# Assignment 4 (Written): Conformal Parameterization

The written part of your next assignment, on conformal surface flattening, is now available below. Conformal flattening is important for (among other things) making the connection between processing of 3D surfaces, and existing fast algorithms for 2D image processing. You’ll have the opportunity to implement one of these algorithms in the coding part of the assignment (to be announced soon).

## 10 thoughts on “Assignment 4 (Written): Conformal Parameterization”

1. peiya_wang says:

Is there no assignment 4 coding assignment or will that be posted later? If it will be posted later, when will it be posted?

2. aremondt says:

Is there a typo in exercise 32? If not: I can prove the identity holds when the normal derivative of $u$ vanishes on the boundary, but when it is the normal derivative of $v$ that vanishes on the boundary I can only get $\Delta v$ to show up on the right-hand side (and not $\Delta u$).

1. Josh says:

Good catch. Assume that the normal derivative of $u$ vanishes.

3. intrepidowl says:

I’m getting a sign error when proving that the inner product is Hermitian in Exercise 31, so I get that $$\langle\langle \alpha,\beta\rangle\rangle=-\overline{\langle\langle\beta,\alpha\rangle\rangle}$$. Is anyone else seeing this, and how might I resolve this?

1. intrepidowl says:

This is probably just notational muck, but what is the cross product of imaginary numbers exactly? It’s possible to reduce the wedge in the inner product down to cross products as we saw earlier in the course, but if we “treat complex numbers as vectors in R2” (as written on the top of page 101/102 in the pdf) then this returns a real number. This doesn’t seem to type-check with the fact that the integral of a complex number should return a complex number, otherwise taking the complex conjugate as in $$\overline{\langle \langle \beta,\alpha\rangle\rangle}$$ wouldn’t make any sense.

1. intrepidowl says:

When I calculate $$\langle\langle \alpha,\alpha\rangle\rangle$$, I keep getting zero because I reduce it to $$\overline{\alpha(\mathcal JX)}\times\alpha(\mathcal JX)+\overline{\alpha(X)}\times\alpha(X)$$ which I think exactly cancel each other. Is that the correct expression?

2. intrepidowl says:

Never mind, all my concerns were cleared up by the post on complex differential forms.

4. yyyy says:

For question 41, is it also possible for the y to converge to -x_n ?

1. Josh says:

Yes, good catch! It should say that $y$ converges to either $-x_n$ or $x_n$.