 Invariants (continued) Summary

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 tarix 17.11.2018 ölçüsü 158 Kb. #80056 • Planar Object, remove column 3. Planar Invariants  • Solution: Homogenous coordinates.

• Represent points in plane as (x,y,w)
• (x,y,w), (kx, ky, kw), (x/w, y/w, 1) represent same point.
• If we think of these as points in 3D, they lie on a line through origin. Set of 3D points that project to same 2D point. Perspective motion and projection For Planar Objects • Form a group.

• They can be composed
• They have inverses.
• Projective transformations equivalent to set of images of images. • Strategy.

• Suppose P represents five points. V1 transforms P so that first 4 to canonical position, and fifth to (a,b,c).
• Next, suppose we are given TP, with T unknown. Find V2 to transform first 4 points of TP to canonical position.
• V2 = V1*T-1. V2P has fifth point = (a,b,c).
• For this to work, V1, V2 must be uniquely determined. Transform to Canonical Position • Note that this is equivalent to what we did in the affine case. Affine coordinates are coordinates of 4th point after first three are transformed to (0,0), (1,0), (0,1). • This cross-ratio is invariant to projective transformations. • p=(x,y,1)T is on line if =0. • This follows from the fact that the cross product is orthogonal to both lines. • Note that a projective transformation can map this to another line, the horizon, which we see. • So parallel lines are affine invariants, since they continue to intersect at infinity. • Invariance isn’t captured by mathematical definition of invariance because 3D to 2D transformations don’t form a group.

• You can’t compose or invert them.

• f is a non-trivial invariant if there exist two image I1 and I2 such that f(I1)~=f(I2). • Theorem: Valid objects are any 3D point sets of size k, for some k. There are no non-trivial invariants of the images of these objects under perspective projection. • So for any pair of images, I, J, from any two objects, f(I) = f(J). • If two objects are identical except for one point, they produce the same image when viewed along a line joining those two points.

• Along that line, those two points look the same.
• The remaining points always look the same. • If I and J are images of the same object with the same pose, but different lighting, f(I)=f(J). • light Lambertian + Point Source • i(x,y)/i(x’,y’)= a(x,y)/a(x’,y’) is invariant to lighting changes. • a(x,y)ny(x,y) = J(x,y). • 3-D objects have no invariants.

• We can deal with this by focusing on planar portions of objects.
• Or special restricted classes of objects.
• Or by relaxing notion of invariants.
• However, invariants have become less popular in computer vision due to these limitations. Yüklə 158 Kb.

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