Tree Spanners for Bipartite Graphs and Probe Interval Graphs 1

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1 Algorthmca (2007) 47: DOI: /s y Algorthmca 2006 Sprnger Scence+Busness Meda, Inc. Tree Spanners for Bpartte Graphs and Probe Interval Graphs 1 Andreas Brandstädt, 2 Feodor F. Dragan, 3 Hoang-Oanh Le, 2 Van Bang Le, 2 and Ryuhe Uehara 4 Abstract. A tree t-spanner T n a graph G s a spannng tree of G such that the dstance between every par of vertces n T s at most t tmes ther dstance n G. The tree t-spanner problem asks whether a graph admts a tree t-spanner, gven t. We frst substantally strengthen the known results for bpartte graphs. We prove that the tree t-spanner problem s NP-complete even for chordal bpartte graphs for t 5, and every bpartte ATE-free graph has a tree 3-spanner, whch can be found n lnear tme. The prevous best known results were NP-completeness for general bpartte graphs, and that every convex graph has a tree 3-spanner. We next focus on the tree t-spanner problem for probe nterval graphs and related graph classes. The graph classes were ntroduced to deal wth the physcal mappng of DNA. From a graph theoretcal pont of vew, the classes are natural generalzatons of nterval graphs. We show that these classes are tree 7-spanner admssble, and a tree 7-spanner can be constructed n O(m log n) tme. Key Words. Chordal bpartte graph, Interval bgraph, NP-completeness, Probe nterval graph, Tree spanner. 1. Introducton. A tree t-spanner T n a graph G s a spannng tree of G such that the dstance between every par of vertces n T s at most t tmes ther dstance n G. The tree t-spanner problem asks whether a graph admts a tree t-spanner, gven t. The noton s ntroduced by Ca and Cornel [1], [2], who found numerous applcatons n dstrbuted systems and communcaton networks; for example, t was shown that tree spanners can be used as models for broadcast operatons [3] (see also [4]). Moreover, tree spanners were used n the area of bology [5], and approxmatng the bandwdth of graphs [6]. We refer to [7] [9] for more background nformaton on tree spanners. The tree t-spanner problem s NP-complete n general [2] for any t 4. However, t can be solved effcently for some partcular graph classes. Especally, the complexty of the tree t-spanner problem s well nvestgated for the class of chordal graphs and ts subclasses. For t 4, the problem s NP-complete for chordal graphs [9], strongly chordal graphs are tree 4-spanner admssble [10] (.e., every strongly chordal graph has 1 An extended abstract of ths paper appeared n the Proceedngs of the 29th Workshop on Graph Theoretc Concepts n Computer Scence (WG 2003), June 19 21, 2003, Elspeet, The Netherlands, pp Lecture Notes n Compter Scence 2880, Sprnger, Berln, The research of H.-O. Le was supported by DFG, Project No. Br1446-4/1. Ths work was done whle R. Uehara was vstng the Unversty of Waterloo. 2 Insttut für Theoretsche Informatk, Fachberech Informatk, Unverstät Rostock, Rostock, Germany. {ab,hoang-oanh.le,le}@nformatk.un-rostock.de. 3 Department of Computer Scence, Kent State Unversty, Kent, OH 44242, USA. dragan@cs.kent.edu. 4 School of Informaton Scence, Japan Advanced Insttute of Scence and Technology, Asahda 1-1, Nom, Ishkawa , Japan. Receved June 17, 2003; revsed August 22, Communcated by H. Gabow. Onlne publcaton July 25, 2006.

2 28 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara a tree 4-spanner), and the followng graph classes are tree 3-spanner admssble: nterval graphs [11], drected path graphs [12], splt graphs [6] (see also [9] for other known results). We frst focus on the tree t-spanner problem for bpartte graphs and ts subclasses. The class of bpartte graphs s a wde and mportant class from both practcal and theoretcal ponts of vew. However, the known results for the complexty of the tree t-spanner problem for bpartte graphs and ther subclasses are few compared wth the chordal graphs and ther subclasses. NP-completeness s only known for general bpartte graphs (ths result can be deduced from the constructon n [2]), and the problem can be solved for regular bpartte graphs and convex graphs as follows: a regular bpartte graph s tree 3-spanner admssble f and only f t s complete [11]; and any convex graph s tree 3-spanner admssble [6]. (Convex graphs were ntroduced by Brandstädt et al. [13]; refer to Secton 2 for a defnton, and see the Appendx for further detals.) We substantally strengthen the known results for bpartte graph classes, and reduce the gap. We show that the tree t-spanner problem s NP-complete even for chordal bpartte graphs for t 5. The class of chordal bpartte graphs s a bpartte analog of chordal graphs, ntroduced by Golumbc and Goss [14], and has applcatons to nonsymmetrc matrces (see [15]). We also show that every bpartte asterodal-trple-edge-free (ATEfree) graph has a tree 3-spanner, and such a tree spanner can be found n lnear tme. The class of ATE-free graphs was ntroduced by Müller [16] to characterze nterval bgraphs. The class of nterval bgraphs s a bpartte analog of nterval graphs and was ntroduced by Harary et al. [17]. Our results reduce the gap between the upper and lower bounds of the complexty of the tree t-spanner problem for bpartte graph classes snce the followng proper nclusons are known [16], [18]: convex graphs nterval bgraphs bpartte ATE-free graphs chordal bpartte graphs bpartte graphs. We next focus on the tree t-spanner problem on probe nterval graphs and related graph classes. The class of probe nterval graphs was ntroduced by Zhang to deal wth the physcal mappng of DNA, whch s a problem arsng n the sequencng of DNA (see [19] [22] for the background). A probe nterval graph s obtaned from an nterval graph by desgnatng a subset P of vertces as probes, and removng the edges between pars of vertces n the remanng set N of nonprobes. In the orgnal papers [19], [22], Zhang ntroduced two varatons of probe nterval graphs. An enhanced probe nterval graph s the graph obtaned from a probe nterval graph by addng the edges jonng two nonprobes f they are adjacent to two ndependent probes. The class of STS-probe nterval graphs s a subset of the probe nterval graphs; n those graphs all probes are ndependent. From the graph theoretcal pont of vew, t has been shown that all probe nterval graphs are weakly chordal [20], and enhanced probe nterval graphs are chordal [19], [22]. In the Appendx we show that (1) the class of STS-probe nterval graphs s equvalent to the class of convex graphs (hence the class s tree 3-spanner admssble), and (2) the class of the (enhanced) probe nterval graphs s ncomparable wth the classes of strongly chordal graphs and rooted drected path graphs. We also menton that, for any gven

3 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 29 probe nterval graph, the graph obtaned by removng all edges jonng probe vertces s an nterval bgraph. Hence, from both vewponts of graph theory and bology, the tree t-spanner problem for (enhanced) probe nterval graphs s worth nvestgatng. Especally, t s natural to ask f those graph classes are tree t-spanner admssble for fxed nteger t. We gve the postve answer to that queston: The classes of probe nterval graphs and enhanced probe nterval graphs are tree 7-spanner admssble. A tree 7-spanner of a (enhanced) probe nterval graph can be constructed n O(m + n log n) tme f t s gven wth an nterval model. Recently, Johnson and Spnrad showed that the recognton problem for the class of probe nterval graphs can be solved n O(n 2 ) tme f each vertex s gven wth nformaton whether t s probe or nonprobe [23], and the tme complexty was mproved to O(m log n) tme by McConnell and Spnrad [24]. Those recognton algorthms also construct wthn the same tme bounds an ntersecton model of a probe nterval graph. Therefore, usng ther algorthms, we can construct a tree 7-spanner for a gven (enhanced) probe nterval graph G = ( P, N, E) n O(m log n) tme. 2. Prelmnares. Gven a graph G = (V, E) and a subset U V, the subgraph of G nduced by U s the graph (U, F), where F ={{u,v} {u,v} E for u,v U}, and s denoted by G[U]. For a subset F of E, we sometmes unfy the edge set F and ts edge nduced subgraph (U, F) wth U ={v {u,v} F for some u V }. A sequence of the vertces v 0,v 1,...,v l s a path, denoted by (v 0,v 1,...,v l ),f{v j,v j+1 } E for each 0 j l 1. The length of a path s the number of edges on the path. For two vertces u and v on G, the dstance of the vertces s the mnmum length of the paths jonng u and v, and s denoted by d G (u,v).acycle s a path begnnng and endng wth the same vertex. The dsk of radus k centered at v s the set of all vertces wth dstance at most k to v, D k (v) ={w V : d G (v, w) k}, and the kth neghborhood N k (v) of v s defned as the set of all vertces at dstance k to v, that s N k (v) ={w V : d G (v, w) = k}. By N(v) we denote the neghborhood of v,.e., N(v) := N 1 (v). More generally, for a subset S V let N(S) = v S N(v) denote the neghborhood of S. (We note that S N(S) may be nonempty.) A connected acyclc edge set s called a tree. A tree jonng all vertces s called a spannng tree.atree t-spanner T n a graph G s a spannng tree of G such that for each par u and v n G, d T (u,v) t d G (u,v). We say that G s tree t-spanner admssble f t contans a tree t-spanner. The tree t-spanner problem s to determne, for gven graph and postve nteger t, f the graph admts a tree t-spanner. A class C of graphs s sad to be tree t-spanner admssble f every graph n C s tree t-spanner admssble. On the tree t-spanner problem, the followng result plays an mportant role: LEMMA 1 [2]. A spannng tree T of G s a tree t-spanner f and only f for every edge {u,v} of G, d T (u,v) t.

4 30 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara A graph G = (V, E) s bpartte f V can be dvded nto two sets V 1 and V 2 wth V 1 V 2 = V and V 1 V 2 = such that every edge jons a vertex n V 1 and another one n V 2. It s well known that a graph G s bpartte f and only f G contans no cycle of odd length [25]. Thus, for each postve nteger k, a tree 2k-spanner of a bpartte graph G s also a tree (2k 1)-spanner. Hence we consder a tree t-spanner for each odd number t for bpartte graphs n ths paper. Here we defne the graph classes dealt wth n ths paper. See the Appendx and [18] and [21] for further detals and references. Interval graphs and related classes. A graph (V, E) wth V ={v 1,v 2,...,v n } s an nterval graph f there s a set of ntervals I ={I 1, I 2,...,I n } such that {v,v j } E f and only f I I j for each and j wth 1, j n. We call the set I the nterval representaton of the graph. For each nterval I, we denote by R(I ) and L(I ) the rght and left endponts of the nterval, respectvely (hence we have L(I ) R(I )). A bpartte graph (X, Y, E) wth X ={x 1, x 2,...,x n1 } and Y ={y 1, y 2,...,y n2 } s an nterval bgraph f there are famles of ntervals I X ={I 1, I 2,...,I n1 } and I Y = {J 1, J 2,...,J n2 } such that {x, y j } E f and only f I J j for each and j wth 1 n 1 and 1 j n 2. We also call the famles of ntervals (I X, I Y ) nterval representaton of the graph. We sometmes unfy a vertex v and ts correspondng nterval I ; I v denotes the nterval correspondng to the vertex v, and R(v) and L(v) denote R(I v ) and L(I v ), respectvely. Chordal graphs and related classes. An edge whch jons two vertces of a cycle but s not tself an edge of the cycle s a chord of that cycle. A graph s chordal f each cycle of length at least 4 has a chord. A graph G s weakly chordal f G and Ḡ contan no nduced cycle C k wth k 5. A bpartte graph G s chordal bpartte f each cycle of length at least 6 has a chord. Let the neghborhood N(e) of an edge e ={v, w} be the unon N(v) N(w) of the neghborhoods of the end-vertces of e. Three edges of a graph G form an asterodal trple of edges (ATE) f for any two of them there s a path from the vertex set from one to the vertex set of the other that avods the neghborhood of the thrd edge. Asterodal-trple-edge-free (ATE-free) graphs are those graphs whch do not contan any ATE. Ths class of graphs was ntroduced n [16], where t was also shown that any nterval bgraph s an ATE-free graph, and any bpartte ATE-free graph s chordal bpartte. For a bpartte graph (X, Y, E), an orderng < of X has the adjacency property f for each vertex y Y, N(y) conssts of vertces that are consecutve (an nterval) n the orderng < of X. A bpartte graph s convex f there s an orderng of X or Y that fulflls the adjacency property [13]. Probe nterval graphs and related classes. A graph G = (V, E) s a probe nterval graph f V can be parttoned nto subsets P and N (correspondng to the probes and nonprobes) and each v V can be assgned to an nterval I v such that {u,v} E f and only f both I u I v and at least one of u and v s n P. In ths paper we assume that P and N are gven, and we denote the consdered probe nterval graph by G = (P, N, E). Note that N s an ndependent set, G[P] s an nterval graph, and G[P {v}] s also an nterval graph for any v N. Let G = (P, N, E) be a probe nterval graph. Let E + be a set of edges {u 1, u 2 } wth u 1, u 2 N such that there are two probes v 1 and v 2 n P such that {v 1, u 1 } E,{v 1, u 2 } E,{v 2, u 1 } E,{v 2, u 2 } E, and {v 1,v 2 } E. Intutvely,

5 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 31 nonprobes u 1 and u 2 are joned by the edge n E + f (1) there are two ndependent probes v 1 and v 2, and (2) both v 1 and v 2 overlap u 1 and u 2. In the case we know that ntervals I u1 and I u2 have to overlap n any affrmatve nterval representatons. Each edge n E + s called an enhanced edge, and the resultng graph G + = (P, N, E E + ) s sad to be an enhanced probe nterval graph. See [19] [22] for further detals. 3. NP-Completeness for Chordal Bpartte Graphs. In ths secton we show that, for any t 5, the tree t-spanner problem s NP-complete for chordal bpartte graphs. The proof s a reducton from Monotone 3SAT whch conssts of nstances of 3SAT such that each clause contans ether only negated varables or only non-negated varables (see [LO2] of [26]), for whch the followng famly of chordal bpartte graphs wll play an mportant role. Frst, S 0 [a, b] s an edge {a, b}, and S 1 [a, b] s the 4-cycle (a, b, b, a, a). Next, for a fxed nteger l>1, S l+1 [a, b] s obtaned from one cycle (a, b, b, a, a), S l [a, a ], S l [b, b ], and S l [a, b ] by dentfyng the correspondng vertces (see Fgure 1). We connect the vertces a and b to other graphs, and use S l [a, b] as a subgraph of bgger graphs. Sometmes, when the context s clear, we smply wrte S l for S l [a, b]. In case l>0 we wrte (a, a, b, b, a) for the 4-cycle n S l [a, b] contanng the edge {a, b}. Each of the edges {a, a }, {a, b }, {b, b } belongs to a unque S l 1, the correspondng S l 1 n S l [a, b]to{a, a }, {a, b }, {b, b }, respectvely. The followng observatons collect basc facts on S l used n the reducton later. OBSERVATION 2. For every nteger l 0, S l [a, b] has a tree (2l + 1)-spanner contanng the edge {a, b}. PROOF. By nducton on l. The case l = 0 s clear. Let l>0, and let (a, a, b, b, a) be the 4-cycle n S l [a, b] contanng the edge {a, b}. Let L, M, R be the correspondng S l 1 contanng the edge {a, a }, {a, b }, {b, b }, respectvely. By the nducton hypothess, each of L, M, R has a tree (2l 1)-spanner T L, T M, T R contanng the edge {a, a }, {a, b }, {b, b }, respectvely. Let TM a, T M b be the connected components of T M {a, b } wth a TM a and b TM b. Then T L TM a and T R TM b are two dsjont trees and T := (T L TM a ) (T R TM b ) {a, b} s a spannng tree of S l [a, b]. Moreover, T s a tree (2l + 1)-spanner of S l [a, b]. To see ths we need only consder edges {x, y} M such that x TM a b and y TM. For such edges we have: The (x, y)-path n T conssts of the (x, a )-path n TM a, the (y, b )-path S1 [a; b] S 2 [a; b] S 3 [a; b] a b a b a b a 0 b 0 a 0 b 0 a 0 b 0 Fg. 1. The graph S l [a, b].

6 32 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara H a e b Fg. 2. The graph obtaned from H and S l [a, b] by dentfyng the edge e ={a, b}. n T b M, and the edges {a, a}, {a, b}, {b, b }. Therefore, d T (x, y) = d TM (x, y) 1 + 3, hence, as T M s a tree (2l 1)-spanner n M, d T (x, y) (2l 1) = 2l + 1. OBSERVATION 3. Let H be an arbtrary graph and let e be an arbtrary edge of H. Let K be an S l [a, b] dsjont from H. Let G be the graph obtaned from H and K by dentfyng the edges e and {a, b}; see Fgure 2. Suppose that T s a tree t-spanner n G, t > 2l, such that the (a, b)-path n T belongs to H. Then d T (a, b) t 2l. PROOF. By nducton on l.forl = 0, the statement follows drectly from the fact that T s a tree t-spanner of G. Let l>0, and suppose nductvely that the statement s true for arbtrary H and S l 1. Let (a, a, b, b, a) be the 4-cycle n K contanng the edge {a, b}, and let L, M, R be the correspondng S l 1 n K contanng the edge {a, a }, {a, b }, {b, b}, respectvely. Let P be the (a, b)-path n T. By assumpton, P H. Consder the (a, a )-path Q n T. We dstngush two cases. Case 1: Q L. In ths case, by defnton of G, Q belongs to H R M and P s a subpath of Q. The nducton hypothess appled to H := H R M and L yelds d T (a, a ) t 2(l 1), hence d T (a, b) = d T (a, a ) d T (b, a ) t 2(l 1) 2 = t 2l. Case 1 s settled. Case 2: Q L. Let P be the (a, b )-path n T.IfP M then P Q P s the (b, b )-path n T. The nducton hypothess appled to H := H M L and R yelds d T (b, b ) t 2(l 1), hence d T (a, b) = d T (a, a ) d T (b, a ) t 2(l 1) 2 = t 2l.

7 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 33 If P M then P L H R and Q P s a subpath of P. The nducton hypothess appled to H := H L R and M yelds hence d T (a, b ) t 2(l 1), d T (a, b) = d T (a, b ) d T (a, a) d T (b, b) t 2(l 1) 1 1 = t 2l. In ether case we are done. Observaton 3 ndcates a way to force an edge {x, y} to be a tree edge for gven odd t: choosng l = (t 1)/2 shows that {a, b} must be an edge of any tree t-spanner T. We now descrbe the reducton. Let k 2 be an nteger, and let F be a 3SAT formula wth m clauses C j for 1 j m, over n varables x for 1 n. We construct a chordal bpartte graph G from F such that G has a tree (2k + 1)-spanner f and only f F s satsfable. DEFINITION 4. In a graph G, an edge {a, b} s sad to be forced by an S l f G s obtaned from two dstnct graphs S l [a, b] and the rest by dentfyng the edges {a, b} n S l [a, b] and an edge n the rest. We requre that each two S l [a, b] and S l [c, d] have at most two vertces n {a, b, c, d} n common. An edge {a, b} s sad to be strongly forced f t s forced by two S k [a, b]. Hereafter, we omt by two S k [a, b] for each strongly forced edge snce t s always forced by two S k [a, b] for the fxed k. By Observaton 3, f G has a tree (2k + 1)-spanner T every strongly forced edge must belong to T. For each varable x create the gadget G(x ) as follows: Take 2m + 4 vertces x 1,...,x m, x 1,...,x m, p, q, r, s, and add the edges {x j, x j } for 1 j, j m, {q, x j } for 1 j m, {r, x j } for 1 j m, {p, x j } for 1 j m, {s, x j } for 1 j m, and {p, r }, {r, s }, {s, q }. Furthermore, each of the edges {p, r }, {r, s }, {s, q }, and {x j, x j } wth 1 j m, s a strongly forced edge, force each edge {a, b} {{q, x j }: 1 j m} {{r, x j }: 1 j m} {{p, x j }:1 j m} {{s, x j }:1 j m} {{x j, x j }:1 j, j m, j j } by an S k 1 [a, b]. Thus, the subgraph n G(x ) nduced by the two ndependent sets {x 1,...,x m } {p, s } and {x 1,...,x m } {q, r } plus the edge {p, q } s a complete bpartte graph (see Fgure 3; n the fgure the S k and S k 1 are omtted, and thck edges are strongly forced). The vertex x j (x j, respectvely) wll be connected to the clause gadget of clause C j f x (x, respectvely) s a lteral n C j. All edges {r, x j } (1 j m) or else all edges {s, x j } (1 j m) wll belong to any tree (2k + 1)-spanner (f any) of the graph G whch we are gong to descrbe.

8 34 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara q r x 1 x1 2 x1 m p s x 1 x 2 1 x m 1 Fg. 3. The gadget G(x ). DEFINITION 5. A clause s postve (negatve, respectvely) f t contans only varables (negaton of varables). We note that each clause s ether postve or negatve snce the gven formula s an nstance of Monotone 3SAT. For each clause C j, G(C j ) s the 4-cycle (c + j, d+ j, d j, c j, c+ j ) where {c + j, d+ j }, {d + j, d j }, and {d j, c j } are strongly forced edges (see Fgure 4). Fnally, the graph G = G(F) s obtaned from all G(v ) and G(C j ) by dentfyng all vertces p, q, r, and s to a sngle vertex p, q, r, and s, respectvely (thus, {p, r}, {r, s} and {s, q} are edges n G), and addng the followng addtonal edges: Connect every x j wth every x j ( ). (Thus, the subgraph nduced by the two ndependent sets {x j :1 n, 1 j m} {p, s} and {x j :1 n, 1 j m} {q, r} plus the edge {p, q} s a complete bpartte graph.) For every postve clause C j :Ifx s n C j then connect x j wth c + j and force the edge {x j, c+ j } by an S k 2[x j, c+ j ]. Connect c j wth r and force the edge {c j, r} by an S k 2 [c j, r]. j For every negatve clause C j :Ifx s n C j then connect x wth c j and force the edge {x j, c j } by an S k 2[x j, c j ]. Connect c+ j wth s and force the edge {c + j, s} by an S k 2 [c + j, s]. The descrpton of the graph G = G(F) s complete. Clearly, G can be constructed n polynomal tme. See Fgure 5 for an example. LEMMA 6. G s chordal bpartte. PROOF. Frst note that each S l s a chordal bpartte graph. By constructon, {x j :1 n, 1 j m} {p, s} {c j :1 j m} {d+ j :1 j m} c + j c j d + j d j Sk Sk Fg. 4. The gadget G(C j ).

9 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 35 q r x1 2 x2 2 x4 2 p s x 1 1 x 1 2 x 1 3 c + 1 c 1 c + 2 c 2 d + 1 d 1 d + 2 d 2 Fg. 5. The reducton gven C 1 = (x 1, x 2, x 3 ) and C 2 = (x 1, x 2, x 4 ). and {x j :1 n, 1 j m} {q, r} {c + j :1 j m} {d j :1 j m} are ndependent sets. Ths partton can be extended n a natural way to a bpartton of V (G) nto two ndependent sets. So, G s bpartte. Next, let G be the subgraph of G nduced by and A := {x j :1 n, 1 j m} {p, s} B := {x j :1 n, 1 j m} {q, r}. Snce G +{{p, q}} s a complete bpartte graph wth the bpartton (A, B), G s a chordal bpartte graph. On the other hand, snce t s the dsjont unon of the clause gadgets, G G s a chordal bpartte graph. We consder an nduced cycle Z n G contanng vertces from both G and G G. By constructon, Z (G G ) {c + j, c j :1 j m}. Let C j be a postve clause. If c j Z then (r, c j, c+ j ) must be a subpath of Z, therefore Z = (r, c j, c+ j, x j, r) for some. Ifc+ j Z (and c j Z) then, for some 1, 2, (x j 1, c + j, x j 2 ) s a subpath of Z. Snce c + j s the neghbor outsde G of x j 1 and of x j 2, Z = (v, x j 1, c + j, x j 2,v)for a vertex v {q, r} {x j :1 n}. Smlarly, Z s a 4-cycle f C j s a negatve clause. Thus, G s chordal bpartte as clamed.

10 36 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara LEMMA 7. Suppose G admts a tree (2k + 1)-spanner. Then F s satsfable. PROOF. Let T be a tree (2k +1)-spanner of G. By constructon of G and Observaton 3, the followng edges of G belong to T : {p, r}, {r, s}, {s, q}, and {x j, x j } for 1 n, 1 j m, {c + j, d+ j }, {d + j, d j }, and {d j, c j } for 1 j m. CLAIM 1. For every and j, {q, x j } E(T ) and {p, x j } E(T ). PROOF OF CLAIM 1. If, for some, j, {q, x j j } E(T) then (p, r, s, q, x, x j ) s the (p, x j )-path n T, hence d T (p, x j ) = 5. However, by Observaton 3, d T (p, x j ) (2k + 1) 2(k 1) = 3, a contradcton. By symmetry, we have {p, x j } E(T ). CLAIM 2. For every and j, exactly one of {r, x j } and {s, x j } belongs to T. PROOF OF CLAIM 2. Both edges {r, x j } and {s, x j } cannot belong to T, otherwse they would form, together wth {r, s} and {x j, x j }, a cycle n T. Now, assume to the contrary that, for some, j, nether {r, x j } nor {s, x j } belongs to T. Then by Observaton 3, d T (r, x j ) = 3 and d T (s, x j ) = 3. Note that by Clam 1, {q, x j } E(T ) and {p, x j } E(T ). Hence by Observaton 3, d T (q, x j ) = 3 and d T (p, x j ) = 3, too. Let P be the (r, x j )-path n T. If P contans {r, s} and {x j, x j } then clearly d T (r, x j ) 5, a contradcton. If P contans {r, s} but not {x j, x j } then wrte P = (r, s,v,x j ). By assumpton, v x j, and as s and p are nonadjacent, v p. Thus, (p, r, s,v,x j, x j ) s the (p, x j )- path n T, hence d T (p, x j ) = 5, a contradcton. If P contans {x j, x j } but not {r, s} then wrte P = (r,v,x j, x j ). By assumpton, v s, and as q and r are nonadjacent, v q. Thus (q, s, r,v,x j, x j j ) s the (q, x )-path n T, hence d T (q, x j ) = 5, a contradcton. If P does not contan {x j, x j } and {r, s} then wrte P = (r, u,v,x j ). In ths case, u,v {s, x j }. Thus (s, r, u,v,x j, x j ) s the (s, x j )-path n T, hence d T (s, x j ) = 5, a contradcton. CLAIM 3. For each, ether all edges {r, x j }, wth 1 j m, belong to T, or all edges {s, x j }, wth 1 j m, belong to T. PROOF OF CLAIM 3. Assume to the contrary that there exst j 1 j 2 such that {r, x j 1 } E(T ) but {r, x j 2 j } E(T ). By Clam 2, {s, x 2 } E(T). Thus, (x j 2 j, x 2, s, r, x j 1 j, x 1 ) s the (x j 2 j, x 1 )-path n T, hence d T (x j 2 j, x 1 )=5. However, by Observaton 3, d T (x j 2 j, x 1 ) (2k + 1) 2(k 1) = 3, a contradcton. Thus, all or none of the edges {r, x j }, wth 1 j m, belong to T. By symmetry, all or none of the edges {s, x j } wth 1 j m, belong to T. Clam 3 follows.

11 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 37 Now, defne a truth assgnment f for varables x,1 n, as follows: { j true f, for some j, {r, x f (x ) = } E(T), false otherwse. By Clam 3, f s well-defned. We are gong to show that f (F) = true. Frst, consder a postve clause C j = (x 1, x 2, x 3 ) and assume to the contrary that f (x 1 ) = f (x 2 ) = f (x 3 ) = false. That s, {r, x j 1 }, {r, x j 2 }, and {r, x j 3 } do not belong to T. By Clam 2, {s, x j 1 }, {s, x j 2 }, and {s, x j 3 } are edges of T. Recall that the edges {c + j, d+ j }, {d + j, d j }, and {d j, c + j } are edges of T, too. Now, snce T s a tree, exactly one of the edges {c + j, x j 1 }, {c + j, x j 2 }, {c + j, x j 3 }, and {c j, r} belongs to T.If{c j, r} E(T ) then (c+ j, d+ j, d j, c j, r, s, x 1 j, x j 1 ) s the (c + j, x j 1 )-path n T, hence d T (c + j, x j 1 ) = 7. However, by Observaton 3, d T (c + j, x j 1 ) (2k + 1) 2(k 2) = 5, a contradcton. If {c + j, x j } E(T) for one { 1, 2, 3 } then (c j, d j, d + j, c + j, x j, x j, s, r) s the (c j, r)-path n T, hence d T (c j, r) = 7, contradctng Observaton 3 agan. Thus, all postve and, smlarly, all negatve clauses C j are satsfed by the assgnment f. Thus each clause C j of F s satsfed by the assgnment f, provng Lemma 7. DEFINITION 8. If x C j (x C j ) then we say, for convenence, that the vertex x j (x j, respectvely) s the correspondng vertex of the varable x (lteral x, respectvely). Note that the correspondng vertex s not shared by two clauses. LEMMA 9. Suppose F s satsfable. Then G admts a tree (2k + 1)-spanner. PROOF. Let f be a truth assgnment for varables x that satsfy F. We frst construct a spannng tree T of G, the subgraph of G nduced by p, q, r, s, x j, x j wth 1 n, 1 j m, c + j, c j, d+ j, d j wth 1 j m. Take {p, r}, {r, s}, {s, q}, {x j, x j } wth 1 n, 1 j m, {c + j, d+ j }, {d + j, d j }, {d j, c j } wth 1 j m, {r, x j } wth 1 n,1 j m, where f (x ) = true, {s, x j } wth 1 n,1 j m, where f (x ) = false nto T. Next, for each clause C j choose a true lteral l C j and let l j {x j : 1 n} {x j :1 n} be the correspondng vertex of l. Then take the edge connectng l j and ts neghbor n {c + j, c j } nto T. So far, T s a tree. Moreover, by case analyss, the followng holds: CLAIM 1. T s a tree 5-spanner of G such that f {a, b} s an edge forced (n G) by an S k 1 then d T (a, b) = 3.

12 38 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara Fnally, extend T at each forced edge usng Observaton 2 n an obvous way to obtan a spannng tree T of G. More precsely, f {a, b} T and s forced by an S l [a, b] (l {k 2, k 1, k}) then take a tree (2l + 1)-spanner T n that S l [a, b] contanng the edge {a, b} nto T ; such a tree spanner T exsts by Observaton 2. Clearly, after takng T nto T, T remans a tree. If {a, b} T and s forced by an S l [a, b] (l {k 2, k 1}) then let T be a tree (2l + 1)-spanner n that S l [a, b] contanng the edge {a, b}. Take the two connected components of T {a, b} nto T. Snce there s an (a, b)-path n T, T remans a tree after takng T {a, b} nto T. Now we show that T s a tree (2k + 1)-spanner of G.As2k and by Clam 1, we only have to check for edges {x, y} n an S l [a, b], l {k 2, k 1, k}. Let T be the tree (2l + 1)-spanner n that S l [a, b] whch has been chosen n extendng T to T. If {a, b} T, then by defnton of T, d T (x, y) = d T (x, y) 2l + 1 2k + 1. If {a, b} T, then by defnton of T, l k. Wrte T a, T b for the connected components of T {a, b} contanng a, respectvely, b. If{x, y} T a (or {x, y} T b ), we agan have d T (x, y) = d T (x, y) 2l + 1 < 2k + 1. a Thus, let x T a, y T b, say. Now, the (x, y)-path n T conssts of the (x, a)-path n T a, the (y, b)-path n T b, and the (a, b)-path n T. Hence d T (x, y) = d T (x, y) 1 + d T (a, b). As T s a (2l + 1)-spanner n S l [a, b] and by Clam 1, f l = k 1 then and f l = k 2 then d T (x, y) (2(k 1) + 1) = 2k + 1, d T (x, y) (2(k 2) + 1) = 2k + 1. Thus, T s a tree (2k + 1)-spanner of G as clamed. Lemmas 6, 7, and 9 mmedately mply the man theorem of ths secton: THEOREM 10. For every fxed k 2, the tree (2k +1)-spanner problem s NP-complete for chordal bpartte graphs. 4. Tree 3-Spanners for Bpartte ATE-Free Graphs. In ths secton we show that any bpartte ATE-free graph admts a tree 3-spanner. We say that a vertex u of a graph G has a maxmum neghbor f there s a vertex w n G such that N(N(u)) = N(w). We wll need the followng result from [27].

13 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 39 LEMMA 11 [27]. Any chordal bpartte graph G has a vertex wth a maxmum neghbor. It s easy to deduce from results [28, Lemma 4.4], [27, Corollary 5], [29, Corollary 1] that a vertex wth a maxmum neghbor of a chordal bpartte graph can be found n lnear tme by the followng procedure: PROCEDURE NICE-VERTEX. Fnd a vertex wth a maxmum neghbor Input: A chordal bpartte graph G = (X Y, E). Output: A vertex wth a maxmum neghbor. Method: ntally all vertces v X Y are unmarked; repeat among unmarked vertces of X select a vertex x such that N(x) contans the maxmum number of marked vertces; mark x and all ts unmarked neghbors; untl all vertces n Y are marked; output the vertex of Y marked last. Now let G = (V, E) be a connected bpartte ATE-free graph and let u be a vertex of G whch has a maxmum neghbor (recall that G s chordal bpartte and therefore such a vertex u exsts). LEMMA 12. Let S be a connected component of a subgraph of G nduced by set V \D k 1 (u) (k 1). Then there s a vertex w N k 1 (u) such that N(w) S N k (u). PROOF. Snce u has a maxmum neghbor, we have N 2 (u) N(w) for some vertex w N(u). Consder now a connected component S of a subgraph of G nduced by set V \D k 1 (u) (k 3). Let w be a vertex of N k 1 (u) such that N(w) S N k (u) s maxmal. Assume that there s a vertex x n S N k (u) whch s not adjacent to w. Then, by maxmalty, for any neghbor z of x n N k 1 (u), there must exst a vertex y n S N k (u) such that {y,w} E and {y, z} / E. Snce vertces x and y both belong to S, they are connected by a path P of G consstng only of vertces from V \D k 1 (u). Let y, x be the neghbors on P of y and x, respectvely. Clearly, snce G s bpartte, x, y N k+1 (u) and {y, x} / E. Consder also shortest paths P(w, u) and P(z, u) of G connectng vertex u wth w and z, respectvely. Vertex x cannot be adjacent wth y snce otherwse a subgraph of G formed by edges {y, x }, {x, x}, {y,w}, {x, z} and paths P(w, u), P(z, u) wll contan an nduced cycle of length at least 6, whch s mpossble. Analogously, vertex y s not adjacent wth x. We clam now that edges a ={y, y }, c ={x, x }, and e ={u,v}, where v s a neghbor of u on P(w, u), form an ATE n G. Indeed, P avods the neghborhood of e snce P V \D k 1 and k > 2, path (y,w) P(w, u) avods the neghborhood of c and path (x, z) P(z, u) avods the neghborhood of a. A contradcton obtaned proves that N(w) S N k (u).

14 40 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara Ths lemma suggests the followng algorthm for constructng a spannng tree of G: PROCEDURE SPAN-ATEG. Tree 3-spanners for bpartte ATE-free graphs Input: A bpartte ATE-free graph G = (V, E) and a vertex u of G wth a maxmum neghbor. Output: A spannng tree T = (V, E ) of G (rooted at u). Method: set E := ; set q := max{d G (u,v): v V }; let s q, {1,...,p q}, be the vertces of N q (u); for every {1,...,p q } do pck a neghbor w of s q n N q 1 (u); add edge {s q,w} to E ; for k := q 1 downto 1 do compute the connected components S1 k,...,sk p k of G[N k (u) {s k+1, {1,...,p k+1 }}]; for every {1,...,p k } do set S := S k N k (u); pck a vertex w n N k 1 (u) such that N(w) S; for each v S add the edge {v, w} to E ; shrnk component S k to a vertex s k and make s k adjacent n G to all vertces from N(S k) N k 1(u). It s easy to see that the graph T = (V, E ) constructed by ths procedure s a spannng tree of G and ts constructon takes only lnear tme. Moreover, T s a shortest path tree of G rooted at u snce, for any vertex x V, d G (x, u) = d T (x, u) holds. THEOREM 13. Let T = (V, E ) be a spannng tree of a bpartte ATE-free graph G = (V, E) output by PROCEDURE SPAN-ATEG. Then, for any x, y V, we have d T (x, y) 3 d G (x, y) and d T (x, y) d G (x, y) + 2. PROOF. Frst we show that d T (x, y) 3 holds for any edge {x, y} of G. Snce G s bpartte, d G (x, u) d G (y, u) =1must hold. Wthout loss of generalty, assume that x N k (u) and y N k 1 (u). Let x be the father of x n T. If x = y we are done; d T (x, y) = 1. Otherwse, x and y are from N k 1 (u) and belong to a common connected component of the graph G[V \D k 2 (u)]. Accordng to the algorthm, x and y share a common father n T. Hence, d T (x, y) = d T (x, y) + 1 = 3. Now consder two arbtrary vertces v and w of G and a shortest (v, w)-path. Applyng the nequalty d T (x, y) 3 to every edge {x, y} of ths path, we wll get d T (v, w) 3 d G (v, w). That d T (x, y) d G (x, y) + 2 already follows from the prevous part of our proof and from Lemma 1 of [30]. For the sake of completeness, we present another proof here. Snce T s a shortest path tree of G rooted at u, the dstances n G and T between a vertex and any of ts ancestors are the same. We wll prove that d T (x, y) d G (x, y) + 2 by nducton on d G (v, w). If v and w are adjacent, then we

15 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 41 are done, because then d T (v, w) 3. Now suppose that d G (v, w) = s 2 and let z be a neghbor of v on a shortest path between v and w. From the nducton assumpton we have d T (z,w) s = s + 1 and d T (v, z) 3. Let a = nca(v, z) be the nearest common ancestor of v and z n the tree T. Snce d T (a,v) = d G (a,v), d T (a, z) = d G (a, z), and {v, z} E we obtan that d T (v, a) d T (z, a) =1. We can addtonally assume that d T (z,w) < d T (v, w) 1, snce otherwse we mmedately conclude d T (v, w) d G (v, w) + 2. From ths and the prevous nequalty we deduce that the vertex nca(w, z) les on the path of T between the vertces a and z. Therefore, a s an ancestor of w, and thus d T (a,w) = d G (a,w). Notce that the dstance sums d T (v, w) + d T (a, z) and d T (v, z) + d T (a,w)are equal. Hence, d T (v, w) = d T (a,w) d T (a, z) + d T (v, z) = d G (a,w) d G (a, z) + d T (v, z) d G (w, z) + 3 = d G (v, w) + 2, concludng the proof. Any nterval bgraph s a bpartte ATE-free graph, and any convex graph s an nterval bgraph. Hence we have the followng corollares: COROLLARY 14. Any nterval bgraph G = (V, E) admts a spannng tree T such that d T (x, y) 3 d G (x, y) and d T (x, y) d G (x, y) + 2 hold for any x, y V. Moreover, such a tree T can be constructed n lnear tme. COROLLARY 15 [6]. Any convex graph G = (V, E) admts a spannng tree T such that d T (x, y) 3 d G (x, y) and d T (x, y) d G (x, y) + 2 hold for any x, y V. Moreover, such a tree T can be constructed n lnear tme. 5. Tree 7-Spanners for (Enhanced) Probe Interval Graphs. In ths secton we show that any (enhanced) probe nterval graph admts a tree 7-spanner. Let G = (P, N, E) be a connected probe nterval graph. We assume that an nterval representaton of G s gven (f not, an nterval model for G can be constructed by a method descrbed n [24] n O(m log n) tme, where n = P + N and m = E ). Let I ={I x : x P} be the ntervals n the nterval model representng the probes and let J ={J y : y N} be the ntervals representng the nonprobes. Frst we dscuss two smple specal cases. If N = then clearly G = (P, E) s an nterval graph. It s known (see [30]) that for any nterval graph G and any vertex u of G there s a shortest path spannng tree T of G rooted at u such that d T (x, y) d G (x, y)+2 holds for any x, y. In fact, a procedure smlar to PROCEDURE SPAN-ATEG produces such a spanner n lnear tme for any nterval graph G and any start vertex u. Evdently, T s a tree 3-spanner of G. To descrbe other specal cases, we need the followng noton. A connected probe nterval graph G = (P, N, E) s superconnected f for any two ntersectng ntervals I v, I w I there always s an nterval J y J such that I v I w J y. For a superconnected probe nterval graph G, a tree 4-spanner can be constructed easly. Frst we gnore all edges n G[P] to get an nterval bgraph G = (X = P, Y = N, E ) and then run PROCEDURE SPAN-ATEG on G. We clam that a spannng tree T of G, produced by that procedure, s a tree 4-spanner of G. Indeed, for any edge {x, y} of G such that x P and y N, d T (x, y) 3 holds by Corollary 14; t s an edge of G, too.

16 42 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara G 0 G G G G u 0 probes u 1 v u 2 u 3 u 4 w nonprobes S 0 L(S 0 ) R(S 0 ) S 1 Fg. 6. Segments and a decomposton of a probe nterval graph. Now consder an edge {v, w} of G wth v, w P. Snce G s superconnected, there s a vertex y N such that I v I w J y,.e., d G (v, w) = 2. Then, by Corollary 14, we have d T (v, w) d G (v, w) + 2 = = 4. Consequently, T s a tree 4-spanner of G. To get a tree 7-spanner for an arbtrary connected probe nterval graph G = (P, N, E), we use the followng strategy. Frst we decompose the graph G nto subgraphs G 0, G 1,...,G k such that G and G j ( j) share at most one common vertex and each G s ether an nterval graph or a superconnected probe nterval graph. Then teratvely, gven a tree 7-spanner T for G 0 G 1 G ( < k) and a tree t-spanner T +1 (t 4) of G +1, we wll extend T to a tree 7-spanner T +1 for G 0 G 1 G G +1 by ether makng all vertces of G +1 adjacent n T +1 to a common neghbor n G 0 G 1 G (f t exsts) or by glung trees T and T +1 at a common vertex. Now we gve a formal descrpton of the decomposton algorthm. Let S 0, S 1,...,S q be segments of the unon y N J y (see Fgure 6 for an llustraton). PROCEDURE DECOMP. A decomposton of a probe nterval graph Input: A probe nterval graph G and ts nterval representaton (I, J ). Output: Subgraphs G 0, G 1,...,G 2q+2 of G, where G 2 ( {0,...,q+1}) s an nterval graph and G 2+1 ( {0,...,q}) s a superconnected probe nterval graph, and specal vertces u j ( j = 1,...,2q + 2), where u j belongs to G j 1 and G j. Method: for = 0 to q do /* defne an nterval graph */ set X := {I x I: L(x) L(S )}; on ntervals X defne an nterval graph G 2 ; let I be an nterval from X wth maxmum R( ) value; set u 2+1 := a vertex of G correspondng to I ; set I := I\(X \{I }); /* defne a superconnected probe nterval graph */ set Y := {I y J : I y S }; set X := {I x I: L(x) R(S )}; defne a probe nterval graph G 2+1 wth probes X and nonprobes Y; let I be an nterval from X wth maxmum R( ) value; set u 2+2 := a vertex of G correspondng to I ; set I := I\(X \{I }); defne on I an nterval graph G 2q+2.

17 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 43 Clearly, all probe nterval graphs G 2+1 ( = 1,...,q) are superconnected and a decomposton of G nto G 0, G 1,...,G 2q+2 can be done n lnear tme f endponts of the ntervals I J are sorted. LEMMA 16. For any = 2,...,2q + 2, R(u ) R(u 1 ) holds. PROOF. When we delete an nterval I v from I, we always leave n I an nterval I u such that R(v) R(u). Now, for an nterval graph G 0 (f t s not empty), we can construct a tree 3-spanner T 0 = T 0 (u 0 ) rooted at any vertex u 0 of G 0. For an nterval graph G 2 ( = 1,...,q + 1), we can construct a tree 3-spanner T 2 = T 2 (u 2 ) rooted at vertex u 2 (see PROCEDURE DECOMP). Snce all those trees are shortest path trees, the neghborhoods of vertex u 2 n G 2 and T 2 concde. Let G 2+1 be an nterval bgraph obtaned from a superconnected probe nterval graph G 2+1 by gnorng all edges between probes and deletng all probes I v such that I v I u2+1. LEMMA 17. For any = 0,...,q, vertex u 2+1 has a maxmum neghbor n G 2+1. PROOF. Accordng to PROCEDURE DECOMP, nterval I correspondng to u 2+1 belongs to I. Let J y be an nterval of J such that L(J y ) R(I ) and R(J y ) s maxmum. We show that vertex y of G 2+1 s a maxmum neghbor of u 2+1 n G 2+1. Consder a vertex w of G 2+1 whch s at dstance 2 from u 2+1 n G 2+1 and assume that ntervals J y, I w do not ntersect. If R(I w )<L(J y ), then necessarly I w s a subnterval of I and w s not a vertex of G 2+1. Hence, we may assume that R(J y)<l(i w ). However, then, by maxmalty of R(J y ), there cannot exst an nterval n J whch ntersects both I and I w. The latter contradcts our assumpton that the dstance n G 2+1 between u 2+1 and w s 2. Let T 2+1 = T 2+1 (u 2+1) be a tree 3-spanner of an nterval bgraph G 2+1 constructed startng at vertex u 2+1, {0,...,q} (see PROCEDURE SPAN-ATEG). Clearly, the neghborhoods of vertex u 2+1 n G 2+1 and T 2+1 concde. We can extend tree T2+1 to a spannng tree T 2+1 = T 2+1 (u 2+1 ) of G 2+1 by addng, for each probe I v of G 2+1 such that I v I u2+1, a pendant vertex v adjacent to u 2+1. LEMMA 18. T 2+1 (u 2+1 ) s a tree 4-spanner for G 2+1, {0,...,q}. Moreover, for any edge {w, u 2+1 } of G 2+1, d T2+1 (w, u 2+1 ) 2 holds. PROOF. Let A ={v: v s a vertex of G 2+1 such that I v I u2+1 } and let H be a superconnected probe nterval graph obtaned from G 2+1 by elmnatng vertces of A. Snce G 2+1 s the nterval bgraph counterpart of H, tree T2+1 s a tree 4-spanner for H. Consder now an edge {v, w} of G 2+1. We may assume that at least one of these vertces (say, v) s from A. If also w A then, by constructon, d T2+1 (v, w) = 2. If w/ A then, snce I v ntersects I w and I v I u2+1, I w must ntersect I u2+1 too. If w s a nonprobe, then

18 44 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara {w, u 2+1 } s an edge of G 2+1 and hence of T2+1.Ifws a probe, then d G (u 2+1,w)= = d T (u 2+1,w)snce T s a shortest path spannng tree (rooted at u 2+1)ofG 2+1. Consequently, n both cases we have d T2+1 (v, w) = 1 + d T (u 2+1,w) Now we are ready to construct a spannng tree T for the orgnal probe nterval graph G = (P, N, E). We say that a vertex v of G domnates a subgraph G k of G f every vertex of G k, dfferent from v, s adjacent to v n G. PROCEDURE SPAN-PIG. Tree 7-spanner for probe nterval graphs Input: A probe nterval graph G = (P, N, E), ts nterval representaton (I, J ) and a decomposton of G nto graphs G 0, G 1,...,G 2q+2. Output: A spannng tree T = (P N, E ) of G. Method: set E = and k := 0; whle k 2q + 2 do f there s an ndex j such that k j and u k domnates G j then do fnd the largest ndex j wth that property; for each v n G k... G j (v u k ) add edge {v, u k } to E ; set k := j + 1; else do f k s even then do fnd a tree 3-spanner T k (u k ) of an nterval graph G k ; add all edges of T k (u k ) to E ; f k s odd then do fnd a tree 4-spanner T k (u k ) of a superconnected probe nterval graph G k ; add all edges of T k (u k ) to E ; set k := k + 1. It s easy to see that the tree T constructed by PROCEDURE SPAN-PIG s a spannng tree of G and ts constructon takes only lnear tme. LEMMA 19. If for graph G k (k {0,...,2q+2}) there exsts a vertex u {u 0,...,u k } whch domnates G k, then there s a vertex u s {u 0,...,u k } such that d T (x, u s ) 1 holds for any x n G k. Otherwse, f such a vertex u does not exst, then for any vertces x, yofg k, d T (x, y) = d Tk (x, y) holds. PROOF. Assume that such a vertex u exsts, but for any u s there s a vertex x n G k such that d T (x, u s )>1. By Lemma 16, R(u k ) R(u ). Hence, vertex u k s also adjacent to all vertces of G k (except tself). On the other hand, snce for any u s there s a vertex x n G k such that d T (x, u s )>1, there was an teraton of the whle loop n PROCEDURE SPAN-PIG where the edges of tree T k (u k ) were added to T. That s, t was detected that vertex u k does not domnate G k. A contradcton obtaned proves the frst part of the lemma. The second part s evdent.

19 Tree Spanners for Bpartte Graphs and Probe Interval Graphs 45 COROLLARY 20. For any vertces x, yofg k (k {0,...,2q + 2}), d T (x, y) max{2, d Tk (x, y)} holds. Such a vertex u s descrbed n Lemma 19 s called the focus of G k n T. LEMMA 21. T s a tree 7-spanner for G. PROOF. Consder an edge {v, w} of G. If both vertces v and w belong to the same graph G k (k = 0,...,2q + 2) then ether d T (v, w) 2ord T (v, w) = d Tk (v, w) 4. Hence, we may assume that they are from dfferent graphs. Clearly, v and w cannot both belong to N. Case 1: v P and w N. In ths case there s a segment S {S 0, S 1,...,S q } such that L(S ) L(w) R(w) R(S ). Clearly, w s a vertex of G 2+1 and, by PROCEDURE DECOMP, no neghbor of w dfferent from u 2+2 can belong to G k (k > 2 + 1). Hence, L(v) L(S ) L(w) R(v) must hold. Moreover, snce L(u 2+1 ) L(S ) L(w) R(v) R(u 2+1 ), vertces w and u 2+1 are adjacent n G and therefore n T 2+1. So, d T2+1 (w, u 2+1 ) = 1. The latter means that ether vertces w and u 2+1 are adjacent n T or they are both adjacent to the focus of G 2+1 n T (see Lemma 19). If v belongs to G 2 then, snce u 2+1 s also n G 2 and d G (v, u 2+1 )=1, d T2 (v, u 2+1 ) 3 must hold. Hence, we have d T (v, w) d T (v, u 2+1 ) + d T (w, u 2+1 ) max{2, d T2 (v, u 2+1 )}+max{2, d T2+1 (w, u 2+1 )} = 5. Now assume that v belongs to G j wth j < 2. Then vertex u j+1 domnates G j+1 snce R(u j+1 ) R(v) L(S ). Let u s be the focus of G j+1 n T and let r be the largest ndex such that graph G r s stll domnated by u s. By PROCEDURE SPAN-PIG, u s s the focus n T of all graphs G j+1,...,g r. Therefore, vertces u j+1,...,u r+1 are all at dstance at most 1 from u s n T. We also have, by Corollary 20, d T (v, u j+1 ) max{2, d Tj (v, u j+1 )} 4. If d T (v, u j+1 )>2, then necessarly s = j + 1 and r 2 (recall that R(u j+1 ) L(S )). Hence, 2 < d T (v, u s ) 4, d T (u s, u 2+1 ) 1, and vertex w s adjacent n T ether to u s or to u 2+1, dependng on whether u s domnates G 2+1 or not. Thus, we have d T (v, w) d T (v, u s ) + d T (u s, u 2+1 ) = 6. Let now d T (v, u j+1 ) 2. If u s domnates G 2, then agan d T (u s, u 2+1 ) 1. Otherwse, r < 2 and vertex u r+1 domnates G 2 snce R(u r+1 ) R(u j+1 ) R(v) L(S ). By PROCEDURE SPAN-PIG, u r+1 s the focus of G 2 n T. Hence, d T (u r+1, u 2+1 ) 1 and therefore d T (u s, u 2+1 ) = d T (u s, u r+1 ) + d T (u r+1, u 2+1 ) = 2. Snce w s adjacent n T ether to u 2+1 or to the focus of G 2+1 n T,wegetd T (v, w) d T (v, u j+1 ) + d T (u j+1, u s ) + d T (u s, u 2+1 ) = 6. Case 2: v, w P. Snce w (as well as v) can be a vertex from {u 1,...,u 2q+2 },tcan belong to few consecutve graphs G,...,G +a. Therefore, let and j be the smallest ndces such that w belongs to G and v belongs to G j. Wthout loss of generalty, assume also that j <. We have R(v) L(w). Snce v n G j s adjacent to w n G, vertex u must be adjacent to v. Snce G can be a proper nterval graph (f s odd), by Lemma 18, we

20 46 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara have d T (w, u ) 2. Recall that, f G s an nterval graph (.e., s even), then we would have d T (w, u ) 1. If j = 1, then both vertces v and u are n G 1 and, therefore, d G (v, u ) = 1 mples d T 1 (v, u ) 4. Hence, we have d T (v, w) d T (v, u ) + d T (w, u ) max{2, d T 1 (v, u )}+max{2, d T (w, u )} = 6. Now assume that j < 1. Then vertex u j+1 domnates G j+1 snce R(u j+1 ) R(v) L(w). Let agan u s be the focus of G j+1 n T and let r be the largest ndex such that graph G r s stll domnated by u s. Snce u s s the focus n T of all graphs G j+1,...,g r, vertces u j+1,...,u r+1 are all at dstance at most 1 from u s n T.By Corollary 20, we also have d T (v, u j+1 ) max{2, d Tj (v, u j+1 )} 4. If d T (v, u j+1 )>2, then agan s = j + 1 and r 1. Hence, 2 < d T (v, u s ) 4, d T (u s, u ) 1 and therefore d T (v, w) d T (v, u s ) + d T (u s, u ) + d T (u,w) = 7. Let now d T (v, u j+1 ) 2. If u s domnates G 1, then agan d T (u s, u ) 1. If u s does not domnate G 1, then vertex u r+1 must domnate t (snce R(u r+1 ) R(u j+1 ) R(v)). Therefore, by PROCEDURE SPAN-PIG, u r+1 s the focus of G 2 n T and d T (u r+1, u ) 1 must hold. Thus, d T (v, w) d T (v, u j+1 )+d T (u j+1, u s )+d T (u s, u )+ d T (u,w) = 7. The man theorem n ths secton s the followng: THEOREM 22. Any probe nterval graph G admts a tree 7-spanner. Moreover, such a tree 7-spanner can be constructed n O(m log n) tme, or n O(m + n log n) tme f the ntersecton model of G s gven n advance. Now let G = (P, N, E) be an enhanced probe nterval graph wth probes P and nonprobes N. COROLLARY 23. Any enhanced probe nterval graph G = (P, N, E) admts a tree 7-spanner. Moreover, such a tree spanner can be constructed n O(m log n) tme. PROOF. By gnorng n G edges between nonprobes, we get a probe nterval graph G. Let T be a tree 7-spanner of G constructed by PROCEDURE SPAN-PIG. We show that T s a tree 7-spanner of G, too. One needs to check the dstance n T only between nonprobes x, y N whch are adjacent n G,.e., {x, y} s an enhanced edge. By the defnton of an enhanced edge, there must exst two nonntersectng probes v, w such that {v, x}, {v, y}, {w, x}, {w, y} are edges n G. Wthout loss of generalty, assume that R(v) < L(w). Then, clearly, vertces x, y,ware all from some superconnected probe nterval graph G 2+1 (see PROCEDURE DECOMP). Let G 2+1 be the nterval bgraph counterpart of G 2+1 and let T 2+1 be a tree 3-spanner of G 2+1. For edges {w, x} and {w, y} of graph G 2+1 we have d T (w, x) max{2, d T2+1 (w, x)} 3 and d T (w, y) max{2, d T2+1 (w, y)} 3 (recall that edges {w, x} and {w, y} are connectng a probe wth nonprobes and hence they are both edges of G 2+1 ). Thus, d T (x, y) d T (x,w)+ d T (y,w) = 6.