This document discusses linear programming approaches for solving casting problems in computational geometry. It first introduces the problem of determining if a polyhedron can be cast based on the angles between facets. This leads to representing the problem as finding the intersection of half-planes, which can be solved using linear programming. The document then provides algorithms for intersecting half-planes and using incremental linear programming to find a solution in linear expected time by randomly ordering the half-plane constraints.

1. LINEAR PROGRAMMING IN
COMPUTATIONAL GEOMETRY
•PROBLEM OF MOLDING
•HALF PLANE INTERSECTION SOLUTION
•INCREMENTAL LINEAR PROGRAMMING SOLUTION
•RANDOMIZATION

2. MANUFACTURING WITH MOLDS
a real life problem of computational geometry
What is mold?
a cavity with same shape as that of object
Restriction
object must have a horizontal top facet
Castable object

3. LEMMA
A polyhedra P can be removed from it’s mold if angle
b/w d and n(f) is at least 90 for each ordinary facet ‘f’ of P
where d is the direction of translation of object
and n(f) is the outward normal of facet f
It leads to the consequence that P can be removed by single
translation if it can be removed by small translations.

4. If we consider the direction of movement as upward from
origin then the d and n can be considered as
d=(dx, dy,1) &
n=(nx, ny, nz).
Now acc. to Lemma, we have
dx nx + dyny+ nz <0
which is nothing but the eqn. of a half plane in plane
Z=1

5. N facet polyhedra will have n-1 such half planes
Common intersection of these planes gives rise to the
region in which the value of
dx and dy lies
empty intersection implies object as uncastable
problem reduces to solving n-1 half plane eqns to get the
common intersection

6. ALGORITHM FOR HALF PLANE INTERSECTION
INPUT := a set H of half planes in the plane
OUTPUT := the convex polygonal region C
ALGORITHM
INTERSECTHALFPLANES(H)
if card(H) = 1
then C← the unique half-plane h ∈ H
←
else Split H into sets H1 and H2 of size n/2 and n/2.
C1 ←INTERSECTHALFPLANES(H1)
C2 ←INTERSECTHALFPLANES(H2)
C←INTERSECTCONVEXREGIONS(C1,C2)
Store C as left and right boundary with sorted list of half planes

7. INTERSECTCONVEXREGIONS(C1,C2)
USE PLANE SWEEP ALGORITHM
ystart <= min(y1,y2)
where y1 and y2 upper end point of C1 and C2
at every event point, new edge e having p as upper
end point appears on boundary
following cases to be tested when e lies on left boundary of C1
continue

8. Case 1

9. Case 2

10. Case 3

11. LINEAR PROGRAMMING:
In case of casting problem, however, we don’t need to know
all solutions to the set of linear constraints; just one solution
will do fine. This allows for a faster algorithm, expected time
is linear.
Finding a solution to a set of linear constraints is LINEAR
PROGRAMMING or LINEAR OPTIMIZATION.

12. BASIC CONSTRUCT OF A LP:
Maximize{Objective function}: c1x1+c2x2+・ ・ ・+cdxd
Subject to {Linear contraints}:
a1,1x1+・ ・ ・+a1,dxd ≤ b1
a2,1x1+・ ・ ・+a2,dxd ≤ b2
...
an,1x1+・ ・ ・+an,dxd ≤ bn

13. Our solution will be the point that maximizes the Objective
Function.
Objective Function can be viewed as a direction c(c1, c2,…, cd )
in the d-dimension plane.
The set of points satisfying the constraints is called feasible
region else infeasible region.
Hence our soln. is the point in the feasible region that is
extreme in the direction c.

15. In case of molding we have n linear constraints in two variables
and we want to find one solution to the set of constraints. We
can do this by taking an arbitrary objective function, and then
solving the linear program defined by the objective function
and the linear constraints.
We use following conventions:
H is the set of n linear constraints i.e, half-planes: h1 , ... , hn
Vector defining objective function: c(cx , cy)
Our goal is to find out point (px , py ) such that cx px +cy py
is max.

16. Possible case of intersection:

17. To make sure that we get a unique soln. we have to impose
restrictions on case ii & iii.
Case ii) we add to our linear program two additional
constraints that will guarantee that the linear program is
bounded. For example, if cx > 0 and cy > 0 we add the
constraints px ≤M and py ≤ M, for some large M ∈ R. let these
constraints be m1 & m2 .
Case iii) we take the lexicographically smallest value.

18. Basis of the algorithm:
Let 1≤ i≤ n, and let Ci and vi be feasible region & optimal point of
each step.Then we have
(i) If vi−1 ∈ hi, then vi = vi−1.
(ii) If vi−1 ∈ hi, then either Ci = 0 or vi ∈ li, where li is the line
bounding hi.

19. Case ii): finding p on li
It can be reduced to 1-D LP problem of finding p on li that
maximizes the OF subject to constraints p ∈ Hi-1 .

20. The interval [ xleft : xright ] is our feasible region and xleft or xright the
optimal soln. .
It take linear time to calculate this ,i.e: O(n).

21. ACTUAL ALGORITHM:

23. RANDOMIZED LINEAR PROGRAMMING:
Worst case time complexity of Increamental LP is O(n2 ) this
can be improved if the order of half planes are suitably changed.
We use a randomize algorithm to get a random sequence of
these half planes. Finally, we get an expected time of O(n).