findLexCompatiblyGVDOrders I
An ideal $I$ is $<$compatibly geometrically vertex decomposable if there exists a (lexicographic) order $<$ such that $I$ is geometrically vertex decomposable and for every (onestep) geometric vertex decomposition, we pick $y$ to be the most expensive indeterminate remaining in the ideal according to $<$ [KR, Definition 2.11]. For the definition of a (onestep) geometric vertex decomposition, see oneStepGVD.
This method computes all possible lex orders $<$ for which the ideal $I$ is $<$compatibly geometrically vertex decomposable.



The ideal in the following example is not squarefree with respect to any indeterminate, so no onestep geometric vertex decomposition exists.



[KR] Patricia Klein and Jenna Rajchgot. Geometric vertex decomposition and liaison. Forum Math. Sigma, 9 (2021) e70:123.
In the ring $k[x_1, \ldots, x_n]$, there are $n!$ possible lexicographic monomial orders, so this function can be computationally expensive.
The object findLexCompatiblyGVDOrders is a method function with options.