EECE/CS 4353 Image Processing - Internet Archive

EECE/CS 4353 Image Processing

Richard Alan Peters IIDepartment of Electrical and Computer Engineering

Fall Semester 2021

Lecture Notes:

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Lecture Notes: Basic Point Processing of Images

Point Processing of Images

In a digital image, a point = a pixel. Point processing transforms a pixel’s

value as function of its value alone. It does not depend on the values of

the pixel’s neighbors.

6 September 2021 1999-2021 by Richard Alan Peters II 2

Examples include: Brightness and contrast adjustment Gamma correction Histogram equalization Histogram matching Color correction.

Point Processing of Images


Point Processing

original + gamma- gamma + brightness- brightness

original + contrast- contrast histogram EQhistogram mod


Wallace and Gromit are characters created and owned byAardman Animation. Seehttps://www.aardman.com/work/#filter=.wallace-gromit

https://www.aardman.com/work/

Point Processing: Pixel Values


A point process transforms one intensity level (or color) intoanother as a function of that one alone. So a point process is

out in .fp p

out in, , .r c f r cp p

That is, the pixel value output is dependent on only the pixel valueinput. That implies

In words, the output at one location is dependent only the value ofthe input image at that same location. Other locations don’t matter.

If , ,andthen , , .

r c b gf g k

r c b k

I

J

Point Ops via Functional Mappings

Input Output

I , pointoperator JImage:

I(r,c) function, f J(r,c)Pixel:

J I

The transformation of image I into image J isaccomplished by replacing each input intensity, g, witha specific output intensity, k, at every location (r,c,b)where I(r,c,b) = g.

The rule that associates k with g is usuallyspecified with a function, f, so that f (g) = k.




One-band Image

Three-band Image

J(r,c) = f ( I(r,c) ),for all pixel locations, (r,c).

J(r,c,b) = f ( I(r,c,b) ), orJ(r,c,b) = fb ( I(r,c,b) ), orfor b = 1, 2, 3, and all (r,c).

J(r,c) = f ( I(r,c) ),for all pixel locations, (r,c).

J(r,c,b) = f ( I(r,c,b) ), orJ(r,c,b) = fb ( I(r,c,b) ), orfor b = 1, 2, 3, and all (r,c).

One-band Image

Three-band Image


Either all 3 bandsare mapped throughthe same function,f, or …

… each band ismapped througha separate func-tion, fb.


Look-Up Tables


Lookup Tables


A lookup table1 is an indexed list of numbers – a vector – that canbe used to implement a discrete function – a mapping from a setof integers, {gin,1, gin,2, ... , gin,n}, to a set of numbers (integers ornot), {gout,1, gout,2, ... , gout,n}. A lookup table can implement afunction such as:

out in in out out, 1

in, out,

if , where 0, , 1 and

then define LUT 1 LUT , for 1, n .

nk k

k k

g f g g n g g

g k g k

We’ve already seen one of these, the colormap.2

1 https://en.wikipedia.org/wiki/Lookup_table ; 2 Lecture 2b, slides 15-34.In MATLAB the indexis the input value + 1.

https://en.wikipedia.org/wiki/Lookup_table

J = LUT(I+1)

Point Operations using Lookup Tables

A lookup table (LUT)can implement afunctional mapping.

,255,,0

,

g

gfkforIf

invaluesontakesifand

,255,,0k

… then the LUTthat implements fis a 256x1 arraywhose (g +1)th

value is k = f (g).

To remap animage, I, to J :


LUT is 256x1.But I may beRxC or RxCx3.

In MATLAB the indexis the input value + 1.

Point Operations = Lookup Table Ops

0 127 255

012

725

5

input value

outp

ut v

alue

index value...

101102103104105106...

...646869707071...

E.g.:

input output


In MATLAB the indexis the input value + 1.

How to Generate a Lookup Table

32/)127(1255;

255,,0.2

xaeax

xa

LetLet

For example, a sigmoid:

a = 2;x = 0:255;LUT = 255 ./ (1+exp(-a*(x-127)/32));

Or in Matlab:

This is justone example.


In MATLAB, LUT isindexed from 1:256.

If I is 3-band, thena) each band is mapped separately using the

same LUT for each band orb) each band is mapped using different LUTs –

one for each band.

Point Ops on RGB Images using Lookup Tables


a) J = LUT(I+1),

b) J(:,:,b) = LUTb(I(:,:,b) +1), for b = 1, 2, 3.

cell

inde

x

cont

ents

0 0

64 32

128 128

192 224

255 255

...

...

...

...

...

...

...

...

1-Band Lookup Table for 3-Band Image

input output

a pixel withthis value

is mapped tothis value


In MATLAB thevalue is index - 1

Photo by Alan Peters,Kyoto, Japan, 1993

R

G

B

Example 3-Band Image


Lightning at Ramasse, Rhone-Alpes, France by Flickr user, Regarde là-bas,https://www.flickr.com/photos/marcel_s_s/8624344496/in/pool-tbasab/

R, G, B Bands

https://www.flickr.com/photos/marcel_s_s/8624344496/in/pool-tbasab/

R

G

B



with 3-Band LUT.


R

G

B



with 3-Band LUT. R is mappedthrough asigmoid.

G is mapped toits negative

The dark values of B are contraststretched while the light valuesare negated and stretched.




3-Band Image with 3-Band LUT.

Resultant bands recom-bined into one image.




3-Band Image with 3-Band LUT.

A silly example to demon-strate some possibilities.


Basic Point Processing:Brightness, Contrast, and Gamma


original + gamma- gamma + brightness- brightness

original + contrast- contrast histogram EQhistogram mod


Point Processing

Point Processes: Original Image

Kinkaku-ji (金閣寺, Temple of theGolden Pavilion), also known asRokuon-ji (鹿苑寺, Deer GardenTemple), is a Zen Buddhist templein Kyoto, Japan.Photo by Alan Peters, August 1993.

Luminance Histogram


For more information on this fascinating, uniqueplace read the historical novel by Mishima,Yukio,The Temple of the Golden Pavilion, translated byIvan Morris, Shinchosha Publishing Co, Ltd., 1956.

if , , 256, , ,, ,

if , , 255255,r c b gr c b g

r c br c b g

IIJ

I

Point Processes: Increase Brightness

0 127 255

012

725

5

g

LUT Mapping 0 and 1 2 3 is the band index.g b , ,

saturation point


Point Processes: Decrease Brightness

0 127 255

012

725

5LUT Mapping

255-g

0 and 1 2 3 is the band index.g b , ,

0, if , , 0, ,

, , , if , , 0r c b g

r c br c b g r c b g

IJ

I I

zero point


Point Processes: Decrease Contrast

( , , ) , , ,r c b a r c b s s T I

0 127 255

012

725

5LUT Mapping

where 0 1.0, s 0,1,2, , 255 , and 1, 2,3 .

a

b

Here, s = 127

s is thecenter ofthe contrastfunction.


Point Processes: Increase Contrast

0, if , , 0,, , , , , if 0 , , 255,

255, if , , 255.

r c br c b r c b r c b

r c b

TJ T T

T

( , , ) , ,r c b a r c b s s T I

0 127 255

012

725

5LUT Mapping 1, s 0, ,255 , 1 2 3a b , ,

zero point

sat. point

Here, s = 127


Point Processes: Increase Contrast

0, if , , 0,, , , , , if 0 , , 255,

255, if , , 255.

r c br c b r c b r c b

r c b

TJ T T

T

( , , ) , ,r c b a r c b s s T I

0 127 255

012

725

5LUT Mapping 1, s 0, ,255 , 1 2 3a b , ,

zero point

sat. point

Here, s = 127


Steps:1. Convert image to class double2. Subtract s from image I3. Multiply result by contrast slope4. Add s back to I5. Convert to class uint8

Steps to Make A Contrast LUT


Double thecontrast of a 3-bitimage centered onintensity level 4

array index

(3) add 4

(1) subtract 4

(4) result

center (2) times 2

(4) 2-to-1 map8 input levels

5 output levels

8 input levels

(0) 1-to-1 map

8 output levels

Point Processes: Contrast Stretch

, ,

, ,

Let min , , max , ,

min , , max x , .

, supp , , measured, , given.,

Then, , .

r c r c

r c r c

m r c M r c

m r c M r c

r c m M m Mr c m

r c M m mM m

I I

J J

I I J J

IJ J J

I I

I I

J J

II

J0 127 255

012

725

5LUT MappingmI MI

mJ

MJ zero point

sat. point

center point


Center Point of Contrast Stretch

, ,

, ,

I J

min , , max , ,

min , , max , ,

, supp ,

, measured, , given.

, , ,m

c , c , slope .2 2

r c r c

r c r c

m r c M r c

m r c M r c

r c

m M m MM mr c r c m mM

M m M mM mM m

I I

J J

I I J J

J JI J

I I

J J J JI I

I I

I I

J J

I

J I


Information Loss from Contrast Adjustment

orig

lo-c

hi-c

histograms


Information Loss from Contrast Adjustmentorig

orig

orig

lo-c hi-c

lo-c

hi-c

rest

rest

lo-c

hi-c

diff

diff

abbreviations:originallow-contrasthigh-contrastrestoreddifference

difference betweenoriginal and restoredlow-contrast

difference betweenoriginal and restoredhigh-contrast


Point Processes: Increased Gamma

0 127 255

012

725

5LUT mapping

1

,, 255 for 1.0

255r c

r c

IJ


Point Processes: Decreased Gamma

0 127 255

012

725

5LUT mappingm M

1,

, 255 for 0 < 1.0255r c

r c

IJ


Point Processes: Decreased Gamma

0 127 255

012

725

5LUT mappingm M

1,

, 255 for 0 < 1.0255r c

r c

IJ


Rather than rather than 255[I/255]1/, gamma correction can also bedefined as 255[I/255]. The effects are opposite. That definition isoften used for image processing as a precursor to coding fortransmission. (https://en.wikipedia.org/wiki/Gamma_correction.)

https://en.wikipedia.org/wiki/Gamma_correction

Gamma Correction: Effect on Histogram


Are LUTs Faster than Direct Computation?


For most computational applications, LUTs are faster than directcomputation, especially when the LUTs are implemented inhardware – e.g. in the frame buffer of a device driver.

In a compiled language, indirect addressing is, I believe, fasterthan arithmetic.1 If so LUTs should be faster than directcomputation.

What about for Matlab?

1I think this is true. But I have not tested it. If anyone would like to test it, I would like to know if I am right or wrong!



Example: Increase brightness of a 24-bit image.

% brightness increase by 32LUTbi = LUT + 32; % since class(LUT) == uint8, LUT(n) == 255 for n > 223J = I + 32; % since class(I) == uint8, J(r,c,b) == 255 for I(r,c,b) > 222K = LUTbi(I+1); % K has the same number of rows columns, and bands as I

I = imread(‘24-bit image file name’);% initialize LUTLUT = uint8(0:255);

Matlab averages over 100 trialsDirect computation: 0.00054486sMap through LUT: 0.057614sDirect/LUT ratio: 0.0094569Direct computation is 105.7425 faster than LUT.

Not in this case, using MATLAB.



Example: Increase contrast of a 24-bit image.

% linear slope in intensity from 0 to 255 over 31 to 223% (LUT indices 32 to 224)LUTct = (LUT-31)*(255/(223-31)); % contrast increase in LUTJ = (I - 31)*(255/(223-31)); % contrast increase of IK = LUTct(I+1); % both J and K are clipped at intensities 0 and 255

I = imread(‘24-bit image file name’);% initialize LUTLUT = uint8(0:255);





% sigmoid function see slide 13a = 2;x = 0:255;LUTsi = uint8(255 ./ (1+exp(-a*(x-127)/32))); % no actual clipping in …J = uint8(255 ./ (1+exp(-a*(double(I)-127)/32))); % … J or K but …K = LUTsi(I+1); % … intensities < 28 are mapped to 0 and > 226 map to 255



Example: Map a 24-bit image through a sigmoid function.I = imread(‘24-bit image file name’);% initialize LUTLUT = uint8(0:255);

EECE/CS 4353 Image Processing - Internet Archive

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