Approximations for Steiner Trees with Minimum Number of Steiner Points

Transcription

1 Journal of Global Optmzaton 18: 17 33, Kluwer Academc ublshers. rnted n the Netherlands. Approxmatons for Stener Trees wth Mnmum Number of Stener onts 1, 1,,,,3, DONGHUI CHEN *, DING-ZHU DU *, XIAO-DONG HU, 4, 4, GUO-HUI LIN, LUSHENG WANG and GUOLIANG XUE 1 Department of Computer Scence and Engneerng, Unversty of Mnnesota, Mnneapols, MN 55455, USA (E-mal: dzd@cs.umn.edu) Department of Computer Scence, Cty Unversty of Hong Kong, Kowloon Tong, Hong Kong, Chna 3 Insttute of Appled Mathematcs, Chnese Academy of Scences, Beng , Chna 4 Department of Computer Scence, The Unversty of Vermont, Burlngton, VT 05405, USA (Accepted n orgnal form 19 January 000) Abstract. Gven n termnals n the Eucldean plane and a postve constant, fnd a Stener tree nterconnectng all termnals wth the mnmum number of Stener ponts such that the Eucldean length of each edge s no more than the gven postve constant. Ths problem s N-hard wth applcatons n VLSI desgn, WDM optcal networks and wreless communcatons. In ths paper, we show that (a) the Stener rato s 1/ 4, that s, the mnmum spannng tree yelds a polynomal-tme approxmaton wth performance rato exactly 4, (b) there exsts a polynomaltme approxmaton wth performance rato 3, and (c) there exsts a polynomal-tme approxmaton scheme under certan condtons. Key words: Stener trees; Approxmaton algorthms; VLSI desgn; WDM optcal networks 1. Introducton Gven a set of n termnals X p 1, p,..., pn n the Eucldean plane, and a postve constant R, the Stener tree problem wth mnmum number of Stener ponts, denoted by ST-MS for short, asks for a tree T spannng a superset of X such that each edge n the tree has a length no more than R and the number C(T )of ponts other than those n X, called Stener ponts [3, 6, 8, 9], s mnmzed. In the classcal Eucldean Stener tree problem (whch asks for a tree spannng a superset of X such that the total length of the tree, that s the sum of lengths of edges n the tree, s mnmzed), a Stener pont always has a degree of 3. In the ST-MS problem, however, degree- Stener ponts are possble. For example, when n and p p, the Eucldean dstance between p 1 1 and p, s larger than R, then the * Supported n part by the Natonal Scence Foundaton under grant CCR Supported n part by Natonal Natural Scence Foundaton of Chna under grant and 973 fundamental research proect of Chna. Supported n part by the Army Research Offce grant DAAH and by the Natonal Scence Foundaton grants ASC and OSR

2 18 DONGHUI CHEN ET AL. optmal tree s a path contanng p1p/r 1 Stener ponts, each of whch has a degree of. The ST-MS problem has an mportant applcaton n wavelength-dvson multplexng (WDM) optcal network desgn [11, 14]. Suppose we need to connect n stes located at p 1, p,..., pn wth WDM optcal network. Due to the lmt n transmsson power, sgnals can only travel a lmted dstance (say R) for guaranteed correct transmsson. If some of the nter-ste dstances are greater than R, we need to provde some amplfers or recevers/ transmtters at some locatons n order to break t nto shorter peces. The ST-MS problem also fnds applcatons n VLSI desgn [, 7, 15], and the evolutonary/ phylogenetc tree constructons n computatonal bology [9]. Recently, Ln and Xue [1] showed that the ST-MS problem s N-hard. They also showed that the approxmaton obtaned from the mnmum spannng tree by smply breakng each edge nto small peces wthn the upper bound has a worst-case performance rato at most 5. In ths paper, we show that ths approxmaton has a performance rato exactly 4. We also present a new polynomal-tme approxmaton wth a performance rato at most 3 and a polynomal-tme approxmaton scheme under certan condtons.. relmnary Any shortest optmal soluton T for the problem ST-MS must have the followng propertes. (a1) No two edges cross each other. (a) Two edges meetng at a vertex form an angle of at least 60. (a3) If two edges form an angle of exactly 60, then they have the same length. To see (a1), consder two edges ac and bd n T. By contradcton, suppose ac and bd cross at e. Note that quadrangle abcd must have an nner angle of at least 90. Wthout loss of generalty, assume abc 90. Then bca 90 and cab 90. Hence ab ac and bc ac, where ab denotes the length of edge ab. When edge ac s removed from T, T would be broken nto two parts contanng vertces a and c, respectvely. One of the parts, say the one contanng a, contans vertex b. Addng edge bc results n a shorter tree stll optmal for ST-MS. Ths contradcts the length-mnmalty of T. Therefore, (a1) holds. To see (a), consder two edges ab and bc n T. By contradcton, suppose abc 60. Then ether cab 60 or bca 60 and hence ether bc ac or ab ac. Usng ac to replace ether bc or ab would reduce the total length of the tree preservng the vertex set, contradctng the length-mnmalty of T among optmal solutons for ST-MS. Therefore, (a) holds. (a3) can be proved by a smlar argument. The followng lemma follows from (a) and (a3). LEMMA 1. There exsts a shortest optmal Stener tree T* for ST-MS such that every vertex n T* has degree at most fve.

3 AROXIMATIONS FOR STEINER TREES 19 roof. It follows mmedately from (a) that every vertex n a shortest optmal tree T for ST-MS has degree at most sx. Consder a vertex u wth degree sx n T. By (a), every angle at u equals 60. By (a3), all edges ncdent to u have the equal length. Next, consder any vertex v wth degree d n T. We clam that f v s adacent to k vertces wth degree sx, then d 6 k. In fact, suppose u s adacent to v wth degree sx. Then u has two degrees uw and ux such that wuv vux 60 and uv uw ux. Thus, vw uw and vx ux. Replacng uw and ux by vw and vx results n stll a shortest optmal tree for ST-MS. But, v gets two more edges. For all vertces wth degree sx and adacent to v, perform the same operaton. We wll obtan a shortest optmal tree for ST-MS such that v has degree d k. Hence, d k 6. Now, for each vertex u wth degree sx, we move only one edge from u to ts adacent vertex. Then every vertex wll have degree at most fve and the resultng tree s stll a shortest optmal tree for ST-MS. Usually, a spannng tree s a tree nterconnectng the gven termnals wth edges between gven termnals. The shortest spannng tree s called the mnmum spannng tree. Spannng trees may not be feasble solutons for the problem ST-MS snce some edges may be too long. To make t feasble, we add ab/r Stener ponts to break each edge ab nto small peces of lengths at most R. The resultng tree wll be called a stenerzed spannng tree. The followng s an nterestng fact. LEMMA. Every stenerzed mnmum spannng tree has the mnmum number of Stener ponts among stenerzed spannng trees. roof. Every mnmum spannng tree can be obtaned from a spannng tree by a sequence of operatons that each replaces an edge by another shorter edge. Snce the shorter edge needs Stener ponts no more than the longer edge needs when we stenerze them. Therefore, the lemma holds. It follows easly from the above two lemmas that the stenerzed mnmum spannng tree s an approxmaton wth performance rato 5 (see [1]). 3. Stenerzed mnmum spannng tree We show the followng tght result n ths secton. THEOREM 1. The stenerzed mnmum spannng tree s a polynomal-tme approxmaton wth performance rato exactly 4. The lower bound can be shown by presentng an example as follows. Consder fve vertces v 1, v,...,v5 of a regular pentagon wth each edge of length 1 where s a small postve real number such that the dstance from the center to

4 0 DONGHUI CHEN ET AL. Fgure 1. each vertex s wthn R (Fgure 1). The stenerzed mnmum spannng tree on v 1, v,...,v5 contans four Stener ponts. However, every optmal tree for ST- MS on v 1, v,...,v5 contans only one Stener pont. Therefore, the performance rato of stenerzed mnmum spannng tree s at least four. To show the tght upper bound, we need frst to study a property of convex path n any shortest optmal tree T for ST-MS. A path q1 q... qm n T s called a convex path f for every 1,,...,m3, qq ntersects q1q 3. An angle of degree more than 10 wll play an mportant role n the proof of Theorem 1. For smplcty, we call such angles bg angles. LEMMA 3. Let q1q1qm be a convex path and m. Suppose there are t bg angles among the m angles q1qq, 3 qqq,...,q 3 4 mqm1q m. Then q q (t )R. 1 m roof. We prove t by nducton on m. For m 3, t s trval that q q 1 3 q q q q R (t )R. Now, suppose m 4. Consder the convex hull H 1 3 of ponts q, q,...,q. If at least one of q and q does not le on the boundary of 1 m 1 H, then by the nducton hypothess, any dstance between two vertces of the convex hull H s at most (t )R and hence any two ponts lyng n H have dstance at most (t )R. Therefore, q q (t )R. 1 m Next, we may assume that both q1 and qm le on the boundary of H. It follows mmedately that whole path qqq les on the boundary of H (Fgure (a)). 1 1 m If q1qq m m1 90, then q1q m q1qm1 and by the nducton hypothess q q (t )R. Hence, q q(t )R. Smlarly, f q qq 90, then 1 m1 1 m 1 m q1qm1 (t )R. Therefore, we may assume q1qq m m1 90 and qqq 1 m 90. It follows that (m ) 18090(m t ) 10t180. Hence, m t 3. Ths means that the path qqq 1 1 m has at most two angles of degrees not more than 10. If qmqm1qm s a bg angle, then by the nducton hypothess, q1qm1 ((t 1) )R. Therefore, q1q m q1qm1 qm1qm (t )R. Smlarly, f q qq s a bg angle, then q q (t )R. Therefore, we may assume m q q q 10 and q qq 10. They are the only two angles not bg m m1 m 1 3

5 AROXIMATIONS FOR STEINER TREES 1 Fgure. on the path qqq 1 1 m. Now, draw a parallelogram qqq 1 m1 pas shown n Fgure (b). Snce q1qq m1 qqq , we have qqm1 p 60. Moreover, qqm1qm qmqm1qm 10. Thus, pqm1qm 60. It follows that Therefore, pq max( pq, q q ) max(q q, q q ) R. m m1 m1 m 1 m1 m q q q p pq q q pq (t 1)R R (t )R. 1 m 1 m m1 m LEMMA 4. In a shortest optmal tree T for ST-MS, there are at most two bg angles at a vertex wth degree three, there s at most one bg angle at a vertex wth degree four, and there s no bg angle wth degree fve roof. Suppose 1,,...,d are all angles at a vertex wth degree d and k ( 0) of them are bg angles. Snce each angle s of at least 60, we have 3601d (d k) 60k10. Thus, 6 (d k) k d k,.e., 5 d k. The lemma follows mmedately from ths nequalty. Note that every leaf n a Stener tree s a termnal. A Stener tree s full f every termnal s a leaf. If a Stener tree s not full, then we can always fnd a termnal wth degree more than one whch enable us to break the tree at ths termnal. In ths way, every Stener tree can be broken nto several small full Stener trees. Those small full Stener trees are called full components of a Stener tree. LEMMA 5. Consder a shortest optmal tree T for ST-MS. Suppose T s a full Stener tree. Let s denote the number of Stener ponts wth degree n T. Then 3s s s n where n s the number of termnals. roof. Snce T has totally s5 s4 s3 s n 1 edges, we have 5s5 4s4 3s s n (s s s s n 1). Hence, 3s s s n

6 DONGHUI CHEN ET AL. Consder a shortest optmal tree T for ST-MS. Suppose T s a full Stener tree on n termnals. The followng fact s easly seen. (b1) T has exactly n convex paths; each connects two termnals, (b) each termnal appears n exactly two convex paths n T, and (b3) each angle at a Stener pont appears n those n convex paths exactly once. Now, we are ready to show Theorem 1. roof of Theorem 1. By Lemma 1, there exsts a shortest optmal tree T* for ST-MS n whch every vertex has degree at most fve. Frst, we assume that T* s a full Stener tree. Let s denote the number of Stener ponts wth degree n T*. By Lemma 5, n 3s5 s4 s3. Consder a spannng tree T consstng of n 1 edges each connectng two termnals at S endponts of a convex path n T* (Fgure 3). By Lemma 3, each edge ab n TS has length upper-bounded by (t )R where t s the number of bg angles on the convex path connectng a and b. Hence, we need at most (t 1) Stener ponts to stenerze edge ab. By Lemma 4, the spannng tree TS can be stenerzed by at most s4 s3 s n 1 Stener ponts. By Lemma, any stenerzed mnmum spannng tree contans at most s4 s3 s n 1 Stener ponts. Clearly, s s s n 1 3s 3s 3s s (s s s s ) If s s s s 0, then s s s n 1 4(s s s s ). If s s s s 0, then T T*. Therefore, n ether case, every stenerzed mnmum 4 3 S spannng tree contans at most 4(s s s s )( 4 C(T*)) Stener ponts Now, suppose T* s not a full Stener tree. Then T* can be decomposed nto several full components T, T,...,T. For each full component T, by the above 1 k argument, we know that the stenerzed mnmum spannng tree on termnals n T contans at most 4 C(T ) Stener ponts. Note that the unon of stenerzed mnmum spannng trees each for termnals n a full component s a stenerzed spannng tree Fgure 3.

7 AROXIMATIONS FOR STEINER TREES 3 for all termnals. By Lemma, the number of Stener ponts n T* s at most k 4 C(T ) 4 C(T*) Approxmaton Let T* be a shortest optmal tree for ST-MS wth Stener ponts of degrees at most fve. Suppose T 1, T,...,Tk are all full components of T*. In the proof of Theorem 1, we showed that the stenerzed mnmum spannng tree on termnals n T contans at most 3 C(T ) 1 Stener ponts. Now, we study when ths upper bound can be mproved. LEMMA 6. Let T* be a shortest optmal tree for ST-MS wth property that every Stener pont has degree at most fve. Let T be a full component of T*. Then the followng hold: (c1) The stenerzed mnmum spannng tree on termnals n T contans at most 3 C(T ) 1 Stener ponts. (c) If T contans a Stener pont wth degree at most four, then the stenerzed mnmum spannng tree on termnals n T contans at most 3 C(T ) Stener ponts. (c3) If the stenerzed mnmum spannng tree on termnals n T contans an edge between two termnals, then t contans at most 3 C(T ) Stener ponts. roof. (c1) and (c3) follow mmedately from the proof of Theorem 1. Next, we show (c). Let s be the number of Stener ponts wth degree n T. Let n be the number of termnals n T. Note that there are exactly n convex paths n T. Choose any n 1 of them and connect two endponts of each path. We wll obtan a spannng tree. Its stenerzaton s denoted by T. Now, assume u s the Stener pont S wth degree at most four. If there s a bg angle at u, then we choose n 1 convex paths not contanng the bg angle. If there s no bg angle at u, then we can choose any n 1 convex path. Wth ths choce, we would have C(T ) s s s S (n 1) 3(s s s ) 3C(T ). 4 3 To desgn a new approxmaton, we need to study how to fnd whether three or four termnals can be connected to a common Stener pont. Note that an angle of less than 90 s acute and an angle of more than 90 s obtuse. A trangle s acute f ts three angles are all acute. A trangle s obtuse f t has one obtuse angle. A trangle s rght f t has one rght angle. LEMMA 7. If a trangle abc s acute or rght, then the mnmum dsk (.e., the dsk of mnmum radus) for coverng the trangle abc s the one bounded by the crcle crcumscrbng abc. If a trangle abc s obtuse or rght, then the mnmum dsk for coverng the trangle abc s the one whose dameter s the longest edge of trangle abc. roof. Suppose ab s the longest edge of the trangle abc. When a dsk covers

8 4 DONGHUI CHEN ET AL. abc, we can always arrange the boundary of the dsk passng through a and b. Ifab s not a dameter of the dsk and c s not on ts boundary, then we can shrnk the dsk stll coverng abc. LEMMA 8. Four termnals a, b, c, d can be covered by a dsk of radus R f and only f each of four trangles abc, bcd, cda, dab can be covered by a dsk of radus R. roof. The only f part s trval. It suffces to show the f part. Wthout loss of generalty, we may assume that abcd form a convex quadrlateral. In fact, f abcd does not form a convex quadrlateral, then one of them must le n the trangle of other three, whch can be covered by a dsk of radus R. Consder the longest edge of complete quadrlateral abcd. (Note: A complete quadrlateral has sx edges.) If ths longest edge s not a dagonal, say ab, then compare acb wth adb. Wthout loss of generalty assume acb adb. Then, the mnmum dsk coverng trangle abc also covers pont d (Fgure 4(a)). Next, we may assume that the longest edge of complete quadrlateral abcd s a dagonal, say ac, and consder followng cases. Case 1. Trangles abc and acd are obtuse or rght. In ths case, abc and cda are obtuse or rght. Therefore, the dsk wth a dameter ac covers a, b, c, d (Fgure 4(b)). Case. Ether trangle abc or acd s acute, say trangle abc. If trangle abd s also acute, then compare acb wth adb. Wthout loss of generalty assume acb adb. Then, the mnmum dsk coverng trangle abc also covers pont d (Fgure 4(a)). Smlar argument may apply to the subcases that trangle bcd s acute and that bdc 90 or adb 90. Moreover, cbd cba 90 and dba cba 90. Therefore, the remander s that bad 90 and dcb 90. In ths subcase, the dsk wth a dameter bd covers a, b, c, d. The proof of Lemma 8 s constructve. We can actually use the proof to fnd the Stener pont to connect the four termnals when t exsts. Now, we present the followng approxmaton algorthm. Fgure 4.

9 AROXIMATIONS FOR STEINER TREES 5 ALGORITHM A. For nput set X of n termnals, sort all n(n 1)/ possble edges between the n termnals n length ncreasng order e 1, e,...,e n(n1)/. Intally, set TA (X, 5) and 1. Then do the followng: Step 1 Step Step 3 whle e R do begn f e connects two dfferent connected components of TA then put e nto T A; : 1; end-whle for each subset of four termnals b, b, c, d respectvely n four connected components of TA do f there exsts a pont s wthn dstance R from a, b, c and d then put the 4-star, consstng of four edges sa, sb, sc, sd, nto T A; whle n(n 1)/ do begn f e connects two dfferent connected components of TA then put e nto T A; : 1; end-whle return T A 4 Clearly, ths algorthm runs n O(n ) tme. THEOREM. Let T* be optmal tree for ST-MS and TA an approxmaton produced by Algorthm A. Then C(T A) 3C(T*). ( ) roof. Denote by T the TA at the begnnng of Step n the algorthm A. (3) () Suppose T T contans k 4-stars. Then C(T A) C(T S) k where TS s a stenerzed mnmum spannng tree on all gven termnals. Let T* bea shortest optmal tree for ST-MS wth Stener ponts of degrees at most fve. Suppose T* has g full components T 1, T,...,T g. We construct a stenerzed () spannng tree T as follows: Intally, put T nto T. For each full component T (1 g), add to T the stenerzed mnmum spannng tree H for termnals n T.If T has a cycle, then destroy the cycle by deletng some edges and Stener ponts of H. An mportant observaton s that f H does not contan an edge between two () termnals, then a Stener pont must be deleted for destroyng a cycle n H T. From ths observaton and by Lemma 6, we have C(T ) 3C(T*) h S where h s the number of full components T s wth propertes that every Stener () pont n T has degree fve and T T has no cycle. Hence, C(T ) 3C(T*) h k. A

10 6 DONGHUI CHEN ET AL. Fgure 5. It suffces to show h k. () (3) Suppose T has p connected components. Then, T has p 3k connected components C 1, C,...,C p3k. Now, we construct a graph H wth vertex set X and () the followng edges: Frst, we put all edges of T nto H. Then consder every full component T (1 h) wth propertes that every Stener pont n T has degree () fve and T T has no cycle. If T has only one Stener pont, then ths Stener pont connects to fve termnals whch must le n at most three C s. Hence, among them there are two pars of termnals; each par le n the same C. Connect the two pars wth two edges and put the two edges nto H. IfT has at least two Stener ponts, then there must exst at least two Stener ponts each connectng to four termnals. We can also fnd two pars of termnals among them such that each par les n the same C. Connect the two pars wth two edges and put the two edges nto H. Clearly, H has at most p h connected components. Snce every connected component of H s contaned by a C, we have p 3k p h. Therefore, h 3k/. What s the exact value of the performance rato of Algorthm A? It s stll open. What we know s that ths value s between.5 and 3. The lower bound.5 can be shown by the nstance n Fgure olynomal-tme approxmaton scheme In ths secton, we consder a varaton of ST-MS. The nput and the constrant are the same. Instead of mnmzng the number of Stener ponts n the tree, we mnmze the number of total ponts (both Stener ponts and gven termnals) n the tree. Obvously, the decson versons of the two problems are dentcal. The new verson s called the Stener tree problem wth mnmum number of total ponts (ST-MT). We construct a polynomal tme approxmaton scheme when the gven set of termnals satsfes certan condtons.

11 AROXIMATIONS FOR STEINER TREES 7 Fgure 6. The rectangle wth partton of sze k., A set X of termnals s c-local f n the mnmum spannng tree of X the length of the longest edge s at most c tmes of the length of the shortest edge. Wthout loss of generalty, we assume that the dstance between any par of termnals n X s at least 1 and c 1. We are nterested n the case where R c THE BASIC IDEA The basc dea of our algorthm s to combne the shftng technque n [4] wth a local optmzaton method. We desgn a set of parttons, each of them parttons the whole area enclosng all termnals nto many rectangular cells (mostly squares) of some constant sze. (See Fgure 6.) Each cell s further dvded nto nteror and boundary areas as n Fgure 7. Then, wth respect to each partton, we organze the termnals contaned n the nteror area of each cell nto several groups such that the Fgure 7. The nteror and boundary areas. The wdth of the boundary areas s l ( 3 log k)c.

12 8 DONGHUI CHEN ET AL. dstance between any two groups s greater than c, and construct an optmal soluton (a local Stener tree) for each group. The collecton of all the local Stener trees n a cell form a local Stener forest for the cell. After that, we connect all the local Stener forests and the termnals contaned n the boundary areas usng the spannng tree approach. Fnally, we select a partton whch yelds an optmal global soluton among all the parttons. 5.. ARTITION STRATEGY Frst, we focus on the parttons. Wthout loss of generalty, assume that the set of termnals X s contaned n a rectangle Rec wth corners (0, 0) (s, 0), (0, t), and (s, t), as shown n Fgure 6. For any postve nteger k, apartton of sze k s a grd n whch adacent horzontal/ vertcal lnes are separated by a dstance k. Clearly, there are k dfferent parttons of sze k, dependng on the postons of the top horzontal lne and the leftmost vertcal lne. We use, where 0, k, to denote the, partton n whch the top horzontal lne and the leftmost vertcal lne are y and x, respectvely. The grd parttons the rectangle Rec nto many cells, most of whch are squares of sze k k. Thus, each cell contans at most k termnals n X. Each cell s dvded nto an nteror area and a boundary area, wth a boundary of wdth l ( 3 log k)c. (See Fgure 7.) 5.3. THE AROXIMATION SCHEME Let X be the set of termnals n the plane, be a partton, and X X be the set of termnals n the nteror areas. An edge s a crossng edge f t s not completely contaned n any nteror area of a cell. A stem n a Stener tree T s a path n T such that every vertex n the path s degree- Stener ponts except that the two vertces at the ends are termnals. A stem s a crossng stem f at least one of the termnals s n mn the boundary area. Let T be an optmal soluton of ST-MT for X. T denotes an optmal soluton of ST-MT for X and C(T ) denotes the total number of ponts n the tree T,.e. C(T ) C(T ) n when T contans n termnals. Snce X s a subset of X, we have p mn C(T ) C(T ). (1) In our algorthm, we deal wth one cell at a tme. Recall that the termnals n the nteror area of a cell are dvded nto several groups and an optmal soluton s constructed for each group. In order to show how to correctly group the termnals n an nteror area, let us consder an optmal soluton of ST-MT T for X. We need to modfy T nto a forest F such that each tree n F s completely ncluded n the nteror areas of some cell. Note that each nteror area of a cell may contan more than one tree n F. Defne the dstance between two trees to be the shortest dstance between any par of termnals n the two trees. We further requre that the dstance between any par of trees n F s greater than c.

13 AROXIMATIONS FOR STEINER TREES 9 LEMMA 9. Let be a partton and T be an optmal soluton of ST-MT for X. T can be modfed nto a forest F such that each tree n F s completely n an nteror area of a cell for and the dstance between any par of trees n F s at least c. Moreover, the total cost C(F ), whch s the sum of the costs of all the trees n F, s at most C(T ). Thus, mn C(F ) C(T ) C(T ). roof. Frst, we elmnate the stems wth length greater than c from T. The dstance between any par of resultng trees s greater than c snce T s optmal. For each tree T n the forest obtaned above, we reconstruct an optmal tree connectng the termnals n T. Wthout loss of generalty, we can assume that each stem n the reconstructed trees has length at most c. (Otherwse, we can repeat the procedure and further decompose the forest.) Now we prove that each tree n the forest obtaned above s completely n an nteror area of a cell. It suffces to show that there s no Stener pont n the boundary area. Suppose there are Stener ponts n the boundary area. Call a Stener pont wth degree greater than a real Stener pont. Note that the dstance between two cells s ( 3 log k)c. It s easy to see that no termnals n dstnct cells are connected n the above resultng forest. Otherwse, there must be a Stener pont s whch s at least 1.5 c log k away from any boundary lne. To reach any boundary lne, s has to 1.5logk create at least k real Stener ponts and k ( c/r 1) degree- Stener 1.5 ponts. Now, we remove all those k c/r Stener ponts n the boundary areas and use them to connect the dsconnected subtrees wth dstance less than c n the correspondng 4 neghbor cells. (See Fgure 8. s s n the shadowed area. At most k 1/r Stener ponts are requred to be added n each of the eght boundary segments of the four cells.) Fgure 8. The eght boundary segments.

14 30 DONGHUI CHEN ET AL. Snce no two cells are connected, we can move the Stener ponts n the boundary areas back to the nteror areas. In ths way, all Stener ponts n the boundary area can be elmnated. It s dffcult to compute the forest F, snce T s unknown. Nevertheless, we can construct a forest whch s smlar to F. Consder the termnals n the nteror area of some fxed cell. By Lemma 9, f the dstance between two termnals s at most c, then they must belong to the same tree of F. Thus, we can group the termnals by formng a mnmum-cost spannng tree of these termnals and then deletng the edges longer than c. Therefore, we get a set of (spannng) trees S,...,S, 1 m consstng of degrees of length at most c. We call these trees the c-spannng trees. Let Y, 1,...,m, be the set of termnals contaned n the c-spannng tree S. Clearly, the termnals n the same group Y belong to the same tree of the forest F. The converse s not necessarly true. Namely, termnals n dfferent groups Y s may also belong to the same tree of F. In other words, to fnd the best way of groupng the termnals, we have to consder all possble ways mergng the groups Y 1,...,Y m. After each such possble merge, we obtan a local Stener forest by constructng an optmal soluton for every new group. We are nterested n a local Stener forest wth the mnmum cost among all possble merges for each cell. Let forest Fˆ denote the collecton of the mnmum-cost local Stener forests, one for each cell. Fˆ has the followng propertes. ˆ LEMMA 10. () Each tree n F s completely contaned n the nteror area of a cell; () The dstance between any par of trees T and T n Fˆ s greater than c; ˆ and () The total cost of the forest F s at most C(F ). Thus, ˆ mn C(F ) C(F ) C(T ). Suppose that there are m groups n a cell. Usng the method n [16], we can m compute a mnmum-cost local Stener forest n O( M(Y)) tme, where M(Y) s the tme to construct an optmal soluton for the set of termnals Y, whch s exponental n the sze of Y AN EXACT ALGORITHM FOR ST-MT Let Y be a set of Y termnals. Wthout loss of generalty, we can assume that the termnals n Y are leaves n the tree. The number of possble topologes (the degree can be unbounded) for Y s at most Y!. Consder a fxed topology T for Y. If the number of canddate ponts for each nternal vertex n T s at most m, then a modfcaton of a standard dynamc programmng algorthm fnds an optmal soluton for the fxed topology T n O(Ym) tme [5]. LEMMA 11. The number of canddate ponts for each nternal vertex s at most 3 Y1 (Y k/r) f termnals n Y are n a square of sze k by k.

15 AROXIMATIONS FOR STEINER TREES 31 mn roof. Let T be an optmal soluton for the fxed topology T. Consder an nternal vertex v at the bottom whose chldren are leaves n T. Wthout ncreasng the number of Stener ponts, we can move the pont assgned to v such that the dstance between v and v ( 1 and ) s Rh, where h s are ntegers, and v1 and v are some chldren of v. Thus, the number of canddate ponts for v s at most Y ( k/r). The heght of T s at most Y 1. For a vertex of heght, the number of canddate ponts s denoted as f(). Then f() f( 1) Y ( k/ 3 R) f( 1). Therefore, for any nternal node, the number of canddate ponts s Y 3 1 at most (Y k/r). Y From the above dscusson, t s easy to see that M(Y) O(Y!(Y k/r) ) CONNECTING THE LOCAL FORESTS AND BOUNDARY OINTS We can construct a Stener tree for X from the forest F as follows. Fx a mnmum-cost spannng tree TS for X and add degree- Stener ponts to ensure that the length of each edge s at most R. Note that each stem n T has length at most c snce X s c-local. Let E denote the set of crossng edges n T S. Construct a graph G by addng all the crossng edges n E to Fˆ and addng degree- Stener ponts to ensure that the length of each edge s at most R. It s easy to see that ˆ S LEMMA 1. G s connected. Now, we are ready to ntroduce our algorthm, whch n fact computes G, for every possble partton,, selects a G, wth the smallest cost, and prunes the selected G, nto a tree. See Fgure 9. THEOREM 3. The performance rato of the algorthm n Fgure 9 s 1 [16(4 3 log k)c/k]. roof. Consder the stems n the mnmum spannng tree for X. Snce the boundary area of each cell conssts of at most 4( 3 log k)ck termnals, each termnal of a crossng stem can be nsde a boundary area at most 4( 3 log k)ck tmes under the k parttons. Snce the length of a stem s at most c, a stem can be a crossng stem at most 4(4 3 log k)ck tmes. Therefore, the total cost of the kg s s bounded as follows:, k1 k1 k1 k1 mn C(G ) kc(t ) C(E ) 0 0,, 0 0 mn kc(t ) 4(4 3 log k)ck C(T ). From Theorem 1, we know that at least one partton yelds a soluton wth cost less than or equal to 1 [16(4 3 log k)c/k] tmes of the optmum. s

16 3 DONGHUI CHEN ET AL. Fgure 9. Algorthm 1. COROLLARY 1. There exsts a polynomal tme approxmaton scheme for ST- MT when the set of termnals s c-local. COROLLARY. Suppose TB s produced by the algorthm n Fgure 9 and T* s an optmal tree for ST-MS. Then C(T B) 16(4 3 log k)c 16(4 3 log k)c 4n 1 C(T*) k k C(T ) where TS s a stenerzed mnmum spannng tree for the same set of n termnals. That s, there exsts a polynomal tme approxmaton scheme for ST-MS when the gven set of termnals s c-local and the mnmum spannng tree on n termnals has length at least (1 )nr for some postve constant. S 6. Dscusson One of the reasons that we are so nterested n the problem ST-MS s that no geometrc optmzaton problem has been found to be MAX-SN-hard. ST-MS may be the one. In fact, Arora s approach [1] does not work for ST-MS. It s an open problem whether ST-MS has a polynomal-tme approxmaton scheme.

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