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It is well known that finding a maximum cardinality (or weight) common independent set in the intersection of 3 matroids is APX-Hard.

Question: Does this problem remain NP-Hard if one of the matroids is a partition matroid with only two sets in the partition*? (no restrictions on the other matroids).

*By "two sets in the partition" I mean that this partition matroid is a matroid $(E,I)$ such that there is a partition $E_1,E_2$ of $E$ and bounds $c_1,c_2 \in \mathbb{N}$ such that $I = \{ S \subseteq E~|~ |S \cap E_1| \leq c_1, |S \cap E_2| \leq c_2 \}$.

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Your problem is a multi-budgeted matroid intersection with two budgets. There is a PTAS for that [1]. However, your case is so special, better algorithm exists.

You can assume there is only a single budget by truncating the rank of one of the other 2 matroids to $c_1+c_2$. Also, the weights are 0-1 weights. Observe the problem is equivalent to the following problem.

Given matroid $M_1$ and $M_2$ and $R, c$, find a common independent set $B$ of $M_1$ and $M_2$, such that $|R\cap B|\leq c$ and $|B|$ is maximized.

The above problem can be solved in polynomial time. One can reduce it to a sequence of maximum weighted matroid intersection. Indeed, we give each element outside of $R$ a weight of $1$, and elements in $R$ has weight $\epsilon << 1$, and look for a maximum weight common independent set of $M_1$ and $M_2$. This will try to find a common independent set $B$ such that $|B|$ is maximized, and under that constraint, $R$ is minimized. However, it is possible that $|R\cap B|>c$, therefore we can truncate the matroid $M_1$ to a matroid with one fewer rank, and repeat the process. We truncate $M_1$ repeatedly until eventually we obtain $|R\cap B|\leq c$. $B$ would be the desired common independent set.

I have no idea what to do if you want maximum weight common independent set while requiring $|R\cap B|\leq C$.

[1] Chekuri, Chandra; Vondrák, Jan; Zenklusen, Rico, Multi-budgeted matchings and matroid intersection via dependent rounding, Randall, Dana (ed.), Proceedings of the 22nd annual ACM-SIAM symposium on discrete algorithms, SODA 2011, San Francisco, CA, USA, January 23–25, 2011. Philadelphia, PA: Society for Industrial and Applied Mathematics (SIAM); New York, NY: Association for Computing Machinery (ACM). 1080-1097 (2011). ZBL1377.90071.

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  • $\begingroup$ Indeed, this shows that the problem is not APX-hard. What about NP-Hardness? $\endgroup$
    – John
    Apr 13, 2023 at 10:42
  • $\begingroup$ I made an update. If you just want a maximum cardinality common independent set, then it is solvable in polynomial time. $\endgroup$
    – Chao Xu
    Apr 13, 2023 at 16:37

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