Tree Spanners for Bipartite Graphs and Probe Interval Graphs 1

Algorthmca (2007) 47: 27 51 DOI: 10.1007/s00453-006-1209-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. 106 118. Lecture Notes n Compter Scence 2880, Sprnger, Berln, 2003. 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, 18051 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 923-1292, Japan. Receved June 17, 2003; revsed August 22, 2005. Communcated by H. Gabow. Onlne publcaton July 25, 2006.

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

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.

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,

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].

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) 1 + 3 = 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.

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.

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 ).

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.

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.

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.

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 + 1 5 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) 1 + 3 = 2k + 1, d T (x, y) (2(k 2) + 1) 1 + 5 = 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].

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).

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

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 1 + 2 = 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.

42 A. Brandstädt, F. F. Dragan, H.-O. Le, V. B. Le, and R. Uehara G 0 G G G G 1 2 3 4 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 = 2 + 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.

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

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)= 2+1 2 = d T (u 2+1,w)snce T 2+1 2+1 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) 3. 2+1 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.

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 )} 3 + 2 = 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 ) + 1 4 + 1 + 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 ) 1 + 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 ) + 1 2 + 1 + 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

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 )} 4 + 2 = 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) 4 + 1 + 2 = 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) 2 + 1 + 2 + 2 = 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) 3 + 3 = 6.