! Solve problem to optimality. ! Solve problem in polytime. ! Solve arbitrary instances of the problem. !approximation algorithm.


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1 Approximation Algorithms Chapter Approximation Algorithms Q Suppose I need to solve an NPhard problem What should I do? A Theory says you're unlikely to find a polytime algorithm Must sacrifice one of three desired features Solve problem to optimality Solve problem in polytime Solve arbitrary instances of the problem approximation algorithm Guaranteed to run in polytime Guaranteed to solve arbitrary instance of the problem Guaranteed to find solution within ratio of true optimum Slides by Kevin Wayne 005 PearsonAddison Wesley All rights reserved Challenge Need to prove a solution's value is close to optimum, without even knowing what optimum value is Load Balancing Load Balancing Input m identical machines; n jobs, job j has processing time t j Job j must run contiguously on one machine A machine can process at most one job at a time Def Let J(i) be the subset of jobs assigned to machine i The load of machine i is L i = " j # J(i) t j Def The makespan is the maximum load on any machine L = max i L i Load balancing Assign each job to a machine to minimize makespan
2 Load Balancing: List Scheduling Load Balancing: List Scheduling Analysis Listscheduling algorithm Consider n jobs in some fixed order Assign job j to machine whose load is smallest so far Theorem [Graham, 966] Greedy algorithm is a approximation First worstcase analysis of an approximation algorithm Need to compare resulting solution with optimal makespan L* ListScheduling(m, n, t,t,,t n ) { for i = to m { L i % 0 load on machine i J(i) % & jobs assigned to machine i for j = to n { i = argmin k L k J(i) % J(i) ' {j L i % L i + t j machine i has smallest load assign job j to machine i update load of machine i Lemma The optimal makespan L* $ max j t j Pf Some machine must process the most timeconsuming job Lemma The optimal makespan Pf The total processing time is " j t j L * " m # j t j One of m machines must do at least a /m fraction of total work Implementation O(n log n) using a priority queue 5 6 Load Balancing: List Scheduling Analysis Load Balancing: List Scheduling Analysis Theorem Greedy algorithm is a approximation Pf Consider load L i of bottleneck machine i Let j be last job scheduled on machine i When job j assigned to machine i, i had smallest load Its load before assignment is L i  t j ( L i  t j ) L k for all ) k ) m Theorem Greedy algorithm is a approximation Pf Consider load L i of bottleneck machine i Let j be last job scheduled on machine i When job j assigned to machine i, i had smallest load Its load before assignment is L i  t j ( L i  t j ) L k for all ) k ) m Sum inequalities over all k and divide by m: blue jobs scheduled before j Lemma L i " t j # m $ k L k = m $ k t k # L * machine i j Now L i = (L i " t j ) 3 + t j # L * { # L* # L* Lemma 0 L i  t j L = L i 7 8
3 Load Balancing: List Scheduling Analysis Load Balancing: List Scheduling Analysis Q Is our analysis tight? A Essentially yes Ex: m machines, m(m) jobs length jobs, one job of length m Q Is our analysis tight? A Essentially yes Ex: m machines, m(m) jobs length jobs, one job of length m m = 0 machine idle machine 3 idle machine idle machine 5 idle machine 6 idle machine 7 idle machine 8 idle machine 9 idle machine 0 idle m = 0 list scheduling makespan = 9 optimal makespan = Load Balancing: LPT Rule Load Balancing: LPT Rule Longest processing time (LPT) Sort n jobs in descending order of processing time, and then run list scheduling algorithm Observation If at most m jobs, then listscheduling is optimal Pf Each job put on its own machine LPTListScheduling(m, n, t,t,,t n ) { Sort jobs so that t t t n for i = to m { L i % 0 J(i) % & for j = to n { i = argmin k L k J(i) % J(i) ' {j L i % L i + t j load on machine i jobs assigned to machine i machine i has smallest load assign job j to machine i update load of machine i Lemma 3 If there are more than m jobs, L* $ t m+ Pf Consider first m+ jobs t,, t m+ Since the t i 's are in descending order, each takes at least t m+ time There are m+ jobs and m machines, so by pigeonhole principle, at least one machine gets two jobs Theorem LPT rule is a 3/ approximation algorithm Pf Same basic approach as for list scheduling L i = (L i " t j ) 3 + t j { # 3 L * # L* # L* Lemma 3 ( by observation, can assume number of jobs > m )
4 Load Balancing: LPT Rule Q Is our 3/ analysis tight? A No Center Selection Theorem [Graham, 969] LPT rule is a /3approximation Pf More sophisticated analysis of same algorithm Q Is Graham's /3 analysis tight? A Essentially yes Ex: m machines, n = m+ jobs, jobs of length m+, m+,, m and one job of length m 3 Center Selection Problem Center Selection Problem Input Set of n sites s,, s n Center selection problem Select k centers C so that maximum distance from a site to nearest center is minimized Input Set of n sites s,, s n Center selection problem Select k centers C so that maximum distance from a site to nearest center is minimized r(c) k = Notation dist(x, y) = distance between x and y dist(s i, C) = min c # C dist(s i, c) = distance from s i to closest center r(c) = max i dist(s i, C) = smallest covering radius Goal Find set of centers C that minimizes r(c), subject to C = k center site Distance function properties dist(x, x) = 0 (identity) dist(x, y) = dist(y, x) (symmetry) dist(x, y) ) dist(x, z) + dist(z, y) (triangle inequality) 5 6
5 Center Selection Example Greedy Algorithm: A False Start Ex: each site is a point in the plane, a center can be any point in the plane, dist(x, y) = Euclidean distance Remark: search can be infinite Greedy algorithm Put the first center at the best possible location for a single center, and then keep adding centers so as to reduce the covering radius each time by as much as possible Remark: arbitrarily bad r(c) greedy center k = centers center site center site 7 8 Center Selection: Greedy Algorithm Center Selection: Analysis of Greedy Algorithm Greedy algorithm Repeatedly choose the next center to be the site farthest from any existing center GreedyCenterSelection(k, n, s,s,,s n ) { C = & repeat k times { Select a site s i with maximum dist(s i, C) Add s i to C site farthest from any center return C Theorem Let C* be an optimal set of centers Then r(c) ) r(c*) Pf (by contradiction) Assume r(c*) < r(c) For each site c i in C, consider ball of radius r(c) around it Exactly one c i * in each ball; let c i be the site paired with c i * Consider any site s and its closest center c i * in C* dist(s, C) ) dist(s, c i ) ) dist(s, c i *) + dist(c i *, c i ) ) r(c*) Thus r(c) ) r(c*) r(c) *inequality ) r(c*) since c i * is closest center r(c) Observation Upon termination all centers in C are pairwise at least r(c) apart Pf By construction of algorithm C* sites r(c) s c i c i * 9 0
6 Center Selection Theorem Let C* be an optimal set of centers Then r(c) ) r(c*) Theorem Greedy algorithm is a approximation for center selection problem The Pricing Method: Vertex Cover Remark Greedy algorithm always places centers at sites, but is still within a factor of of best solution that is allowed to place centers anywhere eg, points in the plane Question Is there hope of a 3/approximation? /3? Theorem Unless P = NP, there no approximation for centerselection problem for any < Weighted Vertex Cover Weighted Vertex Cover Weighted vertex cover Given a graph G with vertex weights, find a vertex cover of minimum weight Pricing method Each edge must be covered by some vertex i Edge e pays price p e $ 0 to use vertex i Fairness Edges incident to vertex i should pay ) w i in total for each vertex i : " p e w e= ( i, j) i 9 9 weight = + + weight = 9 9 Claim For any vertex cover S and any fair prices p e : + e p e ) w(s) Proof " pe " " pe " wi = w( S) e# E i# S e= ( i, j) i# S each edge e covered by at least one node in S sum fairness inequalities for each node in S 3
7 Pricing Method Pricing Method Pricing method Set prices and find vertex cover simultaneously WeightedVertexCoverApprox(G, w) { foreach e in E p e = 0 p = w e e = ( i, j) i while ( edge ij such that neither i nor j are tight) select such an edge e increase p e without violating fairness S % set of all tight nodes return S price of edge ab vertex weight Figure Pricing Method: Analysis Theorem Pricing method is a approximation Pf Algorithm terminates since at least one new node becomes tight after each iteration of while loop 6 LP Rounding: Vertex Cover Let S = set of all tight nodes upon termination of algorithm S is a vertex cover: if some edge ij is uncovered, then neither i nor j is tight But then while loop would not terminate Let S* be optimal vertex cover We show w(s) ) w(s*) w(s) = # w i = # # p e $ i" S i" S e=(i, j) # # p e = # p e $ w(s*) i"v e=(i, j) e" E all nodes in S are tight S, V, prices $ 0 each edge counted twice fairness lemma 7
8 Weighted Vertex Cover Weighted Vertex Cover: IP Formulation Weighted vertex cover Given an undirected graph G = (V, E) with vertex weights w i $ 0, find a minimum weight subset of nodes S such that every edge is incident to at least one vertex in S Weighted vertex cover Given an undirected graph G = (V, E) with vertex weights w i $ 0, find a minimum weight subset of nodes S such that every edge is incident to at least one vertex in S 0 A 6F 9 Integer programming formulation Model inclusion of each vertex i using a 0/ variable x i 6 B 7G 0 x i = " # $ 0 if vertex i is not in vertex cover if vertex i is in vertex cover 6 3 3C D H I 9 33 Vertex covers in  correspondence with 0/ assignments: S = {i # V : x i = Objective function: maximize " i w i x i 7 E total weight = 55 0 J 3 Must take either i or j: x i + x j $ 9 30 Weighted Vertex Cover: IP Formulation Integer Programming Weighted vertex cover Integer programming formulation INTEGERPROGRAMMING Given integers a ij and b i, find integers x j that satisfy: (ILP) min # w i x i i " V s t x i + x j $ (i, j) " E x i " {0, i " V max c t x s t Ax " b x integral n " a ij x j # b i $ i $ m j= x j # 0 $ j $ n x j integral $ j $ n Observation If x* is optimal solution to (ILP), then S = {i # V : x* i = is a min weight vertex cover Observation Vertex cover formulation proves that integer programming is NPhard search problem even if all coefficients are 0/ and at most two variables per inequality 3 3
9 Linear Programming LP Feasible Region Linear programming Max/min linear objective function subject to linear inequalities Input: integers c j, b i, a ij Output: real numbers x j (P) max c t x s t Ax " b x " 0 (P) max " c j x j n j= n s t " a ij x j # b i $ i $ m j= x j # 0 $ j $ n LP geometry in D x = 0 Linear No x, xy, arccos(x), x(x), etc Simplex algorithm [Dantzig 97] Can solve LP in practice Ellipsoid algorithm [Khachian 979] Can solve LP in polytime x = 0 x + x = 6 x + x = Weighted Vertex Cover: LP Relaxation Weighted Vertex Cover Weighted vertex cover Linear programming formulation (LP) min # w i x i i " V s t x i + x j $ (i, j) " E x i $ 0 i " V Theorem If x* is optimal solution to (LP), then S = {i # V : x* i $ is a vertex cover whose weight is at most twice the min possible weight Pf [S is a vertex cover] Consider an edge (i, j) # E Since x* i + x* j $, either x* i $ or x* j $ ( (i, j) covered Observation Optimal value of (LP) is ) optimal value of (ILP) Pf LP has fewer constraints Note LP is not equivalent to vertex cover Pf [S has desired cost] Let S* be optimal vertex cover Then w i i " S* * # $ # w i x i $ w # i i " S LP is a relaxation x* i $ i " S Q How can solving LP help us find a small vertex cover? A Solve LP and round fractional values 35 36
10 Weighted Vertex Cover Theorem approximation algorithm for weighted vertex cover * 7 Load Balancing Reloaded Theorem [DinurSafra 00] If P NP, then no approximation for < 3607, even with unit weights 0 /5  Open research problem Close the gap 37 Generalized Load Balancing Generalized Load Balancing: Integer Linear Program and Relaxation Input Set of m machines M; set of n jobs J Job j must run contiguously on an authorized machine in M j, M Job j has processing time t j Each machine can process at most one job at a time Def Let J(i) be the subset of jobs assigned to machine i The load of machine i is L i = " j # J(i) t j Def The makespan is the maximum load on any machine = max i L i Generalized load balancing Assign each job to an authorized machine to minimize makespan ILP formulation x ij = time machine i spends processing job j LP relaxation (IP) min (LP) min L s t " x i j = t j for all j # J i " $ L for all i # M x i j j x i j # {0, t j for all j # J and i # M j x i j = 0 for all j # J and i % M j L s t " x i j = t j for all j # J i " $ L for all i # M x i j j x i j % 0 for all j # J and i # M j x i j = 0 for all j # J and i & M j 39 0
11 Generalized Load Balancing: Lower Bounds Generalized Load Balancing: Structure of LP Solution Lemma Let L be the optimal value to the LP Then, the optimal makespan L* $ L Pf LP has fewer constraints than IP formulation Lemma The optimal makespan L* $ max j t j Pf Some machine must process the most timeconsuming job Lemma 3 Let x be solution to LP Let G(x) be the graph with an edge from machine i to job j if x ij > 0 Then G(x) is acyclic Pf (deferred) can transform x into another LP solution where G(x) is acyclic if LP solver doesn't return such an x x ij > 0 G(x) acyclic job machine G(x) cyclic Generalized Load Balancing: Rounding Generalized Load Balancing: Analysis Rounded solution Find LP solution x where G(x) is a forest Root forest G(x) at some arbitrary machine node r If job j is a leaf node, assign j to its parent machine i If job j is not a leaf node, assign j to one of its children Lemma Rounded solution only assigns jobs to authorized machines Pf If job j is assigned to machine i, then x ij > 0 LP solution can only assign positive value to authorized machines Lemma 5 If job j is a leaf node and machine i = parent(j), then x ij = t j Pf Since i is a leaf, x ij = 0 for all j parent(i) LP constraint guarantees " i x ij = t j Lemma 6 At most one nonleaf job is assigned to a machine Pf The only possible nonleaf job assigned to machine i is parent(i) job machine job machine 3
12 Generalized Load Balancing: Analysis Generalized Load Balancing: Flow Formulation Theorem Rounded solution is a approximation Pf Let J(i) be the jobs assigned to machine i By Lemma 6, the load L i on machine i has two components: leaf nodes t j j " J(i) j is a leaf Lemma 5 # = # x ij $ # x ij $ L $ L * j " J(i) j is a leaf Lemma j " J LP Lemma (LP is a relaxation) optimal value of LP Flow formulation of LP x i j i x i j j " = t j for all j # J " $ L for all i # M x i j % 0 for all j # J and i # M j x i j = 0 for all j # J and i & M j 0 parent(i) t parent(i) " L * Thus, the overall load L i ) L* Observation Solution to feasible flow problem with value L are in onetoone correspondence with LP solutions of value L 5 6 Generalized Load Balancing: Structure of Solution Conclusions Lemma 3 Let (x, L) be solution to LP Let G(x) be the graph with an edge from machine i to job j if x ij > 0 We can find another solution (x', L) such that G(x') is acyclic Pf Let C be a cycle in G(x) Augment flow along the cycle C At least one edge from C is removed (and none are added) Repeat until G(x') is acyclic 3 3 flow conservation maintained 3 3 Running time The bottleneck operation in our approximation is solving one LP with mn + variables Remark Can solve LP using flow techniques on a graph with m+n+ nodes: given L, find feasible flow if it exists Binary search to find L* Extensions: unrelated parallel machines [LenstraShmoysTardos 990] Job j takes t ij time if processed on machine i approximation algorithm via LP rounding No 3/approximation algorithm unless P = NP G(x) augment along C G(x') 7 8
13 Polynomial Time Approximation Scheme 8 Knapsack Problem PTAS ( + )approximation algorithm for any constant > 0 Load balancing [HochbaumShmoys 987] Euclidean TSP [Arora 996] Consequence PTAS produces arbitrarily high quality solution, but trades off accuracy for time This section PTAS for knapsack problem via rounding and scaling 50 Knapsack Problem Knapsack is NPComplete Knapsack problem Given n objects and a "knapsack" Item i has value v i > 0 and weighs w i > 0 Knapsack can carry weight up to W we'll assume w i ) W Goal: fill knapsack so as to maximize total value Ex: { 3, has value 0 Item Value 6 Weight W = KNAPSACK: Given a finite set X, nonnegative weights w i, nonnegative values v i, a weight limit W, and a target value V, is there a subset S, X such that: # $ W w i i"s v i i"s # % V SUBSETSUM: Given a finite set X, nonnegative values u i, and an integer U, is there a subset S, X whose elements sum to exactly U? Claim SUBSETSUM ) P KNAPSACK Pf Given instance (u,, u n, U) of SUBSETSUM, create KNAPSACK instance: v i = w i = u i # u i $ U i"s V = W = U # u i % U i"s 5 5
14 Knapsack Problem: Dynamic Programming Knapsack Problem: Dynamic Programming II Def OPT(i, w) = max value subset of items,, i with weight limit w Case : OPT does not select item i OPT selects best of,, i using up to weight limit w Case : OPT selects item i new weight limit = w w i OPT selects best of,, i using up to weight limit w w i # 0 if i = 0 % OPT(i, w) = $ OPT(i ", w) if w i > w % & max{ OPT(i ", w), v i + OPT(i ", w " w i ) otherwise Running time O(n W) W = weight limit Not polynomial in input size Def OPT(i, v) = min weight subset of items,, i that yields value exactly v Case : OPT does not select item i OPT selects best of,, i that achieves exactly value v Case : OPT selects item i consumes weight w i, new value needed = v v i OPT selects best of,, i that achieves exactly value v $ 0 if v = 0 & " if i = 0, v > 0 OPT(i, v) = % & OPT(i #, v) if v i > v '& min{ OPT(i #, v), w i + OPT(i #, v # v i ) otherwise V* ) n v max Running time O(n V*) = O(n v max ) V* = optimal value = maximum v such that OPT(n, v) ) W Not polynomial in input size 53 5 Knapsack: FPTAS Intuition for approximation algorithm Round all values up to lie in smaller range Run dynamic programming algorithm on rounded instance Return optimal items in rounded instance Knapsack: FPTAS Knapsack FPTAS Round up all values: v max = largest value in original instance = precision parameter = scaling factor = v max / n # v i = v i % $ " & ", v ˆ # i = v i % $ " & Item Value Weight 3, 656,3 3,80,03 5,7, ,33,99 7 W = original instance Item Value Weight W = rounded instance Observation Optimal solution to problems with or are equivalent Intuition close to v so optimal solution using is nearly optimal; small and integral so dynamic programming algorithm is fast v ˆ v Running time O(n 3 / ) Dynamic program II running time is O(n v ˆ max ), where v ˆ max = # v max $ " % & = # n % $ ' & v v v ˆ 55 56
15 Knapsack: FPTAS Knapsack FPTAS Round up all values: # v i = v i % $ " & " Extra Slides Theorem If S is solution found by our algorithm and S* is any other feasible solution then (+") % v i # % v i i $ S i $ S* Pf Let S* be any feasible solution satisfying weight constraint v i i " S* # $ # v i i " S* $ # v i i " S $ # (v i + %) i " S always round up solve rounded instance optimally never round up by more than $ # v i + n% i" S S ) n DP alg can take v max $ (+&) # v i i" S n = v max, v max ) " i#s v i 57 Load Balancing on Machines Center Selection: Hardness of Approximation Claim Load balancing is hard even if only machines Pf NUMBERPARTITIONING ) P LOADBALANCE Theorem Unless P = NP, there is no approximation algorithm for metric kcenter problem for any < machine machine e a NPcomplete by Exercise 86 b a d Machine f b c Machine e g f length of job f c g d yes Pf We show how we could use a (  ) approximation algorithm for k center to solve DOMINATINGSET in polytime Let G = (V, E), k be an instance of DOMINATINGSET Construct instance G' of kcenter with sites V and distances d(u, v) = if (u, v) # E d(u, v) = if (u, v) 3 E Note that G' satisfies the triangle inequality Claim: G has dominating set of size k iff there exists k centers C* with r(c*) = see Exercise 89 Thus, if G has a dominating set of size k, a (  )approximation algorithm on G' must find a solution C* with r(c*) = since it cannot use any edge of distance 0 Time L 59 60
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