  In a recent work of I.\,Dynnikov and M.\,Prasolov a new method of comparing
Legendrian knots is proposed. In general, to apply the method requires a lot of
technical work. In particular, one needs to search all rectangular diagrams of
surfaces realizing certain dividing configurations. In this paper, it is shown
that, in the case when the orientation-preserving symmetry group of the knot is
trivial, this exhaustive search is not needed, which simplifies the procedure
considerably. This allows one to distinguish Legendrian knots in certain cases
when the computation of the known algebraic invariants is infeasible or is not
informative. In particular, it is disproved here that when~$A\subset\mathbb
R^3$ is an annulus tangent to the standard contact structure along~$\partial
A$, then the two components of~$\partial A$ are always equivalent Legendrian
knots. A candidate counterexample was proposed recently by I.\,Dynnikov and
M.\,Prasolov, but the proof of the fact that the two components of~$\partial A$
are not Legendrian equivalent was not given. Now this work is accomplished. It
is also shown here that the problem of comparing two Legendrian knots having
the same topological type is algorithmically solvable provided that the
orientation-preserving symmetry group of these knots is trivial.
