Four Shortest Vertex-Disjoint Paths in Planar Graphs Jeff Erickson and Yipu Wang University of Illinois at Urbana-Champaign {jeffe,ywang}@illinois.edu 0 Abstract. Let G be an edge-weighted planar graph with k terminal vertices s, t,..., s k, t k. The minimum-sum vertex-disjoint paths problem asks for a set of pairwise vertex-disjoint simple paths of minimum total length, where the ith path connects s i to t i. Even when all terminals lie on a single face, efficient algorithms for this problem are known only for fixed k. We describe the first polynomialtime algorithm for the case of four arbitrary terminal pairs on a single face. 0 Introduction In the vertex-disjoint paths problem, we are given a graph G along with k vertex pairs (s, t ),..., (s k, t k ), and we want to find k pairwise vertex-disjoint paths connecting each node s i to the corresponding node t i. The vertices s,..., s k, t,..., t k are called terminals. The vertex-disjoint paths problem is a special case of multi-commodity flows with applications in VLSI design [, ] and network routing [, ]. This problem is NP-hard [] and remains so even if G is undirected planar [] or if G is directed and k = []. On the other hand, it can be solved in polynomial time if G is undirected and k is bounded [, ] or if G is directed acyclic and k is bounded []. Furthermore, the problem is fixed-parameter tractable with respect to the parameter k in directed planar graphs [, ]. We focus on an optimization version of the vertex-disjoint paths problem, where the goal is to minimize the total length of the paths. This version of the problem has also been considered in the context of network routing, where we want to minimize the amount of energy required to send packets [, ]. In the k-min-sum problem, we are given a graph G, in which every edge e has a non-negative real length l(e), and k pairs of vertices (s, t ),..., (s k, t k ). We want to find k vertex-disjoint paths P,..., P k where each path P i is a path from s i to t i and the total length k i= l(p i ) is as small as possible. (Here l(p i ) = e P i l(e).) Middendorf and Pfeiffer [] proved that the k-min-sum problem is NP-hard when the parameter k is part of the input, even in undirected -regular plane graphs. Not much is known about the complexity of the planar k-min-sum problem for fixed k. In fact, no non-trivial algorithms or hardness results are known for either the -min-sum problem in directed planar graphs or the -min-sum problem in undirected planar graphs, even when all terminals are required to lie on a single face. Polynomial-time algorithms for the planar k-min-sum problem are known for arbitrary k when all k terminals terminals lie on a single face, in one of two patterns. This work was partially supported by NSF grant CCF-0.
Jeff Erickson and Yipu Wang 0 0 0 In a parallel instance, the terminals appear in cyclic order s,..., s k, t k,..., t, and an a serial instance, the terminals appear in cyclic order s, t, s, t,..., s k, t k. Even in directed planar graphs, parallel instances of k-min-sum can be solved using a straightforward reduction to minimum-cost flows []; in fact, this special case can be solved in O(kn) time. A recent algorithm of Borradaile, Nayyeri, and Zafarani [] solves any serial instance of k-min-sum in an undirected planar graph in O(kn ) time. If we allow arbitrary patterns of terminals, fast algorithms are known for only very small values of k. Kobayashi and Sommer [] describe two algorithms, one running in O(n log n) time when k = and all four terminals are covered by at most two faces, the other running in O(n log n) time when k = when all terminals are incident to a single face. Colin de Verdière and Schrijver [] describe an O(kn log n)-time algorithm for directed planar graphs where all sources s i lie on one face and all targets t i lie on another face. Finally, if k, every planar instance of k-min-sum with all terminals on the same face is either serial or parallel. Zafarani [] proved an important structural result for the planar k-min-sum problem. Consider an undirected edge-weighted plane graph G with vertices s, t,..., s k, t k on its outer face. Let Q, Q,..., Q k be the shortest vertex-disjoint paths in G connecting all k terminal pairs, and let P, P,..., P k be the shortest vertex-disjoint paths in G connecting every pair except s k, t k, where the subscript on each path indicates which terminals it connects. Zafarani s Structure Theorem states that if two paths P i and Q j cross, then i = j. We describe the first polynomial-time algorithm to solve the -min-sum problem in undirected planar graphs with all eight terminals on a common face. If the given instance is parallel or serial, it can be solved using existing algorithms; otherwise, the terminals can be labeled s, s, s, t, s, t, t, t in cyclic order around their common face. To solve these instances, our algorithm first computes a solution to the -minsum problem for the terminal pairs s t, s t, s t, using the serial-instance algorithm of Borradaile et al. We identify a small set of key anchor vertices where the -minsum solution intersects the -min-sum solution we want to compute. For each possible choice of anchor vertices, we connect these vertices to the terminals using parallel minsum problems in three carefully constructed subgraphs of G. Overall, our algorithm runs in O(n ) time. Our characterization of the interaction between the -min-sum and - min-sum solutions, which both relies on and extends Zafarani s Structure Theorem [], is the main technical contribution of the paper. Preliminaries For any integer N, let [N] denote the set {,,..., N}. For any plane graph G, we write G to denote the boundary of the outer face of G; we also informally call G the boundary of G. Without loss of generality, we assume that G is a simple cycle. Our algorithms search for pairwise disjoint walks with minimum total length that connect corresponding terminals, rather than explicitly seeking simple paths. Because all edge lengths are non-negative, the shortest set of walks will of course consist of simple paths. The length of a walk w in an edge-weighted graph, which we denote l(w),
Four Shortest Vertex-Disjoint Paths in Planar Graphs 0 0 0 is the sum of the lengths of its edges, with appropriate multiplicity of w is not a simple path. The total length of any set of walks W, which we denote l(w) = w W l(w), is just the sum of their lengths. Two walks meet or touch if they have at least one vertex in common. For any path P and any vertices u and v on that path, we write P[u, v] to denote the subpath of P from u to v. Similarly, let P[u, v) denote the subpath of P from u to the predecessor of v, let P(u, v] denote the subpath of P from the successor of u to v, and let P(u, v) denote the subpath of P from the successor of u to the predecessor of v; these subpaths could be empty. The reversal of any path P is denoted rev(p). The concatenation of two paths P and P is denoted P P. Our -min-sum algorithm relies on a black-box subroutine to solve parallel instances of -min-sum and -min-sum. As observed by van der Holst and de Pina, any parallel instance of k-min-sum can be solved in polynomial time by reduction to minimum-cost flow problem []. In fact, these instances can reduced in O(n) time to a planar instance of minimum-cost flow, by replacing each vertex with a clockwise directed unit-capacity cycle, as described by Colin de Verdiére and Schrijver [] and Kaplan and Nussbaum []. The resulting minimum-cost flow problem can then be solved O(kn) time by performing k iterations of the classical successive shortest path algorithm [,0,], using the O(n)-time shortest-path algorithm of Henzinger et al. [] at each iteration. We omit further details from this version of the paper. To simplify our presentation, we assume that our given instance of -min-sum and every instance of -min-sum and -min-sum considered by our algorithm has a unique solution. If necessary, these uniqueness assumptions can be enforced with high probability using the isolation lemma of Mulmuley, Vazirani, and Vazirani []. We omit further details from this version of the paper. Structure Let G be an undirected plane graph with non-negative edge lengths, and let s, s, s, t, s, t, t, t be eight distinct vertices in cyclic order around the outer face, as illustrated in Figure. Let Q = {Q,..., Q } denote the unique optimal solution to this -minsum instance, where each path Q i connects s i to t i, and let P = {P, P, P } denote the unique optimal solution to the induced -min-sum problem that omits the demand pair s t, where again each path P i connects s i to t i. We can compute P in O(n log n) time using the algorithm of Kobayashi and Sommer [], or in O(n ) time using the more general algorithm of Borradaile et al. []. The paths in P divide G into four regions, as shown in Figure (a). Let X be the unique region adjacent to all the paths in P. For each index i, let C i denote the subpath of G from s i to t i that shares no edges with X, let R i denote the closed region bounded by P i and C i, and let R denote the half-open region R i i \ P i.. How P intersects Q We begin by proving several simple structural properties of the -min-sum solution Q that will help us compute it quickly once we know the -min-sum solution P. Figure (b) shows a typical structure for Q.
Jeff Erickson and Yipu Wang t s t s s R R P P t s P P Q Q t X t s Q t s P s t P s t R (a) Q (b) Fig.. (a) Terminals, paths in P, and the regions they define. (b) Typical structure of Q. Lemma.. If P i crosses Q j, then i = j. In particular, Q does not cross any path in P. Proof. This is an immediate consequence Zafarani s structure theorem []. Lemma.. Q and Q do not meet P. Similarly, Q does not meet P or P. Proof. Q separates s, t, s, t from s and t. Thus, Lemma. implies that Q separates Q and Q from P, and Q separates P and P from Q. Lemma.. Q lies entirely in R. Proof. For the sake of argument, suppose Q leaves R. Let q be any component of Q \ R, as shown at the bottom of Figure. The endpoints x and y of q must lie on P ; let p denote the subpath P [x, y]. Define two new paths P = P \ p q and Q = Q \ q p. Both P and Q are walks from s to t. Let P = {P, P, P } and Q = {Q, Q, Q, Q }. Lemma. implies that q Q does not meet P or P, so P does not meet P or P, which implies that the walks in P 0 are pairwise vertex-disjoint. On the other hand, subpath p lies inside the disk enclosed by P C, so Lemmas. and. imply that Q does not meet Q, Q, or Q. It follows that the walks in Q are also pairwise vertex-disjoint. The unique optimality of P implies l(p) < l(p ), and the unique optimality of Q implies l(q) < l(q ). But l(p) + l(q) = l(p ) + l(q ), so we have a contradiction. Lemma.. Every component of Q \ R meets P, and every component of Q \ R meets P. Proof. For the sake of argument, suppose some component q of of Q \ R does not meet P, as shown at the top of Figure. We derive a contradiction using a similar exchange argument to Lemma.. The endpoints x and y of q must lie on P ; let p denote the subpath P [x, y]. Define two new paths P = P \ p q and Q = Q \ q p. Clearly P and Q are both walks from s to t. Let P = {P, P, P } and Q = {Q, Q, Q, Q }. Lemma. and our assumption that q does not meet P imply that the walks in P are pairwise vertexdisjoint. On the other hand, p lies in the disk enclosed by P C, which implies
Four Shortest Vertex-Disjoint Paths in Planar Graphs t s s p t q s q t s p t Fig.. An impossible configuration of optimal paths, for the proofs of Lemmas. and.. that the walks in Q are also pairwise vertex-disjoint. The optimality of P implies that l(p) < l(p ), and the optimality of Q implies that l(q) < l(q ), but clearly l(p)+l(q) = l(p ) + l(q ), so we have a contradiction. A symmetric argument proves that every component of Q \ R meets P. Lemma.. Either Q R or Q R or both. Proof. For the sake of argument, suppose Q leaves R and Q leaves R. Let S be the closed region bounded by Q C and let S be the closed region bounded by Q C. We call each component of S \ R a left finger, and each component of S \ R a right finger. Lemma. and the Jordan curve theorem imply that each finger is a topological 0 disk that intersects both P and P. Thus, the fingers can be linearly ordered by their intersections with P from s to t (from bottom to top in Figure ). Because Q is a simple path, the fingers intersect Q in the same order. Without loss of generality, suppose the last finger in this order is a right finger. Let s be the last left finger, and let s be the right finger immediately after s. Let w be the last node of P (closest to t ) that lies in s, and let y be the last node of P (closest to t ) that that lies in s. We define four subpaths p = P [w, t ], q = Q [w, t ], p = P [s, y], and q = Q [s, y], as shown on the left of Figure. (Paths p and q could enclose more than one right finger.) t s s q w q p p y z s 0 s x t Fig.. Another impossible configuration of optimal paths, for the proof of Lemma.. Now exchange the subpaths p q and p q to define four new walks P = P \ p q, Q = Q \ q p, P = P \ p q, and Q = Q \ q p. Finally, let P = {P, P, P } and Q = {Q, Q, Q, Q }. As in previous lemmas, we argue that P and Q are sets of vertex-disjoint walks.
Jeff Erickson and Yipu Wang Lemma. implies that Q does not cross P or P, and trivially Q does not cross Q. Thus, none of the paths Q, P, Q touches any of the paths p, q, p, q. It follows that P does not touch either P or P, and similarly, Q and Q does not touch either Q or Q. We define two more auxiliary nodes x and z, as shown on the right in Figure. Let x be the first vertex of P also on Q. Vertex y must precede x on P, because x s and y s. Let z be the first vertex of P also on q. Vertex w must precede z on P, because w s and z s. Trivially, q does not meet q, and P \ p does not meet P \ p. Any left finger 0 formed from q must succeed s. Because s is the last left finger, q does not form any left fingers and does not touch P. By definition, z is the first node of P also on q. On the other hand, all vertices of P \ p (except w) precede w on P, which in turn strictly precedes z on P, so P \ p is disjoint from q. We conclude that P does not meet P, implying that the walks in P are vertex-disjoint: Trivially, p does not meet p, and Q \ q does not meet Q \ q. Since q does not meet P, x is the first vertex of P also on Q \ q. On the other hand, all vertices of p (except y) precede y on P, which in turn strictly precedes x on P, so Q \ q is disjoint from p. Any right finger whose boundary contains a subpath of Q \ q must precede s, and any right finger that meets p must succeed s. Because no right fingers lie strictly between s and s, the path Q \ q does not form any right fingers that meet p. We conclude that Q does not meet Q, which implies that the walks in Q are vertex-disjoint. Finally, as usual, the optimality of P implies that l(p ) < l(p ), and the optimality of Q implies that l(q ) < l(q ), but clearly l(p) + l(q) = l(p ) + l(q ). Without loss of generality, in light of Lemma., we assume for the rest of the paper that P and Q are disjoint, and thus Q R. 0 0. Subgraph Solutions Our algorithm solves several parallel instances of k-min-sum inside certain subgraphs of G. To prove that our algorithm is correct, we need to argue that the subgraph solutions coincide exactly with portions of the desired global solution. As an intermediate step, we first show that the subgraph solutions interact with the global solution in a limited way. Lemma.. Let (G, {s i, t i i [k]}) be a planar instance of k-min-sum, with all terminals s i and t i on G, whose unique solution is Q = {Q,..., Q k }. Let S be a subset of [k] such that the induced planar min-sum instance (G, {s i, t i i S }) is parallel. Let H be a subgraph of G such that () Q i H if and only if i S, and () for all distinct i, j S, no component of Q i H separates components of Q j H from each other in H. For each index i S, let u i and v i be vertices of Q i H such that Q i [u i, v i ] H. Finally, suppose (H, {u i, v i i S }) is a parallel planar min-sum instance, whose unique solution is Π = {π i i S }. Then for all indices i, j S, if i j, then π i does not cross Q j.
Four Shortest Vertex-Disjoint Paths in Planar Graphs Proof. First we establish some notation and terminology. Let κ = S, and re-index the terminals so that S = [κ] and the counterclockwise order of terminals around the outer face of H is u,..., u κ, v κ,..., v. Fix an index i such that i < κ, and consider the paths Q i and π i+. Let C ( ceiling ) denote the path in G from s i to t i that does not contain s i+ or t i+, and let A be the closed region bounded by C and Q i. A point in G is above Q i if it lies in A \ Q i and below Q i if it does not lie in A. Similarly, let F ( floor ) denote the path in H from u i+ to v i+ that does not con- tain u i or v i, and let B be the closed region bounded by F and π i+. A point in H is be- 0 low π i+ if it lies in B \ π i+ and above π i+ if it does not lie in B. Paths Q i and π i+ also divide the interior of G into connected regions, exactly one of which has the entire path C on its boundary; call this region U. Finally, let Q i denote the unique path in G from s i to t i such that C Q i is the boundary of U. Every point on Q i lies on or above Q i, and our assumption () implies that every point in Q i H lies on or above π i+. Thus, intuitively, Q i is the upper envelope of Q i and π i+. In particular, Q i = Q i if and only if Q i and π i+ are disjoint. Similarly, paths Q i and π i+ divide the interior of H into connected regions, exactly one of which contains F on its boundary; call this region L. Let π i+ denote the unique path in H from u i+ to v i+ such that D π i+ is the boundary of L. Assumption () implies that every point on π i+ lies on or below both π i+ and Q i. Thus, intuitively, π i+ is the lower envelope of Q i and π i+. In particular, π i+ = π i+ if and only if Q i and π i+ are disjoint. Each component of Q i \ Q i is an open subpath of π i+ that lies entirely above Q i and therefore is not contained in π i+. Similarly, every component of π i+ \ π i+ is an open subpath of Q i H that lies entirely below π i+ and therefore is not contained in Q i. It follows that l(q i ) + l(π i+ ) l(q i) + l(π i+ ). Finally, let Q = {Q,..., Q κ, Q κ,..., Q k } and Π = {π, π,..., π κ}; see Figure for an example of our construction. (a) (b) Fig.. Proof of Lemma.. The inner red circle is H. (a) The original disjoint paths Q (solid blue) and Π (dashed red). (b) The transformed disjoint paths Q (solid blue) and Π (dashed red). 0 Now suppose for the sake of argument that Q i crosses π i+ for some index i, or equivalently, that Q Q and Π Π. As usual, to derive a contradiction, we need to
Jeff Erickson and Yipu Wang show that Q and Π are sets of disjoint walks. The following case analysis implies that the walks in Q are pairwise disjoint: None of the paths Q κ+,..., Q k intersect H. On the other hand, for all i < κ, Q i \ Q i is a subset of π i+ and therefore lies in H. Trivially, Q κ+,..., Q k are disjoint from Q,..., Q κ. Thus, paths Q..., Q κ, Q κ are disjoint from paths Q κ+,..., Q k. Q κ lies entirely below Q κ and therefore entirely below Q κ. Now consider any point x Q i, for any index i < κ. Point x lies on or above Q i (because every point in Q i lies on or above Q i), and therefore lies above Q i+. So we must have x π i+ and therefore x H. But because x Q i H, x lies either on or above π i+, and therefore lies above π i+. So x cannot lie on Q i+. We 0 conclude that Q i and Q i+ are disjoint. Similar case analysis implies that the walks in Π are pairwise disjoint: π lies entirely above π and therefore entirely above π. Now consider any point x π i+, for any index < i < κ. Point x lies on or below Q i, and therefore below Q i. On the other hand, x lies on or below π i+, and therefore lies below π i. So x cannot lie in π i. We conclude that π i and π i+ are disjoint. The unique optimality of Π and Q implies l(π) < l(π ) and l(q) < l(q ). On the other hand, we immediately have κ ( l(π) + l(q) = l(π ) + l(qi ) + l(π i+ ) ) + i= κ ( l(π ) + l(q i ) + l(π i+ )) + i= = l(π ) + l(q ), k l(q i ) i=κ k l(q i ) giving us a contradiction. We conclude that π i does not cross Q i for any index i. It follows immediately that π i does not cross (in fact, does not touch) any Q j such that j < i. A symmetric argument implies that π i does not cross any Q j such that j > i. i=κ 0 Algorithm Now we are finally ready to describe our algorithm for computing Q given P. We define five anchor vertices as follows; see Figure. If Q meets P, then a is the first vertex of Q that is also on P, and b is the first vertex in the suffix P (a, t ] that is also on Q ; otherwise, a = t and b = s. If Q meets P, then c is the first vertex in their intersection; otherwise, c = t. If P meets the prefix Q [s, c), then d is the last vertex in their intersection; otherwise, d = s. Finally, e is the first vertex of the suffix P (d, t ] that is also on Q.
Four Shortest Vertex-Disjoint Paths in Planar Graphs We also split each path Q i into a prefix Qi s and a suffix Q t i that meet at a single vertex. Specifically, we split Q at a, we split Q at b, we split Q at c, and we split Q at e. Thus, for example, Q s = Q [s, a] and Q t = Q [a, t ]. t s s a c b t s t e s d t Fig.. Anchor vertices a, b, c, d, e. Now suppose that we know the locations of the anchor vertices a, b, c, d, and e. Our algorithm computes Q in three phases; each phase solves a parallel instance of the k-min-sum problem (with k = or k = ) in a subgraph of G in O(n) time, via minimum-cost flows. The subpaths of Q computed in each phase are shown in Figure. t s t s t s s α a c b t s δ ϵ a c b t s a c b ζ t s β t s t s η t e s d t γ e s d t e s d t θ Fig.. Subpaths of Q computed by the three phases of our algorithm. 0. Phase : α, β, and γ Let H be the subgraph of G obtained by deleting every vertex in R except a and c, every edge incident to s or e outside of R, and every vertex of P (d, t ] except e. In the first phase, we compute the shortest set of vertex-disjoint paths in H from s to a, from s to c, and from s to e. Call these paths α, β, and γ, respectively. Lemma.. α = Q s, β = Qs, and γ = Qs. Proof. For the sake of argument, suppose (α, β, γ) (Q s, Qs, Qs ), and define a new set of walks Q := {α Q t, Q, β Q t, γ Qt }. The following exhaustive case analysis implies that the walks in Q are vertex-disjoint.
0 Jeff Erickson and Yipu Wang Paths α, β, and γ are disjoint by definition. Similarly, Q t, Q, Q t, Qt are subpaths of paths in Q and thus are disjoint by defi- nition. P separates Q from α, β, and γ. Lemma. implies that β and γ do not cross Q s, and therefore do not touch Qt. Lemma. also implies that α does not cross Q s, and therefore does not touch Qt. Lemma. also implies that α and β do not cross Q s, and therefore do not touch Qt. Finally, if d = s, then the definition of H implies that γ does not leave R except at s and e, so Lemma. implies that γ is disjoint from Q t. On the other hand, if d s, then Lemma. implies that γ does not cross Q [s, d]; on the other hand, 0 Q t does not meet Q [s, d]. The definition of H implies that γ does not cross the path P [d, t ] and only meets it at d or e; on the other hand, neither d not e are on Q t. Since Q [s, d] P [d, t ] separates γ from Q t, we conclude that Qt and γ are disjoint. Because the walks in Q are vertex-disjoint, the unique optimality of Q implies that l(q) < l(q ). On the other hand, the lemmas in Section. and the definitions of the anchor vertices imply that Q s, Qs, and Qs are indeed paths in H between the appropriate terminals. Moreover, Q s, Qs, and Qs are vertex-disjoint, because they are subpaths of the disjoint paths in Q. Thus, the unique optimality of {α, β, γ} implies that l(α) + l(β) + l(γ) < l(q s ) + l(qs ) + l(qs ). It follows that l(q ) < l(q), giving us the desired contradiction.. Phase : δ and ε If Q and P are disjoint, let δ = t and ε = s. Otherwise, let H be the subgraph of G obtained by deleting every vertex of P (a, t ] except b, all edges incident to b that leave R, and every vertex of α except a. In the second phase, our algorithm computes the shortest vertex-disjoint paths in H from t to a and from s to b. Call these paths δ and ε, respectively. Lemma.. rev(δ) = Q t and ε = Qs. Proof. The lemma is obvious if Q and P are disjoint, so assume otherwise. For the sake of argument, suppose (rev(δ), ε) (Q t, Qs ), and let Q = {Q s rev(δ), ε Q t, Q, Q }. The following exhaustive case analysis implies that the walks in Q are pairwise disjoint. δ and ε are disjoint by definition. Q s, Qt, Q, and Q are disjoint by definition of Q. Lemma. implies that δ does not cross Q s, and therefore does not touch Qt. 0 The path α P [a, t ] separates δ and ε from Q and therefore from Q. Lemma. implies that Q s V(H ) = {a}. It follows that ε does not touch Q s. The unique optimality of Q now implies that l(q) < l(q ). On the other hand, the lemmas in Section. and the definitions of the anchor vertices imply that Q t and Qs are vertex-disjoint paths in H between the appropriate terminals. Thus, the unique optimality of {δ, ε} implies that l(q t ) + l(qs ) > l(δ) + l(ε), and therefore l(q) > l(q ), giving us the desired contradiction.
Four Shortest Vertex-Disjoint Paths in Planar Graphs. Phase : ζ, η, and θ Finally, let H be the subgraph of G obtained by deleting all vertices in α rev(δ), all vertices in β[s, b), all vertices in γ[s, e), and all vertices in ε[s, b). The last phase of our algorithm computes the shortest vertex-disjoint paths in H from b to t, from c to t, and from e to t. Call these paths ζ, η, and θ, respectively. Lemma.. ζ = Q t, η = Qt, and θ = Qt. Proof. Suppose, for the sake of argument, that (ζ, η, θ) (Q t, Qt, Qt ), and let Q := {Q, Q s ζ, Qs η, Qs θ}. As usual, exhaustive case analysis implies that the walks in Q are pairwise disjoint. Several cases rely on Lemmas. and., which imply that 0 α rev(δ) = Q, β = Q s, γ = Qs, and ε = Qs. ζ, η, and θ are disjoint by definition. Q, Q s, Qs, and Qs are disjoint by definition of Q. Q is disjoint from H and thus disjoint from ζ, η, and θ. Q s H = {b}, so Q s is disjoint from η and θ. Q s H = {c}, so Q s is disjoint from ζ and θ. Q s H = {e}, so Q s is disjoint from ζ and η. The unique optimality of Q now implies that l(q) < l(q ). On the other hand, Q t, Qt, and Qt are paths between appropriate terminals in H. Thus, the unique optimality of {ζ, η, θ} implies that l(q t )+l(qt )+l(qt ) > l(ζ)+l(η)+ l(θ), and therefore l(q) > l(q), giving us the desired contradiction.. Summary Finally, we summarize our overall -min-sum algorithm. First, in a preprocessing phase, we compute P using the algorithm of Kobayashi and Sommer []. Then for all possible choices for the anchor vertices a, b, c, d, e, we compute the paths α, β, γ, δ, ε, ζ, η, θ as described in the previous sections, first under the assumption that Q R, and then under the symmetric assumption that Q R (mirroring the definitions of the anchor vertices and the paths). Altogether, we compute the solutions to O(n ) parallel instances of -min-sum and -min-sum, each in O(n) time via minimum-cost flows. Thus, the overall running time of our algorithm is O(n ). 0 References. Ravindra K. Ahuja, Thomas L. Magnanti, and James Orlin. Network Flows: Theory, Algorithms, and Applications. Prentice Hall,.. Glencora Borradaile, Amir Nayyeri, and Farzad Zafarani. Towards single face shortest vertex-disjoint paths in undirected planar graphs. Proc. rd Ann. Europ. Symp. Algorithms,,. Lecture Notes Comput. Sci., Springer.. Robert G. Busacker and Paul J. Gowen. A procedure for determining minimal-cost network flow patterns. Technical Paper ORO-TP-, Oper. Res. Office, Johns Hopkins U., November 0. http://www.dtic.mil/docs/citations/ad0.
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