangulation. To the right: The particular 3-coloring of the zig-zag triangulation of C uing color {0,1,2} = {black,blue,red}. Thi contruction can alway be achieed and ha the following propertie: (1) All point of P \{} lie in the interior of Q, (2) In the zig-zag triangulation of C, eery ertex of CH(P), and eery ertex of Q, except for q 2,q k+1, ha een degree, (3) In the 3-coloring of the triangulation of C that tart with color 0 at q 2 Q, and color 1 at q 2 CH(P), we hae that q k CH(P) receie color 0, ertex q k+1 Q receie color 1, and receie color 2. See to the right in Figure 8. Now, pre-proce Q P a explained in 3. Thi pre-proceing will contruct in a firt tep the polygon P ued in the algorithm of Theorem 1. The reader hould ee that thi time P i the polygon whoe upper part U(P) i Q\{q 2,q k+1}, and whoe lower part L(P) i formed by the adjacencie p i p i+1, for 1 i n 2, along with the adjacencie q 2 q 2, q k+1 q k. In a econd tep, the ear formed by conecutie conex 12

13 ertice of L(P) will be computed. Recall that thi ear are the conex polygon Q j in the pre-proceing phae. Here, by uitably locating q 2,q k+1 Q we can aume that both q 2,q k CH(P) are reflex ertice of P, ee Figure 9. Moreoer, obere that any other ertex q i CH(P), 2 < i < k, mut alo be a reflex ertex of P. Thi implie that the conex polygon Q j are contained in CH(P). Thi i important becaue in a third tep of the pre-proceing we compute a arbitrary triangulation of P minu thoe Q j. At thi point we will chooe the zig-zag triangulation of C a part of thi arbitrary triangulation, adding other arbitrary edge inide CH(P) if the zig-zag triangulation of C doe not complete a triangulation of P minu the Q j, the important thing here i that the zig-zag triangulation of C, which contain CH(P), appear. P P q 3 q k q 3 q k q 2 q k+1 q 2 q k+1 = q 1 = q 1 = q 1 = q 1 Figure 9: Polygon P hown in gray. The dah line delimit the ear that are contructed by the algorithm. They are contained in CH(P) ince eery ertex of CH(P) i a reflex ertex of P. To the left we can ee a whole 3-colored triangulation of P, where the zig-zag triangulation of C appear. The 3-coloring i extended from the 3-coloring of C. If we now execute the algorithm of Theorem 1, the firt thing it will do i to compute a fan triangulation of each Q j, if any exit, ee to the right in Figure 9. Thi will complete a triangulation T(P) of polygon P. The econd thing it will do i to compute a 3-coloring of T(P). Clearly we can extend the particular 3-coloring of the zig-zag triangulation of C, explained before, to a 3-coloring of T(P). Here again the important thing i that thi particular 3-coloring of the triangulation of C appear. The third thing the algorithm will do i to re-color with the color of the mallet chromatic cla of the 3-coloring of T(P). Here we hae two cae: Vertex tay of color 2. Obere that ince q 2,q k+1 Q,q 2,q k CH(P), the neighbor of in Q and CH(P) repectiely, hae color 0 and 1, then doe not create any conflict with the 3-coloring of T(P). Thu we will keep the adjacencie 13

14 q 2,q 2,q k and q k+1, and we will execute the algorithm to the end. Since the color of q 2,q k+1,q 2,q k, are neer changed by the algorithm, we end up haing a 3-colored triangulation, where alo ha een degree, that ue at mot n interior Steiner point. Note that thee interior Steiner point are alo interior w.r.t. CH(P), ince the algorithm of Theorem 1 would introduce the Steiner point inide Q, but trictly below the lower part L(P) of P. Since the adjacencie q 2,q 2,q k and q k+1 are preent in the output triangulation, the Steiner point fall trictly inide CH(P). The reader can erify that if we now remoe Q\{}, along with all the adjacencie that it take with it, we are left with the deired een triangulation. Vertex change color. Then we will aume without lo of generality that get color 0. If got color 1 we would hae a ymmetric coneration. Remember that by the particular 3-coloring of C we hae that q 2,q k+1 Q hae color 0,1 repectiely, and q 2,q k CH(P) hae color 1,0 repectiely, ee Figure 8. Thu the algorithm will introduce the Steiner point of color 0, to take the place of, and will ymbolically delete. Thi time we can charge to q 2 which i one of the n+1 3 point of L(P) of color 0. Now introduce the adjacency p n 2 and execute the algorithm until it finihe. There, point p n 2 would bethe lat proceed point. So we arrie at the configuration hown in the left upper corner of Figure 10. In thi configuration we necearily hae that the degree of q 2 = p 1 CH(P) i odd, ince it i adjacent to q 2, Q which hae both color 0. The degree of p n 2 i alo odd ince it i adjacent to,q k = p n 1 CH(P), and both are colored 0 a well. So the only option we hae now are whether ha een or odd degree between q 2 and p n 2. Thi i equialent to conider whether p n 2 i of color 2 or 1 repectiely. Thoe two configuration are hown in the right upper corner, and in the middle repectiely, in Figure 10 in olid. Now let u put back and we will finih the contruction a hown in Figure 10 with dahed line. We might requirethe ueof another Steiner point, a hown to theright in theuppercorner of theame figure, which can becharged to q k = p n 1. In the end get een degree a well. It i not hard to erify that the total number of interior Steiner point i again at mot n , that all Steiner point are interior to CH(P), and that the remoal of Q \ {} doe not detroy any parity. Hence Theorem 2 follow. 5 Peudo-odd and odd triangulation Working locally with (peudo-)odd triangulation i lightly eaier. The following oberation wa already pointed out in [7]: 3 It i not n thi time ince there i no conflict in CH(P). 14

15 q 3 q k = p n 1 q k 1 q 3 q k = p n 1 q k 1 q 3 q 2 = p 1 q 2 p n 2 q k q 3 q 2 = p 1 q 2 p n 2 q k q k+1 q k+1 = q 1 = q 1 = q 1 = q 1 q 3 q k = p n 1 q k 1 p n 2 q 3 q 2 = p 1 q 2 q k q k+1 = q 1 = q 1 Figure 10: Polygon P i hown in light gray. Uing color {0,1,2} = {black,blue,red}, if we made a if point p n 2 wa the lat point, we arrie at the configuration hown in the left upper corner. The white color of point p n 2 mean that we do not care about it real color at thi time. If we put back, and we color it with 3, we can add the dahed adjacencie hown in the middle and the right upper corner, depending on the actual color of p n 2. The color of will conflict with the color of q 3,q k Q, but thi i not a problem ince in the end we will remoe Q\{}. Oberation 2. Let be a triangle in a triangulation. Then at mot een interior Steiner point uffice to obtain an odd-triangulation of. The way thi odd triangulation of i obtained i hown in Figure 11. Obere that, in general, no imilar tatement can be done for (peudo-)een triangulation. For implicity we will right now focu on peudo-odd triangulation only. The algorithm to contruct them i eentially the ame a for peudo-een triangulation, howeer, thi time we will not be able to ue a 3-coloring a guide. That 3-coloring played an important role in the upper bound on the number of interior Steiner point that we ued: We did not hae to explicitly track whether we introduced at mot one Steiner point eery three interior point, but we jut had to take care of oling conflict with the coloring a they appeared. The 3-coloring then enured uch conflict to appear at mot one-third of the time. So for (peudo-)odd triangulation we will hae to explicitly take care of introducing at mot one Steiner point per eery three interior point. The pre-proceing of P will be again a explained in 3. Let P be again the unique ertex with mallet y-coordinate, the one we ue to ort the interior point 15

16 Figure 11: All interior point are Steiner point. Gray ertice are of een parity before the introduction of Steiner point, and all black ertice on the boundary of are of odd parity. The middle cae pawn two ub-problem, of the kind hown to the right, by introducing one Steiner point. Thi gie in total = 7 Steiner point for an odd-triangulation of. angularly around. The pre-proceing phae end with the contruction of the conex polygon Q j, the big ear hanging from the lower part L(P) of the therein created polygon P. For peudo-een triangulation we would fan-triangulate thoe Q j, if any. For peudo-odd we will chooe different triangulation, the main idea i the following: If eery ertex of a conex polygon Q j i of een degree in a triangulation T(P) of P, then we could jut directly make them adjacent to and leae them all odd. Unfortunately thi might not alway be poible. We will try nonethele to achiee omething imilar, and the following reult hown in [5] will be ery ueful: Lemma 2 (O. Aichholzer et al.). Let Q be a conex polygon where each of it ertice ha a parity aigned. Let p,q,r be any three conecutie ertice of Q. Then there exit a triangulation of Q that make all ertice of Q happy with the poible exception of p,q,r. After the pre-proceing there i an arbitrary triangulation of P minu the Q j. We will complete a triangulation T(P) of P a follow: for each one of the Q j et the parity of it ertice appropriately o that we can apply Lemma 2 and make them all een in T(P), with the poible exception of it lat three ertice, w.r.t. the angular order of the interior point around. The triangulation of thoe Q j, howeer, might change during the algorithm depending on what conflict we encounter. Haing thi particular triangulation T(P) of P we will again can L(P) from the left to right, following the angular order p 1,...,p k of the k interior point of P around, o again p 0 and p k+1 are the neighbor of in CH(P). We will again aume that by the time we are proceing point, the adjacency i already preent and eery point p r, r < j, i of odd degree in the current contruction. Neethele, obere that ince we want to add roughly k 3 interior Steiner point, proceing actually mean proceing,+1,+2, o we hae different cae depending on the local ituation. That i, if we 16

17 are currently tuck at it i becaue it current degree i een, otherwie we could jut continue. Sothecaewehaetotudyarethetripleofparitie: (e,e,e),(e,e,o),(e,o,e) and (e,o,o), where they correpond entry-wie to the current paritie of (,+1,+2 ), and e,o tand for een and odd repectiely. It i ery important to keep in mind that in the triple (,+1,+2 ), the only point adjacent to i. Alo, in all cae we will aume that +3 exit. If that i not the cae then one of,+1,+2 i p k+1, and we would find ourele with a problem of contant ize that can be oled uing a contant number of interior Steiner point, due to Oberation 2. We will now jump to the cae ditinction. (1) (e,e,e). Regardle of whether,+1,+2 are reflex or conex ertice of P, the configuration i hown in olid to the left in Figure 12 and it olution i hown dahed (e,e,e) (e,o,e) (e,e,o) Figure 12: In the figure, the color repreent the parity of the ertice before the adjacencie of the olution are added. Gray color mean een degree, black color mean odd degree, and white mean that we do not necearily take care of that point at thi tep. The original configuration are hown in olid black while their olution are hown dahed. Point i a Steiner point. (2) (e,o,e). Regardle of whether,+1,+2 are reflex or conex ertice of P, look the middle configuration of Figure 12. (3) (e,e,o). If +2 i a reflex ertex of P, then the ituation i hown to the right in Figure 12. If +2 i a conex ertex of P, then +2 i part of ome conex polygon Q we applied Lemma 2 on. Obere that Q cannot be a triangle, otherwie +2 would be the middle ertex and therefore +2 would hae een degree, ince it two neighbor in Q would be adjacent. Thu Q mut hae ize at leat four. Thi impliethat cannotbepartofqeither, becaueotherwiewecouldafelyaume that, a any other ertex of Q to the left of, i happy due to Lemma 2. Thi mean that +1 i necearily a reflex ertex of P, and thu the leftmot ertex of 17

18 Q. Finally, thi all mean that Q cannot actually hae ize larger than four, i.e., Q mut necearily be a conex quadrilateral. If Q had ize larger than four we could afely aume that the degree of +2 in the triangulation of Q i een due to Lemma 2 and the fact that neither +1 nor +2 hae been modified by the algorithm before. Therefore the ituation i a pictured in the upper left corner of Figure 13 and it olution i hown right below pj (e,e,o) (e,o,o) (e,o,o) Figure 13: The color repreent the parity of the ertice before the adjacencie of the olution are added. Gray color mean een degree, black color mean odd degree, and white mean that we do not necearily take care of that point at thi tep. The original configuration are hown in olid black while their olution are hown dahed. Point i a Steiner point. (4) (e,o,o). Thi cae i the hardet of all. If +1 i a reflex ertex of P, then the ituation i hown in the econd figure in the upper part of Figure 13, and whoe olution i hown right there dahed. If +1 i a conex ertex of P, then obere that +2 cannot be a conex ertex a well, ince otherwie,...,+3 are part of the ame conex polygon, and would be happy due to Lemma 2. So +2 i a reflex ertex of P. More, if the degree of +3 in T(P) i odd then the olution i like hown to the right in Figure 13. Thu we enter a cae where +2 i a reflex ertex of P and +3 i of een degree in T(P). Thi cae cannot be oled locally, conidering only,...,+3, and there i more than one way we can deal with it. The way we will do it here i the following: We will go ahead until +6 and we will ole,...,p 6 with at mot two Steiner point, one of them will be charged to the triple,+1,+2, and the other to the triple +3,+4,+5. At the end of the contruction we will be left with the adjacency +6, which i where the algorithm will continue. So for all what follow we will aume that +4,...,+6 exit, otherwie +3 = p k+1, and we find ourele with the lat four point the algorithm would proce. Thi can be oled with a contant number, larger than one, of Steiner point, which turn into a contant oerhead oerall. 18

19 Since we are auming that +3 i currently een, we hae four main ub-cae that depend on the pair of paritie (, ) of (+4,+5 ) in T(P). The configuration and olution are pictured in Figure (e,o,o,e)+(e,e) 1 2 (e,o,o,e)+(e,o) 1 2 (e,o,o,e)+(o,o) (e,o,o,e)+(o,e) 1 (e,o,o,e)+(o,e) 2 Figure 14: The color repreent the parity of the ertice before the adjacencie of the olution are added. Gray color mean een degree, black color mean odd degree, and white mean that we do not necearily take care of that point at thi tep. The original configuration are hown in olid black while their olution are hown dahed. Point 1, 2 are Steiner point. (4.1) (e,e). The configuration i hown in the left upper corner of Figure 14 in olid and it olution i hown dahed. Obere that for the olution it doe not play a role whether +3,+4,+5 are reflex or conex ertice of P. (4.2) (e,o). The configuration i hown in the econd figure on the upper part of Figure 14 in olid and it olution i hown dahed. Again, it doe not matter whether +3,+4,+5 are reflex or conex ertice of P. (4.3) (o,o). If +4 i a reflex ertex of P, the configuration i hown in olid to the right on the upper part of Figure 14, and it olution i hown dahed. Now let u argue that +4 cannot be a conex ertex. If +4 were a conex ertex it would be part of a conex polygon Q we applied Lemma 2 on in the beginning. Thi polygon Q cannot be a triangle, becaue then +4 would be the middle point and it would then hae een degree in T(P). Alo, Q cannot hae at leat fie ide becaue by Lemma 2, again, +4 could be 19

20 made of een degree in a triangulation of Q, without affecting the parity of +3 in the ame triangulation. Thu the only cae that kind of make ene i that Q i a conex quadrilateral. So, there are two poible option for Q. Either Q = +2,+3,+4,+5 or Q = +3,+4,+5,+6. The former i not poible becaue +2 would be part of Q and the algorithm would hae choen in the beginning to leae +2 of een degree, intead of odd, uing a triangulation of Q. Thi i achieable uing Lemma 2. The latter i alo not poible becaue although +3 i currently of een degree, +4,+5 are of odd degree in the current triangulation of Q. Since +4,+5 are the two middle ertice of Q, and Q ha only two triangulation, then at leat one of them would get een degree, and again, we would be dicuing a different cae. Therefore +4 mut necearily be a reflex ertex of P. (4.4) (o,e). If +4 happen to be a reflex ertex of P, then the configuration i hown to the left on the lower part of Figure 14, and it olution i hown dahed. If +4 i a conex ertex of P, then it part of a conex polygon Q we applied Lemma 2 on in the beginning. By the ame argument a in cae (o,o) we know that Q can be neither a triangle nor a conex polygon larger than four. Neerthele, thitimeqcanindeedbethequadrilateral+3,+4,+5,+6. The other poibility for Q i in thi cae not poible either due to the ame argument. Since Q i a conex quadrilateral and +4 i of odd degree in the current triangulation of Q, then the only poibility i that the diagonal i preent. The configuration i hown to the right on the lower part of Figure 14 in olid and it olution i hown dahed. Thi conclude the cae analyi. From all cae dicued it i clear that the algorithm introduce at mot one interior Steiner point per triple of interior point of P, and at the end it might require to bruteforce a configuration of contant ize, that due to Oberation 2 can be ole uing a contant number c of interior Steiner point a well. Hence oerall the algorithm make ue of at mot k 3 +c interior Steiner point and Theorem 3 follow. 5.1 Extenion to odd triangulation The extenion of the algorithm of Theorem 3 to odd triangulation i now ery eay. We again encloe the et of point P in a bigger polygon, jut a we did for een triangulation. What the configuration look like can be een to the left in Figure 8 on page 12. We then run the algorithm for peudo-odd triangulation, and at the end the only thing that can happen i that the piot P get een degree. Howeer, by Oberation 2, piot can be made odd by locally adding a contant number of Steiner point in one of it adjacent triangle. Thu, in total, the number of ued interior Steiner point i at mot n c, for ome contant c. The odd triangulation i finally obtained by remoing the bigger polygon encloing P. Therefore Theorem 4 follow. 20

21 6 Concluion In thi paper we hae preented algorithm that contruct, with help of Steiner point, (peudo-)een and (peudo-)odd triangulation of a gien et of point P. The number of Steiner point that the algorithm ue i roughly k 3 for the peudo ariant, where k i the number of interior point of P, and roughly n 3 for the other cae. It i important to obere that our algorithm do not modify the conex hull of P, and therefore they preere extent meaure of P, uch a diameter, width, among other. If we do not care about modifying CH(P), or about the poition of the Steiner point, then the tak i in general ignificantly eaier. For example, only two Steiner point far away from CH(P) would uffice to contruct a peudo-een triangulation, ay one at and the other at. Albeit being thi contruction poible, we do not know why it would be intereting to ue it, ince the output et of point doe not look anything like the one that wa gien a the input. We could beliee that the technique hown here could be puhed further to improe the oerall number of Steiner point, ay to go from one-third to one-ixth, but thi will imply a ignificantly larger number of cae to analyze, we ee thi really a a econdary intereting improement. What we really ee a the primary open problem are the following: I it poible to alway contruct (peudo-)een or (peudo-)odd triangulation of a gien et of point P uing only a contant number of (interior) Steiner point? In other word, how big i the lower bound on the number of (interior) Steiner point that are required to alway contruct uch triangulation? We hae failed o far trying to proe a (ub-)linear lower bound, which i the natural gue when working on thee kind of problem. Moing away from uing Steiner point, would it be poible to contruct een or odd triangulation of a gien et of point P where at mot a contant number of point remain unhappy? Thi quetion wa poed by Aichholzer et al. in [5]. In that ame paper they proed a lower bound of roughly n 108 unhappy ertice when the aignment of paritie i not uniform. Thu, a they pointed out, the intereting cae are all een and all odd. Note that although thee two quetion look imilar, they might not be equialent. If we are intereted in an een triangulation of P and we contruct one where a contant number of ertice remain unhappy, thoe unhappy ertice might be far from each other. Thi mean that we might hae to bring them together, at leat in pair, uing Steiner point. We hae to get rid of them of them in pair, at leat, ince the number of odd ertice i alway een. But to get them cloe to each other we might require more than a contant number of Steiner point. Thu the technique to ole thoe two open problem might be rather different. We ee a real challenge there. 21

22 7 Acknowledgement The author would like to thank Jorge Urrutia and Marco Heredia for uggeting the problem and for helpful comment in the initial tage of thi work, and to Raimund Seidel for aluable feedback. Thi reearch wa partially upported by CONACYT-DAAD of México. Reference [1] P. J. Heawood, On the four-color map theorem, Quart. J. Pure Math., ol. 29, pp , , 2 [2] R. Steinberg, The tate of the three color problem, in Quo Vadi, Graph Theory? A Source Book for Challenge and Direction (J. W. K. John Gimbel and L. V. Quinta, ed.), ol. 55 of Annal of Dicrete Mathematic, pp , Eleier, , 2 [3] K. Dik, L. Kowalik, and M. Kurowki, A new 3-color criterion for planar graph, inwg (L.Kucera, ed.), ol of Lecture Note in Computer Science, pp , Springer, , 2, 3 [4] O. Aichholzer, T. Hackl, C. Huemer, F. Hurtado, and B. Vogtenhuber, Large bichromatic point et admit empty monochromatic 4-gon, SIAM J. Dicrete Math., ol. 23, no. 4, pp , , 2 [5] O. Aichholzer, T. Hackl, M. Hoffmann, A. Pilz, G. Rote, B. Speckmann, and B. Vogtenhuber, Plane graph with parity contraint, in WADS (F. K. H. A. Dehne, M. L. Gariloa, J.-R. Sack, and C. D. Tóth, ed.), ol of Lecture Note in Computer Science, pp , Springer, , 2, 16, 21 [6] S. Fik, A hort proof of chátal watchman theorem, J. Comb. Theory, Ser. B, ol. 24, no. 3, p. 374, [7] A. Pilz, Parity propertie of geometric graph, Mater thei, Graz Unierity of Technology,