PTAS FOR GRAPHS OF BOUNDED TREEWIDTH - Planar Graphs and Graphs of Bounded Treewidth

The purpose of this section is to prove Theorem 1.5 by presenting a PTAS for Steiner forest on graphs of bounded treewidth.

6.1 Groups

We define a notion of group that will be crucial in the description of the algorithm.

A group is defined by a setS of center vertices, a setX of “interesting” vertices, and a maximum distancer; thegroupGG(X, S, r) contains S and those vertices of X that are at distance at mostrfrom some vertex inS.

Lemma 6.1. Let T be a Steiner tree of X ⊆ V(G) with length W. For every >0, there is a set S⊆X ofO(1 + 1/)vertices such thatX =GG(X, S, W).

Proof. Let us select vertices s1, s2, . . . from X as long as possible, with the requirement that the distance of si is more than W from every sj, 1 ≤ j < i.

Suppose that st is the last vertex selected this way. We claim that t ≤ 1 + 2/.

Consider a shortest closed tour inGthat visits the verticess₁, . . . , s_t(not necessarily in the order of their indices). As the distance between any two such vertices is more thanW, the total length of the tour is more thantW (assuming thatt >1). On the other hand, all these vertices are on the treeT and it is well known that there is a closed tour that visits every vertex of the tree in such a way that every edge of the tree is traversed exactly twice and no other edge of the graph is used. Hence the shortest tour has length at most 2W andt≤2/follows.

The following consequence of the definition of group is easy to see.

Proposition 6.2. If S₁, S₂, X₁, X₂ are subsets of vertices of G and r₁, r₂ are real numbers, then

GG(X1, S1, r1)∪ GG(X2, S2, r2)⊆ GG(X1∪X2, S1∪S2,max{r1, r2}).

6.2 Conforming solutions

Let (T,B) be a rooted nice tree decomposition of widthk, letIbe the nodes ofT, and let B ={Bi | i∈ I} be the bags of the decomposition. For every i ∈I, let V_i be the set of vertices appearing inB_i or in the bag of a descendant of B_i. Let A_i be the set ofactive verticesat bag B_i: those verticesv ∈V_i for which there is

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Fig. 2. The 4 components K1, K2, K3, K3 of F partition Ai into 4 classes. Note that the restriction ofF toVihas 6 components.

a demand {v, w} ∈ D with w6∈ V_i. Let G_i := G[V_i]. A Steiner forest F induces a partitionπ_i(F) of A_i for everyi∈I: let two vertices ofA_i be in the same class of π_i(F) if and only if they are in the same component of F. Note that if F is restricted toG_i, then a component ofF can be split into up tok+ 1 components, thusπi(F) is a coarser partition than the partition defined by the components of the restriction ofF toGi. See Figure 4 for an example.

Let Π = (Πi)_i∈I be a collection such that Πi is a set of partitions ofAi. If some Steiner forestF satisfiesπi(F)∈Πifor everyi∈I, then we say thatF conformsto Π. The aim of this subsection is to give an algorithm for bounded treewidth graphs that finds a minimum-length solution conforming to a given Π. For fixed k, the running time is polynomial in the size of the graph and the size of the collection Π on a graph with treewidth at mostk. In Section 6.3, we construct a polynomial-size collection Π such that there is a (1 +)-approximate solution that conforms to Π.

Putting together these two results, we get a PTAS for the Steiner forest problem on bounded treewidth graphs.

Lemma 6.3. For every fixedk, there is a polynomial time algorithm that, given a graph Gwith treewidth at most k and a collectionΠ, finds the minimum-length Steiner forest conforming toΠ.

The proof of Lemma 6.3 follows the standard dynamic programming approach, but it is not completely trivial. First, we use a technical trick that makes the presentation of the dynamic programming algorithm simpler. We can assume that every terminal vertex v has degree 1: otherwise, moving the terminal to a new degree 1 vertexv⁰ attached to vwith an edgevv⁰ having length 0 does not change the problem and does not increase treewidth. Thus by Lemma 2.1, it can be assumed that we have a nice tree decomposition (T,B) of width at most k where

α2

α1

S α1∨α2

Fig. 3. ForestF1 induces partitionα1 onS, forestF2 induces partitionα2, the union of the two forests induces the partitionα1∨α2.

no terminal vertex is introduced and the join nodes contain no terminal vertices;

this assumption simplifies the presentation. For the rest of the section, we fix such a tree decomposition and notationVi, Ai, etc. refer to this fixed decomposition.

Let us introduce terminology and notation concerning partitions. A partitionα of a setS can be considered as an equivalence relation onS. Hence we use notation (x, y)∈αto say thatxandyare in the same class ofα. We denote byx^αthe class ofαthat contains elementx.

If F is a subgraph of G and S ⊆ V(G), then F induces a partition α of S:

(x, y) ∈ α if and only if x and y are in the same component of F (and every x∈S\V(F) forms its own class). We say that partitionαisfinerthan partitionβ if (x, y)∈αimplies (x, y)∈β; in this case,β iscoarserthanα. Ifα=β, thenαis both finer and coarser thanβ. We denote byα₁∨α₂ the unique finest partitionα coarser than bothα₁andα₂. This definition is very useful in the following situation.

Let F₁, F₂ be subgraphs ofG, and suppose that F₁ and F₂ induce partitions α₁ andα₂ of a setS⊆V(G), respectively. IfF₁ andF₂ intersect only inS, then the partition induced by the union ofF₁ andF₂is exactlyα₁∨α₂ (see Figure 3). Let β_i be a partition ofB_i for somei∈I and letF_i be a subgraph of G[V_i]. Then we denote by Fi+βi the graph obtained from Fi by adding a new edgexy for every (x, y)∈βi. Note that Fi+βi is not necessarily a subgraph ofG[Vi].

Following the usual method of designing algorithms for bounded-treewidth graphs, we define several subproblems for each node i∈ I. A subproblem at node i cor-responds to finding a subgraph Fi in Gi satisfying certain properties: informally speaking,Fiis supposed to be the restriction of a Steiner forestF toVi. The prop-erties defining a subproblem prescribe howFishould look from the “outside world”

(i.e., from the part ofGoutsideVi) and they contain all the information necessary for deciding whether Fi can be extended, by edges outside Vi, to a conforming so-lution. Let us discuss briefly and informally what information these prescriptions should contain. Clearly, the edges ofF_iinB_iand the wayF_i connects the vertices ofB_i(i.e., the partitionαofB_iinduced byF_i) is part of this information.

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more, the way F_i partitionsA_i should also be part of this information. However, there is a subtle detail that makes the description of our algorithm significantly more technical. The definition of π_i(F) = π means that the components of F partition A_i in a certain way. But the restriction F_i of F to V_i might induce a finer partition of Ai than π: it is possible that two components of Fi are in the same component ofF (see Figure 4). This means that we cannot require that the partition ofAiinduced byFibelongs to Π. We avoid this problem by “imagining”

the partitionβ ofBi induced by the full solutionF, and require thatFi partition Ai according to πif each class of β becomes connected somehow. In other words, instead of requiring that Fi itself partitionsAi in a certain way, we require that Fi+β induce a certain partition.

Formally, each subproblemP is defined by a tuple (i, H, π, α, β, µ), where (S1) i∈I is a node ofT,

(S2) H is a spanning subgraph ofG[Bi] (i.e., contains all vertices ofG[Bi]), (S3) π∈Πi is a partition ofAi,

(S4) α,β are partitions ofBi,βis coarser thanα, andαis coarser than the partition induced by the components ofH, and

(S5) µis an injective mapping from the classes ofπto the classes ofβ.

The solution c(i, H, π, α, β, µ) of a subproblemP is the minimum length of a sub-graphF_i ofG[V_i] satisfying all of the following requirements:

(C1) F_i[B_i] =H (which implies B_i⊆V(F_i)).

(C2) αis the partition ofB_i induced byF_i. (C3) The partition of A_i induced byF_i+β isπ.

(C4) For every descendant dof i(including d=i), the partition ofA_d induced by Fi+β belongs to Πd.

(C5) If there is a terminal pair (x1, x2) withx1, x2∈Vi, then they are connected in Fi+β.

(C6) Every x∈Ai is in the component ofFi+β containingµ(x^π).

We solve these subproblems by bottom-up dynamic programming. Let us discuss how to solve a subproblem depending on the type ofi.

Leaf nodes i. Ifiis a leaf node, then the value of the solution is trivially 0.

Join node i having children i1, i2. Note that Ai₁ andAi₂ are disjoint: the vertices of a join node are not terminal vertices. The setAi is a subset ofAi₁∪Ai₂

and it may be a proper subset: if there is a pair (x, y) withx∈Ai₁,y ∈Ai₂, then xory might not be inAi.

We show that the value of the subproblem is c(i, H, π, α, β, µ) =

min

(J1),(J2),(J3),(J4)

(c(i₁, H, π¹, α¹, β, µ¹) +c(i₂, H, π², α², β, µ²)−`(H)), (8) where the minimum is taken over all tuples satisfying, for p = 1,2, all of the following conditions:

(J1) α¹∨α²=α.

(J2) πandπ^p are the same onA_i_p∩A_i.

(J3) For every v ∈ Ai_p ∩Ai, µ(v^π) = µ^p(v^π^p) (note that the classes of β are the domain of bothµandµ^p).

(J4) If there is a terminal pair (x₁, x₂) withx₁∈A₁ andx₂∈A₂, thenµ¹(x^π₁¹) = µ²(x^π₂²).

We will use the following observation repeatedly. Let F be a subgraph of G_i and letF^p =F[V_i_p]. Suppose that F induces partitionαonB_i andβ is a coarser partition thanα. Then two vertices ofVi_pare connected inF+β if and only if they are connected inF^p+β. Indeed,F^3−pdoes not provide any additional connectivity compared toF^p+β: asβ is coarser thanα_3−p, if two vertices ofBiare connected by a path inF^3−p, then they are already adjacent inF^p+β.

Proof of (8)left ≤(8)right:

LetP1= (i1, H, π¹, α¹, β, µ¹) andP2= (i2, H, π², α², β, µ²) be subproblems min-imizing the right-hand side of (8), and letF¹ and F² be optimal solutions of P1

andP2, respectively. LetF be the union of subgraphsF¹and F². It is clear that the length ofF is exactly the right-hand side of (8): the common edges ofF¹ and F²are exactly the edges ofH. We show thatF is a solution of P, i.e.,F satisfies requirements (C1)–(C6).

(C1): Follows fromF¹[Bi] =F²[Bi] =F[Bi] =H.

(C2): Follows from (J1) and from the fact thatF¹ andF²intersect only inB_i. (C3): First consider two vertices x, y ∈ Ai_p ∩Ai. Vertices x and y are connected

in F +β if and only if they are connected in F^p+β. By (C3) for F^p, this is equivalent to (x, y)∈π^p, which is further equivalent to (x, y)∈ π by (J2).

Now suppose thatx∈ A_i₁ ∩A_i and y ∈A_i₂∩A_i. In this case, xand y are connected inF +β if and only if there is a vertex of B_i reachable from x in F¹+β and fromy in F²+β, or in other words,µ¹(x^π¹) =µ²(y^π²). By (J3), this is equivalent toµ(x^π) =µ(y^π), or (x, y)∈π(asµis injective).

(C4): Ifdis a descendant ofip, then the statement follows using that (C4) holds for solutionF^p of P^p and the fact that for every descendant dof ip, Fi+β and F+β induce the same partition of Ad. Ford=i, the statement follows from the previous paragraph, i.e., from the fact thatF+βinduces partitionπ∈Π_i onA_i.

(C5): Consider a pair (x1, x2). If x1, x2 ∈ Vi₁ or x1, x2 ∈ Vi₂, then the statement follows from (C5) on F¹ or F². Suppose now that x1 ∈ Vi₁ and x2 ∈ Vi₂; in this case, we havex1 ∈ Ai₁ and x2 ∈ Ai₂. By (C6) on F¹ and F², xp is connected toµ^p(x^π_p^p) inF^p+β. By (J4), we haveµ¹(x^π₁¹) =µ²(x^π₂²), hencex₁ andx₂ are connected to the same class ofβ in F+β.

(C6): Consider anx∈Ai that is inAi_p. By condition (C6) onF^p, we have thatxis connected inF^p+β (and hence inF+β) toµ^p(v^π^p), which equalsµ(v^π) by (J3).

Proof of (8)left ≥(8)right:

LetF be a solution of subproblem (i, H, π, α, β, µ) and let F^p be the subgraph ofF induced byV_i_p. To prove the inequality, we need to show three things. First,

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we have to define two tuples (i₁, H, π¹, α¹, β, µ¹) and (i₂, H, π², α², β, µ²) that are subproblems, i.e., they satisfy (S1)–(S5). Second, we show that (J1)–(J4) hold for these subproblems. Third, we show that F¹ and F² are solutions for these subproblems (i.e., (C1)–(C6)), hence they can be used to give an upper bound on the right-hand side that matches the length ofF.

Letα^p be the partition of Bi induced by the components ofF^p; as F¹ and F² intersect only inBi, we haveα=α¹∨α², ensuring (J1). Sinceβ is coarser thanα, it is coarser than bothα¹andα². Letπ^pbe the partition ofAi_pdefined byF+β;

we haveπp∈Πi_p by (C4) forF. Furthermore, by (C3) forF,πis the partition of Ai induced by F+β, hence it is clear thatπ andπ^p are the same onAi_p∩Ai, so (J2) holds. This also means thatF+β (or equivalently,F^p+β) connects a class of π^p to exactly one class ofβ; letµ^p be the corresponding mapping from the classes of π^p to β. Now (J4) is immediate. It is clear that the tuple (ip, H, π^p, α^p, β, µ^p) satisfies (S1)–(S5).

We show thatF^p is a solution of subproblem (i_p, H, π^p, α^p, β, µ^p). As the edges ofH are shared byF¹ andF², it will follow that the right-hand side of (8) is not greater than the left-hand side.

(C1): Obvious from the definition ofF¹ andF². (C2): Follows from the wayα^pis defined.

(C3): Follows from the definition of π^p, and from the fact that F+β and F^p+β induce the same partition onA_i_p.

(C4): Follows from (C4) onF and from the fact thatF +β andF^p+β induces the same partition onAd.

(C5): Suppose that x₁, x₂ ∈V_i_p. Then by (C5) for F, x₁ and x₂ are connected in F+β, hence they are connected inF_i+β as well.

(C6): Follows from the definition ofµ^p.

Introduce node iof vertex v. Letj be the child ofi. Sincev is not a terminal vertex, we haveAj =Ai. LetF⁰be a subgraph ofG[Vj] and letFS be obtained from F⁰by adding vertexvtoF⁰and makingvadjacent toS⊆Bj. Ifα⁰is the partition ofBj induced by the components ofF⁰, then we define the partitionα⁰[v, S] ofBi

to be the partition obtained by joining all the classes of α⁰ that intersect S and addingv to this new class (ifS=∅, then{v} is a class ofα⁰[v, S]). It is clear that α⁰[S, v] is the partition ofB_i induced byF_S.

We show that the value of a subproblem is given by c(i, H, π, α, β, µ) = min

(I1),(I2),(I3)c(j, H[Bj], π, α⁰, β⁰, µ⁰) + X

xv∈E(H)

`(xv), (9) where the minimum is taken over all tuples satisfying all of the following:

(I1) α=α⁰[v, S], whereS is the set of neighbors ofvin H.

(I2) β⁰ isβ restricted toBj.

(I3) For everyx∈A_i,µ(x^π) is the class ofβ containingµ⁰(x^π).

Proof of (9)left ≤(9)right:

LetF⁰ be an optimal solution of subproblemP⁰ = (j, H[B_j], π, α⁰, β⁰, µ⁰). LetF be the graph obtained fromF⁰ by adding to it the edges ofH incident to v; it is

clear that the length ofF is exactly the right-hand side of (9). Let us verify that (C1)–(C6) hold forF.

(C1): Immediate.

(C2): Holds because of (I1) and the wayα⁰[v, S] was defined.

(C3)–(C5): Observe that F +β connects two vertices of Vj if and only if F⁰+β⁰ does.

Indeed, if a path in F +β connects two vertices via vertex v, then the two neighborsx, yofvon the path are in the same class ofβ asv(using thatαand βare coarser than the partition induced byH), hence (I2) implies thatx, yare in the same class ofβ⁰ as well. In particular, for every descendantd ofi, the components ofF+β and the components ofF⁰+β give the same partition of A_d.

(C6): Follows from (C6) forF⁰ and from (I3).

Proof of (9)left ≥(9)right:

Let F be a solution of subproblem (i, H, π, α, β, µ) and let F⁰ be the subgraph ofF induced byV_j. We define a tuple (j, H[B_j], π, α⁰, β⁰, µ⁰) that is a subproblem, show that it satisfies (I1)–(I3), and thatF⁰ is a solution of this subproblem.

Let α⁰ be the partition of Vj induced by F⁰ and let β⁰ be the restriction of β onBj; these definitions ensure that (I1) and (I2) hold. Letµ⁰(x^π) =µ(x^π)\ {v}, which is a class ofβ⁰; clearly, this ensures (I3). Note that this is well defined, as it is not possible thatµ(x^π) is a class ofβ consisting of onlyv: by (C6) forF, this would mean thatv is the only vertex ofBi reachable fromxinF. Sincevis not a terminal vertex,v 6=x, thus ifv is reachable fromx, then at least one neighbor of v has to be reachable fromxas well.

Let us verify that (S1)–(S5) hold for the tuple (j, H[Bj], π, α⁰, β⁰, µ⁰). (S1) and (S2) clearly hold. (S3) follows from the fact that (C4) holds for F and Ai =Aj. To see that (S4) holds, observe that (x, y)∈ α⁰ implies (x, y)∈ α, which implies (x, y)∈β, which implies (x, y)∈β⁰. (S5) is clear from the definition of µ⁰.

The difference between the length ofFand the length ofF⁰is exactlyP

xv∈E(H)`(xv).

Thus to show that the left-hand side of (9) is at most the right-hand side of (9), it is sufficient to show thatF⁰ is a solution of subproblem (j, H[Bj], π, α⁰, β⁰, µ⁰).

(C1)–(C2): Obvious.

(C3)–(C5): As in the other direction, follow from the fact that F⁰+β⁰ induces the same partition ofVj asF+β.

(C6): By the definition of µ⁰, it is clear thatµ⁰(x^π) is exactly the subset of B_j that is reachable fromxin F+β and hence inF⁰+β⁰.

Forget nodei of vertexv. Letj be the child ofi. We haveVi=Vj and hence A_i=A_j. We show that the value of a subproblem is given by

c(i, H, π, α, β, µ) = min

(F1),(F2),(F3),(F4)c(j, H⁰, π, α⁰, β⁰, µ⁰), (10) where the minimum is taken over all tuples satisfying all of the following:

(F1) H⁰[Bi] =H.

(F2) αis the restriction of α⁰ to B_i.

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(F3) β is the restriction ofβ⁰ toB_i and (x, v)∈β⁰ if and only if (x, v)∈α⁰. (F4) For everyx∈Ai,µ(x^π) is the (nonempty) setµ⁰(x^π)\ {v}(which implies that

µ⁰(x^π) contains at least one vertex of Bi).

Proof of (10)left≤(10)right:

Let F be a solution of (j, H⁰, π, α⁰, β⁰, µ⁰). We show that F is a solution of (j, H, π, α, β, µ) as well.

(C1): Clear because of (F1).

(C2): Clear because of (F2).

(C3)–(C5): We only need to observe thatF +β and F+β⁰ have the same components:

since by (F3), (x, v) ∈ β⁰ implies (x, v) ∈ α⁰, the neighbors of v in F +β⁰ are reachable fromv in F, thusF+β⁰ does not add any further connectivity compared toF+β.

(C6): Observe that if µ⁰(x^π) are the vertices of Bj reachable fromxin F+β⁰, then µ(x^π) = µ⁰(x^π)\ {v} are the vertices of Bi reachable from x in F +β⁰. We have already seen that F +β and F +β⁰ have the same components, thus the nonempty setµ(x^π) is indeed the subset ofBi reachable fromxinF +β.

Furthermore, by (F3),β is the restriction ofβ⁰ onBi, thus ifµ⁰(x^π) is a class ofβ⁰, thenµ(x^π) is a class ofβ.

Proof of (10)left≥(10)right:

LetF be a solution of (j, H, π, α, β, µ). We define a tuple (j, H⁰, π, α⁰, β⁰, µ⁰) that is a subproblem, we show that (F1)–(F3) hold, and that F is a solution of this subproblem.

Let us define H⁰ = F[Bj] and let α⁰ be the partition of Bj induced by the components of F; these definitions ensure that (F1) and (F2) hold. We define β⁰ as the partition obtained by extendingβ to Bj such that v belongs to the class of β that contains a vertexx∈Bi with (x, v)∈α⁰ (asβ is coarser than the partition induced by H, there is at most one such class; if there is no such class, then we let{v} be a class of β⁰). It is clear that (F3) holds for this β⁰. Let us note that F+β andF+β⁰ have the same connected components: if (x, v)∈β⁰, thenxand v are connected in F. Letµ⁰(x^π) be the subset ofB_j reachable from xin F+β⁰ (or equivalently, inF+β). It is clear thatµ(x^π) =µ⁰(x^π)\ {v⁰} holds, hence (F4) is satisfied.

Let us verify first that (S1)–(S5) hold for (j, H⁰, π, α⁰, β⁰, µ⁰). (S1) and (S2) clearly holds. (S3) follows from the fact that (S3) holds for (i, H, π, α, β, µ) and Ai=Aj. To see that (S4) holds, observe that ifx, y∈Bi, then (x, y)∈α⁰ implies (x, y)∈α, which implies (x, y)∈β, which implies (x, y)∈β⁰. Furthermore, if (x, v)∈α⁰, then (x, v)∈β⁰ by the definition ofβ. (S5) is clear from the definition of µ⁰.

We show thatF is a solution of (j, H⁰, π, α⁰, β⁰, µ⁰).

(C1): Clear from the definition ofH⁰. (C2): Clear from the definition ofα⁰.

(C3)–(C5): Follow from the fact thatF+βandF+β⁰have the same connected components.

(C6): Follows from the definition ofµ⁰.

6.3 Constructing the partitions

Recall that the collection Π = (Πi)i∈I contains a set of partitions Πifor eachi∈I.

We construct these sets Π_i in the following way. Each partition in Π_i is defined by a sequence ((S₁, r₁), . . . ,(S_p, r_p)) of at most k+ 1 pairs and a partition ρ of {1, . . . , p}. The pair (S_j, r_j) consists of a setS_jofO((k+ 1)(1 + 1/)) vertices ofG_i and a nonnegative real numberr_j, which equals the distance between two vertices of G. This means that there are at most |V(G)|O((k+1)(1+1/))· |V(G)|² possible pairs (S_j, r_j) and hence at most |V(G)|^O((k+1)²^(1+1/)) different sequences. The number of possible partitions ρis O(k^k). Thus if we construct Πi by considering all possible sequences constructed from every possible choice of (Sj, rj), the size of Πi is polynomial in|V(G)| for every fixedkand.

We construct the partition πcorresponding to a particular sequence and ρthe following way. Each pair (Sj, rj) can be used to define a groupRj =GG(Ai, Sj, rj) ofAi. Roughly speaking, for each class P ofρ, there is a corresponding class of π that contains the union of Rj for every j ∈ P. However, the actual definition is somewhat more complicated. We wantπ to be a partition, which means that the subsets ofAicorresponding to the different classes ofρshould be disjoint. In order to ensure disjointness, we define R⁰_j :=Rj\Sj−1

j⁰=1Rj⁰. The partition π of Ai is constructed as follows: for each classP ofρ, we letS

j∈PR⁰_j be a class ofπ. Note that these classes are disjoint by construction. If these classes fully coverAi, then we put the resulting partitionπinto Π_i; otherwise, the sequence does not define a partition. This finishes the construction of Π_i.

Before showing that there is a near-optimal solution conforming to the collection Π defined above, we need a further definition. For two verticesuandv, we denote byu < vthe fact that the topmost bag containinguis a proper descendant of the topmost bag containing v. Note that each bag is the topmost bag of at most one vertex in a nice tree decomposition (recall that we can assume that the root bag contains only a single vertex). Thus ifuandv appear in the same bag, thenu < v orv < uholds, i.e., this relation defines a total ordering of the vertices in a bag. We can extend this relation to connected subsets of vertices: for two disjoint connected sets K1, K2, K1 < K2 means that K2 has a vertex v such that u < v for every vertex u∈ K1, or in other words, K1 < K2 means that the topmost bag where vertices fromK1 appear is a proper descendant of the topmost bag where vertices from K2 appear. If there is a bag containing vertices from both K1 andK2, then eitherK1< K2or K2< K1holds. The reason for this is that the bags containing vertices fromK1∪K2 form a connected subtree of the tree decomposition, and if the topmost bag in this subtree contains vertexv∈K1∪K2, thenu < v for every other vertexuinK1∪K2.

Lemma 6.4. There is a(1 +k)-approximate solution conforming toΠ.

Proof. Let F be a minimum-length Steiner forest. We describe a procedure that adds further edges toF to transform it into a Steiner forestF⁰ that conforms to Π and has length at most (1 +k)`(F). We need a delicate charging argument to show that the total increase of the length is at mostk·`(F) during the procedure.

In each step, we charge the increase of the length to an ordered pair (K₁, K₂) of components ofF. We are charging only to pairs (K₁, K₂) having the property that

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In document Planar Graphs and Graphs of Bounded Treewidth (Pldal 21-31)