# Optimal Matroid Partitioning Problems

## Abstract

This paper studies optimal matroid partitioning problems for various objective functions. In the problem, we are given k weighted-matroids on the same ground set. Our goal is to find a feasible partition that minimizes (maximizes) the value of an objective function. A typical objective is the maximum over all subsets of the total weights of the elements in a subset, which is extensively studied in the scheduling literature. Likewise, as an objective function, we handle the maximum/minimum/sum over all subsets of the maximum/minimum/total weight(s) of the elements in a subset. In this paper, we determine the computational complexity of the optimal partitioning problem with the above-described objective functions. Namely, for each objective function, we either provide a polynomial time algorithm or prove NP-hardness. We also discuss the approximability for the NP-hard cases.

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1. 1.

We remark that the condition $$I_i\ne \emptyset ~(\forall i\in [k])$$ is imposed to make the objective function well-defined. Moreover, if we define $$\max _{e\in \emptyset }w_i(e)=0$$, $$\min _{e\in \emptyset }w_i(e)=\infty$$, and $$\sum _{e\in \emptyset }w_i(e)=0$$, then we can reduce the problem where empty sets are allowed to our problem by adding dummy elements.

2. 2.

In order to make the paper self-contained, we give a proof in Appendix.

3. 3.

More precisely, we prove that it is NP-hard to distinguish whether the minimum $$(\max ,\min )$$ and $$(\sum ,\min )$$-value of a given problem instance is positive or zero.

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## Acknowledgements

We thank the anonymous reviewer for giving us comments which improves the presentation of the manuscript. The first author is supported by JSPS KAKENHI Grant Number JP16K16005. The second author is supported by JSPS KAKENHI Grant Number JP15H06286. The third author is supported by JSPS KAKENHI Grant Number JP24106002, JP25280004, JP26280001, and JST CREST Grant Number JPMJCR1402, Japan. The forth author is supported by JST ERATO Grant Number JPMJER1201, Japan, and JSPS KAKENHI Grant Number JP17K12646.

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Correspondence to Kei Kimura.

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## Omitted Proof in Theorem 2

### Lemma 2

() Let $$(E, \mathcal {I})$$ be a partition matroid defined by $$\mathcal {I} = \{ I \,:\,|I \cap S_i| \le \eta _i \ (i\in [p]) \}$$ with weight w, where $$(S_1,\dots ,S_p)$$ is a partition of E, $$\eta _1,\dots ,\eta _p$$ are positive integers, and $$|S_i|=\eta _i\cdot k$$ for each $$i\in [p]$$. Let $$I_{i,j}$$ consist of $$\eta _i$$ elements with the $$\eta _i$$ largest weights in $$S_i\setminus (\bigcup _{h=1}^{j-1}I_{i,h})$$. Then $$(\bigcup _{i\in [p]} I_{i,1},\dots ,\bigcup _{i\in [p]} I_{i,k})$$ is an optimal solution to the minimum $$(\sum ,\max )$$-value matroid partitioning problem $$(E, (\mathcal {I}, w), k)$$.

### Proof

Let $$(I_1^*,\dots ,I_k^*)$$ be an optimal partition. Without loss of generality, we may assume that $$\max _{e\in I_1^*}w(e)\ge \dots \ge \max _{e\in I_k^*}w(e)$$. Let j be any index in [k]. In addition, let $$(i', e_j)$$ be the pair of an index and an element attaining $$\max _{i \in [p]} \max _{e \in I_{i,j}}w(e)$$. We claim that $$\max _{e \in I^*_j} w(e) \ge w(e_j)$$. To show this, we suppose the contrary. We denote $$S = \bigcup _{h < j} I_{i', h} \cup \{e_j\}$$. Note that $$|S| = (j-1)\eta _{i'}+1$$ and $$w(e) \ge w(e_j)$$ for all $$e \in S$$. Since $$(E, \mathcal {I})$$ is a partition matroid, at most $$(j-1)\eta _{i'}$$ elements in S are contained in $$I^*_1, \ldots , I^*_{j-1}$$. By assumption $$\max _{e \in I^*_j} w(e) < w(e_j)$$, there is an index $$\ell > j$$ such that $$I^*_{\ell }$$ has some element $$e' \in I^*_{\ell } \cap S$$. Then we have $$\max _{e \in I^*_{\ell }}w(e) \ge w(e') \ge w(e_j) > \max _{e\in I^*_j} w(e)$$, which contradicts the assumption $$\max _{e\in I^*_j} w(e) \ge \max _{e \in I^*_{\ell }}w(e)$$.

Thus, we have

\begin{aligned} \max _{e\in I_j^*}w(e) \ge w(e_j)=\max _{i\in [p]}\max _{e\in I_{i,j}}w(e)=\max _{e\in \bigcup _{i\in [p]}I_{i,j}}w(e). \end{aligned}

Therefore, $$(\bigcup _{i\in [p]} I_{i,1},\dots ,\bigcup _{i\in [p]} I_{i,k})$$ is also an optimal solution. $$\square$$

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