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COMP 546 Lecture 12 Illumination and Reflectance Tues. Feb. 20, 2018 1 Illumination and Reflectance Shading Brightness versus Lightness Color constancy Shading on a sunny day L N(x) Lamberts (cosine)

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1

COMP 546

Lecture 12

Illumination and Reflectance

  • Tues. Feb. 20, 2018
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Illumination and Reflectance

  • Shading
  • Brightness versus Lightness
  • Color constancy
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SLIDE 3

N(x) L

 

Shading on a sunny day

𝑂 𝑀 Lambert’s (cosine) Law:

𝐽 π‘Œ = 𝑂(π‘Œ) βˆ™ 𝑀

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Unit Surface Normal

1 πœ–π‘Ž πœ–π‘Œ

2

+ πœ–π‘Ž πœ–π‘

2

+ 1 πœ–π‘Ž πœ–π‘Œ , πœ–π‘Ž πœ–π‘ , βˆ’1

𝑂 ≑

𝑂

4

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Shading on a sunny day

5

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Cast and Attached Shadows

cast : 𝑂(π‘Œ) βˆ™ 𝑀 > 0 but light is

  • ccluded

attached : 𝑂(π‘Œ) βˆ™ 𝑀 < 0 𝑀

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SLIDE 7

Shading models

  • sunny day (last lecture)
  • sunny day + low relief
  • cloudy day
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SLIDE 8

Examples of low relief surfaces

(un)crumpled paper

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Low relief surface

1 πœ–π‘Ž πœ–π‘Œ

2

+ πœ–π‘Ž πœ–π‘

2

+ 1 πœ–π‘Ž πœ–π‘Œ , πœ–π‘Ž πœ–π‘ , βˆ’1

𝑂 ≑

9

πœ–π‘Ž πœ–π‘Œ

β‰ˆ 0

β‰ˆ0 β‰ˆ0

πœ–π‘Ž πœ–π‘

β‰ˆ 0

Thus,

Suppose and are both small.

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Linear shading model for low relief

𝐽 π‘Œ, 𝑍 β‰ˆ

πœ–π‘Ž πœ–π‘Œ , πœ–π‘Ž πœ–π‘ , βˆ’1 βˆ™ π‘€π‘Œ, 𝑀𝑍, π‘€π‘Ž

10

  • shadows can still occur
  • ne equation per point but two unknowns
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SLIDE 11

Example: curtains

π‘Ž π‘Œ, 𝑍 = π‘Ž0 + 𝑏 sin( π‘™π‘Œ π‘Œ )

𝑀 𝑂 π‘Œ π‘Ž

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SLIDE 12

Example: curtains

π‘Ž π‘Œ, 𝑍 = π‘Ž0 + 𝑏 sin( π‘™π‘Œ π‘Œ ) πœ–π‘Ž πœ–π‘Œ = 𝑏 π‘™π‘Œ cos π‘™π‘Œ π‘Œ πœ–π‘Ž πœ–π‘ = ,

π‘Œ π‘Ž

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Example: curtains

𝐽 π‘Œ, 𝑍 β‰ˆ

πœ–π‘Ž πœ–π‘Œ , πœ–π‘Ž πœ–π‘ , βˆ’1 βˆ™ π‘€π‘Œ, 𝑀𝑍, π‘€π‘Ž

π‘Ž π‘Œ, 𝑍 = π‘Ž0 + 𝑏 sin( π‘™π‘Œ π‘Œ ) πœ–π‘Ž πœ–π‘Œ = 𝑏 π‘™π‘Œ cos π‘™π‘Œ π‘Œ 𝐽 π‘Œ, 𝑍 = 𝑏 π‘™π‘Œ cos π‘™π‘Œ π‘Œ π‘€π‘Œ βˆ’ π‘€π‘Ž πœ–π‘Ž πœ–π‘ = ,

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Example: curtains

π‘Ž π‘Œ, 𝑍 = π‘Ž0 + 𝑏 sin( π‘™π‘Œ π‘Œ ) 𝐽 π‘Œ, 𝑍 = βˆ’ π‘€π‘Ž + 𝑏 π‘™π‘Œ π‘€π‘Œcos π‘™π‘Œ π‘Œ

𝑀 = (π‘€π‘Œ, 𝑀𝑍, π‘€π‘Ž)

π‘Œ π‘Ž 𝑂

Q: where do the intensity maxima and minima occur?

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Shading on a cloudy day

(my Ph.D. thesis)

15

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 (x)

Shading on a Cloudy Day

Shadowing effects cannot be ignored.

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Shading on a Cloudy Day

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Shading is determined by shadowing and surface normal.

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Shading on a Sunny Day

Shading determined by surface normal only.

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Shape from shading

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Q: What is the task ? What problem is being solved? A: Estimate surface slant, tilt, curvature. How to account for (or estimate) the lighting ?

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Illumination and Reflectance

  • Shading

[Shading and shadowing models assume that intensity variations

  • n a surface are entirely due to illumination. But surfaces have

reflectance variations too. ]

  • Brightness versus Lightness
  • Color constancy
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Which paper is lighter ?

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Paper on the left is in shadow. It has lower physical intensity and it appears

  • darker. But do both papers seem to be of same (white) material?
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Which paper is lighter ?

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Image is processed so that the right paper is given same image intensities as left

  • paper. Now, right paper appears to be made of different material. Why?
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𝐽 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦, 𝑧)

Abstract version β€œReal” example

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𝐽 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦, 𝑧) shading & shadows material luminance

Physical quantities

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low reflectance, high illumination low reflectance, low illumination high reflectance, low illumination

𝐽 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦, 𝑧)

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𝐽 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦, 𝑧) β€œlightness” β€œbrightness”

Perceptual quantities

(no standard term for perceived illumination)

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All four indicated β€œsquares” have same intensity. What is the key difference between the two configurations ?

Adelson’s corrugated plaid illusion.

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β€œLightness” perception

Q: What is the task ? What problem is being solved? A:

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β€œLightness” perception

Q: What is the task ? What problem is being solved? A: Estimate the surface reflectance, by discounting the illumination. 𝐽 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦, 𝑧)

?

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β€œLightness” perception:

solution sketch

Q: What is the task ? What problem is being solved? A: Estimate the surface reflectance, by discounting illumination effects. Compare points that have same illumination. 𝐽 𝑦1, 𝑧1 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦1, 𝑧1) 𝐽 𝑦2, 𝑧2 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ (𝑦2, 𝑧2)

=

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Illumination and Reflectance

  • Shading
  • Brightness versus Lightness
  • Color constancy
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Recall lecture 3 - color

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Illumination Surface Reflectance (fraction) Absorption by photoreceptors (fraction)

There are three different spectra here.

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LMS cone responses

𝐽 𝑦, 𝑧, πœ‡ = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧, πœ‡ βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ 𝑦, 𝑧, πœ‡

Cone response

𝐽 𝑦, 𝑧, πœ‡ 𝐷𝑀𝑁𝑇(πœ‡) π‘’πœ‡

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Surface Color Perception

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Q: What is the task? What is the problem to be solved? A:

illumination Surface Reflectance

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Surface Color Perception

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Q: What is the task? What is the problem to be solved? A: Estimate the surface reflectance, by discounting the illumination. 𝐽 𝑦, 𝑧, πœ‡ = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œ 𝑦, 𝑧, πœ‡ βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“ 𝑦, 𝑧, πœ‡

illumination Surface Reflectance

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SLIDE 36

𝐽𝑆𝐻𝐢 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œπ‘†π»πΆ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“π‘†π»πΆ (𝑦, 𝑧)

For simplicity, let’s ignore the continuous wavelength πœ‡ and just consider 𝑆𝐻𝐢 (LMS).

πœ‡

Cone response

𝐽 𝑦, 𝑧, πœ‡ 𝐷𝑀𝑁𝑇(πœ‡) π‘’πœ‡

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β€œColor Constancy”

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Task: estimate the surface reflectance, by discounting the illumination

𝐽𝑆𝐻𝐢 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œπ‘†π»πΆ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“π‘†π»πΆ (𝑦, 𝑧)

illumination Surface Reflectance

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Why we need color constancy

  • object recognition
  • skin evaluation (health, emotion, …)
  • food quality
  • …
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Example 1: spatially uniform illumination

= *

𝐽𝑆𝐻𝐢 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œπ‘†π»πΆ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“π‘†π»πΆ (𝑦, 𝑧)

Given this, how to estimate this?

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Solution 1: β€˜max-RGB’ Adaptation

= *

𝐽𝑆𝐻𝐢 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œπ‘†π»πΆ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“π‘†π»πΆ (𝑦, 𝑧)

Divide each 𝐽𝑆𝐻𝐢 channel by the max value of 𝐽𝑆𝐻𝐢 in each channel. When does this give the correct answer?

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Example 2: non-uniform illumination

= *

𝐽𝑆𝐻𝐢 𝑦, 𝑧 = π‘—π‘šπ‘šπ‘£π‘›π‘—π‘œπ‘π‘’π‘—π‘π‘œπ‘†π»πΆ 𝑦, 𝑧 βˆ— π‘ π‘“π‘”π‘šπ‘“π‘‘π‘’π‘π‘œπ‘‘π‘“π‘†π»πΆ 𝑦, 𝑧 Solution: See Exercises.

sun + blue sky

  • nly

blue sky

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Illumination and Reflectance

  • Shape from Shading
  • Brightness versus Lightness
  • Color constancy

Different solutions require different underlying

  • assumptions. This is separate issue from how we can

code up a solution using neural circuits.