Linear Programming- Manufacturing with

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2 Casting process 2 / 36

3 The of Casting top facet ordinary facet Every ordinary facet f has a corresponding facet in the mold, which we denote by ˆf. castable object top facet 3 / 36

4 The of Casting 4 / 36 Question: Given an object P, can it be removed from its mold by a single translation? In other words, we want to decide whether a direction d exists such that P can be translated to infinity in direction d without intersecting the interior of the mold during the translation. Observation: Every ordinary facet f must move away from, or slide along, its corresponding facet ˆf of the mold.

5 The of Casting Angle between two vectors in 3d: The angle of the vectors is the smaller of the two angles measured in the plane determined by them. 5 / 36

6 Lemma 4.1 P can be removed from its mold in direction d if and only if d makes an angle of at least π/2 with the outward normal of all ordinary facets of P. f p d η( ˆf ) P 6 / 36

7 1-1 corresponding between direction and points z z = 1 x y 7 / 36

8 By Lemma 4.1 η = ( η x, η y, η z ): outward normal of an ordinary facet d = (d x, d y, 1): removal direction ( η, d) π/2 η. d 0 η x d x + η y d y + η z 0 This inequality describe a half-plane in the plane z = 1. 8 / 36

9 An object P is castable, for a fixed top facet the of half-planes is non-empty. If the is solvable in O(A) time, then the castability can be solved in O(An) time. We will solve the intersect5ion in O(n) expected time. Theorem 4.2: In O(n 2 ) expected time and using O(n) storage it can be decided whether a polyhydron with n facets is castable. 9 / 36

10 Problem: Note: Given a set H = {h 1, h 2,..., h n } of half-planes, find the of them. Given a set of n linear constraints in two variables, is there a point in the plane, find all points that satisfy all the constraints. For casting, we do not need to find the of half-planes. Here we consider a more general. 10 / 36

11 Observations: Since a half-plane is convex, the of half-planes is convex. Since the is convex, every half-plane bounding line can contribute at most one edge. A few examples of s of half-planes: (i) (ii) (iii) (iv) (v) 11 / 36

12 We give a rather straightforward divide-and-conquer algorithm to compute the of a set of n half-planes. It is based on a routine INTERSECTCON- VEXREGIONS to compute the convex polygonal regions. We Anfirst straightforward give the overalldivide-and-conquer algorithm. algorithm: the of n half-planes Algorithm INTERSECTHALFPLANES(H) Input. A set H of n half-planes in the plane. Output. The convex polygonal region C := h H h. 1. if card(h)=1 2. then C the unique half-plane h H 3. else Split H into sets H 1 and H 2 of size n/2 and n/2. 4. C 1 INTERSECTHALFPLANES(H 1 ) 5. C 2 INTERSECTHALFPLANES(H 2 ) 6. C INTERSECTCONVEXREGIONS(C 1,C 2 ) What remains is to describe the procedure INTERSECTCONVEXREGIONS. But Time wait didn t complexity: we see this before, in Chapter 2? Indeed, Corollary 2.7 states that we can { compute the polygons in O(nlogn + k logn) time, where O(1) n is the total number of vertices if n = in 1 T (n) = the two polygons. We must be a bit careful 2T in(n/2) applying + O(n this result log n) to our if, n > 1 because the regions we have can be unbounded, or degenerate to a segment or a point. Hence, T (n) the regions O(n log are 2 not n). necessarily polygons. But it is not difficult to modify the algorithm from Chapter 2 so that it still works. 12 / 36

13 We give a rather straightforward divide-and-conquer algorithm to compute the of a set of n half-planes. It is based on a routine INTERSECTCON- VEXREGIONS to compute the convex polygonal regions. We Anfirst straightforward give the overalldivide-and-conquer algorithm. algorithm: the of n half-planes Algorithm INTERSECTHALFPLANES(H) Input. A set H of n half-planes in the plane. Output. The convex polygonal region C := h H h. 1. if card(h)=1 2. then C the unique half-plane h H 3. else Split H into sets H 1 and H 2 of size n/2 and n/2. 4. C 1 INTERSECTHALFPLANES(H 1 ) 5. C 2 INTERSECTHALFPLANES(H 2 ) 6. C INTERSECTCONVEXREGIONS(C 1,C 2 ) What remains is to describe the procedure INTERSECTCONVEXREGIONS. But Time wait didn t complexity: we see this before, in Chapter 2? Indeed, Corollary 2.7 states that we can { compute the polygons in O(nlogn + k logn) time, where O(1) n is the total number of vertices if n = in 1 T (n) = the two polygons. We must be a bit careful 2T in(n/2) applying + O(n this result log n) to our if, n > 1 because the regions we have can be unbounded, or degenerate to a segment or a point. Hence, T (n) the regions O(n log are 2 not n). necessarily polygons. But it is not difficult to modify the algorithm from Chapter 2 so that it still works. 12 / 36

14 the of n half-planes Question: Can we compute the convex polygons in o(n log n) time? Answer: Yes, we can. v e 1 e2 13 / 36

15 the of n half-planes Question: Can we compute the convex polygons in o(n log n) time? Answer: Yes, we can. v e 1 e2 13 / 36

16 convex polygons h 3 h 2 h 1 right boundary h 4 h 5 left boundary L left (C) = h 3,h 4,h 5 L right (C) = h 2,h 1 14 / 36

17 convex polygons We use a plane sweep algorithm: Note: At most 4 line segments intersect the sweep line. left edge C2 left edge C1 C 2 C 1 right edge C1 = nil right edge C2 15 / 36

18 convex polygons At each event point some new edge e appears on the boundary. To handle the edge e: several cases: e belongs to C1 or to C2, and whether it is on the left or the right boundary. We consider the case when e is on the left boundary of C1. p: the upper endpoint of e. Three case (based on C): (1) the edge with p as upper endpoint, (2) the edge with e left_edge_c2 as upper endpoint, and (3) the edge with e right_edge_c2 as upper endpoint. 16 / 36

19 convex polygons It performs the following actions. (i) p (ii) p right edge C2 e right edge C2 e 17 / 36

20 convex polygons p e left edge C2 18 / 36

21 convex polygons Theorem 4.3 The convex polygonal regions in the plane can be computed in O(n) time. T (n) = T (n) O(n log n). Corollary 4.4 { O(1) if n = 1 2T (n/2) + O(n) if n > 1 The common of a set of n half-planes in the plane can be computed in O(n log n) time and linear storage. 19 / 36

22 convex polygons Theorem 4.3 The convex polygonal regions in the plane can be computed in O(n) time. T (n) = T (n) O(n log n). Corollary 4.4 { O(1) if n = 1 2T (n/2) + O(n) if n > 1 The common of a set of n half-planes in the plane can be computed in O(n log n) time and linear storage. 19 / 36

23 convex polygons Theorem 4.3 The convex polygonal regions in the plane can be computed in O(n) time. T (n) = T (n) O(n log n). Corollary 4.4 { O(1) if n = 1 2T (n/2) + O(n) if n > 1 The common of a set of n half-planes in the plane can be computed in O(n log n) time and linear storage. 19 / 36

24 convex polygons Theorem 4.3 The convex polygonal regions in the plane can be computed in O(n) time. T (n) = T (n) O(n log n). Corollary 4.4 { O(1) if n = 1 2T (n/2) + O(n) if n > 1 The common of a set of n half-planes in the plane can be computed in O(n log n) time and linear storage. 19 / 36

25 Incremental Previous Section: We computed the of n half-planes in O(n log n) time. The has Ω(n log n) lower bound. So, we cannot solve the casting faster in this way. Note that, for casting we only need to know if the is empty or not. In this section: An "expected" linear time algorithm presented to solve the using linear programming. 20 / 36

26 Incremental optimization : Maximize c 1 x 1 + c 2 x c d x d (objective function) Subject to 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 (constraints) 21 / 36

27 Incremental optimization : Objective function: f(x 1,..., x d ) = c 1 x 1 + c 2 x c d x d = c. x. c = (c 1,..., c d ), x = (x 1,..., x d ). feasible region c solution 22 / 36

28 in R 2 LP in the plane: H: n half-plane or in other words n linear constraints f c (p): Objective function = c x p x + c y p y, c = (c x, c y ), p = (p x, p y ). Goal: find p R 2 s.t. p H and f c (p) is maximized. Cases: (i) (ii) ρ (iii) (iv) e v 23 / 36

29 in R 2 Unique Answer: solution 24 / 36

30 in R 2 Bound the feasible region: we add to half-plane to the set of half-planes { px M if c m 1 = x > 0 p x M Otherwise { py M if c m 2 = y > 0 p y M Otherwise Notations: (H, c): a linear program. H i = {m 1, m 2, h 1, h 2,..., h i }. C i = m 1 m 2 h 1 h 2 h i. Clearly C 0 C 1 C 2 C n = C. v i : the optimal solution for C i. If C i =, for some i, then the is infeasible ( is empty). 25 / 36

31 in R 2 Bound the feasible region: we add to half-plane to the set of half-planes { px M if c m 1 = x > 0 p x M Otherwise { py M if c m 2 = y > 0 p y M Otherwise Notations: (H, c): a linear program. H i = {m 1, m 2, h 1, h 2,..., h i }. C i = m 1 m 2 h 1 h 2 h i. Clearly C 0 C 1 C 2 C n = C. v i : the optimal solution for C i. If C i =, for some i, then the is infeasible ( is empty). 25 / 36

32 in R 2 incremental algorithm Lemma 4.5 We have (i) If v i 1 h i, then v i = v i 1. (ii) If v i 1 h i, then either C i = or v i l i, where l i is the line bounding h i. Proof. (i): Clear, since C i C i 1. (ii): q C i 1 v i v i 1 26 / 36

33 in R 2 incremental algorithm Lemma 4.5 We have (i) If v i 1 h i, then v i = v i 1. (ii) If v i 1 h i, then either C i = or v i l i, where l i is the line bounding h i. Proof. (i): Clear, since C i C i 1. (ii): q C i 1 v i v i 1 26 / 36

34 in R 2 incremental algorithm 27 / 36 (i) h 5 h 3 h 4 (ii) h 6 h 5 h 3 h 4 c v 4 = v 5 h 1 h 2 h 1 v 5 v 6 h 2

35 in R 2 incremental algorithm New Simpler : Find the point p on l i that maximizes f c (p), subject to the constraints p h, for h H i 1. Maximize f c (x) subject to x σ(h, l i ), h H i 1 and l i h is bounded to the left x σ(h, l i ), h H i 1 and l i h is bounded to the right x σ(h,l i ) h σ(h,l i ) l i 28 / 36

36 in R 2 incremental algorithm 1-dimensional linear program: Maximize f c (x) subject to x σ(h, l i ), h H i 1 and l i h is bounded to the left x σ(h, l i ), h H i 1 and l i h is bounded to the right x left = max {σ(h, l i ) : l i h is bounded to the left} h H i 1 x right = min {σ(h, l i ) : h H i 1 l i h is bounded to the right} Lemma 4.6 A 1-dimensional linear program can be solved in linear time. Hence, if case (ii) of Lemma 4.5 arises, then we can compute v i, in O(i) time. 29 / 36

37 in R 2 incremental algorithm 4 G 2DBOUNDEDLP Algorithm: 1. Let v 0 be the corner of C Let h 1,...,h n be the half-planes of H. 3. for i 1 to n 4. do if v i 1 h i 5. then v i v i 1 6. else v i the point p on l i that maximizes f c (p), subject to the constraints in H i if p does not exist 8. then Report that the linear program is infeasible and quit. 9. return v n We now analyze the performance of our algorithm. Lemma 4.7 Algorithm 2DBOUNDEDLP computes the solution to a bounded linear program with n constraints and two variables in O(n 2 ) time and linear storage. Lemma 4.7 Algorithm 2DBOUNDEDLP computes the solution to a bounded linear program with n constraints and two variables in O(n 2 ) time and linear storage. Proof. To prove that the algorithm correctly finds the solution, we have to show that after every stage whenever we have added a new half-plane h i the point v i is still the optimum point for C i. This follows immediately from Lemma / 36

38 in R 2 incremental algorithm 4 G 2DBOUNDEDLP Algorithm: 1. Let v 0 be the corner of C Let h 1,...,h n be the half-planes of H. 3. for i 1 to n 4. do if v i 1 h i 5. then v i v i 1 6. else v i the point p on l i that maximizes f c (p), subject to the constraints in H i if p does not exist 8. then Report that the linear program is infeasible and quit. 9. return v n We now analyze the performance of our algorithm. Lemma 4.7 Algorithm 2DBOUNDEDLP computes the solution to a bounded linear program with n constraints and two variables in O(n 2 ) time and linear storage. Lemma 4.7 Algorithm 2DBOUNDEDLP computes the solution to a bounded linear program with n constraints and two variables in O(n 2 ) time and linear storage. Proof. To prove that the algorithm correctly finds the solution, we have to show that after every stage whenever we have added a new half-plane h i the point v i is still the optimum point for C i. This follows immediately from Lemma / 36

39 in R 2 incremental algorithm A bad case: h 1 h 2 c v 2 v 3 v n v 5 v 4 h 5 h 3 h 4 h n 31 / 36

40 32 / 36 the of the half-planes yet. We now meet a quite intriguing phenomenon. in R 2 Although we have no way to Randomized determine an ordering of H that is guaranteed to lead to a good running time, we have a very simple way out of our. We simply pick a random ordering of H. Of course, we could have bad luck and pick an order that leads Randomized to a quadratic running algorithms time. But with some luck, we pick an order that makes it run much faster. Indeed, we shall prove below that most orders lead to a fast Use algorithm. randomness For completeness, to avoid we first bad repeat cases. the algorithm. Algorithm 2DRANDOMIZEDBOUNDEDLP(H, c,m 1,m 2 ) Input. A linear program (H {m 1,m 2 }, c), where H is a set of n half-planes, c R 2, and m 1, m 2 bound the solution. Output. If (H {m 1,m 2 }, c) is infeasible, then this fact is reported. Otherwise, the lexicographically smallest point p that maximizes f c (p) is reported. 1. Let v 0 be the corner of C Compute a random permutation h 1,...,h n of the half-planes by calling RANDOMPERMUTATION(H[1 n]). 3. for i 1 to n 4. do if v i 1 h i 5. then v i v i 1 6. else v i the point p on l i that maximizes f c (p), subject to the constraints in H i if p does not exist 8. then Report that the linear program is infeasible and quit. 9. return v n

41 in R 2 Lemma 4.8 The 2-dimensional linear programming with n constraints can be solved in O(n) randomized expected time using worst-case linear storage. Proof. X i = { 1 if vi 1 h i 0 otherwise Total time = n (O(i) X i ) i=1 ( n ) E(Total time) = E (O(i) X i ) = i=1 n (O(i) E(X i )). i=1 33 / 36

42 in R 2 Lemma 4.8 The 2-dimensional linear programming with n constraints can be solved in O(n) randomized expected time using worst-case linear storage. Proof. X i = { 1 if vi 1 h i 0 otherwise Total time = n (O(i) X i ) i=1 ( n ) E(Total time) = E (O(i) X i ) = i=1 n (O(i) E(X i )). i=1 33 / 36

43 in R 2 Lemma 4.8 The 2-dimensional linear programming with n constraints can be solved in O(n) randomized expected time using worst-case linear storage. Proof (Cont.). Any upper bound for E(x i )? E(X i ) = P r(v i 1 h i ) 2 (by backward analysis) i E(Total time) = n (O(i) E(X i )) i=1 n i=1 ( O(i) 2 ) O(n). i 34 / 36

44 in R 2 Lemma 4.8 The 2-dimensional linear programming with n constraints can be solved in O(n) randomized expected time using worst-case linear storage. Proof (Cont.). Any upper bound for E(x i )? E(X i ) = P r(v i 1 h i ) 2 (by backward analysis) i E(Total time) = n (O(i) E(X i )) i=1 n i=1 ( O(i) 2 ) O(n). i 34 / 36

45 in R 2 Lemma 4.8 The 2-dimensional linear programming with n constraints can be solved in O(n) randomized expected time using worst-case linear storage. Proof (Cont.). Any upper bound for E(x i )? E(X i ) = P r(v i 1 h i ) 2 (by backward analysis) i E(Total time) = n (O(i) E(X i )) i=1 n i=1 ( O(i) 2 ) O(n). i 34 / 36

46 in R 2 Backward Analysis: Why P r(v i 1 h i ) 2 i? c v n half-planes defining v n 35 / 36

47 in R 2 Backward Analysis: Why P r(v i 1 h i ) 2 i? c v n 35 / 36

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