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On the azulenoids with large ring of 5-gons

On the azulenoids with large ring of 5-gons. Roman Soták (joint work with R. Hajduk and F. Kardoš) Institute of Mathematics Faculty of Science P. J. Šafárik University, Košice. Marseille 2007. Examples:. G  M n ( n ,4). G  M 4 (4,5). G  M 2 (3,6) (only 2-connected). G  M 3 (5,8).

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On the azulenoids with large ring of 5-gons

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  1. On the azulenoids with large ring of 5-gons Roman Soták (joint work with R. Hajduk and F. Kardoš) Institute of Mathematics Faculty of Science P.J.Šafárik University, Košice Marseille 2007

  2. Examples: GMn(n,4) GM4(4,5) GM2(3,6) (only 2-connected) GM3(5,8) Mn(p,q) – set of all 3-regular plane graphs having only p-gonal and q-gonal faces such that q-gonal faces form a ringRn of nq-gons. (p≥3, q≥4) (Deza, Grishukhin ’02)

  3. Problem:For which (n,p,q) there exists a map from Mn(p,q)? Denote by xI and xO the number of vertices in inner and outer part of the graph determined by the ring Rn. From Euler formula we can derive the following: ((4 - p) (q - 4) + 4) n + (6 - p) (xI + xO) = 4p For example: Case q=4; then xI = xO = 0 and therefore n = p. Moreover, there exist such maps.

  4. From this equation following results are derived: (results of Deza and Grishukhin)

  5. Case Mn(5,q) for q ≥ 8: Malkevitch (’70?): There are graphs from Mn(5,8) for infinitely many odd numbers n. There is no graph from Mn(5,10k) for odd n. Theorem (Madaras and R.S.): The graph from Mn(5,q) exists for the following particular values a) n = 6, q≡ 0, 1, 4 (mod 5), q ≥ 5, q ≠ 9 b) n = 8, q ≡ 1, 4 (mod 5), q ≥ 5 c) n = 10, q ≡ 0 (mod 10), q ≥ 20 d) n = 12 + 4k, k ≥ 0, q ≡ 2, 3 (mod 5), q ≥ 13 e) n = 14 + 4k, k ≥ 0, q = 10 Other cases are still mostly open. From the proof – idea for the case n = 6, q ≡ 1 (mod 5)

  6. I for r = 1 O for r = 1 GM6(5,11) We use following basic configurations (with some half-edges): A A* Br for r = 4 H = H* By concatenation of basic configurations we have inner and outer parts I and O of our graph: I = H ○ A○ … ○ A ○ B2 ○ A* ○ … ○ A* ○ H (r copies of A, r copies of A*) O = H ○ B5r+2 ○ H

  7. Case Mn(7,5) for n ≥ 28: • In this case we turn our attention to the dual: triangulation with vertices of degree from {5,7} such that 5-vertices induce Cn. • Denote by Dnhalf(7,5) the set of all graphs with following properties: • the outer face has size n, all other faces are triangles; • vertices incident with the outer face have degree from {3,4}, all other vertices have degree 7; • there are no chords between vertices from the outer face. Then Mn(7,5) is nonempty if and only if there are corresponding interior and exterior graphs in Dnhalf(7,5). By corresponding we mean, that one can identify the vertices of degrees 3 and 4 to form vertices of degree 5. For example, wheel W7 is from D7half(7,5), but there is no corresponding exterior.

  8. To each graph G from Dnhalf(7,5) we can assign a circular sequence s(G) from {3,4}n in a natural way . Now we can construct other sequences from a sequence s(G) by replacing: • some subseqence 333 by 433334 • some 343 by 43334 • some 3443 by 4334334 • some 34k3 by 433(43)k-134 Each of them is realizable by a graph from Dmhalf(7,5), where m is the length of the new sequence. • Moreover, we can prove following observations: • If a circular sequence is realizable, then it can be constructed recursively from 37 (realizable by W7) using only first three replacements. • If a circular sequence is realizable and contains 433334, then it is also realizable after replacing this subsequence by 333 (inverse operation) • the same for 43334 by 343 • the same for 4334334 by 3443

  9. Algorithm for searching for maps from Mn(7,5): • Start (add to set D) with the sequence 37 • Take the next unmarked sequence, mark it, construct new sequences by first three replacements, add them to D, for each of the added sequences test realizability of the corresponding exterior sequence. • Repeat the previous step Outputs of algorithm: 1. Cardinality of Dnhalf(7,5) 2. There is no map from Mn(7,5) for odd n < 52 3. There is no map from Mn(7,5) for n = 34 4. We found all maps from Mn(7,5) for even n < 52 (see Examples 1, 2, 3, 4)

  10. Conjectures: • The set Mn(7,5) is nonempty for all even n > 34. • There is an odd n with nonempty Mn(7,5). • There is a constant k, such that the set Mn(7,5) is nonempty for all n > k. • There is a map from Mn(7,5) with automorphism group Γ for each Γ from {D7d, D7, S14, C7v, C7, D3d, D3, S6, C3v, C3, D2d, D2, S4, C2v, C2, Ci, Cs, C1} • For example, there is a map from M260(7,5) with automorphism group S4. • Theorem: • The set Mn(7,5) is nonempty for • n = 28 + 4k, k ≥ 0 • n = 42 + 12k, k ≥ 0 From the proof – we find the dual of G Mn(7,5) for n = 32 + 8k, k ≥ 1

  11. corresponding sequence of degrees for I: (33343444344344334334)2 for O (shifted and reverted) (44434333433433443443)2 corresponding sequences: (333443344434434434334334)2 (444334433343343343443443)2

  12. Thanks for your attention.

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