Topological sorting and NP-hard proof

Question

I meet a problem. I can find a sub-optimal solution, but cannot find an optimal one and cannot prove its NPC hardness.

The problem can also be described as follows. Given a sequence $X=\{x_1,x_2,...,x_n\}$, a partial order $<$, where $x_i <x_j$ means that $x_i$ must be in front of $x_j$, and a new item $x_k$ as well as related statements $x_k<x_i$ for some $i$'s and $x_j<x_k$ for some $j$'s to be added to the partial order, how to find the least replacement solution to insert $x_k$ into the original sequence $X$ without violating the resulting partial order?

It can be ensured that the partial order forms no cycle such as $x_i<\cdots<x_j<\cdots <x_i$.

Answer 1

I guess this optimality depends on how you define "the least replacement solution". I'm taking it to mean minimum insertions / deletions. If you prefer minimum number of swaps necessary, take a look at step 4.1 and 4.2 and you may be able to optimize it for such, though I don't show it here.

If it is ensured that no additional $x_k$ and its partial constraints will cause a cycle, then this becomes relatively easy. Consider a topological ordering $\{x_1, x_2, \ldots, x_i, \ldots, x_j, \ldots\}$ where $x_i$ and $x_j$ have no transitive relation. We now have five options for additional constraints to be added:

1. $x_k$ with no constraints: put $x_k$ anywhere in the ordering.
2. $x_k > x_i$: put $x_k$ anywhere after $x_i$ in the ordering.
3. $x_k < x_j$: put $x_k$ anywhere before $x_i$ in the ordering.
4. $x_i < x_k < x_j$: put $x_k$ anywhere in between $x_i$ and $x_j$ in the ordering.
5. $x_j < x_k < x_i$: this is the only case where we will need to rearrange existing nodes because $x_i$ and $x_j$ are out of order currently.

For case 5 we will first worry about the fact that we are transitively adding an edge $x_j < x_i$, then after re-ordering, this will resolve to case 4. Do the following:

1. Let $A_i$ be all elements reachable ("after") in the DFS starting at $x_i$ of the current partial ordering.
2. Let $B_j$ be all elements on any path from the root to $x_j$. This can be computed in linear time by doing a DFS on the reverse partial ordering, starting at $x_j$.
3. Note that if $x_j \in A_i$ then this violates our original assumption. If $x_i \in B_j$, this also violates our original assumption. So we know $x_j \not\in A_i$ and $x_i \not\in B_j$.
4. Greedily move all elements in $B_j$ before all elements in $A_i$. This can be done by doing one of the following which has minimum insertions:
   1. Move all elements in $B_j$ that are after $x_i$ in the ordering, before $x_i$.
   2. Move all elements in $A_i$ that are before $x_j$ in the ordering, after $x_j$.
5. You can show that the minimum of these two will be a lower bound for the number of insertions / moves necessary.

After you have separated $B_j$ into the first half of the order and $A_i$ into the second half, this case reduces to case 4 as described above. Overall this will take $O(n + m)$.