Solidness of amoebas of maximally sparse polynomials

Let $f(z)=\sum_{\alpha\in A}a_\alpha z^\alpha$, with $A$ a finite subset of the integer lattice ${\bf Z}^n$, be a complex Laurent polynomial. Its amoeba is the subset of ${\bf R}^n$ obtained as the image of $\{f(z)=0\}$ under the mapping $(z_1,\dots,z_n)\mapsto(\log\vert z_1\vert,\dots,\log\vert z_n\vert)$. The amoeba is said to be solid if the number of connected components of its complement is minimal, that is, equal to the number of vertices of the Newton polytope $\Delta_f$ of $f$. Solid amoebas are particularly well adapted to tropical geometry. The polynomial $f$ is said to be maximally sparse if the support of summation $A$ is minimal, that is, equal to the set of vertices of $\Delta_f$. When $n=1$ a maximally sparse polynomial is a binomial.

Question: Does every maximally sparse polynomial have a solid amoeba?

The conjecture is mainly based on empirical data (=computer pictures). I did prove with Hans Rullgård that if the number of vertices is less than or equal to $n+2$, then the tropical spine is contained in the amoeba. (So it would seem very plausible that the number of complement components is minimal for maximally sparse polynomials with at most $n+2$ terms.)

(contributed by Mikael Passare)

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