Transcript

Graphs of Non-Crossing Perfect Matchings*

C. Hernando1, F. Hurtado2, and Marc Noy3

1 Department de Matematica Aplicada I, Universitat Politecnica de Catalunya, Diagonal647, 08028 Barcelona, Spain. e-mail: [email protected];3 Department de Matematica Aplicada II, Universitat Politecnica de Catalunya, PauGargallo 5, 08028 Barcelona, Spain. e-mails: 2hurtado, [email protected]

Abstract. Let Pn be a set of n ¼ 2m points that are the vertices of a convex polygon, and letMm be the graph having as vertices all the perfect matchings in the point set Pn whose edgesare straight line segments and do not cross, and edges joining two perfect matchings M1 andM2 if M2 ¼ M1 � ða; bÞ � ðc; dÞ þ ða; dÞ þ ðb; cÞ for some points a; b; c; d of Pn. We prove thefollowing results about Mm: its diameter is m� 1; it is bipartite for every m; the connec-tivity is equal to m� 1; it has no Hamilton path for m odd, m > 3; and finally it has aHamilton cycle for every m even, m � 4.

Key words. Perfect matching, Non-crossing configuration, Gray code

1. Introduction

Given a graph G, one can consider an associated graph MðGÞ whose vertices arethe perfect matchings of G and where two perfect matchings are adjacent if theirsymmetric difference is a cycle C of G. In this case one says that C is an alternatingcycle for the two matchings. This definition is closely related to the matchingpolytope MðGÞ of G, a polytope whose vertices are the incidence vectors of all thematchings in G, since two perfect matchings are adjacent in MðGÞ if, and only if,they are adjacent in the graph of MðGÞ [8].

The graphs MðGÞ have been studied in the past and some general results areknown, like the fact that they are always Hamiltonian [9]. Particular attention hasbeen paid to the case in which G is a plane bipartite graph, a situation of par-ticular interest in the study of chemical compounds [10, 15]. Another noteworthyinstance is when G ¼ Kn;n, and in this case MðKn;nÞ is the graph of the so calledassignment polytope [2, 3].

In this paper we study a geometric version of the problem. Let Pn be a set ofn ¼ 2m points that are the vertices of a convex polygon, and let us consider

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Graphs and Combinatorics (2002) 18:517–532

Graphs andCombinatorics� Springer-Verlag 2002

matchings in the point set Pn whose edges are straight line segments. A perfectmatching in Pn is said to be non-crossing if no two of its edges intersect. The pointsof Pn are labeled, consequently two matchings are considered equal only if theyhave exactly the same set of edges. We define the graph Mm as the graph having asvertices the non-crossing perfect matchings of Pn and edges joining M1 and M2 ifM2 ¼ M1 � ða; bÞ � ðc; dÞ þ ða; dÞ þ ðb; cÞ for some points a; b; c; d of Pn (seeFig. 1). Observe that in this case the symmetric difference of M1 and M2 is a cycleof length four. This definition is adopted so that two adjacent matchings differ aslittle as possible, namely only in two edges.

The graph M4 is depicted in Fig. 2. Some relevant properties can be observed:the graph is bipartite (this is indicated by black and white vertices), it is Hamil-tonian, and every vertex has another vertex at maximum distance 3. Moreover, ithas minimum degree 3 and it can be checked that it is 3-connected. The aim ofthis paper is to establish analogous properties for Mm with m � 6.

In Section 2 we show how to obtain a shortest path in Mm between any twovertices; in fact, our results provide a linear time algorithm for finding shortest

a

b

c

d

a

b

d

c

Fig. 1. Adjacent matchings in M6

Fig. 2. The graph M4

518 C. Hernando et al.

paths. As a corollary, we prove that every vertex of Mm has eccentricity equal tom� 1 and, as a consequence, the diameter of Mm is m� 1. In Section 3 we showthat Mm is a bipartite graph for every m, and in Section 4 we prove that theminimum degree and the connectivity of Mm are equal to m� 1. Finally, we showin Section 5 that Mm has no Hamilton path for m odd, m > 3, and, drawing onprevious work by Ruskey and Proskurowski [11], we show that Mm is Hamilto-nian for every m even, m � 4.

A result we wish to emphasize is that shortest paths can be easily constructed,which gives as a corollary the determination of the exact value of the diameter forevery m. These problems are usually difficult and have been only partly solved inrelated graphs like graphs of triangulations [7, 14], and graphs of non-crossingspanning trees [1, 5].

On the other hand, it is well-known that Mm has Cm vertices, whereCm ¼ 1

mþ12mm

� �is a Catalan number. In the paper we make use of several

bijections between matchings in Mm and other combinatorial objects countedby the Catalan numbers, like plane trees and balanced binary strings. In addi-tion, in our study of Mm we have found a bijection between matchings in Mm

an certain permutations of length m. This fact has an interesting combinatorialconsequence: it produces a new family of permutations counted by the Catalannumbers, not defined in terms of one forbidden subsequence of length three(see [13]).

From now on a non-crossing perfect matching will be called simply amatching. The graph Mm does not depend on the exact position of the points of Pnand, to fix ideas, we take Pn as the set of vertices of a regular n-gon. Edgescorresponding to the sides of the polygon are called boundary edges.

2. Shortest Paths, Eccentricities and Diameter

We recall that the distance dðu; vÞ between two vertices u and v in a graph is thelength r of a shortest path u ¼ w0 � w1 � � � � � wr ¼ v connecting them, where �denotes adjacency. The eccentricity of a vertex v in a graph is the maximum of thedistances between v and any other vertex of the graph, and the diameter is themaximum of the eccentricities.

In this section we show a simple method for finding shortest paths between anytwo vertices in Mm and, as a consequence, we obtain the exact value of thediameter. Before that we need to introduce some definitions and notations. If e isan edge not belonging to a matching M in Mm, but one of its neighbors M

contains e, we say that M is obtained by insertion of e into M , and we use thenotation M e for the matching M. If e belongs to M we simply defineM e ¼ M .

Observe that the insertion of an arbitrary edge e into a given matching isnot always possible, but it is certainly feasible when e is a boundary edge.In this case, if M � M 0 and e is one of the edges exchanged, thenM e ¼ M 0 e. If M � M 0 but e is not exchanged, then it can be checked thatM e � M 0 e.

Graphs of Non-Crossing Perfect Matchings 519

The insertion of a sequence of edges is recursively defined by

M ðe1; e2; . . . ; erÞ ¼ ðM ðe1; e2; . . . ; er�1ÞÞ er:

Lemma 2.1. Let M 0 and M 00 be in Mm and suppose they share the boundary edge e.Then all the matchings in any shortest path between M 0 and M 00 contain e.

Proof. Let dðM 0;M 00Þ ¼ d and let

M 0 ¼ M0 � M1 � � � � � Md ¼ M 00

be a shortest path between M 0 and M 00. Suppose that e does not belong to somematching in the path and let Mi be the first one with this property. Consider nowthe sequence

M 0 ¼ M0 e � M1 e � � � � � Md e ¼ M 00:

By the remarks before Lemma 2.1, consecutive matchings in this sequence areeither equal or adjacent, but Mi�1 e ¼ Mi e, and this gives a path of lengthsmaller than d, a contradiction. (

Let M be in Mm. Let us consider the set E1 of boundary edges in M , andremove the endpoints of E1 both from Pn and from M . This gives a new point setwith a new matching, and we can similarly define the current boundary edges E2

and repeat the process. A disassembly sequence for the matching M is any se-quence ðe1; . . . ; emÞ consisting of all the edges in M , with the property that all theedges in E1 come first, next all the edges in E2, and so on.

Observe that a disassembly sequence ðe1; . . . ; emÞ for a matching M can beinserted into any other matching M 0, and that M 0 ðe1; . . . ; emÞ ¼ M .

Theorem 2.2. Let ðe1; . . . ; emÞ be a disassembly sequence for a matching M, and letM 0 be any other matching. Then

M 0 � M 0 e1 � M 0 ðe1; e2Þ � � � � � M 0 ðe1; . . . ; emÞ ¼ M

is a shortest path between M 0 and M , where it is understood that some of theadjacencies in the above expression can be equalities.

Proof. Let

M 0 ¼ M0 � M1 � � � � � Md ¼ M

be a shortest path P 0 between M 0 and M .Let us assume first that e1 does not belong to M 0, and prove that there is a

shortest path from M 0 to M in which the first step is the insertion of e1. As e1belongs to M but not to M 0, there is an i > 0 such that Mi is the first matching inthe sequence containing e1. In the sequence

520 C. Hernando et al.

M 0 � M0 e1 � M1 e1 � � � � � Md e1 ¼ M ;

the equality M 0 ¼ M0 from the original path has been replaced by the properadjacency M 0 � M0 e1, but the adjacency Mi�1 � Mi has been replaced by theequality Mi�1 e1 ¼ Mi e1, therefore the length remains unchanged and we stillhave a shortest path.

If e1 belongs toM 0, the first step in the path P 0 is simply an equality. Henceforthwe see that in any case there is a path from M 0 to M starting by the insertion of e1.By Lemma 2.1 we know that all matchings in shortest paths from M 0 e1 to Mcontain e1. The same argument shows that there is a path fromM 0 e1 toM startingby the insertion of e2, and the repetition of the process proves the claim. (

Remark that the above result provides in fact a linear time algorithm forfinding shortest paths.

Corollary 2.3. If two matchings M and M 0 in Mm have exactly k edges in common,then

dðM ;M 0Þ � m� k � 1:

Proof. Observe that the path in the statement of Theorem 2.2 involves no ex-change when an edge already in M is to be inserted; this gives dðM ;M 0Þ � m� k.But in the final step, when only two edges have to be inserted, a single exchangeinserts in fact two edges. Hence dðM ;M 0Þ � m� k � 1. (

This implies that the diameter is at most m� 1. We show now that this is infact the exact value of the diameter.

The union M1 [M2 of two matchings in Mm is the union C1 � � � Ck of edgedisjoint cycles, in which the edges from M1 and the edges from M2 alternate ineach cycle (an edge e in both M1 and M2 gives a ‘‘double’’ edge in M1 [M2, whichwe consider as a cycle of length 2). This decomposition is closely related to thedistance between the two matchings:

Theorem 2.4. If M1 and M2 are such that M1 [M2 ¼ C1 � � � Ck then

dðM1;M2Þ ¼1

2

Xk

i¼1

ðlengthðCiÞ � 2Þ:

Proof. The result is obvious for m ¼ 2, for m > 2 we distinguish two cases.

If M1 and M2 share a boundary edge e, we remove the endpoints of e from thepoint set Pn, and from both M1 and M2, then apply induction to the resulting pointset and matchings. Otherwise, let e be a boundary edge in M2 but not in M1. Weknow from Theorem 2.2 that there is a shortest path between M1 and M2 in whichwe first move from M1 to M1 e. The cycle in M1 [M2 containing e becomes in

Graphs of Non-Crossing Perfect Matchings 521

ðM1 eÞ [M2 the double edge e plus a cycle with length equal to lengthðCÞ � 2; wenow apply the previous case to M1 e and M2. (

The preceding result is now applied to a special situation, which gives as aconsequence the lower bound for the diameter. Define the rotation of an edge ði; jÞas the edge ðiþ 1; jþ 1Þ, addition being modulo 2m. Given a matching M 2 Mm,its rotation M is the matching consisting of all the edges of M rotated.

Theorem 2.5. Let M 2 Mm and let M be its rotation. Then dðM ;MÞ ¼ m� 1.

Proof. Let r be an oriented line, not crossed by any edge in M , and leti; iþ 1; . . . ; iþ p be the points in Pn to the left of r. We claim that the union of theedges in M to the left of r, together with its rotations, is an alternating pathstarting at i, going through all the points to the left of r, and ending in iþ p þ 1(Fig. 3a). Observe that applying the claim to the edges to the right of r we get asimilar path from iþ p þ 1 to i. The concatenation of the two paths shows thatM [M is a single cycle of length n ¼ 2m, and Theorem 2.4 implies the result.

ii+1

i+p i+p+1

r

i+1

i+q+1

i+q

i

i+pi+p+1

r’

r’’

(a) (b)

Fig. 3. Edges in M are continuous; edges in the rotate matching M are dashed

12

3

2m

2m-1

ρ

Fig. 4. Plane tree associated to a matching

522 C. Hernando et al.

The claim is obvious when p ¼ 1, and we proceed by induction on p. Ifði; iþ pÞ is an edge in M , we can take a line parallel to r leaving exactly the pointsiþ 1; . . . ; iþ p � 1 to its left; by induction this gives by rotation a path from iþ 1to iþ p, which can be concatenated with the edge ði; iþ pÞ and its rotationðiþ 1; iþ p þ 1Þ. If ði; iþ pÞ is not an edge in M we can find a number q and twooriented lines r0 and r00, non crossed by any edge of M , such that r0 has exactly thepoints i; . . . ; iþ q to its left and r00 the points iþ qþ 1; . . . ; iþ p (Fig. 3b). Now weapply induction to these two portions and concatenate the two resulting paths. (

Corollary 2.6. The eccentricity of every vertex in Mm is equal to m� 1, and thediameter of Mm is equal to m� 1.

3. Mm is a Bipartite Graph

In this section we show that Mm is a bipartite graph for all m. To this end we use aclassical bijection between non-crossing perfect matchings and plane trees. Given amatching M in Mm define a plane tree tM with mþ 1 nodes as follows: there is anode of tM for every edge ofM , plus a root q that is placed outside the edge ð1; 2mÞ.Join q to all the nodes corresponding to edges visible from q and proceed recur-sively (see Fig. 4, where the edges of tM are dashed and the root is in grey).

Recall that the height of a node in a tree is the distance from the root, and thepath length of a tree is the sum of the heights of all its nodes. Let us denote thepath length of a tree t by pðtÞ. For instance, pðtM Þ ¼ 10 in the matching of Fig. 4.

Lemma 3.1. If M and M 0 are adjacent in Mm, then the path lengths of tM and tM 0

have different parity.

Proof. Suppose M 0 ¼ M � ab� cd þ ad þ bc, and suppose the node in tM corre-sponding to ðc; dÞ is a descendant of the node corresponding to ða; bÞ. If we let Cbe the set of vertices of Pn to the right of the edge ðc; dÞ, then it is straightforwardto check that

pðtM Þ � pðtM 0 Þ ¼ 1 þ 2jCj:

The other cases are treated similarly. (

Partition the matchings in Mm according to whether the path length of theassociated tree is even or odd. The previous lemma says that all the edges arebetween the two parts of the partition, and as a corollary we have:

Corollary 3.2. The graph Mm is bipartite for every m.

Let us mention a different proof of this result. Label the points of Pn asf1; 10; 2; 20; . . . ;m;m0g in circular order, and let M be any (non-crossing perfect)matching. Then, because of parity considerations, any point i in f1; . . . ;mg hasto be matched in M with a point j0 in f10; . . . ;m0g. If we set ri ¼ j, thisdefines a permutation r ¼ rðMÞ of n letters. It is easy to check that if M and

Graphs of Non-Crossing Perfect Matchings 523

M 0 are adjacent in Mm then rðMÞ and rðM 0Þ differ in a single transposition(but not conversely), hence they have different parity. This establishes thebipartition.

This proof has an interesting consequence. It is well known that the number ofnon-crossing perfect matchings of Pn is the Catalan number 1

mþ12mm

� �. The above

correspondence then gives a family of permutations that is counted by the Catalannumbers. All the families of permutations we know with this property are definedin terms of one forbidden subsequence of length three [13]. Our family is not ofthis kind since the only forbidden subsequence for m ¼ 3 is 231 but it appears inthe permutation 4231 that does come from a matching. Hence we have foundwhat appears to be a new family of permutations counted by the Catalannumbers.

4. Connectivity

We begin this section with a simple lemma.

Lemma 4.1. The minimum degree of Mm is equal to m� 1.

Proof. Let M be in Mm, and let G be the graph having a vertex for every edge inM , and where two edges uv and u0v0 are adjacent if they see each other, i.e., if thereis a straight-line segment connecting them and touching no other segment. Thenthe degree of M in Mm is equal to the number of edges in G. But G has m verticesand is clearly connected, so that it has at least m� 1 edges. Finally, observe thatthe matchings formed by m parallel edges have degree exactly m� 1. (

We are now ready for the main result.

Theorem 4.2. The connectivity of Mm is equal to m� 1.

Proof. By the above lemma we only need to show that Mm is ðm� 1Þ-connected.The proof is by induction on m, starting with the case m ¼ 3, which is clear sinceM3 is the graph K2;3. Suppose then m � 4.

By Menger’s theorem, given M 0 and M 00 in Mm it is enough to prove that thereare m� 1 paths from M 0 to M 00 internally disjoint. We consider three cases.

1) M 0 and M 00 have a common boundary edge e ¼ ði; iþ 1Þ. Removing theendpoints of e, by induction there are m� 2 internally disjoint paths from M 0 toM 00 all of them containing e. If e0 ¼ ðiþ 1; iþ 2Þ, we know there exists a path Pfrom M 0 e0 to M 00 e0 such that all its vertices contain e0. Since e and e0 cannotboth be contained in a matching, concatenating P with the adjacenciesM 0 � M 0 e0 and M 00 e0 � M 00 we obtain a path from M 0 to M 00 internally disjointwith the previous ones.

2) M 0 and M 00 have no common boundary edge and none of them has all itsedges on the boundary. Then, without of loss of generality, we can assume thatthere exist boundary edges e ¼ ði; iþ 1Þ and e0 ¼ ðiþ 1; iþ 2Þ such that

524 C. Hernando et al.

e 2 M 00; e 62 M 0; e0 62 M 0; e0 62 M 00:

By Lemma 4.1, M 0 has at least m� 1 different neighbors M1; . . . ;Mm�1. Onlyone of them contains e, assume it is M1, and only one of them contains e0, assumeit is M2. A simple check shows that the matchings M2 e; . . . ;Mm�1 e are alldifferent. Observe also that

M1 e0 6¼ Mi; i ¼ 2; . . . ;m� 1;

since Mm does not have triangles (Corollary 3.2), and that

M1 e0 6¼ Mi e; i ¼ 2; . . . ;m� 1;

since e and e0 cannot both be in a matching.By induction, and by a consequence of Menger’s theorem (see [4, p. 82]), there

are m� 2 internally disjoint paths between M 00 and M2 e; . . . ;Mm�1 e, all ofthem containing e. By induction again, there are m� 2 internally disjoint pathsbetween M1 e0 and M 00 e0, all of them containing e0. Take one of them notcontaining M2 (by assumption m� 2 > 1) and concatenate it withM 0 � M1 � M1 e0 and M 00 e0 � M 00; in this way we have an additional path fromM 0 and M 00 internally disjoint with the former ones.

3) The only case left is M 0 ¼ fe1; e3; . . . ; e2m�1g and M 00 ¼ fe2; e4; . . . ; e2mg,where ei ¼ ði; iþ 1Þ, i ¼ 1; . . . ; 2m. Let Mi be the matching obtained from M 0 byexchanging e1 and e2i�1, that is,

Mi ¼ M 0 � ð1; 2Þ � ð2i� 1; 2iÞ þ ð2; 2i� 1Þ þ ð2i; 1Þ; i ¼ 2; . . . ;m� 1:

These are m� 2 matchings adjacents to M 0, none of them containing e1 nor e2m.Now M2 e2m; . . . ;Mm�1 e2m are all different and none of them contains e1; asinthe former case, there are m� 2 internally disjoint paths connecting them toM 00.Now take a path from M 0 to M 00 e1, all of whose vertices contain e1,andconcatenate it with M 00 e1 � M 00, obtaining m� 1 disjoint paths form M 0

to M 00. (

5. Hamiltonian Properties

Let Em be the number of matchings M in Mm such that the path length of tM iseven, and let Om be the number of matchings in Mm such that the path length oftM is odd. We next show that jEm � Omj > 1 for m odd and, as a consequence ofLemma 3.1, that Mm has no Hamilton path. We make use of the fact that thegenerating function CðzÞ ¼

Pm�0 Cmzm of the Catalan numbers satisfies the

quadratic equation

zCðzÞ2 � CðzÞ þ 1 ¼ 0:

Graphs of Non-Crossing Perfect Matchings 525

Lemma 5.1. jEm � Omj > 1 for m odd, m > 3.

Proof. Let tm;k be the number of plane trees with m nodes and path length k, andlet Qðu; zÞ ¼

Ptm;kukzm be the corresponding bivariate generating function. It is a

standard fact [12] that Q satisfies the functional equation

Qðu; zÞ ¼ 1

1 � Qðu; zuÞ :

We are interested in DðzÞ ¼P

dmzm ¼ Qð�1; zÞ, since the coefficient of zm inthis series is the difference between the number of plane trees with even pathlength and odd path length. Taken into account that Em � Om ¼ dmþ1 we onlyneed to consider the even part of DðzÞ

P ðzÞ ¼ DðzÞ þ Dð�zÞ2

:

Straightforward manipulation of the equations Qð1; zÞ ¼ 1=ð1 � Qð1; zÞÞ andQð�1; zÞ ¼ 1=ð1 � Qð�1;�zÞÞ gives

P ðzÞ2 � PðzÞ � z2 ¼ 0:

Comparing this with the equation for CðzÞ we see that P ðzÞ ¼ z2CðzÞ2. This impliesthat

dm ¼ ð�1Þm=2Cm=2�1

for m even, and the result follows. (

As a corollary we have:

Theorem 5.2. The graph Mm has no Hamilton path for m odd, m > 3.

1 2 3 4 5 6 7 8 9 10

M

109

8

7

6

54

3

2

1

s (M) :1 1 1 0 1 0 1 0 0 0

Fig. 5. Binary string associated to a matching

526 C. Hernando et al.

For m even, the same proof as before gives Em � Om ¼ 0, thus the necessarycondition for the existence of a Hamilton cycle is fulfilled. Our last result is thatHamilton cycles do exist in this case.

To this end we use yet another bijection, namely between matchings of Mm

and the set Bm of binary strings consisting of m zeros and m ones, and havingthe prefix property, i.e., in every prefix the number of ones it at least thenumber of zeros. The bijection is illustrated in Fig. 5, where a matching isrepresented linearly with curved arcs, and an arc going up corresponds to a 1and an arc going down to a 0. It is immediate that the resulting string hasthe prefix property and that this defines indeed a bijection between V ðMmÞand Bm.

Let bðMÞ be the binary string associated with the matching M . Then it is easyto check that two matchings M and M 0 are adjacent in Mm if, and only if,

bðMÞ ¼ x1y0z and bðM 0Þ ¼ x0y1z

111011100000

111011010000

111011001000

111011000100

111011000010

111010100010

111010100100

111010101000

111010110000

111010010010

111010010100

111010011000

111010001100

111010001010

111111000000

111110100000

111110010000

111110001000

111110000100

111110000010

111101100000

111101010000

111101001000

111101000100

111101000010111100110000

111100101000

111100100100

111100100010

111100011000

111100010100

111100010010

111100001100

111100001010

Fig. 6. Top: generalized vertices. Bottom: Hamilton path connecting them

Graphs of Non-Crossing Perfect Matchings 527

for some binary string y with k zeros and k ones for some k � 0 and having theprefix property. If we impose that y is empty in the above adjacency rule, that is, ifwe only allow the interchange of a zero with an adjacent one, we obtain aspanning subgraph Sm of Mm. This graph was studied in [11], where the authorsshowed how to construct a Hamilton path when m is even. This path cannot beextended to a Hamilton cycle, since the string 1m0m has degree one in Sm. Our goalis to modify this construction conveniently in order to obtain a Hamilton cycle inthe larger graph Mm.

We sketch the construction in [11] and direct the reader to this reference forfull details. Given s ¼ x10k 2 Bm, let s ¼ x1110kþ2 be the set of all strings in Bmþ2

11001111100000

110101110000

110100111000

1010110111000 110010111000

101101110000

101011110000

101100111000

1100111110000

111000111000

111001110000

1110111100000

111111000000

110111100000A

N

XZ

M

1100111000

111011000

11100100

11100010

11010010

11001110010101100

101011010

11011000

11110000

11010100

10110100

10110010

11001010

X

Fig. 7. A Hamiltonian cycle in M6 obtained from a Hamiltonian cycle in M4

Z

Z

Z

M

-1

N

Fig. 8. The path N

528 C. Hernando et al.

beginning with x1. Let us call s the generalized vertex of s. The basic idea is totake two adjacent vertices s and t in Mm and recursively construct a Hamiltonpath spanning all the vertices in s [ t � Mmþ2. For example, Fig. 6 shows thegeneralized vertices 111011105 and 141106, and a path connecting all the verti-ces.

Generalized vertices always have this grid structure in two dimensions. Thetwo degrees of freedom correspond to the two ‘‘free’’ ones that can be moved. Thepaths connecting pairs of generalized vertices can be seen as a three dimensionalobject.

The construction of the Hamilton cycle for m even is by induction on m,starting with the cycle of M4 appearing in Fig. 7. We start inductively with aHamilton cycle Cm�2 in Mm�2, then we substitute each vertex s ¼ x10t by itsgeneralized vertex s ¼ x1110tþ2 and we build Hamilton paths between con-secutive generalized vertices. In this way we construct a Hamilton cycle Cm inMm. This illustrated in Fig. 7 for m ¼ 6 and in Fig. 11 this cycle has to beread by columns. Among the paths between consecutive generalized vertices,the paths A, X , Y and Z are those appearing in [11] (see Fig. 10). The paths Nand M are shown in Fig. 8 and Fig. 9, respectively. In Figures 8, 9 and 10, thelabel in the vertical and the horizontal direction is always X , unless indicatedotherwise. In the other direction, the label is always A (or A�1) unless indicatedotherwise.

Hence we conclude:

Theorem 5.3. The graph Mm is Hamiltonian for m even, m � 4.

Y

Z

Z

Z

Z

M

Y

Fig. 9. The path M

Graphs of Non-Crossing Perfect Matchings 529

6. Concluding Remarks

We have established a number of significant properties of the graphs Mm. Anopen problem is to compute the group of automorphisms. Without loss of gen-erality we can assume that P2m is the set of vertices of a regular 2m-gon; then everysymmetry of the polygon acts on the set of matchings and induces an automor-phism of Mm. We conjecture that these are the only automorphisms of the graph.

Finally, a natural line of research is to extend the study of graphs ofnon-crossing matchings to the case of sets of points in the plane that are notnecessarily vertices of a convex polygon. Preliminary results have been obtainedrecently in this direction [6].

Fig. 10. The paths A, X, Y and Z

530 C. Hernando et al.

Acknowledgments. We are grateful to the referees for comments that helped to improve thepresentation of the paper.

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Fig. 11. A Hamilton cycle in M6

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Received: October 10, 2000Final version received: January 17, 2002

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