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The Feedback Arc Set Problem with Triangle Inequality Is a Vertex Cover Problem

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Abstract

We consider the (precedence constrained) Minimum Feedback Arc Set problem with triangle inequalities on the weights, which finds important applications in problems of ranking with inconsistent information. We present a surprising structural insight showing that the problem is a special case of the minimum vertex cover in hypergraphs with edges of size at most 3.

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Notes

  1. Recall that a 0∖1 solution δ ∗ is minimal if the removal of any arc (i,j) from its support makes it unfeasible.

  2. When (i,j′)∈M v this is the only possible case, namely case (ii) does not hold. Indeed, see Fig. 6, when (i,j′)∈M v , the following is a valid constraint δ v,j′+δ j,k′+δ k,v ≥1 and we are considering the case with δ v,j′=δ j,k′=0.

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Acknowledgements

To my beloved mom Elsa.

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Authors and Affiliations

  1. IDSIA, USI-SUPSI, Lugano, Switzerland

    Monaldo Mastrolilli

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  1. Monaldo Mastrolilli
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Correspondence to Monaldo Mastrolilli.

Additional information

This research is supported by Swiss National Science Foundation project N. 200020-144491/1 “Approximation Algorithms for Machine Scheduling Through Theory and Experiments” and by Hasler Foundation Grant 11099.

Appendices

Appendix A: Ranking with Probability Inequalities: a Counterexample

The following example shows that probabilities inequalities are not sufficient for (3a)–(3d) to be a proper formulation:

$$w_{(i,j)}+w_{(j,i)}=1\quad \mbox{for all distinct } i,j $$

Consider the instance with 8 nodes with weight zero on the arcs displayed in Fig. 7 (therefore the reversed arcs have weight 1). Moreover, all the arcs in {2,3}×{7,8} have weight 1 (the reversed zero). Finally, all the remaining arcs have weight 0.5, namely those in {1}×{4,5,6} and the reversed ones. A feasible solution for (2a)–(2d) is obtained by picking all the displayed arcs in Fig. 7 and none of the reversed ones (therefore we have to pick also those in {2,3}×{7,8}, {7,8}×{2,3}, {4,5,6}×{1} and {1}×{4,5,6} in order to satisfy the constraints in (2a)–(2d)). This solution has value 7, whereas any total ordering has value not smaller than 7.5 (the best total ordering is (2,3,4,5,6,7,8,1)).

Fig. 7
figure 7

Counterexample for probability inequalities

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Appendix B: A Comment on Formulation (3a)–(3d)

If the poset is not empty the additional constraints that are present in formulation (3a)–(3d) but not in (2a)–(2d) are also necessary. Indeed, in Figure 8 any permutation that complies with the precedence constraints has value larger than the solution suggested in the picture with a cycle.

Fig. 8
figure 8

Solution \(\delta^{*}_{(1,2)}= \delta^{*}_{(2,3)}=\delta^{*}_{(3,4)}= \delta^{*}_{(4,1)}=\delta^{*}_{(1,3)}= \delta^{*}_{(3,1)}=\delta^{*}_{(2,4)}= \delta^{*}_{(4,2)}=1\) has value smaller than any valid permutation

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Mastrolilli, M. The Feedback Arc Set Problem with Triangle Inequality Is a Vertex Cover Problem. Algorithmica 70, 326–339 (2014). https://doi.org/10.1007/s00453-013-9811-2

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