Suppose you have a 2-dimensional polyhedral surface with specified lengths for the edges so that the vertices all have positive curvature. I believe this has a unique isometric embedding into 3-dimensional Euclidean space as the boundary of a convex polyhedron. Could someone confirm this? If so, is there a reasonable algorithm for finding the isometric embedding computationally?



Isometric embedding of a positively curved polyhedral surface
Yes, this is the historically first version of the Alexandrov embedding theorem which was earlier discussed [here](https://mathoverflow.net/questions/22122/on-alexandrov-embedding-theorem). See the original Alexandrov's monograph (reviewed [here](http://www.math.cornell.edu/~connelly/alexandrov.pdf)), or see [my book](http://www.math.ucla.edu/~pak/book.htm) if you like downloadable stuff (at least for now). 


The algorithmic issue is difficult. If you want the usual complexity, in my paper with Fedorchuk we [showed](http://www.math.ucla.edu/~pak/papers/pp16.pdf) that this requires finding roots of polynomials with exponential degrees. The practical algorithm (based on a new completely different proof of the Alexandrov theorem) was found by Bobenko and Izmestiev [here](http://arxiv.org/abs/math/0609447) and further analyzed [here](http://arxiv.org/abs/0812.5030). 


