Spherical Harmonics comparison

You could take the difference between the coefficients of the two, then integrate over the sphere with these new coefficients, but using the absolute value of the function, to get the L1 distance. To integrate I'd say probably just do a basic Monte Carlo integration by having a set of a few dozen or so points on the unit sphere that you can plug into the resulting difference SH function. This should work because the L1 distance between two functions is something like

[eqn]\iint_S | \hat{f}(s) - \hat{g}(s) | ds[/eqn]

where [eqn]\hat{f}(s) = \sum_{i} c_i y_i(s)[/eqn] and [eqn]\hat{g}(s) = \sum_{i} d_i y_i(s)[/eqn]

[eqn]\Rightarrow \hat{f}(s) - \hat{g}(s) = \sum_{i}(c_i - d_i) y_i(s)[/eqn]

Here, c and d are your coefficients, and the y's are the SH basis functions. So a Monte Carlo integration would be something like

[eqn]\frac{1}{N} \sum_{j}^{N} |\sum_{i}(c_i - d_i) y_i(x_j)| w(x_j)[/eqn]

for a set of N points (uniformly distributed) on the unit sphere. Here w(x) is a weight function, which would be equal to 4pi if you use a uniform distribution on the sphere.

You could also replace taking the absolute value with an L2 norm to get the L2 distance. I think I got all that right... hope that helps.

--Buzzy

It helps a lot !
Many thanks.
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