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This problem is a following up question on this one. The only difference is the newly added constraint in the bold font.

Set S, which is an non-empty finite subset of $\{ (i,j) : i, j \in N \land i \neq j \}$ and is also a transitive closure, is given. E.g. $S=\{(1,2), (2,3), (1,3), (2,4), (1,4)\}$. For each element $(i,j)$, we have weight $w_{ij}=c(i)>0$, where $c$ is some benefit function, and a binary decision variable $x_{ij}$. The optimization problem is defined as follows:

$$\text{maximize} \sum_{(i,j)\in S} w_{ij}x_{ij} $$ $$\text{s.t.} x_{ij}+x_{ik} \le 1 $$ $$x_{ij}+x_{jk} \le 1 $$

Note this problem is not the Maximum Weighted Matching problem as edges that share the same end point are allowed.

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1 Answer 1

up vote 9 down vote accepted

This is easy to solve in polynomial time.

To see this, it may be easiest to interpret $S$ as a preorder $\le$ on $N$:

  • $i \le j$ iff $(i,j) \in S$ or $i = j$.

Then we define the equivalence relation $\sim$:

  • $i \sim j$ iff $i \le j$ and $j \le i$.

Now let $W$ consist of all equivalence classes of $\sim$; for each $i \in N$ let $[i] \in W$ be its equivalence class. Put otherwise, $[i]$ consists of all $j \in N$ such that $(j,i) \in S$ and $(i,j) \in S$. Now we can define the natural partial order $\le'$ on $W$:

  • $[i] \le' [j]$ iff $i \le j$ for each $i,j \in N$.

We say that $[i] \in W$ is maximal if $[i] \le' [j]$ implies $[i] = [j]$. We have the following property that is easy to verify:

  • If $[i]$ is maximal, we cannot get the benefit $c(j)$ for all $j \in [i]$.

Hence in any solution, for every maximal class $[i]$, we have to choose at least one node $j \in [i]$ such that we do not get the benefit of $c(j)$.

We now construct a solution as follows: For each maximal class $[i]$, choose a node $j \in [i]$ that minimises $c(j)$; we say that $j$ is a sink node. Now assume that $k \in N$ is not a sink node. There are too cases:

  • $k \in [i]$, where $[i]$ is maximal. Now we can simply choose the edge $(k,j) \in S$, where $j \in [i]$ is a sink node, and collect the benefit of $c(k)$.

  • $k \in [i]$, where $[i]$ is not maximal. Then there is a maximal set $[\ell]$ with $[i] \le [\ell]$. Let $j \in [\ell]$ be the sink node, choose the edge $(k,j) \in S$, and collect the benefit of $c(k)$.

That is, we can get the benefit of $c(k)$ for all nodes $k \in N$ that are not sink nodes; the solution is optimal.

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