Journal of Discrete Algorithms. Approximability of partitioning graphs with supply and demand

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1 Journal of Discree Algorihms ) Conens liss available a ScienceDirec Journal of Discree Algorihms Approximabiliy of pariioning graphs wih supply dem Takehiro Io a,,erikd.demaine b, Xiao Zhou a, Takao Nishizeki a a raduae School of Informaion Sciences, Tohoku Universiy, Aoba-yama , Sendai, , Japan b MIT Compuer Science Arificial Inelligence Laboraory, 32 Vassar S., Cambridge, MA 02139, USA aricle info absrac Aricle hisory: Received 28 March 2007 Acceped 14 March 2008 Available online 29 March 2008 Keywords: Approximaion algorihm Dem FPTAS raph pariion problem MAXSNP-hard Series-parallel graph Supply Suppose ha each verex of a graph is eiher a supply verex or a dem verex is assigned a posiive real number, called he supply or he dem. Each dem verex can receive power from a mos one supply verex hrough edges in. Onehuswisheso pariion ino conneced componens by deleing edges from so ha each componen C eiher has no supply verex or has exacly one supply verex whose supply is a leas he sum of dems in C, wishes o maximize he fulfillmen, ha is, he sum of dems in all componens wih supply verices. This maximizaion problem is known o be NP-hard even for rees having exacly one supply verex srongly NP-hard for general graphs. In his paper, we focus on he approximabiliy of he problem. We firs show ha he problem is MAXSNP-hard hence here is no polynomial-ime approximaion scheme PTAS) for general graphs unless P = NP. We hen presen a fully polynomial-ime approximaion scheme FPTAS) for series-parallel graphs having exacly one supply verex Elsevier B.V. All righs reserved. 1. Inroducion Consider a graph such ha each verex is eiher a supply verex or a dem verex. Each verex v is assigned a posiive real number; he number is called he supply of v if v is a supply verex; oherwise, i is called he dem of v. Each dem verex can receive power from a mos one supply verex hrough edges in. One hus wishes o pariion ino conneced componens by deleing edges from so ha each componen C has exacly one supply verex whose supply is a leas he sum of dems of all dem verices in C. However, such a pariion does no always exis. So we wish o obain a pariion of ino conneced componens so ha each componen C eiher has no supply verex or has exacly one supply verex whose supply is a leas he sum of dems of all dem verices in C, wish o maximize he fulfillmen, ha is, he sum of dems of he dem verices in all componens wih supply verices. We call his problem he maximum pariion problem [7]. Fig. 1a) illusraes a soluion of he maximum pariion problem for a graph, whose fulfillmen is 2 + 7) ) ) ) = 45. In Fig. 1a) each supply verex is drawn as a recangle each dem verex as a circle, he supply or dem is wrien inside, he deleed edges are drawn by hick doed lines, each conneced componen wih a supply verex is shaded. The maximum pariion problem has some applicaions o he power supply problem for power delivery neworks [3,7, 10,14]. Le be a graph of a power delivery nework. Each supply verex represens a feeder, which can supply elecrical power. Each dem verex represens a load, which requires elecrical power supplied from exacly one of he feeders hroughanework.eachedgeof represens a cable segmen, which can be urned off by a swich. Then he maximum * Corresponding auhor. addresses: akehiro@ecei.ohoku.ac.jp T. Io), edemaine@mi.edu E.D. Demaine), zhou@ecei.ohoku.ac.jp X. Zhou), nishi@ecei.ohoku.ac.jp T. Nishizeki) /$ see fron maer 2008 Elsevier B.V. All righs reserved. doi: /j.jda

2 628 T. Io e al. / Journal of Discree Algorihms ) Fig. 1. a) Pariion of a graph wih maximum fulfillmen, b) pariion of a series-parallel graph having exacly one supply verex, c) a sar S wih a supply verex a he cener. pariion problem represens he power supply swiching problem o maximize he sum of all loads ha can be supplied powers in a nework reconfigured by urning off some cable segmens. iven a se A of inegers an upper bound ineger) b, hemaximum subse sum problem [4,5] asks o find a subse C of A such ha he sum of inegers in C is no greaer han he bound b is maximum among all such subses C. The maximum subse sum problem can be reduced in linear ime o he maximum pariion problem for a paricular ree, called a sar, wih exacly one supply verex a he cener, as illusraed in Fig. 1c) [7]. Since he maximum subse sum problem is NP-hard, he maximum pariion problem is also NP-hard even for sars. Thus i is very unlikely ha he maximum pariion problem can be exacly solved in polynomial ime even for rees. However, here is a fully polynomial-ime approximaion scheme FPTAS) for he maximum pariion problem on rees [7]. One may hus expec ha he FPTAS for rees can be exended o a larger class of graphs, for example series-parallel graphs parial k-rees, ha is, graphs wih bounded reewidh [1,2]. In his paper, we sudy he approximabiliy of he maximum pariion problem. We firs show ha he maximum pariion problem is MAXSNP-hard, hence here is no polynomial-ime approximaion scheme PTAS) for he problem on general graphs unless P = NP. We hen presen an FPTAS for series-parallel graphs having exacly one supply verex. The FPTAS for series-parallel graphs can be exended o parial k-rees. Fig. 1b) depics a series-parallel graph ogeher wih a conneced componen C found by our FPTAS. One migh hink ha i would be sraighforward o exend he FPTAS for he maximum subse sum problem in [5] o an FPTAS for he maximum pariion problem wih a single supply verex. However, his is no he case since we mus ake a graph srucure ino accoun. For example, he verex v of dem 2 drawn by a hick circle in Fig. 1b) canno be supplied power even hough he supply verex w has marginal power ) = 2, while he verex v in Fig. 1c) can be supplied power from he supply verex w in he sar having he same supply dems as in Fig. 1b). Indeed, we no only exend he scaling rounding echnique bu also employ many new ideas o derive our FPTAS. An early version of he paper has been presened a [6]. The res of he paper is organized as follows. In Secion 2 we show ha he maximum pariion problem is MAXSNPhard. In Secion 3 we presen a pseudo-polynomial-ime algorihm for series-parallel graphs. In Secion 4 we presen an FPTAS based on he algorihm in Secion MAXSNP-hardness Assume in his secion ha a graph has one or more supply verices. See Figs. 1a) 2b).) The main resul of his secion is he following heorem. Theorem 1. The maximum pariion problem is MAXSNP-hard for biparie graphs.

3 T. Io e al. / Journal of Discree Algorihms ) Fig. 2. a) Variable gadge x j, b) biparie graph Φ corresponding o an insance Φ wih hree clauses C 1 = x 1 x 2 x 3 ), C 2 = x 1 x 2 x 3 ) C 3 = x 1 x 2 x 3 ). A varian of he MAXSAT problem, called he 3-occurrence MAX3SAT problem, is MAXSNP-hard [11,12]. An insance Φ of he problem consiss of a collecion of m clauses C 1, C 2,...,C m on n variables x 1, x 2,...,x n such ha each clause has exacly hree lierals each variable appears a mos hree imes in he clauses. The 3-occurrence MAX3SAT problem is o find a ruh assignmen for he variables which saisfies he maximum number of clauses. Since each clause has exacly hree lierals, we have n 3m. 1) In order o prove Theorem 1, we use he concep of L-reducion which is a special kind of reducion ha preserves approximabiliy [11,12]. Suppose ha boh A B are maximizaion problems. Then we say ha AcanbeL-reducedoB if here exis wo polynomial-ime algorihms Q R wo posiive consans α β which saisfy he following wo condiions 1) 2) for each insance I A of A: 1) he algorihm Q reurns an insance I B = Q I A ) of B such ha OPT B I B ) α OPT A I A ), where OPT A I A ) OPT B I B ) denoe he maximum soluion values of I A I B, respecively; 2) for each feasible soluion of I B wih value c B, he algorihm R reurns a feasible soluion of I A wih value c A such ha OPT A I A ) c A β OPT ) B I B ) c B. Noe ha, by condiion 2) of he L-reducion, R mus reurn he opimal soluion of I A for he opimal soluion of I B. We now prove Theorem 1. Proof of Theorem 1. I suffices o show ha he 3-occurrence MAX3SAT problem can be L-reduced o he maximum pariion problem for biparie graphs. We firs show ha condiion 1) of he L-reducion holds for α = 26. I suffices o show ha, from each insance Φ of he 3-occurrence MAX3SAT problem, one can consruc in polynomial ime a biparie graph Φ as an insance of he maximum pariion problem such ha OPT MPP Φ ) 26 OPT SAT Φ), 2) where OPT MPP Φ ) is he maximum soluion value of he maximum pariion problem for Φ OPT SAT Φ) is he maximum soluion value of he 3-occurrence MAX3SAT problem for Φ. We firs make a variable gadge x j for each variable x j,1 j n; x j is a binary ree wih hree verices as illusraed in Fig. 2a); he roo is a supply verex of supply 7, he wo leaves x j x j are dem verices of dems 4. Then he graph Φ is consruced as follows. For each variable x j,1 j n, pu he variable gadge x j o he graph, for each clause C i,1 i m, pu a dem verex C i of dem 1 o he graph. Finally, for each clause C i,1 i m, joina dem verex x j or x j )in x j,1 j n, wih he dem verex C i if only if he lieral x j or x j )isinheclause C i, as illusraed in Fig. 2b). Clearly, Φ can be consruced in polynomial ime, is a biparie graph. I should be noed ha, since each variable x j,1 j n, appears a mos hree imes in he clauses, he supply verex in x j has enough power o supply all dem verices C i whose corresponding clauses have x j or x j. We hen verify Eq. 2). One can easily have OPT MPP Φ ) = 4n + OPT SAT Φ). 3) Noe ha, for each j, 1 j n, exacly one of he wo dem verices x j x j is supplied power in he maximum soluion of he maximum pariion problem for Φ hence he firs erm 4n of he righ side of Eq. 3) represens he sum of he dems in x j,1 j n, which are supplied power. Since OPT SAT Φ) m, by Eqs. 1) 3) we have OPT MPP Φ ) 12m + m = 13m. 4)

4 630 T. Io e al. / Journal of Discree Algorihms ) On he oher h, we have OPT SAT Φ) m/2, because if a ruh assignmen saisfies only less han half of clauses of Φ, hen he negaion of he ruh assignmen saisfies a leas half of he clauses of Φ. Therefore, by Eq. 4) we have OPT MPP Φ ) 13m = 26 m 2 26 OPT SATΦ). We have hus verified Eq. 2). We nex show ha condiion 2) of he L-reducion holds for β = 1. One can give a ruh assignmen in polynomial ime from a pariion P of Φ, as follows: se a variable x j o TRUE if he dem verex x j in x j is supplied power in P ; oherwise, se x j o FALSE. I suffices o show ha OPT SAT Φ) c Φ OPT MPP Φ ) f P), 5) where c Φ is he number of clauses of Φ saisfied by he ruh assignmen f P) is he fulfillmen of P, ha is, he sum of dems of all dem verices in componens wih supply verices. Eiher he dem verex x j or he dem verex x j is no supplied power in P for each variable gadge x j,1 j n. Therefore, we have c Φ f P) 4n. 6) By Eqs. 3) 6) we have OPT SAT Φ) c Φ OPT SAT Φ) f P) ) 4n = 4n + OPT SAT Φ) ) f P) = OPT MPP Φ ) f P). We have hus verified Eq. 5). 3. Pseudo-polynomial-ime algorihm Since he maximum pariion problem is srongly NP-hard [8], here is no pseudo-polynomial-ime algorihm for general graphs unless P = NP. However, Io e al. presened a pseudo-polynomial-ime algorihm for he maximum pariion problem on series-parallel graphs having one or more supply verices [8]. In his secion we presen anoher pseudo-polynomial-ime algorihm on series-parallel graphs having exacly one supply verex, which is suied o an FPTAS presened in Secion 4. More precisely, we have he following heorem. Theorem 2. The maximum pariion problem for a series-parallel graph wih a single supply verex can be solved in ime OF 2 n) if he dems he supply are inegers, where n is he number of verices in F is an arbirary upper bound on he maximum soluion value for. A rivial example of he upper bound F is he supply of he supply verex. Anoher example is he sum of dems of all dem verices in. A beer upper bound will be given in Secion 4. In he remainder of his secion we give an algorihm o solve he maximum pariion problem in ime OF 2 n) as a proof of Theorem 2. In Secion 3.1 we give a definiion of a series-parallel graph. In Secion 3.2 we define some erms presen ideas of our algorihm. We hen presen our algorihm in Secion 3.3. We finally show, in Secion 3.4, ha our algorihm akes ime OF 2 n) Terminologies definiions Awo-erminal) series-parallel graph is defined recursively as follows [13]: 1) A graph wih a single edge is a series-parallel graph. The ends of he edge are called he erminals of denoed by v s ) v ). See Fig. 3a).) 2) Le 1 be a series-parallel graph wih erminals v s 1 ) v 1 ), le 2 be a series-parallel graph wih erminals v s 2 ) v 2 ). a) A graph obained from 1 2 by idenifying v 1 ) wih v s 2 ) is a series-parallel graph, whose erminals are v s ) = v s 1 ) v ) = v 2 ). Such a connecion is called a series connecion, is denoed by = 1 2. See Fig. 3b).) b) A graph obained from 1 2 by idenifying v s 1 ) wih v s 2 ) idenifying v 1 ) wih v 2 ) is a series-parallel graph, whose erminals are v s ) = v s 1 ) = v s 2 ) v ) = v 1 ) = v 2 ). Such a connecion is called a parallel connecion, is denoed by = 1 2. See Fig. 3c).) The erminals v s ) v ) of are ofen denoed simply by v s v, respecively. Since we deal wih he maximum pariion problem, we may assume wihou loss of generaliy ha is a simple graph hence has no muliple edges.

5 T. Io e al. / Journal of Discree Algorihms ) Fig. 3. a) A series-parallel graph wih a single edge, b) series connecion, c) parallel connecion. Fig. 4. a) A series-parallel graph, b) a binary decomposiion ree T of, c)asubgraph u of. A series-parallel graph can be represened by a binary decomposiion ree [13]. Fig. 4a) illusraes a series-parallel graph, Fig. 4b) depics a binary decomposiion ree T of. Labels s p aached o inernal nodes in T indicae series parallel connecions, respecively. Nodes labeled s p are called s- p-nodes, respecively. Every leaf of T represens a subgraph of induced by a single edge. Each node u of T corresponds o a subgraph u = V u, E u ) of induced by all edges represened by he leaves ha are descendans of u in T. Fig. 4c) depics u for he lef child u of he roo r of T in Fig. 4b). u is a series-parallel graph for each node u of T, = r for he roo r of T. Since a binary decomposiion ree of a given series-parallel graph can be found in linear ime [13], we may assume ha a series-parallel graph is binary decomposiion ree T are given Terms ideas Suppose ha here is exacly one supply verex w in a graph = V, E), as illusraed in Figs. 1b) c). Le supw) be he supply of w. For each dem verex v, we denoe by demv) he dem of v. Ledemw) = 0alhoughw is a supply verex. Then, insead of finding a pariion of, we shall find a se C V such ha a) w C; b) v C demv) supw); c) C induces a conneced subgraph of. Such a se C V is called a supplied se for. Thefulfillmen f C) of a supplied se C is he sum of dems of all dem verices in C, ha is, f C) = v C demv).

6 632 T. Io e al. / Journal of Discree Algorihms ) Fig. 5. a) A supplied se C for a series-parallel graph, b) a conneced se C s for u, c) a separaed pair C s, C ) of ses for u,d)aseparaedpair C s, C ) of isolaed ses for u. A supplied se C is called he maximum supplied se for if f C) is maximum among all supplied ses for. Then he maximum pariion problem is o find a maximum supplied se for a given graph. Themaximum fulfillmen f ) of a graph is he fulfillmen f C) of he maximum supplied se C for. For he series-parallel graph in Fig. 1b), he supplied se C shaded in he figure has he maximum fulfillmen, hence f ) = f C) = 23, while f S) = 25 for he sar S in Fig. 1c). [Main ideas] Le be a series-parallel graph, le u, u u be nodes of a binary decomposiion ree T of, le u = V u, E u ), u = V u, E u ) u = V u, E u ) be he subgraphs of for nodes u, u u, respecively, as illusraed in Fig. 5a). Every supplied se C for naurally induces subses of V u, V u V u. The supplied se C for indicaed by a doed closed curve in Fig. 5a) induces a single subse C s of V u in Fig. 5b) such ha C s = C V u v s u ), v u ) C s.on he oher h, C induces a pair of subses C s C of V u in Fig. 5c) such ha C s C = C V u, C s C =, v s u ) C s v u ) C.AseC s, C s or C is no always a supplied se for u or u, because i may no conain he supply verex w. C s is a conneced se for u, ha is, C s induces a conneced subgraph of u, while he pair C s, C ) is a separaed pair of ses for u, ha is, C s C induce verex-disjoin conneced subgraphs of u.thesec in Fig. 5a) conainsnoerminalsof u. In such a case, we regard ha demv s u )) = demv u )) = 0 C induces a separaed pair of singleon ses C s, C ) such ha C s ={v s u )} C ={v u )}, as illusraed in Fig. 5d). The formal definiions will be given laer.) If a se C s, C s or C conains he supply verex w, hen he se may have he marginal power, he amoun of which is no greaer han supw). If a se does no conain w, hen he se may have he deficien power, he amoun of which should be no greaer han supw). Thus we laer inroduce five funcions g, h 1, h 2, h 3 h 4 ; for a series-parallel graph u a real number x, hevalueg u, x) represens he maximum marginal power or he minimum deficien power of conneced ses for u ; for a series-parallel graph u a real number x, he value h i u, x), 1 i 4, represens he maximum marginal power or he minimum deficien power of separaed pairs of ses for u. Our idea is o compue g u, x) h i u, x), 1 i 4, from he leaves of T o he roo r of T by means of dynamic programming. [Formal definiions of conneced ses separaed pair of ses ] We now formally define he noion of conneced ses separaed pair of ses for a series-parallel graph. Le u = V u, E u ) be a subgraph of for a node u of a binary decomposiion ree T of, le v s = v s u ) v = v u ).We call a se C V u a conneced se for u if C saisfies he following hree condiions see Fig. 5b)): a) v s, v C; b) C induces a conneced subgraph of u ; c) v C demv) supw). ApairofsesC s, C V u is called a separaed pair of ses) for u if C s C saisfy he following hree condiions see Fig. 5c)):

7 T. Io e al. / Journal of Discree Algorihms ) Fig. 6. a) A series-parallel graph, b)asubgraph in u of u,c)asubgraph ou u of. a) C s C =, v s C s v C ; b) C s C induce conneced subgraphs of u ; c) v C s C demv) supw). We hen classify conneced ses separaed pairs furher ino smaller classes. The power flow around a erminal depends on wheher he erminal is a supply verex or a dem verex. Since we wan o deal wih he wo cases uniformly, we inroduce a virual graph u for a subgraph u of ; u is obained from u by regarding each of he wo erminals v s v as a dem verex whose dem is zero. We denoe by dem x) he dem of a dem verex x in u, hence { dem 0 if x is vs or v x) = ; demx) oherwise. Clearly every conneced se for u is a conneced se for u. However, a conneced se C for u is no necessarily a conneced se for u ;forexample,if x C dem x) supw) bu x C demx) = demv s) + demv ) + x C dem x)> supw), henc is no a conneced se for u. Similarly, every separaed pair for u is a separaed pair for u, while no every separaed pair for u is a separaed pair for u.wedenoeby in u he graph obained from u by deleing he wo erminals v s v as illusraed in Fig. 6b), while we denoe by ou u he graph obained from by deleing all he verices of u excep v s v as illusraed in Fig. 6c). Le R w ={x R: x supw)}, where R denoes he se of all real numbers. For each real number i R w,wecalla conneced se C for u an i-conneced se if C saisfies he following wo condiions a) b): a) if i > 0, hen w C i + dem x) supw); x C b) if i 0, hen w / C dem x) i = i. x C An i-conneced se C for u wih i > 0 is a supplied se for u, hence corresponds o some supplied se C r for he whole graph = r such ha w C C r, where r is he roo of T ;anamouni of he remaining power of w can be delivered ouside u hrough v s or v ; hence he margin of C is i. On he oher h, an i-conneced se C for u wih i 0 is no a supplied se for u, bu may correspond o a supplied se C r for = r such ha w / C C r w C r ;anamoun i of power mus be delivered o C from w hrough v s or v, hence he deficiency of C is i. For an i-conneced se C for u,le f C, i) = dem x). x C

8 634 T. Io e al. / Journal of Discree Algorihms ) Then f C, i) = f C) for u if 0 < i supw). On he oher h, if supw) i 0, hen f C, i) represens he fulfillmen of C when an amoun i of power is delivered o C from w in ou u. According o he definiion of an i-conneced se, a conneced se C for u is no a 0-conneced se for u if C conains he supply verex w = v s, v ) dem x) = supw). x C Because he dems of v s v are posiive, we have demx)>supw) x C hence such a conneced se C for u is no a conneced se for u we need no o ake C ino accoun. Thus, if C is a 0-conneced se for u,henc ={v s, v } u has an edge v s, v ). Le σ / R w be a symbol. For each pair of j k in R w {σ }, wecallaseparaedpairc s, C ) for u a j, k)-separaed pair if C s, C ) saisfies he following seven condiions a) g): a) if j R w j > 0, hen w C s j + x C s dem x) supw); b) if j R w j 0, hen w / C s x C s dem x) j; c) if j = σ,henc s ={v s }; d) if k R w k > 0, hen w C k + x C dem x) supw); e) if k R w k 0, hen w / C x C dem x) k; f) if k = σ,henc ={v }; g) if j, k R w j + k 0, hen j 0 k 0. Since here is exacly one supply verex w in, here is no j, k)-separaed pair C s, C ) for u such ha j > 0 k > 0. A j, k)-separaed pair C s, C ) for u wih j > 0 corresponds o a supplied se C r for he whole graph = r such ha w C s C r ;anamoun j of he remaining power of w can be delivered ouside C s hrough v s, hence he margin of C s is j. A j, k)-separaed pair C s, C ) for u wih j 0 may correspond o a supplied se C r for such ha C s C r eiher w C or w C r C s C ;anamoun j of power mus be delivered o C s hrough v s, hence he deficiency of C s is j. A j, k)-separaed pair C s, C ) for u wih j = σ corresponds o a supplied se C r for such ha v s / C r, ha is, v s is never supplied power. See Figs. 5a) d).) Clearly C s ={v s } if C s is a 0, k)-separaed pair for u.a j, k)- separaed pair C s, C ) for u wih k > 0, k 0ork = σ corresponds o a supplied se C r for similarly as above. For a j, k)-separaed pair C s, C ) for u,le x C s C dem x) if j, k R w ; f C s, C, j, k) = x C s dem x) if j R w k = σ ; x C dem x) if j = σ k R w. Le f {v s }, {v }, σ, σ ) = max { f C u ) C u is a supplied se for u such ha v s, v / C u }, le f {v s }, {v }, σ, σ ) = 0if u has no supplied se C u such ha v s, v / C u. [Formal definiions of funcions g h i, 1 i 4] Le denoe he se of all series-parallel graphs. We now formally define a funcion g :, R) R w { } as follows: for a series-parallel graph u a real number x R, g u, x) = { max i R w u has an i-conneced se C such ha f C, i) x}. 7)

9 T. Io e al. / Journal of Discree Algorihms ) If u has no i-conneced se C wih f C, i) x for any number i R w,henleg u, x) =. We hen formally define a funcion h 1 :, R) R w { } as follows: for a series-parallel graph u a real number x R, h 1 u, x ) = max { j + k u has a j, k)-separaed pair C s, C ) such ha j, k R w, j + k supw), f C s, C, j, k) x }. 8) If u has no j, k)-separaed pair C s, C ) wih f C s, C, j, k) x for any pair of numbers j k in R w,henleh 1 u, x) =. I should be noed ha a j, k)-separaed pair C s, C ) for u wih j, k R w corresponds o a supplied se C r for such ha C s C C r, hence we can simply ake he summaion of j k as he marginal power or he deficien power of C s C. We nex formally define a funcion h 2 :, R) R w { } as follows: for a series-parallel graph u a real number x R, h 2 u, x ) = max { j R w u has a j, σ )-separaed pair C s, {v } ) such ha f C s, {v }, j, σ ) x }. 9) If u has no j, σ )-separaed pair C s, {v }) wih f C s, {v }, j, σ ) x for any number j R w,henleh 2 u, x) =.We hen formally define a funcion h 3 :, R) R w { } as follows: for a series-parallel graph u a real number x R, h 3 u, x ) = max { k R w u has a σ, k)-separaed pair {v s }, C ) such ha f {vs }, C, σ, k ) x }. 10) If u has no σ, k)-separaed pair {v s}, C ) wih f {v s }, C, σ, k) x for any number k R w,henleh 3 u, x) =.We finally define a funcion h 4 :, R) {0, } as follows: for a series-parallel graph u a real number x R, h 4 u, ) { 0 if x = u has a σ, σ )-separaed pair {v s }, {v }) such ha f {v s }, {v }, σ, σ ) x; 11) oherwise. Clearly, he five funcions g h i,1 i 4, are non-increasing. For any negaive real number x < 0, we have g u, x) = g u, 0) h i u, x) = h i u, 0), 1 i 4. Our algorihm compues g u, x) h i u, x), 1 i 4, for each node u of a binary decomposiion ree T of a given series-parallel graph from he leaves o he roo r of T by means of dynamic programming Algorihm We firs show how o compue he maximum fulfillmen f ) of a given graph from g, x) h i, x), 1 i 4. [How o compue f )] Suppose ha g r, x) h i r, x), 1 i 4, have been compued for he roo r of T.Since = r, one can easily compue f ) from g, x) h i, x), 1 i 4, as in he following wo cases a) b), where v s = v s ) v = v ). Case a): one of v s v is he supply verex w he oher is a dem verex. One may assume wihou loss of generaliy ha v s is he supply verex w v is a dem verex. Le C be a supplied se for having he maximum fulfillmen. Then here are he following wo cases i) ii), as illusraed in Fig. 7: i) v is supplied power from v s = w), ha is, v s, v C; ii) v is no supplied power, ha is, v s C v / C. For Case i), we compue f 1 ) as follows: f 1 ) = max { x + demv ) x R supw) + g, x) demv ) 0 }. 12) Noe ha g, x) 0foreverynumberx R since has no supply verex. If supw) + g, x) demv )<0forany number x R, henle f 1 ) =. For Case ii), we compue f 2 ) as follows: f 2 ) = max { x R supw) + h 2, x) 0 }. 13) Noe ha h 2, x) 0foreverynumberx R. Ifsupw) + h 2, x)<0foranynumberx R, henle f 2 ) =. Fig. 7. Two cases in Case a).

10 636 T. Io e al. / Journal of Discree Algorihms ) Fig. 8. Four cases in Case b). We hus have f ) = max { f 1 ), f 2 ) }. Case b): boh v s v are dem verices. 14) Le C be a supplied se for having he maximum fulfillmen. In his case, here are he following four cases iii) vi), as illusraed in Fig. 8: iii) v s, v C; iv) v s C v / C; v) v s / C v C; vi) v s, v / C. For Case iii), we compue f 3 ) as follows: f 3 ) = max { x + demv s ) + demv ) x R g, x) demv s ) demv ) 0 }. 15) If g, x) demv s ) demv )<0foranynumberx R, henle f 3 ) =. For Case iv), we compue f 4 ) as follows: f 4 ) = max { x + demv s ) x R h 2, x) demv s ) 0 }. 16) If h 2, x) demv s )<0foranynumberx R, henle f 4 ) =. For Case v), we compue f 5 ) as follows: f 5 ) = max { x + demv ) x R h 3, x) demv ) 0 }. 17) If h 3, x) demv )<0foranynumberx R, henle f 5 ) =. For Case vi), we compue f 6 ) as follows: f 6 ) = max { x R h 4, x) = 0 }. If h 4, x) = for any number x R, henle f 6 ) =. We hus have f ) = max { f 3 ), f 4 ), f 5 ), f 6 ) }. We hen explain how o compue g u, x) h i u, x), 1 i 4, for each node u of T. 18) 19) [How o compue g u, x) h i u, x), 1 i 4] We firs compue g u, x) h i u, x), 1 i 4, for each leaf u of T, for which u conains exacly one edge as illusraed in Fig. 3a). Since he wo erminals of u are dem verices of dems zero, we have { g u, x) = 0 if x 0; oherwise. Similarly, for each index i, 1 i 4, we have { h i u, x) = 0 if x 0; oherwise. We nex compue g u, x) h i u, x), 1 i 4, for each inernal node u of T from he counerpars of he wo children of u in T. However, we show only how o compue h 1 u, x) for a p-node u of T, because one can similarly compue g u, x) h i u, x), 1 i 4, for each p-node s-node of T ; he deails are given in Appendix A. 20) 21)

11 T. Io e al. / Journal of Discree Algorihms ) Fig. 9. Combining a j 1, k 1 )-separaed pair C s1, C 1 ) for 1 a j 2, k 2 )-separaed pair C s2, C 2 ) for 2 o a j, k)-separaed pair C s, C ) for u = 1 2 wih j, k R w. We compue h 1 u, x) for a p-node u of T.Le u = 1 2, le v s = v s u ) v = v u ). See Figs. 3c) 9.) Le C s, C ) be a j, k)-separaed pair for u wih j, k R w such ha f C s, C, j, k) x R j +k = h 1 u, x) =.The j, k)-separaed pair C s, C ) for u can be obained by combining a j 1, k 1 )-separaed pair C s1, C 1 ) for 1 wih a j 2, k 2 )- separaed pair C s2, C 2 ) for 2 such ha f C s, C, j, k) = f C s1, C 1, j 1, k 1 ) + f C s2, C 2, j 2, k 2 ), where j 1, j 2, k 1, k 2 R w such ha j 1 + j 2 = j k 1 + k 2 = k, as illusraed in Fig. 9. Since f C s, C, j, k) = f C s1, C 1, j 1, k 1 ) + f C s2, C 2, j 2, k 2 ) x, we have f C s1, C 1, j 1, k 1 ) y f C s2, C 2, j 2, k 2 ) x y for some number y R. SinceC s1, C 1 ) is a j 1, k 1 )-separaed pair for 1 wih f C s1, C 1, j 1, k 1 ) y, one may assume by Eq. 8) ha j 1 + k 1 = h 1 1, y). Similarly, one may assume ha j 2 + k 2 = h 1 2, x y). Sinceh 1 u, x) = j + k = j 1 + j 2 ) + k 1 + k 2 ) = h 1 1, y) + h 1 2, x y), one can compue h 1 u, x) as follows: h 1 { u, x) = max h1 y 1, y) + h 1 2, x y)}. 22) I should be noed ha he maximum above is aken over all real numbers y R such ha if h 1 1, y) + h 1 2, x y) 0henh 1 1, y) 0 h 1 2, x y) 0. Remember condiion g) of a j, k)-separaed pair.) 3.4. Proof of Theorem 2 We now show ha our algorihm akes ime OF 2 n) for a series-parallel graph as a proof of Theorem 2, where F is an arbirary upper bound on he maximum fulfillmen f ) of. Forexample,F = min{supw), v V demv)}. Since all dems he supply in a given series-parallel graph are inegers, f C u ) is an ineger for any supplied se C u for u. Similarly, f C, i) f C s, C, j, k) are inegers for any i-conneced se C any j, k)-separaed pair C s, C ) for u, respecively. We denoe by Z he se of all inegers. Le Z w ={x Z: x w}. Defineafuncionĝ :, Z) Z w { } similarly as g :, R) R w { } in Eq. 7): foraseries-parallelgraph u an ineger x Z, wedefine ĝ u, x) = max{ i Z w u has an i-conneced se C such ha f C, i) x}. Define funcions ĥ1, ĥ2, ĥ3 :, Z) Z w { } ĥ4 :, Z) {0, } similarly as h 1, h 2, h 3 h 4 in Eqs. 8) 11). Define inegral values ˆfi ), 1 i 6, similarly as f i ), 1 i 6, in Eqs. 12), 13) 15) 18), respecively. Then clearly ˆf i ) = f i ), 1 i 6, since all dems he supply in are inegers. Therefore, by Eqs. 14) 19) we can compue f ) from ˆf i ), 1 i 6. We shall hus compue values ĝ u, x) ĥi u, x), 1 i 4, for all inegers x Z. However, one can easily observe ha i suffices o compue hem only for inegers x Z + F, where Z+ F ={x Z 0 x F }; remember ha F is an upper bound of he maximum fulfillmen f ) of. For each leaf u of T all inegers x Z + F, one can easily compue values ĝ u, x) ĥi u, x), 1 i 4, in ime O Z + F ) = OF ) by he counerpars of Eqs. 20) 21). Since is a series-parallel simple graph of n verices, has a mos 2n 3 edges hence T has a mos 2n 3 leaves. One can hus compue ĝ u, x) ĥi u, x), 1 i 4, for all leaves u of T in ime OFn). For each inernal node u of T all inegers x Z + F, one can compue ĝ u, x) ĥi u, x), 1 i 4, in ime O Z + F 2 ) = OF 2 ) by he counerpars of Eq. 22) in Secion 3.3 Eqs. A.1) A.16) in Appendix A. SinceT has a mos 2n 4 inernal nodes, one can compue ĝ, x) ĥi, x), 1 i 4, in ime OF 2 n). One can compue he maximum fulfillmen f ) of from ĝ, x) ĥi, x), 1 i 4, in ime OF ) by he counerpars of Eqs. 12) 19). Thus he maximum pariion problem can be solved in ime OF 2 n). This complees a proof of Theorem FPTAS Assume in his secion ha he supply all dems are posiive real numbers which are no always inegers. Since he maximum pariion problem is MAXSNP-hard, here is no PTAS for he problem on general graphs unless P = NP. However, using he pseudo-polynomial-ime algorihm in Secion 3, we can obain an FPTAS for series-parallel graphs having exacly one supply verex, have he following heorem.

12 638 T. Io e al. / Journal of Discree Algorihms ) Fig. 10. a) Original problem insance, b) insance scaled by facor = 1. Theorem 3. There is a fully polynomial-ime approximaion scheme for he maximum pariion problem on a series-parallel graph having exacly one supply verex. In he remainder of his secion, as a proof of Theorem 3, we give an algorihm o find a supplied se C for a seriesparallel graph wih f C) 1 ε) f ) in ime polynomial in n 1/ε for any real number ε, 0< ε < 1, where n is he number of verices in. Thus our approximae maximum fulfillmen f ) of is f C), hence he error is bounded by ε f ), ha is, f ) f ) = f ) f C) ε f ). We now ouline our algorihm he analysis. We exend he ordinary scaling rounding echnique for he knapsack problem [5,9] he maximum pariion problem on rees [7] apply i o he maximum pariion problem for a series-parallel graph wih a single supply verex. For some scaling facor, we consider he se {..., 2,, 0,, 2,...} as he range of funcions g h i,1 i 4, find he approximae soluion f ) by using he pseudo-polynomial-ime algorihm in Secion 3. As we will show laer in Lemma 2b), we have f ) f )<4n. Inuiively, Eq. 24) holds because he series parallel connecions are execued no more han 2n imes each connecion adds a mos 2 o he error f ) f ). Choosing an appropriae upper bound F on f ) such ha F /2 f ) F, aking = εf /8n), wehaveeq.23). One may expec ha an FPTAS could be obained simply by using an ordinary scaling rounding echnique he pseudo-polynomial-ime algorihm, as follows: 1) scale down he supply supw) by supw) = supw)/, scale up he dem demv) by demv) = demv)/ for each dem verex v, 2) find a supplied se C for having he maximum fulfillmen for he scaled insance by using he pseudo-polynomialime algorihm, 3) compue he fulfillmen f C) for he original insance, 4) oupu f C) as an approximae maximum fulfillmen f ) for he original one. Alhough such a sraighforward mehod always finds a feasible soluion for he original insance, he error f ) f ) canno be bounded by 4n. Consider an example in Fig. 10, where he supply verex is drawn by a recangle each dem verex by a circle. Fig. 10a) depics an original insance, while Fig. 10b) depics an insance scaled by facor = 1. For he original insance, he supplied se shaded in Fig. 10a) has he maximum fulfillmen f ) = On he oher h, for he scaled one, he supplied se C shaded in Fig. 10b) is found by he pseudo-polynomial-ime algorihm, C has a fulfillmen of f C) = f ) = for he original one. Thus f ) f ) = 100, hence f ) f ) canno be bounded by 4n = 12. Similarly, one can easily observe ha f ) f ) canno be bounded by cn for any fixed consan c. Thus he sraighforward mehod above canno yield an FPTAS. We now give he deails of our algorihm he proof of is correcness. For a posiive real number, ler = {..., 2,, 0,, 2,...} R + F ={x R 0 x F }. The funcions g h i, 1 i 4, in Secion 3 have range R. In his secion we define new funcions ḡ, h1, h2, h3 h 4 which have a sampled range R approximae g, h 1, h 2, h 3 h 4, respecively. I should be noed ha ḡ h i,1 i 4, do no always ake he same value as g h i,1 i 4, respecively, even for x R. More precisely, we i) define ḡ h i,1 i 4, for x R by he counerpars of Eqs. 7) 11), recursively compue ḡ h i,1 i 4, for x R by he counerpars of Eqs. 20) 22) Eqs. A.1) A.16) in Appendix A; ii) define compue values f i ), 1 i 6, by he counerpars of Eqs. 12), 13) 15) 18); iii) define compue f ) as follows: f ) = max { f1 ), f2 ) } 25) if one of v s ) v ) is he supply verex w, f ) = max { f3 ), f 4 ), f 5 ), f 6 ) } 26) if boh v s ) v ) are dem verices. 23) 24)

13 T. Io e al. / Journal of Discree Algorihms ) We will show laer in Lemma 2b) ha f ) is an approximae value of f ) saisfying Eq. 24). I should be noed ha he dems he supply are never scaled rounded when we compue he funcions ḡ h i,1 i 4, as above, hence hese funcions ake real values which are no necessarily in R. Le T be a binary decomposiion ree of. WedenoebynT ) he number of nodes in T. For a node u of T,wedenoe by T u asubreeoft which is rooed a u is induced by all descendans of u in T.WedenoebynT u ) he number of nodes in T u. The funcions ḡ h i,1 i 4, approximae he original funcions g h i,1 i 4, as in he following lemma. Noe ha ḡ u, x) = ḡ u, 0) h i u, x) = h i u, 0), 1 i 4, for any negaive number x R. Lemma 1. For each node u of a binary decomposiion ree T of, he following a) b) hold: a) i) ḡ u, x) g u, x) for any number x R ; ii) ḡ u, x) is non-increasing; iii) for any number x R, here is an ineger α such ha 0 α nt u ) 1 ḡ u, x/ α) g u, x), b) for each index i, 1 i 4, i) h i u, x) h i u, x) for any number x R ; ii) h i u, x) is non-increasing; iii) for any number x R, here is an ineger β i such ha 0 β i nt u ) 1 h i u, x/ β i ) h i u, x ). Proof. See Appendix B. We hen have he following lemma. Lemma 2. The following a) b) hold: a) for each index i, 1 i 6, f i ) nt ) f i ); b) f ) 4n < f ) f ). Proof. a) We prove only for he index i = 1, ha is, f 1 ) nt ) f 1 ), 27) because one can similarly prove for he oher indices. Le v s = v s ) v = v ). One may assume ha v s = w for f 1 ) f 1 ). Lex be a real number such ha x + demv ) = f 1 ) =, 28) hen by Eq. 12) we have supw) + g, x) demv ) 0. 29) By Lemma 1a) here is an ineger α such ha 0 α nt ) 1 30) ḡ, x/ α ) g, x). 31) By Eqs. 29) 31) we have supw) + ḡ, x/ α ) demv ) 0.

14 640 T. Io e al. / Journal of Discree Algorihms ) Therefore, by he counerpar of Eq. 12) we have f1 ) x/ α + demv ). 32) By Eqs. 28), 30) 32) we have f1 ) x/ nt ) 1 ) + demv ) = x/ + ) nt ) + demv ) x nt ) + demv ) f 1 ) nt ). We have hus verified Eq. 27). b) By Lemma 1a), Eq. 12) is counerpar, we have f 1 ) f 1 ). Similarly we have f i ) f i ), 2 i 6. Therefore by Eqs. 14), 19), 25) 26) we have f ) f ). ByLemma 2a) Eqs. 14), 19), 25) 26) we have f ) nt ) f ). Since is a series-parallel simple graph, has a mos 2n 3 edges hence T has a mos 2n 3 leaves. Therefore T has a mos 4n 7 nodes hence nt )<4n. Wehushave f ) 4n < f ). We are now ready o prove Theorem 3. Proof of Theorem 3. One may assume wihou loss of generaliy ha, for each dem verex v of a finie dem, a series-parallel graph has a pah Q v) going from he supply verex w o v such ha he sum of dems on Q v) does no exceed supw), hence v is conained in some supplied se C for. Oherwise,v canno be conained in any supplied se, hence one can regard ha v has an infinie dem. We do no delee such a verex v from, because he resuling graph may no be series-parallel.) One can examine in polynomial ime wheher here exiss such a pah Q v) for each dem verex v in ; his can be done for all dem verices v in in ime On 2 ) simply by applying a single-source shores pah algorihm o a graph similar o a line-graph of. One may assume ha has one or more dem verices of finie dems; oherwise, f ) = 0. Le V be he se consising of he supply verex w all dem verices of finie dems in. Le be a subgraph of induced by V, hen is conneced. Le m d = max{demv) v V }, le v V be a dem verex such ha demv ) = m d.then has a supplied se C conaining v, C is a supplied se also for. Wehushave f ) f C) dem v ) = m d. 33) We now choose an upper bound F on f ) such ha F f ) F. 2 Consider a simple greedy algorihm o find a supplied se for. The algorihm raverses by he breadh-firs search saring from w, includes raversed dem verices in a supplied se as much as possible so ha he se induces a conneced subgraph of he sum of dems in he se does no exceed supw). LeC A be a supplied se for found by he greedy algorihm. If eiher f C A ) = supw) or f C A ) = v V demv), henc A is he maximum supplied se for hence for. One may hus assume wihou loss of generaliy ha f C A )<supw) f C A )< v V demv). Then, here are dem verices in which were raversed bu could no be included in C A.Lev be he verex, among hese verices, ha was firs raversed. Then we have supw)< f C A ) + demv ). 34) 35) We choose F as follows: F = 2 max { f C A ), m d }. 36) Then, since f ) supw) demv ) m d, by Eqs. 35) 36) we have f )< f C A ) + demv ) f C A ) + m d 2 max { f C A ), m d } = F. Since C A is a supplied se for, wehave f ) f C A ) hence by Eqs. 33) 36) f ) max { f C A ), m d } = F 2.

15 T. Io e al. / Journal of Discree Algorihms ) We have hus verified Eq. 34). Le = εf 8n. Then by Lemma 2b) Eqs. 34) 37) we have f )< f ) + 4n εf 8n f ) + ε f ), hence we have Eq. 23). One can observe ha he algorihm akes ime O R + 2 ) n 3 ) F n = O, ε 2 because R + F = F / +1, hence by Eq. 37) we have F / 8n/ε. 37) 5. Conclusions In his paper, we sudied he approximabiliy of he maximum pariion problem. We firs showed ha he maximum pariion problem is MAXSNP-hard. We hen gave an FPTAS for series-parallel graphs having exacly one supply verex. I is easy o modify he FPTAS so ha i acually finds a supplied se for a series-parallel graph. The FPTAS for series-parallel graphs can be exended o ha for parial k-rees alhough i would become much more complicaed. In he ordinary knapsack problem, each iem is assigned a size value, one wishes o choose a subse of iems ha maximizes he sum of values of iems such ha heir oal size does no exceed he size of a bag [5,9]. Consider a slighly modified version of he maximum pariion problem on graphs in which each dem verex is assigned no only a dem bu also a value, one wishes o find a pariion which maximizes he sum of values of all dem verices in componens wih supply verices. This problem is indeed a generalizaion of he ordinary knapsack problem, can be solved for series-parallel graphs parial k-rees using echniques similar o hose for he maximum pariion problem if here is exacly one supply verex. Noe ha he sard approximaion mehods for he knapsack problem in [5,9] canno be applied o he modified maximum pariion problem. Acknowledgemens We hank MohammadTaghi Hajiaghayi for fruiful discussions. We also hank he referees for heir commens, one of which leads us o an improvemen of he ime complexiy of our FPTAS. Appendix A. How o compue g u, x) h i u, x), 1 i 4 In his secion, we explain how o compue g u, x) h i u, x), 1 i 4, for each inernal node u of T from he counerpars of he wo children of u in T. We firs consider a parallel connecion. [Parallel connecion] Le u = 1 2, le v s = v s u ) v = v u ). See Figs. 3c) A.1 A.3.) We have shown in Secion 3 ha one can compue h 1 u, x) in Eq. 8) by Eq. 22). We now show how o compue h 2 u, x) in Eq. 9). For j R w, every j, σ )-separaed pair C s, {v }) for u wih f C s, {v }, j, σ ) x can be obained by combining a j 1, σ )-separaed pair C s1, {v }) for 1 wih a j 2, σ )-separaed pair Fig. A.1. Combining a j 1, σ )-separaed pair C s1, {v }) for 1 a j 2, σ )-separaed pair C s2, {v }) for 2 o a j, σ )-separaed pair C s, {v }) for u = 1 2.

16 642 T. Io e al. / Journal of Discree Algorihms ) Fig. A.2. Combining a σ, σ )-separaed pair {v s }, {v }) for 1 a σ, σ )-separaed pair {v s}, {v }) for 2 o a σ, σ )-separaed pair {v s}, {v }) for u = 1 2. Fig. A.3. Forming an i-conneced se C for u = 1 2. C s2, {v }) for 2 such ha j 1, j 2 R w, j 1 + j 2 = j f C s, {v }, j, σ ) = f C s1, {v }, j 1, σ ) + f C s2, {v }, j 2, σ ), as illusraed in Fig. A.1. We can hus compue h 2 u, x) as follows: h 2 u, ) { x = max h2 y 1, y) + h 2 2, x y)} A.1) where he maximum above is aken over all real numbers y R such ha if h 2 1, y) + h 2 2, x y) 0henh 2 1, y) 0 h 2 2, x y) 0. One can compue h 3 u, x) in Eq. 10) similarly as h 2 u, x). We hen show how o compue h 4 u, x) in Eq. 11). Everyσ, σ )-separaed pair {v s}, {v }) for u wih f {v s}, {v }, σ, σ ) x can be obained by combining a σ, σ )-separaed pair {v s }, {v }) for 1 wih a σ, σ )-separaed pair {v s}, {v }) for 2 such ha f {v s}, {v }, σ, σ ) = f {v s }, {v }, σ, σ ) + f {v s }, {v }, σ, σ ), as illusraed in Fig. A.2. We can hus compue h 4 u, x) as follows: h 4 u, ) x = { max h 4 1, y) + h 4 2, x y) y R}. A.2) We nex show how o compue g u, x) in Eq. 7). There are he following wo cases a) b) where an i-conneced se C for u wih f C, i) x is formed from he counerpars of u s children, as illusraed in Figs. A.3a) b). We define wo funcions g a g b for he wo cases a) b), respecively. Case a): C is obained by combining an i 1 -conneced se C 1 for 1 wih an i 2-conneced se C 2 for 2 such ha f C, i) = f C 1, i 1 ) + f C 2, i 2 ) i 1 + i 2 = i. See Fig. A.3a).) We define g a u, x) for each real number x R, as follows: g a u, x) { = max g y 1, y) + g 2, x y)} A.3) where he maximum above is aken over all real numbers y R such ha if g 1, y) + g 2, x y) 0heng 1, y) 0 g 2, x y) 0.

17 T. Io e al. / Journal of Discree Algorihms ) Fig. A.4. Combining an i 1 -conneced se C 1 for 1 an i 2-conneced se C 2 for 2 o an i-conneced se C for u,where u = 1 2. Case b): C is obained by eiher combining a conneced se for 1 wih a separaed pair for 2 or combining a separaed pair for 1 wih a conneced se for 2. One may assume wihou loss of generaliy ha an i-conneced se C is obained by combining a j 1, k 1 )-separaed pair C s1, C 1 ) for 1 wih an i 2-conneced se C 2 for 2 such ha f C, i) = f C s1, C 1, j 1, k 1 ) + f C 2, i 2 ), where j 1 + k 1 + i 2 = i. See Fig. A.3b).) We define g b u, x) for each real number x R, as follows: g b u, x) = max y { h1 1, y) + g 2, x y)} A.4) where he maximum above is aken over all real numbers y R such ha if h 1 1, y) + g 2, x y) 0henh 1 1, y) 0 g 2, x y) 0. From g a g b above, one can compue g u, x) as follows: g u, x) = max { g a u, x), g b u, x)}. A.5) We nex consider a series connecion. [Series connecion] Le u = 1 2, le v be he verex of idenified by he series connecion, ha is, v = v 1 ) = v s 2 ). See Figs. 3b) A.4 A7.) We define sdv) as follows: { supv) if v is a supply verex, sdv) = demv) if v is a dem verex. Remember ha demw) = 0 for he supply verex w. We firs show how o compue g u, i) in Eq. 7). Fori R w,everyi-conneced se C for u wih f C, i) x can be obained by combining an i 1 -conneced se C 1 for 1 wih an i 2-conneced se C 2 for 2 such ha f C, i) = f C 1, i 1 ) + f C 2, i 2 )+demv) i 1 + i 2 + sdv) = i, as illusraed in Fig. A.4. Therefore g u, x) can be compued for each real number x R, as follows: g u, x) { = max g y 1,y 1, ) y 1 + g 2, ) } y 2 + sdv) A.6) 2 where he maximum above is aken over all real numbers y 1 y 2 such ha a) y 1, y 2 R; b) y 1 + y 2 + demv) = x; c) if g 1, y 1) + g 2, y 2) + sdv) 0, hen g 1, y 1) 0, g 2, y 2) 0 sdv)<0. We nex show how o compue h 1 u, x) in Eq. 8). There are he following wo cases a) b) where a j, k)-separaed pair C s, C ) for u wih f C s, C, j, k) x is formed from he counerpars of u s children, as illusraed in Figs. A.5a) b). We define wo funcions h a 1 hb 1 for he wo cases a) b), respecively. Case a): C s, C ) is obained by eiher combining a conneced se for 1 wih a separaed pair for 2 or combining a separaed pair for 1 wih a conneced se for 2. One may assume wihou loss of generaliy ha a j, k)-separaed pair C s, C ) is obained by combining an i 1 -conneced se C 1 for 1 wih a j 2, k)-separaed pair C s2, C ) for 2 such ha f C s, C, j, k) = f C 1, i 1 ) + f C s2, C, j 2, k) + demv) i 1 + j 2 + sdv) = i. See Fig. A.5a).)

18 644 T. Io e al. / Journal of Discree Algorihms ) Fig. A.5. Forming a j, k)-separaed pair C s, C ) for u,where j, k R w u = 1 2. We define h a 1 u, x) for each real number x R, as follows: h1 a u, ) { x = max g y 1,y 1, ) y 1 + h1 2, ) } y 2 + sdv) 2 where he maximum above is aken over all real numbers y 1 y 2 such ha A.7) a) y 1, y 2 R; b) y 1 + y 2 + demv) = x; c) if g 1, y 1) + h 1 2, y 2) + sdv) 0, hen g 1, y 1) 0, h 1 2, y 2) 0 sdv)<0. Case b): C s, C ) is obained by combining a j, σ )-separaed pair C s, {v}) for 1 wih a σ, k)-separaed pair {v}, C ) for 2 such ha f C s, C, j, k) = f C s, {v}, j, σ ) + f {v}, C, σ, k). See Fig. A.5b).) We define h b 1 u, x) for each real number x R as follows: h1 b u, ) { x = max h2 y 1,y 1, ) y 1 + h3 2, )} y 2 2 where he maximum above is aken over all real numbers y 1 y 2 such ha A.8) a) y 1, y 2 R; b) y 1 + y 2 = x; c) if h 2 1, y 1) + h 3 2, y 2) 0, hen h 2 1, y 1) 0 h 3 2, y 2) 0. If v is he supply verex w, henleh b 1 u, x) = for each real number x R; since u has a supplied se C ={v}, he dem) verices in C s C canno be supplied power; noe ha such a case is regarded as a σ, σ )-separaed pair for u. From h a 1 hb 1 above, one can compue h 1 u, x) as follows: h 1 u, x ) = max { h a 1 u, x ), h b 1 u, x )}. We hen show how o compue h 2 u, x) in Eq. 9). There are he following wo cases a) b) where a j, σ )- separaed pair C s, {v }) for u wih f C s, {v}, j, σ ) x is formed from he counerpars of u s children, as illusraed in Figs. A.6a) b). We define wo funcions h a 2 hb 2 for he wo cases a) b), respecively. Case a): C s, {v }) is obained by combining an i 1 -conneced se C 1 for 1 wih a j 2, σ )-separaed pair C s2, {v }) for 2 such ha f C s, {v }, j, σ ) = f C 1, i 1 ) + f C s2, {v }, j 2, σ ) + demv) i 1 + j 2 + sdv) = j. See Fig. A.6a).) A.9) We define h a 2 u, x) for each real number x R, as follows: h2 a u, ) { x = max g y 1,y 1, ) y 1 + h2 2, ) } y 2 + sdv) 2 A.10)

19 T. Io e al. / Journal of Discree Algorihms ) Fig. A.6. Forming a j, σ )-separaed pair C s, {v }) for u,where j R w u = 1 2. where he maximum above is aken over all real numbers y 1 y 2 such ha a) y 1, y 2 R; b) y 1 + y 2 + demv) = x; c) if g 1, y 1) + h 2 2, y 2) + sdv) 0, hen g 1, y 1) 0, h 2 2, y 2) 0 sdv)<0. Case b): C s, {v }) is obained by combining a j, σ )-separaed pair C s, {v}) for 1 wih a σ, σ )-separaed pair {v}, {v }) for 2 such ha f C s, {v }, j, σ ) = f C s, {v}, j, σ ) + f {v}, {v }, σ, σ ). See Fig. A.6b).) We define h b 2 u, x) for each real number x R as follows: h2 b u, ) { x = max h2 y 1,y 1, ) y 1 + h4 2, )} y 2 2 A.11) where he maximum above is aken over all real numbers y 1 y 2 such ha a) y 1, y 2 R; b) y 1 + y 2 = x; c) if h 2 1, y 1) + h 4 2, y 2) 0, hen h 2 1, y 1) 0 h 4 2, y 2) 0. If v is he supply verex w, henleh b 2 u, x) = for each real number x R. From h a 2 hb 2 above, one can compue h 2 u, x) as follows: h 2 u, x ) = max { h a 2 u, x ), h b 2 u, x )}. A.12) One can compue h 3 u, x) in Eq. 10) similarly as h 2 u, x). We finally show how o compue h 4 u, x) in Eq. 11). There are he following wo cases a) b) where a σ, σ )- separaed pair {v s }, {v }) for u wih f {v s}, {v }, σ, σ ) x is formed from he counerpars of u s children, as illusraed in Figs. A.7a) b). We define wo funcions h a 4 hb 4 for he wo cases a) b), respecively. Case a): {v s }, {v }) is obained by combining a σ, σ )-separaed pair {v s }, {v}) for 1 wih a σ, σ )-separaed pair {v}, {v }) for 2 such ha f {v s}, {v }, σ, σ ) = f {v s }, {v}, σ, σ ) + f {v}, {v }, σ, σ ). See Fig. A.7a).) We define h a 4 u, x) for each real number x R, as follows: h4 a u, ) { x = max h4 y 1,y 1, ) y 1 + h4 2, )} y 2 2 A.13) where he maximum above is aken over all real numbers y 1 y 2 such ha y 1 + y 2 = x.