# ! Solve problem to optimality. ! Solve problem in poly-time. ! Solve arbitrary instances of the problem. #-approximation algorithm.

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1 Approximation Algorithms 11 Approximation Algorithms Q Suppose I need to solve an NP-hard problem What should I do? A Theory says you're unlikely to find a poly-time algorithm Must sacrifice one of three desired features Solve problem to optimality Solve problem in poly-time Solve arbitrary instances of the problem #-approximation algorithm Guaranteed to run in poly-time Guaranteed to solve arbitrary instance of the problem Guaranteed to find solution within ratio # of true optimum Challenge Need to prove a solution's value is close to optimum, without even knowing what optimum value is Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne Load Balancing 111 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 Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne 4

4 Load Balancing: LPT Rule Load Balancing: LPT Rule Observation If at most m jobs, then list-scheduling is optimal Pf Each job put on its own machine Q Is our 3/ analysis tight? A No Lemma 3 If there are more than m jobs, L* ) t m+1 Pf Consider first m+1 jobs t 1,, t m+1 Since the t i 's are in descending order, each takes at least t m+1 time There are m+1 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 ) t j { # 3 L * # L* # 1 L* Theorem [Graham, 1969] LPT rule is a 4/3-approximation Pf More sophisticated analysis of same algorithm Q Is Graham's 4/3 analysis tight? A Essentially yes Ex: m machines, n = m+1 jobs, jobs of length m+1, m+,, m-1 and one job of length m Lemma 3 ( by observation, can assume number of jobs > m ) Center Selection Problem 11 Center Selection Input Set of n sites s 1,, s n Center selection problem Select k centers C so that maximum distance from a site to nearest center is minimized k = 4 r(c) center site Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne 16

5 Center Selection Problem Center Selection Example Input Set of n sites s 1,, s n Center selection problem Select k centers C so that maximum distance from a site to nearest center is minimized 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 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 r(c) 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) center site Greedy Algorithm: A False Start Center Selection: Greedy Algorithm 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 Greedy algorithm Repeatedly choose the next center to be the site farthest from any existing center Remark: arbitrarily bad greedy center 1 Greedy-Center-Selection(k, n, s 1,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 k = centers center site Observation Upon termination all centers in C are pairwise at least r(c) apart Pf By construction of algorithm 19 0

6 Center Selection: Analysis of Greedy Algorithm Center Selection 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*) *-inequality \$ r(c*) since c i * is closest center Theorem Let C* be an optimal set of centers Then r(c) \$ r(c*) Theorem Greedy algorithm is a -approximation for center selection problem 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 r(c) r(c) Question Is there hope of a 3/-approximation? 4/3? C* sites r(c) s c i c i * Theorem Unless P = NP, there no #-approximation for center-selection problem for any # < 1 Weighted Vertex Cover 114 The Pricing Method: Vertex Cover Weighted vertex cover Given a graph G with vertex weights, find a vertex cover of minimum weight weight = weight = 9 Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne 4

7 Weighted Vertex Cover Primal-Dual Algorithm Pricing method Each edge must be covered by some vertex Edge e pays price p e ) 0 to use edge Primal-dual algorithm Set prices and find vertex cover simultaneously Fairness Edges incident to vertex i should pay \$ w i in total for each vertex i : " p e w e= ( i, j) i 4 9 Weighted-Vertex-Cover-Approx(G, w) { foreach e in E p e = 0 p = w e e = ( i, j) while (- edge i-j such that neither i nor j are tight) select such an edge e increase p e without violating fairness i 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 S % set of all tight nodes return S each edge e covered by at least one node in S sum fairness inequalities for each node in S 5 6 Primal Dual Algorithm: Analysis Theorem Primal-dual algorithm is a -approximation Pf Algorithm terminates since at least one new node becomes tight after each iteration of while loop 116 LP Rounding: Vertex Cover Let S = set of all tight nodes upon termination of algorithm S is a vertex cover: if some edge i-j is uncovered, then either i or j is not tight But then while loop would not terminate Let S* be optimal vertex cover We show w(s) \$ w(s*) w( S) = " wi = " " pe " " pe = " pe i# S i# S e= ( i, j) i# V e= ( i, j) e# E w( S*) all nodes in S are tight S + V, prices ) 0 each edge counted twice fairness lemma 7 Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne

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 10 A 6F 9 Weighted vertex cover Integer programming formulation (ILP) min # w i x i i " V s t x i + x j \$ 1 (i, j) " E x i " {0,1 i " V 16 B 7G C H 9 Observation If x* is optimal solution to (ILP), then S = {i " V : x* i = 1 is a min weight vertex cover 3 D I 33 7 E 10 J 3 total weight = Integer Programming Linear Programming INTEGER-PROGRAMMING Given integers a ij and b i, find integers x j that satisfy: n " a ij x j # b i 1\$ i \$ m j=1 x j # 0 1\$ j \$ n x j integral 1\$ j \$ n Observation Vertex cover formulation proves that integer programming is NP-hard search problem (even if all coefficients are 0/1 and at most two nonzeros per inequality) 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 Linear No x, xy, arccos(x), x(1-x), etc (P) max " c j x j n j=1 n s t " a ij x j = b i 1# i # m j=1 x j \$ 0 1# j # n Simplex algorithm [Dantzig 1947] Can solve LP in practice Ellipsoid algorithm [Khachian 1979] Can solve LP in poly-time 31 3

9 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 \$ 1 (i, j) " E x i \$ 0 i " V Observation Optimal value of (LP) is \$ optimal value of (ILP) 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 ) 1, either x* i ) or x* j ) ( (i, j) covered Pf [S has desired cost] Let S* be optimal vertex cover Then Note LP is not equivalent to vertex cover w i i " S* * # \$ # w i x i \$ 1 w # i i " S i " S LP is a relaxation x* i ) Q How can solving LP help us find a small vertex cover? A Solve LP and round fractional values Weighted Vertex Cover Good news -approximation algorithm is basis for most practical heuristics can solve LP with min cut ( faster primal-dual schema ( linear time (see book) PTAS for planar graphs Solvable in poly-time on bipartite graphs using network flow 117 Load Balancing Reloaded Bad news [Dinur-Safra, 001] If P NP, then no #-approximation for # < 13607, even with unit weights 10 / Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne

10 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 Generalized Load Balancing: Lower Bounds Generalized Load Balancing: Structure of LP Solution Lemma 1 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 time-consuming job Lemma 3 Let x be an extreme point 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 (we prove contrapositive) Let x be a feasible solution to the LP such that G(x) has a cycle Define % x ij ±" y ij = & ' x ij (i, j) # C (i, j) \$ C % x ij m " z ij = & ' x ij (i, j) # C (i, j) \$ C The variables y and z are feasible solutions to the LP Observe x = y + z Thus, x is not an extreme point j +0 i

11 Generalized Load Balancing: Rounding Generalized Load Balancing: Analysis Rounded solution Find extreme point LP solution x 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 4 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 non-leaf job is assigned to a machine Pf The only possible non-leaf job assigned to machine i is parent(i) job machine 41 4 Generalized Load Balancing: Analysis Conclusions 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: Running time The bottleneck operation in our -approximation is solving one LP with mn + 1 variables Remark Possible to solve LP using max flow techniques (see text) leaf nodes parent(i) 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 t parent(i) " L * j " J LP Lemma 1 (LP is a relaxation) optimal value of LP Extensions: unrelated parallel machines [Lenstra-Shmoys-Tardos 1990] Job j takes t ij time if processed on machine i -approximation algorithm via LP rounding No 3/-approximation algorithm unless P = NP Thus, the overall load L i \$ L* 43 44

12 Polynomial Time Approximation Scheme 118 Knapsack Problem PTAS (1 + 0)-approximation algorithm for any constant 0 > 0 Load balancing [Hochbaum-Shmoys 1987] Euclidean TSP [Arora 1996] FPTAS PTAS that is polynomial in input size and 1/0 Consequence PTAS produces arbitrarily high quality solution, but trades off accuracy for time This section FPTAS for knapsack problem via rounding and scaling Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne 46 Knapsack Problem Knapsack is NP-Complete 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, 4 has value 40 Item 1 Value 1 6 Weight 1 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 SUBSET-SUM: 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 SUBSET-SUM \$ P KNAPSACK Pf Given instance (u 1,, u n, U) of SUBSET-SUM, create KNAPSACK instance: v i = w i = u i # u i \$ U i"s V = W = U # u i % U i"s 47 48

13 Knapsack Problem: Dynamic Programming 1 Knapsack Problem: Dynamic Programming II Def OPT(i, w) = max value subset of items 1,, i with weight limit w Case 1: OPT does not select item i OPT selects best of 1,, i 1 using up to weight limit w Case : OPT selects item i new weight limit = w w i OPT selects best of 1,, i 1 using up to weight limit w w i # 0 if i = 0 % OPT(i, w) = \$ OPT(i "1, w) if w i > w % & max{ OPT(i "1, w), v i + OPT(i "1, 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 1,, i that yields value exactly v Case 1: OPT does not select item i OPT selects best of 1,, i-1 that achieves exactly value v Case : OPT selects item i consumes weight w i, new value needed = v v i OPT selects best of 1,, i-1 that achieves exactly value v \$ 0 if v = 0 & " if i = 0, v > 0 OPT(i, v) = % & OPT(i #1, v) if v i > v '& min{ OPT(i #1, v), w i + OPT(i #1, 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 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 0 = precision parameter 1 = scaling factor = 0 v max / n # v i = v i % \$ " & ", v ˆ # i = v i % \$ " & Item Value Weight 1 134, ,34 3 1,810, ,17, ,343,199 7 W = 11 original instance Item Value Weight W = 11 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 / 0) Dynamic program II running time is O(n v ˆ max ), where v ˆ max = # v max \$ " % & = # n % \$ ' & v v v ˆ 51 5

14 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 (1+") % 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 1 \$ # v i + n% i" S S \$ n DP alg can take v max \$ (1+&) # v i i" S n 1 = 0 v max, v max \$ i"s v i 53 Algorithm Design by Éva Tardos and Jon Kleinberg Copyright 005 Addison Wesley Slides by Kevin Wayne Center Selection: Hardness of Approximation Knapsack: State of the Art Theorem Unless P = NP, there is no ( - 0) approximation algorithm for k-center problem for any 0 > 0 Pf We show how we could use a ( - 0) approximation algorithm for k- center to solve DOMINATING-SET in poly-time Let G = (V, E), k be an instance of DOMINATING-SET Construct instance G' of k-center with sites V and distances d(u, v) = if (u, v) " E d(u, v) = 1 if (u, v) 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*) = 1 Thus, if G has a dominating set of size k, a ( - 0)-approximation algorithm on G' must find a solution C* with r(c*) = 1 since it cannot use any edge of distance This lecture "Rounding and scaling" method finds a solution within a (1 + 0) factor of optimum for any 0 > 0 Takes O(n 3 / 0) time and space Ibarra-Kim (1975), Lawler (1979) Faster FPTAS: O(n log (1 / 0) + 1 / 0 4 ) time Idea: group items by value into "large" and "small" classes run dynamic programming algorithm only on large items insert small items according to ratio v i / w i clever analysis 55 56

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