1 / 42

Current Jotto standings…

Current Jotto standings…. Sophs. Jrs. Srs. Profs. pluot 1. pluot 2. pluot 1. pluot 2. squid 2. squid 1. squid 0. squid 1. Sophs & "Profs" guessing. Turning direction. (15,30). 3. (20,20). (40,20). 2. 3'. Bertrand Planes' Life Clock. (10,10). 1.

leia
Download Presentation

Current Jotto standings…

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Current Jotto standings… Sophs Jrs Srs Profs pluot 1 pluot 2 pluot 1 pluot 2 squid 2 squid 1 squid 0 squid 1 Sophs & "Profs" guessing...

  2. Turning direction (15,30) 3 (20,20) (40,20) 2 3' Bertrand Planes' Life Clock (10,10) 1 def turningDirection( pt1, pt2, pt3 ): """ returns 1 if pt1 -> pt2 -> pt3 is a CCW turn returns -1 if pt1 -> pt2 -> pt3 is a CW turn returns 0 if they are collinear """ x1, y1 = pt1; x2, y2 = pt2; x3, y3 = pt3; # the signed magnitude of the cross product CP = (x2-x1)*(y3-y1)-(y2-y1)*(x3-x1) if CP > 0: return 1 if CP < 0: return -1 return 0 Python code for CCW turning

  3. long bets...

  4. Geometric algorithms All points on this line are Line Segment intersection… (xA,yA) + s(xB-xA,yB-yA) (xB,yB) dxA dyA (xi,yi) (x2,y2) (xA,yA) (x1,y1) All points on this line are (x1,y1) + r(x2-x1,y2-y1) dx1 dy1 dx1 dxA dy1dyA r s xA - x1 yA- y1 Solving these equations finds the intersection via r and s. =

  5. Geometric algorithms Line Segment intersection… pt2 = (20,20) pt5 = (40, 20) pt3 = (10, 20) pt1 = (10, 10) pt4 = (20, 10) Line segment #1 runs from (10, 10) to (20, 20) Line segment #2 runs from (10, 20) to (20, 10) Intersection result = (15.0, 15.0, 1, 0.5, 0.5) Line segment #1 runs from (10, 10) to (10, 20) Line segment #2 runs from (20, 20) to (20, 10) Intersection result = (0, 0, 0, 0, 0) Line segment #1 runs from (10, 10) to (20, 20) Line segment #2 runs from (20, 10) to (40, 20) Intersection result = (0.0, 0.0, 1, -1.0, -1.0)

  6. Geometric algorithms Java has its advantages! Line2d.linesIntersect(x1, y1, x2, y2, x3, y3, x4, y4); Polygon().contains(x,y); Polygon Line2D

  7. Convex Hull First approach: brute force? the segments surrounding the exterior of a point set.

  8. The Graham Scan "first algorithm in computational geometry" a stack (e.g., python list) 4) push P[0] and P[1] onto S 5) run the scan: i = 2 while i < N: A = the top of stack S B = the second point in S if (P[i] is to the left of B wrt A): push P[i] onto stack S i = i+1 else: pop stack S and discard the top 1) Find extremal point, P[0] 2) Sort all other points in terms of their angles with P[0] - use atan2 ! 3) the sorted list is P[i]

  9. Graham Scan: java Stack grahamScan(Coordinate[] c) { Point p; Stack ps = newStack(); ps.push(c[0]); ps.push(c[1]); ps.push(c[2]); for (int i = 3; i < c.length; i++) { p = ps.pop(); while (computeOrientation(ps.peek(), p, c[i]) > 0)) { ps.pop(); } ps.push(p); ps.push(c[i]); } ps.push(c[0]); return ps; }

  10. Convex Hull #2 Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  11. Jarvis March draw a line to this point (why?) start here Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  12. Jarvis March draw a line to this point q start here draw a line to the point with the LEAST relative angle (q) Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  13. Jarvis March draw a line to the point with the LEAST relative angle (q) start here Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  14. Jarvis March draw a line to the point with the LEAST relative angle (q) q relative to the previous angle! Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  15. Jarvis March continue until you return… Jarvis’s March - shown here convex hull Graham’s Scan - see previous

  16. This week's Problems… • read over these problems: judge easy/hard? • try one or two to finish this afternoon/evening... • what geometric computation is needed? Hopefully everyone can get 1 or 2 of these completed!

  17. Try them out!!

  18. The EE Problem Input # of test cases # of room vertices, # of routers, # of test points example input C + B vertices of the room A test locations + + routers Locations of the routers A Locations of the test points Output B C Data Set 1: 10.81 3.42 0.00 The strength of the signal at each test location == 1.0/(closest visible router ** 2)

  19. W W W W The IE Problem Input Industrial Engineering # of test cases 1 4 4 0.1 0.1 0.0 0.9 1.0 0.05 1.1 -0.1 -0.1 -0.1 0.8 0 1.1 0.5 0.7 0 0.3 0.5 0 0.3 # of stores and possible warehouse locations (x,y) location of stores $.5 (x,y,price) location of warehouses and their cost to build in Mega$ S delivery cost = Euclidean distance S S Output $.3 $.3 S $.8 Data Set 1: 2.32 The minimum total cost to supply all stores from some warehouse(s).

  20. The Superpaint Problem Input one side of the square lattice Col 4 Col 3 Col 2 Col 1 . . . . C . C . . . . . C . . . number of occupied squares Row 1 4 3 2 1 2 3 4 1 Row 2 Locations of the cows (row,col) Row 3 Row 4 Col 4 Output Col 3 Col 2 Col 1 . . . . B . B . . B . . B . B . Row 1 5 Row 2 Row 3 The number of locations that "attack" all occupied squares with a Queen's move Row 4

  21. The Safepens Problem Input Number of rectangular fences 4 1 1 16 16 6 6 11 13 7 7 9 12 3 3 10 5 The fences! (lower left and upper right vertices) (16,16) (11,13) (9,12) (7,7) (6,6) Output (10,5) (3,3) 3 1 (1,1) The deepest nesting level The number of pens at that level

  22. "Sweepline algorithm" http://en.wikipedia.org/wiki/Bentley–Ottmann_algorithm

  23. D Priority Queue C Binary Search Tree A B http://en.wikipedia.org/wiki/Bentley–Ottmann_algorithm

  24. The Screens Problem Input # of test cases # of room vertices # of projector screens (x,y) location of "you" room Vertices of the room oriented screens Output Data Set 1: 90.00% Line segments of the screens The total fraction of the presentation you can observe (all screens' contributions!)

  25. Current Jotto standings… Sophs Jrs Srs Profs Chalk 1 Chalk 0 Chalk 1 Chalk 1 Quine 1 Quine 1 Quine 2 Quine 2 aught 2 aught 1 aught 1 aught 2 jotto 2 jotto 2 jotto 0 jotto 1 savvy 2 savvy 0 savvy 1 savvy 1 clash 2 clash 0 clash 1 clash 1

  26. Current Jotto standings… Sophs Jrs Srs Others icily 0 icily 0 icily 1 icily 1 strep 2 strep 2 strep 2 strep 1 spork 1 spork 3 spork 0 spork 0 spend 2 spend 2 spend 2 spend 2 peeps 2 peeps 1 peeps 2 peeps 1 furls 1 furls 1 furls 0 furls 1 Ghost 2 Ghost 1 Ghost 1 Ghost 0 Tanks 2 Tanks 1 Tanks 2 Tanks 1 Gecko 2 Gecko 1 Gecko 1 Gecko 1

  27. "QuickHull"

  28. "QuickHull" choose L and R pts draw chord

  29. "QuickHull" choose L and R pts draw chord • assign sides • find farthest point on each side • create triangle recurse!

  30. "QuickHull" choose L and R pts draw chord • assign sides • assign sides • find farthest point on each side • create triangle recurse!

  31. The Sweepline Algorithm choose L and R pts draw chord • assign sides • assign sides • find farthest point on each side • create triangle recurse!

  32. The Safepens Problem Input Number of rectangular fences 4 1 1 16 16 6 6 11 13 7 7 9 12 3 3 10 5 The fences! (lower left and upper right vertices) (16,16) (11,13) (9,12) (7,7) (6,6) Output (10,5) (3,3) 3 1 (1,1) The deepest nesting level The number of pens at that level

  33. The Superpaint Problem Input one side of the square lattice Col 4 Col 3 Col 2 Col 1 . . . . C . C . . . . . C . . . number of occupied squares Row 1 4 3 2 1 2 3 4 1 Row 2 Locations of the cows (row,col) Row 3 Row 4 Col 4 Output Col 3 Col 2 Col 1 . . . . B . B . . B . . B . B . Row 1 5 Row 2 Row 3 The number of locations that "attack" all occupied squares with a Queen's move Row 4

  34. The Screens Problem Input # of test cases # of room vertices # of projector screens (x,y) location of "you" room Vertices of the room oriented screens Output Data Set 1: 90.00% Line segments of the screens The total fraction of the presentation you can observe (all screens' contributions!)

  35. The EE Problem Input # of test cases # of room vertices, # of routers, # of test points example input C + B vertices of the room A test locations + + routers Locations of the routers A Locations of the test points Output B C Data Set 1: 10.81 3.42 0.00 The strength of the signal at each test location == 1.0/(closest visible router ** 2)

  36. W W W W The IE Problem Input Industrial Engineering # of test cases 1 4 4 0.1 0.1 0.0 0.9 1.0 0.05 1.1 -0.1 -0.1 -0.1 0.8 0 1.1 0.5 0.7 0 0.3 0.5 0 0.3 # of stores and possible warehouse locations (x,y) location of stores $.5 (x,y,price) location of warehouses and their cost to build in Mega$ S delivery cost = Euclidean distance S S Output $.3 $.3 S $.8 Data Set 1: 2.32 The minimum total cost to supply all stores from some warehouse(s).

  37. Convex Hull Problems… • read over these problems… • which ones are convex hull? • which ones could be convex hull? and the rest?

  38. Current Jotto standings… Quine 5 Win! Sophs Jrs Srs Others icily 0 icily 0 icily 1 icily 1 strep 2 strep 2 strep 2 strep 1 spork 1 spork 3 spork 0 spork 0 spend 2 spend 2 spend 2 spend 2 peeps 2 peeps 1 peeps 2 peeps 1 furls 1 furls 1 furls 0 furls 1 Ghost 2 Ghost 1 Ghost 1 Ghost 0 Tanks 2 Tanks 1 Tanks 2 Tanks 1 Gecko 2 Gecko 1 Gecko 1 Gecko 1

  39. What are these? public int mystery1(Point A, Point B, Point P) { int cp1 = (B.x-A.x)*(P.y-A.y) - (B.y-A.y)*(P.x-A.x); if (cp1>0) return 1; else return -1; } public int mystery2(Point A, Point B, Point C) { int ABx = B.x-A.x; int ABy = B.y-A.y; int num = ABx*(A.y-C.y)-ABy*(A.x-C.x); if (num < 0) num = -num; return num; }

  40. What are these? public int mystery1(Point A, Point B, Point P) { int cp1 = (B.x-A.x)*(P.y-A.y) - (B.y-A.y)*(P.x-A.x); if (cp1>0) return 1; else return -1; } sortOfDistance Hints: whichSide public int mystery2(Point A, Point B, Point C) { int ABx = B.x-A.x; int ABy = B.y-A.y; int num = ABx*(A.y-C.y)-ABy*(A.x-C.x); if (num < 0) num = -num; return num; }

More Related