Reconstruction of the geometric structure of a set of points in the plane from its geometric tree graph
aa r X i v : . [ m a t h . C O ] D ec Reconstruction of the Geometric Structure of a Set of Points inthe Plane from its Geometric Tree Graph
Chaya Keller and Micha A. PerlesEinstein Institute of Mathematics, Hebrew UniversityJerusalem 91904, IsraelAugust 26, 2018
Abstract
Let P be a finite set of points in general position in the plane. The structure of thecomplete graph K ( P ) as a geometric graph includes, for any pair [ a, b ] , [ c, d ] of vertex-disjoint edges, the information whether they cross or not.The simple (i.e., non-crossing) spanning trees (SSTs) of K ( P ) are the vertices of theso-called Geometric Tree Graph of P , G ( P ). Two such vertices are adjacent in G ( P ) if theydiffer in exactly two edges, i.e., if one can be obtained from the other by deleting an edgeand adding another edge.In this paper we show how to reconstruct from G ( P ) (regarded as an abstract graph)the structure of K ( P ) as a geometric graph. We first identify within G ( P ) the verticesthat correspond to spanning stars. Then we regard each star S ( z ) with center z as therepresentative in G ( P ) of the vertex z of K ( P ). (This correspondence is determined onlyup to an automorphism of K ( P ) as a geometric graph.) Finally we determine for anyfour distinct stars S ( a ) , S ( b ) , S ( c ) , and S ( d ), by looking at their relative positions in G ( P ),whether the corresponding segments cross. Graph reconstruction is an old and extensive research topic. It dates back to the
ReconstructionConjecture raised by Kelly and Ulam in 1941 (see [8, 12]), which asserts that every graph on atleast three vertices is uniquely determined by its collection of vertex deleted subgraphs.As a natural extension of the Reconstruction Conjecture, numerous papers considered eitherreconstruction of structures other then graphs (a research topic proposed by Ulam in 1960), orreconstructions of graphs from other information. In the first direction, reconstructed objectsinclude colored graphs, hypergraphs, matroids, relations, and other classes. In the second direc-tion, the “information” may be k -vertex deleted subgraphs, edge-deleted subgraphs, elementarycontractions, spanning trees, etc. In addition, various papers considered reconstruction of pa-rameters of the graph instead of its full structure. Such parameters include the order, the degreesequence, planarity, the types of spanning trees, and many others (see the surveys [2, 10] forreferences).In this paper, we study the problem of reconstructing the geometric structure of a set ofpoints in the plane from its geometric tree graph. Tree graphs were defined in 1966 by Cummins [3] in the context of listing all spanning treesof a given connected graph effectively. The tree graph T ( G ) of a graph G has the spanningtrees of G as its vertices, and two spanning trees are adjacent if one can be obtained from the1ther by deleting an edge and adding another edge. These graphs were studied in a number ofpapers and were shown to be Hamiltonian and to have the maximal possible connectivity (see,e.g., [7, 9]).In 1996, Avis and Fukuda [1] defined the geometric tree graph , as the counterpart of treegraphs in the geometric graph setting. Definition 1.1.
Let P be a finite point set in general position in the plane. The geometrictree graph G ( P ) is defined as follows. The vertices of G ( P ) are the simple (i.e., non-crossing)spanning trees (SSTs) of K ( P ) . Two such vertices are adjacent in G ( P ) if they differ in exactlytwo edges, i.e., if one can be obtained from the other by deleting an edge and adding anotheredge. Geometric tree graphs were shown to be connected [1], and upper and lower bounds on theirdiameter were established [1, 5].We study a reconstruction problem for geometric graphs: Is the geometric tree graph G ( P )sufficient for “reconstructing” the structure of K ( P )? In a sense, this question is a geometriccounterpart of the work of Sedl´aˇcek [11], who studied the question whether a graph can be re-constructed from its spanning trees. As we deal with a geometric setting, we seek to reconstructthe geometric structure of the graph. Definition 1.2.
Let P be a finite set of points in general position in the plane. The geometricstructure of the complete graph K ( P ) as a geometric graph includes, for any pair [ a, b ] , [ c, d ] ofvertex-disjoint edges, the information whether they cross or not. Our main result is the following:
Theorem 1.3.
For any finite set P of points in general position in the plane, the geometricstructure of K ( P ) can be reconstructed from the geometric tree graph G ( P ) . While the proof of the theorem is elementary, it is rather complex, and consists of severalstages:1.
Maximal cliques in G ( P ) . We study thoroughly the structure of maximal cliques in G ( P ). We divide these cliques into two types, called “union max-cliques” and “intersectionmax-cliques”, and show that given a maximal clique in G ( P ), one can determine its type.This study spans Section 2.2. Stars and brushes in G ( P ) . We show how to identify the vertices of G ( P ) that cor-respond to spanning stars and spanning brushes (i.e., spanning trees of diameter 3 witha single internal edge), by examining the max-cliques to which they belong. The starsare determined only up to an automorphism of K ( P ) (obviously, one cannot do better),and once they are fixed, the brushes are determined uniquely. This part of the proof ispresented in Section 3.3. The geometric structure of K ( P ) . We show how the geometric structure of K ( P ) canbe derived from information on the brushes in G ( P ). This part is presented in Section 4.In the last part of the paper, Section 5, we consider abstract (i.e., non-geometric) graphs,and show that a variant of the argument developed in Sections 2 and 3 can be used to provethe following result: 2 heorem 1.4. For any n ∈ N , the automorphism group of the tree graph of K n is isomorphicto Aut( K n ) ∼ = S n . Our treatment of the geometric reconstruction problem (i.e., K ( P ) from G ( P )) falls short ofthis. It leaves open the (quite implausible) possibility that the geometric tree graph G ( P ) hasan automorphism η , other than the identity, that fixes each star and each brush. This leavesopen, for further research, the following question. Question 1.5.
Is this true that for any finite set P of points in general position in the plane,we have Aut( G ( P )) ∼ = Aut( K ( P )) , where G ( P ) is treated as an abstract graph, whereas K ( P ) is treated as a geometric graph? G ( P ) In this section we study the structure of maximal (with respect to inclusion) cliques in thegeometric tree graph G ( P ). We divide the maximal cliques into two types, called U-cliques andI-cliques, and our ultimate goal is to determine, given a maximal clique in G ( P ), what is itstype.We start in Section 2.1 with a few definitions and notations, to be used throughout thepaper. In Sections 2.2 and 2.3 we describe a classification of the maximal cliques into twotypes, presented originally in [13], and discuss basic properties of both types. In order todistinguish between general combinatorial considerations and geometric arguments specific toSSTs, we start in Section 2.2 with a general combinatorial framework, and leave the geometricarguments to Section 2.3.In Sections 2.4 and 2.5 we study degenerate maximal cliques, i.e., maximal cliques of size2. In Section 2.4 we give a geometric characterization of the situation when a maximal cliqueis degenerate, and in Section 2.5 we show how to identify whether a given degenerate maximalclique is a U-clique or an I-clique. Finally, in Section 2.6 we show how to determine whether agiven non-degenerate maximal clique is a U-clique or an I-clique. Notation 2.1.
The following notations and conventions are used throughout the paper. Thestraight line that passes through points x, y is denoted by ℓ ( x, y ) . The vertex and edge sets of agraph G are denoted by V ( G ) and E ( G ) , respectively. Since the set P is fixed, we shall alwaysidentify a spanning subgraph of K ( P ) (and, in particular, a spanning subtree) with its set ofedges. In our study we shall extensively use maximal cliques of G ( P ). These are defined as follows: Definition 2.2. A max-clique in a graph G is a maximal (with respect to inclusion) cliqueincluded in G . Since any max-clique is a complete graph on its vertex set, we shall identify amax-clique with its set of vertices. We shall use the following observation on the structure of G ( P ), proved by Avis andFukuda [1]. Claim 2.3 ( [1], Lemma 3.15) . For any set P of points in general position in the plane, G ( P ) is connected and its diameter is ≤ | P | − . .2 Types of Max-cliques in Tree Graphs A generic combinatorial way to treat a tree graph is to consider a base set X and a graph G whose vertices are q -subsets of X (not necessarily all the q -subsets), such that two vertices A, B ∈ V ( G ) are adjacent if and only if | A △ B | = 2. (In the case of the (geometric) tree graphof a point set P , we have X = E ( K ( P )), q = | P | −
1, and the vertices of G are the sets of edgesof (simple) spanning trees of K ( P ).)Let A, B be two adjacent vertices of G . Denote I = A ∩ B, D = A △ B and U = A ∪ B = I ∪ D. A third vertex C ∈ V ( G ) is a common neighbor of A and B if and only if:1. C ∩ D = ∅ and I ⊂ C . In this case, C is obtained from I by adding a single element.2. D ⊂ C and C ⊂ U . In this case, C is obtained from U by removing a single element.(It is easy to see that in any other case, either | A △ C | 6 = 2 or | B △ C | 6 = 2.)If C, C ′ are both common neighbors of A and B , such that C satisfies (1) and C ′ satisfies(2), then clearly, | C △ C ′ | = 4. Hence, if two common neighbors of A and B are themselvesneighbors, then either both satisfy (1) or both satisfy (2). On the other hand, it is clear thatany two common neighbors satisfying (1) are themselves neighbors, and the same holds for (2).Thus, any pair A, B of adjacent vertices of G is included in at most two max-cliques:1. U ( A, B ) = { C ∈ V ( G ) | C ⊂ A ∪ B } , and2. I ( A, B ) = { C ∈ V ( G ) | C ⊃ A ∩ B } .For any two elements C, C ′ ∈ U ( A, B ), the union C ∪ C ′ is constant (and equal to U ). Likewise,for any two elements C, C ′ ∈ I ( A, B ), the intersection C ∩ C ′ is constant (and equal to I ). Thisis the motivation behind the following definition. Definition 2.4.
A max-clique of the first type will be called a Union max-clique, or a
U-clique ,and a max-clique of the second type will be called an Intersection max-clique or an
I-clique . Remark 2.5.
It is clear that given two vertices
A, B of a max-clique C , we cannot determinewhether C is a U-clique or an I-clique. However, if we are given a third vertex B ′ ∈ C , we candetermine the type, according to which one of the equalities A ∩ B = A ∩ B ′ , A ∪ B = A ∪ B ′ holds. Moreover, once we determine that C is, say, an I-clique, we know that C = I ( A, B ) as this is the unique I-clique that includes A and B . Hence, three vertices of a max-cliquedetermine it uniquely. Definition 2.6. If U ( A, B ) = { A, B } , we say that { A, B } is a degenerate U-clique . Similarly,if I ( A, B ) = { A, B } , we say that { A, B } is a degenerate I-clique . For a pair of adjacent vertices
A, B , there are four possible situations:1. | U ( A, B ) | ≥ | I ( A, B ) | ≥
3. In this case, there are exactly two max-cliques thatcontain A and B .2. | U ( A, B ) | ≥ | I ( A, B ) | = 2. In this case, there is a unique max-clique that contains A and B , namely U ( A, B ). (The I-clique that contains A and B is degenerate.)3. The same as (2), with the roles of U ( A, B ) and I ( A, B ) interchanged.4. U ( A, B ) = I ( A, B ) = { A, B } . In this case, the set { A, B } itself is a max-clique (that is,both a U-clique and an I-clique). As we shall see in the sequel, this situation cannot occurin our geometric setting. 4 .3 Types of Max-Cliques in G ( P ) Now we turn to the geometric graph G ( P ) and show additional properties of max-cliques thatfollow from its geometric structure. In order to make our notation suggestive, we denote nowthe two adjacent vertices of G ( P ) by T , T , and the edges in their symmetric difference by e ∈ T \ T and e ∈ T \ T .In G ( P ), U-cliques and I-cliques have a geometric meaning.1. U ( T , T ) . Consider the graph ¯ T = T ∪ T = T ∪ { e } . Obviously, it is a connectedgraph with a unique cycle. This cycle contains e and e , and is simple if and only if e and e do not cross. (Note that e cannot cross another edge of T , as both these edgesbelong to the SST T .) As shown above, U ( A, B ) consists of all vertices of G ( P ) thatare obtained from ¯ T by removing a single edge. Since the vertices of G ( P ) are the edgesets of SSTs, we can say that removing an edge (other than e , e ) from ¯ T results in anelement of U ( T , T ) if and only if the unique cycle of ¯ T is simple, and the removed edgebelongs to that cycle. Note that if e , e cross, then removal of any edge other than e , e from ¯ T results in a non-simple graph, and thus, the only elements of U ( T , T ) are T and T .2. I ( T , T ) . Consider the graph ˜ T = T ∩ T = T \ { e } . Obviously, it is a simple forestwith two connected components. As shown above, I ( A, B ) consists of all vertices of G ( P )that are obtained from ˜ T by adding a single edge. Since the vertices of G ( P ) are the edgesets of SSTs, we can say that adding an edge to ˜ T results in an element of I ( A, B ) if andonly if that edge makes the forest ˜ T into a simple spanning tree of K ( P ). The geometric interpretation allows us to characterize the cases when U-cliques and I-cliquesare degenerate.
Claim 2.7.
Let T , T be SSTs such that T \ T = { e } and T \ T = { e } . The U-clique U ( T , T ) is degenerate if and only if e and e cross.Proof. By the geometric interpretation, if e and e cross then the only vertices of U ( T , T ) are T and T , and thus, it is degenerate. If e and e do not cross, then ¯ T is a simple connectedgraph on n vertices with n edges. (Note that ¯ T is simple since as T , T are SSTs, the only edgesin ¯ T that may cross each other are e , e .) Thus, ¯ T has a cycle of order at least 3. Removal ofany edge from this cycle gives rise to a vertex in U ( T , T ). Therefore, in this case U ( T , T ) isnon-degenerate. Proposition 2.8.
Let P be a finite set of points in the plane, no three on a line, | P | ≥ . Let T , T be SSTs of K ( P ) such that T \ T = { e } and T \ T = { e } . The I-clique I ( T , T ) isdegenerate only in the following case:The convex polygon conv( P ) has three consecutive vertices x, v, y (i.e., [ x, v ] and [ v, y ] areedges of conv( P ) ), such that:1. The triangle conv( x, v, y ) contains no other points of P .2. e = [ x, v ] and e = [ v, y ] .3. [ x, y ] ∈ T ∩ T . ywx Figure 1: Illustration to the proof of Proposition 2.8.
Proof.
Assume that (1)–(3) hold, and let T ∈ I ( T , T ). T is connected, and thus, has anedge e that emanates from v . Note that as x, v, y are consecutive vertices of conv( P ) and thetriangle conv( x, v, y ) contains no other points of P , any edge of K ( P ) that emanates from v (other than e and e ) must cross [ x, y ]. Thus, either e = e , e = e , or e crosses [ x, y ]. Thelatter is impossible, since [ x, y ] ∈ T ∩ T , and as shown above, T ∩ T is included in any elementof I ( T , T ). If e = e , then T = (( T ∩ T ) ∪ { e } ) ⊂ T , which implies T = T (as both havethe same number of edges). Similarly, if e = e then T = T . Therefore, the only elements of I ( T , T ) are T and T , i.e., I ( T , T ) is degenerate.In the other direction, assume that I ( T , T ) is degenerate. As mentioned above, the graph˜ T = T ∩ T is a simple forest with two connected components. Color the vertices of onecomponent white and the vertices of the other component black, and call an edge colorful if itsendpoints are of different colors.Since ˜ T is planar, it can be extended to a triangulation T of conv( P ) with vertex set P .As T is connected, it contains a colorful edge e . A triangle in T to which e belongs clearlycontains another colorful edge e ′ . Addition of either e or e ′ to ˜ T results in a simple tree, andthus, gives rise to a vertex of I ( T , T ). (Note that e and e ′ cannot cross edges of ˜ T since theybelong to a triangulation that extends ˜ T .) Since I ( T , T ) is degenerate, this implies that allother edges of T are not colorful. We claim that this can happen only in the case described inthe statement of the proposition.Consider the edges e, e ′ . If e is not a boundary edge of conv( P ) then it belongs to anothertriangle in T . The other triangle must contain an additional colorful edge, contradicting theassumption that only e and e ′ are colorful. The same holds for e ′ , and thus, both e and e ′ areboundary edges of conv( P ). Denote their common vertex by v and their other endpoints by x, y , respectively.It is clear that Condition (1) above holds for x, v, y , since T is a triangulation of P andconv( v, x, y ) is one of its triangles. To see that Condition (2) holds, note that ˜ T ∪ { e } and˜ T ∪ { e ′ } are the only vertices of I ( T , T ), and thus, are equal to T and T . As T i = ˜ T ∪ { e i } for i = 1 ,
2, the edges e, e ′ must coincide with e and e .Finally, since | P | ≥
4, [ x, y ] is a diagonal of conv( P ), and thus, in the triangulation T
6t belongs to another triangle conv( x, y, w ) (see Figure 1). This triangle is monochromatic(as otherwise, there are at least four colorful edges in T ), hence w, v are of different colors.We apply a flip to the triangulation T , replacing the triangles conv( x, y, v ) , conv( x, y, w ) byconv( x, v, w ) , conv( y, v, w ). The resulting triangulation includes an additional colorful edge[ v, w ] that does not cross any edge of ˜ T , except possibly for [ x, y ]. Since by assumption, thereare only two colorful edges that do not cross edges of ˜ T , we must have [ x, y ] ∈ ˜ T , which meansthat Condition (3) holds. Corollary 2.9.
Each edge of G ( P ) is contained in at least one non-degenerate max-clique.Proof. Let [ T , T ] ∈ G ( P ) and denote T \ T = { e } and T \ T = { e } . If e , e do not crossthen U ( T , T ) is non-degenerate by Claim 2.7. If e , e cross then I ( T , T ) is non-degenerate,since by the proof of Proposition 2.8, if I ( T , T ) is degenerate, then the only edges that canbe added to T ∩ T to form a simple tree share a vertex, which is not the case for e , e thatcross in an interior point. G ( P ) Lemma 2.10.
Let T be a SST of G ( P ) , and let D be an I-clique that contains T . Denote thecommon intersection of pairs of elements of D by ˜ T , and let { e } = T \ ˜ T . If e is not a leaf edgeof T , then | V ( D ) | ≥ .Proof. We begin the proof with the argument used in the proof of Proposition 2.8. Namely, weconsider the graph ˜ T , which is a simple forest with two components. We color its componentsblack and white, and extend it to a triangulation T of P . The triangulation contains a colorfuledge [ a, b ], and consequently, another colorful edge [ a, b ′ ] in the same triangle. Assume w.l.o.g.that a is black, b and b ′ are white.Now we would like to use the assumption that e is not a leaf edge of T . This assumptionimplies that each connected component of ˜ T has at least two vertices. Consequently, any vertex v ∈ P is an endpoint of at least one monochromatic edge of T (as otherwise, v would be isolatedin ˜ T ).If both [ a, b ] and [ a, b ′ ] are boundary edges of conv( P ), then (since △ a, b, b ′ is a triangle in T ) these are the only edges of T that emanate from a . This is impossible, as they are bothcolorful. Hence, we can assume w.l.o.g. that [ a, b ] is a diagonal of conv( P ), and thus, belongsto two triangles, △ abc and △ abd . (Note that either c or d is equal to the vertex b ′ mentionedabove.) We consider several cases, according to the colors of c and d :1. Case 1: c and d are white. (This case is illustrated in Figure 2.) Consider the neighborsof a in T . Since T is a triangulation, all these vertices lie on a path in T . As statedbefore, since a is black, at least one of its neighbors must be black. On the other hand,some of its neighbors (including b, c, d ) are white. Hence, at least one of the edges inthe path connecting the neighbors of a is colorful (see Figure 2). In addition, the edges[ a, b ] , [ a, c ] , [ a, d ] are colorful. Thus, T contains at least four colorful edges, and each ofthem gives rise to a vertex of D . Therefore, | V ( D ) | ≥ For sake of clarity, we use here and in the sequel the notation [
T, T ′ ] for edges of G ( P ), like is commonlyused for geometric graphs, although G ( P ) is treated as an abstract graph. bcd Figure 2: Illustration to Case 1 of Lemma 2.10. a b c d a bcd k
Figure 3: Illustrations to Case 3a of Lemma 2.102.
Case 2: c and d are black. Since the black vertices a, c, d are all neighbors of the whitevertex b , the argument of Case 1 applies, with the roles of a with b , black and white,interchanged.3. Case 3: c and d have different colors. Assume w.l.o.g. that c is white and d is black.In this case, the edges [ a, b ] , [ a, c ] , and [ b, d ] are colorful. We further divide this case intothree subcases, according to whether [ a, c ] and [ b, d ] are diagonals of conv( P ) or not, andshow that in each case, T contains at least one additional colorful edge.(a) Case 3a: [ b, d ] is a diagonal of conv( P ) . In this case, [ b, d ] belongs to an additionaltriangle of T (i.e., other than △ abd ). If this triangle is △ bdc , then the edge [ c, d ] iscolorful, implying | V ( D ) | ≥ △ bdk forsome k = c , then, as b and d have different colors, one of the edges [ b, k ] and [ d, k ] iscolorful, again implying | V ( D ) | ≥ Case 3b: [ a, c ] is a diagonal of conv( P ) . The argument of Case 3a applies, withthe roles of a and b , c and d , black and white, interchanged.(c) Case 3c: Both [ a, c ] and [ b, d ] are boundary edges of conv( P ) . In this case, (cid:3) abcd is a convex quadrilateral, and thus, its diagonal [ c, d ] lies inside it (see Fig-ure 4). As both △ abc and △ abd are triangles in T , they do not contain points of P ,hence [ c, d ] does not cross any edge of ˜ T . (Note that [ a, b ] is not an edge of ˜ T , sinceit is colorful.) Since [ c, d ] is colorful, the graph ˜ T ∪ { [ c, d ] } is an SST, hence belongsto D . Therefore, | V ( D ) | ≥
4, which completes the proof of the lemma.8 bcd
Figure 4: Illustration to Case 3c of Lemma 2.10.Using the lemma, we can identify whether a given degenerate clique is a U-clique or anI-clique.
Proposition 2.11.
Let [ S, T ] be an edge in G ( P ) that constitutes a degenerate clique. (Thisactually means that there is only one max-clique that includes { S, T } .) If [ S, T ] is included ina max-clique of 3 vertices, then the degenerate clique [ S, T ] is an I-clique. If [ S, T ] is includedin a max-clique of at least 4 vertices, then the degenerate clique [ S, T ] is a U-clique.Proof. By Corollary 2.9, each edge is included in at least one non-degenerate max-clique. Hence,[
S, T ] is included in a max-clique of size at least 3.If the degenerate clique [
S, T ] is an I-clique, then, by Proposition 2.8, S ∩ T is a forest withtwo connected components. One of them is an isolated vertex v that lies on the boundary ofconv( P ), and the two neighbors of v on the boundary of conv( P ), x, y , are adjacent in S ∩ T .In such a case, the unique cycle in the graph S ∪ T is the triangle △ ( x, v, y ). By the geometriccharacterization of Section 2.3, this implies that the size of the U-clique that includes [ S, T ] is3. If the degenerate clique [
S, T ] is a U-clique, then, by Claim 2.7, the unique cycle of S ∪ T includes the edges s ∈ S \ T and t ∈ T \ S , and these edges cross. This implies that s is not aleaf edge of S , and thus, by Lemma 2.10, the size of the I-clique that includes [ S, T ] is at least4. G ( P ) Our next goal is to identify whether a given non-degenerate max-clique is a U-clique or anI-clique. This identification is somewhat more complex, and requires some preparations.
Definition 2.12.
For any SST T ∈ G ( P ) , define a graph D T as follows: V ( T ) is the set ofmax-cliques that contain T (including degenerate cliques). Two max-cliques are adjacent in D T if and only if their intersection is a single edge (that obviously has T as one of its endpoints). D T include T , it follows that any two non-adjacent vertices of D T intersect in T only. Theorem 2.13.
Assume that | P | ≥ . For any T ∈ G ( P ) , the graph D T is connected. Before we present the proof of the theorem, we show how it can be used (along with severalof the previous lemmas) to identify the types of max-cliques in G ( P ).By the discussion in Section 2.2, each edge of G ( P ) belongs to exactly one U-clique andexactly one I-clique (one of them possibly degenerate). This implies that D T is 2-colorable.Indeed, we can color all its vertices that are U-cliques white and all its vertices that are I-cliquesblack. If two vertices are adjacent, both include the same edge [ S, T ] ∈ G ( P ), and thus, one isa U-clique and the other is an I-clique, so they have different colors. Using Theorem 2.13, wecan conclude that D T is connected and 2-colorable, which implies that the 2-coloring is unique,in the sense that fixing the color of any vertex determines the colors of all other vertices. Thiswill allow us to determine the types of all max-cliques, by the following four-step process:1. Pick an SST T that belongs to a degenerate clique { S, T } . (It is easy to show that sucha T always exists. We show this in Lemma 2.14 below.) Using Proposition 2.11, identifywhether { S, T } is a U-clique or an I-clique.2. Consider the vertices of D T and determine for each of them whether it is a U-clique oran I-clique. (This is possible by the explanation above. Since we know the “color” of thevertex { S, T } of D T , we can determine the colors of all other vertices.)3. Consider a neighbor T ′ of T . Note that the edge [ T, T ′ ] ∈ G ( P ) belongs to a max-clique C that is a vertex of both D T and D T ′ . Determine whether C is a U-clique or an I-clique.(This is possible, as by the previous step we can determine the type for all max-cliquesthat are vertices of D T .) Using this information about C ∈ V ( D T ′ ), determine for eachvertex of D T ′ whether it is a U-clique or an I-clique.4. Repeat Step 3 with a “new” vertex of G ( P ) every time until the types of all max-cliquesare determined. (Since G ( P ) is connected by Claim 2.3, we indeed reach all vertices of G ( P ) in this way.)Therefore, in order to determine the types of all max-cliques we have to prove a simplelemma on the existence of degenerate max-cliques, and to prove Theorem 2.13. We providethese two items now. Lemma 2.14.
For any set P , | P | ≥ , of points in general position in the plane, there existsan SST that belongs to a degenerate U-clique in G ( P ) .Proof. By the Erd˝os-Szekeres theorem [4], there exist four points in P in convex position. De-note these points a, b, c, d , such that [ a, b, c, d ] is a convex quadrilateral (in this order). Considerthe tree T that includes the edges [ a, b ] , [ b, d ] , [ d, c ], all edges that connect a to each p ∈ P thatlies above the straight line ℓ ( b, d ) and all edges that connect c to each p ∈ P that lies below ℓ ( b, d ), as shown in Figure 5. Clearly, T is an SST of K ( P ). Addition of the edge [ a, c ] to T creates a self-crossing cycle, and thus, by the discussion in Section 2.3, T ∈ V ( G ) belongs to adegenerate U-clique. 10 b cd Figure 5: An illustration for the proof of Lemma 2.14.
Proof of Theorem 2.13.
Suppose
C, C ′ ∈ V ( D T ). Choose edges [ T, S ] ∈ C , [ T, S ′ ] ∈ C ′ . Inorder to prove that C is connected to C ′ by a path in D T , it suffices to find a sequence S , S , . . . , S ℓ of SST’s, such that S = S , S ℓ = S ′ , and for each 0 ≤ i < ℓ , there is a max-clique C i of G ( P ) that includes both [ T, S i ] and [ T, S i +1 ]. This will be done in four steps.1. Step 1.
Extend the SST T to a triangulation T of conv( P ). Recall that T is a 2-connectedgraph.2. Step 2. If S ⊂ T , leave S as it is. If not, T contains the graph ˜ T = T ∩ S = T \ { e } .Since T is 2-connected, there is another edge e ∗ in T that connects the two componentsof ˜ T . Define S ∗ = ˜ T ∪ { e ∗ } . S ∗ is an SST of K ( P ). Note that S ∗ is included in theI-clique I ( T, S ) (since T ∩ S ∗ = T ∩ S = ˜ T ).Do the same for S ′ : leave it, if S ′ ⊂ T , or replace it by some S ′ ∗ ⊂ T , such that S ′ ∗ ∈ I ( T, S ′ ).Step 2 allows us to restrict our attention to the case where both S and S ′ are included in T (and all intermediate SST’s will be included in T , as well).3. Step 3.
Assume T ∩ S = T \ { e } , T ∩ S ′ = T \ { e ′ } . If e = e ′ , then [ T, S ] and [
T, S ′ ]belong to the same I-clique. Suppose e = e ′ . Since T is a simple tree, there is a uniquesimple path in T whose edges are (in this order) e , e , . . . , e m , with e = e and e m = e ′ .For 1 < i < m , choose an edge e ∗ i of T other than e i that connects the two componentsof T \ { e i } , and define S i = T \ { e i } ∪ { e ∗ i } . In addition, let S = S and S m = S ′ . Nowwe only have to connect S i − with S i for i = 2 , , . . . , m . (Note that the desired sequence S , S , . . . , S ℓ will be the concatenation of the sequences connecting S to S , S to S etc., and ℓ will be the sum of their lengths.)4. Step 4.
Suppose T is a triangulation of conv( P ) with vertex set P . Let T, S, S ′ beSST’s of K ( P ) that are included in T . Assume that both S and S ′ are adjacent to T in G ( P ), S ∩ T = T \ { e } , S ′ ∩ T = T \ { e ′ } , e = e ′ , and e, e ′ share a vertex. Suppose e = [ x, y ] , e ′ = [ x ′ , y ], and let k = deg( y ) −
2. Denote the edges of T that emanate from y by e, e ′ , f , f , . . . , f k . Removal of all edges that emanate from y divides T into k + 311onnected components: D that includes x , D ′ that includes x ′ , F s (1 ≤ s ≤ k ) thatincludes the second endpoint of f s , and F k +1 that consists of the isolated vertex y . Since T is 2-connected, we can extend the forest B = D ∪ D ′ ∪ F ∪ . . . ∪ F k into an SST of K ( P \ { y } ) by adding k + 1 edges of T .Let ¯ S be such an extension. ¯ S includes a unique simple path π from x to x ′ . Thispath starts in the component D of B , and ends in D ′ . It visits some of the intermediatecomponents F i in a particular order (see Figure 6). By appropriately labelling thesecomponents, we may assume that π visits D, F , F , . . . , F l , D ′ in this order (0 ≤ l ≤ k ).Let g be the edge of π that passes from D to F , g i (1 ≤ i ≤ l −
1) the edge that passesfrom F i to F i +1 , and g l be the edge that passes from F l to D ′ . (If l = 0, then there isonly one edge g = g that passes directly from D to D ′ .)Now we can describe the passage from [ T, S ] to [
T, S ′ ]. Let us start with the simple case l = 0. Put S = T \ { e } ∪ { g } , S = T \ { e ′ } ∪ { g } . Then T ∩ S = T ∩ S , T ∪ S = T ∪ S ,and T ∩ S = T ∩ S ′ . Thus, { T, S, S } and { T, S , S ′ } are included in I-cliques, while { T, S , S } is included in a U-clique. Hence, ( S, S , S , S ′ ) is the required sequence.When l >
0, define S = T \ { e } ∪ { g } , S = T \ { f } ∪ { g } ,S = T \ { f } ∪ { g } , S = T \ { f } ∪ { g } ,. . .S l − = T \ { f l − } ∪ { g l − } , S l = T \ { f l } ∪ { g l − } ,S l +1 = T \ { f l } ∪ { g l } , S l +2 = T \ { e ′ } ∪ { g l } . Then each of the triples { T, S, S } , { T, S i , S i +1 } (for 1 ≤ i ≤ l ), and { T, S l +2 , S ′ } isincluded in an I-clique, and each of the triples { T, S i − , S i } (for 1 ≤ i ≤ l + 1) isincluded in a U-clique. Therefore, ( S, S , S , . . . , S l +1 , S l +2 , S ′ ) is the required sequence.This completes the proof of the theorem.As explained after the statement of Theorem 2.13, the Theorem and Lemma 2.14 imply thefollowing corollary. Corollary 2.15.
Assume | P | ≥ . Given G ( P ) , we can determine for each max-clique in itwhether it is a U-clique or an I-clique. In this section we use the results of Section 2 to identify the vertices of G ( P ) that representstars (i.e., SSTs of diameter 2) and brushes (i.e., SSTs of diameter 3). The stars, considered inSection 3.1, are determined only up to an automorphism of K ( P ) as a geometric graph. Thebrushes, considered in Section 3.2, are determined uniquely given a determination of the stars. Definition 3.1. A star is a tree of diameter 2. f e'f f y D g g g F D'F F Figure 6: An illustration for the proof of Theorem 2.13.
Notation 3.2.
For x ∈ P , we call the spanning star whose center is x an x -star , and denoteit by S ( x ) . Theorem 3.3.
A vertex T ∈ G ( P ) is a star if and only if all U-cliques that include T are ofsize 3.Proof. Recall that by the geometric interpretation of max-cliques presented in Section 2.3, if U ( S, T ) is a non-degenerate U-clique then all its elements are obtained from the graph S ∪ T by removing an edge from its unique cycle. In particular, | U ( S, T ) | is the length of the uniquecycle of S ∪ T . If the unique cycle of S ∪ T is self-crossing (which can occur only if its lengthis ≥
4) then U ( S, T ) is degenerate.Assume that T ∈ G ( P ) is an x -star, and let U ( S, T ) be a U-clique that includes T . Since T is a star, S ∪ T is obtained from T by adding an edge that connects two leaves v, w of T .Hence, the unique cycle of S ∪ T , ([ x, v ] , [ v, w ] , [ w, x ]), is of length 3. Thus, | U ( S, T ) | = 3, asasserted.On the other hand, we show that if T ∈ G ( P ) is an SST of diameter ≥
3, then T belongsto a U-clique of size = 3. Since diam( T ) ≥ T contains an internal edge [ a, b ] (i.e., both a and b are not leaves of T ). Consider the graph T \ { [ a, b ] } , that is obviously a forest with twoconnected components. Color the vertices of the connected component that includes a blackand the vertices of the component that includes b white.We would like to show that there exists a colorful edge e that uses neither a nor b and doesnot cross any edge of T \ { [ a, b ] } (see Figure 7). This will conclude the proof, since in such acase, denoting S = T ∪ { e } \ { [ a, b ] } , we find that S is an SST, [ S, T ] ∈ G ( P ), and the uniquecycle of the graph S ∪ T = T ∪ { e } is of length ≥
4. Then, by the geometric interpretationabove, if e crosses [ a, b ] then U ( S, T ) is a degenerate U-clique, and otherwise, | U ( S, T ) | is equalto the length of the unique cycle of S ∪ T , that is ≥
4. Hence, in any case, T lies in a U-cliqueof size = 3.We extend T to a triangulation T of conv( P ), and consider three cases:13 ea b Figure 7: An illustration for the proof of Theorem 3.3 – beginning. a b cd d Figure 8: An illustration for the proof of Theorem 3.3 – case 1.1.
Case 1: [ a, b ] is a boundary edge of conv( P ) . The edge [ a, b ], being colorful, iscontained in a colorful triangle △ abc ∈ T . The other colorful edge of △ abc is not aboundary edge of conv( P ), as otherwise, one of the two connected components of T \{ [ a, b ] } consists of a single vertex, which contradicts the assumption that [ a, b ] is an internaledge of T . Denote that colorful edge [ a, c ]. The neighbors of a in T constitute a path[ b, c, d , d , . . . ]. Since the connected component of a in the graph T \ { [ a, b ] } (i.e., the“black” component) includes more than one vertex, at least one of the d i ’s is black. Thus,the path contains a colorful edge e that uses neither a nor b (see Figure 8). Furthermore,as e belongs to T , it does not cross any edge of T . Hence, e is the edge whose existencewas claimed.2. Case 2: b is an internal vertex of conv( P ) . In this case, [ a, b ] is an internal edge of T , and thus, it is contained in two triangles △ abc, △ abd ∈ T . We further divide this caseinto two sub-cases:(a) Case 2a: Either c or d (or both) are black. Assume w.l.o.g. that c is black.Since b is an internal vertex of conv( P ), the neighbors of b in T constitute a cycle[ d, a, c, d , d , . . . , d k , d ]. Since the connected component of b in the graph T \ { [ a, b ] } contains more than one vertex, at least one of the d i ’s or d is white. Since a, c areblack, this implies that the cycle includes at least two colorful edges, and at leastone of them uses neither a nor b (see left part of Figure 9). As in Case 1, this is thedesired edge e .(b) Case 2b: Both c and d are white. By the same argument as above, a has a black14 dbc d d d d d a dbcd d d Figure 9: An illustration for the proof of Theorem 3.3 – case 2. d d a bc d a bc d Figure 10: An illustration for the proof of Theorem 3.3 – Case 3.neighbor in T . As the neighbors of a in T form a (possibly closed) path, at leastone edge of this path is colorful (see right part of Figure 9). This edge uses neither a nor b (as both edges that use b in this path, [ b, c ] and [ b, d ], are not colorful). Asin the previous cases, this is the desired edge e .3. Case 3: Both a and b are boundary vertices of conv( P ) , and [ a, b ] is a diagonalof conv( P ) . As in Case 2, [ a, b ] lies in two triangles △ abc, △ abd ∈ T . We further dividethis case to two sub-cases:(a) Case 3a: c and d are of the same color. W.l.o.g., c and d are white. By thesame arguments as above, a has a black neighbor, and thus, the path of a ’s neighborsincludes a colorful edge (see left part of Figure 10). This edge does not use b , asboth edges that use b (that are [ b, c ] , [ b, d ]) are not colorful, and it clearly does notuse a .(b) Case 3b: c and d are of different colors. In this case, the edge [ c, d ] is as desired,since it is colorful, does not use a, b , and does not cross edges of T \ { [ a, b ] } (see right15 q Figure 11: A pq -brush.part of Figure 10). Note that since [ c, d ] crosses [ a, b ], the U-clique that includes T in this case is degenerate. This completes the proof of the theorem. Remark 3.4.
It may happen that T ∈ G ( P ) is not a star, but all U-cliques that contain T areeither degenerate or of size 3. G ( P ) Definition 3.5.
Let P be a set of points in general position in the plane, and let p, q ∈ P . A pq -brush is an SST of diameter 3 whose only internal edge is [ p, q ] . Figure 11 shows an example of a pq -brush.In this subsection we aim at identifying the vertices of G ( P ) that represent brushes. Theidentification uses distances in the graph G ( P ), defined (as usual) as the length of the shortestpath between two vertices, and denoted by d G ( P ) ( S, T ). Note that by the structure of G ( P ), itis clear that for any pair of vertices S, T ∈ G ( P ), we have d G ( P ) ( S, T ) ≥ | ∆( S, T ) | , where ∆( S, T ) is the symmetric difference between the edge sets of the graphs S and T . Theorem 3.6.
An SST T ∈ G ( P ) is a pq -brush if and only if it is not a star and d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) = n − . (1)For sake of convenience, we divide the theorem into two propositions. Proposition 3.7.
Assume that T ∈ G ( P ) satisfies d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) = n − . Then T is a pq -brush, or T = S ( p ) , or T = S ( q ) . roof. First, we note that d G ( P ) ( S ( p ) , S ( q )) ≥ | ∆( S ( p ) , S ( q )) | = n − , and thus, by the triangle inequality, d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) ≥ n − T ∈ G ( P ).Assume that T satisfies (1). Let k, ℓ be the numbers of edges of T that emanate from q, p (respectively), and let r be the number of edges of T that use neither p nor q . We consider twocases:1. [ p, q ] T . In this case, we have k + ℓ + r = n − , d G ( P ) ( T, S ( p )) ≥ | ∆( T, S ( p )) | = k + r, and d G ( P ) ( T, S ( q )) ≥ | ∆( T, S ( q )) | = ℓ + r. Hence, d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) ≥ k + ℓ + 2 r = ( n −
1) + r > n − .
2. [ p, q ] ∈ T . In this case, k + ℓ + r = n, d G ( P ) ( T, S ( p )) ≥ | ∆( T, S ( p )) | = k + r − , and D G ( P ) ( T, S ( q )) | ≥ | ∆( T, S ( q )) = ℓ + r − . Hence, d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) ≥ k + ℓ + 2 r − n + r − ≥ n − , and equality can hold only if r = 0, which means that all edges of T emanate either from p or from q and [ p, q ] ∈ T , i.e., T is a pq -brush, or T = S ( p ) or T = S ( q ). Proposition 3.8.
Let T ∈ G ( P ) be a pq -brush. Then d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) = n − . In order to prove Proposition 3.8, we need a lemma.
Definition 3.9.
Let T be a pq -brush (or T = S ( p ) or T = S ( q ) ). We say that T is of type( k, ℓ ) if val( T, p ) = k + 1 and val( T, q ) = ℓ + 1 (i.e., the numbers of edges of T that emanatefrom p, q are k + 1 , ℓ + 1 , respectively). Lemma 3.10. If T is a pq -brush (or a star) of type ( k, ℓ ) , k < n − , then it is adjacent in G ( P ) to some pq -brush (or star) T ′ of type ( k + 1 , ℓ − . (By symmetry, if ℓ < n − then T is adjacent in G ( P ) to some pq -brush (or star) T ′′ of type ( k − , ℓ + 1) .) q a x Figure 12: An illustration to the proof of Lemma 3.10.
Proof of the Lemma.
Assume w.l.o.g. that [ p, q ] is placed horizontally, and consider the half-plane above it. Let [ p, x ] be an edge such that the angle α between [ p, q ] and [ p, x ] is minimalamongst all edges of T that emanate from p (see Figure 12). Let T ′ = ( T \ { [ p, x ] } ) ∪ { [ q, x ] } .Due to the minimality of the angle α , [ q, x ] does not cross any of the edges of T that emanatefrom p . Thus, T ′ is a pq -brush of type ( k + 1 , ℓ −
1) (or T ′ = S ( q ), if ℓ = 1), and [ T, T ′ ] ∈ G ( P )since | ∆( T, T ′ ) | = 2. Proof of Proposition 3.8.
Let T ∈ G ( P ) be a pq -brush. Assume w.l.o.g. that T is of type ( k, ℓ ).Repeated use of Lemma 3.10 enables us to construct a path h T , T , . . . , T n − i in G ( P ) such that T = S ( q ), T k = T , and T n − = S ( p ). This implies d G ( P ) ( T, S ( p )) ≤ ℓ and d G ( P ) ( T, S ( q )) ≤ k ,hence, d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) ≤ ℓ + k = n −
2. Since we have shown above that d G ( P ) ( T, S ( p )) + d G ( P ) ( T, S ( q )) ≥ n − T , this completes the proof. K ( P ) In this section we achieve a complete reconstruction of the geometric structure of K ( P ), basedon the identification of stars and brushes presented in Section 3. Most of the effort is devotedto obtaining a complete identification of the brushes, in the sense that given a pq -brush T anda vertex x = p, q , we determine whether [ x, p ] ∈ T or [ x, q ] ∈ T . This step is presented inSection 4.1. The finalization of the proof of Theorem 1.3, presented in Section 4.2, is easy. G ( P ) So far, we know which vertices of G ( P ) are brushes. Furthermore, if we identify the points of P with the stars in G ( P ) (an identification that is determined only up to an automorphism of K ( P )), we can say for each brush T , what are the vertices p, q that are its “centers”, and howmany edges of T emanate from each of the central vertices p, q .Our goal now is to gain full information on the brushes. Namely, for a pq -brush T and avertex x = p, q , we would like to determine whether [ p, x ] ∈ T or [ q, x ] ∈ T .As an intermediate step, we would like to determine, for given x, y = p, q , whether both x and y are connected in T to the same vertex (either p or q ), or one of them is connected to p and the other to q . 18 roposition 4.1. Let T be a pq -brush, and let x, y ∈ P be different from each other and from p and q . The leaf edges of T whose endpoints are x and y emanate from the same internalvertex of T if and only if for any xy -brush S , we have d G ( P ) ( T, S ) ≥ n − . One direction of the proposition is immediate. If the leaf edges emanate from the samevertex, e.g., [ p, x ] , [ p, y ], then at most one (and actually, exactly one) of these edges can belongto an xy -brush (as ([ x, y ] , [ y, p ] , [ p, x ]) form a cycle). Since every edge of an xy -brush S emanatesfrom either x or y , we have | ∆( S, T ) | ≥ n − d G ( P ) ( T, S ) ≥ | ∆( S, T ) | = n − xy -brush S .On the other hand, if the leaf edges emanate from different vertices, e.g., [ p, x ] , [ q, y ], itis possible that an xy -brush S include both these edges, and then | ∆( S, T ) | = n −
3. Wewill construct an xy -brush that satisfies this condition, and furthermore, satisfies the strongercondition d G ( P ) ( S, T ) = n −
3. Before we show this construction, we need a few preparations.
Definition 4.2.
Let G be a geometric graph, and let O be a point in the plane. We say that O sees a point P if the open segment ( O, P ) does not meet any edge or vertex of G . We say that O sees an edge e ∈ E ( G ) if it sees every point X ∈ e , including the endpoints. Lemma 4.3.
Let G = ( V, E ) be a crossing-free geometric graph, with no isolated vertices.Suppose V is a disjoint union V = V ∪ W , where | V | = 2 (say, V = { p, q } ), and each edge of G connects a vertex of V with a vertex of W . Suppose O ∈ R \ S { aff( e ) : e ∈ E ( G ) } . Then O sees some vertex w ∈ W . We note that a similar lemma was proved in [6]. The assumption on O in [6] is O conv( V ( G )), and the assertion is the same as in our lemma. Proof.
Draw a ray R that emanates from O , crosses some edge of G , and does not meet anyvertex of G . (It is clear that such rays exist.) Denote by C the first crossing point of R with anedge of G . Then C is an interior point of an edge, say [ p, w ], of G , and O sees C . Now rotate R around O towards w , until it hits w . If the triangle △ OCw does not contain any vertex of G ,except w , then O sees w . Otherwise, there is a first position R ′ of the rotated ray that meets V . Let v be the point of R ′ ∩ V closest to O . Then O sees v . If v ∈ W , we are done. Assume,therefore, that v ∈ V . Clearly, v = p since 0 < ∠ vOp < ∠ wOp < π . Hence, v = q .Among the edges that emanate from q , let [ q, w ′ ] be the edge such that ∠ Oqw ′ is minimal.As before, rotate R ′ around O towards w ′ , until it hits w ′ . If the triangle △ Oqw ′ does notcontain any vertex of G , except q and w ′ , then O sees w ′ . Otherwise, there is a first position R ′′ of the rotated ray that meets V . Let v ′ be the point of R ′′ ∩ V closest to O . Then O sees v ′ .Now, we observe that v ′ V . Indeed, v ′ = q since 0 < ∠ v ′ Oq < ∠ qOw ′ < π , and v ′ = p since0 < ∠ v ′ Op = ∠ v ′ Oq + ∠ qOp < ∠ w ′ Oq + ∠ wOp < π . Therefore, v ′ ∈ W , which completes theproof.Now we are ready to prove Proposition 4.1. Proof of Proposition 4.1.
We already proved above that if the two leaf edges of T whose end-points are x, y emanate from the same vertex, then for any xy -brush S , d G ( P ) ( T, S ) ≥ n − p, x ] , [ q, y ]. We consider two cases, according to the placement of p, q, x, y in the plane. In each19 p x y Figure 13: An illustration to the proof of Proposition 4.1: Case 1. The regions are numbered bythe order of their consideration, where the missing number 3 corresponds to Phase 3 in which[ p, q ] is replaced by [ x, y ]. The notation ( i ) − ( v ) where i ∈ { , , } and v ∈ { x, y } means thatwe are going to connect all points in Region i to v .case, we show that we can pass from T to a suitable xy -brush S in n − d G ( P ) ( T, S ) = n −
3, and thus complete the proof of the proposition. Note that inall the steps of the path connecting T to S , the edges [ p, x ] , [ q, y ] remain untouched. Case 1: x, y are on the same side of ℓ ( p, q ) . In this case, at least one of the edges [ p, x ] , [ q, y ] is included in a line that supports the set { p, x, q, y } (which means that all points in the set are on the same side of the line). We assumew.l.o.g. that [ p, x ] has this property.The passage from T to an appropriate S is performed by a 4-phase procedure, illustrated inFigure 13. In each phase (except for phase 3 that will be described below), we consider one ofthe regions of the plane denoted in the figure: 1 , ,
4, and deal with all points of P that belongto that region.1. Region 1 (Reg1).
This region is the open half-plane to the left of the line ℓ ( p, x ).Assume that | P ∩ Reg | = k . We are going to perform k steps: in each step, we takeone of these points, remove the edge that connects it to either p or q , and add an edge20hat connects it to x . Of course, we must maintain the simplicity during all steps, andthis is achieved using Lemma 4.3.Let G be the geometric graph whose edges are all edges of T of the form [ p, w ] or [ q, w ],where w lies in Reg
1. The graph G and the point O = x satisfy the assumptionsof Lemma 4.3, and thus, by the Lemma, x sees one of its vertices w ∈ Reg
1, say w .Assume, for example, that [ q, w ] ∈ E ( G ). Define T = T \ { [ q, w ] } ∪ { [ x, w ] } . Since x sees w , the edge [ x, w ] does not cross any other edge of T . Thus, T is an SST and | T △ T | = 2, which implies that [ T, T ] ∈ G ( P ).Now, we repeat the first step with the SST T in place of T . That is, we define G whoseedges are all edges of T of the form [ p, w ] or [ q, w ] where w lies in Reg
1, except for [ q, w ].As before, we apply Lemma 4.3 with G and O = x and obtain a vertex w that is seenfrom x . Then, we define T by removing from T the edge that connects w to either p or q and adding the edge [ x, w ]. Note that the edge [ x, w ] that was not included in G cannot cross [ x, w ], as they both emanate from x .By continuing in the same fashion, we obtain a sequence T , T , . . . , T k such that T = T ,[ T i , T i +1 ] ∈ G ( P ) for all i , and in T k , all points in P ∩ Reg x .It should be noted that the parts of T i that are not included in the auxiliary graph G i ,i.e., the edge [ p, q ] and the edges [ p, w ] , [ q, w ], w ∈ R \ Reg
1, are all disjoint from theconvex set
Reg
1, and thus cannot cross the new edge [ x, w i +1 ] (as x ∈ bdry( Reg
Region 2 (Reg2).
This region contains all points that lie above ℓ ( p, q ) and on the rightside of ℓ ( p, x ). Assume that | P ∩ Reg | = k . We start with T k and perform k steps: ineach step, we consider one of these points, remove the edge that connects it to either p or q , and add an edge that connects it to y . As before, the simplicity is maintained duringall steps, by using Lemma 4.3.Let G k be the geometric graph whose edges are all edges of T k of the form [ p, w ] or[ q, w ], where w lies in Reg
2. The graph G k and the point O = y satisfy the assumptionsof Lemma 4.3, and thus, by the Lemma, y sees one of the vertices w ∈ Reg
2, call it w k +1 .Without loss of generality, [ p, w k +1 ] ∈ G k . Define T k +1 = T k \{ [ p, w k +1 ] }∪{ [ y, w k +1 ] } .Since y sees w k +1 , the edge [ y, w k +1 ] does not cross any other edge of T k . Thus,[ T k , T k +1 ] ∈ G ( P ).By continuing in the same fashion, we obtain a sequence T k +1 , T k +2 , . . . , T k + k suchthat [ T i , T i +1 ] ∈ G ( P ) for all i , and in T k + k , all points in P ∩ Reg x and all points in P ∩ Reg y .3. Phase 3.
In this phase, we add the edge [ x, y ] and remove the edge [ p, q ] (that otherwisecloses a cycle ([ x, y ] , [ y, q ] , [ q, p ] , [ p, x ])). Formally, we define T k + k +1 = T k + k \ { [ p, q ] } ∪{ [ x, y ] } . Note that the edge [ x, y ] does not cross any edge of T k + k , as in T k + k , allpoints of P ∩ Reg y .4. Region 4 (Reg4).
This region contains all points that lie below ℓ ( p, q ) and on the right of ℓ ( p, x ). Assume | P ∩ Reg | = k . As all points of P except for p, q, x, y belong to one of theregions: Reg , Reg , Reg
4, we have k + k + k = n −
4. We construct a sequence of SSTs T k + k +2 , . . . , T k + k + k +1 such that in T k + k + k +1 , all points in P ∩ Reg x and all points in P ∩ ( Reg ∪ Reg
4) are connected to y . Hence, T k + k + k +1 = T n − is an xy -brush that satisfies d G ( P ) ( T, T n − ) = n −
3, as desired.21 p x y
Figure 14: An illustration to the proof of Proposition 4.1: Case 2. The notation ( i ) − ( v ) where i ∈ { , } and v ∈ { x, y } means that the points in Region i are connected to v .Let G k + k +1 be the geometric graph whose edges are all edges of T k + k +1 of the form[ p, w ] or [ q, w ], where w lies in Reg
4. The graph G k + k +1 and the point O = y satisfythe assumptions of Lemma 4.3, and thus, by the Lemma, y sees one of the vertices w ∈ Reg
4, say w k + k +1 . Without loss of generality, [ p, w k + k +1 ] ∈ G k + k +1 . Define T k + k +2 = T k + k +1 \ { [ p, w k + k +1 ] } ∪ { [ y, w k + k +1 ] } . Since y sees w k + k +1 , the edge[ y, w k + k +1 ] does not cross any other edge of T k . (It should be noted that the factthat y lies outside Reg y sees Reg y , and P is in general position.) Thus,[ T k + k +1 , T k + k +2 ] ∈ G ( P ).By continuing in the same fashion, we obtain a sequence T k + k +2 , T k + k +3 , . . . , T k + k + k +1 such that [ T i , T i +1 ] ∈ G ( P ) and T k + k + k +1 = T n − is the desired xy -brush. Case 2: x, y are on different sides of ℓ ( p, q ) . This case is treated in a fashion similar to Case 1. We divide all points of P \ { p, q } intotwo regions, where Region 1 (Reg1) consists of the points above ℓ ( p, q ) and Region 2 (Reg2)consists of the points below ℓ ( p, q ) (see Figure 14). In the first phase, we consider the points of Reg x ), disconnect them from p or q and connect them to x instead. The procedureis identical to the procedure of the first phase of Case 1. In the second phase, we consider thepoints of Reg y ), disconnect them from p or q and connect them to y instead. Theprocedure is, again, similar. Finally, in the third phase we remove the edge [ p, q ] and insert theedge [ x, y ] instead. (As at this stage, all points in P \ { p, q, x, y } are connected to either x or y ,this step does not create crossings.) As a result, we obtain a sequence T = T , T , T , . . . , T n − ,such that [ T i , T i +1 ] ∈ G ( P ) for all i and T n − is an xy -brush, as desired.As Cases 1,2 include all possible placements of x, y, p, q , the proof is complete.Now we are ready to identify every brush completely. Corollary 4.4.
Let T ∈ V ( G ( P )) be a pq -brush and let x ∈ P , x = p, q . Given G ( P ) , we candetermine whether [ p, x ] ∈ T or [ q, x ] ∈ T .Proof. It follows from the proof of Proposition 3.8 that T belongs to a path h S ( p ) = T , T , T , . . . , T n − , T n − = S ( q ) i p x y l Figure 15: An illustration to the proof of Proposition 4.5.in G ( P ) such that in T i , deg( p ) = n − − i and deg( q ) = i + 1. Consider T . Since it has onlyone vertex x = p, q that is connected to q , we can use Proposition 4.1 to determine it. (Herewe use the assumption that n ≥ T and use Proposition 4.1 again todetermine the additional vertex x connected to q in T . (Note that [ x , q ] ∈ E ( T ), and thus, x is identified as the unique vertex x such that the leaf edge of T that emanates from it hasthe same second endpoint as the leaf edge that emanates from x .) We can continue in thesame fashion and get a complete identification of T , T , . . . , T n − , including T . Our last step toward the identification of the geometric structure of K ( P ) is the following easyproposition. Proposition 4.5.
Let p, q, x, y be four different points in P . The segments [ p, x ] and [ q, y ] donot cross if and only if there exists a pq -brush that includes the edges [ p, x ] and [ q, y ] .Proof. It is clear that if [ p, x ] and [ q, y ] cross then no pq -brush can contain both edges [ p, x ]and [ q, y ], as a brush is a simple tree. If [ p, x ] and [ q, y ] do not cross, then they are strictlyseparated by some line ℓ . In such a case, we can define a pq -brush in which all vertices thatlie on the same side of ℓ as p are connected to p , and all other vertices are connected to q (seeFigure 15). This pq -brush includes both [ p, x ] and [ q, y ].Now we are ready to prove our main theorem. Proof of the Main Theorem.
Consider the geometric tree graph G ( P ). The vertices x, y, p, q are identified with the stars S ( x ) , S ( y ) , S ( p ) , S ( q ) ∈ G ( P ). By Theorem 3.6, we can identify all23 q -brushes in G ( P ). By Corollary 4.4, we can check for each of them whether it includes both[ p, x ] and [ q, y ] or not. By Proposition 4.5, if none of the pq -brushes contains both [ p, x ] and[ q, y ], then these segments cross, and otherwise, they do not cross. This completes the proof ofthe theorem. K n In this section we consider the abstract (i.e., non-geometric) graph K n . Recall that, as definedin the introduction, the vertices of the tree graph G ( K n ) are all the spanning trees of K n , andtwo spanning trees are adjacent if they differ in exactly two edges. We prove Theorem 1.4,stating that the automorphism group of G ( K n ) is isomorphic to Aut( K n ) = S n .It turns out that the theorem can be proved by roughly the same methodology as theproof of Theorem 1.3, as shown below. Altogether, the proof in the abstract setting turns outconsiderably simpler than its geometric counterpart. Identification of stars in G ( K n ) . Denote G = G ( K n ), and let V ( K n ) = { v , . . . , v n } . Asin the geometric case, our first step is identification of the vertices of G that represent stars.Unlike the geometric case, here the identification is immediate. Claim 5.1.
Let T ∈ V ( G ) . Then T represents a star if and only if for any T ′ ∈ V ( G ) , d G ( T, T ′ ) ≤ n − .Proof. We observe that in the abstract case, d G ( S , S ) = | ∆( S , S ) | for any S , S ∈ V ( G ).(In the geometric case, we could only say that d G ( P ) ( S , S ) ≥ | ∆( S , S ) | .)Assume that T is a star. Since any spanning tree T ′ of K n shares at least one edge with T ,we have d G ( T, T ′ ) = | ∆( T, T ′ ) | ≤ n − T is not a star then it is easy to see that the graph T c = K n \ T is connected, and thus, there exists T ′ ∈ V ( G ) that does not share an edge with T . Hence, d G ( T, T ′ ) = | ∆( T, T ′ ) | = n − G between any pair of stars in n −
2, it follows that thequantity max { T ′ ∈ V ( G ): T ′ = T } d G ( T, T ′ ) equals n − T represents a star and n − center of the graph G .These vertices can be identified with the vertices of K n in an arbitrary way (as any auto-morphism of K n clearly induces an automorphism of G ). So, we call the n vertices in G thatrepresent stars S ( v ) , . . . , S ( v n ) (in some arbitrary order). Valences of vertices in G ( K n ) . The next simple step is identifying, for any T ∈ V ( G ) andany vertex v , what is the valence of v in T . Note that we were not able to obtain such anidentification in the geometric setting. Claim 5.2.
Let T ∈ V ( G ) and v ∈ V ( K n ) . The valence δ T ( v ) of v in T is n − − d G ( T, S ( v )) .Proof. Since all edges in S ( v ) emanate from v , it is clear that | ∆( T, S ( v )) | = n − − δ T ( v ).As d G ( T, S ( v )) = | ∆( T, S ( v )) | , the assertion follows. Max-cliques in G ( K n ) . Our next step is examination of max-cliques in G . As in the geometriccase, we would like to determine whether a given max-clique is a U -clique or an I -clique. Proposition 5.3.
Given a max-clique C of G , we can determine whether it is a U -clique or an I -clique. roof. As the discussion in Section 2.2 is purely combinatorial, it applies without change tothe abstract setting. In particular, all vertices in a U -clique U ( S, T ) are obtained from S ∪ T by removing an edge from its unique cycle, and all vertices in an I -clique I ( S, T ) are obtainedfrom the two-component forest S ∩ T by adding an edge that connects its two components. Asthere are no geometric restrictions in our case, it follows that | U ( S, T ) | is equal to the size of theunique cycle in S ∪ T (and, in particular, is between 3 and n ), and | I ( S, T ) | = k ( n − k ), where k is the number of vertices in one of the connected components of S ∩ T . Hence, determinationwhether C is a U -clique or an I -clique is non-trivial only if |C| = n − U -clique U ( S, T ) is of size n − C of S ∪ T is of size n −
1, which meansthat S ∪ T consists of C plus a single additional edge. Each element of U ( S, T ) is obtainedfrom S ∪ T by removing one edge from C . Assume w.l.o.g. that C = h v , v , . . . , v n − , v i , andthe additional edge is [ v n , v ]. It is clear that v is never a leaf in a tree of U ( S, T ), v n is a leafin all n − U ( S, T ), and each of the vertices v , . . . , v n − is a leaf in exactly two treesof U ( S, T ). In addition, U ( S, T ) has two trees that are paths. (These are the trees obtainedby removing [ v , v ] and [ v n − , v ].) These two trees can be recognized by checking that theirsequence of valences is 1 , , , . . . , , I -clique I ( S, T ) is of size n − S ∩ T , one componentconsists of a single vertex x . Assume, in addition, that I ( S, T ) has two elements that are paths.(Otherwise, we can determine that I ( S, T ) is an I -clique by the previous paragraph.) This ispossible only if the second component of S ∩ T is a path P . In such a case, each endpoint of P is a leaf in n − n −
1) trees of I ( S, T ). As in U ( S, T ), all vertices except one areleaves in at most two trees of U ( S, T ), this property allows to determine that I ( S, T ) is indeedan I -clique. The automorphism group of G ( K n ) . Our last step is to show that the information on G obtained so far is sufficient for determining uniquely the spanning tree represented by eachvertex of G . Namely, given T ∈ V ( G ) and two vertices p, q ∈ V ( K n ), we would like to determinewhether [ p, q ] ∈ E ( T ) or not. If this is possible, it implies that Aut( G ) ∼ = Aut( K n ) ∼ = S n , sinceour determination is unique up to the arbitrary identification of the vertices of G that representstars with the vertices of K n . Hence, this will complete the proof of Theorem 1.4.First, we consider the case when neither p nor q is a leaf in T . Claim 5.4.
Let T ∈ V ( G ) and suppose p, q ∈ V ( K n ) , p = q , δ T ( p ) , δ T ( q ) ≥ . Then [ p, q ] ∈ E ( T ) if and only if T has a neighbor T ′ in G in which the valences of both p and q are smallerby than in T .Proof. If [ p, q ] E ( T ) then no removal of an edge from E ( T ) can reduce the valences of both p and q , and thus, T ′ as described in the claim does not exist. On the other hand, if [ p, q ] ∈ E ( T )then the graph ˜ T = T \ { [ p, q ] } is a two-component forest in which both components are of size ≥
2. Hence, there exists an edge [ p ′ , q ′ ] that connects the two components of ˜ T and uses neither p nor q . The tree T ′ = T \ { [ p, q ] } ∪ { [ p ′ , q ′ ] } is a neighbor of T as described in the claim.Now we can assume w.l.o.g. that p is a leaf in T . We perform a four-step procedure:1. Find a leaf p ′ of T such that d T ( p, p ′ ) > T ′ of T in G ( K n ) such that δ T ′ ( p ) , δ T ′ ( p ′ ) ≥ U -clique U ( T, T ′ ), and find a tree S ∈ U ( T, T ′ ) such that δ S ( p ) = 1 and δ S ( p ′ ) = 2. 25. We find a vertex p ′′ such that δ S ( p ′′ ) = δ T ( p ′′ ) −
1. We claim that if p ′′ = q then[ p, q ] ∈ E ( T ), and otherwise, [ p, q ] E ( T ).We show below that the four steps can indeed be performed, and that they allow to determinewhether [ p, q ] ∈ E ( T ) or not, as claimed. Step 1.
First, we note that if d T ( p, p ′ ) = 2 for all leaves p ′ of T , then T is a star, and thus,[ p, q ] ∈ E ( T ) for the unique q whose valence in T is greater than 1 and [ p, q ] E ( T ) for anyother q . Hence, we may assume that there exists a leaf p ′ such that d T ( p, p ′ ) >
2, and we onlyhave to detect it.Consider the set of leaves of T other than p : A = { p i ∈ V ( K n ) : p i = p, δ T ( p i ) = 1 } .(Note that we can recognize this set, as we are able to determine valences of vertices.) Weclaim that d T ( p, p ′ ) > T ′ of T in G ( K n ) such that δ T ′ ( p ) = δ T ′ ( p ′ ) = 2. This allows to detect the desired p ′ by going over the elements of A , andfor each of them, going over the neighbors of T in G ( K n ) and checking whether the claimedneighbor exists.To see that the claim holds, note that a neighbor T ′ of T satisfies δ T ′ ( p ) = δ T ′ ( p ′ ) = 2, ifand only if it is of the form T ′ = T \ ∪{ [ p, p ′ ] } \ { e } , for an edge e that belongs to the uniquecycle C of T ∪ { [ p, p ′ ] } and uses neither p nor p ′ . If d ( p, p ′ ) = 2, then C is of length 3, andthus, it has no edges that use neither p nor p ′ . Thus, no such neighbor T ′ exists. If d ( p, p ′ ) > C is of length >
3, and thus, it includes an edge e that uses neither p nor p ′ . The tree T ′ = T \ { e } ∪ { [ p, p ′ ] } is the desired neighbor of T . Step 2.
This step is immediate, as the required neighbor T ′ was already found in Step 1. Step 3.
The required neighbor S is the tree obtained from T ∪ { [ p, p ′ ] } by removing theunique edge of the cycle C that uses p but not p ′ (call it [ p, p ′′ ]). The U -clique U ( T, T ′ ) can berecognized using Proposition 5.3, since there exist only two max-cliques of G ( K n ) that includeboth T and T ′ – a U -clique and an I -clique – and Proposition 5.3 allows us to determine, whichof them is the U -clique. Then, S can be recognized as the unique element of U ( T, T ′ ) in whichthe valences of p, p ′ are 1 and 2, respectively. Step 4.
It is clear that the unique vertex whose valence in S is smaller by one than its valencein T is p ′′ , as defined in Step 3. By the construction of C , [ p, p ′′ ] is the unique edge of E ( T )that emanates from p , i.e., p ′′ is the unique neighbor of p in T . Hence, [ p, q ] ∈ E ( T ) if and onlyif q = p ′′ , as asserted. The vertex p ′′ is detected by comparing the valences of the vertices in S with their respective valences in T .This completes the proof of Theorem 1.4. References [1] D. Avis and K. Fukuda, Reverse Search for Enumeration, Discrete Applied Mathematics , pp. 21–46, 1996.[2] J. A. Bondy and R. L. Hemminger, Graph reconstruction – a survey, J. Graph Theory l (1977), pp. 227-268.[3] R. L. Cummins, Hamilton circuits in tree graphs, IEEE Trans. Circuit Th., (1966),pp. 82–90.[4] P. Erd˝os and G. Szekeres, A Combinatorial Problem in Geometry, Compositio Math. ,pp. 463-470, 1935. 265] M. C. Hernando, F. Hurtado, A. M´arquez, M. Mora and M. Noy, Geometric Tree Graphsof Points in Convex Position, Discrete Applied Mathematics (1972), pp. 187–193.[8] P. J. Kelly, A congruence theorem for trees, Pacific J. Math. (1957), pp. 961-968.[9] G. Liu, On connectivities of tree graphs, J. Graph Theory (1988), pp. 453–459.[10] S. Ramachandran, Graph reconstruction – some new developments, AKCE J. Graphs.Combin., (2004), pp. 51–61.[11] J. Sedl´aˇcek, The reconstruction of a connected graph from its spanning trees, Mat. ˇCasopisSloven. Akad. Vied.24