11. APPROXIMATION ALGORITHMS
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1 Copng wth NP-completeness 11. APPROXIMATION ALGORITHMS load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Q. Suppose I need to solve an NP-hard problem. What should I do? A. Sacrfce one of three desred features.. Solve arbtrary nstances of the problem.. Solve problem to optmalty.. Solve problem n polynomal tme. ρ-approxmaton algorthm. Guaranteed to run n poly-tme. Guaranteed to solve arbtrary nstance of the problem Guaranteed to fnd soluton wthn rato ρ of true optmum. Lecture sldes by Kevn Wayne Copyrght 005 Pearson-Addson Wesley Challenge. Need to prove a soluton's value s close to optmum, wthout even knowng what optmum value s Last updated on /3/16 9:0 AM Load balancng 11. APPROXIMATION ALGORITHMS load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Input. m dentcal machnes; n jobs, job j has processng tme t j. Job j must run contguously on one machne. A machne can process at most one job at a tme. Def. Let J() be the subset of jobs assgned to machne. The load of machne s L = Σ j J() t j. Def. The makespan s the maxmum load on any machne L = max L. Load balancng. Assgn each job to a machne to mnmze makespan. machne 1 a dmachne 1 f machne b c Machne e g 0 L1 L tme
2 Load balancng on machnes s NP-hard Load balancng: lst schedulng Clam. Load balancng s hard even f only machnes. Pf. NUMBER-PARTITIONING P LOAD-BALANCE. NP-complete by Exercse 8.6 Lst-schedulng algorthm. Consder n jobs n some fxed order. Assgn job j to machne whose load s smallest so far. Lst-Schedulng(m, n, t 1,t,,t n ) { e a b f c g d for = 1 to m { L 0 } J() load on machne jobs assgned to machne length of job f for j = 1 to n { = argmn k L k J() J() {j} machne has smallest load assgn job j to machne L L + t j update load of machne machne 1 machne a dmachne 1 f b c Machne e g yes } } return J(1),, J(m) Implementaton. O(n log m) usng a prorty queue. 0 L tme 5 6 Load balancng: lst schedulng analyss Beleve t or not Theorem. [Graham 1966] Greedy algorthm s a -approxmaton. Frst worst-case analyss of an approxmaton algorthm. Need to compare resultng soluton wth optmal makespan L*. Lemma 1. The optmal makespan L* max j t j. Pf. Some machne must process the most tme-consumng job. Lemma. The optmal makespan Pf. The total processng tme s Σ j t j. L * m 1 j t j. One of m machnes must do at least a 1 / m fracton of total work. L * m 1 j t j. 7 8
3 Load balancng: lst schedulng analyss Load balancng: lst schedulng analyss Theorem. Greedy algorthm s a -approxmaton. Pf. Consder load L of bottleneck machne. Let j be last job scheduled on machne. When job j assgned to machne, had smallest load. Its load before assgnment s L t j L t j L k for all 1 k m. Theorem. Greedy algorthm s a -approxmaton. Pf. Consder load L of bottleneck machne. Let j be last job scheduled on machne. When job j assgned to machne, had smallest load. Its load before assgnment s L t j L t j L k for all 1 k m. Sum nequaltes over all k and dvde by m: blue jobs scheduled before j L t j 1 m k L k = 1 m k t k L * Lemma machne j Now L = (L t j )!# "# $ + t j % L* L* L *. Lemma 1 0 L - tj L = L tme 9 10 Load balancng: lst schedulng analyss Load balancng: lst schedulng analyss Q. Is our analyss tght? A. Essentally yes. Q. Is our analyss tght? A. Essentally yes. Ex: m machnes, m (m 1) jobs length 1 jobs, one job of length m. Ex: m machnes, m (m 1) jobs length 1 jobs, one job of length m. lst schedulng makespan = 19 optmal makespan = 10 m = 10 machne dle machne 3 dle machne dle machne 5 dle machne 6 dle machne 7 dle machne 8 dle machne 9 dle machne 10 dle m =
4 Load balancng: LPT rule Load balancng: LPT rule Longest processng tme (LPT). Sort n jobs n descendng order of Observaton. If at most m jobs, then lst-schedulng s optmal. processng tme, and then run lst schedulng algorthm. Pf. Each job put on ts own machne. Lemma 3. If there are more than m jobs, L* t m+1. LPT-Lst-Schedulng(m, n, t 1,t,,t n ) { Sort jobs so that t 1 t for = 1 to m { } L 0 J() t n load on machne jobs assgned to machne Pf. Consder frst m+1 jobs t 1,, t m+1. Snce the t 's are n descendng order, each takes at least t m+1 tme. There are m + 1 jobs and m machnes, so by pgeonhole prncple, at least one machne gets two jobs. for j = 1 to n { = argmn k L k machne has smallest load Theorem. LPT rule s a 3/-approxmaton algorthm. Pf. Same basc approach as for lst schedulng. } J() J() {j} L L + t j } return J(1),, J(m) assgn job j to machne update load of machne 13 L = (L t j )!# "# $ + t j % 3 L *. L* 1 L* Lemma 3 ( by observaton, can assume number of jobs > m ) 1 Load Balancng: LPT rule Q. Is our 3/ analyss tght? A. No. 11. APPROXIMATION ALGORITHMS Theorem. [Graham 1969] LPT rule s a /3-approxmaton. Pf. More sophstcated analyss of same algorthm. Q. Is Graham's /3 analyss tght? A. Essentally yes. load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Ex: m machnes, n = m + 1 jobs, jobs of length m, m + 1,, m 1 and one more job of length m. 15
5 Center selecton problem Center selecton problem Input. Set of n stes s 1,, s n and an nteger k > 0. Input. Set of n stes s 1,, s n and an nteger k > 0. Center selecton problem. Select set of k centers C so that maxmum dstance r(c) from a ste to nearest center s mnmzed. Center selecton problem. Select set of k centers C so that maxmum dstance r(c) from a ste to nearest center s mnmzed. r(c) k = centers Notaton. dst(x, y) = dstance between stes x and y. dst(s, C) = mn c C dst(s, c) = dstance from s to closest center. r(c) = max dst(s, C) = smallest coverng radus. Goal. Fnd set of centers C that mnmzes r(c), subject to C = k. center ste Dstance functon propertes. dst(x, x) = 0 [ dentty ] dst(x, y) = dst(y, x) [ symmetry ] dst(x, y) dst(x, z) + dst(z, y) [ trangle nequalty ] Center selecton example Greedy algorthm: a false start Ex: each ste s a pont n the plane, a center can be any pont n the plane, dst(x, y) = Eucldean dstance. Remark: search can be nfnte! Greedy algorthm. Put the frst center at the best possble locaton for a sngle center, and then keep addng centers so as to reduce the coverng radus each tme by as much as possble. Remark: arbtrarly bad! k = centers k = centers r(c) greedy center 1 center ste center ste 19 0
6 Center selecton: greedy algorthm Center selecton: analyss of greedy algorthm Repeatedly choose next center to be ste farthest from any exstng center. GREEDY-CENTER-SELECTION (k, n, s1, s,, sn) C. REPEAT k tmes Select a ste s wth maxmum dstance dst(s, C). C C s. RETURN C. ste farthest from any center Lemma. Let C* be an optmal set of centers. Then r(c) r(c*). Pf. [by contradcton] Assume r(c*) < ½ r(c). For each ste c C, consder ball of radus ½ r(c) around t. Exactly one c * n each ball; let c be the ste pared wth c *. Consder any ste s and ts closest center c * C*. dst(s, C) dst(s, c ) dst(s, c *) + dst(c *, c ) r(c*). Thus, r(c) r(c*). Δ-nequalty ½ r(c) r(c*) snce c * s closest center ½ r(c) ½ r(c) c Property. Upon termnaton, all centers n C are parwse at least r(c) apart. Pf. By constructon of algorthm. C* ste s c * 1 Center selecton Domnatng set reduces to center selecton Lemma. Let C* be an optmal set of centers. Then r(c) r (C*). Theorem. Greedy algorthm s a -approxmaton for center selecton problem. Remark. Greedy algorthm always places centers at stes, but s stll wthn a factor of of best soluton that s allowed to place centers anywhere. Queston. Is there hope of a 3/-approxmaton? /3? e.g., ponts n the plane Theorem. Unless P = NP, there no ρ-approxmaton for center selecton problem for any ρ <. Pf. We show how we could use a ( ε) approxmaton algorthm for CENTER-SELECTION selecton to solve DOMINATING-SET n poly-tme. Let G = (V, E), k be an nstance of DOMINATING-SET. Construct nstance G' of CENTER-SELECTION wth stes V and dstances dst(u, v) = 1 f (u, v) E dst(u, v) = f (u, v) E Note that G' satsfes the trangle nequalty. G has domnatng set of sze k ff there exsts k centers C* wth r(c*) = 1. Thus, f G has a domnatng set of sze k, a ( ε)-approxmaton algorthm for CENTER-SELECTION would fnd a soluton C* wth r(c*) = 1 snce t cannot use any edge of dstance. 3
7 Weghted vertex cover 11. APPROXIMATION ALGORITHMS load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Defnton. Gven a graph G = (V, E), a vertex cover s a set S V such that each edge n E has at least one end n S. Weghted vertex cover. Gven a graph G wth vertex weghts, fnd a vertex cover of mnmum weght. 9 9 weght = + + weght = 11 6 Prcng method Prcng method Prcng method. Each edge must be covered by some vertex. Edge e = (, j) pays prce p e 0 to use both vertex and j. Set prces and fnd vertex cover smultaneously. Farness. Edges ncdent to vertex should pay w n total. WEIGHTED-VERTEX-COVER (G, w) for each vertex : p e e=(, j) w S. FOREACH e E pe 0. 9 Farness lemma. For any vertex cover S and any far prces pe : e pe w(s). Pf. WHILE (there exsts an edge (, j) such that nether nor j s tght) Select such an edge e = (, j). Increase pe as much as possble untl or j tght. S set of all tght nodes. RETURN S. each edge e covered by at least one node n S sum farness nequaltes for each node n S 7 8
8 Prcng method example Prcng method: analyss Theorem. Prcng method s a -approxmaton for WEIGHTED-VERTEX-COVER. Pf. Algorthm termnates snce at least one new node becomes tght after each teraton of whle loop. Let S = set of all tght nodes upon termnaton of algorthm. S s a vertex cover: f some edge (, j) s uncovered, then nether nor j s tght. But then whle loop would not termnate. vertex weght prce of edge a-b Let S* be optmal vertex cover. We show w(s) w(s*). w(s) = w = S S p e e=(, j) V p e e=(, j) = p e w(s*). e E all nodes n S are tght S V, prces 0 each edge counted twce farness lemma 9 30 Weghted vertex cover 11. APPROXIMATION ALGORITHMS Gven a graph G = (V, E) wth vertex weghts w 0, fnd a mn weght subset of vertces S V such that every edge s ncdent to at least one vertex n S. load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem total weght = = 57 3
9 Weghted vertex cover: IP formulaton Weghted vertex cover: IP formulaton Gven a graph G = (V, E) wth vertex weghts w 0, fnd a mn weght subset of vertces S V such that every edge s ncdent to at least one vertex n S. Integer programmng formulaton. Model ncluson of each vertex usng a 0/1 varable x. x = " # $ Vertex covers n 1 1 correspondence wth 0/1 assgnments: S = { V : x = 1}. 0 f vertex s not n vertex cover 1 f vertex s n vertex cover Objectve functon: mnmze Σ w x. Weghted vertex cover. Integer programmng formulaton. Observaton. If x* s optmal soluton to (ILP), then S = { V : x * = 1} s a mn weght vertex cover. (ILP) mn w x V s. t. x + x j 1 (, j) E x {0,1} V Must take ether vertex or j (or both): x + x j Integer programmng Lnear programmng Gven ntegers a j, b, and c j, fnd ntegers x j that satsfy: Gven ntegers a j, b, and c j, fnd real numbers x j that satsfy: max c t x s. t. Ax b x ntegral n a j x j b 1 m j=1 x j 0 1 j n x j ntegral 1 j n (P) max c t x s. t. Ax b x 0 (P) max c j x j n j=1 n s. t. a j x j b 1 m j=1 x j 0 1 j n Observaton. Vertex cover formulaton proves that INTEGER-PROGRAMMING s an NP-hard search problem. Lnear. No x, xy, arccos(x), x(1 x), etc. Smplex algorthm. [Dantzg 197] Can solve LP n practce. Ellpsod algorthm. [Khachan 1979] Can solve LP n poly-tme
10 LP feasble regon Weghted vertex cover: LP relaxaton LP geometry n D. Lnear programmng relaxaton. x 1 = 0 (LP) mn w x V s. t. x + x j 1 (, j) E x 0 V Observaton. Optmal value of (LP) s optmal value of (ILP). Pf. LP has fewer constrants. Note. LP s not equvalent to vertex cover. ½ ½ x 1 + x = 6 x = 0 x 1 + x = 6 Q. How can solvng LP help us fnd a small vertex cover? A. Solve LP and round fractonal values. ½ Weghted vertex cover: LP roundng algorthm Weghted vertex cover napproxmablty Lemma. If x* s optmal soluton to (LP), then S = { V : x * ½} s a vertex cover whose weght s at most twce the mn possble weght. Theorem. [Dnur-Safra 00] If P NP, then no ρ-approxmaton for WEIGHTED-VERTEX-COVER for any ρ < (even f all weghts are 1). Pf. [S s a vertex cover] Consder an edge (, j) E. Snce x * + x j * 1, ether x * ½ or x j * ½ (, j) covered. Pf. [S has desred cost] Let S* be optmal vertex cover. Then w S* * w x 1 w S LP s a relaxaton x * ½ S Theorem. The roundng algorthm s a -approxmaton algorthm. Pf. Lemma + fact that LP can be solved n poly-tme. On the Hardness of Approxmatng Mnmum Vertex Cover Irt Dnur Samuel Safra May 6, 00 Abstract We prove the Mnmum Vertex Cover problem to be NP-hard to approxmate to wthn a factor of , extendng on prevous PCP and hardness of approxmaton technque. To that end, one needs to develop a new proof framework, and borrow and extend deas from several felds. Open research problem. Close the gap. 39 0
11 Generalzed load balancng 11. APPROXIMATION ALGORITHMS load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Input. Set of m machnes M; set of n jobs J. Job j J must run contguously on an authorzed machne n M j M. Job j J has processng tme t j. Each machne can process at most one job at a tme. Def. Let J() be the subset of jobs assgned to machne. The load of machne s L = Σ j J() t j. Def. The makespan s the maxmum load on any machne = max L. Generalzed load balancng. Assgn each job to an authorzed machne to mnmze makespan. Generalzed load balancng: nteger lnear program and relaxaton Generalzed load balancng: lower bounds ILP formulaton. x j = tme machne spends processng job j. (IP) mn L s. t. x j = t j for all j J L for all M j x j x j {0, t j } for all j J and M j Lemma 1. The optmal makespan L* max j t j. Pf. Some machne must process the most tme-consumng job. Lemma. Let L be optmal value to the LP. Then, optmal makespan L* L. Pf. LP has fewer constrants than IP formulaton. x j = 0 for all j J and M j LP relaxaton. (LP) mn L s. t. x j = t j for all j J L for all M j x j x j 0 for all j J and M j x j = 0 for all j J and M j 3
12 Generalzed load balancng: structure of LP soluton Generalzed load balancng: roundng Lemma 3. Let x be soluton to LP. Let G(x) be the graph wth an edge between machne and job j f x j > 0. Then G(x) s acyclc. Pf. (deferred) can transform x nto another LP soluton where G(x) s acyclc f LP solver doesn't return such an x Rounded soluton. Fnd LP soluton x where G(x) s a forest. Root forest G(x) at some arbtrary machne node r. If job j s a leaf node, assgn j to ts parent machne. If job j s not a leaf node, assgn j to any one of ts chldren. Lemma. Rounded soluton only assgns jobs to authorzed machnes. Pf. If job j s assgned to machne, then x j > 0. LP soluton can only assgn postve value to authorzed machnes. x j > 0 G(x) acyclc job G(x) cyclc job machne machne 5 6 Generalzed load balancng: analyss Generalzed load balancng: analyss Lemma 5. If job j s a leaf node and machne = parent(j), then x j = t j. Pf. Snce s a leaf, x j = 0 for all j parent(). LP constrant guarantees Σ x j = t j. Theorem. Rounded soluton s a -approxmaton. Pf. Let J() be the jobs assgned to machne. By LEMMA 6, the load L on machne has two components: Lemma 6. At most one non-leaf job s assgned to a machne. Pf. The only possble non-leaf job assgned to machne s parent(). leaf nodes: t j j J() j s a leaf = x j x j L L * Lemma 1 Lemma 5 j J() j s a leaf j J LP Lemma (LP s a relaxaton) optmal value of LP parent: t parent() L * job machne Thus, the overall load L L*. 7 8
13 Generalzed load balancng: flow formulaton Generalzed load balancng: structure of soluton Flow formulaton of LP. x j x j j = t j for all j J L for all M x j 0 for all j J and M j x j = 0 for all j J and M j Lemma 3. Let (x, L) be soluton to LP. Let G(x) be the graph wth an edge from machne to job j f x j > 0. We can fnd another soluton (x', L) such that G(x') s acyclc. Pf. Let C be a cycle n G(x). Augment flow along the cycle C. flow conservaton mantaned At least one edge from C s removed (and none are added). Repeat untl G(x') s acyclc. 3 3 augment flow along cycle C Observaton. Soluton to feasble flow problem wth value L are n 1-to-1 correspondence wth LP solutons of value L. 1 3 G(x) 5 1 G(x') Conclusons Runnng tme. The bottleneck operaton n our -approxmaton s solvng one LP wth m n + 1 varables. Remark. Can solve LP usng flow technques on a graph wth m + n + 1 nodes: gven L, fnd feasble flow f t exsts. Bnary search to fnd L*. Extensons: unrelated parallel machnes. [Lenstra-Shmoys-Tardos 1990] Job j takes t j tme f processed on machne. -approxmaton algorthm va LP roundng. If P NP, then no no ρ-approxmaton exsts for any ρ < 3/. 11. APPROXIMATION ALGORITHMS load balancng center selecton prcng method: vertex cover LP roundng: vertex cover generalzed load balancng knapsack problem Mathematcal Programmng 6 (1990) North-Holland APPROXIMATION ALGORITHMS FOR SCHEDULING UNRELATED PARALLEL MACHINES Jan Karel LENSTRA Endhoven Unversty of Technology, Endhoven, The Netherlands, and Centre for Mathematcs and Computer Scence, Amsterdam, The Netherlands Davd B. SHMOYS and l~va TARDOS Cornell Unversty, Ithaca, NY, USA 51
14 Polynomal-tme approxmaton scheme Knapsack problem PTAS. (1 + ε)-approxmaton algorthm for any constant ε > 0. Load balancng. [Hochbaum-Shmoys 1987] Eucldean TSP. [Arora, Mtchell 1996] Consequence. PTAS produces arbtrarly hgh qualty soluton, but trades off accuracy for tme. Knapsack problem. Gven n objects and a knapsack. Item has value v > 0 and weghs w > 0. Knapsack has weght lmt W. Goal: fll knapsack so as to maxmze total value. Ex: { 3, } has value 0. we assume w W for each Ths secton. PTAS for knapsack problem va roundng and scalng. tem value weght orgnal nstance (W = 11) 53 5 Knapsack s NP-complete Knapsack problem: dynamc programmng I KNAPSACK. Gven a set X, weghts w 0, values v 0, a weght lmt W, and a target value V, s there a subset S X such that: W w S v S V SUBSET-SUM. Gven a set X, values u 0, and an nteger U, s there a subset S X whose elements sum to exactly U? Theorem. SUBSET-SUM P KNAPSACK. Pf. Gven nstance (u1,, un, U) of SUBSET-SUM, create KNAPSACK nstance: v = w = u V = W = U u S u S U U Def. OPT(, w) = max value subset of tems 1,..., wth weght lmt w. Case 1. OPT does not select tem. OPT selects best of 1,, 1 usng up to weght lmt w. Case. OPT selects tem. New weght lmt = w w. OPT selects best of 1,, 1 usng up to weght lmt w w. # 0 f = 0 % OPT(, w) = $ OPT( 1, w) f w > w % & max{ OPT( 1, w), v + OPT( 1, w w )} otherwse Theorem. Computes the optmal value n O(n W) tme. Not polynomal n nput sze. Polynomal n nput sze f weghts are small ntegers
15 Knapsack problem: dynamc programmng II Knapsack problem: dynamc programmng II Def. OPT(, v) = mn weght of a knapsack for whch we can obtan a soluton of value v usng a subset of tems 1,...,. Note. Optmal value s the largest value v such that OPT(n, v) W. Case 1. OPT does not select tem. OPT selects best of 1,, 1 that acheves value v. Case. OPT selects tem. Consumes weght w, need to acheve value v v. OPT selects best of 1,, 1 that acheves value v v. Theorem. Dynamc programmng algorthm II computes the optmal value n O(n vmax) tme, where vmax s the maxmum of any value. Pf. The optmal value V* n vmax. There s one subproblem for each tem and for each value v V*. It takes O(1) tme per subproblem. Remark 1. Not polynomal n nput sze! Remark. Polynomal tme f values are small ntegers. OPT(, v) = 0 v 0 =0 v>0 mn {OPT( 1,v), w + OPT( 1,v v )} Knapsack problem: polynomal-tme approxmaton scheme Knapsack problem: polynomal-tme approxmaton scheme Intuton for approxmaton algorthm. Round all values up to le n smaller range. Run dynamc programmng algorthm II on rounded/scaled nstance. Return optmal tems n rounded nstance. tem value weght tem value weght Round up all values: 0 < ε 1 = precson parameter. v max = largest value n orgnal nstance. θ = scalng factor = ε v max / n. v = Observaton. Optmal solutons to problem wth v are equvalent to optmal solutons to problem wth v ˆ. Intuton. v close to v so optmal soluton usng v s nearly optmal; v ˆ small and ntegral so dynamc programmng algorthm II s fast. v, ˆv = v orgnal nstance (W = 11) rounded nstance (W = 11) 59 60
16 Knapsack problem: polynomal-tme approxmaton scheme Knapsack problem: polynomal-tme approxmaton scheme Theorem. If S s soluton found by roundng algorthm and S* s any other feasble soluton, then Pf. Let S* be any feasble soluton satsfyng weght constrant. S v S v (1 + ) always round up S v S v subset contanng only the tem of largest value Theorem. For any ε > 0, the roundng algorthm computes a feasble soluton whose value s wthn a (1 + ε) factor of the optmum n O(n 3 / ε) tme. Pf. We have already proved the accuracy bound. Dynamc program II runnng tme s O(n v ˆ max ), where S S v (v + ) solve rounded nstance optmally never round up by more than θ choosng S* = { max } v S v + 1 v ˆv max = v max = n = v S S + n S n v + 1 v θ = ε v max / n v + 1 v S thus v S v (1 + ) S v v max Σ S v 61 6
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