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S May 16, 2021 at 14:26 history suggested Matthieu Latapy
added "lattices" and "graph-theory" tags
May 16, 2021 at 11:54 review Suggested edits
S May 16, 2021 at 14:26
May 12, 2021 at 16:19 answer added possiblywrong timeline score: 8
May 9, 2021 at 10:42 vote accept Joseph O'Rourke
May 8, 2021 at 23:26 history edited Joseph O'Rourke CC BY-SA 4.0
deleted 1 character in body
May 8, 2021 at 23:10 answer added Jukka Kohonen timeline score: 5
May 8, 2021 at 22:25 history edited Joseph O'Rourke CC BY-SA 4.0
added 207 characters in body
May 5, 2021 at 6:25 comment added Jukka Kohonen You could try small instances with a SAT solver. The standard reduction of Hamiltonian cycle to SAT has variables $x_{ij}$ encoding "the $i$th element of the cycle is vertex $j$", and appropriate constraints. It would seem straightforward to add constraints to enforce orthogonality, namely, after given two vertices you cannot next visit the third vertice that would continue the straight line.
May 4, 2021 at 21:28 comment added Joseph O'Rourke @FrançoisBrunault: Good catch! Fixed now.
May 4, 2021 at 21:28 history edited Joseph O'Rourke CC BY-SA 4.0
Fixed incorrect figure.
May 4, 2021 at 20:34 comment added François Brunault Ah, but this is easily fixed by adding edges.
May 4, 2021 at 20:26 comment added François Brunault It seems that 4 vertices are not visited in $C_4$, looking near two of the yellow sticks (or I need to change my glasses).
May 4, 2021 at 14:27 comment added Joseph O'Rourke @FedorPetrov: Nice! Thanks. That settles Q1.
May 4, 2021 at 13:45 comment added Fedor Petrov chess coloring proves that for odd $n$ there is no Hamiltonian cycle in $C_n$ (orthogonal or not)
May 4, 2021 at 13:21 history asked Joseph O'Rourke CC BY-SA 4.0