### VaibhavAgarwala's blog

By VaibhavAgarwala, history, 6 weeks ago, I have polygon and its vertices represented with randomly shuffled 2d points, how can I find the correct order?

In correct order I mean order so that segment between P[i] and P[(i + 1) % n] is a side of polygon.

Or if this is hard, how to solve it for 4 sides?

I'm looking for an efficient solutions. Many thanks. Comments (11)
 »
 » You can find the upper hull of the points, add it to the list, then add the points sorted by their x coordinates (lower points first)
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   Simple and useful.I actually used comparator based on this idea, didn't split the points
 »
 » Is the polygon guaranteed to be convex?
•  » » It's not stated anywhere, so I assume not. It will also be a very obvious / known question in that case.
•  » » » why are people downvoting this? I answered the question in another comment and the author of the blog seemed fine with it. The convex case would just be finding a convex hull, and there are many known algorithms for that
•  » » » » If you are given a set of points in $R^2$, then there isn't a unique way of making a non-convex polygon out of them. Take for example the points $(1,0),(0,1),(-1,0),(0,-1),(0,0)$. There are 4 natural ways you could make a non-convex polygon out of them.Only when you assume the points should form a convex polygon will the answer be unique.
•  » » » » » of course it wouldn't be unique, but we can't make assumptions to make our life simpler. and also a convex polygon of n vertices can have 2n orders :/ (rotation, flip)That said, I understand why there's the confusion, but my answer seemed to satisfy VaibhavAgarwala, so I guess he meant any order (correct me if I'm wrong)
•  » » » » » » Yeah, I meant any order.
 » wouldn't it be the right thing to do,to simply sort them in clockwise direction, using the formula of cos between 2 vectors,however,you have to fing the center of that polygon to form the vectors