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 $M$. 



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> 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 $M$?
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Or alternatively,



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> 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 $M$?
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Simplifying triangulations of 3-manifolds
There are many such examples, depending upon what you mean by "triangulation". If a triangulation is just a glueing of tetrahedra along faces, then the simplest one is probably the following: the 3-sphere has a triangulation with 1 tetrahedron, and a triangulation with 2 tetrahedra (it is a nice excercise to find them). However, you cannot apply any $4 \to 1$ or $3 \to 2$ move to a triangulation having only 2 tetrahedra.


If by "triangulation" you mean a "simplicial triangulation", then it suffices to construct a triangulation of the 3-sphere such that every edge meets at least 4 distinct tetrahedra. In this case, no $4 \to 1$ and no $3 \to 2$ moves can apply. 


For instance, you can take a triangulation of the 2-sphere such that every vertex is incident to at least 4 triangles (as an example, the octahedron). By doing a suspension of this triangulation you get a triangulation of the 3-sphere such that every edge is incident to at least 4 tetrahedra.


