Simplifying triangulations of 3-manifolds

Throughout, by finite triangulation I mean a triangulation consisting of a finite number of triangles.

Suppose $T$ and $T'$ are finite triangulations of a 3-manifold $M$. We will say that $T'$ is simpler than $T$ iff $T'$ consists of the same number or fewer triangles than $T$ and that $T'$ is a simplest triangulation of $M$ iff $forall$ triangulation $T$ of $M$, $T'$ is simpler than $T$.

Note: If a 3-manifold $M$ has a finite triangulation, then clearly it has a simplest triangulation.

By a theorem of Pachner (Theorem A.1.1. in 'The geometry of dynamical triangulations') any two triangulations of a manifold can be transformed from one to another by a finite number of stellar subdivisions. As we are only dealing with 3-manifolds, there are only 4 stellar subdivisions; known as the $1 to 4$, $2 to 3$, $3 to 2$ and $4 to 1$ moves as described in http://at.yorku.ca/t/a/i/c/45.pdf and hereafter called the Pachner moves. So clearly, there exists a finite sequence of Pachner moves from any finite triangulaiton $T$ of $M$ to $T'$, a simplest triangulation of $M$.

If $T$ is a finite triangulation of $M$, does the greedy algorithm of just applying as many $4 to 1$ and $3 to 2$ Pachner moves to $T$ as possible always result in a simplest triangulation of $M$?

Or alternatively,

Is there a finite triangulation $T$ of a 3-manifold $M$ such that repeatedly applying only the $4 to 1$ and $3 to 2$ Pachner moves does not eventually result in a simplest triangulation of $M$?

• Next lo.logic - A book explaining power and limitations of Peano Axioms?
• Previous ag.algebraic geometry - Zariski open sets are dense in analytic topology