Wikipedia:Reference desk/Archives/Mathematics/2021 April 30

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April 30[edit]

6 Cycles on a cube with diagonals.[edit]

Consider a cube with diagonals (so 24 edges). How many distinct ways are there to have 6 cycles of 4 edges? The ones that I have found are

  • 6 equators (containing opposite edges and the diagonals connecting them.
  • 6 hourglasses (containing the diagonals on one face with opposite edges of the face)
  • 6 parallelograms (view two adjoining faces, top edge of left face, diagonal from Top left to bottom of line between them, bottom edge of right face and diagonal back to start, three go around one vertex, three around the opposite vertex)
  • mirror of the 6 parallelograms in previous entry.

Do the 6 cycles have to all be of the same shape to cover all of the edges? Are there any others were the 6 all are of the same shape?Naraht (talk) 12:21, 30 April 2021 (UTC)[reply]

No, and yes, as shown: --116.86.4.41 (talk) 17:44, 30 April 2021 (UTC)[reply]
Some ways
A somewhat brute force search produced 853 decompositions into 4-cycles, which, after division modulo the cubical symmetries, fall into 44 equivalence classes, or orbits. Each orbit has a size that divides 48, the number of cubical symmetries. There is one supersymmetric decomposition, that of the parallelograms – I have not treated mirror symmetries as different from the other symmetries – while there are six bloated orbits of fully asymmetric decompositions. The full account (where n × s means n orbits of size s) is: 1 × 1; 1 × 2; 2 × 3; 0 × 4; 4 × 6; 4 × 8; 9 × 12; 2 × 16; 15 × 24; 6 × 48.  --Lambiam 15:55, 2 May 2021 (UTC)[reply]

Followup[edit]

Conversely, can there be 4 cycles of 6 edges each?Naraht (talk) 12:21, 30 April 2021 (UTC)[reply]

Yes. Number the vertices 0 through 7 so that x and 7-x are always opposite corners. (There are 8 essentially different ways of doing this.) Then one example is (012765)(023754)(031746)(153624). According to Cycle decomposition (graph theory), Kn-I can be decomposed into cycles of length m, m and n both even, 4≤m≤n, iff m divides the number of edges in Kn-I (=m(m-2)/2). In this case the graph is K8-I with 24 edges, so there are decompositions into 4-cycles, 6-cycles, and 8-cycles. --RDBury (talk) 23:34, 30 April 2021 (UTC)[reply]