# Balanced partition for trees

Let T be a tree with non-negative weights on the edges. Partition the vertices so each side have n/2 vertices and minimize the sum of the weights of the edges crossing the partition.

This problem can be solved in O(n^3) time, and it generalizes to graphs with constant treewidth [1].

Here we describe the dynamic programming solution for tree and also show the running time is O(n^2). This is probably the solution in [1] when the input is a tree, so this might lead to O(2^kn^2) time algorithm for graphs with treewidth k.

This problem can be reduced to the following problem by arbitrarily root the tree at a leaf, and replace each vertex with degree d+1 into a binary tree with d leaves, and assign weight 0 to the internal nodes (except the root) of the replacing binary tree, and infinite edge weight for the edges connecting internal nodes inside the binary tree.

Let T be a rooted binary tree with non-negative weights on the edges, and 0-1 weights on the vertices. Partition the vertices so each side have the same weight and minimize the sum of the weights of the edges crossing the partition.

Let W(v) to be the sum of the vertex weights of the subtree rooted at v. Let D(v,k) to be the subproblem of minimum cost partition of the the vertices of the tree rooted at v into vertices (A,B) such that v\in A and \sum_{a\in A} {w(a)}=k. Assume v have two children u and w. There are 4 cases to consider, depending on which of \{u,w\} is on the same side of the partition as v.

\displaystyle D(v,k) = \min \begin{cases} \min_{i} D(u,i) + D(w,k-w(v)-i)\\ \min_{i} w(vu) + D(u,W(u)-i) + D(w,k-w(v)-i)\\ \min_{i} w(vw) + D(u,i) + D(w,W(w) - (k-w(v)-i))\\ \min_{i} w(vu) + w(vw) + D(u,W(u)-i) + D(w,W(w)-(k-w(v)-i))\\ \end{cases}

Where i taken over all the numbers that make sense. Other cases are similar and simpler. Note that D(v,k) for a particular k can be computed O(k \min\{W(u),W(w)\}) time. So computing D(v,k) for all 1\leq k\leq W(v) takes O(W(u)W(w)) time.

Hence the running time would obey T(n) \leq \max_{a+b=n} T(a)+T(b)+O(ab). One can show T(n)=O(n^2) is a solution by induction.

# References

[1] K. Jansen, M. Karpinski, A. Lingas, E. Seidel, **Polynomial time approximation schemes for max-bisection on planar and geometric graphs**, SIAM Journal on Computing. 35 (2005) 110–119 10.1137/S009753970139567X.