The first week of 2018 was quite calm, with the most devoted taking stabs at the still-ongoing Prime New Year contest by snarknews. Only one problem remains unsolved there, and it looks like it might take some team effort to crack it. I think it's within the spirit of the contest to discuss its problems publicly (after all, many even have analysis posted online), so I propose to do just that. Please skip the following paragraphs if you don't want spoilers!

The problem goes like this: you are given an undirected graph with at most 200000 vertices and edges. Your goal is to split its vertices into four disjoint groups A, B, C, D in such a way that:

I have been thinking about the problem for some time, but only have the following not very helpful observations:

The problem goes like this: you are given an undirected graph with at most 200000 vertices and edges. Your goal is to split its vertices into four disjoint groups A, B, C, D in such a way that:

- The subgraph formed by groups A+B is connected and has at least 2 vertices.
- The subgraph formed by groups C+D is connected and has at least 2 vertices.
- Each vertex in A has an edge to each vertex in C.
- There are no other edges between subgraphs A+B and C+D.

I have been thinking about the problem for some time, but only have the following not very helpful observations:

- If the graph has an articulation point, then we can take this point as set A, any of the components it separates as set B, and the rest as C+D. So we can assume the graph is vertex-biconnected.
- The answer is not always "Yes" - the smallest counterexample is the graph with 4 vertices and 5 edges. This example also shows that if a biconnected graph has a vertex of degree 2, then this vertex and both adjacent vertices must be in the same half of our splitting, since both ends of a splitting edge must have degree of at least 3 (two splitting edges plus one inside the half).
- There are also counterexamples without vertices of degree 2.
- There might be some hashing trick involved. I.e., for example suppose we assign a random number
*a*to each vertex, and compute for each vertex the value sum(_{i}*a*_{i}*-a*), where the summation goes over all adjacent vertices_{j}*j*. Then if we add up those values within one half of the graph (A+B in the above terminology), then almost everything will cancel out, leaving us |C|**a*_{A}-|A|**a*_{C}, where*a*_{A}is the sum of*a*in A. But where to go from here?.._{i }

In my last post, I have held a vote for the best problem of 2017. And the winner is this AtCoder problem by maroonrk: you are given two rooted trees on the same set of 10

Thanks for reading, and check back next week.

^{5}vertices. You need to assign integer weights to all vertices in such a way that for*each*subtree of*each*of the trees the sum of weights is either 1 or -1, or report that it's impossible. The solution is explained in this post. Congratulations to the problemsetter and to AtCoder, and thanks to all problemsetters of 2017!Thanks for reading, and check back next week.

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