Geometry Algorithms

7/25/2019 Geometry Algorithms

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Geometry

Tmas Ken MagnssonBjarki gst Gumundsson

School of Computer Science

Reykjavk University

rangursrk forritun og lausn verkefna
http://ru.is/http://ru.is/td


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Today were going to cover

Geometry Computational geometry

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Points and vectors

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Points and vectors

(3, 6)

(6, 3)

(1, 3)

Points are representedby a pair of numbers,(x,y).

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Points and vectors

(3, 6)

(6, 3)

(1, 3)

Points are representedby a pair of numbers,(x,y).

Vectors are representedin the same way.

Thinking of points asvectors us to do manythings.

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Points and vectors

uv

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Points and vectors

uv

Simplest operation,addition is dened asx0y0

+

x1y1

=

x0+x1y0+y1

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Points and vectors

uv

u + v


+

x1y1

=

x0+x1y0+y1

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Points and vectors

uv


+

x1y1

=

x0+x1y0+y1

Subtraction is dened inthe same manner

x0y0x

1y1

=x

0x

1y0 y1

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Points and vectors

uv

u v


+

x1y1

=

x0+x1y0+y1

Subtraction is dened inthe same manner

x0y0x

1y1

=x

0x

1y0 y1

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Points and vectors

struct point {

double x, y;point(double _x, double _y) {

x = _x, y = _y;

}

point operator+(const point &oth){

return point(x + oth.x, y + oth.y);

}

point operator-(const point &oth){return point(x - oth.x, y - oth.y);

}

};

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Points and vectors

or we could use thecomplexclass.typedef complex point;

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Points and vectors

or we could use thecomplexclass.typedef complex point;

Thecomplexclass inC++and Javahas methods denedfor Addition Subtraction Multiplication by a scalar

Length Trigonometric functions

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Points and vectors

Complex numbers have the real part and the imaginarypart. Can be thought of as vectors or points on thecomplex plane.

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Points and vectors

Complex numbers have the real part and the imaginarypart. Can be thought of as vectors or points on thecomplex plane. double real(p)returns the real part, or in our case,

thexvalue ofp

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Points and vectors


thexvalue ofp double imag(p)returns the imaginary part,yvalue of

p.

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Points and vectors



p. double abs(p)returns the absolute value of the

complex number, the length of the vector.

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Points and vectors



p. double abs(p)returns the absolute value of the

complex number, the length of the vector. double sin(p),double cos(p),double tan(p),

trigonometric functions.

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Lines and line segments

Line segments arerepresented by a pair ofpoints,((x0,y0), (x1,y1)).

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Li d li


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p0

p1p1 p0


Distance between twopoints is the length of the

line segment or vectorbetween the points.

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Li d li t


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p1 p0


Distance between twopoints is the length of the

line segment or vectorbetween the points.

d((x0,y0), (x1,y1))

= |(x1 x0,y1 y0)|=

(x1 x0)2 + (y1 y0)2

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Li d li t


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struct point {

...

double distance(point oth = point(0,0)) const {

return sqrt(pow(x - oth.x, 2.0)

+ pow(y - oth.y, 2.0));}

...

}

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Li d li t


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struct point {

...

double distance(point oth = point(0,0)) const {

return sqrt(pow(x - oth.x, 2.0)

+ pow(y - oth.y, 2.0));}

...

}

Or use theabsfunction withcomplex.

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Line representation same asline segments.

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q1

p1

q0

p0


Treat them as lines passingthrough the two points.

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p1

p0

r0

r1



Or as a point and a directionvector.

p+t r

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p1

p0

r0

r1



Or as a point and a directionvector.

p+t r

Either way

pair

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Circles


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Circles

Circles are very easy torepresent.

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Circles


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Circles

p0

p1


Center pointp= (x,y).

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Circles


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Circles

p0

p1

r0r1



And the radiusr.

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Circles


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Circles

p0

p1

r0r1



And the radiusr.pair

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Dot product


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Dot product

Given two vectors

u=

x0y0

v=

x1y1

the dot product of uand vis dened as

x0y0

x1y1

=x0 x1+y0 y1

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Dot product


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Dot product

Given two vectors

u=

x0y0

v=

x1y1

the dot product of uand vis dened as

x0y0

x1y1

=x0 x1+y0 y1

Which in geometric terms isu v= |u||v| cos

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Dot product


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Dot product

u

v

Allows us to calculate the

angle between uand v.

=arccos

u v

|u||v|

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Dot product


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Dot product

u

v

vu

Allows us to calculate the

angle between uand v.

=arccos

u v

|u||v|

And the projection of vonto u.

vu=

u v

|u|

u

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Dot product


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Dot product

u

v

vu

p

q

The closest point on uto pisq.

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Dot product


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Dot product

u

v

vu

p

q


The distance frompto uis thedistance fromptoq.

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Dot product


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p

u

v

vu

p

q


The distance frompto uis thedistance fromptoq.

Unlessqis outside u, then theclosest point is either of theendpoints.

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Dot product


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p

Rest of the code will use the complexclass.#define P(p) const point &p

#define L(p0, p1) P(p0), P(p1)

double dot(P(a), P(b)) {

return real(a) * real(b) + imag(a) * imag(b);

}

double angle(P(a), P(b), P(c)) {return acos(dot(b - a, c - b) / abs(b - a) / abs(c - b));

}

point closest_point(L(a, b), P(c), bool segment = false) {

if (segment) {

if (dot(b - a, c - b) > 0) return b;if (dot(a - b, c - a) > 0) return a;

}

double t = dot(c - a, b - a) / norm(b - a);

return a + t * (b - a);

}14

Cross product


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p

Given two vectors

u=

x0y0

v=

x1y1

the cross product of uand vis dened as

x0y0

x1y1

=x0 y1 y0 x1

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Cross product


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p

Given two vectors

u=

x0y0

v=

x1y1

the cross product of uand vis dened as

x0y0

x1y1

=x0 y1 y0 x1

Which in geometric terms is|u v| = |u||v| sin

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Cross product


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p

u

v

Allows us to calculate the area

of the triangle formed by uand v.

|u v|

2

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Cross product


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u

v

12|u v|

u

v



|u v|

2

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Cross product


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u

v

12|u v|

u

v



|u v|

2

And can tell us if the anglebetweenuand vis positive ornegative.

|u v| 0 i >

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Counterclockwise


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A

B

C

v

u

Given three pointsA,Band C, we

want to know if they form acounter-clockwise angle in thatorder.

A B C

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Counterclockwise


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A

B

C

v

u

+

Given three pointsA,Band C, we

want to know if they form acounter-clockwise angle in thatorder.

A B C

We can examine the cross productof and the area of the triangleformed by

u=B C

v= B A

u v>0

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Counterclockwise


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A

B

C

u

v

The points in the reverse order do

not form a counter clockwise angle.

C B A

In the reverse order the vectorsswap places

u=B A v= B C

u v


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v

u

A

B

C

The points in the reverse order do

not form a counter clockwise angle.

C B A

In the reverse order the vectorsswap places

u=B A v= B C

u v


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double cross(P(a), P(b)) {return real(a)*imag(b) - imag(a)*real(x);

}

double ccw(P(a), P(b), P(c)) {

return cross(b - a, c - b);

}

bool collinear(P(a), P(b), P(c)) {

return abs(ccw(a, b, c)) < EPS;

}

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Intersections


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Very common task is to nd the intersection of two lines orline segments.

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Intersections


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Very common task is to nd the intersection of two lines orline segments. Given a pair of points(x0,y0),(x1,y1), representing a

line we want to start by obtaining the formAx+By= C.

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Intersections


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We can do so by setting

A=y1 y0

B=x0 x1

C=A x0+B y1

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Intersections


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We can do so by setting

A=y1 y0

B=x0 x1

C=A x0+B y1

If we have two lines given by such equations, wesimply need to solve for the two unknowns,xandy.

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Intersections


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For two lines

A0x+B0y=C0A1x+B1y=C1

The intersection point is

x=(B1 C0 B0 C1)

D

y=(A0 C1 A1 C0)

D

WhereD=A0 B1 A1 B0

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IntersectionsQ l bl d h f


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Quite similar problem is to nd the intersections of twocircles.

rA rB

A Bd

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IntersectionsQ it i il bl i t d th i t ti f t


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rA rB

A Bd

Ifd> r0+r1the circles donot intersect.

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IntersectionsQ it i il bl i t d th i t ti f t


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rA rB

A Bd


Ifd< |r0 r1|, one circles iscontained within the other.

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IntersectionsQ ite similar problem is to nd the intersections of t o


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rA rB

A Bd



Ifd= 0and r0=r1, thecircles are the same.

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IntersectionsQuite similar problem is to nd the intersections of two


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rA rB

A Bd



Ifd= 0and r0=r1, thecircles are the same.

Lets look at the last case.

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rA rB

A B

h

a b

We can solve for the vectorsaand hfrom the equations

a2 +h2 =r20 b2 +h2 =r21

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rA rB

A B

h

a b

We can solve for the vectorsaand hfrom the equations

a2 +h2 =r20 b2 +h2 =r21

We get

a= r2A r

2

B+d2)

2 dh2 =r2A a

2

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Intersections


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#define C(p, r) const point &p, double r

int intersect(C(A, rA), C(B, rB), point & res1, point & res2) {

double d = abs(B - A);if ( rA + rB < d - EPS || d < abs(rA - rB) - EPS){

return 0;

}

double a = (rA*rA - rB*rB + d*d) / 2*d;

double h = sqrt(rA*rA - a*a);point v = normalize(B - A, a);

u = normalize(rotate(B-A), h);

res1 = A + v + u;

res2 = A + v - u;

if (abs(u) < EPS){

return 1;}

return 2;

}

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Polygons


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Polygons are represented by

a list of points in the orderrepresenting the edges.

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Polygons


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a list of points in the orderrepresenting the edges. To calculate the area

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Polygons


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We pick one starting point.

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Polygons


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We pick one starting point. Go through all the other

adjacent pair of points and sumthe area of the triangulation.

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Polygons


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Polygons


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Polygons


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Polygons


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Even if we sum up area outsidethe polygon, due to the crossproduct, it is subtracted later.

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Polygons


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Even if we sum up area outsidethe polygon, due to the crossproduct, it is subtracted later.

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Polygons


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double polygon_area_signed(const vector &p) {

double area = 0;

int cnt = size(p);

for (int i = 1; i + 1 < cnt; i++){

area += cross(p[i] - p[0], p[i + 1] - p[0])/2;

}

return area;}

double polygon_area(vector &p) {

return abs(polygon_area_signed(p));

}

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Convex hull Given a set of points, we want to nd the convex hull


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G e a set o po ts, e a t to d t e co e uof the points.

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p ,of the points.

The convex hull of points can be visualized as theshape formed by a rubber band around the set ofpoints.

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p ,of the points.


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pof the points.


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Convex hull


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Graham scan:

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Convex hull


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Graham scan: Pick the pointp0with the lowestycoordinate.

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Convex hull


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Graham scan: Pick the pointp0with the lowestycoordinate. Sort all the points by polar angle withp0.

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Convex hull


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Graham scan: Pick the pointp0with the lowestycoordinate. Sort all the points by polar angle withp0. Iterate through all the points

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Convex hull


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Graham scan: Pick the pointp0with the lowestycoordinate. Sort all the points by polar angle withp0. Iterate through all the points If the current point forms a clockwise angle with the

last two points, remove last point from the convex set.

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Convex hull


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last two points, remove last point from the convex set. Otherwise, add the current point to the convex set.

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Convex hull


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last two points, remove last point from the convex set. Otherwise, add the current point to the convex set.

Time complexityO(Nlog N).

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Convex hull


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Convex hull


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Convex hull


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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hull

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p01

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Convex hullpoint hull[MAXN];

int convex_hull(vector p) {

int n = size(p), l = 0;


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sort(p.begin(), p.end(), cmp);

for (int i = 0; i < n; i++) {if (i > 0 && p[i] == p[i - 1])

continue;

while (l >= 2 && ccw(hull[l - 2], hull[l - 1], p[i]) >= 0)

l--;

hull[l++] = p[i];

}

int r = l;

for (int i = n - 2; i >= 0; i--) {

if (p[i] == p[i + 1])

continue;

while (r - l >= 1 && ccw(hull[r - 2], hull[r - 1], p[i]) >= 0)

r--;

hull[r++] = p[i];}

return l == 1 ? 1 : r - 1;

}

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Convex hull


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Many other algorithms exist

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Convex hull


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Many other algorithms exist Gift wrapping aka Jarvis march.

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Convex hull


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Many other algorithms exist Gift wrapping aka Jarvis march. Quick hull, similar idea to quicksort.

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Convex hull


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Many other algorithms exist Gift wrapping aka Jarvis march. Quick hull, similar idea to quicksort. Divide and conquer.

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Convex hull


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Many other algorithms exist Gift wrapping aka Jarvis march. Quick hull, similar idea to quicksort. Divide and conquer.

Some can be extended to three dimensions, or higher.

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Point in convex polygonSimple algorithm to check if a point is in a convex polygon.


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We start by calculating the


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We start by calculating thearea of the polygon.

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To check if our point iscontained in the polygon wesum up the area of the

triangles formed the pointand every two adjacentpoints.

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To check if our point iscontained in the polygon wesum up the area of thetriangles formed the pointand every two adjacentpoints.

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y garea of the polygon.


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The total area of the trianglesis equal to the area of the

polygon i the point is insidethe polygon.

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area of the polygon.




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area of the polygon.




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area of the polygon. To check if our point is

contained in the polygon wesum up the area of thetriangles formed the pointand every two adjacentpoints.



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Point in concave polygon


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How about non convex polygon?

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How about non convex polygon? Theeven-odd rulealgorithm.

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How about non convex polygon? Theeven-odd rulealgorithm. We examine a ray passing through the polygon to the

point.

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H b l ?


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How about non convex polygon? Theeven-odd rulealgorithm. We examine a ray passing through the polygon to the

point.

If the ray crosses the boundary of the polygon, then italternately goes from outside to inside, and outside toinside.

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Ray from the outside ofthe polygon to the point.


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C t th b f


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Count the number ofintersection points.

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C t th b f


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If odd, then the point isinside the polygon.

If even, then the point isoutside the polygon.

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Count the number of


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Does not matter whichray we pick.

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Point in concave polygon Ray from the outside of

the polygon to the point. Count the number of


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Closest pair of points

Given a set of points, we want to nd the pair of pointswith the smallest distance between them.

Di id d l ith


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Divide and conquer algorithm;

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Divide and conquer algorithm; Sort points by the x-coordinate.

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Divide and conquer algorithm; Sort points by the x-coordinate. Split the set into two equal sized sets by the vertical

line of the medialxvalue.

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line of the medialxvalue. Solve the problem recursively in the left and right

subset.

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subset. Sort the two subsets by they-coordinate.

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subset. Sort the two subsets by they-coordinate.

Find the smallest distance among the pair of pointswhich lie on dierent sides of the line.

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Takk fyrir nnina!

G i kk l k t !


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Gangi ykkur vel og ga skemmtun prnu morgun!

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Geometry Algorithms