Nonexistence of a Weakly Neighbourly Polyhedral
Map of Type {6, 6}
Nandini Nilakantan
CONTENTS
1. Introduction
2. Proofs
Acknowledgments
References
For the existence of an n-vertex polyhedral map of type {p, p},
it is known that n must be ≥ (p − 1)2 and equality holds if and
only if K is weakly neighbourly. We have seen in [Brehm et al.
02] that there is a unique polyhedral map of type {5, 5} on 16
vertices. In [Brehm 90], Brehm constructed a polyhedral map of
type {6, 6} with 26 vertices. In this article, we prove that there
do not exist any polyhedral maps of type {6, 6} on 25 vertices.
As a consequence, we show that the minimum number of edges
in polyhedral maps of Euler characteristic -25 is > 75.
1.
2000 AMS Subject Classification: Primary 52B70, 51M20, 57M20
Keywords: Polyhedral maps, polyhedral 2-manifolds,
regular graph design
INTRODUCTION
A finite collection K of cycles, edges, and vertices of a
complete graph is called a polyhedral complex (of dimension 2) if (i) each edge of a cycle in K is in K, (ii) each
vertex of each edge in K is in K, and (iii) any two cycles
have at most one common edge. If u and v are vertices of
a face F and uv is not an edge of F , then uv is called a diagonal. We denote a face u1 · · · um u1 either by u1 · · · um
or (u1 , . . . , um ) and a diagonal (or edge) uv by (u, v).
The vertex-set of a complex K is denoted by V (K).
For a complex K, the edge graph EG(K) of K is the
graph whose vertices and edges are the vertices and edges
of K, respectively.
A polyhedral complex K is called a polyhedral 2manifold if EG(K) is connected and for each vertex v of K, the faces containing v are of the form
(v, u1 , . . . , u2 ), . . . , (v, um−1 , . . . , um ), (v, um , . . . , u1 ) for
some m ≥ 3.
A polyhedral 2-manifold K defines a decomposition
of a 2-manifold M into abstract polygonal discs whose
interiors are open discs which are pairwise distinct. Such
a 2-manifold M is unique up to homeomorphism and is
called the geometric carrier of K.
A polyhedral 2-manifold is called a polyhedral map if
the intersection of any two faces is empty, a vertex, or
c A K Peters, Ltd.
1058-6458/2003 $ 0.50 per page
Experimental Mathematics 12:3, page 257
258
Experimental Mathematics, Vol. 12 (2003), No. 3
an edge. Hence, a diagonal in a polyhedral map is the
diagonal of a unique face and is not an edge.
If f0 (K), f1 (K), and f2 (K) are the number of vertices,
edges, and faces of a complex K, then χ(K) := f0 (K) −
f1 (K) + f2 (K) is called the Euler characteristic of K.
If the number of diagonals in a polyhedral map K on
n
nvertices is d(K), then it is clear that d(K) + f1 (K) ≤
2 . A polyhedral map is called a weakly neighbourly
polyhedral map (in short, wnp map) if any pair of vertices
is in a face. So, an n-vertex polyhedral map K is weakly
neighbourly if and only if d(K) + f1 (K) = n2 .
A polyhedral 2-manifold K is said to be a polyhedral
2-manifold of type {p, q} or a {p, q} equivelar polyhedral
2-manifold, if each face is a p-cycle and the degree of each
vertex is q (see [Brehm and Schulte 97, Brehm and Wills
93, Datta and Nilakantan 01]).
If K is a polyhedral map of type {p, q} on n vertices,
then d(K) = nq(p−3)/2 and f1 (K) = nq/2. Here, we are
interested in those cases where p = q. For a polyhedral
map K of type {p, p}, we see that np(p − 3)/2 + np/2 ≤
n(n − 1)/2 and therefore n ≥ (p − 1)2 . Thus, if K is a
wnp map of type {p, p}, then n = (p − 1)2 . Clearly, the
4-vertex 2-sphere is the unique wnp map of type {3, 3}.
In [Brehm 90], Brehm proved that there exist exactly
three wnp maps of type {4, 4} (the geometric carrier of
two of these is the torus and that of the third is the Klein
bottle). In [Brehm 90], Brehm constructed the following :
Example 1.1. Two maps :
F
= {(i, i + 1, i + 2, i + 5, i + 8) : 1 ≤ i ≤ 16}.
(Additions are modulo 16.)
S
= {(i, i + 1, i + 2, i + 5, i + 8, i + 17) : 1 ≤ i ≤ 26}.
(Additions are modulo 26.)
F is a polyhedral map of type {5, 5} and S is a polyhedral
map of type {6, 6}. F is weakly neighbourly, but S is not
(e.g., vertices 1 and 14 are not in any common face). The
geometric carriers of F and S are the orientable surface
of Euler characteristic −8 and the nonorientable surface
of Euler characteristic −26, respectively.
For the existence of an n-vertex polyhedral map M of
type {6, 6}, it is obvious that n is ≥ 7. Here, we present
such a polyhedral 2-manifold with 12 vertices.
Example 1.2. A polyhedral 2-manifold of type {6, 6} with
12 vertices :
P
= {ui ui+1 ui+2 ui+4 ui+6 ui+9 : 1 ≤ i ≤ 12}.
(Additions are modulo 12.)
Clearly, the number of edges in P is 36 and hence χ(P ) =
−12. It is not difficult to see that the geometric carrier
of P is a nonorientable surface.
In [Brehm et al. 02], we have shown that F is the
unique 16-vertex polyhedral map of type {5, 5}. Here we
prove:
Theorem 1.3. There does not exist any polyhedral map
of type {6, 6} on 25 vertices.
For χ ≤ 1, let E(χ) be the smallest number E for
which there exists a polyhedral map of Euler characteristic χ with E edges. From a result (Proposition 2 in
[Brehm 90]), it is seen that if M is a polyhedral map of
Euler characteristic −25, then the number of edges of M
is ≥ 75. So, E(−25) ≥ 75. Here we prove :
Corollary 1.4. E(−25) > 75.
Remark 1.5. For an n-vertex polyhedral map K of type
{p, p}, the following are equivalent: (i) K is weakly neighbourly, (ii) n = (p−1)2 , and (iii) χ(K) = (p−1)2 (4−p)/2.
So, we can replace the assumption “weakly neighbourly”
by “25-vertex” or by “of Euler characteristic −25” in the
theorem.
Remark 1.6. From the theorem, it follows that for the
existence of an n-vertex polyhedral map of type {6, 6}, n
is ≥ 26, where as a polyhedral 2-manifold of type {6, 6}
exists on 12 vertices (namely, P in Example 1.2).
Remark 1.7. In design theory, a regular graph design
(RGD) with parameters b, v, r, k, λ1 , λ2 (where λ1 − λ2 =
±1) is an incidence system with v points and b blocks in
which each point is incident with r blocks, each block is
incident with k points, and any two distinct points are
together incident with λ1 or λ2 blocks. The point graph
of an RGD is the graph whose vertices are the points
of the RGD and two vertices are adjacent if and only if
the corresponding points are together incident with λ2
blocks.
Now, given a wnp map of type {p, q}, one can associate
an RGD with parameters r = q, k = p, λ1 = 1, λ2 = 2 by
simply declaring the vertices of the map to be the points
and the vertices lying on a face to form a block. The
RGD thus obtained has the following special property:
(P) The induced subgraph of the point graph on each
block is a cycle.
Nilakantan: Nonexistence of a Weakly Neighbourly Polyhedral Map of Type {6, 6}
Conversely, given an RGD with r = q, k = p, λ1 =
1, λ2 = 2 satisfying (P), one can uniquely recover the wnp
map of type {p, q}. However, as far as we know, RGDs
with property (P) have not been studied in the design
theory literature.
2.
PROOFS
In this section, we give proofs of the results. We first
state two propositions proved by Brehm in [Brehm 90].
We need these two propositions to prove the corollary.
Proposition
2.1. If M is a polyhedral map, then f1 (M ) ≤
√
Y ( 2Y + 2)/8, where Y = f0 (M ) + f2 (M ). Equality
holds if and only if M is a wnp map of type {k, k} with
f0 (M ) = f2 (M ) = (k − 1)2 for some k.
Proposition 2.2. If M is a polyhedral map with Euler
characteristic χ, then
√ f1 (M ) ≥ G(χ) − χ, where G(χ) :=
min{m ∈ N : m( 2m − 6) ≥ −8χ and m ≥ 8}.
Proof of Theorem 1.3: If possible, let there exist a 25vertex polyhedral map M of type {6, 6}. In this case,
f2 (M ) = f0 (M ) = 25 and M is weakly neighbourly.
Thus any two faces of M have a common vertex. Let the
vertices of M be 1, . . . , 25. Without loss of generality, we
can assume that the faces containing the vertex 25 are
(25, 24, 23, 22, 21, 20), (20, 1, 2, 3, 4, 25), (25, 4, 5, 6, 7, 8),
(25, 8, 9, 10, 11, 12), (25, 12, 13, 14, 15, 16), and (25, 16, 17,
18, 19, 24). The remaining 19 faces are of the form (24,
19, j1 , k1 , j2 , i1 ), (24, i1 , j3 , k2 , j4 , i2 ), (24, i2 , j5 , k3 , j6 , i3 ),
(24, i3 , j7 , j8 , i4 , 23), (23, i4 , j9 , k4 , j10 , i5 ), (23, i5 , j11 , k5 ,
j12 , i6 ), (23, i6 , j13 , k6 , j14 , i7 ), (23, i7 , j15 , j16 , i8 , 22), (22,
i8 , j17 , k7 , j18 , i9 ), (22, i9 , j19 , k8 , j20 , i10 ), (22, i10 , j21 , k9 ,
j22 , i11 ), (22, i11 , j23 , j24 , i12 , 21), (21, i12 , j25 , k10 , j26 ,
i13 ), (21, i13 , j27 , k11 , j28 , i14 ), (21, i14 , j29 , k12 , j30 , i15 ),
(21, i15 , j31 , j32 , i16 , 20), (20, i16 , j33 , k13 , j34 , i17 ), (20, i17 ,
j35 , k14 , j36 , i18 ), and (20, i18 , j37 , k15 , j38 , 1) where i1 , . . . ,
i18 , j1 , . . . , j38 , k1 , . . . , k15 ∈ {1, . . . , 19}.
The unknown vertices i1 , . . . , i18 denote the vertices
in level 1, j1 , . . . , j38 denote the vertices in level 2, and
k1 , . . . , k15 denote the vertices in level 3 (see Figure 1).
These vertices have to be suitably identified in order to
arrive at a feasible solution. By adopting a method similar to that in [Brehm et al. 02], it will be virtually
impossible to arrive at a solution. Hence, a computer
program had to be developed. The number of combinations of the vertices to be considered was immense and
demanded very high computer storage and time.
259
The following algorithm was adopted and was implemented in C++. The unknown vertices in level
1 are grouped as the blocks {i1 , i2 , i3 }, {i4 , i5 , i6 , i7 },
{i8 , i9 , i10 , i11 }, {i12 , i13 , i14 , i15 }, {i16 , i17 , i18 } corresponding to those vertices in the links of the vertices
24, 23, 22, 21, and 20, respectively. Clearly, each of the
elements in each block has to be different. For each of
the above sets, the possible combinations of the vertices
are evaluated by checking on the initial condition matrix.
This matrix is a 25 × 25 symmetric matrix. The ij-th entry is 1, if the vertices i and j are joined by an edge, -1,
if ij is a diagonal, and 0 if it is neither. Clearly, all the
diagonal elements of the matrix are 0 and the maximum
entry is 2 (as each edge can belong to exactly two triangles). With the available data, the elements of the initial
condition matrix T are as given below.
T1,2 = T1,20 = 1, T1,3 = T1,4 = T1,25 = −1,
T2,1 = T2,3 = 1, T2,4 = T2,20 = T2,25 = −1,
T3,2 = T3,4 = 1, T3,1 = T3,20 = T3,25 = −1,
T4,25 = 2, T4,3 = T4,5 = 1, T4,1 = T4,2 = T4,6 = T4,7 =
T4,8 = T4,20 = −1,
T5,4 = T5,6 = 1, T5,7 = T5,8 = T5,25 = −1,
T6,5 = T6,7 = 1, T6,4 = T6,8 = T6,25 = −1,
T7,6 = T7,8 = 1, T7,4 = T7,5 = T7,25 = −1,
T8,25 = 2, T8,7 = T8,9 = 1, T8,4 = T8,5 = T8,6 = T8,10 =
T8,11 = T8,12 = −1,
T9,8 = T9,10 = 1, T9,11 = T9,12 = T1,25 = −1,
T10,9 = T10,11 = 1, T10,8 = T10,12 = T10,25 = −1,
T11,10 = T11,12 = 1, T11,8 = T11,9 = T11,25 = −1,
T12,25 = 2, T12,11 = T12,13 = 1, T12,8 = T12,9 = T12,10 =
T12,14 = T12,15 = T12,16 = −1,
T13,12 = T13,14 = 1, T13,15 = T13,16 = T13,25 = −1,
T14,13 = T14,15 = 1, T14,12 = T14,16 = T14,25 = −1,
T15,14 = T15,16 = 1, T15,12 = T15,13 = T15,25 = −1,
T16,25 = 2, T16,15 = T16,17 = 1, T16,12 = T16,13 = T16,14 =
T16,18 = T16,19 = T16,24 = −1,
T17,16 = T17,18 = 1, T17,19 = T17,24 = T17,25 = −1,
T18,17 = T18,19 = 1, T18,16 = T18,24 = T18,25 = −1,
T19,18 = T19,24 = 1, T19,16 = T19,17 = T19,25 = −1,
T20,25 = 2, T20,1 = T20,21 = 1, T20,2 = T20,3 = T20,4 =
T20,22 = T20,23 = T20,24 = −1,
T21,20 = T21,22 = 1, T21,23 = T21,24 = T21,25 = −1,
T22,21 = T22,23 = 1, T22,20 = T22,24 = T22,25 = −1,
T23,22 = T23,24 = 1, T23,20 = T23,21 = T23,25 = −1,
T24,25 = 2, T24,19 = T24,23 = 1, T24,16 = T24,17 = T24,18 =
T24,20 = T24,21 = T24,22 = −1,
T25,4 = T25,8 = T25,12 = T25,16 = T25,20 = T25,24 = 2,
T25,1 = T25,2 = T25,3 = T25,5 = T25,6 = T25,7 = T26,9 =
T25,10 = T25,11 = T25,13 = T25,14 = T25,15 = T25,17 =
T25,18 = T26,19 = T25,21 = T25,21 = T25,22 = T25,23 = −1.
260
Experimental Mathematics, Vol. 12 (2003), No. 3
k
3
k
2
j
4
k
1
j
3
j
7
j
2
j
1
18
14
j
5
k
4
j
6
i
1
19
17
15
j
8
i
2 i
3
24
13
k
5
j
k
j
9 j
6
10 j11 12
j
13 j
i
14
i
5
i
j
4
6
15
i
7
j
23
16
16
i
8
12
11
10
25
i
9
i
10
22
8
9
i
11
7
4
6
j
17
5
3
i
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1
2
j
38
k
15
i
18
i
17
i
16
j
18
j
19 k8
j
j 20
21
j
22
j
23
21
20
k
7
k
9
j
i
24
j
13
25
i
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j
26
j
k
j
28 27
10
j
j
29
k
30
11
k
12
i
15
j
37
j
j
35 34 j
j
33
36
j
32
j
31
k
13
k
14
k
1
k
2
Level 3
j
3
j
2
j
4
j
1
Level 2
i
2
i
1
i
3
Level 1
FIGURE 1. The first three levels of a possible polyhedral map in 25 vertices.
All the other entries of the initial condition matrix are
0. For a feasible solution to exist, in each of the 25 rows,
there should be exactly 6 entries which are 2 and the
remaining 18 entries should be -1 (all the diagonal entries
of the matrix are 0). At each stage in the program, the
matrix is checked for the symmetry condition and for the
Nilakantan: Nonexistence of a Weakly Neighbourly Polyhedral Map of Type {6, 6}
existence of entries which are different from 0, 1, 2, and
-1.
From the above initial condition matrix, some of the
pertinent data necessary for the vertex numbering are as
follows:
Number of zeros in rows 1 to 25 are 19, 19, 19, 15,
19, 19, 19, 15, 19, 19, 19, 15, 19, 19, 19, 15, 19, 19, 19,
15, 19, 19, 19, 15, and 0, respectively and the number of
ones in rows 1 to 25 are 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, and 0, respectively.
The total number of combinations possible for
the sets {i1 , i2 , i3 }, {i4 , i5 , i6 , i7 }, {i8 , i9 , i10 , i11 },
{i12 , i13 , i14 , i15 }, and {i16 , i17 , i18 } are 1314, 39776,
39776, 39776, and 1314, respectively. This is further
reduced to 1014, 22776, 22776, 22776, and 1014 combinations by eliminating all those combinations where the
adjacent sets do not satisfy the condition matrix. Thus,
1.0866 × 1020 cases have been reduced to 1.2148 × 1019
cases.
The further elimination is done as follows:
Step 1: Consider the first element in the first block.
Step 2: If i1 i2 = i2 i3 = 0, then proceed further. Otherwise, go to Step 20.
Step 3: Consider the first element of the second block.
Step 4: : If i3 i4 = i4 i5 = i5 i6 = i6 i7 = 0 and i4 = i1 ,
i2 , i3 and i3 = i5 , i6 , i7 , then proceed. Otherwise,
go to Step 19.
261
Step 11: Check for the conditions to be satisfied in the
condition matrix for the different combinations of
vertices in the first two blocks. If satisfied, then
proceed further. Otherwise, go to Step 17.
Step 12: Consider the first element of the fifth block
/
Step 13: If i15 i16 = i16 i17 = i17 i18 = 0 and i16 ∈
{i11 , i12 , i13 , i14 , i18 }, i17 = i15 , then proceed. Otherwise, go to Step 16.
Step 14: Check for the conditions to be satisfied in the
condition matrix for the different combinations of
vertices in the first two blocks. If satisfied, then
proceed further. Otherwise, go to Step 16.
Step 15: If a possible solution exists in level 1 with vertices i1 , i2 , i3 , i4 , i5 , i6 , i7 , i8 , i9 , i10 , i11 , i12 , i13 ,
i14 , i15 , i16 , i17 , i18 , then the selection of vertices
in the levels 2 and 3 are made in a very simplified
way. This is performed by evaluating for a particular
vertex in level 1 the following:
(i) the number of −1s in level 1 and in the adjacent levels for which this particular vertex is a
common vertex.
(ii) the number of 1s required for each vertex. If
the sum of these is less than the total number
of 0s and 1s available for all the vertices in level
1, only then will a possible solution exist. Otherwise, the set arrived at is rejected.
Step 5: Check for the conditions to be satisfied in the
condition matrix for the different combinations of
vertices in the first two blocks. If satisfied, then
proceed further. Otherwise, go to Step 19.
As an example, consider the set
Step 6: Consider the first element of the third block.
This corresponds to the 1st element in block
1, 5555th element in block 2, 3121th element
in block 3, 1746th element in block 4, and
294th element in block 5 of level 1.
The
set {2, 5, 3, 6, 2, 7, 9, 4, 10, 2, 8, 3, 7, 10, 5, 9, 6, 10} corresponds to the vertices in level 1. In this set, the
link of vertex 2 requires 6 -1s from the adjacent elements in level 1 (i.e., there have to be 6 vertices from
level 1, each of which is a diagonal with 2), 12 -1s
from the adjacent elements in the other levels, 6 additional 1s and 6 vertices from the interior vertices
20, 21, 22, 23, 24, 25. This totals up to 30. This
is impossible as only 19 0 and 2 1s are available as
seen from the condition matrix. Hence, this set is
rejected. If it is not, then a possible solution exists.
Step 7: If i8 i9 = i9 i10 = i10 i11 = i11 i12 = 0 and
/ {i3 , i4 , i5 , i6 , i7 }, i7 ∈
/ {i9 , i10 , i11 }, then proi8 ∈
ceed. Otherwise, go to Step 18.
Step 8: Check for the conditions to be satisfied in the
condition matrix for the different combinations of
vertices in the first two blocks. If satisfied, then
proceed further. Otherwise, go to Step 18.
Step 9: Consider the first element of the fourth block.
Step 10: If i11 i12 = i12 i13 = i13 i14 = i14 i15 = 0 and
/ {i7 , i8 , i9 , i10 , i11 }, i11 ∈
/ {i13 , i14 , i15 }, then
i12 ∈
proceed. Otherwise, go to Step 17.
rej 0 5554 3120 1745 293 -1 * 2 5 3 6 2 7 9 4 10 2 8
3 7 10 5 9 6 10 *1 1 -1 -1 * 2 6 12 6 6 30 19 2 21 *
21 2 30
262
Experimental Mathematics, Vol. 12 (2003), No. 3
This procedure is carried out for all the combinations
in each block of level 1. Proceed further.
Step 16: Consider the next set of {i16 , i17 , i18 } in the
fifth block and go to Step 13.
Step 17: Consider the next set of {i12 , i13 , i14 , i15 } in
the fourth block and go to Step 10.
Step 18: Consider the next set of {i8 , i9 , i10 , i11 } in the
third block and go to Step 7.
ACKNOWLEDGMENTS
I thank Basudeb Datta for suggesting the problem and also
for many useful discussions. I thank my father and also Hariharan for helping me a lot with the programming. I thank
Bhaskar Bagchi for explaining the relation between equivelar
wnp maps and regular graph designs. Remark 1.7 is due to
him. I thank the anonymous referee for many useful comments which have led to substantial improvements in the presentation of the paper. I am very grateful to the National
Board for Higher Mathematics, DAE, India for providing me
with financial support.
Step 19: Consider the next set of {i4 , i5 , i6 , i7 } in the
second block and go to Step 4.
REFERENCES
Step 20: Consider the next set of {i1 , i2 , i3 } in the first
block and go to Step 2.
[Brehm 90] U. Brehm. “Polyhedral Maps with Few Edges.”
In Topics in Comb. and Graph Theory (RingelFetstschrift), edited by R. Bodendiek and R. Henn, pp.
153–162. Heidelberg: Physica-Verlag, 1990.
It is observed from the computations performed that
no possible solution exists to this problem. Hence, the
theorem follows.
[Brehm et al. 02] U. Brehm, B. Datta, and N Nilakantan.
“The Edge-Minimal Polyhedral Maps of Euler Characteristic −8.” Beiträge zur Algebra und Geometrie 43
(2002), 583–596.
Proof of Corollary 1.4: Clearly, G(−25) = 50. So, by
Proposition 2.2, E(−25) ≥ 75. If possible, let M be
a polyhedral map with χ(M ) = −25 and f1 (M ) = 75.
Then f0 (M ) + f2 (M ) = 50. Hence, by Proposition 2.1,
M is a {k, k}-equivelar wnp map with f0 (M ) = f2 (M ) =
(k − 1)2 for some k. Then (k − 1)2 = 25 and hence k = 6.
The corollary now follows from the theorem.
[Brehm and Schulte 97] U. Brehm and E. Schulte. “Polyhedral Maps.” In Handbooks of Discrete and Computational
Geometry, edited by J. E. Goodman and J. O’Rourke,
pp. 345–358. Orlando, FL: CRC Press, 1997.
Remark 2.3. Due to the vast number of calculations involved, we believe that the algorithm is in the most optimal form.
[Datta and Nilakantan 01] B. Datta and N. Nilakantan.
“Equivelar Polyhedra with Few Vertices.” Discrete and
Comput Geom. 26 (2001), 429–461.
[Brehm and Wills 93] U. Brehm amd J. M. Wills. “Polyhedral Manifolds.” In Handbook of Convex Geometry,
edited by P. M. Gruber and J. M. Wills, pp. 535–554.
New York: Elsevier Publishers, 1993.
Nandini Nilakantan, Mathematics Section, The Abdus Salam, International Center for Theoretical Physics, Trieste, Italy
(nandini@ictp.trieste.it)
Received April 16, 2002; accepted in revised form March 11, 2003