It's well known that any oriented closed 3-manifold (topological or smooth) can be obtained by surgerizing along a (framed oriented) link $L$ in the 3-sphere $S^{3}$. Even better, Kirby found a complete set of relations that classify all 3-manifold in terms of links. In short, we have the map


$$\{\mbox{framed oriented link in }S^3\}/\{\mbox{kirby move}\} \xrightarrow{\sim} \{3\mbox{-manifold (oriented and closed)}\}.$$


A natural question arise.


1. Given two (oriented and framed) link presentations, is there an algorithm (that halts in finite time) determines if they give rise to the same $3$-manifold?


Or even better, one may ask


2. Given a link presentation $l$, is there any sort of normal form $[l]$ for it (in the sense that $l$ and $l'$ are related by Kirby moves if and only if $[l] = [l']$?


### Related


* A program [KLO](https://community.middlebury.edu/%7Emathanimations/klo/) lets you play with link presentations under Kirby moves. But it doesn't seem to determine if two link presentations produce the same $3$-manifold.
* The most popular programs **snappea** and **regina** present a $3$-manifold in terms of triangulation. It does not seem that they provide a way to determine from their Kirby diagram.



Normal form of framed links under Kirby moves
1. It is a folklore result that geometrisation solves the homeomorphism problem for (compact, connected, oriented) three-manifolds. [Kuperberg](https://arxiv.org/abs/1508.06720) discusses this, and improves the running time to elementary recursive. So, use Snappy to convert the given Kirby diagrams to triangulations and appeal to the above.
2. Sure. First place a reasonable complexity on diagrams. Say, count the number of crossings, add the bit-size of the framings, and break ties using some lexicographic complexity on encodings of graphs (as Burton does using isoSigs of triangulations). Now generate all diagrams up to the complexity of the given framed link. Use (1) to decide homeomorphism and take the smallest. This algorithm to compute normal forms is basically as fast as your solution to the homeomorphism problem (times the number of Kirby diagrams up to the given complexity, of course!).


And to repeat a remark from above - snappy can produce a triangulation from a framed link.


