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Bang-Jensen, J.

Bellitto, Thomas

Lochet, William

Yeo, A.

English

arXiv

University of Bergen

The directed 2-linkage problem with length constraints∗J.Bang-Jensen†, T. Bellitto‡, W. Lochet§, A. Yeo¶AbstractThe weak 2-linkage problem for digraphs asks for a given digraph and verticess1, s2, t1, t2 whether D contains a pair of arc-disjoint paths P1, P2 such that Pi isan (si, ti)-path. This problem is NP-complete for general digraphs but polynomiallysolvable for acyclic digraphs [8]. Recently it was shown [3] that if D is equipped witha weight function w on the arcs which satisfies that all edges have positive weight,then there is a polynomial algorithm for the variant of the weak-2-linkage problemwhen both paths have to be shortest paths in D. In this paper we consider the unitweight case and prove that for every pair constants k1, k2, there is a polynomialalgorithm which decides whether the input digraph D has a pair of arc-disjointpaths P1, P2 such that Pi is an (si, ti)-path and the length of Pi is no more thand(si, ti) + ki, for i = 1, 2, where d(si, ti) denotes the length of the shortest (si, ti)-path. We prove that, unless the exponential time hypothesis (ETH) fails, there isno polynomial algorithm for deciding the existence of a solution P1, P2 to the weak2-linkage problem where each path Pi has length at most d(si, ti) + c log1+ǫ n forsome constant c.Keywords: (arc)-disjoint paths, shortest disjoint paths, acyclic digraph,linkage1 IntroductionNotation throughout this paper follows [2, 1]. We use [0, i] to denote the set {0, 1, 2, . . . , i}.Problems concerning disjoint paths with prescribed end vertices in graphs and digraphsplay an important role in many combinatorial problems. Among the most important such∗Research supported by the Danish research council under grant number DFF-7014-00037B.†Department of Mathematics and Computer Science, University of Southern Denmark, Odense, Den-mark (email: jbj@imada.sdu.dk).‡Department of Mathematics and Computer Science, University of Southern Denmark, Odense, Den-mark (email: bellitto@imada.sdu.dk).§University of Bergen, Norway (email: william.lochet@gmail.com).¶Department of Mathematics and Computer Science, University of Southern Denmark, Odense, Den-mark (email: yeo@imada.sdu.dk).1problems are the k-linkage problem and the weak k-linkage problem which we for-mulate below for digraphs.k-linkageInput: A digraph D = (V,A) and distinct vertices s1, s2, . . . , sk, t1, t2, . . . , tkQuestion: Does D contain k vertex-disjoint paths P1, P2, . . . , Pk such that Pi is an(si, ti)-path for i ∈ [k]?weak k-linkageInput: A digraph D = (V,A) and not necessarily distinct verticess1, s2, . . . , sk, t1, t2, . . . , tkQuestion: Does D contain k arc-disjoint paths P1, P2, . . . , Pk such that Pi is an(si, ti)-path for i ∈ [k]?It is an easy and well-known fact that the k-linkage problem and the weak k-linkage problems are polynomially equivalent in the sense that one can easily reduceone to the other by a polynomial reduction see e.g. [1, Chapter 10].A famous and very important result by Robertson and Seymour [13] shows that thecorresponding linkage problems for undirected graphs are polynomially solvable for fixedk and that the problems are in fact FPT, meaning that there is an algorithm for eachproblem whose running time is of the form O(f(k)nc) for some computable function fand a constant c. It was shown in [13] that c = 3 will do and this has been improved toc = 2 in [10].For directed graphs the situation is quite different: Fortune, Hopcroft and Wyllie [8]proved that already the 2-linkage and the weak 2-linkage problems are NP-complete.They also showed that if the input is an acyclic digraph, then both linkage problems arepolynomially solvable when the number of terminals is fixed (not part of the input).Theorem 1.1. [8] The weak k-linkage problem in acyclic digrpahs is solvable in timeO(k!nk+2).Eilam-Tzoreff [7] proved that for undirected graphs the 2-linkage problem is alsopolynomially solvable if each edge of the input graph is equipped with a positive lengthand the goal is to check whether there is a solution P1, P2 such that Pi is a shortest(si, ti)-path for i = 1, 2. This result has recently been extended by Gottschau et al. in[9] and by Kobayashi and Sako in [11] to the case where edges of length zero are allowed.The problem has also been studied on digraphs by Bérczi and Kobayashi [3] whorecently proved the following:2Theorem 1.2. There exists a polynomial algorithm that runs in time |V |O(a1+a2) forthe following problem as well as its arc-version: Given a digraph D = (V,A), ver-tices s1, s2, t1, t2 ∈ V , a weight function w on A such that the weight of every di-rected cycle is positive and numbers a1 and a2; decide whether D has disjoint pathsP 11 , . . . , P1a1, P 21 , . . . , P2a2such that P ij is a shortest (si, ti)-path for i = 1, 2 and 1 6 j 6 ai.There are several other papers dealing with shortest path version of the k-linkageproblem, see e.g. [4, 6, 12].In this paper we consider the following variant of the weak-2-linkage problem wherethe paths do not have to be shortest paths (in terms of number of arcs) but there is abound on how far from being shortest they can be. Throughout the paper, we denote byd(u, v) the length of the shortest (u, v)-path.short weak 2-linkage SW2L(D, s1, s2, t1, t2, k1, k2)Input: A digraph D = (V,A), vertices s1, s2, t1, t2 ∈ V and natural numbers k1, k2Question: Is there a pair of arc-disjoint paths P1, P2 such that Pi is an (si, ti)-pathand |A(Pi)| ≤ d(si, ti) + ki?Clearly this problem is NP-complete when k1, k2 = n−1 since that puts no restrictionon P1, P2 in a solution. The main result of our paper is that when k1, k2 are bothconstants the short weak 2-linkage problem can be solved in polynomial time. Wealso prove that the problem is NP-complete when there is no restriction on the length ofone of the paths. Finally, we show that under the exponential time hypothesis, there isno polynomial algorithm for the short weak 2-linkage when k1, k2 ∈ O(log1+ǫ n) nomatter how small the value of ǫ is as long as it is positive.2 2-linkage with almost shortest pathsLet D be a digraph and s a vertex. The reach of s is the set of vertices x such that thereexists a path from s to x in D. Using breadth-first-search we can partition the reach of avertex s into levels, such that Lis denotes the set of vertices x such that the shortest pathfrom s to x is of length i. We say that an arc uv is between two levels if, d(s, u) < ∞and d(s, v) = d(s, u) + 1.Suppose s1 and s2 are fixed, let A1 denote the set of arcs between two consecutivelevels from s1 and A2 the set of arcs between two consecutive levels from s2. Note thatboth A1 and A2 induce acyclic digraphs. Furthermore, an arc uv is in Ai if and onlyif some shortest (si, v)-path uses the arc uv. We will use the following lemma and itsanalogous for A2:Lemma 2.1. If P is a path from s1 to t1 of length at most d(s1, t1) + k, then P uses atmost k arcs not belonging to A1.3Proof. Every path from s1 to t1 must visit every level with index smaller than d(s1, t1)at least once. Moreover, it must use an arc of A1 to go from one level to the next, whichends the proof. Theorem 2.2. For every fixed choice of positive integers k1, k2 the problem short weaklinkage problem with input [D, s1, s2, t1, t2, k1, k2] is polynomially solvable.Proof: Let k = max{k1, k2}, n = |V | and m = |A|. We shall describe an al-gorithm that runs in time nO(k) for the problem. Let E1 = (v1, u1), . . . , (vi, ui) andE2 = (z1, w1), . . . , (zj , wj) be two ordered subsets of A of at most k arcs each and thatavoid A1 and A2 respectively. Recall that A1 denotes the set of arcs between two consec-utive levels from s1 and A2 the set of arcs between two consecutive levels from s2. LetdAℓ(x, y) denote the distance from x to y in the digraph induced by the arcs in Aℓ. Wecall E1 and E2 feasible if the following holds.• dA1(s1, v1)+1+dA1(u1, v2)+1+dA1(u2, v3)+ · · ·+dA1(ui−1, vi)+1+dA1(ui, t1) 6dD(s1, t1) + k1• dA2(s2, z1)+1+dA2(w1, z2)+1+dA2(w2, z3)+ · · ·+dA2(wj−1, zi)+1+dA2(wj , t2) 6dD(s2, t2) + k2We will describe an O(nC) algorithm for some constant C, which decides if there existsa solution P1, P2 to the problem such that for ℓ = 1, 2, Pℓ only uses arcs of Aℓ and Eℓ. Tosolve the general question, we only need to run this algorithm for all feasible choices ofE1 and E2. We note that there are less than (mk)2 ≤ n4k (as m ≤ n2) ways of choosingE1 and E2. So the algorithm only needs to be run at most O(n4k) times.Let now E1 and E2 be fixed. We create the digraph D′ by adding the verticess′1, t′1, s′2, t′2 to D and the following paths:• A path from s′1 to every vertex x ∈ {s1, u1, u2, . . . , ui} of length dD(s1, x) + 1.• A path from every vertex x ∈ {t1, v1, v2, . . . , vi} to t′1 of length dD(x, t1) + 1.• A path from s′2 to every vertex x ∈ {s2, w1, w2, . . . , wj} of length dD(s2, x) + 1.• A path from every vertex x ∈ {t2, z1, z2, . . . , zj} to t′2 of length dD(x, t2) + 1.All the internal vertices on all the above added paths are distinct and new vertices.However, the length of all these paths cannot be grater than n and the number of thesepaths is 2i + 2j + 4 = O(n). Hence, the number of vertices we add is still polynomial inthe size of the initial graph. Let P ∗1 be a (s′1, t′1)-path in D′ and let x be the first vertexon P ∗1 from {s1, u1, u2, . . . , ui} and let y be the last vertex on P∗1 from {t1, v1, v2, . . . , vi}.4Then the subpath from s′1 to x has length dD(s1, x)+ 1 and the subpath from y to t′1 haslength dD(y, t1) + 1. This implies that the length of P∗1 is the following.(dD(s1, x)+1)+(dD(y, t1)+1)+dD(x, y) = dD(s1, x)+dD(x, y)+dD(y, t1)+2 ≥ dD(s1, t1)+2As there exists an (s′1, t′1)-path of length dD(s1, t1) + 2 in D′ (using the arcs s′1s1 andt1t′1 and a shortest (s1, t1)-path in D), we note that the shortest (s′1, t′1)-path in D′ haslength exactly dD(s1, t1) + 2. Furthermore if the subpath of P∗1 from x to y only usesarcs from A1 then it has length dD(x, y) and we have equality everywhere in the aboveequation, which implies that the length of P ′1 is dD(s1, t1) + 2 = dD′(s′1, t′1). Analogouslyif the length of P ∗1 is dD′(s′1, t′1) then the subpath from x to y only uses arcs from A1.Clearly the analogous result also holds for a shortest path from s′2 to t′2. By Theorem1.2, we know that we can determine in polynomial time if there exist i+1 shortest pathsfrom s′1 to t′1 and j + 1 shortest paths from s′2 to t′2 such that all i + j + 2 paths arearc-disjoint.We claim that if such paths exist, then the answer to our instance of the short weaklinkage problem is true and if there is no such i + j + 2 arc-disjoint paths for anyfeasible choice of E1 and E2, then the answer to our instance is false.First assume that we found i+ j + 2 arc-disjoint paths for some feasible choice of E1and E2. Now remove all vertices in V (D′)\V (D) from the (s′1, t′1)-paths and add the arcsE1. Note that the outdegree of s1 will be one (as it belongs to one of the paths) and theindegree of s1 will be zero. Analogously the indegree of t1 will be one and the outdegreewill be zero. All other vertices will have indegree and out degree equal to each other(if they belong to k paths then the indegree and outdegree will both be k). Thereforethe arcs in the resulting subdigraph form a path from s1 to t1 plus possibly a numberof cycles. As the total number of arcs in the subdigraph is less than d(s1, t1) + k1 thepath from s1 to t1 (after discarding any cycles) also has length less than d(s1, t1) + k1.Indeed, the graph contains all the arcs of E1 (i arcs) and all the arcs of the i+1 paths oflength d(s1, t1) + 2 except those of D′ \D. The number of arcs of D′ \D in those pathsis N1 =∑x∈{s1,u1,...,ui}(d(s1, x) + 1) +∑x∈{v1,...,vi,t1}(d(x, t1) + 1). Let us set u0 = s1and vi+1 = t1. The total number of arcs in our graph is:i+ (d(s1, t1) + 2)× (i + 1)−N1 = i+i∑ℓ=0[d(s1, t1)− d(s1, uℓ)− d(vℓ+1, t1)]6 i+i∑ℓ=0d(uℓ, vℓ+1)6 d(s1, t1) + k1 by definition of the feasibility of E1Analogously we find a path from s2 to t2 of length less than d(s2, t2) + k2. By our5construction these paths are arc-disjoint, completing the proof of one direction.Now assume that there exist arc-disjoint paths P1 and P2 in D, such that Pℓ is a(sℓ, tℓ)-path of length less than d(sℓ, tℓ)+kℓ. Let Eℓ be the arcs on Pℓ that do not belongto Aℓ (ℓ ∈ [2]). Hence, E1 and E2 are feasible. Using these E1 and E2 and the subpathsof P1 and P2 after removing the arcs in E1 and E2 we note that we can obtain the desiredi+ j + 2 arc-disjoint paths in D′. This completes the proof.D′ has size O(n2), so the existence of this path Q can be checked in polynomial time,and the overall problem can be solved in time nO(k).3 Non-polynomial casesThis section is devoted to the proof of the NP-completeness of the problem of semi-shortweak 2-linkage:semi-short weak 2-linkage SSW2L(D, s1, s2, t1, t2, k)Input: A digraph D = (V,A), vertices s1, s2, t1, t2 ∈ V and a natural number kQuestion: Is there a pair of arc-disjoint paths P1, P2 such that P1 is an (s1, t1)-pathof length |A(P1)| ≤ d(s1, t1) + k and P2 is an (s2, t2)-path?Eilam-Tzoreff proved in [7] that the problem of SSW2L is NP-complete if P1 has to bea shortest (s1, t1)-path, which is the case k = 0. We observe that this can be generalizedto all values of k.Theorem 3.1. The semi-short weak 2-linkage problem is NP-complete for all values ofk.Proof. The proof is by reducing SSW2L for k = 0 to SSW2L with any value of k. Letk be fixed and let D, s1, s2, t1, t2, 0 be an instance of SSW2L. Let us now create D′ byreplacing every arc of D by a directed path of length k+1. Let P ′1 and P′2 be (s1, t1) and(s2, t2)-paths respectively in D′.Note that every path P ′ from s1 to t1 in D′ is thus the image of a path P from s1 to t1in D by subdivision of the edges of P , and P ′ is (k+1) times longer than P . Let P1 andP2 be the preimage of P′1 and P′2. Hence, if |A(P′1)| 6 dD′(s1, t1)+k < dD′(s1, t1)+k+1,then |A(P1)| < dD(s1, t1) + 1, which means that P1 is a shortest (s1, t1)-path in D.It is also easy to see that P ′1 and P′2 arc disjoint if and only if P1 and P2 are. Thus,(P ′1, P′2) is a solution of SSW2L(D′, s1, s2, t1, t2, k) if and only if P1 and P2 are solutionsof SSW2L(D, s1, s2, t1, t2, 0). This proves the NP-completeness of SSW2L for all k. Note that our algorithm in Theorem 2.2 is only polynomial for constant k, so it isnatural to ask whether we could replace constant k by some function of n.6Recall the so-called Exponential Time Hypothesis (ETH) which in one of many for-mulations says that there exist a real number δ > 0 so that no algorithm can solve 3-SATinstances with m clauses in time O(2δm). This modification of the commonly knownversion of ETH is in fact equivalent to that, see e.g. [5, Theorem 14.4].Theorem 3.2. Assuming that ETH is true, then for every ǫ > 0 there is no polynomialalgorithm for weak short 2-linkage problem when the input D is a digraph on nvertices and k1, k2 = Θ(log1+ǫ n).Proof: We give the proof when k1, k2 = log1+ǫN , where N is the number of verticesin the input digraph. Let [D, s1, s2, t1, t2] be an instance of the weak 2-linkage problemand let n be the number of vertices of D. Let ǫ′ be defined such that 2n11+ǫ′=⌈2n11+ǫ⌉and note that ǫ′ ≤ ǫ and that ǫ′ > 0 when n is large enough. Construct a new digraphD′ by adding an independent set of size 2n11+ǫ′− n so that the resulting digraph hasN = 2n11+ǫ′vertices, implying that we have (logN)1+ǫ′= n. Clearly every pair of arc-disjoint (s1, t1)-, (s2, t2)-paths, P1, P2 in D′ use only vertices from D and hence each oftheir lengths is at most n = log1+ǫ′N ≤ log1+ǫN so Pi is at most ki longer that theshortest (si, ti)-path for i = 1, 2.Suppose there is an algorithm for the weak short 2-linkage problem that runs intime O(N c) for inputs on N vertices when k1, k2 = log1+ǫ N for some fixed constanctc > 0. Then we haveN c = (2n11+ǫ′)c= 2c·n11+ǫ′< 2δ·nfor every fixed constant δ > 0 provided that n is large enough. This means that we cansolve the general weak 2-linkage problem in time O(2δ·n) for every δ > 0.To see that this contradicts the ETH, we just have to observe that the reduction from3-SAT to the 2-linkage problem in [8] (see also [1, Section 10.2]) converts a 3-SAT formulawith n variables and m clauses into an instance of 2-linkage with at most dm verticeswhere d is a constant (it is at most 61). Furthermore, the 2-linkage problem for a digraphon n vertices reduces to the weak 2-linkage problem on a digraph with twice as manyvertices.74 RemarksSlivkins [14] proved that the weak k-linkage problem is W [1]-hard for acyclic digraphs.We can prove that the same holds for short weak k-linkage in acyclic digraphs.Indeed, consider an instance of weak k-linkage on an acyclic digraph D and considera topological ordering v1, ..., vn of the vertices of D, i.e. an ordering such that for everyarc vivj , we have j > i. Let us build D′ from D by replacing every arc vivj in D by adirected path of length (j − i) in D′. Hence, D′ is still acyclic and every walk between avertex vi and a vertex vj in D is now replaced by a walk of length j − i in D′ and is thusa shortest walk. Therefore, a solution of short weak k-linkage in D′ immediately providesa solution of weak k-linkage in D.References[1] J. Bang-Jensen and G. Gutin. Digraphs: Theory, Algorithms and Applications.Springer-Verlag, London, 2nd edition, 2009.[2] J. Bang-Jensen and G Gutin. Classes of Directed Graphs. Springer Monographs inMathematics. Springer Verlag, London, 2018.[3] K. Bérczi and Y. Kobayashi. The directed disjoint shortest paths problem. In25th Annual European Symposium on Algorithms, ESA 2017, September 4-6, 2017,Vienna, Austria, pages 13:1–13:13, 2017.[4] A. Björklund and T. Husfeldt. Shortest two disjoint paths in polynomial time. InAutomata, Languages, and Programming - 41st International Colloquium, ICALP2014, Copenhagen, Denmark, July 8-11, 2014, Proceedings, Part I, pages 211–222,2014.[5] M. Cygan, F.V. Fomin, L. Kowalik, D. Lokshtanov, D. Marx, M. Pilipczuk,M. Pilipczuk, and S. Saurabh. Parameterized Algorithms. Springer, 2015.[6] É. Colin de Verdière and A. Schrijver. Shortest vertex-disjoint two-face paths inplanar graphs. ACM Trans. Algorithms, 7(2):19:1–19:12, 2011.[7] T. Eilam-Tzoreff. The disjoint shortest paths problem. Discrete Applied Mathemat-ics, 85(2):113–138, 1998.[8] S. Fortune, J.E. Hopcroft, and J. Wyllie. The directed subgraph homeomorphismproblem. Theor. Comput. Sci., 10:111–121, 1980.[9] M. Gottschau, M. Kaiser, and C. Waldmann. The undirected two disjoint shortestpaths problem. Oper. Res. Lett., 47(1):70–75, 2019.8[10] K. Kawarabayashi, Y. Kobayashi, and B. A. Reed. The disjoint paths problem inquadratic time. J. Comb. Theory, Ser. B, 102(2):424–435, 2012.[11] Y. Kobayashi and R. Sako. Two disjoint shortest paths problem with non-negativeedge length. Oper. Res. Lett., 47(1):66–69, 2019.[12] Y. Kobayashi and C. Sommer. On shortest disjoint paths in planar graphs. DiscreteOptimization, 7(4):234–245, 2010.[13] N. Robertson and P.D. Seymour. Graph minors. XIII: The disjoint paths problem.J. Combin. Theory Ser. B, 63:65–110, 1995.[14] A. Slivkins. Parameterized tractability of edge-disjoint paths on directed acyclicgraphs. SIAM J. Discrete Math., 24(1):146–157, 2010.9

The directed 2-linkage problem with length constraints

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The directed 2-linkage problem with length constraints

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