Wraps and Loops—Which Sequences are Admissible?

Question

If you take a non-intersecting closed loop on a torus (that is to say, a path which ends where it starts and does not cross over itself, drawn inside a square whose edges "wrap" left to right and top to bottom), you can use it to infinitely tile the plane in a lined pattern:

Example loop (w/ basepoint), and the corresponding tiling in the plane

Moreover, if we take our basepoint and follow the path produced on the plane diagram until we hit a basepoint once again, we find that we move through a series of tiles, perhaps not ending up where we started:

Extracting the arrow sequence from the loop (w/ given basepoint and travel direction)

In this way, we say that the sequence ↑←↓↓←↑→↓ is admissible. However, not all sequences of arrows can be produced this way; a diagram producing the sequence simply cannot be drawn. Those sequences are called inadmissible. Can you find an effective means for determining if a provided sequence is admissible or inadmissible?

Here are some example cases to get you started (which your algorithm should correctly label as admissible or inadmissible):

1. ↓←→↓↑→←↓
2. ↑↑↑→↑←
3. ↑↑↓→←↑→↑→

Answer 1

UPDATE: As a reference I've written a super dumb brute force algo which just tries all possible permutations of the horizontal and vertical boundary crossings.

Here are the results for the four test cases:

End of UPDATE.

Note: The core of the algorithm is still rather sketchy, so this may be regarded as a partial answer.

Sketch of decision "algorithm":

     input: ULDDLURD
     1) form pairs, wrapping around: UL,LD,DD,DL,LU,UR,RD,DU
     2) discard direction: UL,LD,UU,RU,DR,LD,UL,UD
     if both UU and LL occur return FALSE
     pool U,D->V and L,R->H and for each try to find an
         ordering such that
     for all pairs made in (1,2) the following is satisfied
     if there are two pairs U1L1,U2L2 then U1<U2 <=> L1<L2
         and similar for DR
     if there are two pairs L1D1,L2D2 then L1<L2 <=> D1>D2
         and similar for RU
     if there are two pairs U1U2,U2U3 then U1<U2 <=> U2<U3
         and similar for LL
     for any pair U1D1 (or D1U1) all V such that U1<V<D1 are
         also part of a UD (or DU) pair
         and similar for LR (or RL)
     for any two pairs U1L1,L2D2 we have L2>L1
         and similar for R1U1,U2L2, for R1U1,U2U3,
         for U1U3,U2L2, for U1L1,L2L3 and for L1L3,L2D2
     for any two pairs D1R1,R2U2 we have R2<R1
         and similar for L1D1,D2R2, for L1D1,U3U2,
         for U3U1,D2R2, for D1R1,R3R2 and for R3R1,R2U2
     return TRUE if such two orderings exists otherwise
         return FALSE
 

Remarks:

1. The orderings in the algorithm, of course, directly correspond to the (spatial) order of crossings along tile boundaries
2. The rules spelt out are just a rundown of what needs to hold so no segments cross
3. Admittedly, the actual algorithmic test for/construction of the orderings is a challenge in itself.