1. Iteration Solution of the Global Illumination Problem László Szirmay-Kalos

2. Solution by iteration l Expansion uses independent samples –no resuse of visibility and illumination information l Iteration may use the complete previous information – L n = L e +  L n-1 –  pixel =  L n

3. Storage of the temporary radiance: finite elements l FEM: l Projecting to an adjoint base:  L j (n) b j (p) =  L j e b j (p) +   L j (n-1) b j (p) L(p)   L j b j (p) L i (n) = L i e +  L j (n-1) p= (x,  )

4. FEM iteration l Matrix form: l Jacobi iteration –Complexity: O(steps · 1 step) = O(c · N 2 ) l Other iteration methods: –Gauss-Seidel iteration: O(c · N 2 ) –Southwell iteration: O(N · N) –Successive overrelaxation: O(c · N 2 ) L (n) = L e + R  L (n-1)

5. Problems of classical iteration l storage complexity of the finite-element representation –4 variate radiance: very many basis functions l error accumulation –   L/(1-q), q  contraction l The error is due to the drastic simplifications of the form-factor computation

6. Stochastic iteration Use instantiations of random operator  * – L n = L e +  * n L n-1 l which behaves as the real operator in average –   * n L   L

7. Example: x = 0.1 x + 1.8 Solution by stochastic iteration l Random transport operator: –x n = T n x n + 1.8, T is r.v. in [ 0, 0.2] l n : 1 2 3 4 5 l T n sequence: 0 0.1 0.2 0.15 0.05 l x n sequence: 0  1.8 1.91 2.18 2.13 1.9 l Not convergent! l Averaging: 1.8 1.85 1.9 2.04 1.96

8. Iteration with a single ray Transfer the whole power from x into  selected with probability:L(x,  ) cos  x   x   x   L e (x,  ) 1 2 3 

9. Making it convergent L n = L e +  * n L n-1  n =  L n is not convergent  pixel =(  L 1 +  L 2 +...+  L m )/m

10. Stochastic iteration with FEM Diffuse case l Projected transport operator: –directional integral of the transport operator –surface integral of the projection l Alternatives: –both explicitely: classical iteration, stochastic radiosity –surface integral explicitely: transillumination radiosity –both implicitely: stochastic ray-radiosity

11. Stochastic radiosity P = P e + HP Random transport operator: H* P| i = H ij  Expected value: E[H*P| i ]=  j H ij   P i /  = H P| i Selects a single (a few) patch with the probability of its relative power and transfers all power from here

12. Transillumination radiosity Selects a single (a few) directions and transfers all power into these directions Projected rendering equation: L = L e + R  L Transport operator: R ij = = f i /A i    Ai  b j (h(x,-  ’  ) cos  ’ dxd  ’ Random transport operator: R ij * = 4  f i /A i  Ai  b j (h(x,-  ’  ) cos  ’dx

13.  Ai  b j (h(x,-  ’  ) cos  ’dx AiAi ’’ A(i,j,  ’) AjAj  Transillumination plane A(i,j,  ’)= projected area of path j, which is visible from path i in direction  ’

14. Stochastic ray radiosity Selects a single (a few) rays (points+dirs) with a probability proportional to the power  cos  /area and transfers all power by these rays P = P e + HP Random transport operator: if y and  are selected: H* P| i = f i  b i (h(y,  )  Expected value of the random transport operator: E[H*P| i ]=  j f i   Aj b i (h(y,  )  cos  /  dy/A j P j /  =  j f i /A j  Aj b i (h(y,  ) cos  dy P j = H P| i

15. Stochastic iteration for the non-diffuse case L n = L e +  * n L n-1 l Reduce the storage requirements of the finite-element representation Search  * which require L not everywhere L n (p n+1 ) = L e (p n+1 )+  * n (p n+1,p n ) L n-1 (p n )

16. Stochastic integration l Projected transport operator: –directional integral of the transport operator –directional-surface integrals of the projection l Alternatives: –all integrals explicitely: classical iteration –all integrals implicitely: iteration with a single ray –directional integral of the transport operator implicitely, integral of the projection explicitely

17. Ray-bundle based iteration L e pixel Storage requirement: 1 variable per patch

18. Finite elements for the positional variation l FEM: l Projected rendering equation: –L(  ’) = L e (  ’) +   F(  ’,  ) A(  ’) L(  ’)d  ’ l Random transport operator: –Select a global direction  ’ randomly: –  * L(  ’) = 4  F(  ’,  ) A(  ’) L(  ’) L(x,  )   L j (  ) b j (x)

19. Ray-bundle iteration Generate the first random direction   FOR each patch i L[i] = L e (  1 ) FOR m = 1 TO M Reflect incoming radiance L to the eye and add contribution/M to Image Generate random global direction  m+1 L = L e (  m+1 )+ 4  F(  m,  m+1 ) A(  m ) L(  m ) ENDFOR Display Image

20. Ray-bundle images 10k patches 500 iterations 9 mins 60k patches 600 iterations 45 mins 60k patches 300 iterations 30 mins

21. Can we use quasi-Monte Carlo samples in iteration? 1/(M-1)   *(p i )  *(p i-1 ) L e    2 L e 1/(M-1)  f(p i,p i-1 )   f(x,y) dxdy p i must be infinite-distribution sequence!

22. Future improvements ? l Problem formulation –Monte-Carlo integral –Expansion versus iteration l Same accuracy with fewer samples –importance sampling –very uniform sequences, stratification l Making the samples cheaper –fast visibility computations –global methods: coherence principle