Computational_Geometry_Lec1.pdf

Elective III: Computational

Geometry Code: CSEB 604

Pritha Banerjee

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Resources

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Books:

Computational Geometry : Algorithms and

Applications, Mark De Berg, Marc van Kreveld, Mark

Overmars, Otfried Schwarzkopf, Springer

Computational Geometry in C, Joseph O. Rourke,

Cambridge

Lecture notes of David Mount from the web

Convex Hull

Input : a set of points P = {p1, p2, p3pn} on

a plane

Output : CH(P) , Find an unique convex

polygon whose vertices are points from P &

contains all points of P.

CH(P): List the vertices of convex polygon in

clockwise order.

Convex Hull: Naive Algo

Elective III : CG

Consider all nC2 line segments and check if rest of

the other n-2 points are to the right of the chosen line segment

Convex Hull: Naive Method

Input : P = {p1, p2, p3pn} on a plane

Output : List L ,the vertices of convex polygon in

clockwise order.

1. For all ordered pairs (p,q) PxP ( p q)

2. valid = true; {(p,q) is edge of CH}

3. for all points r P p or q

4. if r lies left of line (p,q) (ordered) , edge (p,q)

is not CH edge, valid = false

5. If valid then add (p,q) to E

6. Construct list L of vertices , sorted in clock wise order

Elective III : CG

Naive Algo:Slow Convex Hull All the nC2 line segments => n(n-1) = n

2 n pair of

points

For each pair, check all n-2 (O(n)) points are left/right

(constant time operation)

=> O(n3 )

Last step: edges in E are directed, start from any

edge, put both points in L, take end point, find edge

starting with this in E, put in L..=> n+ n-1 + n-

2=> O(n2 ) { can do O(nlogn) but of no use here)

Total: O(n3 ) + O(n2 ) => O(n3 )

Elective III : CG

Incremental algorithm

Take help of Incremental algorithm : std.

design technique

Add points one at a time in the Polygon ,

updating solution after each addition

Construct upper and lower hull and join them

Sort on x coordinate

Take points from left to right constructing

upper hull

Take points right to left to construct lower

hull

Graham Scan: Incremental algorithm

Elective III : CG

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1, 2

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Elective III : CG

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1, 2, 3,

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14

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1,2,3,

Elective III : CG

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1, 2, 4

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1, 2, 3, 4

Elective III : CG

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1, 2, 4, 5,

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1,2,4

1,2,5

Elective III : CG

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1, 2, 5, 6

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1, 2, 5

1, 2, 5, 6, 7

Elective III : CG

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1, 2, 5,6

1, 2, 5, 6, 7

1, 2, 5, 7

Elective III : CG

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1, 2, 5, 7

1, 2, 5, 7, 8

1, 2, 5, 8

Elective III : CG

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1, 2, 5, 8

1, 2, 5, 8, 9

1, 2, 5, 8, 9, 10

1, 2, 5, 8, 9

1, 2, 5, 8, 10

Elective III : CG

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2

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1, 2, 5, 8, 10

1, 2, 5, 10

Elective III : CG

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1, 2, 5, 10

1, 2, 5, 10, 11

1, 2, 5, 10, 11

1, 2, 5, 10, 11, 12

1, 2, 5, 10, 11, 12, 13

Elective III : CG

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2

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1, 2, 5, 10, 11, 13

Elective III : CG

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1, 2, 5, 10, 11, 13

1, 2, 5, 10, 13

1, 2, 5, 10, 13, 14

Elective III : CG

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1, 2, 5, 10, 14

1, 2, 5, 10, 14, 15

1, 2, 5, 10, 14, 15 Upper Hull constructed

Elective III : CG

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Lower Hull constructed by moving

from right to left in similar fashion


Sort points based on x coordinate: O(n log n)

Append left most points p1, p2 in L_upper

For i = 2 to n

While L_upper contains more than 2 points

Append pi to L_upper if last two points in L_upper and

pi makes right turn at last point of L_upper

Else delete the last point from L_upper

Similarly for Lower Hull L_lower

Start with pn and pn-1 in L_lower

Graham Scan: complexity

Complexity:

Sorting : O(n log n)

What data structure we use ? Stack

How many times an element is inserted or

deleted from stack for constructing L_upper?

Only once=> over the for loop : O(n)

Similarly for L_lower

Total : O(n) + O(n log n) => O(n log n)

Graham Scan: Proof of correctness

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1. Consider p1.. p(i-1) are CH

vertices, adding pi

2. p1.. p(i-1) are hull edges=> all

points lie below the chain

3. When pi is added, this new chain

must be above the old chain

;otherwise, there must be some

point in the slab between p(i-1)

and pi that is violating hull

properties and it does not

represent a correct hull edges

p(i-1)

pi

Jarvis March (Gift Wrapping) :

Output Sensitive Algorithm

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Start with one of the left-most, right-most, top-most,bottom-most point consider a vertical line through this point Rotate clockwise until it hits a point , that becomes next point in CH

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Complexity : Finding one of the extreme point : O(n) Finds next Hull vertex by rotation=> O(n) There are h hull vertex=> O(hn) Time complexity : O(nh), If h is O(log n) => same as Graham Scan

Convex Hull

Input : a set of points P = {p1, p2, p3pn} on

a plane

Output : CH(P) , Find an unique convex

polygon whose vertices are points from P &

contains all points of P.

CH(P): List the vertices of convex polygon in

clockwise order.

Given two CH: how to Find

Combined Hull?

Resources

32

Books:

Computational Geometry in C, Joseph O. Rourke,

Cambridge, Pages: 91-95, Section 3.8 and Pages: 69-

72, Section 3.4

Lecture notes of David Mount from the web, Pages: 14-

17


Combined Hull?

Find upper and lower tangent for both HA and HB disregarding all other

points lying between lower and upper tangents

HA HB

Upper

tangent

Lower

tangent


Combined Hull?

Tangent : if p1 and p3 are on the same side

of the tangent line at p2

p2 p1 p3

HA

HB

ab is Lower Tangent: if point (a-1) , (a+1) are on the left

(Orient() is +ve) of segment ab and point (b-1), (b+1) are also on the left

a

a-1 a+1 b

b-1 b+1

ab is Upper Tangent: if point (a-1) , (a+1) are on the right

(Orient() is -ve) of segment ab and point (b-1), (b+1) are also on the right

How to find lower tangent for two hulls: Check all possible pairs? O(n2)


Combined Hull?

a = right most point of left Hull A

b = left most point of right Hull B

While T = ab is not lower tangent to both A and B

while T not lower tangent to A then a = a-1

while T not lower tangent to B then b = b + 1

Return ab

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HA

HB

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Complexity: finding lower tangent

Finding the leftmost and rightmost point in

HA and HB is : O(n)

Each vertex on the hull vertices are visited

only once: O(n)

Convex hull: divide and conquer

Applicable to any dimension

Like merge sort

Sort points P by x coordinate : O(n log n)

CH(S)

If |P| 3, CH(P) is a triangle

Else partition P into 2 sets A ( lowest x

coordinates) and B ( highest x coordinates)

Recursively compute HA= CH(A) , HB= CH(B)

Merge two hulls into CH by computation of upper and lower tangents of HA and HB

Time complexity

Sorting : O(n log n)

T(n): time complexity for hull algo

Merge step : O(n)

T(n) = 2T(n/2) + O(n) = O(n log n)

Time complexity is O(n log n)

Elective III : CG

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