S TEINER O RIENTATION has a Solution ⇒ G RID T ILING

has a Solution

Since the instance (G,T) of Steiner Orientation has a solution, let G^∗ be the orientation which satisfies all pairs from T. Note that the set of vertices B∪D∪F∪H has no incoming edges. Similarly, the set of verticesA∪C∪E∪G has no outgoing edges.

Lemma 2. No yellow edge can be on a path inG^∗ which satisfies any terminal pair of Type I or II.

Proof. Fixj∈[k]. We just prove the lemma for the terminal pair (bⁿ_j, a¹_j) since the proof for other terminal pairs is similar. Suppose there is a yellow edge on some pathP satisfying the terminal pair (bⁿ_j, a¹_j). This yellow edge cannot be of Category I sinceDhas no incoming edges, and hence we could not have reached Din the first place starting frombⁿ_j. However, this yellow edge also cannot be of Category II sinceH has no incoming edges and hence we could not have reached

H in the first place starting frombⁿ_j.

The next lemma restricts the orientations of the undirected paths of green edges.

74 R. Chitnis and A. E. Feldmann

Lemma 3. In the orientation G^∗, for each i∈[k] we have that

– there exists an integer λ_i ∈[n] such that the paths A_i, B_i are oriented away from, and towardsλ_i, respectively,

– there exists an integer μ_i ∈[n] such that the paths C_i, D_i are oriented away from, and towardsμ_i, respectively,

– there exists an integer δ_i ∈[n] such that the paths E_i, F_i are oriented away from, and towardsδ_i, respectively,

– there exists an integer _i ∈[n] such that the paths G_i, H_i are oriented away from, and towards_i, respectively.

Proof. Fix i ∈ [k]. We just prove the lemma for the paths A_i, B_i since the proof for other cases is similar. By Lemma2, we know that the paths satisfying the terminal pairs (bⁿ_i, a¹_i) and (b¹_i, aⁿ_i) cannot contain any yellow edges. Since the only non-yellow edges incoming to A are blue edges from B (and B has no incoming edges), it follows that the terminal pairs (bⁿ_i, a¹_i) and (b¹_i, aⁿ_i) of Type I are satisfied by edges from the graph G^∗[A_i∪B_i]. The path satisfying the terminal pair (bⁿ_i, a¹_i) has to travel downwards along B_i, use a blue edge and then finally travel downwards along A_i. Similarly, the path satisfying the terminal pair (b¹_i, aⁿ_i) has to travel upwards alongB_i, use a blue edge and then finally travel upwards along A_i. Since we can only orient each green edge in exactly one direction, it follows that both these paths must use the same blue edge, i.e., there exists an integerλ_i∈[n] such that the pathsA_i, B_i are oriented

away from, and towardsλ_i, respectively.

Lemma 4. For each i∈[k] and integers λ_i,μ_i, δ_i,_i as given by Lemma3we have that

– λ_i=μ_i, – δ_i=_i.

Proof. Fixi∈[k]. We just prove that λ_i =μ_i since the proof for the other case is similar. Consider the terminal pairs (dⁿ_i, a¹_i) and (d¹_i, aⁿ_i) of Type III. The only outgoing edges from D are to A∪C. However, A∪C has not outgoing edges.

Hence, the aforementioned terminal pairs are satisfied by edges fromG^∗[D∪A].

By Lemma3, we know that A_i is oriented away from λ_i and D_i is oriented towards μ_i. Hence, if μ_i > λ_i then the pair (dⁿ_i, a¹_i) is not satisfied, and if μ_i< λ_i then the pair (d¹_i, aⁿ_i) is not satisfied. Thus we haveλ_i=μ_i. Lemma 5. No yellow edge can be on a path satisfying any terminal pair of Type V or VI.

Proof. Fixj∈[k]. We just prove the lemma for the terminal pair (b¹_j, cⁿ_j) since the proof for other terminal pairs is similar. Suppose there is a yellow edge on some pathP satisfying the terminal pair (b¹_j, cⁿ_j). This yellow edge cannot be of Category I sinceDhas no incoming edges, and hence we could not have reached Din the first place starting fromb¹_j. However, this yellow edge also cannot be of Category II sinceH has no incoming edges and hence we could not have reached

H in the first place starting fromb¹_j.

A Tight Lower Bound for Steiner Orientation 75 Lemma 6. For each 1≤i, j≤k, we have that

– any path which satisfies the terminal pair(b¹_j, cⁿ_j)mustcontain the horizontal canonical pathQ^λ_j^j,

– any path which satisfies the terminal pair(f_iⁿ, g¹_i) must contain the vertical canonical pathP_i^δi.

Proof. Fix j ∈[k]. Consider the terminal pair (b¹_j, cⁿ_j), and letP be any path satisfying it. By Lemma5, we know that P cannot have any yellow edges. By Lemmas3 and4, we know thatB_j, C_j are oriented towards, and away fromλ_j, respectively. We claim that the first edge on P which leavesB_j is b^λ_j^j →v_1,j^1,λj. Clearly,P cannot have any edge fromB_j toA_j, sinceA_j has no outgoing edges.

Hence, the pathP is of the following type: the vertical upwards pathb¹_j →b²_j → . . . b^τ_j followed by the blue edge b^τ_j → v¹_1,j^,τ. Since B_j is oriented towards λ_j it follows that λ_j ≥τ. Ifλ_j > τ then by orientation of the black grid edges and orange edges (note that the splitting doesn’t really change the rows/columns level) it follows that P reachesC_j at a vertexc^ψ_j where λ_j > τ ≥ψ. However, C_j is oriented away from λ_j which contradicts that P is a path fromb¹_j to cⁿ_j. Hence, we have thatλ_j =τ=ψ. Therefore,P contains a subpath which starts at b^λ_j^j and ends at c^λ_j^j and all edges of this subpath (except the first and last blue edges) are contained in the graph G^{∗ k}_i=1V(G_i,j)

, i.e., P contains the canonical horizontal pathQ^λ_j^j.

The proof of the second part of the lemma is similar, and we omit the details

here.

Lemma 7. The instance(k, n,{S_i,j}1≤i,j≤k)of Grid Tiling has a solution.

Proof. We show that (δ_i, λ_j) ∈S_i,j for each 1≤i, j ≤k. This will imply that Grid Tiling has a solution.

Fix any 1 ≤i, j ≤k. By Lemma6, we know that the orientation G^∗ must contain the horizontal canonical pathQ^λ_j^j (to satisfy the pair (b¹_j, cⁿ_j)) and also the vertical canonical path P_i^δi (to satisfy the pair (f_iⁿ, g_i¹)). We now claim that the vertex v_i,j^δi,λj cannot be split: suppose to the contrary that it is split.

By Definition3, the path P_i^δi orients the edge v^δ_i,j,LB^i,λj −v_{i,j,T R}^δi,λj as v^δ_i,j,LB^i,λj ← v^δ_{i,j,T R}^i,λj . However, by Definition2, the pathQ^λ_j^j orients the edgev^δ_i,j,LB^i,λj −v_{i,j,T R}^δi,λj as v^δ_i,j,LB^i,λj → v_{i,j,T R}^δi,λj , which is a contradiction. Hence, the vertex v^δ_i,j^i,λj is not split, i.e., (δ_i, λ_j)∈S_i,j for each 1≤i, j≤k.

2.4 Obtaining thef(k)·N^o(k) Lower Bound

It is easy to see that the graphGhasO(n²k²) vertices and can be constructed in poly(n+k) time. Combining the two directions from Subsects.2.2and2.3, we get

76 R. Chitnis and A. E. Feldmann

a paramterized reduction fromGrid TilingtoSteiner Orientation. Hence, the W[1]-hardness of Steiner Orientationfollows from the W[1]-hardness of Grid Tiling[11]. Chen et al. [2] showed that, for any functionf, the existence of anf(k)·n^o(k)algorithm fork-Cliqueviolates ETH. There is a simple reduction (see Theorem 14.28 from [6]) fromk-Cliquetok×kGrid Tilingimplying the same runtime lower bound for the latter problem. Our reduction transforms the problem ofk×kGrid Tilinginto an instance of Steiner Orientationwith O(k) demand pairs. Composing the two reductions, we obtain that under ETH there is nof(k)·n^o(k)time algorithm for Steiner Orientation. Recall from Remark1that the graphGconstructed in theSteiner Orientation instance has genus 1, and hence thef(k)·n^o(k)lower bound holds for genus 1 graphs too.

This concludes the proof of Theorem1.

References

1. Arkin, E.M., Hassin, R.: A note on orientations of mixed graphs. Discret. Appl.

Math.116(3), 271–278 (2002)

2. Chen, J., Huang, X., Kanj, I.A., Xia, G.: Strong computational lower bounds via parameterized complexity. J. Comput. Syst. Sci.72(8), 1346–1367 (2006) 3. Chitnis, R., Esfandiari, H., Hajiaghayi, M.T., Khandekar, R., Kortsarz, G., Sed-

dighin, S.: A tight algorithm for strongly connected steiner subgraph on two ter- minals with demands. Algorithmica77(4), 1216–1239 (2017)

4. Chitnis, R., Feldmann, A.E., Manurangsi, P.: Parameterized approximation algo- rithms for directed Steiner network problems. CoRR, abs/1707.06499 (2017) 5. Chitnis, R.H., Hajiaghayi, M., Marx, D.: Tight bounds for planar strongly con-

nected Steiner subgraph with fixed number of terminals (and extensions). In:

SODA, pp. 1782–1801 (2014)

6. Cygan, M., Fomin, F.V., Kowalik, L., Lokshtanov, D., Marx, D., Pilipczuk, M., Pilipczuk, M., Saurabh, S.: Parameterized Algorithms. Springer, Cham (2015).

https://doi.org/10.1007/978-3-319-21275-3

7. Cygan, M., Kortsarz, G., Nutov, Z.: Steiner forest orientation problems. SIAM J.

Discret. Math.27(3), 1503–1513 (2013)

8. Hassin, R., Megiddo, N.: On orientations and shortest paths. Linear Algebra Appl.

114, 589–602 (1989). Special Issue Dedicated to Alan J. Hoffman

9. Impagliazzo, R., Paturi, R.: On the complexity of k-SAT. J. Comput. Syst. Sci.

62(2), 367–375 (2001)

10. Impagliazzo, R., Paturi, R., Zane, F.: Which problems have strongly exponential complexity? J. Comput. Syst. Sci.63(4), 512–530 (2001)

11. Marx, D.: On the optimality of planar and geometric approximation schemes. In:

FOCS, pp. 338–348 (2007)

12. Marx, D.: Can you beat treewidth? Theory Comput.6(1), 85–112 (2010) 13. Marx, D.: A tight lower bound for planar multiway cut with fixed number of

terminals. In: Czumaj, A., Mehlhorn, K., Pitts, A., Wattenhofer, R. (eds.) ICALP 2012. LNCS, vol. 7391, pp. 677–688. Springer, Heidelberg (2012).https://doi.org/

10.1007/978-3-642-31594-7 57

14. Marx, D., Pilipczuk, M.: Everything you always wanted to know about the param- eterized complexity of Subgraph Isomorphism (but were afraid to ask). In: STACS, pp. 542–553 (2014)

A Tight Lower Bound for Steiner Orientation 77 15. Marx, D., Pilipczuk, M.: Optimal parameterized algorithms for planar facility loca- tion problems using voronoi diagrams. In: Bansal, N., Finocchi, I. (eds.) ESA 2015.

LNCS, vol. 9294, pp. 865–877. Springer, Heidelberg (2015). https://doi.org/10.

1007/978-3-662-48350-3 72

16. Pilipczuk, M., Wahlstr¨om, M.: Directed multicut is W[1]-hard, even for four ter- minal pairs. CoRR, abs/1507.02178 (2015)

17. Pilipczuk, M., Wahlstr¨om, M.: Directed multicut is W[1]-hard, even for four ter- minal pairs. In: SODA, pp. 1167–1178 (2016)

Can We Create Large k -Cores by Adding