A 1.5 approximation algorithm for augmenting edge-connectivity of a graph from 1 to 2. ∗ Guy Even†

Guy Kortsarz‡

Zeev Nutov

§

June 5, 2009

Abstract We consider the following NP-hard problem: given a connected graph G = (V, E) and an edge set E on V disjoint to E, find a minimum size subset of edges F ⊆ E such that (V, E ∪ F ) is 2-edge-connected. In [4] we presented a 1.8 approximation for the problem. In this paper we improve the ratio to 1.5.

∗

Preliminary version appeared in APPROX 2001, LNCS 2129, pp. 90-101, 2001. Dept. of Electrical Engineering-Systems, Tel-Aviv University, Tel-Aviv 69978, Israel. E-mail:[email protected] ‡ Rutgers University, Camden. E-mail:[email protected] Partially suported by NSF grant number 0728787. § The Open University of Israel, Raanana, Israel. E-mail:[email protected]

†

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1

Introduction

A graph (possibly with parallel edges) is k-edge-connected if there are k pairwise edge-disjoint paths between every pair of its nodes. We consider the following problem: Tree Augmentation Problem (TAP) Instance: A tree T = (V, E) and a set of links E on V disjoint to E. Objective: Find a minimum size subset F ⊆ E of edges such that G + F is 2-edge-connected. TAP is sometimes posed as the problem of covering a laminar family (see e.g., [1]). Namely, given a laminar family E on a groundset V , and an edge set E on V , find a minimum size edge subset F ⊆ E such that for every S ∈ E, there is an edge in F with one endpoint in S and the other in V − S. TAP is also equivalent to the problem of finding a minimum size augmenting edge set to increase the edge-connectivity from k to k + 1 for any odd k; this is since the family of minimum cuts of a k-connected graph with k odd is laminar. TAP is APX-hard even if the set E of links forms a cycle on the leaves of T [1]. For previous work on the problem and discussion see [4]. In the conference version [2] of this paper we presented and sketched a proof of a 1.5-approximation algorithm for the problem, improving the ratio 1.875+ε of Nagamochi [5]. Since the full paper was too long and complicated we presented in [4] an easier 1.8-approximation algorithm for TAP. In this paper we present some additional ideas to prove: Theorem 1.1: TAP admits a 1.5-approximation algorithm. Our ratio matches the recently obtained lower bound on integrality gap of the natural LPrelaxation for TAP, see [3].

2

Preliminaries

Here is some notation used in the paper. Let T = (V, E) be a tree. For u, v ∈ V let (u, v) denote the edge in T and uv the link in E between u and v. Let p(uv) denote the path between u and v in T . A link uv covers all the edges (and nodes) along the path p(uv). We designate a node r of T as the root, and refer to the pair T, r as a rooted tree (we do not mention the root when it is clear from the context). The choice of r defines a partial order on V : u is a descendant of v (or v is an ancestor of u) if v belongs to p(ru); if, in addition, (u, v) ∈ T , then u is a child of v, and v is the parent of u. The leaves of T are the nodes in V − r that have no descendants. We denote the leaf set of T by L(T ), or simply by L, when the context is clear. The rooted subtree of T induced by v and its descendants is denoted by Tv (v is the root of Tv ). A subtree T 0 of T is called a rooted subtree of T if T 0 = Tv for some v ∈ V . For a node set U and a link set M , U ∩ T 0 is the set of nodes in U that belong to T 0 , and M ∩ T 0 is the set of links in M with both endnodes in T 0 . To contract a (not necessarily rooted) subtree T 0 of T /I is to combine all nodes in T 0 into a single node v, named after the least common ancestor of the nodes in T 0 . The edges and links with both endpoints in T 0 are deleted. The edges and links with one endpoint in T 0 now have v as their new endpoint, but retain their correspondence with the edge or link of the original graph. For simplicity, we say that the new node v obtained by contracting T 0 “contains” a node u if u ∈ T 0 . If we add a link uv to a partial solution I, then the nodes along the path p(uv) belong to the same 2-edge-connected component of the augmented graph (V, E ∪ I). Hence, we may contract

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the nodes along p(uv). For a set of links I ⊆ E, let T /I denote the tree obtained by contracting every 2-edge-connected component of T ∪ I into a single node. Since all contractions are induced by subsets of links, we refer to the contraction of every 2-edge-connected component of T ∪ I into a single node simply as the contraction of the links in I. A link u0 v 0 is a shadow of a link uv if p(u0 v 0 ) ⊆ p(uv). Adding all shadows of existing links does not affect the optimal solution size, since every shadow can be replaced by some existing link covering all edges covered by the shadow. Thus: Assumption 2.1: The set of links E is closed under shadows. The following reductions allow us to make some simplifying assumptions about TAP instances. Definition 2.2: A cover F of T is shadow minimal if for every link uv ∈ F , replacing uv by a proper shadow of uv results in a set of links that does not cover T . Claim 2.3: ([4, Claim 2.6]) Let F be a shadow minimal cover of T . Then there is exactly one link in F incident to every leaf a of T . In particular, the leaf-to-leaf links in F form a matching on L. Further simplifying assumptions can be made when we consider the relationship between pairs of edges in T . Definition 2.4: An edge e of T is reducible if all links in E that cover e are shadows of the same link, or if there exists an edge e0 6= e such that every link that covers e0 and is not a proper shadow of any other link, also covers e. It is not hard to see that we may preprocess the instance so that: Assumption 2.5: ([4, Assumption 2.8]) T has no reducible edges. The following statement shows that if there are no reducible edges, then paths in T consisting of degree 2 nodes can not end at leaves. Proposition 2.6: ([4, Proposition 2.9]) If T has no reducible edges, then every parent of a leaf has at least two children. For X, Y ⊆ V and a link set F , let F (X, Y ) = {xy ∈ F : x ∈ X, y ∈ Y } denote the set of links in F that have one endpoint in X and the other in Y ; for x ∈ V let deg F (x) = |F (x, V )| be the degree of x w.r.t. F . For the rest of the paper fix F to be some optimum shadow minimal solution for an instance T = (V, E), E of TAP.

3 3.1

The lower bound and the algorithm The Improved Lower Bound

Definition 3.1: A node s ∈ V − r is a stem of T if it has exactly two children and both of them are leaves. The link between the two children of a stem, if exists, is called a twin-link.

2

Claim 3.2: ([4, Claim 3.2]) If T has no reducible edges, then there is a twin-link in E between the children of every stem. The following lower bound was established and used in [4]: Lemma 3.3: (The Leaf-Stem Lower Bound) ([4, Lemma 3.3]) Let L be the set of leaves and S be the set of stems of T , and let X = V \ (L ∪ S). Let F be a shadow minimal optimal solution and let MF be the set of leaf-to-leaf non twin links in F . Then: 2 1 1X degF (x). (1) |F | ≥ |L| − |MF | + 3 3 3 x∈X

The following definition was not used in [4] and is one of the main three new ideas we use in order to improve the ratio to 1.5.

x

r

v b2

s b1

a1

Figure 1: Illustration to Definition 3.4. Links are shown by dashed lines and paths by wiggly lines. Definition 3.4: A link b1 b2 is a locking link of a leaf a1 if the following holds (see Figure 3.1). The nodes b1 , b2 , a1 are the leaves of a rooted proper subtree Tv of T (so v 6= r) so that b2 is a child of v, the leaves b1 , a1 are children of a stem s that is a proper descendant of v, and any link in E incident to a1 and distinct from the twin link a1 b1 goes to b2 or to an ancestor x of a1 . The importance of locking links comes from the following simple observation. Observation 3.5: If b1 b2 ∈ F is a locking link of a1 then F contains a link (an F -locked link) from a1 to an ancestor x of a1 . Proof: The only links in E incident to the locked leaf a1 are the links to b1 , b2 and to ancestors of a1 . As b1 , b2 have degree 1 in F , by Claim 2.3, the statement follows. 2 Lemma 3.6: [The Improved Lower Bound] Let F, MF , X be as in the Leaf-Stem Lower Bound. Let M be a maximum matching among all matchings in E without twin links and without locking links. Let M0 be the set of links in M that have no common endpoint with a link in MF . Let J be the set of non-F -locked links in F . Then: 1 1 1X 2 |F | ≥ |L| − |M | + |M0 | + degJ (x). (2) 3 3 3 3 x∈X

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Proof: Let MF0 be the set of locking links in MF and Note that (MF \ MF0 ) ∪ M0 is a matching without locking links and without twin links of size |MF | − |MF0 | + |M0 |. As M is a maximum matching of this type, we have |M | ≥ |MF | − |MF0 | + |M0 |. Thus |MF | ≤ |M | − |M0 | + |MF0 |. Substituting the last inequality in (1) gives 1X 2 1 |M | − |M0 | + |MF0 | + |F | ≥ |L| − degF (x) 3 3 3 x∈X ! 1 1 1 1 X = |L| − |M | + |M0 | + degF (x) − |MF0 | . 3 3 3 3 x∈X

Now we observe that the last term in the obtained bound equals the last term in (2). Indeed, by Observation 3.5, for any link b1 b2 ∈ MF0 there is a unique link a1 x ∈ F with x ∈ X. 2 The term 32 |L| − 13 |M | in (2) can be computed in polynomial time. The other terms depend on F which is not known to us, and are partly revealed during the run of the algorithm.

3.2

The credit scheme and the Coupons Invariant

We explain how we use the Improved Lower Bound. Let U denote the set of leaves unmatched by M , so |U | = |L| − 2 · |M |. By multiplying both sides of (2) by 1.5 and rearranging terms, we obtain 3 1 1X 3 · |F | ≥ |U | + · |M | + · |M0 | + degJ (x) . (3) 2 2 2 2 x∈X

We prove the following statement that implies Theorem 1.1.

Theorem 3.7: There exists a polynomial time algorithm that for an instance of TAP computes a solution I of size at most the right-hand size of (3). Thus |I| ≤ 1.5 · |F |. Initially, the algorithm assigns coupons to unmatched leaves and to links in M so that the total number of coupons assigned equals the sum of the first two terms in the right hand side of (3). For technical reasons, we also assign 1 coupon to the root r. Initial assignment of coupons: • Every unmatched leaf gets 1 coupon, and r gets 1 coupon. • Every link in M gets 3/2 coupons. Compound nodes. Our algorithm iteratively contracts certain subtrees of T . We refer to the nodes created by contraction as compound nodes. Compound nodes always own 1 coupon. Noncompound nodes are referred to as original nodes (of T ). Each time a contraction takes place, the new compound node gets 1 coupon, which together with the links added is paid by the total credit of the contracted subtree. For technical reasons, r is also considered as a compound node. Our algorithm is designed to maintain the following invariant: Invariant 3.8: [Coupons Invariant] 1. Every unmatched leaf and every compound node of T /I (including r) owns 1 coupon. 2. Every link in M owns 3/2 coupons. 4

P 0 Tickets. To charge the last two terms 21 |M0 | and 12 x∈X degF (x) − |MF | in (3), we introduce tickets. A ticket worth 1/2 coupon. We will have 2 types of tickets, according to the term we want to charge: M0 -tickets are used to charge the term 12 |M0 |. If we can prove that some link e ∈ M belongs to M0 , namely, that e has no endnode in common with a link in MF , then we may claim an M0 -ticket (1/2 coupon) at e. P X-tickets are used to charge the last term 12 x∈X degJ (x) in (3). If we can prove that some x ∈ X is not linked by F to any locked leaf a1 so that its locking link b1 b2 ∈ F , then we may claim an X-ticket (1/2 coupon) at x. In fact, we will often claim tickets in a subtree T 0 of T without specifying exactly the link e ∈ M or the node x ∈ X on which the ticket is claimed, nor its type. All we care about is that T 0 contains a ticket. No ticket is claimed twice, if we follow the following simple rule. Rule for claiming tickets: Immediately after claiming tickets in T 0 , the whole tree T 0 is contracted into a compound node. This indeed implies that no ticked is claimed twice. If an M0 -ticket was claimed on a link e ∈ M ∩T 0 , then e disappears after T 0 is contracted. If an X-ticket is claimed at a node x, then x is contracted into a compound node, while X-tickets are never claimed on compound nodes. Summarizing, we use the following notation for the credit distributed in a (not necessarily rooted) subtree T 0 of T /I. Let coupons(T 0 ) denote the total number of coupons owned by T 0 ; this includes the coupons owned by nodes of T 0 (every unmatched leaf and every compound node of T 0 owns 1 coupon), and link in M with both endnodes in T 0 . Let P 3/2 coupons for every 0 0 0 tickets(T ) = |M0 ∩ T | + x∈X∩T 0 degF (x) − |MF ∩ T 0 | denote the number of tickets in T 0 . Let credit(T 0 ) = coupons(T 0 ) + 21 tickets(T 0 ). The high level idea of our algorithm is to repeatedly find a tree T 0 in T /I and a cover B 0 of T 0 so that credit(T 0 ) ≥ |B 0 | + 1. In such a case we have enough credit in T 0 to pay for the |B 0 | links to cover and contract T 0 , and to leave a coupon on the resulting compound node. Leaving a coupon is crucial in order to mantain Part 1 of the Coupon Invariant. If we iteratively cover trees in this way until T is completely covered, then the number of links we use is at most the right hand side of Inequality (3), as claimed in Theorem 3.7.

3.3

Greedy contractions and the Matching Invariant

One simple way of finding a subtree T 0 and its cover B 0 as above, that may not use rooted subtrees, and relies on coupons only, is as follows. Definition 3.9: If there exists a set I 0 of α links that covers a subtree T 0 of T , and if coupons(T 0 ) ≥ |I 0 | + 1, then adding I 0 to the partial solution, and assigning one coupon to the obtained compound node is a greedy α-contraction. A greedy α-contraction, if exists, can be found in polynomial time if α is a constant. One of the steps of the algorithm is to apply all possible greedy α-contractions exhaustively for α = 1, 2, 3, 4. We now list two properties of the instance after all greedy 1, 2-contractions are exhausted. Definition 3.10: A matching on the leaves of T /I is proper if its contraction does not create a new leaf.

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Lemma 3.11: ([4, Lemma 3.10]) If all greedy 1, 2-contractions are exhausted, then M is a proper matching. Lemma 3.12: If all greedy 1, 2-contractions are exhausted then every stem has one unmatched child and one matched child. Proof: If the two children of the stem are unmatched then adding the twin link is a greedy 1-contraction. If both children of the stem are matched then adding the two matched links is a greedy 2-contraction. 2 Invariant 3.13: [Matching Invariant] 1. M is a proper matching on the original leaves. 2. Every stem has one matched child and one unmatched child.

3.4

The algorithm

Our approach is to iteratively exhaust greedy α-contractions for α ≤ 4, and then to find a rooted subtree T 0 of T /I with its cover B 0 , to contract T 0 with B 0 , and to leave a coupon on the created compound node (in order to maintain Part 1 of the Coupons Invariant). In the next section we will prove the following statement: Lemma 3.14: Suppose that no greedy α-contraction applies for α ≤ 4. Then there exists a polynomial time algorithm that finds a rooted subtree T 0 of T /I and a cover B 0 ⊆ E of T 0 so that credit(T 0 ) ≥ |B 0 | + 1 holds. Algorithm 1 Approx(T = (V, E), E) (A 1.5-approximation algorithm) 1: I ← ∅. 2: Contract reducible edges of T and update I accordingly. 3: M ← maximum matching of leaf-to-leaf links among all matchings in E not containing twin-links or locking links. 4: while T /I has more than one node do 5: Exhaust greedy α-contractions for α ≤ 4 and update I accordingly. 6: Find a rooted subtree T 0 of T /I and a cover B 0 of T 0 as in Lemma 3.14. 7: Contract T 0 , give 1 coupon on the new compound leaf, and set I ← I ∪ B 0 . 8: end while 9: Return I. Algorithm Approx initiates I ← ∅ as a partial cover. It applies the reductions from Section 2 (specifically, we need Assumptions 2.1 and 2.5), computes a maximum matching M on the leaves among the non-twin links and non-locking links, and initiates the credit scheme as described in Section 3.2. In the main loop, the algorithm iteratively exhausts greedy α-contractions for α ≤ 4, computes T 0 , B 0 as in Lemma 3.14, adds B 0 to I, contracts T 0 , and leaves a coupon on the created compound leaf. The stopping condition is when I covers T , namely, when T /I is a single node.

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It is easy to verify that all actions needed to be taken by the algorithm can be implemented in polynomial time. The credit scheme used implies that the algorithm computes a solution I as in Theorem 3.7, namely |I| is at most 1.5 times the right-hand size of (3). Hence the approximation ratio of 1.5 follows from the Improved Lower Bound, and the proof of Theorem 1.1 will be complete, if we prove Lemma 3.14. For the rest of the paper we deal with proving Lemma 3.14 via various concepts and claims.

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Proof of Lemma 3.14

4.1

Semi-closed trees and their basic covers

The up-link incident to a node u is defined as the link uv such that v is as close as possible to the root; under Assumption 2.1, v is an ancestor of u. We denote the highest link incident to u by up(u). For a node set U ⊆ V , we let up(U ) = {up(u) : u ∈ U }. Definition 4.1: A rooted subtree T 0 of T is leaf-closed if every link incident to a leaf of T 0 has its both endnodes in T 0 . A subtree T 0 is minimally leaf closed if it is leaf-closed but any rooted proper subtree of T 0 is not leaf closed. Claim 4.2: ([5]) If T 0 is minimally leaf-closed then up(L(T 0 )) covers T 0 . The following definitions and claims hold for any proper matching M . We will also assume that every link in M owns 3/2 coupons, Definition 4.3: A semi-closed tree (w.r.t. a proper matching M ) is defined and found as follows. (i) Contract every link in M − I to get T /(I ∪ M ). (ii) Find a minimally leaf-closed rooted subtree T 00 = (T /(I ∪ M ))v of T /(I ∪ M ). (iii) T 0 = (T /I)v , namely T 0 is obtained from T 00 by “decontracting” every link in M − I. We note that T 0 may not be closed with respect to matched leaves and in addition it may not be closed with respect to stems; however, T 0 is closed w.r.t. its unmatched leaves. Definition 4.4: Let T 0 be a semi-closed subtree of T /I. The basic cover B(T 0 ) of T 0 consists of M ∩ T 0 (namely, links in M with both endnodes in T 0 ) plus the up-links of the unmatched leaves of T 0 . As we assume that M is proper, B(T 0 ) indeed covers T 0 , by Claim 4.2. It is not hard to see that all semi-closed trees and their basic covers can be computed in polynomial time, see [4]. In what follows, let T 0 be a semi-closed tree, and let v be the root of T 0 . Claim 4.5: ([4]) Let C(T 0 ) denote the set of non-leaf compound nodes of T 0 (this includes r, if r ∈ T 0 ). Then 1 (4) coupons(T 0 ) − |B(T 0 )| = · |M ∩ T 0 | + |C(T 0 )| . 2 In particular, coupons(T 0 ) ≥ |B(T 0 )|. Definition 4.6: A tree is called deficient if credit(T 0 ) < |B(T 0 )| + 1. 7

4.2

Some non-deficient semi-closed trees

Claim 4.7: ([4, Claims 4.4 and 4.6]) If T 0 is deficient then C(T 0 ) = ∅ (in particular, r ∈ / T 0 ) and T 0 has exactly one matched pair. Claim 4.8: If T 0 has at least 3 stems then T 0 is not deficient. Proof: By the Matching Invariant each stem has a child that belongs to a matched pair. If T 0 has at least 3 stems then T 0 has at least 2 matched pairs. Thus T 0 is not deficient, by Claim 4.7. 2 Corollary 4.9: If T 0 has a ticket then T 0 is not deficient. Proof: By Claim 4.7 we may assume that T 0 has exactly one matched pair. Observing that then coupons(T 0 ) = |L(T 0 )| − 0.5 while |B(T 0 )| = |L(T 0 )| − 1, the statement follows. 2 Claim 4.10: If T 0 has a locking link b1 b2 ∈ F then T 0 is not deficient. Proof: Say first that T 0 has an additional unmatched leaf a0 apart from a. Let b1 b2 ∈ F be the unique matched pair in T 0 (See Claim 4.7). By shadow minimality degF (s) = 0. Let x be the node covering a0 . As T 0 is closed with respect to a0 , x ∈ T 0 . It is easy to see (see later in Figure 4.3) that T 0 contains exactly one stem if it chose a locking link. Thus x 6∈ {b1 , b2 , s} because b1 b2 ∈ F and degF (b1 ) = degF (b2 ) = 1. Thus x ∈ T 0 ∩ X and we may assume that x is not compound (see Claim 4.7). Thus T 0 has a ticket and is not deficient by Claim 4.9. Hence to try and avoid a ticket, the tree T 0 must have the shape as in Definition 3.4. But then there must be a link in F covering the edge between v and its parent, and this link has an endnode x in T . The node x 6= a1 , as the tree is a1 closed. x ∈ / {b1 , b2 } as degF (b1 ) = degF (b2 ) = 1 and b1 b2 ∈ F . Finally, x 6= s as degF (s) = 0. Hence x ∈ X ∩ T 0 and so T 0 has an X-ticket at x hence it is not deficient. 2 Lemma 4.11: If T 0 has no locking links then tickets(T 0 ) ≥ 1 + |U ∩ T 0 | − (2|M ∩ T 0 | + |S ∩ T 0 |) + |M0 ∩ T 0 | .

(5)

Proof: Note that there are no links between nodes in U as greedy 1-contraction does not apply. In order to avoid a ticket every leaf in U ∩ T 0 has to be covered either by a node that is a part of a matched link or by a stem (this follows as we assume that the tree has no locking links, and all compound nodes are leaves). Indeed, if a ∈ U is covered by x ∈ X, since T 0 is a-closed x ∈ T 0 ∩ X and a ticket at x ∈ T 0 follows. In addition, a node is required to cover the root v of T 0 . Thus it takes |U ∩ T 0 | + 1 nodes to cover the non-matched leaves and root of the tree (without tickets) and the covering nodes must be nodes that are part of matched pairs or stems. Every stem has degree 1 (by shadow minimality) and so does any node that belongs to a matched pair. The number of vertices that are part of in matched pairs plus the number of stems is |S ∩ T 0 | + 2|M ∩ T 0 |. Let b = |U ∩ T 0 | + 1 − |S ∩ T 0 | + 2|M ∩ T 0 |. If b is positive, b X-tickets follow. Besides this we have to add the M0 ∩ T 0 M0 -tickets. 2 Claim 4.12: If T 0 has at least 5 leaves and no locking links in F then T 0 has an M0 -ticket and thus it is not deficient. 8

Proof: Note that T 0 has at least 3 unmatched leaves. By Inequality (5), tickets(T 0 ) ≥ 4 − 2 − s, where s is the number of stems in T 0 . Consequently, the only case to consider is 5 leaves and s = 2 (because T 0 has at most 2 stems by Claim 4.8) for otherwise a ticket follows. The unique matched pair can not be a twin pair as M contains no twin links. Let a1 , b1 be the children of s1 and a2 , b2 the children of s2 , where b1 , b1 is the matched pair. Let a3 be the additional unmatched leaf of T 0 . Now, if for example b1 the F open node with respect to T 0 then by shadow minimality degF (s1 ) = 0. This leaves only b2 , s2 to cover a1 , a2 , a3 , and a ticket follows. A ticket follows similarly similarly if b2 is T 0 -open. Finally if, say, s1 is T 0 -open then by shadow minimality a1 b1 ∈ F . To avoid a ticket a2 b2 ∈ F (as a2 can not be connected to s1 by shadow minimality). In this case for the first time we claim an M0 -ticket. Note that twin links do not belong to MF . Thus b1 , b2 are matched in M but as neither b1 nor b2 is matched in MF . Thus an M0 -ticket follows. 2 Claim 4.13: If T 0 has one matched pair, at least 4 leaves, and no stems, then T 0 has an X-ticket, and thus it is not deficient. Proof: As |U | ≥ 2 the claim follows from Lemma 4.11 (because S ∩ T 0 = ∅).

2

Claim 4.14: If T 0 has 3 leaves then it has no stem. Proof: If T 0 has a stem, then the unique matched pair b1 b2 ∈ M is a locking link which is a contradiction 2 We summarize... Corollary 4.15: If T 0 is deficient then all its compound nodes are leaves, it has exactly one matched pair and 3 or 4 leaves. Furthermore, if T 0 has 4 leaves then it has 1 or 2 stems, and if T 0 has 3 leaves then it has no stems.

4.3

Trees with 4 leaves and 2 stems are not deficient

Note that if the root v of T 0 is not r, at least one node in T 0 must be T 0 -open (to cover the root v). And to avoid a ticket if must be either a stem or one of b1 or b2 . The following notion is the third central idea we need (apart from locking links and matching tickets) in order to prove the 1.5 ratio. Definition 4.16: A node x ∈ T 0 is a T 0 -jumper (or just a jumper for short) if there exists a link (x, y) such that: (i) y 6∈ T 0 and (ii) y is active or y is a part of a matched pair. Figure 4.3 depicts two examples of jumper nodes. The root of T 0 is v. Lemma 4.17: If T 0 has 4 leaves and 2 stems then it is not deficient. The theorem will be proved via 4 claims. Before we state these claims we give a few observations. Suppose T 0 contains exactly two stems and 4 leaves. Let the stems be s1 and s2 . By the Matching Invariant, s1 has two twin children, a1 and b1 so that a1 is active and b1 is a matched leaf. Similarly, s2 has two twin children: a matched leaf b2 and an active child a2 . Since there is only one matching link in the tree, the link (b1 , b2 ) must be the matching link. Thus, the tree is as in Figure 4.3 (any other shape will imply at least 5 leaves). 9

v

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s a0

b3

b1

a2

a1

b4

b2

b2 b1

a2

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(A)

(B)

Figure 2: A jumper node b2 : (A) b2 linked to an active node a0 (B) b2 linked to a matched leaf b3 .

v

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Figure 3: 4 leaves and two stems. The edges between v and u can be a path

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Claim 4.18: It can not be that a1 s2 ∈ E is a link and b1 is a jumper. Also, it can not be that a2 s1 ∈ E and b2 is a jumper. Proof: This would imply a greedy 3 or 4 contraction step. For example if a2 s1 ∈ E and b2 is linked to a leaf a0 , then adding a2 s1 , a1 b1 and b2 a0 is a 3 greedy step. 2 Claim 4.19: If a1 s2 6∈ E and a2 s1 6∈ E then T 0 owns an M0 -ticket. Proof: Note that in this case a1 b2 , a2 b1 6∈ E as otherwise by shadow completion a1 s2 ∈ E or a2 s1 ∈ E. Hence, to avoid an X-ticket for covering a1 and a2 , we need to assume that a1 b1 , a2 b2 ∈ F . However, this implies that b1 , b2 are both unmatched in MF (twin links do not belong to MF ) and an M0 -ticket follows. 2 Claim 4.20: If a2 s1 6∈ E and (a1 , s2 ) ∈ E then T 0 owns an M0 -ticket. Proof: The link a2 b1 can not exist for otherwise its shadow a2 s1 would exist. Hence, to avoid an X-ticket for covering a2 , it must be that (a2 , b2 ) ∈ F . As b1 is not a jumper (by Claim 4.18), it follows that (b1 , b2 ) are both unmatched in MF (recall that even if a1 b1 ∈ F a1 b1 6∈ MF because it is a twin link) and an M0 -ticket follows. 2 The case that (a1 , s2 ) 6∈ E and (a2 , s1 ) ∈ E follows by symmetry. Claim 4.21: If both (a1 , s2 ) and (a2 , s1 ) are links then T 0 owns a ticket. Proof: By Claim 4.18, both b1 and b2 are not jumpers. To avoid an M0 -ticket F must contain both link b1 a2 and b1 a2 . However, this implies a 2-greedy step, a contradiction. 2 Since we have covered all possibilities this finishes the proof of Lemma 4.17

Deficient trees with 4 leaves

4.4

By Lemma 4.17 and Claim 4.13 a deficient tree with 4 leaves must have one stem. In Fig. 4.4 we depict all shapes of trees with 4 leaves and one stem. The node t is not a stem and a 2 may be a compound node but each of a1 , b1 , b2 is an original leaf.

v

v s

a1

b1

b2 (A)

t

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t s

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b2

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(B)

v

t s

b1 (C)

b2

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b2

(D)

Figure 4: Possible configurations of 4 leaves, 1 stem, and 1 matched pair. Recall that no greedy 3, 4-contractions are possible. This discards some sub-cases as explained by the following observation. 11

Observation 4.22: If (a2 , s) ∈ E then b2 is not a jumper. For example, if b2 a0 ∈ E and a0 is a leaf then the contraction of the 3 links a2 s, b2 a0 and a1 b1 is a greedy 3-contraction. Thus we may assume that either b2 is not a jumper or that (a2 , s) 6∈ E. Claim 4.23: If (a1 , b1 ) 6∈ F the tree owns a ticket. Proof: If a1 b1 6∈ F then to cover a1 without a ticket, a1 needs to be linked to b2 (recall, t is not a stem). But then a2 has to be linked in F to b1 (recall that by shadow-minimality degF (s) = 0). This gives a 2-greedy operation, a contradiction. 2 Claim 4.24: If a2 s ∈ F then T 0 owns an M0 -ticket. Proof: We need to assume a1 b1 ∈ MF (to avoid a ticket). The node b2 can not be a jumper (Observation 4.22). Thus, b2 can not be matched to a leaf of T outside T 0 . As a2 s, a1 b1 ∈ F neither b1 , nor b2 are matched in MF . An M0 -ticket follows. 2 Thus, to avoid a ticket, the only node of L ∪ S that can cover a2 is b2 . Thus we may assume that a1 b1 , a2 b2 ∈ F by Claims 4.23 and 4.24. This avoids a ticket as a2 may be a compound node. We now can show that the tree in part (A) is not deficient. Claim 4.25: The tree in Figure 4.4 part (A) is not deficient. Proof: By the above we may assume that a2 b2 ∈ F . A leaf is created when b2 a2 is contracted. As this degF (b2 ) = degF (b1 ) = 1 and t is not a stem, the node that covers this leaf, belongs to X 2 Claim 4.26: If T 0 is deficient then a1 b1 , a2 b2 ∈ F and s is F -open w.r to T 0 Proof: Note that b1 a1 , b2 a2 ∈ F and thus b1 , b2 , a1 can not cover the root v as they have degree 1 in F . The node a2 is T 0 -closed. Thus s has to cover the root to avoid a ticket. 2 This pins down all the (possibly) deficient trees with 4 leaves. The shape of the tree has to be as (B), (C) or (D) in Figure 4.4, a1 b1 and a2 b2 belong to F , and s is F -open with respect to T 0 .

4.5

Deficient trees with 3 leaves

In [4] the stems have been contracted by the first 1, 2-greedy operations. It was established in [4, Claim 4.9] what is the shape of all deficient trees with 3 leaves, as long T 0 has no stem. But by Claim 4.14, T 0 indeed has no stems. Thus by [4, Claim 4.9] we get the following identification of all deficient trees with 3 leaves. Lemma 4.27: [4, Claim 4.9] A tree with 3 leaves is deficient if there is an ordering of b1 b2 so that: 1. The contraction of b2 a does not create a leaf. 2. b1 is T 0 open. See Figure 4.5 for the description of all deficient trees with 3 leaves. Note that x is not a stem. In case (A) if b1 a exists as well we assume that up(b1 ) is at least as high reaching as up(b2 ) 12

x b1

b2 b2

a

(A)

b1

a

(B)

Figure 5: The thick edges are links. The broken edges are links to vertices outside T 0 . The contraction of ab2 does not create a leaf

4.6

Finding a non-deficient tree when all semi-closed tree are deficient

We now explain a specified method to deal with the case all the trees are deficient. First, find all deficient trees. These trees are clearly disjoint (as they are rooted, if they are not disjoint one of the trees would have to be a subtree of the other which is not possible. This follows by the definition of semi-closed trees). We note that the set of deficient trees may cover only a small part of T . For example take a path r = v1 , . . . , v2k , and attach to every vi a leaf ui . Then attach to v2k any deficient tree so that v2k = v. The matching M is b1 b2 plus the appropriate ui , vi edges. The tree T 0 is a unique semi-closed subtree in this example, and T 0 is deficient. In particular, many matched pairs may not belong to any deficient tree. After finding all deficient trees we operate as follows. For every deficient tree with 4 leaves first momentarily add the edge a1 b1 into I and pay for that with the coupon of a1 . For convenience, let us still call the new compound node by b1 . (Note that for the moment the Matching Invariant does not hold as b1 is a compound node. But we will restore this invariant once the operation of covering a new tree ends). We still consider b1 b2 as a matched link (and this matched link still has 1.5 credit) and a2 as an unmatched node (and indeed it has credit 1). Observe that after this transformation, all deficient trees have 3 leaves. ˜ as follows. The matchings M and M ˜ contains the same matched Now define a new matching M links that do not belong to any deficient tree. But in a deficient tree we replace the match link b 1 b2 by b2 a in the case of a 3 leaves tree, and replace b1 b2 by b2 a2 in the case of a 4 leaves tree. In any ˜ (formally, these changes also mean switching case, make b1 an unmatched leaf with respect to M coupons thus the new matched pair gets 1.5 coupons and the new unmatched leaf gets 1 coupon). ˜ is Note that as the contraction of all the links of the form a2 b2 or ab2 does not create a leaf, M ˜ . An example of this process is given in Figure proper. Find a semi-closed tree T˜ with respect to M ˜ 6. By claim 4.9 if T has a ticket it is not deficient. The way we prove the existence of a ticket is as follows. Assuming its root u is not r (in which case we are done) who can cover the root in T˜? Claim 4.28: If T˜ resulted from a deficient tree with 4 leaves it has a ticket. Proof: Recall that avoid a ticket, we need to assume that a1 b1 ∈ F and a2 b2 ∈ F . Thus the nodes a1 , b1 , b2 can not cover the root as they are original leaves in T and so degF (a1 ) = degF (b1 )) = degF (b2 ) = 1. The stem s can not cover the root since s is contained in b1 , and b1 is an unmatched ˜ , and therefore T˜ is s-closed. Finally, a2 was T 0 closed so it is certainly leaf with respect to M ˜ T -closed. Hence the node covering u belongs to X and a ticket follows 2 13

r

(a)

b3

b4

a2

r

(b)

b1

b3 b2

b4

a2

b1 b2

a1

a1

Figure 6: Illustration of switching the matched pairs and finding T˜. Tree edges are shown by bold lines, matching links by dashed lines. (a) An example when all minimally semi-closed trees are ˜ and the minimally semi-closed trees w.r.t. M ˜. deficient. (b) The matching M Claim 4.29: If the tree T˜ resulted from a deficient tree with 3 leaves then T˜ has a ticket Proof: Clearly T˜ is a-closed as T 0 was. It is also b1 closed as b1 is an unmatched node with ˜ . If we are in case (B) of Figure (4.5) then b2 can not cover u as degF (b2 ) = 1. In case respect to M (A) of the figure, as T˜ is b1 -closed and up(b1 ) is at least as high as up(b2 ), T˜ is b2 closed. Hence the node that covers u belongs to X and T˜ has a ticket 2 Returning to the invariants: It is immediate to see that if T˜ contains nodes from two deficient trees, both trees are contracted into the new compound node. Therefore, the deficient trees remaining after T˜ is contracted are disjoint. We reverse the ˜ . Namely we restore b1 b1 as the matched pairs and the a or a2 vertices operations done to define M as the unmatched leaves and reset the credit. Now we perform all 1 and 2 greedy operation. By Claim ([4, 3.9]) and Claim 3.12 the Matching Invariant is restored and all the other invariants clearly hold as we left a coupon in the compound node created by contracting T˜. This shows that we can find in any case a non-deficient tree and cover it, and then and return to a situation where all the invariants hold. This finished the proof of Lemma 3.14. This also end the proof of Theorem 3.7. Acknowledgement:

We thank Jon Feldman for some useful discussions.

References [1] J. Cheriyan, T. Jord´an, and R. Ravi, On 2-coverings and 2-packing of laminar families. ESA, LNCS 1643 pp. 510–520, 1999. [2] G. Even, J. Feldman, G. Kortsarz and Z. Nutov, A 3/2-approximation for augmenting a connected graph into a two-connected graph. APPROX, pp. 90–101, 2001.

14

[3] N. Garg, R. Khandekar, and K. Talwar, Covering a laminar family. Manuscript, 2000. [4] G. Even, J. Feldman, G. Kortsarz and Z. Nutov, A 1.8 approximation algorithm for augmenting edge-connectivity of a graph from 1 to 2. Manuscript, 2007. [5] H. Nagamochi, An approximation for finding a smallest 2-edge connected subgraph containing a specified spanning tree. Discrete Applied Mathematics, 126:83-113, 2003.

15

Guy Kortsarz‡

Zeev Nutov

§

June 5, 2009

Abstract We consider the following NP-hard problem: given a connected graph G = (V, E) and an edge set E on V disjoint to E, find a minimum size subset of edges F ⊆ E such that (V, E ∪ F ) is 2-edge-connected. In [4] we presented a 1.8 approximation for the problem. In this paper we improve the ratio to 1.5.

∗

Preliminary version appeared in APPROX 2001, LNCS 2129, pp. 90-101, 2001. Dept. of Electrical Engineering-Systems, Tel-Aviv University, Tel-Aviv 69978, Israel. E-mail:[email protected] ‡ Rutgers University, Camden. E-mail:[email protected] Partially suported by NSF grant number 0728787. § The Open University of Israel, Raanana, Israel. E-mail:[email protected]

†

1

1

Introduction

A graph (possibly with parallel edges) is k-edge-connected if there are k pairwise edge-disjoint paths between every pair of its nodes. We consider the following problem: Tree Augmentation Problem (TAP) Instance: A tree T = (V, E) and a set of links E on V disjoint to E. Objective: Find a minimum size subset F ⊆ E of edges such that G + F is 2-edge-connected. TAP is sometimes posed as the problem of covering a laminar family (see e.g., [1]). Namely, given a laminar family E on a groundset V , and an edge set E on V , find a minimum size edge subset F ⊆ E such that for every S ∈ E, there is an edge in F with one endpoint in S and the other in V − S. TAP is also equivalent to the problem of finding a minimum size augmenting edge set to increase the edge-connectivity from k to k + 1 for any odd k; this is since the family of minimum cuts of a k-connected graph with k odd is laminar. TAP is APX-hard even if the set E of links forms a cycle on the leaves of T [1]. For previous work on the problem and discussion see [4]. In the conference version [2] of this paper we presented and sketched a proof of a 1.5-approximation algorithm for the problem, improving the ratio 1.875+ε of Nagamochi [5]. Since the full paper was too long and complicated we presented in [4] an easier 1.8-approximation algorithm for TAP. In this paper we present some additional ideas to prove: Theorem 1.1: TAP admits a 1.5-approximation algorithm. Our ratio matches the recently obtained lower bound on integrality gap of the natural LPrelaxation for TAP, see [3].

2

Preliminaries

Here is some notation used in the paper. Let T = (V, E) be a tree. For u, v ∈ V let (u, v) denote the edge in T and uv the link in E between u and v. Let p(uv) denote the path between u and v in T . A link uv covers all the edges (and nodes) along the path p(uv). We designate a node r of T as the root, and refer to the pair T, r as a rooted tree (we do not mention the root when it is clear from the context). The choice of r defines a partial order on V : u is a descendant of v (or v is an ancestor of u) if v belongs to p(ru); if, in addition, (u, v) ∈ T , then u is a child of v, and v is the parent of u. The leaves of T are the nodes in V − r that have no descendants. We denote the leaf set of T by L(T ), or simply by L, when the context is clear. The rooted subtree of T induced by v and its descendants is denoted by Tv (v is the root of Tv ). A subtree T 0 of T is called a rooted subtree of T if T 0 = Tv for some v ∈ V . For a node set U and a link set M , U ∩ T 0 is the set of nodes in U that belong to T 0 , and M ∩ T 0 is the set of links in M with both endnodes in T 0 . To contract a (not necessarily rooted) subtree T 0 of T /I is to combine all nodes in T 0 into a single node v, named after the least common ancestor of the nodes in T 0 . The edges and links with both endpoints in T 0 are deleted. The edges and links with one endpoint in T 0 now have v as their new endpoint, but retain their correspondence with the edge or link of the original graph. For simplicity, we say that the new node v obtained by contracting T 0 “contains” a node u if u ∈ T 0 . If we add a link uv to a partial solution I, then the nodes along the path p(uv) belong to the same 2-edge-connected component of the augmented graph (V, E ∪ I). Hence, we may contract

1

the nodes along p(uv). For a set of links I ⊆ E, let T /I denote the tree obtained by contracting every 2-edge-connected component of T ∪ I into a single node. Since all contractions are induced by subsets of links, we refer to the contraction of every 2-edge-connected component of T ∪ I into a single node simply as the contraction of the links in I. A link u0 v 0 is a shadow of a link uv if p(u0 v 0 ) ⊆ p(uv). Adding all shadows of existing links does not affect the optimal solution size, since every shadow can be replaced by some existing link covering all edges covered by the shadow. Thus: Assumption 2.1: The set of links E is closed under shadows. The following reductions allow us to make some simplifying assumptions about TAP instances. Definition 2.2: A cover F of T is shadow minimal if for every link uv ∈ F , replacing uv by a proper shadow of uv results in a set of links that does not cover T . Claim 2.3: ([4, Claim 2.6]) Let F be a shadow minimal cover of T . Then there is exactly one link in F incident to every leaf a of T . In particular, the leaf-to-leaf links in F form a matching on L. Further simplifying assumptions can be made when we consider the relationship between pairs of edges in T . Definition 2.4: An edge e of T is reducible if all links in E that cover e are shadows of the same link, or if there exists an edge e0 6= e such that every link that covers e0 and is not a proper shadow of any other link, also covers e. It is not hard to see that we may preprocess the instance so that: Assumption 2.5: ([4, Assumption 2.8]) T has no reducible edges. The following statement shows that if there are no reducible edges, then paths in T consisting of degree 2 nodes can not end at leaves. Proposition 2.6: ([4, Proposition 2.9]) If T has no reducible edges, then every parent of a leaf has at least two children. For X, Y ⊆ V and a link set F , let F (X, Y ) = {xy ∈ F : x ∈ X, y ∈ Y } denote the set of links in F that have one endpoint in X and the other in Y ; for x ∈ V let deg F (x) = |F (x, V )| be the degree of x w.r.t. F . For the rest of the paper fix F to be some optimum shadow minimal solution for an instance T = (V, E), E of TAP.

3 3.1

The lower bound and the algorithm The Improved Lower Bound

Definition 3.1: A node s ∈ V − r is a stem of T if it has exactly two children and both of them are leaves. The link between the two children of a stem, if exists, is called a twin-link.

2

Claim 3.2: ([4, Claim 3.2]) If T has no reducible edges, then there is a twin-link in E between the children of every stem. The following lower bound was established and used in [4]: Lemma 3.3: (The Leaf-Stem Lower Bound) ([4, Lemma 3.3]) Let L be the set of leaves and S be the set of stems of T , and let X = V \ (L ∪ S). Let F be a shadow minimal optimal solution and let MF be the set of leaf-to-leaf non twin links in F . Then: 2 1 1X degF (x). (1) |F | ≥ |L| − |MF | + 3 3 3 x∈X

The following definition was not used in [4] and is one of the main three new ideas we use in order to improve the ratio to 1.5.

x

r

v b2

s b1

a1

Figure 1: Illustration to Definition 3.4. Links are shown by dashed lines and paths by wiggly lines. Definition 3.4: A link b1 b2 is a locking link of a leaf a1 if the following holds (see Figure 3.1). The nodes b1 , b2 , a1 are the leaves of a rooted proper subtree Tv of T (so v 6= r) so that b2 is a child of v, the leaves b1 , a1 are children of a stem s that is a proper descendant of v, and any link in E incident to a1 and distinct from the twin link a1 b1 goes to b2 or to an ancestor x of a1 . The importance of locking links comes from the following simple observation. Observation 3.5: If b1 b2 ∈ F is a locking link of a1 then F contains a link (an F -locked link) from a1 to an ancestor x of a1 . Proof: The only links in E incident to the locked leaf a1 are the links to b1 , b2 and to ancestors of a1 . As b1 , b2 have degree 1 in F , by Claim 2.3, the statement follows. 2 Lemma 3.6: [The Improved Lower Bound] Let F, MF , X be as in the Leaf-Stem Lower Bound. Let M be a maximum matching among all matchings in E without twin links and without locking links. Let M0 be the set of links in M that have no common endpoint with a link in MF . Let J be the set of non-F -locked links in F . Then: 1 1 1X 2 |F | ≥ |L| − |M | + |M0 | + degJ (x). (2) 3 3 3 3 x∈X

3

Proof: Let MF0 be the set of locking links in MF and Note that (MF \ MF0 ) ∪ M0 is a matching without locking links and without twin links of size |MF | − |MF0 | + |M0 |. As M is a maximum matching of this type, we have |M | ≥ |MF | − |MF0 | + |M0 |. Thus |MF | ≤ |M | − |M0 | + |MF0 |. Substituting the last inequality in (1) gives 1X 2 1 |M | − |M0 | + |MF0 | + |F | ≥ |L| − degF (x) 3 3 3 x∈X ! 1 1 1 1 X = |L| − |M | + |M0 | + degF (x) − |MF0 | . 3 3 3 3 x∈X

Now we observe that the last term in the obtained bound equals the last term in (2). Indeed, by Observation 3.5, for any link b1 b2 ∈ MF0 there is a unique link a1 x ∈ F with x ∈ X. 2 The term 32 |L| − 13 |M | in (2) can be computed in polynomial time. The other terms depend on F which is not known to us, and are partly revealed during the run of the algorithm.

3.2

The credit scheme and the Coupons Invariant

We explain how we use the Improved Lower Bound. Let U denote the set of leaves unmatched by M , so |U | = |L| − 2 · |M |. By multiplying both sides of (2) by 1.5 and rearranging terms, we obtain 3 1 1X 3 · |F | ≥ |U | + · |M | + · |M0 | + degJ (x) . (3) 2 2 2 2 x∈X

We prove the following statement that implies Theorem 1.1.

Theorem 3.7: There exists a polynomial time algorithm that for an instance of TAP computes a solution I of size at most the right-hand size of (3). Thus |I| ≤ 1.5 · |F |. Initially, the algorithm assigns coupons to unmatched leaves and to links in M so that the total number of coupons assigned equals the sum of the first two terms in the right hand side of (3). For technical reasons, we also assign 1 coupon to the root r. Initial assignment of coupons: • Every unmatched leaf gets 1 coupon, and r gets 1 coupon. • Every link in M gets 3/2 coupons. Compound nodes. Our algorithm iteratively contracts certain subtrees of T . We refer to the nodes created by contraction as compound nodes. Compound nodes always own 1 coupon. Noncompound nodes are referred to as original nodes (of T ). Each time a contraction takes place, the new compound node gets 1 coupon, which together with the links added is paid by the total credit of the contracted subtree. For technical reasons, r is also considered as a compound node. Our algorithm is designed to maintain the following invariant: Invariant 3.8: [Coupons Invariant] 1. Every unmatched leaf and every compound node of T /I (including r) owns 1 coupon. 2. Every link in M owns 3/2 coupons. 4

P 0 Tickets. To charge the last two terms 21 |M0 | and 12 x∈X degF (x) − |MF | in (3), we introduce tickets. A ticket worth 1/2 coupon. We will have 2 types of tickets, according to the term we want to charge: M0 -tickets are used to charge the term 12 |M0 |. If we can prove that some link e ∈ M belongs to M0 , namely, that e has no endnode in common with a link in MF , then we may claim an M0 -ticket (1/2 coupon) at e. P X-tickets are used to charge the last term 12 x∈X degJ (x) in (3). If we can prove that some x ∈ X is not linked by F to any locked leaf a1 so that its locking link b1 b2 ∈ F , then we may claim an X-ticket (1/2 coupon) at x. In fact, we will often claim tickets in a subtree T 0 of T without specifying exactly the link e ∈ M or the node x ∈ X on which the ticket is claimed, nor its type. All we care about is that T 0 contains a ticket. No ticket is claimed twice, if we follow the following simple rule. Rule for claiming tickets: Immediately after claiming tickets in T 0 , the whole tree T 0 is contracted into a compound node. This indeed implies that no ticked is claimed twice. If an M0 -ticket was claimed on a link e ∈ M ∩T 0 , then e disappears after T 0 is contracted. If an X-ticket is claimed at a node x, then x is contracted into a compound node, while X-tickets are never claimed on compound nodes. Summarizing, we use the following notation for the credit distributed in a (not necessarily rooted) subtree T 0 of T /I. Let coupons(T 0 ) denote the total number of coupons owned by T 0 ; this includes the coupons owned by nodes of T 0 (every unmatched leaf and every compound node of T 0 owns 1 coupon), and link in M with both endnodes in T 0 . Let P 3/2 coupons for every 0 0 0 tickets(T ) = |M0 ∩ T | + x∈X∩T 0 degF (x) − |MF ∩ T 0 | denote the number of tickets in T 0 . Let credit(T 0 ) = coupons(T 0 ) + 21 tickets(T 0 ). The high level idea of our algorithm is to repeatedly find a tree T 0 in T /I and a cover B 0 of T 0 so that credit(T 0 ) ≥ |B 0 | + 1. In such a case we have enough credit in T 0 to pay for the |B 0 | links to cover and contract T 0 , and to leave a coupon on the resulting compound node. Leaving a coupon is crucial in order to mantain Part 1 of the Coupon Invariant. If we iteratively cover trees in this way until T is completely covered, then the number of links we use is at most the right hand side of Inequality (3), as claimed in Theorem 3.7.

3.3

Greedy contractions and the Matching Invariant

One simple way of finding a subtree T 0 and its cover B 0 as above, that may not use rooted subtrees, and relies on coupons only, is as follows. Definition 3.9: If there exists a set I 0 of α links that covers a subtree T 0 of T , and if coupons(T 0 ) ≥ |I 0 | + 1, then adding I 0 to the partial solution, and assigning one coupon to the obtained compound node is a greedy α-contraction. A greedy α-contraction, if exists, can be found in polynomial time if α is a constant. One of the steps of the algorithm is to apply all possible greedy α-contractions exhaustively for α = 1, 2, 3, 4. We now list two properties of the instance after all greedy 1, 2-contractions are exhausted. Definition 3.10: A matching on the leaves of T /I is proper if its contraction does not create a new leaf.

5

Lemma 3.11: ([4, Lemma 3.10]) If all greedy 1, 2-contractions are exhausted, then M is a proper matching. Lemma 3.12: If all greedy 1, 2-contractions are exhausted then every stem has one unmatched child and one matched child. Proof: If the two children of the stem are unmatched then adding the twin link is a greedy 1-contraction. If both children of the stem are matched then adding the two matched links is a greedy 2-contraction. 2 Invariant 3.13: [Matching Invariant] 1. M is a proper matching on the original leaves. 2. Every stem has one matched child and one unmatched child.

3.4

The algorithm

Our approach is to iteratively exhaust greedy α-contractions for α ≤ 4, and then to find a rooted subtree T 0 of T /I with its cover B 0 , to contract T 0 with B 0 , and to leave a coupon on the created compound node (in order to maintain Part 1 of the Coupons Invariant). In the next section we will prove the following statement: Lemma 3.14: Suppose that no greedy α-contraction applies for α ≤ 4. Then there exists a polynomial time algorithm that finds a rooted subtree T 0 of T /I and a cover B 0 ⊆ E of T 0 so that credit(T 0 ) ≥ |B 0 | + 1 holds. Algorithm 1 Approx(T = (V, E), E) (A 1.5-approximation algorithm) 1: I ← ∅. 2: Contract reducible edges of T and update I accordingly. 3: M ← maximum matching of leaf-to-leaf links among all matchings in E not containing twin-links or locking links. 4: while T /I has more than one node do 5: Exhaust greedy α-contractions for α ≤ 4 and update I accordingly. 6: Find a rooted subtree T 0 of T /I and a cover B 0 of T 0 as in Lemma 3.14. 7: Contract T 0 , give 1 coupon on the new compound leaf, and set I ← I ∪ B 0 . 8: end while 9: Return I. Algorithm Approx initiates I ← ∅ as a partial cover. It applies the reductions from Section 2 (specifically, we need Assumptions 2.1 and 2.5), computes a maximum matching M on the leaves among the non-twin links and non-locking links, and initiates the credit scheme as described in Section 3.2. In the main loop, the algorithm iteratively exhausts greedy α-contractions for α ≤ 4, computes T 0 , B 0 as in Lemma 3.14, adds B 0 to I, contracts T 0 , and leaves a coupon on the created compound leaf. The stopping condition is when I covers T , namely, when T /I is a single node.

6

It is easy to verify that all actions needed to be taken by the algorithm can be implemented in polynomial time. The credit scheme used implies that the algorithm computes a solution I as in Theorem 3.7, namely |I| is at most 1.5 times the right-hand size of (3). Hence the approximation ratio of 1.5 follows from the Improved Lower Bound, and the proof of Theorem 1.1 will be complete, if we prove Lemma 3.14. For the rest of the paper we deal with proving Lemma 3.14 via various concepts and claims.

4

Proof of Lemma 3.14

4.1

Semi-closed trees and their basic covers

The up-link incident to a node u is defined as the link uv such that v is as close as possible to the root; under Assumption 2.1, v is an ancestor of u. We denote the highest link incident to u by up(u). For a node set U ⊆ V , we let up(U ) = {up(u) : u ∈ U }. Definition 4.1: A rooted subtree T 0 of T is leaf-closed if every link incident to a leaf of T 0 has its both endnodes in T 0 . A subtree T 0 is minimally leaf closed if it is leaf-closed but any rooted proper subtree of T 0 is not leaf closed. Claim 4.2: ([5]) If T 0 is minimally leaf-closed then up(L(T 0 )) covers T 0 . The following definitions and claims hold for any proper matching M . We will also assume that every link in M owns 3/2 coupons, Definition 4.3: A semi-closed tree (w.r.t. a proper matching M ) is defined and found as follows. (i) Contract every link in M − I to get T /(I ∪ M ). (ii) Find a minimally leaf-closed rooted subtree T 00 = (T /(I ∪ M ))v of T /(I ∪ M ). (iii) T 0 = (T /I)v , namely T 0 is obtained from T 00 by “decontracting” every link in M − I. We note that T 0 may not be closed with respect to matched leaves and in addition it may not be closed with respect to stems; however, T 0 is closed w.r.t. its unmatched leaves. Definition 4.4: Let T 0 be a semi-closed subtree of T /I. The basic cover B(T 0 ) of T 0 consists of M ∩ T 0 (namely, links in M with both endnodes in T 0 ) plus the up-links of the unmatched leaves of T 0 . As we assume that M is proper, B(T 0 ) indeed covers T 0 , by Claim 4.2. It is not hard to see that all semi-closed trees and their basic covers can be computed in polynomial time, see [4]. In what follows, let T 0 be a semi-closed tree, and let v be the root of T 0 . Claim 4.5: ([4]) Let C(T 0 ) denote the set of non-leaf compound nodes of T 0 (this includes r, if r ∈ T 0 ). Then 1 (4) coupons(T 0 ) − |B(T 0 )| = · |M ∩ T 0 | + |C(T 0 )| . 2 In particular, coupons(T 0 ) ≥ |B(T 0 )|. Definition 4.6: A tree is called deficient if credit(T 0 ) < |B(T 0 )| + 1. 7

4.2

Some non-deficient semi-closed trees

Claim 4.7: ([4, Claims 4.4 and 4.6]) If T 0 is deficient then C(T 0 ) = ∅ (in particular, r ∈ / T 0 ) and T 0 has exactly one matched pair. Claim 4.8: If T 0 has at least 3 stems then T 0 is not deficient. Proof: By the Matching Invariant each stem has a child that belongs to a matched pair. If T 0 has at least 3 stems then T 0 has at least 2 matched pairs. Thus T 0 is not deficient, by Claim 4.7. 2 Corollary 4.9: If T 0 has a ticket then T 0 is not deficient. Proof: By Claim 4.7 we may assume that T 0 has exactly one matched pair. Observing that then coupons(T 0 ) = |L(T 0 )| − 0.5 while |B(T 0 )| = |L(T 0 )| − 1, the statement follows. 2 Claim 4.10: If T 0 has a locking link b1 b2 ∈ F then T 0 is not deficient. Proof: Say first that T 0 has an additional unmatched leaf a0 apart from a. Let b1 b2 ∈ F be the unique matched pair in T 0 (See Claim 4.7). By shadow minimality degF (s) = 0. Let x be the node covering a0 . As T 0 is closed with respect to a0 , x ∈ T 0 . It is easy to see (see later in Figure 4.3) that T 0 contains exactly one stem if it chose a locking link. Thus x 6∈ {b1 , b2 , s} because b1 b2 ∈ F and degF (b1 ) = degF (b2 ) = 1. Thus x ∈ T 0 ∩ X and we may assume that x is not compound (see Claim 4.7). Thus T 0 has a ticket and is not deficient by Claim 4.9. Hence to try and avoid a ticket, the tree T 0 must have the shape as in Definition 3.4. But then there must be a link in F covering the edge between v and its parent, and this link has an endnode x in T . The node x 6= a1 , as the tree is a1 closed. x ∈ / {b1 , b2 } as degF (b1 ) = degF (b2 ) = 1 and b1 b2 ∈ F . Finally, x 6= s as degF (s) = 0. Hence x ∈ X ∩ T 0 and so T 0 has an X-ticket at x hence it is not deficient. 2 Lemma 4.11: If T 0 has no locking links then tickets(T 0 ) ≥ 1 + |U ∩ T 0 | − (2|M ∩ T 0 | + |S ∩ T 0 |) + |M0 ∩ T 0 | .

(5)

Proof: Note that there are no links between nodes in U as greedy 1-contraction does not apply. In order to avoid a ticket every leaf in U ∩ T 0 has to be covered either by a node that is a part of a matched link or by a stem (this follows as we assume that the tree has no locking links, and all compound nodes are leaves). Indeed, if a ∈ U is covered by x ∈ X, since T 0 is a-closed x ∈ T 0 ∩ X and a ticket at x ∈ T 0 follows. In addition, a node is required to cover the root v of T 0 . Thus it takes |U ∩ T 0 | + 1 nodes to cover the non-matched leaves and root of the tree (without tickets) and the covering nodes must be nodes that are part of matched pairs or stems. Every stem has degree 1 (by shadow minimality) and so does any node that belongs to a matched pair. The number of vertices that are part of in matched pairs plus the number of stems is |S ∩ T 0 | + 2|M ∩ T 0 |. Let b = |U ∩ T 0 | + 1 − |S ∩ T 0 | + 2|M ∩ T 0 |. If b is positive, b X-tickets follow. Besides this we have to add the M0 ∩ T 0 M0 -tickets. 2 Claim 4.12: If T 0 has at least 5 leaves and no locking links in F then T 0 has an M0 -ticket and thus it is not deficient. 8

Proof: Note that T 0 has at least 3 unmatched leaves. By Inequality (5), tickets(T 0 ) ≥ 4 − 2 − s, where s is the number of stems in T 0 . Consequently, the only case to consider is 5 leaves and s = 2 (because T 0 has at most 2 stems by Claim 4.8) for otherwise a ticket follows. The unique matched pair can not be a twin pair as M contains no twin links. Let a1 , b1 be the children of s1 and a2 , b2 the children of s2 , where b1 , b1 is the matched pair. Let a3 be the additional unmatched leaf of T 0 . Now, if for example b1 the F open node with respect to T 0 then by shadow minimality degF (s1 ) = 0. This leaves only b2 , s2 to cover a1 , a2 , a3 , and a ticket follows. A ticket follows similarly similarly if b2 is T 0 -open. Finally if, say, s1 is T 0 -open then by shadow minimality a1 b1 ∈ F . To avoid a ticket a2 b2 ∈ F (as a2 can not be connected to s1 by shadow minimality). In this case for the first time we claim an M0 -ticket. Note that twin links do not belong to MF . Thus b1 , b2 are matched in M but as neither b1 nor b2 is matched in MF . Thus an M0 -ticket follows. 2 Claim 4.13: If T 0 has one matched pair, at least 4 leaves, and no stems, then T 0 has an X-ticket, and thus it is not deficient. Proof: As |U | ≥ 2 the claim follows from Lemma 4.11 (because S ∩ T 0 = ∅).

2

Claim 4.14: If T 0 has 3 leaves then it has no stem. Proof: If T 0 has a stem, then the unique matched pair b1 b2 ∈ M is a locking link which is a contradiction 2 We summarize... Corollary 4.15: If T 0 is deficient then all its compound nodes are leaves, it has exactly one matched pair and 3 or 4 leaves. Furthermore, if T 0 has 4 leaves then it has 1 or 2 stems, and if T 0 has 3 leaves then it has no stems.

4.3

Trees with 4 leaves and 2 stems are not deficient

Note that if the root v of T 0 is not r, at least one node in T 0 must be T 0 -open (to cover the root v). And to avoid a ticket if must be either a stem or one of b1 or b2 . The following notion is the third central idea we need (apart from locking links and matching tickets) in order to prove the 1.5 ratio. Definition 4.16: A node x ∈ T 0 is a T 0 -jumper (or just a jumper for short) if there exists a link (x, y) such that: (i) y 6∈ T 0 and (ii) y is active or y is a part of a matched pair. Figure 4.3 depicts two examples of jumper nodes. The root of T 0 is v. Lemma 4.17: If T 0 has 4 leaves and 2 stems then it is not deficient. The theorem will be proved via 4 claims. Before we state these claims we give a few observations. Suppose T 0 contains exactly two stems and 4 leaves. Let the stems be s1 and s2 . By the Matching Invariant, s1 has two twin children, a1 and b1 so that a1 is active and b1 is a matched leaf. Similarly, s2 has two twin children: a matched leaf b2 and an active child a2 . Since there is only one matching link in the tree, the link (b1 , b2 ) must be the matching link. Thus, the tree is as in Figure 4.3 (any other shape will imply at least 5 leaves). 9

v

v

s

s a0

b3

b1

a2

a1

b4

b2

b2 b1

a2

a1

(A)

(B)

Figure 2: A jumper node b2 : (A) b2 linked to an active node a0 (B) b2 linked to a matched leaf b3 .

v

u

s1

a

1

s2

b1

b2

a

2

Figure 3: 4 leaves and two stems. The edges between v and u can be a path

10

Claim 4.18: It can not be that a1 s2 ∈ E is a link and b1 is a jumper. Also, it can not be that a2 s1 ∈ E and b2 is a jumper. Proof: This would imply a greedy 3 or 4 contraction step. For example if a2 s1 ∈ E and b2 is linked to a leaf a0 , then adding a2 s1 , a1 b1 and b2 a0 is a 3 greedy step. 2 Claim 4.19: If a1 s2 6∈ E and a2 s1 6∈ E then T 0 owns an M0 -ticket. Proof: Note that in this case a1 b2 , a2 b1 6∈ E as otherwise by shadow completion a1 s2 ∈ E or a2 s1 ∈ E. Hence, to avoid an X-ticket for covering a1 and a2 , we need to assume that a1 b1 , a2 b2 ∈ F . However, this implies that b1 , b2 are both unmatched in MF (twin links do not belong to MF ) and an M0 -ticket follows. 2 Claim 4.20: If a2 s1 6∈ E and (a1 , s2 ) ∈ E then T 0 owns an M0 -ticket. Proof: The link a2 b1 can not exist for otherwise its shadow a2 s1 would exist. Hence, to avoid an X-ticket for covering a2 , it must be that (a2 , b2 ) ∈ F . As b1 is not a jumper (by Claim 4.18), it follows that (b1 , b2 ) are both unmatched in MF (recall that even if a1 b1 ∈ F a1 b1 6∈ MF because it is a twin link) and an M0 -ticket follows. 2 The case that (a1 , s2 ) 6∈ E and (a2 , s1 ) ∈ E follows by symmetry. Claim 4.21: If both (a1 , s2 ) and (a2 , s1 ) are links then T 0 owns a ticket. Proof: By Claim 4.18, both b1 and b2 are not jumpers. To avoid an M0 -ticket F must contain both link b1 a2 and b1 a2 . However, this implies a 2-greedy step, a contradiction. 2 Since we have covered all possibilities this finishes the proof of Lemma 4.17

Deficient trees with 4 leaves

4.4

By Lemma 4.17 and Claim 4.13 a deficient tree with 4 leaves must have one stem. In Fig. 4.4 we depict all shapes of trees with 4 leaves and one stem. The node t is not a stem and a 2 may be a compound node but each of a1 , b1 , b2 is an original leaf.

v

v s

a1

b1

b2 (A)

t

s

a2

a1

v

t s

b1

b2

a2

a1

(B)

v

t s

b1 (C)

b2

a2

a1

b1

a2

b2

(D)

Figure 4: Possible configurations of 4 leaves, 1 stem, and 1 matched pair. Recall that no greedy 3, 4-contractions are possible. This discards some sub-cases as explained by the following observation. 11

Observation 4.22: If (a2 , s) ∈ E then b2 is not a jumper. For example, if b2 a0 ∈ E and a0 is a leaf then the contraction of the 3 links a2 s, b2 a0 and a1 b1 is a greedy 3-contraction. Thus we may assume that either b2 is not a jumper or that (a2 , s) 6∈ E. Claim 4.23: If (a1 , b1 ) 6∈ F the tree owns a ticket. Proof: If a1 b1 6∈ F then to cover a1 without a ticket, a1 needs to be linked to b2 (recall, t is not a stem). But then a2 has to be linked in F to b1 (recall that by shadow-minimality degF (s) = 0). This gives a 2-greedy operation, a contradiction. 2 Claim 4.24: If a2 s ∈ F then T 0 owns an M0 -ticket. Proof: We need to assume a1 b1 ∈ MF (to avoid a ticket). The node b2 can not be a jumper (Observation 4.22). Thus, b2 can not be matched to a leaf of T outside T 0 . As a2 s, a1 b1 ∈ F neither b1 , nor b2 are matched in MF . An M0 -ticket follows. 2 Thus, to avoid a ticket, the only node of L ∪ S that can cover a2 is b2 . Thus we may assume that a1 b1 , a2 b2 ∈ F by Claims 4.23 and 4.24. This avoids a ticket as a2 may be a compound node. We now can show that the tree in part (A) is not deficient. Claim 4.25: The tree in Figure 4.4 part (A) is not deficient. Proof: By the above we may assume that a2 b2 ∈ F . A leaf is created when b2 a2 is contracted. As this degF (b2 ) = degF (b1 ) = 1 and t is not a stem, the node that covers this leaf, belongs to X 2 Claim 4.26: If T 0 is deficient then a1 b1 , a2 b2 ∈ F and s is F -open w.r to T 0 Proof: Note that b1 a1 , b2 a2 ∈ F and thus b1 , b2 , a1 can not cover the root v as they have degree 1 in F . The node a2 is T 0 -closed. Thus s has to cover the root to avoid a ticket. 2 This pins down all the (possibly) deficient trees with 4 leaves. The shape of the tree has to be as (B), (C) or (D) in Figure 4.4, a1 b1 and a2 b2 belong to F , and s is F -open with respect to T 0 .

4.5

Deficient trees with 3 leaves

In [4] the stems have been contracted by the first 1, 2-greedy operations. It was established in [4, Claim 4.9] what is the shape of all deficient trees with 3 leaves, as long T 0 has no stem. But by Claim 4.14, T 0 indeed has no stems. Thus by [4, Claim 4.9] we get the following identification of all deficient trees with 3 leaves. Lemma 4.27: [4, Claim 4.9] A tree with 3 leaves is deficient if there is an ordering of b1 b2 so that: 1. The contraction of b2 a does not create a leaf. 2. b1 is T 0 open. See Figure 4.5 for the description of all deficient trees with 3 leaves. Note that x is not a stem. In case (A) if b1 a exists as well we assume that up(b1 ) is at least as high reaching as up(b2 ) 12

x b1

b2 b2

a

(A)

b1

a

(B)

Figure 5: The thick edges are links. The broken edges are links to vertices outside T 0 . The contraction of ab2 does not create a leaf

4.6

Finding a non-deficient tree when all semi-closed tree are deficient

We now explain a specified method to deal with the case all the trees are deficient. First, find all deficient trees. These trees are clearly disjoint (as they are rooted, if they are not disjoint one of the trees would have to be a subtree of the other which is not possible. This follows by the definition of semi-closed trees). We note that the set of deficient trees may cover only a small part of T . For example take a path r = v1 , . . . , v2k , and attach to every vi a leaf ui . Then attach to v2k any deficient tree so that v2k = v. The matching M is b1 b2 plus the appropriate ui , vi edges. The tree T 0 is a unique semi-closed subtree in this example, and T 0 is deficient. In particular, many matched pairs may not belong to any deficient tree. After finding all deficient trees we operate as follows. For every deficient tree with 4 leaves first momentarily add the edge a1 b1 into I and pay for that with the coupon of a1 . For convenience, let us still call the new compound node by b1 . (Note that for the moment the Matching Invariant does not hold as b1 is a compound node. But we will restore this invariant once the operation of covering a new tree ends). We still consider b1 b2 as a matched link (and this matched link still has 1.5 credit) and a2 as an unmatched node (and indeed it has credit 1). Observe that after this transformation, all deficient trees have 3 leaves. ˜ as follows. The matchings M and M ˜ contains the same matched Now define a new matching M links that do not belong to any deficient tree. But in a deficient tree we replace the match link b 1 b2 by b2 a in the case of a 3 leaves tree, and replace b1 b2 by b2 a2 in the case of a 4 leaves tree. In any ˜ (formally, these changes also mean switching case, make b1 an unmatched leaf with respect to M coupons thus the new matched pair gets 1.5 coupons and the new unmatched leaf gets 1 coupon). ˜ is Note that as the contraction of all the links of the form a2 b2 or ab2 does not create a leaf, M ˜ . An example of this process is given in Figure proper. Find a semi-closed tree T˜ with respect to M ˜ 6. By claim 4.9 if T has a ticket it is not deficient. The way we prove the existence of a ticket is as follows. Assuming its root u is not r (in which case we are done) who can cover the root in T˜? Claim 4.28: If T˜ resulted from a deficient tree with 4 leaves it has a ticket. Proof: Recall that avoid a ticket, we need to assume that a1 b1 ∈ F and a2 b2 ∈ F . Thus the nodes a1 , b1 , b2 can not cover the root as they are original leaves in T and so degF (a1 ) = degF (b1 )) = degF (b2 ) = 1. The stem s can not cover the root since s is contained in b1 , and b1 is an unmatched ˜ , and therefore T˜ is s-closed. Finally, a2 was T 0 closed so it is certainly leaf with respect to M ˜ T -closed. Hence the node covering u belongs to X and a ticket follows 2 13

r

(a)

b3

b4

a2

r

(b)

b1

b3 b2

b4

a2

b1 b2

a1

a1

Figure 6: Illustration of switching the matched pairs and finding T˜. Tree edges are shown by bold lines, matching links by dashed lines. (a) An example when all minimally semi-closed trees are ˜ and the minimally semi-closed trees w.r.t. M ˜. deficient. (b) The matching M Claim 4.29: If the tree T˜ resulted from a deficient tree with 3 leaves then T˜ has a ticket Proof: Clearly T˜ is a-closed as T 0 was. It is also b1 closed as b1 is an unmatched node with ˜ . If we are in case (B) of Figure (4.5) then b2 can not cover u as degF (b2 ) = 1. In case respect to M (A) of the figure, as T˜ is b1 -closed and up(b1 ) is at least as high as up(b2 ), T˜ is b2 closed. Hence the node that covers u belongs to X and T˜ has a ticket 2 Returning to the invariants: It is immediate to see that if T˜ contains nodes from two deficient trees, both trees are contracted into the new compound node. Therefore, the deficient trees remaining after T˜ is contracted are disjoint. We reverse the ˜ . Namely we restore b1 b1 as the matched pairs and the a or a2 vertices operations done to define M as the unmatched leaves and reset the credit. Now we perform all 1 and 2 greedy operation. By Claim ([4, 3.9]) and Claim 3.12 the Matching Invariant is restored and all the other invariants clearly hold as we left a coupon in the compound node created by contracting T˜. This shows that we can find in any case a non-deficient tree and cover it, and then and return to a situation where all the invariants hold. This finished the proof of Lemma 3.14. This also end the proof of Theorem 3.7. Acknowledgement:

We thank Jon Feldman for some useful discussions.

References [1] J. Cheriyan, T. Jord´an, and R. Ravi, On 2-coverings and 2-packing of laminar families. ESA, LNCS 1643 pp. 510–520, 1999. [2] G. Even, J. Feldman, G. Kortsarz and Z. Nutov, A 3/2-approximation for augmenting a connected graph into a two-connected graph. APPROX, pp. 90–101, 2001.

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[3] N. Garg, R. Khandekar, and K. Talwar, Covering a laminar family. Manuscript, 2000. [4] G. Even, J. Feldman, G. Kortsarz and Z. Nutov, A 1.8 approximation algorithm for augmenting edge-connectivity of a graph from 1 to 2. Manuscript, 2007. [5] H. Nagamochi, An approximation for finding a smallest 2-edge connected subgraph containing a specified spanning tree. Discrete Applied Mathematics, 126:83-113, 2003.

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