Tutorial 9 - Fast Marching Methods

Transcription

1 Numerical Geometry of Images Tutorial 9 Fast Marching Methods c 2012

2 Why measure distances? Shape analysis Image analysis Registration Morphology Navigation And many other uses.. There are many (mostly 2D) fast Euclidean distance transforms - see the site links for more examples.

3 How to measure distances? Given a domain Ω (for a moment assume that Ω R 2 ) and a set of source points X 0, for each point x Ω, measure its distance from X 0, T (x). Formally, the problem boils down to solving the eikonal equation (from Greek εικων= image). Ω T (x, y) = 1, T (X 0 ) = 0.

4 In optics and acoustics this equation describes the propagation of electromagnetic or pressure waves through some (non-homogenous) medium. It is responsible for the fact that the light (sound) traverses the shortest path between two points (Fermat s principle) and as consequence it describes light refraction and Snell s law.

5 Fermat s principle - a simple example..

6 Fermat s principle - a not-so-simple example..

7 The Fast Marching algorithm Imagine that Ω is a uniformly distributed forest. At time T = 0, a bad guy sets fire simultaneously in all the source points X 0. The fire front advances outwards from these initial points. Firemen register the time T (x) at which the fire arrives to the location x. Once a point is touched by the fire, the trees burn out and the front never visits again the locations where it had already passed. The fast marching algorithm is a numerical tool, which simulates the described scenario.

8 The Fast Marching algorithm A few observations when computing a solution to the Eikonal equation: Our boundary is also our zero level set - the set of source points. This uses one property of distances. The construction of the solution should be made monotonous, going from closer points to farther points - we are validating the triangle inequality on several stencils. This is another property of distances. The third property of distances (symmetry) is implicitly assumed.

9 The equation, as a solution to the minimum distance problem, is well defined in terms of narrow band around a level set.

10 The Fast Marching algorithm Initialization 1. Set T (X 0 ) = 0 and mark the points from X 0 as Black. 2. Set the rest of the points T (Ω \ X 0 ) = and mark them X 0 as Green. Iteration 1. Green points adjacent to Black points become Red. 2. Red points are updated, i.e. their T is estimated from the arrival times of the neighboring Black points. 3. The Red point with the smallest T becomes Black. 4. The process reiterates until all points are Black.

11 Note that a point is updated only by adjacent points (4-neighbor adjacency) with smaller value of T, and we demand that the new T value is larger than that of the updating points. A Black point has the T value smaller than that of all non-black points. Therefore, once a point becomes Black, it need not be updated anymore. This can also be thought of in terms of updating an upper and a lower bound until they (almost) meet. The maximum number of point updates is the number of neighbors on the grid, which is O(N).

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13 The update step T 1 T 3 =? T 2 When we say update x 3 from x 1, x 2, we mean: Given two points x 1, x 2 with known front arrival times T 1, T 2, estimate the arrival time T 3 to the point x 3.

14 Assume that the front is planar (a good approximation for T h), i.e. propagating from some plane ˆn x + p = 0. The points x i are distant T i from the plane. The knowledge of T 1, T 2 allows to determine the parameters of the plane ˆn, p. (note: there are algorithms which do not assume a planar wavefront).

15 Assume w.l.o.g. that x 1 = (0, 0), x 2 = ( 2h, 0), x 3 = ( h, 2 h) x 3 x 1 x 2

16 We have: T 1 = ˆn x 1 + p = p T 2 = ˆn x 2 + p = 2hn x + T 1 from where n x = T 2 T 1 2h n y = ± 1 nx 2 = ± 1 2h 2 (T 2h 1 T 2 ) 2.

17 Since x 3 is above x 1 and x 2 w.r.t. the y-axis, the solution n y = 1 2h 2 (T 2h 1 T 2 ) 2 corresponds to the source, from which the front arrives to x 3 before it arrives to x 1 and x 2, yielding T 3 < T 1, T 2. We discard this solution. T 3 = ˆn x 3 + p = 2 2 hn x hn y + T 1 = 1 2 (T 2 T 1 ) h 2 (T 2 T 1 ) 2 + T 1 = 1 2 (T 2 + T 1 ) h 2 (T 2 T 1 ) 2 (T 1 + T 2 + ) 2h 2 (T 2 T 1 ) 2. = 1 2

18 If T 1 T 2 > h, the wavefront propagation direction n comes outside of the triangle x 1 x 2 x 3 and arrives to x 3 before arriving to x 1 or x 2. This can be see from the algebra. Assume T 1 > T 2 and T 1 > T 2 + h, T 3 = 1 2 (T 1 + T 2 ± ) 2h 2 (T 2 T 1 ) 2 < 1 2 (T 1 + T 2 + h) < 1 2 (2T 2 + 2h) = T 2 + h < T 1. In this case, we have no choice but to update x 3 according to T 3 = min{t 1, T 2 } + h.

19 Extension to the weighted Euclidean case When the wavefront has non-uniform propagation velocity V (x, y), a weighted distance map is computed by solving the eikonal equation or V (x, y) T (x, y) = 1 T (x, y) = F (x, y), where F (x, y) = V 1 (x, y) > 0 is the slowness field.

20 This equation can be reformulated as g T (x, y) = 1, where the front arrival time is measured according to the weighted Euclidean metric, dt = dx 2 + dy 2 F (x, y). In the weighted case, the time gained by traversing the distance h when updating x 3 from the neighboring points x 1, x 2 is not h but hf (x, y). Hence, T 3 = 1 2 (T 1 + T 2 + ) 2F 2 h 2 (T 2 T 1 ) 2, valid for T 1 T 2 hf.

21 Extension to (acute) triangulated manifolds The eikonal equation can be solved on general surfaces, represented as triangulated meshes (polyhedral or piecewise-linear approximation). The algorithm is very similar, yet now each vertex on the mesh is updated from the triangles built upon it. We first deal with triangles with acute angles.

22 E t C h F D H u B G A C θ a h φ G b B D A

23 We are looking for t = EC, where t > u Note: CD = b t u t Using the law of cosines: Using the law of sines: BD 2 = CD 2 + a 2 2aCDcosθ sinϕ CD = sinθ BD

24 Looking at the triangle CBG, we have = setting h in the equation h = asinϕ = a CD BD sinθ acdsinθ CD 2 + a 2 2aCDcosθ t u h = F We obtain: (t u) CD 2 + a 2 2aCDcosθ acdsinθ (t u) CD 2 + a 2 2aCDcosθ ab (t u) t sinθ = F = F

25 CD 2 + a 2 2aCDcosθ ab 1 t sinθ = F ( ) (t u) 2 ( b + a t 2 2a b ( b (t u) t ) 2 + a 2 2a ( b (t u) t (t u) t ) cosθ = Fab 1 t sinθ ) cosθ = F 2 a 2 b 2 1 t 2 sin2 θ

26 (b (t u)) 2 + a 2 t 2 2tab (t u) cosθ = F 2 a 2 b 2 sin 2 θ ( b 2 t 2 2tu + u 2) ( ) +a 2 t 2 2 t 2 tu abcosθ = F 2 a 2 b 2 sin 2 θ ( ) t 2 a 2 + b 2 2abcosθ + t (2uabcosθ 2ub 2) ( ) +b 2 u 2 F 2 a 2 sin 2 θ = 0

27 Note that t > u according to our assumptions, and CD should stay within the segment CA. By changing the direction of h we obtain: acosθ < b (t u) t < a cosθ

28 The discussion so far has been for non-obtuse triangulations. If obtuse triangles exist, they can be split into 2 by adding an extra edge (and 2 virtual triangles) to the triangulation graph (which now becomes non-planar, but still works for our purposes) Obtuse angle and splitting section A Triangulated surface patch A B A B Unfolded triangulated patch