The diameter of the ridge-graph of a cyclic polytope

Discrete Math. Appl., Vol. 19, No. 1, pp. 47–53 (2009) © de Gruyter 2009

DOI 10.1515/DMA.2009.003

The diameter of the ridge-graph of a cyclic polytope A. M. MAKSIMENKO

Abstract — It is shown that the diameter of the the ridge-graph, that is, the graph of the polytope C .d; n/ dual to the given polytope, where d is the dimension and n is the number of facets of the polytope, is equal to n d maxf0; d.n 2d /=.bd=2c C 1/eg:

Following the generally accepted terminology [1, 2, 10], we call a d -dimensional polytope a d -polytope and its k-dimensional faces, k-faces. Similarly, 0-faces are vertices, 1-faces are edges, and .d 1/-faces of a d -polytope are called facets (or hyperfaces), and .d 2/faces are called subfacets or ridges. Two vertices of a polytope are called adjacent if they are connected by an edge of the polytope. The graph of a polytope is the sets of vertices and edges of the polytope. The set of vertices of the ridge-graph of a polytope coincides with the set of facets, and two facets are adjacent if they have a common ridge. Thus, the ridge-graph of a polytope is, in fact, the graph of the polytope dual to the initial polytope [1, 2]. A path in a graph is a sequence of vertices .v0 ; v1 ; : : : ; vl / where each two neighbouring vertices vi 1 and vi , i 2 f1; 2; : : : ; lg, are adjacent, and the number of vertices l is the length of the path. The distance between two vertices in a graph is the length of the shortest path containing these vertices. The diameter of a graph is the maximal distance between two vertices of the graph. We denote by .d; n/ the maximal diameter of the graphs of polytopes in the class of all d -polytopes with n facets. In 1957, V. Hirsh (see [5]) proposed the hypothesis that .d; n/ n d (more precisely, the hypothesis concerned all polytopes). In the general case, the validity of this hypothesis is still an open question. The existing upper bounds for .d; n/ are even not polynomial [12]. It is known [7, 11] that for n > d 8 the inequality .d; n/ n d is true. It is also known [2, 10] that it is sufficient to prove the hypothesis for simple polytopes (that is, for such d -polytopes each vertex of which belongs exactly to d facets). In addition, there are several results related to calculation of the diameters of polytopes of some special forms [2, 12]. One of the important classes of polytopes is the class of cyclic [4, 8] and dual to them polytopes. A cyclic polytope C.d; n/ is a polytope with n vertices x i D x.i /, i 2 f1; 2; : : : ; ng, which are located on a curve x./ D .; 2 ; : : : ; d / in the space Rd . These polytopes Originally published in Diskretnaya Matematika (2009) 21, No. 1 (in Russian). Received November 30, 2007. Revised August 8, 2008.

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A. M. Maksimenko

stand out by a series of their properties [2]. They are simplicial, bd=2c-adjacent, and have the maximal number of faces of all dimensionalities in the class of d -polytopes with a fixed number of vertices [17]. The polytopes C .d; n/ dual to the cyclic polytopes are simple and have the maximal number of vertices in the class of d -polynomials with the fixed number n of facets. These facts raised hopes that, among other things, the polytopes C .d; n/ give an example of polytopes with maximal diameter of the graph. Unfortunately, even for d D 4 and n D 9 their diameter is equal to 4, whereas in [15] an example of a 4-polytope with the same number of facets and the diameter of the graph equal to 5 is given (later it was shown [3] that for d D 4 and n D 9 this is a unique, up to combinatorial equivalency, example of a polytope with diameter 5). We introduce the notation c .d; n/ D diam C .d; n/: In 1964, V. Klee showed [13] that the Hirsh hypothesis is true for C .d; n/: c .d; n/ n

d;

(1)

and (1) becomes equality for d < n 2d . In that paper, the hypothesis was formulated that for n > 2d the equality c .d; n/ D bn=2c is true (slightly later it was shown [15] that this is false). But due to inaccessibility of paper [13] these results are absolutely unknown, so that some 30 years after their appearance attempts to find the value c .d; n/ for n D 2d with the use of computer calculations were undertook [6], and in paper [16] recalling fact (1) the author gave reference to paper [14] where there is no mention of this result. In order to close this problem, we give below the exact value of c .d; n/. Theorem 1. The diameter c .d; n/ of the ridge-graph of the cyclic polytope C.d; n/ is calculated by the formula ( n d for d < n 2d; c .d; n/ D n d d.n 2d /=.bd=2c C 1/e for n > 2d: Proof. Validity of the theorem for d < n 2d was proved in [13], therefore in what follows we assume that n > 2d: We denote the set of vertices of the polytope C.d; n/ by X D fx 1 ; x 2 ; : : : ; x n g, supposing that the vertices are numbered in order of increasing parameter , that is, 1 < 2 < : : : < n . The following Gale criterion [9] is valid. All faces of the polytope C.d; n/ are simplexes, and the subset Y X consisting of d vertices determines some facet if and only if for any two x i ; x j 2 X n Y , i < j , the number of vertices x k 2 Y , for which i < k < j , is even. In particular, it follows from simpliciality of a cyclic polytope that two its facets are adjacent if they have exactly d 1 common vertices. Let us prove the theorem for the case of even dimensionality. Let d D 2k, show that c .d; n/ n

d

.n

2d /=.k C 1/;

(2)

and give an example of two facets of a cyclic polytope such that the distance between these facets is equal to the integer part of the right-hand side of (2). Brought to you by | Bibliotheek TU Delft (Bibliotheek TU Delft) Authenticated | 172.16.1.226 Download Date | 6/20/12 11:14 AM

49

The diameter of the ridge-graph

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17 16

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13 14

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Figure 1. Two adjacent facets of the polytope C.10; 18/

It follows from the Gale criterion that the set of vertices of each facet can be always partitioned uniquely into k disjoint pairs of the form fx i ; x i C1 g; where i 2 f1; 2; : : : ; ng and the addition in i C 1 is the addition modulo n (if i D n, then i C 1 D 1). Conversely, any subset Y X of the form Y D fx i1 ; x i1 C1 ; x i2 ; x i2 C1 ; : : : ; x ik ; x ik C1 g is the set of some facet. For the sake of clearness of the further representation we put in correspondence to each vertex x i the point vi on a circle of unit radius: vi D .cos.2 i=n/; sin.2 i=n/: Two points vi and vi C1 is called a pair and is denoted by pi , the set of all such pairs for i 2 f1; 2; : : : ; ng is denoted by P . Suppose that the vertices of some facet are represented by a set of disjoint pairs. Then the transition to the adjacent facet is a shift of one or several adjacent to each other pairs by one point along the circle (see Fig. 1). In other words, in order to go to any adjacent facet it is sufficient to choose one of k pairs and the shift direction (clockwise or anticlockwise), in this case, if on the path of the pair there is another pair, then it is shifted in the same direction as the first pair and so on. Further we use the following rule of shifting. During the shifting the pair can simultaneously move several adjacent neighbours locating on its path, but cannot have an intersection with another pair. Thus, the problem of finding the distances between facets in a ridge-graph of a cyclic polytope reduces to the following problem. On the circle with n points, k pairs of points are chosen, they determine the vertices of the first facet, further they will be referred to as pieces, and k pairs are marked, they correspond to the vertices of the second facet, further they will be referred to as cells (see Fig. 2). It is required to find the minimum number of shifts of pieces sufficient to fill all cells by particles. We denote this number by l.F1 ; F2 /, here F1 is the set of pieces and F2 is the set of cells, F1 ; F2 P . To each pair of the set P n .F1 [ F2 / we put in correspondence an arc of the circle between the points vi and vi C1 which does not contain these points. This arc will be denoted by ai and referred to as an interval (see Fig. 2). We denote the set of all intervals by A. It is obvious that jAj > 0. A subset W A is called a lacuna if there exist i; j 2 f1; 2; : : : ; ng such that W D fai ; ai C1 ; : : : ; aj g and ai 1 and aj C1 do not belong to A. Here, as before, the sums of indices are calculated modulo n. It is clear that A is uniquely representable in Brought to you by | Bibliotheek TU Delft (Bibliotheek TU Delft) Authenticated | 172.16.1.226 Download Date | 6/20/12 11:14 AM

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A. M. Maksimenko

6

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1 18

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Figure 2. The example of facets F1 D f1; 2; 3; 5; 6; 11; 12; 16; 17; 18g (pieces) and F2 D f3; 4; 5; 6; 8; 9; 10; 11; 12; 13g (cells) of the polytope C.10; 18/

the form A D W1 [ W2 [ : : : [ Ws ;

0 < s d;

(3)

where Wi , i D 1; : : : ; s, are lacunas. We define the length l.W / of the lacuna W as the number of points vi , 1 i n, contained in W , that is, l.W / D jW j 1: Then the total length of lacunas is equal to the number of vertices which do not belong to the chosen facets and s X

l.Wi / n

2d:

i D1

Lemma 1. Suppose that a lacuna W is removed from the circle together with all points contained in it and all pieces of F1 cannot move through this lacuna. Then l.F1 ; F2 / n

d

l.W /:

Proof. For each right shift exactly one point becomes free from pieces and exactly one point becomes occupied. Note that if the pieces move in an optimal way, then the point which became non-occupied cannot be occupied again. In order to prove this, we assume the contrary. Then two following variants are possible. (1) The piece which freed a point occupied it again. Then non-optimality of the shift is obvious. (2) The piece occupied a point was shifted in the same direction as the piece freed this point. Then these two shifts (freeing and occupying of this point) could be realised by one step. Thus, the number of shifts cannot be greater than the number of all points not occupied by pieces. It follows directly from Lemma 1 that l.F1 ; F2 / n

d

max l.Wi /:

1i s


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The diameter of the ridge-graph

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S1

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Figure 3. Three cuts for Fig. 2: l.S1 / D 1, l.S2 / D 0, l.S3 / D 2

In particular, with the use of this inequality it is easy to prove the theorem for d < n 2d . It is sufficient to give an example of two facets which have 2d n common vertices. It is clear that the number of shifts approaching these facets is equal to n d . By the part of the circle between the lacunas Wi and Wj is meant the part which will be gone in the moving along the circle in the anticlockwise direction from Wi to Wj . We introduce an auxiliary function f .i; j / equal to the number of pieces located between Wi and Wj without of the number of cells on this part of the circle. It is easy to see that this function possesses the following properties. Property 1. The equality f .i; j / D

f .j; i / is true.

Property 2. The equality f .i; j / D f .i; m/ C f .m; j / is true. Property 3 (continuity). If f .i; j / > 1, then between Wi and Wj there exists Wm such that f .i; m/ D 1. If f .i; j / D 0 for some i and j , then the lacunas Wi and Wj are called adjacent. Property 4 (transitivity). If Wi and Wj are adjacent and Wj and Wm are adjacent, then Wi and Wm are also adjacent. Let Wi and Wj be adjacent. Then removing them from the circle and applying Lemma 1 for each of the appearing arcs, we obtain the inequality l.F1 ; F2 / n

d

l.Wi /

l.Wj /:

We can treat similarly the situation with several pairwise adjacent lacunas. Using this possibility, we partition the set of all lacunas into subsets Si , 1 i t s, in such a way that all lacunas belonging to one subset are pairwise adjacent and any two lacunas of different subsets are not adjacent. The subsets Si will be referred to as cuts (see Fig. 3). By the length l.S / of a cut S is meant the total lengths of lacunas contained in it. Then, obviously, l.F1 ; F2 / n d max l.Si /; 1i t

where t is the number of cuts. Since t X

j D1

l.Sj / D

s X

l.Wi / n

2d;

i D1


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C.6; 19/

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C.4; 18/

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C.6; 21/ 14

15 16 17

1 21 20 19 18

Figure 4. Examples of diametrically opposite facets.

we see that the inequality l.F1 ; F2 / n

d

n

2d t

is true. It remains to prove that t k C 1;

(4)

where k D d=2. According to (3), the number of lacunas obeys the inequality s d . Therefore, in order to prove inequality (4), it is sufficient to show that among all cuts there are at most two cuts containing exactly one lacuna each. Let a cut S 0 contain exactly one lacuna Wi 0 . Hence it follows that for any j ¤ i 0 f .i 0 ; j / ¤ 0:

(5)

Then by property 3 (continuity) f .i 0 ; j /f .i 0 ; m/ > 0 for any m ¤ i 0 that is, the function g 0 .j / D f .i 0 ; j / takes the values of the same sign for all j . Consider one more cut S 00 containing only one lacuna Wi 00 , i 00 ¤ i 0 . Let us show that for any j ¤ i 0 and m ¤ i 00 f .i 0 ; j /f .i 00 ; m/ < 0: Suppose the contrary, that is, suppose that for some j and m f .i 0 ; j /f .i 00 ; m/ > 0; the equality here is impossible according to (5) and the similar assertion for i 00 . Then this inequality is true for j D i 00 and m D i 0 , but this is impossibly by virtue of property 1. Thus, among all cuts there are at most two cuts containing exactly one lacuna each, so inequality (4), and together with it inequality (2), are proved. Fig. 4 contains an example of facets the distance between which is the maximal integer not exceeding the right-hand side of (2). In [15] there is an example of such kind for d D 6 and n D 23. The algorithm of construction of such example for arbitrary d and n is seen from the figure. Thus, in the case of even dimensionality the theorem is proved. Consider the case d D 2k C 1. It follows from the Gale criteria that in the case of odd dimensionality each facet has to contain at least one of the points v1 or vn . Let F1 and F2 be some facets. Two cases are possible. In the first case, F1 and F2 contain the same point of the set fv1 ; vn g. Then this point can be removed from consideration, and therefore we find ourselves in the case of an even dimensionality. Thus, l.F1 ; F2 / .n

1/

.d

1/

.n

1/ 2.d kC1

1/