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$UPDATE:$ UPDATE: I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.

$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2&-&6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2& &6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$

$HOW THIS THING WORKS:$ HOW THIS THING WORKS:

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $$\begin{array}{ccccccccccccc} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

$UPDATE:$ I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.

$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2&-&6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$

$HOW THIS THING WORKS:$

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

UPDATE: I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.

$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2& &6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$

HOW THIS THING WORKS:

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{ccccccccccccc} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

added 2136 characters in body
Source Link
domotorp
  • 18.7k
  • 3
  • 57
  • 125

$UPDATE:$ I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.

$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2&-&6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$

$HOW THIS THING WORKS:$

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

$UPDATE:$ I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.

$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2&-&6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$

$HOW THIS THING WORKS:$

I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.

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domotorp
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I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.

Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.

One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.

$\begin{array}{clrr} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array} $

Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.