Rendering: Monte Carlo Integration II Bernhard Kerbl Research - PowerPoint PPT Presentation

න Rendering: Monte Carlo Integration II Bernhard Kerbl Research Division of Computer Graphics Institute of Visual Computing & Human-Centered Technology TU Wien, Austria With slides based on material by Jaakko Lehtinen, used with permission

Today’s Goal Integrating the cosine-weighted radiance 𝑀 𝑗 (𝑦, 𝜕) at a point 𝑦 Integral of the light function over the hemisphere, w.r.t. direction/solid angle at 𝜕 Let’s find a solution! How do we integrate over the hemisphere? How do we do it smartly ? Rendering – Monte Carlo Integration I 2

Sampling a Unit Disk Imagine we have a disk-shaped surface with radius 𝑠 = 1 that registers incoming light (color) from directional light sources As an exercise, we want to approximate the total incoming light over the disk’s surface area r = 1 2 We integrate over an area of size 𝜌 We will use the Monte Carlo integral for that 2 Rendering – Monte Carlo Integration II 3

Uniformly Sampling the Unit Disk If we can manage to uniformly sample the disk, then we can compute the Monte Carlo integral as a simple average × 𝜌 By drawing uniform samples in 𝑦 and 𝑧 , we cannot cover the area precisely Inscribed square: information lost Circumscribed square: unnecessary samples Rendering – Monte Carlo Integration II 4

Uniformly Sampling the Unit Disk If we can manage to uniformly sample the disk, then we can compute the Monte Carlo integral as a simple average By drawing uniform samples in 𝑦 and 𝑧 , we cannot cover the area precisely Inscribed square: information lost Circumscribed square: unnecessary samples Rendering – Monte Carlo Integration II 5

Uniformly Sampling the Unit Disk If we can manage to uniformly sample the disk, then we can compute the Monte Carlo integral as a simple average By drawing uniform samples in 𝑦 and 𝑧 , we cannot cover the area precisely Inscribed square: information lost Circumscribed square: unnecessary samples Rendering – Monte Carlo Integration II 6

Uniformly Sampling the Unit Disk If we can manage to uniformly sample the disk, then we can compute the Monte Carlo integral as a simple average By drawing uniform samples in 𝑦 and 𝑧 , we cannot cover the area precisely Inscribed square: information lost Circumscribed square: unnecessary samples This is actually somewhat ok! Rendering – Monte Carlo Integration II 7

Rejection Sampling Requires a PDF 𝑞 𝑌 (𝑦) and a constant 𝑑 such that 𝑔 𝑦 < 𝑑𝑞 𝑌 (𝑦) Draw 𝜊 𝑗 and 𝑌 𝑗 from their respective distributions. If the point (𝑌 𝑗 , 𝜊 𝑗 𝑑𝑞 𝑌 𝑌 𝑗 ) lies under 𝑔 𝑦 , then the sample is accepted loop forever: sample 𝑌 from 𝑞 𝑌 ’s distribution if 𝜊 𝑗 ⋅ 𝑑𝑞 𝑌 𝑌 𝑗 < 𝑔 𝑌 𝑗 then return 𝑌 𝑗 Rendering – Monte Carlo Integration II 8

Rejection Sampling Requires a PDF 𝑞 𝑌 (𝑦) and a constant 𝑑 such that 𝑔 𝑦 < 𝑑𝑞 𝑌 (𝑦) Draw 𝜊 𝑗 and 𝑌 𝑗 from their respective distributions. If the point (𝑌 𝑗 , 𝜊 𝑗 𝑑𝑞 𝑌 𝑌 𝑗 ) lies under 𝑔 𝑦 , then the sample is accepted loop forever: sample 𝑌 from 𝑞 𝑌 ’s distribution if 𝜊 𝑗 ⋅ 𝑑𝑞 𝑌 𝑌 𝑗 < 𝑔 𝑌 𝑗 then return 𝑌 𝑗 𝑗𝑔 ( 𝑦 2 + 𝑧 2 ≤ 1) Unit disk: 𝑔 𝑦, 𝑧 = ൝1 , 𝑞 𝑦, 𝑧 = 1 4 , 𝑑 = 4 0 𝑝𝑢ℎ𝑓𝑠𝑥𝑗𝑡𝑓 Rendering – Monte Carlo Integration II 9

Back to the Unit Disk We do not want to waste samples if we can avoid it Instead, find a way to generate uniform samples on the disk Second attempt: draw from 2D polar coordinates Polar coordinates defined by radius 𝑠 ∈ [0,1) and angle 𝜄 ∈ [0,2𝜌) Transformation to cartesian coordinates: 𝑦 = 𝑠 sin 𝜄 y = 𝑠 cos 𝜄 Rendering – Monte Carlo Integration II 10

Uniformly Sampling the Unit Disk? Convert two 𝜊 to ranges 0, 1 , [0,2𝜌) for polar coordinates Convert to cartesian coordinates void sampleUnitDisk() { std::default_random_engine r_rand_eng(0xdecaf); std::default_random_engine theta_rand_eng(0xcaffe); std::uniform_real_distribution<double> uniform_dist(0.0, 1.0); for (int i = 0; i < NUM_SAMPLES; i++) { auto r = uniform_dist(r_rand_eng); auto theta = uniform_dist(theta_rand_eng) * 2 * M_PI; auto x = r * sin(theta); auto y = r * cos(theta); samples2D[i] = std::make_pair(x, y); } } Rendering – Monte Carlo Integration II 11

Clumping We successfully sampled the unit disk in the proper range 1 However, the distribution is not uniform with respect to the area 0,5 0 Samples clump together at center -0,5 Averaging those samples will give us a skewed result for the integral! -1 -1 -0,5 0 0,5 1 Rendering – Monte Carlo Integration II 12

Uniformly Sampling the Unit Disk: A Solution The area of a disk is proportional to 𝑠 2 , times a constant factor 𝜌 If we see the disk as concentric rings of width Δ𝑠 , the 𝑘 inner rings 2 𝑠 𝑘 up to radius 𝑠 𝑘 = 𝑘Δ𝑠 should contain 𝑂 out of 𝑂 total samples 𝑠 𝑗 Conversely, the 𝑗 𝑢ℎ sample should lie in the ring at radius 𝑠 𝑗 = 𝑠 𝑂 Since 𝜊 is uniform in [0, 1) , we can switch 𝑘 𝑂 for 𝜊 to get 𝑠 𝑗 = 𝑠 𝜊 𝑗 Rendering – Monte Carlo Integration II 13

Uniformly Sampling the Unit Disk: A Solution 1 It works, and it is not even a bad way to arrive at the correct solution 0,5 0 -0,5 However, for more complex scenarios, we -1 might struggle to find the solution so easily -1 -0,5 0 0,5 1 1 With the tools we introduced earlier, we can 0,5 formalize this process for arbitrary setups 0 -0,5 -1 -1 -0,5 0 0,5 1 Rendering – Monte Carlo Integration II 14

Polar To Cartesian Coordinates Let’s transform a regular grid from polar to cartesian coordinates 𝜄 𝑠 𝑌 = 𝑠 cos(𝜄), 𝑍 = 𝑠 sin(𝜄) Rendering – Monte Carlo Integration II 15

First Attempt to Learn the PDF Take 100k samples, transform and see which box they end up in ξ 1 , ξ 2 𝑌 = 𝜊 1 cos(2𝜌𝜊 2 ), 𝑍 = 𝜊 1 sin(2𝜌𝜊 2 ) Rendering – Monte Carlo Integration II 16

Knowing the PDF If we know the effect of a transformation 𝑈 on the PDF, we can Use it in the Monte Carlo integral to weight our samples, or Compensate to get a uniform sampling method after transformation 𝐽𝑜𝑞𝑣𝑢 (𝜊 1 , 𝜊 2 ) 𝐷𝑏𝑠𝑢𝑓𝑡𝑗𝑏𝑜 (𝑦, 𝑧) 𝑄𝑝𝑚𝑏𝑠 (r, 𝜄) 𝑈(𝑠, 𝜄) Rendering – Monte Carlo Integration II 17

Computing the PDF after a Transformation Assume a random variable 𝐵 and a bijective transformation 𝑈 that yields another variable 𝐶 = 𝑈 𝐵 Bijectivity dictates that 𝑐 = 𝑈(𝑏) must be either monotonically increasing or decreasing with 𝑏 This implies that there is a unique 𝐶 i for every 𝐵 i , and vice versa In this case, the CDFs for the two variables fulfill 𝑄 𝐶 𝑈(𝑏) = 𝑄 𝐵 (𝑏) Rendering – Monte Carlo Integration II 18

Computing the PDF after a Transformation 𝑒𝑄 𝐶 (𝑐) = 𝑒𝑄 𝐵 (𝑏) If 𝑐 = 𝑈(𝑏) and 𝑐 increases with 𝑏 , we have: 𝑒𝑏 𝑒𝑏 𝑒𝑄 𝐶 (𝑐) = 𝑒𝑄 𝐵 (𝑏) If 𝑐 decreases with 𝑏 (e.g. 𝑐 = −𝑏) , we have: − 𝑒𝑏 𝑒𝑏 Since 𝑞 𝐶 is the non-negative derivative of 𝑄 𝐶 , we can rewrite as: 𝑒𝑐 𝑞 𝐶 𝑐 𝑒𝑏 = 𝑞 𝐵 𝑏 , 𝑉𝑡𝑗𝑜𝑕: 𝑒𝑄 𝑌 𝑦 = 𝑞 𝑌 𝑦 𝑒𝑦 𝑒𝑧 𝑒𝑧 −1 𝑞 𝐶 𝑐 = 𝑒𝑐 𝑞 𝐵 𝑏 𝑒𝑏 Rendering – Monte Carlo Integration II 19

Computing the PDF after a Transformation −1 𝑒𝑐 Let’s interpret 𝑞 𝐶 𝑐 = 𝑞 𝐵 𝑏 𝑒𝑏 −1 𝑒𝑐 It is the probability density of 𝐵 , multiplied by 𝑒𝑏 −1 𝑒𝑐 has two intuitive interpretations: 𝑒𝑏 the change in sampling density at point 𝑏 if we transform 𝑏 by 𝑈 or , the inverse change in the volume of an infinitesimal hypercube at point 𝑏 if we transform 𝑏 by 𝑈 Rendering – Monte Carlo Integration II 20

Multidimensional Transformations If we try to apply the above to the unit disk, we fail at 𝑦 = 𝑠 sin 𝜄 −1 𝑒𝑦 We can’t evaluate : the transformation that produces one 𝑒𝑠 target variable is dependent on both input variables and vice-versa We cannot compute the change in the PDF between individual variables, we must take them all into account simultaneously It’s matrix time Rendering – Monte Carlo Integration II 21

Multidimensional Transformations We write the set of 𝑂 values from a multidimensional variable Ԧ 𝐵 as a vector Ԧ 𝑏 and the 𝑂 outputs of transformation 𝑈 as a vector 𝑐 : 𝑏 1 𝑐 1 𝑈 1 ( Ԧ 𝑏) ⋮ ⋮ 𝑏 = Ԧ , 𝑐 = = ⋮ = 𝑈( Ԧ 𝑏) 𝑏 𝑂 𝑐 𝑂 𝑈 𝑂 ( Ԧ 𝑏) Instead of quantifying the change in volume incurred by 𝑈 𝑏 , 𝑒𝑈 𝑏 , our goal is now to quantify the change incurred by 𝑈 Ԧ 𝑏 𝑒𝑏 Rendering – Monte Carlo Integration II 22