informally: in an art gallery with n paintings what is the optimal position for a camera? formally:...
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Informally: In an art gallery with n paintings what is the optimal position for a camera?
Formally: Given a set of n points in the Euclidean space, find a point in their convex hull that maximizes the product of the distances to the n given points› Convex hull = minimal convex set that
contains all the points
Simple isosceles triangle with vertices (-1,0), (1,0), and (0,5)
F(x,y) = prod of the distances (squared) from a point (x,y) to the three vertices = [(x+1)2 + (y-0)2] [(x-1)2 + (y-0)2] [(x-0)2 + (y-5)2]
Maximum Modulus Principle tells us that such a point is on the boundary
Let’s call the point that maximizes F(x,y) i.e. the product of the square of the distances, the “art gallery point”
2D – specifically, the case of a triangle› Why is the art gallery point on a boundary?› How do you find it?› Where is it on any given triangle?
3D› Is the art gallery point still on a boundary?› Is everything the same as in 2D?
Start on one side of a triangle and pick a random point on it
Put this random point’s coordinates in terms of one variable, alpha
Create the product of distances function (will be a function dependent solely on alpha)
Maximize it: find the derivative and set it equal to 0› After some cancellation, this will be a cubic equation
Hence, finding the AG point along one side of a triangle amounts to solving for the roots of a cubic polynomial
Repeat the steps on the previous slide for the other two edges
Now we will have 3 possible AG points Compare the actual product of the
distances returned by each point The biggest one(s) wins
Easiest case to study and to check Observation: an equilateral triangle has
3 AG points, which are at the midpoints of each side of the triangle
Proof: verifies that the maximizer of the product of the distances along an edge is at the midpoint
This is where it gets interesting The big question: Is there one AG point
on the base (i.e. at the origin, like we saw with equilaterals), or are there 2 symmetrical ones on the legs?
Fix the height of a triangle on the y-axis
Drop legs to the x-axis symmetrically around the y-axis
Run it up to h = 70 and a = 100 (lots of triangles!)
Question: Per each height, what is the range of a’s for which the origin is the AG?Horizontal distances from y-axis
Fix
ed h
eigh
t
2 0 2 4 6 8 10a
0 .5
1 .0
1 .5
2 .0
2 .5
3 .0
h
0 20 40
20
40
60
80
Observations: Bottom range
adds one to it every 15
Top of upper range is almost linear, goes up by 2’s, except on every 10 it jumps by 3
Bottom of upper range starts at first point where base length exceeds leg lengthFixed height values
Hor
izon
tal d
ista
nces
fro
m y
-axi
s
0 20 40 60
20
40
60
80
100
120
Observations: Bottom point
hovers around 60, except for in the beginning
Upper point hovers around 129 or so
Good news is if your angle is within say 70 to 120, you’re guaranteed that the AG is on the origin
Fixed height values
Non
-isos
cele
s an
gle
of t
he t
riang
le
Stationary point = point where the derivative or gradient = 0
With a triangle, the stationary point of the function p(z) is a saddle point
In 3D, the stationary point is a minimum!
Shows a break between 2D and 3D› Otherwise, Kalantari proves the MMP, under
certain conditions, for the 3D case as well
Still don’t know how to tell when looking at an isosceles triangle where the AG will be…› What explains those observations I listed?› Will there eventually be three ranges? What
might they look like? Is there a way to know when looking at any
triangle, not just isosceles or equilateral? Is there a geometric characterization of the
AG point? What about with more than 3 points? What about 3D (say, a tetrahedron)?
Convex hull Complex numbers Maximum Modulus Principle Gauss-Lucas Theorem Stationary points Newton’s Method Fundamental Theorem of Algebra Good old algebra and calculus
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