Monte Carlo Integration COS 323

Transcription

1 Monte Carlo Integration COS 323

2 Integration in d Dimensions? One option: nested 1-D integration f(x,y) g(y) y f ( x, y) dx dy ( ) = g y dy x Evaluate the latter numerically, but each sample of g(y) is itself a 1-D integral, done numerically

3 Integration in d Dimensions? Midpoint rule in d dimensions? In 1D: (b-a)/h points In 2D: (b-a)/h 2 points In general: O(1/h d ) points Required # of points grows exponentially with dimension, for a fixed order of method Curse of dimensionality Other problems, e.g. non-rectangular domains

4 Rethinking Integration in 1D 1 f ( x) dx =? 0 f(x) x=1 Slide courtesy of Peter Shirley

5 We Can Approximate 1 1 = 0 f ( x) dx g( x) dx 0 f(x) g(x) x=1 Slide courtesy of Peter Shirley

6 Or We Can Average 1 f ( x) dx = E( f ( x)) 0 f(x) E(f(x)) x=1 Slide courtesy of Peter Shirley

7 Estimating the Average 1 0 f (x)dx 1 N N i=1 f (x i ) f(x) E(f(x)) Slide courtesy of Peter Shirley

8 Other Domains b a f (x)dx b a N N i=1 f (x i ) f(x) < f > ab x=a x=b Slide courtesy of Peter Shirley

9 Monte Carlo Integration No exponential explosion in required number of samples with increase in dimension (Some) resistance to badly-behaved functions Le Grand Casino de Monte-Carlo

10 Variance b a Var N b a N i= 1 f ( x) dx f ( x i ) = = b a N N ( b a) N i= 1 2 b a N 2 f ( x N i= 1 i ) Var[ Var[ f ( x f ( x i )] * 1 i )] N with a correction of N- (consult a statistician for details) Variance decreases as 1/N Error of E decreases as 1/sqrt(N)

11 Variance Problem: variance decreases with 1/N Increasing # samples removes noise slowly E(f(x))

12 Variance Reduction Techniques Problem: variance decreases with 1/N Increasing # samples removes noise slowly Variance reduction: Stratified sampling Importance sampling

13 Stratified Sampling Estimate subdomains separately E k (f(x)) M 1 M k Can do this recursively!

14 Stratified Sampling This is still unbiased E k (f(x)) E = k vol(m ) j f (x ) jn j =1 N j N j n =1 M 1 M k

15 Stratified Sampling Less overall variance if less variance in subdomains Var[ E] = k j =1 vol(m j ) 2 N j Var[ f (x)] M j E k (f(x)) M 1 M k Total variance minimized when number of points in each subvolume M j proportional to error in M j.

16 Importance Sampling Put more samples where f(x) is bigger E(f(x)) f (x)dx = 1 N Y i Ω N i=1 where Y i = f (x i ) p(x i ) and x i drawn from P(x)

17 Importance Sampling This is still unbiased E(f(x)) [ Y ] E i = = Ω Ω Y ( x) p( x) dx f ( x) p( x) dx p( x) = f ( x) dx Ω for all N

18 Importance Sampling Variance depends on choice of p(x): E(f(x)) Var(E) = 1 N N n =1 f (x n ) p(x n ) 2 E 2

19 Importance Sampling Zero variance if p(x) ~ f(x) E(f(x)) p( x) = cf ( x) Y i = f ( x p( x Var( Y ) i i = ) 1 = ) c 0 Less variance with better importance sampling

20 Random number generation

21 True random numbers

22 Generating Random Points Uniform distribution: Use pseudorandom number generator 1 Probability 0 Ω

23 Pseudorandom Numbers Deterministic, but have statistical properties resembling true random numbers Common approach: each successive pseudorandom number is function of previous

24 Desirable properties Random pattern: Passes statistical tests (e.g., can use chi-squared) Long period: As long as possible without repeating Efficiency Repeatability: Produce same sequence if started with same initial conditions (for debugging!) Portability

25 Linear Congruential Methods x n+ Choose constants carefully, e.g. a = b = c = 2 32 Results in integer in [0, c) ( ax b) mod c = + 1 n Simple, efficient, but often unsuitable for MC: e.g. exhibit serial correlations

26 Problem with LCGs

27 Lagged Fibonacci Generators Takes form x n = (x n-j «x n-k ) mod m, where operation «is addition, subtraction, or XOR Standard choices of (j, k): e.g., (7, 10), (5,17), (6,31), (24,55), (31, 63) with m = 2 32 Proper initialization is important and hard Built-in correlation! Not totally understood in theory (need statistical tests to evaluate)

28 Seeds Why? Approaches: Ask the user (for debugging) Time of day True random noise: from radio turned to static, or thermal noise in a resistor, or

29 Seeds Lava lamps! o_rand.html?id=ou0gaaaaebaj

30 Pseudorandom Numbers Most methods provide integers in range [0..c) To get floating-point numbers in [0..1), divide integer numbers by c To get integers in range [u..v], divide by c/(v u+1), truncate, and add u Better statistics than using modulo (v u+1) Only works if u and v small compared to c

31 Generating Random Points Uniform distribution: Use pseudorandom number generator 1 Probability 0 Ω

32 Sampling from a non-uniform distribution Specific probability distribution: Function inversion Rejection f (x)

33 Sampling from a non-uniform distribution Inversion method Integrate f(x): Cumulative Distribution Function f (x) 1 f ( x) dx

34 Sampling from a non-uniform distribution Inversion method Integrate f(x): Cumulative Distribution Function Invert CDF, apply to uniform random variable f (x) 1 f ( x) dx

35 Sampling from a non-uniform distribution Specific probability distribution: Function inversion Rejection f (x)

36 Sampling from a non-uniform distribution Rejection method Generate random (x,y) pairs, y between 0 and max(f(x))

37 Sampling from a non-uniform distribution Rejection method Generate random (x,y) pairs, y between 0 and max(f(x)) Keep only samples where y < f(x) Doesn t require cdf: Can use directly for importance sampling.

38 Example: Computing pi

39 With Stratified Sampling

40 Monte Carlo in Computer Graphics

41 or, Solving Integral Equations for Fun and Profit

42 or, Ugly Equations, Pretty Pictures

43 Computer Graphics Pipeline Modeling Animation Rendering Lighting and Reflectance

44 Rendering Equation L o Light ( x, ω') = L ( x, ω') + L ( x, ω) f ( x, ω, ω')( ω n) dω e dω ω Ω i Surface n x r ω ' Viewer Ω [Kajiya 1986]

45 Rendering Equation L o ( x, ω') = L ( x, ω') + L ( x, ω) f ( x, ω, ω')( ω n) dω e Ω i r This is an integral equation Hard to solve! Can t solve this in closed form Simulate complex phenomena Heinrich

46 Rendering Equation L o ( x, ω') = L ( x, ω') + L ( x, ω) f ( x, ω, ω')( ω n) dω e Ω i r This is an integral equation Hard to solve! Can t solve this in closed form Simulate complex phenomena Jensen

47 Monte Carlo Integration f(x) 1 N 1 f ( x) dx N 0 i= 1 f ( x i ) Shirley

48 Estimate integral for each pixel by random sampling Monte Carlo Path Tracing

49 Global Illumination From Grzegorz Tanski, Wikipedia

50 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics

51 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Eye Pixel Caustics x Surface LP = L(x S e)da

52 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics Eye Light x x Surface L(x,w) = L e(x,x e) + S f r ( x, x x, x e) L(x x)v ( x, x ) G( x, x ) da

53 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics Herf L(x,w) = L e(x,x e) + S f r ( x, x x, x e) L(x x)v ( x, x ) G( x, x ) da

54 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics Eye ω Surface ω Light L (x,w) o = L e(x,w) + Ω f r ( x, w, w) L (x,w )( w n) dw i x Surface

55 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics L (x,w) o = L e(x,w) + Ω f r ( x, w, w) L (x,w )( w n) dw i Debevec

56 Monte Carlo Global Illumination Rendering = integration Antialiasing Specular Surface Light Soft shadows Indirect illumination Caustics Eye ω ω L (x,w) o = L e(x,w) + Ω f r ( x, w, w) L (x,w )( w n) dw i x Diffuse Surface

57 Monte Carlo Global Illumination Rendering = integration Antialiasing Soft shadows Indirect illumination Caustics L (x,w) o = L e(x,w) + Ω f r ( x, w, w) L (x,w )( w n) dw i Jensen

58 Challenge Rendering integrals are difficult to evaluate Multiple dimensions Discontinuities Partial occluders Highlights Caustics L(x,w) = L e(x,x e) + S f r ( x, x x, x e) L(x Drettakis x)v ( x, x ) G( x, x ) da

59 Challenge Rendering integrals are difficult to evaluate Multiple dimensions Discontinuities Partial occluders Highlights Caustics L(x,w) = L e(x,x e) + S f r ( x, x x, x e) L(x Jensen x)v ( x, x ) G( x, x ) da

60 Monte Carlo Path Tracing Big diffuse light source, 20 minutes Jensen

61 Monte Carlo Path Tracing 1000 paths/pixel Jensen

62 Monte Carlo Path Tracing Drawback: can be noisy unless lots of paths simulated 40 paths per pixel: Lawrence

63 Monte Carlo Path Tracing Drawback: can be noisy unless lots of paths simulated 1200 paths per pixel: Lawrence

64 Reducing Variance Observation: some paths more important (carry more energy) than others For example, shiny surfaces reflect more light in the ideal mirror direction Idea: put more samples where f(x) is bigger

65 Importance Sampling Idea: put more samples where f(x) is bigger 1 f ( x) dx 0 i= 1 Y i = f ( x p( x i i N 1 = Y N ) ) i

66 Effect of Importance Sampling Less noise at a given number of samples Uniform random sampling Importance sampling Equivalently, need to simulate fewer paths for some desired limit of noise