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Guarding Terrains though the Lens of Parameterized Complexity
- 1. Ben-Gurion University of the Negev, Israel Guarding Terrains through the Lens of Parameterized Complexity Akanksha Agrawal PARAMETERIZED COMPLEXITY SEMINAR Based on a joint work with Sudeshna Kolay and Meirav Zehavi, SWAT 2020 (and an invited survey with Meirav Zehavi, CSR 2020)
- 2. * A Short Walk Through the Terrains A Brief History Of Terrain Guarding A Kernel and an XP Algorithm *Artistic icons in this presentation are from Iconfinder.
- 3. * A Short Walk Through the Terrains v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Terrains Sequence of points in the plane, such that any vertical line is intersected continuously
- 4. Terrains A Short Walk Through the Terrains v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Sequence of points in the plane, such that any vertical line is intersected continuously
- 5. A Short Walk Through the Terrains Not a Terrain! v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Sequence of points in the plane, such that any vertical line is intersected continuously
- 6. Vertices and Edges of a Terrains A Short Walk Through the Terrains v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 edge vertex Vertex: Points in the sequence that defines the terrain Edge: Line segment connecting consecutive points (vertices) of the terrain
- 7. Points of a Terrain A Short Walk Through the Terrains v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 point point All points (including the vertices) on the edges of the terrain are its points
- 8. v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 >180o Reflex: angle > 180o Convex: angle <= 180o <=180o A Short Walk Through the Terrains Reflex/Convex Vertices
- 9. A Short Walk Through the Terrains Visibility Two points on the terrain see each other if the line segment between them is contained on or above the terrain v1 v2 v3 v4 v5 v6 v7 v8 v9 v10
- 10. A Short Walk Through the Terrains Orthogonal Terrains All the edges of the terrain are horizontal/vertical lines
- 11. QUESTION Terrain T Integer k & Is there is a set of at most k points in T that sees all the points of T? v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Terrain Guarding NPUT
- 12. QUESTION Terrain T Integer k & Is there is a set of at most k points in T that sees all the points of T? v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Terrain Guarding NPUT Where can we place the guards? What all points do we need to guard? DIFFERENT VERSIONS Discrete: Vertices guarding vertices Continuous: Points guarding points
- 13. QUESTION Terrain T Integer k & Is there is a set of at most k points in T that sees all the points of T? v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Terrain Guarding NPUT Where can we place the guards? What all points do we need to guard? DIFFERENT VERSIONS Discrete: Vertices guarding vertices Continuous: Points guarding points
- 14. A Brief History Of Terrain Guarding Classical Complexity Terrain Guarding is a close cousin of the Art Gallery problem, where the objective is to guard a polygon, instead of a terrain.
- 15. A Brief History Of Terrain Guarding Terrain Guarding is a close cousin of the Art Gallery problem, where the objective is to guard a polygon, instead of a terrain. An NP-hardness for Terrain Guarding was claimed in 1995, but a proof was for it was only obtained after 15 years. [King and Krohn] no algorithm2⌦(n/3) Classical Complexity
- 16. A Brief History Of Terrain Guarding Terrain Guarding is a close cousin of the Art Gallery problem, where the objective is to guard a polygon, instead of a terrain. An NP-hardness for Terrain Guarding was claimed in 1995, but a proof was for it was only obtained after 15 years. [King and Krohn] Orthogonal Terrain Guarding was shown to be NP-hard only in 2018. [Bonnet and Giannopoulos] no algorithm2⌦(n/3) Classical Complexity
- 17. A Brief History Of Terrain Guarding Approximation Algorithms In 2005, the first constant factor approximation was obtained for Discrete Terrain Guarding. [Ben-Moshe et al.]
- 18. A Brief History Of Terrain Guarding Approximation Algorithms In 2005, the first constant factor approximation was obtained for Discrete Terrain Guarding. [Ben-Moshe et al.] Later, Discrete Terrain Guarding was shown to admit a PTAS. [Gibson et al.]
- 19. A Brief History Of Terrain Guarding Approximation Algorithms By discretization for Continuous Terrain Guarding, the following results were obtained: 1) NP-completeness and 2) a PTAS. [Friedrichs et al.]
- 20. A Brief History Of Terrain Guarding Approximation Algorithms By discretization for Continuous Terrain Guarding, the following results were obtained: 1) NP-completeness and 2) a PTAS. [Friedrichs et al.] Discretization v1 v2 v3 v4 v5 v6 v7 v8 v9 v10
- 21. A Brief History Of Terrain Guarding Approximation Algorithms v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Discretization Allowed Guard Set Points to be Guarded FINITE By discretization for Continuous Terrain Guarding, the following results were obtained: 1) NP-completeness and 2) a PTAS. [Friedrichs et al.]
- 22. A Brief History Of Terrain Guarding Parameterized Complexity An XP-algorithm for Terrain Guarding running in time . Art Gallery: has no algorithm running in time: or [Ashok et al.] nO( p k) 2o(n) f(k) · nO(k/ log k)
- 23. A Brief History Of Terrain Guarding Parameterized Complexity An XP-algorithm for Terrain Guarding running in time . Art Gallery: has no algorithm running in time: or A -time FPT algorithm for Orthogonal Terrain Guarding. [Ashok et al.] nO( p k) 2o(n) f(k) · nO(k/ log k) kO(k) nO(1)
- 24. A Brief History Of Terrain Guarding Parameterized Complexity An XP-algorithm for Terrain Guarding running in time . Art Gallery: has no algorithm running in time: or A -time FPT algorithm for Orthogonal Terrain Guarding. [Ashok et al.] FPT algorithms for Terrain Guarding for structural parameters like onion peeling number and guard range. [Khodakarami et al.] nO( p k) 2o(n) f(k) · nO(k/ log k) kO(k) nO(1)
- 25. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices
- 26. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices XP algorithm for Orthogonal Terrain Guarding parameterized by the number of minima
- 27. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices
- 28. v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 >180o <=180o A Quick Recap Reflex/Convex Vertices Reflex: angle > 180o Convex: angle <= 180o
- 29. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices Why #reflex vertices?
- 30. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices Why #reflex vertices? v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Placing guards at all reflex vertices guards the whole terrain
- 31. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices Why #reflex vertices? v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Placing guards at all reflex vertices guards the whole terrain k r
- 32. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices Restricted to Vertices Guarding Vertices
- 33. Overview of our Kernelization MARK 1: Mark enough solution candidates. MARK 2: Mark enough vertices to be guarded. Short Circuit: Remove unmarked vertices.
- 34. Overview of our Kernelization MARK 1: Mark enough solution candidates. MARK 2: Mark enough vertices to be guarded. Short Circuit: Remove unmarked vertices.
- 35. A Claim to Remember Order Claim x yp q x sees y & p sees q x sees q
- 36. Region of (no) Chaos Convex Regions Sequence of maximal consecutive convex vertices
- 37. Region of (no) Chaos Convex Regions Sequence of maximal consecutive convex vertices Any two vertices of a convex region see each other
- 38. MARK 1: Mark enough solution candidates. • Mark all reflex vertices • Mark endpoints of convex regions Kernel for Terrain Guarding: Mark 1
- 39. Kernel for Terrain Guarding: Mark 1 MARK 1: Mark enough solution candidates. • Mark all reflex vertices • Mark endpoints of convex regions • For pair of convex regions C,C’, mark the smallest and the largest vertices of C’, that sees all of C (also do the above for a pair of reflex vertex and a convex region) C C’u v w
- 40. Construction of Mark 1 is Correct! For a convex region C, if a vertex v to the right of it, sees the right endpoint, then v sees everything in C v C If there is a solution, then there is a solution from Mark 1
- 41. Construction of Mark 1 is Correct! v=q C For a convex region C, if a vertex v to the right of it, sees the right endpoint, then v sees everything in C Order Claim x y p If there is a solution, then there is a solution from Mark 1
- 42. Construction of Mark 1 is Correct! More generally, once v starts seeing, it sees up to the left vertex v C Order Claim p seen If there is a solution, then there is a solution from Mark 1
- 43. Construction of Mark 1 is Correct! v (unmarked guard) C Consider a solution S If there is a solution, then there is a solution from Mark 1
- 44. Construction of Mark 1 is Correct! v (unmarked guard) Consider the closest marked guard, u1 and u2, to the left and right of v, respectively u1 u2 One of them works as a replacement for v?C Consider a solution S If there is a solution, then there is a solution from Mark 1
- 45. Construction of Mark 1 is Correct! v (unmarked guard) Consider the closest marked guard, u1 and u2, to the left and right of v, respectively u1 u2 One of them works as a replacement for v?C Consider a solution S If there is a solution, then there is a solution from Mark 1 Case 1: v sees a convex region partially Case 2: v sees all convex regions fully
- 46. Construction of Mark 1 is Correct! C v (unmarked guard) u1 u2 C’ p Consider a solution S If there is a solution, then there is a solution from Mark 1 Case 1: v sees a convex region partially, say to its left
- 47. Construction of Mark 1 is Correct! C v (unmarked guard) u1 u2 C’ z p Consider a solution S If there is a solution, then there is a solution from Mark 1 Case 1: v sees a convex region partially, say to its left
- 48. Construction of Mark 1 is Correct! C v (unmarked guard) u1 u2 C’ z’ (sees z; after C’) z’ sees all of C’! z p Consider a solution S If there is a solution, then there is a solution from Mark 1 Case 1: v sees a convex region partially, say to its left
- 49. Construction of Mark 1 is Correct! z’ C v (unmarked guard) u1 u2 C’ z p Consider a solution S If there is a solution, then there is a solution from Mark 1 u2 is a replacement! • vertex to the left of v seen by u2 Case 1: v sees a convex region partially, say to its left
- 50. Construction of Mark 1 is Correct! z’ C v (unmarked guard) u1 u2 C’ z p Consider a solution S If there is a solution, then there is a solution from Mark 1 u2 is a replacement! • vertex to the left of v seen by u2 • vertex to the right of v seen by z’ Case 1: v sees a convex region partially, say to its left
- 51. Construction of Mark 1 is Correct! Consider a solution S If there is a solution, then there is a solution from Mark 1 z’ C v (unmarked guard) u1 u2 C’ z u2 is a replacement! • vertex to the left of v seen by u2 • vertex to the right of v seen by z’ p Case 1: v sees a convex region partially, say to its left
- 52. Construction of Mark 1 is Correct! Consider a solution S If there is a solution, then there is a solution from Mark 1 z’ C v (unmarked guard) u1 u2 C’ z u2 is a replacement! • vertex to the left of v seen by u2 • vertex to the right of v seen by z’ p Case 1: v sees a convex region partially, say to its left
- 53. Construction of Mark 1 is Correct! Consider a solution S If there is a solution, then there is a solution from Mark 1 z’ C v (unmarked guard) u1 u2 C’ z u2 is a replacement! • vertex to the left of v seen by u2 • vertex to the right of v seen by z’ p w Case 1: v sees a convex region partially, say to its left
- 54. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! Consider a solution S
- 55. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! C’ Consider a solution S
- 56. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! C’ Consider a solution S
- 57. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! C’ Consider a solution S
- 58. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! C’ someone sees C’ fully Consider a solution S
- 59. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1 C v (unmarked guard) u1 u2 Case 2: v sees all convex regions fully u2 is a replacement! C’ someone sees C’ fully Consider a solution S
- 60. Construction of Mark 1 is Correct! v (unmarked guard) u1 u2 C Consider a solution S If there is a solution, then there is a solution from Mark 1 Case 1: v sees a convex region partially Case 2: v sees all convex regions fully Replacement for v exists!
- 61. Construction of Mark 1 is Correct! If there is a solution, then there is a solution from Mark 1
- 62. Overview of our Kernelization MARK 1: Mark enough solution candidates. MARK 2: Mark enough vertices to be guarded. Short Circuit: Remove unmarked vertices. Poly(r) vertices
- 63. Short Circuit to Complete the Algorithm Terrain T, integer k Terrain T’, integer k NPUT x y p q x y
- 64. Overview of our Kernelization MARK 1: Mark enough solution candidates. MARK 2: Mark enough vertices to be guarded. Short Circuit: Remove unmarked vertices. Terrain Guarding admits a polynomial kernel, when parameterized by the number of reflex vertices Poly(r) vertices +NP-completeness of Terrain Guarding
- 65. Overview of our Kernelization +the Known Discretization => Continuous Terrain Guarding Terrain Guarding admits a polynomial kernel, when parameterized by the number of reflex vertices
- 66. This Talk Polynomial kernel for Terrain Guarding parameterized by the number of reflex vertices XP algorithm for Orthogonal Terrain Guarding parameterized by the number of minima Each vertex is adjacent to at most one horizontal and at most one vertical edge
- 67. Minima of an Orthogonal Terrain A horizontal edge uv, where the two neighbors of u and v are both at a higher elevation them. minimum Reflex vertices are circles and convex vertices are squares
- 68. Maxima of an Orthogonal Terrain A horizontal edge uv, where the two neighbors of u and v are both at a lower elevation than u and v. maximum
- 69. Valleys Roughly, a valley is a maximal regions between a maxima and a minima W1 W2 W3 W4 #Valleys <= #minima + 2
- 70. Minima vs. Sol. Size Solution Size = 2 |Minima| >>2 Solution Size >> 1 |Minima| =1
- 71. W1 W2 W3 W4 (T,k) is a yes-instance A subset of reflex vertices of size at most k, guarding all the convex vertices A Useful Result
- 72. A Useful Result W1 W2 W3 W4 (T,k) is a yes-instance A subset of reflex vertices of size at most k, guarding all the convex vertices Focus only on reflex solutions guarding convex vertices!
- 73. DP States XP Algorithm Integer k’<= k and a height h Wi pi h bpi to be seen to be seen allowed guards allowed guards
- 74. DP States XP Algorithm For each valley Wi, we have: • What we must necessarily be seen from the left and right side of the valley Integer k’<= k and a height h Wi pi h bpi to be seen to be seen allowed guards allowed guards
- 75. DP States 1 if and only if there is a set of reflex vertices of size at most k’, at height at least h, such that for each Wi, we have: • All the necessary vertices are seen XP Algorithm For each valley Wi, we have: • What we must necessarily be seen from the left and right side of the valley Integer k’<= k and a height h Wi pi h bpi to be seen to be seen allowed guards allowed guards
- 76. DP States 1 if and only if there is a set of reflex vertices of size at most k’, at height at least h, such that for each Wi, we have: • All the necessary vertices are seen XP Algorithm For each valley Wi, we have: • What we must necessarily be seen from the left and right side of the valley Integer k’<= k and a height h Wi pi h bpi to be seen to be seen allowed guards allowed guards #Table Entries is bounded by = #Valleys⌫ n2⌫+1 · (k + 1)
- 77. W1 W2 W3 W4 At height h, at most 2#Valleys many reflex vertices height h A Fact
- 78. Recursive Formula W1 W2 W3 W4 lowest height h* • Go over all subsets of guards to be placed at h* At height h, at most 2#Valleys many reflex
- 79. Recursive Formula W1 W2 W3 W4 lowest height h* • Go over all subsets of guards to be placed at h* • Increment what needs to be seen for each valley Updation Safe!
- 80. Recursive Formula W1 W2 W3 W4 lowest height h* • Go over all subsets of guards to be placed at h^* • Increment what needs to be seen for each valley Take OR over all answers Updation Safe?
- 81. DP States XP Algorithm Wi pi h bpi to be seen to be seen allowed guards allowed guardsContinuous Continuous Safe Updatation always possible!
- 82. XP Algorithm Orthogonal Terrain Guarding admits an algorithm running in timen2⌫+c = #Valleys⌫
- 83. Conclusion & Open Problems We designed a polynomial kernel for Terrain Guarding when parameterized by the number of reflex vertices. Is the problem FPT parameterized by the solution size?
- 84. Conclusion & Open Problems We designed a polynomial kernel for Terrain Guarding when parameterized by the number of reflex vertices. Is the problem FPT parameterized by the solution size? We designed an XP-algorithm for Orthogonal Terrain Guarding when parameterized by the number of minima. Is the problem FPT parameterized by the number of minimas?
- 85. Conclusion & Open Problems Thanks! We designed a polynomial kernel for Terrain Guarding when parameterized by the number of reflex vertices. Is the problem FPT parameterized by the solution size? We designed an XP-algorithm for Orthogonal Terrain Guarding when parameterized by the number of minima. Is the problem FPT parameterized by the number of minimas?