8
$\begingroup$

We would like to know whether there are any known approximation results for the cardinality constrained minimum $s$-$t$-cut on directed graphs. We weren't able to find any such result in literature.

The problem is defined as follows:

Instance: A directed graph $G=(V,E),ドル a cost function $w : E \to \mathbb{R_0^+},ドル two vertices $s,t \in V$ and an integer $k$.

Solution: An $s$-$t$-cut, i.e. a partition of $V$ into two sets $V_1, V_2$ such that $s \in V_1,ドル $t \in V_2$ and the number of edges that cross the cut is at most $k,ドル i.e. $|\{ (u,v) \in E: u \in V_1, v \in V_2 \}| \le k$.

Measure (to minimize): The cost of the cut: $$ \sum_{ (u,v) \in E : u \in V_1, v \in V_2 } w(u,v) $$

In "Cardinality constrained and multicriteria (multi)cut problems" the autors prove that this problem is strongly NP-Hard even for undirected graphs.

We are mainly interested in approximation algorithms for directed graphs, but approximation results for the undirected case might also be useful.

Thank you for any insights.

asked Mar 25, 2013 at 16:41
$\endgroup$
1
  • $\begingroup$ Sorry it is not an answer, actually I want to ask how to transfer the bi-criteria approximation into the mono-criteria approximation? please forgive me. $\endgroup$ Commented Aug 10, 2019 at 21:36

1 Answer 1

11
$\begingroup$

We can get a $(2,2)$ bi-criteria approximation as follows (or more generally $(1+\varepsilon, 1 + 1/\varepsilon)$ bi-criteria approximation).

We may assume that we know the cost of the optimal solution. Denote it by $OPT$. Let $$w'(u,v) = \frac{w(u,v)}{OPT} + \frac{1}{k}.$$ Consider the optimal solution $(V_1, V_2)$. Then $$\sum_{(u,v) \in E(V_1, V_2)} w'(u,v) = \sum_{(u,v) \in E(V_1, V_2)} \left(\frac{w(u,v)}{OPT} + \frac{1}{k}\right) = 1 + \frac{|E(V_1,V_2)|}{k} \leq 2.$$

Our algorithm finds the minimum directed $s$-$t$ cut $(V_1', V_2')$ in $G$ with edge weights $w'$. The cost of this cut is at most 2ドル$. Therefore, the cut $(V_1', V_2')$ cuts at most $$E(V_1', V_2') = \sum_{(u,v)\in E(V_1',V_2') } 1 \leq k \sum_{(u,v)\in E(V_1',V_2')} w'(u,v) \leq 2k$$ edges. The cost of the cut is at most $$\sum_{(u,v)\in E(V_1',V_2')} w(u,v) \leq OPT \sum_{(u,v)\in E(V_1',V_2')} w'(u,v) \leq 2OPT.$$

answered Mar 25, 2013 at 18:30
$\endgroup$
3
  • $\begingroup$ Thanks a lot. Your algorithm is very interesting and insightful. Unfortunately it seems that a bicriteria approximation algorithm is not enough for us, as we need every approximate solution to be feasible w.r.t the cardinality constraint. Do you know if any such result is known in literature? Thanks again! $\endgroup$ Commented Mar 27, 2013 at 11:48
  • $\begingroup$ Using this approach, you can get $k$-approximation. Probably, you can also get $(k+1)/2$ approximation using LP. But getting anything better than that seems to be pretty difficult. In particular, there is an LP integrality gap for the natural LP relaxation. Let $V={s,t,x}$. Connect $s$ to $x$ with $(k+1)/2$ edges of weight 1 each, connect $x$ to $t$ with $k+1$ edges of weight 0. The optimal combinatorial solution has cost $(k+1)/2$. The optimal LP solution assign $x_e=2/(k+1)$ to edges from $s$ to $x,ドル and $x_e=1-2/(k+1)$ to edges from $x$ to $t$. Its cost is 1ドル$. The gap is $(k+1)/2$. $\endgroup$ Commented Mar 27, 2013 at 20:39
  • $\begingroup$ Thanks for your answer, what you say hints that it might be hard to develop a "good" monocreteria approximation for the problem as even the $(k+1)/2$-apx can be fairly bad. Thanks again for your insights! :) $\endgroup$ Commented Apr 5, 2013 at 15:20

Your Answer

Draft saved
Draft discarded

Sign up or log in

Sign up using Google
Sign up using Email and Password

Post as a guest

Required, but never shown

Post as a guest

Required, but never shown

By clicking "Post Your Answer", you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.