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I'm reading Advanced Global Illumination.

Here is the part confusing me:

Advanced Global Illumination 2nd Edition, page 34

  1. What do the second equation and $\delta$-function mean?

  2. Why the third equation is a sufficient condition even though a reason is given?

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  • $\begingroup$ I'm working through the same book and coincidentally went through that chapter (again) in the last few days. Are you aware of the UC Davis lectures on YouTube that use this book? Unfortunately the lecturer doesn't address your questions specifically in the BRDF lecture. I was formulating an answer for you but I'm not fully understanding it either, so I'll not reply as I don't want to misinform. I will try to write my thoughts about it when I have time though... $\endgroup$ – PeteUK Feb 24 '17 at 23:12
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You are right to be confused. What I think they should have written: $$ L( x \leftarrow \Psi ) = L_{in} \delta(\Psi - \alpha) $$ using $ \alpha $ instead of $ \Theta $, which is already used as a dummy variable in the integral.

You should look up the Dirac delta function to learn its meaning and properties. In this context, you can imagine the $ L$ above as representing a very concentrated beam (a laser) coming from the angle $ \alpha $. Practically, to do the integral over $ \Psi $ with $ \delta (\Psi - \alpha) $ present in the integrand, remove the integration and replace all occurrences of $ \Psi $ with $ \alpha $. Then it should be clear how they arrive at the next line, which should read, for all $ \alpha $: $$ \int f_r(x, \alpha \rightarrow \Theta) \; \cos(N_x, \Theta)\; d \omega_\Theta \leq 1 $$.

The fact that this condition is sufficient follows from two facts: 1. that any function (e.g. $ L $) can be approximated by a sum of many $ \delta $ functions, and 2. that everything is linear. In other words, if I write $ N(L) $ and $ D(L) $ for the numerator and denominator of the left hand side of (2.21), then you can see that if $ L = L_1 + L_2 $, then $ N(L_1+L_2) = N(L_1) + N(L_2) $ and $ D(L_1+L_2) = D(L_1) + D(L_2) $. So if I know $ N(L_1) \leq D(L_1) $ and $ N(L_2) \leq D(L_2) $, then I know $ N(L) \leq D(L) $ since $$ N(L) = N(L_1) + N(L_2) \leq D(L_1) + D(L_2) = D(L) $$.

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