The converse contraction conjecture for non-exceptional Hurwitz bisets

Let $B$ be a non-exceptional Hurwitz biset, and let $B^*$ denote its associated semigroup of elements. An element of $B^*$ is unobstructed if it has no Thurston obstruction. Converse contraction conjecture. If every element of $B^*$ is unobstructed, then $B$ is contracting. The converse to the implication that contraction makes every element of $B^*$ unobstructed is unresolved; the difficulty is that individual spectral radii may all be less than one while the joint spectral radius is equal to one, a phenomenon known for nonnegative matrices.