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Introduction

Hi, I am quite sure I made an ignorant misinterpertation. Hence, I will apologize in advance.

The question is about the paper AAM Revisited in respect to equation 30 of section 4.1.2. In it, each (factor of the)Jacobian(delta W/delta x and y) is computed as (1-a-b, 0) and (0, 1-a-b) where a and b are given by equation 25 and 26 in section 4.1.1.

Also, the jacobian is evaluated in respect to vertex xi, assuming there is no alternation to the mesh(given in the paper in page 34 figure 13, precomputation step 4 in which the author states that the jacobians are evaluated in respect to (x;0) where the 0 portion is the variation).

Essentially the xio and xi in equation 25 and 26 should be equivalent as well as yio and yi which is written in the paper as p = 0 as well as q = 0.

Question

Here, I will get to my question. If xi and xio as well as yi and yio are equivalent when computing the factor of a jacobian given in equation 30, given the fact that in such situation a and b will evaluate to 0(as the numerator will become 0) as is evident from equation 25 and 26, shouldn't the values in equation 30 always be (1,0) and (0,1)? Thanks in advance!

Note

As I have not seen any other research paper related questions in stackoverflow(possibly due to my lack of experience), I suspect that this kind of question is best asked on a different site. If that is the case, may I ask any recommandations?

alkasm
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Isamu Isozaki
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  • Can you give a little better explanation of why "xio and xi in equation 25 and 26 should be equivalent as well as yio and yi" because I don't understand why. To be clear, `dW(x;0)/dt` means "the Jacobian of the identity warp" in other words, how each vertex `x` moves under small perturbations of the identity warp. The point of precomputing the Jacobian of the identity warp is that if you know how small perturbations in the identity warp affect the vertices, you know how perturbations affect the vertices under *any* warp by composing them. – alkasm Sep 09 '17 at 23:15
  • Oh I see---yes, you are correct at the vertices, which makes sense---the warp that maps a point to itself is the identity warp. It does say on pg 20 too that "the Jacobian is only non-zero in the triangles around xi. It takes the maximum value of 1 at the vertex xi and decays away linearly." I'm just unsure if, in the actual algorithm on page 34, you're supposed to use the vertices of the base mesh *s0* (hence your question) or of another mesh *s* or of the pixels inside the mesh (and whether those should be pixels inside *s0* or *s*) since I've never implemented AAM. – alkasm Sep 09 '17 at 23:59
  • I think that we use s0 instead of s because as can be seen from equation 2, s is equal to eigenvectors of s times p which is added to so, the mean shape. I am extremely sorry for my late response – Isamu Isozaki Sep 10 '17 at 00:06
  • It appears that it is clear for you, sir, now, but since I did lack explanation, I will do that here. In regards to your first comment, you are correct. Yet they, the components of the Jacobians, can be evaluated by equation 30 where they are respectively (1-a-b, 0) and (0, 1-a-b) where as a and b becomes 0 at p=0, I think the Jacobians are always 1. I think that is the gist of my question. – Isamu Isozaki Sep 10 '17 at 00:09
  • @AlexanderReynolds Thanks for your support! Do you think that all in all, I am not mistaken in thinking that equation 30 always evaluates to 1 when there is no variation? – Isamu Isozaki Sep 10 '17 at 00:13
  • @AlexanderReynolds I am sorry, maybe I misunderstood you. In a step before equation 29, as it is said that s contains x1, y1, x2 and so on, explicitly stating they are vertices, I think they are points on the mesh(face) s. In regards it being s or so, I think it is so because as I mentioned in my first comment, s is given by the addition of so with the eigenvectors of s times p. Since p = 0, I think s = so. I am again sorry for the inconvenience. – Isamu Isozaki Sep 10 '17 at 00:23
  • It's unnecessary to apologize, this site is for asking questions and this stuff can indeed be confusing :). In the algorithm stated, step 1 is to compute the warp, which the section 4.3.4 address. At the end of 4.3.4 is the updating of `p` and `q`. Then in step 4 of the algorithm is when you're finding the Jacobian, so aren't `p` and `q` nonzero at that point? – alkasm Sep 10 '17 at 02:22
  • @AlexanderReynolds I do not think so. The reason is that on page 34 of the paper in figure 13, the Jacobians(both delta N/delta q and delta W/delta p) are mentioned to be pre-computed at (x;0) which means at p = 0 and q = 0. I think this is because the Jacobian is used for the Taylor expansion on page 16 equation 21 where once it is computed at p=0, it is multiplied by the gradient and delta p where only delta p is not a constant(I think). To summarize, I think it is evaluated at p and q = 0 because the Jacobians are computed with p=0 and q=0. – Isamu Isozaki Sep 10 '17 at 02:36
  • These p's and q's are confusing to me since I've never seen the AAM stuff. Is the boldface **p** just the "shape parameters" which **s_i** gets multiplied by like in the first equations in the paper? Because in the literature for Lucas-Kanade by the same authors, **W(x;p)** means the pixel locations **x** transformed by a transformation with parameters **p** (i.e. a homography matrix with elements **p**); are these the same thing? – alkasm Sep 10 '17 at 03:54
  • @AlexanderReynolds they are the same thing. In the Lucas-Kanade algorithm, the x represents the initial shape so where p, as you nicely clarified, is just the coefficient of the shape's eigenvectors. – Isamu Isozaki Sep 10 '17 at 05:30
  • To be more specific, the ps account for facial motion while the qs account for the global transformation of the face(rotation, scaling and translation). – Isamu Isozaki Sep 10 '17 at 05:41

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