Approximation and Randomized Algorithms (ARA) Lecture 2, September 1, 2010

Transcription

1 Approximation and Randomized Algorithms (ARA) Lecture 2, September 1, 2010

2 Last time Algorithm Revision Algorithms for the stable matching problem Five illustrative algorithm problems Computatibility

3 Today Short tour of NP concepts We start the approximation algorithms topic Load balancing algorithms Center selection problem

4 Independent Set Input. Graph. Goal. Find maximum cardinality independent set. subset of nodes such that no two joined by an edge

5 Competitive Facility Location Input. Graph with weight on each node. Game. Two competing players alternate in selecting nodes. Not allowed to select a node if any of its neighbors have been selected. Goal. Select a maximum weight subset of nodes Second player can guarantee 20, but not 25. 5

6 Decision Problems Decision problem. X is a set of strings. Instance: string s. Algorithm A solves problem X: A(s) = yes iff s X. Polynomial time. Algorithm A runs in poly-time length of s if for every string s, A(s) terminates in at most p( s ) "steps", where p( ) is some polynomial. 6

7 Definition of P P. Decision problems for which there is a polytime algorithm. Problem Description Algorithm Yes No MULTIPLE Is x a multiple of y? Grade school division 51, 17 51, 16 RELPRIME Are x and y relatively prime? Euclid (300 BCE) 34, 39 34, 51 PRIMES Is x prime? AKS (2002) EDIT- DISTANCE Is the edit distance between x and y less than 5? Dynamic programming niether neither acgggt ttttta LSOLVE Is there a vector x that satisfies Ax = b? Gauss-Edmonds elimination , ,

8 NP Certification algorithm intuition. Certifier views things from "managerial" viewpoint. Certifier doesn't determine whether s X on its own; rather, it checks a proposed proof t that s X. Def. Algorithm C(s, t) is a certifier for problem X if for every string s, s X iff there exists a string t such that C(s, t) = yes. "certificate" or "witness" NP. Decision problems for which there exists a poly-time certifier. C(s, t) is a poly-time algorithm and t p( s ) for some polynomial p( ). Remark. NP stands for nondeterministic polynomial-time. 8

9 P, NP P. Decision problems for which there is a poly-time algorithm. NP. Decision problems for which there is a polytime certifier. Claim. P NP. Pf. Consider any problem X in P. By definition, there exists a poly-time algorithm A(s) that solves X. Certificate: t = ε, certifier C(s, t) = A(s). 9

10 The Main Question: P Versus NP Does P = NP? [Cook 1971, Edmonds, Levin, Yablonski, Gödel] Is the decision problem as easy as the certification problem? Clay $1 million prize. EXP P NP EXP P = NP If P NP If P = NP If yes: Efficient algorithms for 3-COLOR, TSP, FACTOR, SAT, If no: No efficient algorithms possible for 3-COLOR, TSP, SAT, would break RSA cryptography (and potentially collapse economy) Consensus opinion on P = NP? Probably no. 10

11 Proof of P NP is checked NOW! August 6, 2010 Vinay Deolalikar (HP research labs): claims he has solved P NP problem Claim = currently a 102-page proof, still expanding Still to be completely checked Wikipedia (link to the 102-page manuscript) Blog of Richard Lipton (Georgia Tech):

12 NP-Complete NP-complete. A problem Y in NP with the property that for every problem X in NP, X p Y. Theorem. Suppose Y is an NP-complete problem. Then Y is solvable in poly-time iff P = NP. Pf. If P = NP then Y can be solved in poly-time since Y is in NP. Pf. Suppose Y can be solved in poly-time. Let X be any problem in NP. Since X p Y, we can solve X in poly-time. This implies NP P. We already know P NP. Thus P = NP. Fundamental question. Do there exist "natural" NP-complete problems? 12

13 Circuit Satisfiability CIRCUIT-SAT. Given a combinational circuit built out of AND, OR, and NOT gates, is there a way to set the circuit inputs so that the output is 1? The "First" NP-Complete Problem: CIRCUIT-SAT is NP-complete. [Cook 1971, Levin 1973] output yes: ??? hard-coded inputs inputs 13

14 Establishing NP-Completeness Remark. Once we establish first "natural" NP-complete problem, others fall like dominoes. Recipe to establish NP-completeness of problem Y. Step 1. Show that Y is in NP. Step 2. Choose an NP-complete problem X. Step 3. Prove that X p Y. Justification. If X is an NP-complete problem, and Y is a problem in NP with the property that X P Y then Y is NP-complete. Pf. Let W be any problem in NP. Then W P X P Y. By transitivity, W P Y. Hence Y is NP-complete. by definition of NP-complete by assumption 14

15 NP-Completeness Observation. All problems below are NP-complete and polynomial reduce to one another! CIRCUIT-SAT by definition of NP-completeness 3-SAT INDEPENDENT SET DIR-HAM-CYCLE GRAPH 3-COLOR SUBSET-SUM VERTEX COVER HAM-CYCLE PLANAR 3-COLOR SCHEDULING SET COVER TSP 15

16 Extent and Impact of NP- Completeness Extent of NP-completeness. [Papadimitriou 1995] Prime intellectual export of CS to other disciplines. 6,000 citations per year (title, abstract, keywords). more than "compiler", "operating system", "database" Broad applicability and classification power. "Captures vast domains of computational, scientific, mathematical endeavors, and seems to roughly delimit what mathematicians and scientists had been aspiring to compute feasibly. 16

17 More Hard Computational Problems Aerospace engineering: optimal mesh partitioning for finite elements. Biology: protein folding. Chemical engineering: heat exchanger network synthesis. Civil engineering: equilibrium of urban traffic flow. Economics: computation of arbitrage in financial markets with friction. Electrical engineering: VLSI layout. Environmental engineering: optimal placement of contaminant sensors. Financial engineering: find minimum risk portfolio of given return. Game theory: find Nash equilibrium that maximizes social welfare. Genomics: phylogeny reconstruction. Mechanical engineering: structure of turbulence in sheared flows. Medicine: reconstructing 3-D shape from biplane angiocardiogram. Operations research: optimal resource allocation. Physics: partition function of 3-D Ising model in statistical mechanics. Politics: Shapley-Shubik voting power. Pop culture: Minesweeper consistency. Statistics: optimal experimental design. 17

18 A Note on Terminology Knuth's original suggestions. Hard. Tough. Herculean. Formidable. Arduous. but Hercules known for strength not time "creative research workers are as full of ideas for new terminology as they are empty of enthusiasm for adopting it." -Don Knuth Some English word write-ins. Impractical. Bad. Heavy. Tricky. Intricate. Prodigious. Difficult. Intractable. Costly. Obdurate. Obstinate. Exorbitant. Interminable. 18

19 A Note on Terminology Hard-boiled. [Ken Steiglitz] In honor of Cook. Hard-ass. [Al Meyer] Hard as satisfiability. Sisyphean. [Bob Floyd] Problem of Sisyphus was time-consuming. but Sisyphus never finished his task Ulyssean. [Don Knuth] Ulysses was known for his persistence. and finished! 19

20 A Note on Terminology: Consensus NP-complete. A problem in NP such that every problem in NP polynomial reduces to it. NP-hard. [Bell Labs, Steve Cook, Ron Rivest, Sartaj Sahni] A decision problem such that every problem in NP reduces to it. not necessarily in NP NP-hard search problem. A problem such that every problem in NP reduces to it. not necessarily a yes/no problem 20

21 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. ρ-approximation algorithm. Guaranteed to run in poly-time. Guaranteed to solve arbitrary instance of the problem Guaranteed to find solution within ratio ρ of true optimum. Challenge. Need to prove a solution's value is close to optimum, without even knowing what optimum value is! 21

22 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 = Σ 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. 22

23 Load Balancing: List Scheduling List-scheduling algorithm. Consider n jobs in some fixed order. Assign job j to machine whose load is smallest so far. List-Scheduling(m, n, t 1,t 2,,t n ) { for i = 1 to m { L i 0 load on machine i J(i) φ jobs assigned to machine i } } for j = 1 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 Implementation. O(n log n) using a priority queue. 23

24 Load Balancing: List Scheduling A B C D E F G H I J Machine 1 Machine 2 Machine 3 0 Time 24

25 Load Balancing: List Scheduling B C D E F G H I J A Machine 1 Machine 2 Machine 3 0 Time 25

26 Load Balancing: List Scheduling C D E F G H I J A Machine 1 B Machine 2 Machine 3 0 Time 26

27 Load Balancing: List Scheduling D E F G H I J A Machine 1 B Machine 2 C Machine 3 0 Time 27

28 Load Balancing: List Scheduling E F G H I J A Machine 1 B D Machine 2 C Machine 3 0 Time 28

29 Load Balancing: List Scheduling F G H I J A E Machine 1 B D Machine 2 C Machine 3 0 Time 29

30 Load Balancing: List Scheduling G H I J A E Machine 1 B D Machine 2 C F Machine 3 0 Time 30

31 Load Balancing: List Scheduling H I J A E Machine 1 B D Machine 2 G C F Machine 3 0 Time 31

32 Load Balancing: List Scheduling I J A E HMachine 1 B D Machine 2 G C F Machine 3 0 Time 32

33 Load Balancing: List Scheduling J A E HMachine 1 I B D Machine 2 G G C F Machine 3 0 Time 33

34 Load Balancing: List Scheduling A E HMachine 1 I B D Machine 2 G C F Machine 3 J 0 Time 34

35 Load Balancing: List Scheduling D E H Machine 1 C B A Machine G 2 I F Machine 3 J 0 Optimal Schedule A E HMachine 1 I B D Machine 2 G C F Machine 3 J 0 List schedule 35

36 Load Balancing: List Scheduling Analysis Theorem. [Graham, 1966] Greedy algorithm is a 2- approximation. First worst-case analysis of an approximation algorithm. Need to compare resulting solution with optimal makespan L*. Lemma 1. The optimal makespan L* max j t j. Pf. Some machine must process the most time-consuming job. Lemma 2. The optimal makespan Pf. The total processing time is Σ j t j. L * m 1 j t j. One of m machines must do at least a 1/m fraction of total work. 36

37 Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-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 1 k m. blue jobs scheduled before j machine i j 0 L i - t j L = L i 37

38 Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-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 1 k m. Sum inequalities over all k and divide by m: L i t j 1 m k L k = 1 m k t k Lemma 1 L * Now L i = (L i t j ) + L* t j L* 2L *. Lemma 2 38

39 Load Balancing: List Scheduling Analysis Q. Is our analysis tight? A. Essentially yes. Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m m = 10 machine 2 idle machine 3 idle machine 4 idle machine 5 idle machine 6 idle machine 7 idle machine 8 idle machine 9 idle machine 10 idle list scheduling makespan = 19 39

40 Load Balancing: List Scheduling Analysis Q. Is our analysis tight? A. Essentially yes. Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m m = 10 optimal makespan = 10 40

41 Load Balancing: LPT Rule Longest processing time (LPT). Sort n jobs in descending order of processing time, and then run list scheduling algorithm. LPT-List-Scheduling(m, n, t 1,t 2,,t n ) { Sort jobs so that t 1 t 2 t n for i = 1 to m { L i 0 J(i) φ } load on machine i jobs assigned to machine i } for j = 1 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 41

42 Load Balancing: LPT Rule Observation. If at most m jobs, then list-scheduling is optimal. Pf. Each job put on its own machine. Lemma 3. If there are more than m jobs, L* 2 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 3/2 approximation algorithm. Pf. Same basic approach as for list scheduling. L i = (L i t j ) + L* t j 2 3 L *. 1 2 L* Lemma 3 ( by observation, can assume number of jobs > m ) 42

43 Load Balancing: LPT Rule Q. Is our 3/2 analysis tight? A. No. 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 = 2m+1 jobs, 2 jobs of length m+1, m+2,, 2m-1 and one job of length m. 43

44 Center Selection Problem 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 44

45 Center Selection Problem 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. 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. Distance function properties. dist(x, x) = 0 dist(x, y) = dist(y, x) dist(x, y) dist(x, z) + dist(z, y) (identity) (symmetry) (triangle inequality) 45

46 Center Selection Example 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! r(c) center site 46

47 Greedy Algorithm: A False Start 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! greedy center 1 k = 2 centers center site 47

48 Center Selection: Greedy Algorithm Greedy algorithm. Repeatedly choose the next center to be the site farthest from any existing center. Greedy-Center-Selection(k, n, s 1,s 2,,s n ) { } C = φ 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 Observation. Upon termination all centers in C are pairwise at least r(c) apart. Pf. By construction of algorithm. 48

49 Center Selection: Analysis of Greedy Algorithm Theorem. Let C* be an optimal set of centers. Then r(c) 2r(C*). Pf. (by contradiction) Assume r(c*) < ½ r(c). For each site c i in C, consider ball of radius ½ 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 ) 2r(C*). Thus r(c) 2r(C*). -inequality r(c*) since c i * is closest center ½ r(c) ½ r(c) C* sites ½ r(c) s c i c i * 49

50 Center Selection Theorem. Let C* be an optimal set of centers. Then r(c) 2r(C*). Theorem. Greedy algorithm is a 2-approximation for center selection problem. Remark. Greedy algorithm always places centers at sites, but is still within a factor of 2 of best solution that is allowed to place centers anywhere. e.g., points in the plane Question. Is there hope of a 3/2-approximation? 4/3? Theorem. Unless P = NP, there no ρ-approximation for center-selection problem for any ρ < 2. 50