acm schedule… 2/23 geometry problems 3/2 lab problems (geometry!) 3/30 "mock"...

ACM Schedule…

2/23 Geometry problems

3/2 Lab problems (geometry!)

3/30 "Mock" competition and lab session (4:15)

3/9 Final algorithm discussion…

3/16 [Spring break]

3/23 [Extended programming practicum break]

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

?? 2 ?? 2 ?? 0 ?? 1

Last time: the Muddy Problem

Input

Output

4 4*.*..******...*.

# of rows and columns

The environment

*

4The smallest number of boards needed to cover all the muddy patches!

represents a muddy patch

. represents a grassy patch

Turning direction

Python code for CCW turning

1

2

3

def turningDirection( pt1, pt2, pt3 ): """ returns 1 if pt1 -> pt2 -> pt3 is a turn returns -1 if pt1 -> pt2 -> pt3 is a 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

Bertrand Planes' Life Clock

(10,10)

(20,20)

(15,30)

3'

(40,20)

Geometric algorithms

Line Segment intersection…

(x1,y1)

(x2,y2)

(xB,yB)

(xA,yA)

rs

xA - x1

yA - y1

dx1 dxA

dy1 dyA

=

(xi,yi)

All points on this line are (x1,y1) + r(x2-x1,y2-y1)

(xA,yA) + s(xB-xA,yB-yA)All points on this line are

dx1 dy1

dxA dyA

Solving these equations finds the intersection via r and s.


Line Segment intersection…

pt1 = (10, 10)

pt2 = (20,20)

pt3 = (10, 20)

pt4 = (20, 10)

pt5 = (40, 20)

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)


Java has its advantages!

Line2d.linesIntersect(x1, y1, x2, y2, x3, y3, x4, y4);

Polygon().contains(x,y);

Line2D Polygon

This week's Problems…

• read over these problems…

• rank them in "difficulty" order

• what geometric computation is needed?

The EE ProblemInput

Output

# of test cases

Data Set 1:10.813.420.00The strength of the signal at each test location ==

1.0/(closest visible router ** 2)

# of room vertices, # of routers, # of test points

vertices of the room

Locations of the routers

test locations

routers

example input

Locations of the test points

+

+

+

A

B

C

C

A B

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

The IE ProblemInput

Output

# of test cases

Data Set 1:2.32

The minimum total cost to supply all stores from some warehouse(s).

# of stores and possible warehouse locations

(x,y) location of stores

(x,y,price) location of warehouses and their cost to build in Mega$

S

S

S

SW

W

WW

$.3$.3

$.5

$.8

delivery cost = Euclidean distance

Industrial Engineering

The Superpaint ProblemInput

Output

4 32 12 34 1

one side of the square lattice

Locations of the cows (row,col)

5

The number of locations that "attack" all occupied squares

with a Queen's move

Row 1 . . . .C . C . . . . .C . . .

number of occupied squares

Row 2

Row 3

Row 4

Col 1

Col 2

Col 3

Col 4

Row 1 . . . .B . B . . B . .B . B .

Row 2

Row 3

Row 4

Col 1

Col 2

Col 3

Col 4

The Safepens Problem

Input

Output

41 1 16 166 6 11 137 7 9 123 3 10 5

Number of rectangular fences

The fences!

(lower left and upper right vertices)

3 1

The deepest nesting level

The number of pens at that level

(1,1)

(16,16)

(6,6)

(11,13)

(7,7)

(9,12)

(3,3)

(10,5)

http://en.wikipedia.org/wiki/Bentley–Ottmann_algorithm

"Sweepline algorithm"

http://en.wikipedia.org/wiki/Bentley–Ottmann_algorithm

A

B

C

D Priority Queue

Binary Search Tree

The Screens ProblemInput

Output

# of test cases

Vertices of the room

Data Set 1:90.00%

The total fraction of the presentation you can observe (all screens' contributions!)

# of projector screens

# of room vertices

(x,y) location of "you"

Line segments of the screens

room

oriented screens


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

?? 2 ?? 2 ?? 0 ?? 1


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

Convex Hull

the segments surrounding the

exterior of a point set.

First approach: brute force?

The Graham Scan

"first algorithm in computational geometry"

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]

4) push P[0] and P[1] onto S

a stack (e.g.,

python list)

i = 2

while i < N:

A = the top of stack S B = the second point on S

if (P[i] is to the left of B to A): push P[i] onto stack S i = i+1 else: pop stack S and discard the top

5) run the scan:

Graham Scan: java

Stack grahamScan(Coordinate[] c) { Point p; Stack ps = new Stack(); 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;}

Convex Hull #2

convex hull Jarvis’s March - shown hereGraham’s Scan - see previous

Jarvis March

convex hull

start here

draw a line to this point (why?)

Jarvis’s March - shown hereGraham’s Scan - see previous

Jarvis March

convex hull

start here

draw a line to this point

draw a line to the point with the LEAST relative angle ()


Jarvis March

convex hull

start here



Jarvis March

convex hull


but bigger than the previous angle!


Jarvis March

convex hull

continue until you return…


"QuickHull"

"QuickHull"

choose L and R pts

draw chord

"QuickHull"

choose L and R pts

draw chord

• assign sides

• find farthest point on each side

• create triangle

recurse!

"QuickHull"

choose L and R pts

draw chord

• assign sides


• create triangle

recurse!

• assign sides

Java's Polygon class

contains

constructors

intersects (a rect)

getBounds (get a bbox)

uses the WIND_EVEN_ODD rule

The Sweepline Algorithm

choose L and R pts

draw chord

• assign sides


• create triangle

recurse!

• assign sides

The Safepens Problem

Input

Output

41 1 16 166 6 11 137 7 9 123 3 10 5

Number of rectangular fences

The fences!

(lower left and upper right vertices)

3 1

The deepest nesting level

The number of pens at that level

(1,1)

(16,16)

(6,6)

(11,13)

(7,7)

(9,12)

(3,3)

(10,5)

The Superpaint ProblemInput

Output

4 32 12 34 1

one side of the square lattice

Locations of the cows (row,col)

5

The number of locations that "attack" all occupied squares

with a Queen's move

Row 1 . . . .C . C . . . . .C . . .

number of occupied squares

Row 2

Row 3

Row 4

Col 1

Col 2

Col 3

Col 4

Row 1 . . . .B . B . . B . .B . B .

Row 2

Row 3

Row 4

Col 1

Col 2

Col 3

Col 4

The Screens ProblemInput

Output

# of test cases

Vertices of the room

Data Set 1:90.00%

The total fraction of the presentation you can observe (all screens' contributions!)

# of projector screens

# of room vertices

(x,y) location of "you"

Line segments of the screens

room

oriented screens

The EE ProblemInput

Output

# of test cases

Data Set 1:10.813.420.00The strength of the signal at each test location ==

1.0/(closest visible router ** 2)

# of room vertices, # of routers, # of test points

vertices of the room

Locations of the routers

test locations

routers

example input

Locations of the test points

+

+

+

A

B

C

C

A B

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

The IE ProblemInput

Output

# of test cases

Data Set 1:2.32

The minimum total cost to supply all stores from some warehouse(s).

# of stores and possible warehouse locations

(x,y) location of stores

(x,y,price) location of warehouses and their cost to build in Mega$

S

S

S

SW

W

WW

$.3$.3

$.5

$.8

delivery cost = Euclidean distance

Industrial Engineering

Convex Hull Problems…

• read over these problems…

• which ones are convex hull?

• which ones could be convex hull?

and the rest?


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

Win! Quine 5

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; }

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; }

Hints:sortOfDistancewhichSide

acm schedule… 2/23 geometry problems 3/2 lab problems (geometry!) 3/30 "mock"...

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