CS 372: Computational Geometry Lecture 10 Linear Programming in Fixed Dimension

Transcription

1 CS 372: Computational Geometry Lecture 10 Linear Programming in Fixed Dimension Antoine Vigneron King Abdullah University of Science and Technology November 7, 2012 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

2 1 Introduction 2 One dimensional case 3 Two-dimensional linear programming 4 Generalization to higher dimension Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

3 Outline This lecture is on linear programming. Special case: fixed dimension. It means d = O(1) variables. Even without this restriction, there are polynomial-time algorithms. For any fixed dimension, we give a simple linear-time algorithm. Reference: Textbook Chapter 4. Dave Mount s lecture notes, lectures Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

4 Example A factory can make two types of products: X and Y A product of type X requires 10 hours of manpower, 4l of oil, and 5m 3 of storage. A product of type Y requires 8 hours, 2l of oil and 10m 3 storage. A product X can be sold $200 and a product Y can be sold $250. You have 168 hours of manpower available, as well as 60l of oil and 150m 3 of storage. How many products of each type should you make so as to maximize their total price? Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

5 Formulation x and y denote the number of products of type X and Y, respectively. Maximize the price under the constraints f (x, y) = 200x + 250y x 0 y 0 10x + 8y 168 4x + 2y 60 5x + 10y 150. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

6 Geometric Interpretation 4x + 2y = 60 f (x, y) = constant Feasible region 10x + 8y = 168 5x + 10y = 150 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

7 Geometric Interpretation 4x + 2y = 60 optimal (x, y) 10x + 8y = 168 5x + 10y = 150 f (x, y) = constant Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

8 Solution From previous slide, at the optimum: x = 8 y = 11. Luckily these are integers. So it is the solution to our problem. If we add the constraint that all variables are integers, we are doing integer programming. We do not deal with it in CS 372. We consider only linear inequalities, no other constraint. Our example was a special case where the linear program has an integer solution, hence it is also a solution to the integer program. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

9 Problem Statement Maximize the objective function under the constraints f (x 1, x 2... x d ) = c 1 x 1 + c 2 x c d x d a 1,1 x a 1,d x d b 1 a 2,1 x a 2,d x d b 2... a n,1 x a n,d x d b n This is linear programming in dimension d. Equivalently, the goal could be to minimize f. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

10 Geometric Interpretation Each constraint represents a half-space in R d. The intersection of these half-spaces forms the feasible region. The feasible region is a convex polyhedron in R d. feasible region a constraint Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

11 Convex Polyhedra Definition (Convex polyhedron) A convex polyhedron is an intersection of half spaces in R d. May also be called convex polytope. A convex polyhedron is not necessarily bounded. Special case: A convex polygon is a a bounded, convex polytope in R 2. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

12 Convex Polyhedra in R 3 a tetrahedron a cube a cone Faces of a convex polyhedron in R 3 : Vertices, edges and facets. Example: A cube has 8 vertices, 12 edges and 6 facets. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

13 Geometric Interpretation Let c = (c 1, c 2,... c d ). We want to find a point v opt of the feasible region such that c is an outer normal at v opt, if there is one. c v opt Feasible region Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

14 Infeasible Linear Programs The feasible region may be empty. In this case there is no solution to the linear program. The program is said to be infeasible. We would like to know when it is the case. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

15 Unbounded Linear Programs The feasible region may be unbounded in the direction of c. In this case, we say that the linear program is unbounded. Then we want to find a ray ρ in the feasible region along which f takes arbitrarily large values. ρ Feasible region c Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

16 Degenerate Cases A linear program may have an infinite number of optimal solutions. f (x, y) = opt c In this case, we report only one solution. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

17 Background Many optimization problems in engineering, operations research, and economics are linear programs. A practical algorithm: The simplex algorithm. Exponential time in the worst case. There are polynomial time algorithms. Interior point methods. Integer programming is NP-hard. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

18 Background Computational geometry techniques give good algorithms in low dimension. Running time is O(n) when d is constant. But exponential in d: Time O(3 d2 n). This lecture: Seidel s algorithm. Simple, randomized. Expected running time O(d!n). This is O(n) when d = O(1) In practice, very good for low dimension. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

19 One dimensional case Maximize the objective function f (x) = cx under the constraints a 1 x b 1 a 2 x b 2.. a n x b n Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

20 Interpretation If a i > 0 then constraint i corresponds to the interval (, b ] i. a i If a i < 0 then constraint i corresponds to the interval [ ) bi,. a i The feasible region is an intersection of intervals. So the feasible region is an interval. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

21 Interpretation a 1 > 0 b 1 /a 1 a 2 < 0 b 2 /a 2 L feasible region R R Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

22 Algorithm Case 1: (i 1, i 2 ) such that a i1 < 0 < a i2. Compute Compute It takes O(n) time. b i R = min. a i >0 a i b i L = max. a i <0 a i If L > R then the program in infeasible. Otherwise If c > 0 then the solution is x = R. If c < 0 then the solution is x = L. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

23 Algorithm Case 2: a i > 0 for all i. Compute R = min b i a i. If c > 0 then the solution is x = R. If c < 0 then the program is unbounded and the ray (, R] is a solution. Case 3: a i < 0 for all i. Compute L = max b i a i. If c < 0 then the solution is x = L. If c > 0 then the program is unbounded and the ray [L, ) is a solution. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

24 Two-Dimensional Linear Programming First approach: Compute the feasible region. In O(n log n) time by divide and conquer+plane sweep. Other method: see later, lecture on duality. The feasible region is a convex polyhedron. Find an optimal point. Can be done in O(log n) time. Overall, it is O(n log n) time. This lecture: An expected O(n)-time algorithm. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

25 Preliminary We only consider bounded linear programs. We make sure that our linear program is bounded by enforcing two additional constraints m 1 and m 2. Objective function: f (x, y) = c1 x + c 2 y Let M be a large number. If c1 0 then m 1 is x M. If c 1 0 then m 1 is x M. If c 2 0 then m 2 is y M. If c2 0 then m 2 is y M. In practice, it often comes naturally. For instance, in our first example, it is easy to see that M = 30 is sufficient. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

26 New constraints y M m 2 c m 1 M x Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

27 Notation The i th constraint is: It defines a half-plane h i. a i,1 x + a i,2 y b i. l i h i Let l i denote the line that bounds h i. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

28 Algorithm A randomized incremental algorithm. We first compute a random permutation of the constraints (h 1, h 2... h n ). We denote H i = {m 1, m 2, h 1, h 2... h i }. We denote by v i a vertex of H i that maximizes the objective function. In other words, v i is a solution to the linear program where we only consider the first i constraints. v0 is simply the vertex of the boundary of m 1 m 2. Idea: knowing v i 1, we insert h i and find v i. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

29 Example m 1 v 0 m 2 c f (x, y) = constant Feasible region Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

30 Example v 1 m 1 m 2 c f (x, y) = f (v 1 ) Feasible region h 1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

31 Example v 2 = v 1 m 1 m 2 c Feasible region h 2 f (x, y) = f (v 2 ) h 1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

32 Example m 1 h 3 m 2 v 3 c h 2 Feasible region f (x, y) = f (v 3 ) h 1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

33 Example m 1 h 3 m 2 v 4 = v 3 c h 4 h 2 Feasible region f (x, y) = f (v 4 ) h 1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

34 Algorithm Randomized incremental algorithm. Before inserting h i, we only assume that we know v i 1. How can we find v i? Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

35 First Case First case: v i 1 h i. h i v i 1 c Feasible region Then v i = v i 1. Proof? Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

36 Second Case Second case: v i 1 h i. v i 1 c Feasible region for H i 1 l i h i Then v i 1 is not in the feasible region of H i. So v i v i 1. What do we know about v i? Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

37 Second Case There exists an optimal solution v i l i. c v i Feasible region for H i 1 l i h i Proof? How can we find v i? Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

38 Second Case Assume that a i,2 0, then the equation of l i is We replace y with b i a i,1 x a i,2 objective function. y = b i a i,1 x a i,2. in all the constraints of H i and in the We obtain a one dimensional linear program, with variable x. If it is feasible, its solution gives us the x-coordinate of v i. We obtain the y-coordinate using the equation of l i. If this linear program is infeasible, then the original 2D linear program is infeasible too and we are done. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

39 Analysis First case is done in O(1) time: Just check whether v i 1 h i. Second case in O(i) time: One dimensional linear program with i + 2 constraints. So the algorithm runs in O(n 2 ) time. Is there a worst case example where it runs in Ω(n 2 ) time? What is the expected running time? We need to know how often the second case happens. We define the random variable X i. Xi = 0 in first case (v i = v i 1 ). Xi = 1 in second case (v i v i 1 ). Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

40 Analysis When X i = 0 we spend O(1) time at i-th step. When X i = 1 we spend O(i) time. So the expected running time E[T (n)] is ( n ) E[T (n)] = O 1 + i.e[x i ]. Note: E[X i ] is simply the probability that X i = 1 in our case. i=1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

41 Analysis We denote C i = H i. In other words, C i is the feasible region at step i. v i is adjacent to two edges of C i, these edges correspond to two constraints h and h. C i h v i h If v i v i 1, then h i = h or h i = h. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

42 Analysis What is the probability that h i = h or h i = h? We use backwards analysis. We assume that H i is fixed. So h i is chosen uniformly at random in H i \ {m 1, m 2 }. So the probability that h i = h or h i = h is at most 2/i. It could be 1/i or 0 if vi is defined by m 1 and/or m 2. So E[X i ] 2/i So ( n ) E[T (n)] = O 1 + i. 2 = O(n). i i=1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

43 Generalization to higher dimension First attempt: Each constraint is a half-space. Can we compute their intersection and get the feasible region? In R 3 it can be done in O(n log n) time. (Not covered in CS 372.) In higher dimension, the feasible region has Ω(n d 2 ) vertices in the worst case. So computing the feasible region requires Ω(n d 2 ) time. Here, we will give a O(n) expected time algorithm for finding one optimal point in the feasible region, when d = O(1). Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

44 Preliminary A hyperplane in R d has equation where α 1 x 1 + α 2 x α d x d = β d. (α 1, α 2,..., α d ) R d \ {0} d. In general position, d hyperplanes intersect at one point. Each constraint h i is a half-space, bounded by a hyperplane h i. We assume general position in the sense that: Any d hyperplanes h i1,..., h id intersect at exactly one point. The intersection of any d + 1 such hyperplanes is empty. No such hyperplane is orthogonal to c. Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

45 Algorithm We generalize the 2D algorithm. We first find d constraints m 1, m 2,... m d that make the linear program bounded: If ci 0 then m i is x i M. If ci < 0 then m i is x i M. We pick a random permutation (h 1, h 2,... h n ) of H. Then H i is {m 1, m 2,... m d, h 1, h 2,... h i }. We maintain v i, the solution to the linear program with constraints H i and objective function f. v 0 is the vertex of d m i. i=1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

46 Algorithm We assume d = O(1). Inserting h i is done in the same way as in R 2 : If v i 1 h i then v i 1 = v i. Otherwise, v i h i. We find v i by solving a linear program with i + d constraints in R d 1. If this linear program is infeasible, then the original linear program is infeasible too, so we are done. It can be done in expected O(i) time. (By induction.) Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

47 Analysis What is the probability that v i v i 1? By our general position assumption, v i belongs to exactly d hyperplanes that bound constraints in H i. The probability that vi v i 1 is the probability that one of these d constraints was inserted last. By backwards analysis, it is d/i. So the expected running time of our algorithm is ( n ) E[T (n)] = O 1 + i. d = O(dn) = O(n). i i=1 Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48

48 Conclusion This algorithm can be made to handle unbounded linear programs and degenerate cases. A careful implementation of this algorithm runs in O(d!n) time. So it is only useful in low dimension. It can be generalized to other types of problems. See textbook: Smallest enclosing disk. Sometimes we can linearize a problem and use a linear programming algorithm. We will see such cases in homework or tutorial. (Example: Finding the enclosing annulus with minimum area.) Antoine Vigneron (KAUST) CS 372 Lecture 9 November 7, / 48