$f(a,c)$ are either equal or collinear.

\[hyper\] A central arrangement ${{\mathcal A}}$ is called [*hypersolvable*]{} if it admits a composition series of solvable extensions: $$\{H\}={{\mathcal A}}_{1}\subset {{\mathcal A}}_{2}\subset \cdots {{\mathcal A}}_{i}\subset {{\mathcal A}}_{i+1} \cdots \subset{{\mathcal A}}_{l}={{\mathcal A}}$$

For example, for generic arrangements, known to be hypersolvable, a composition series can be easily constructed starting with an arrangement containing just one hyperplane, and adding one more hyperplane for each extension.

As in the fiber-type case, the fundamental group $\pi_1(M_{{{\mathcal A}}})$, for ${{\mathcal A}}$ hypersolvable, is an iterated almost direct product of finitely generated free groups $\mathbb{F}_{m_i}, \; m_i=|{{\mathcal A}}_i \setminus {{\mathcal A}}_{i-1}|$.

The length $l=l({{\mathcal A}})$ of a composition series, called [*the length of ${{\mathcal A}}$*]{}, is a combinatorial invariant: it coincides with the degree of the Poincar' e polynomial of the quadratic OS-algebra $\overline{A^*}({{\mathcal A}})$, see [@JP1 Theorem B].

\[thm:lr\] Let ${{\mathcal A}}$ be a hypersolvable arrangement. Then ${{\mathcal A}}$ is supersolvable if and only if $l({{\mathcal A}})=r({{\mathcal A}})$.

Hence comparing the length of a hypersolvable arrangement to its rank is a simple combinatorial test for the asphericity of the complement.

As a general rule, $l({{\mathcal A}}) \geq r({{\mathcal A}})$ for ${{\mathcal A}}\subset V$ hypersolvable. Each solvable extension ${{\mathcal B}}\subset_{\neq} 