02502 image analysis week4 -neighbourhood...
TRANSCRIPT
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DTU Compute24. September 2019 02502 - week 4DTU Compute24. September 2019 02502– week 4
Tim B. Dyrby ([email protected])Associate Professor, DTU Compute
&Rasmus R. Paulsen ([email protected])Associate Professor, DTU Compute
02502Image AnalysisWeek 4 - Neighbourhood Processing
1
http://courses.compute.dtu.dk/02502/
Plenty of slides adapted from Thomas Moeslunds lectures
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DTU Compute24. September 2019 02502 - week 4
Lecture 4Neighbourhood Processing
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DTU Compute24. September 2019 02502 - week 4
What can you do after today?• Describe the difference between point processing and neighbourhood processing• Compute a rank filtered image using the min, max, median, and difference rank filters• Compute a mean filtered image• Decide if median or average filtering should be used for noise removal• Choose the appropriate image border handling based on a given input image• Implement and apply template matching• Compute the normalised cross correlation and explain why it should be used• Apply given image filter kernels to images• Use edge filters on images• Describe finite difference approximation of image gradients including the magnitude and
the direction• Compute the magnitude of the gradient• Describe the concept of edge detection
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DTU Compute24. September 2019 02502 - week 4
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DTU Compute24. September 2019 02502 - week 4
Point processing
0 2
1 2
1 2 1
2 5 3
1 3
2 2
0 1
1 2 0
2 1 4
1 0 1912
Input Output
• The value of the output pixel is only dependent on the value of one input pixel
• A global operation – changes all pixels
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DTU Compute24. September 2019 02502 - week 4
Point processing
n Grey level enhancement– Process one pixel at a time independent of all other pixels– For example used to correct Brightness and Contrast
Too lowcontrastCorrect
Too highcontrast
Too highbrightness
Too lowbrightness
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DTU Compute24. September 2019 02502 - week 4
Neighbourhood processing
0 2
1 2
1 2 1
2 5 3
1 3
2 2
0 1
1 2 0
2 1 4
1 0 1912
Input Output
• Several pixels in the input has an effect on the output
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DTU Compute24. September 2019 02502 - week 4
Use of filtering
Noise removal Enhance edges Smoothing
• Image processing
• Typically done before actual image analysis
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DTU Compute24. September 2019 02502 - week 4
Salt and pepper noise
• Pixel values that are very different from their neighbours
• Very bright or very dark spots• Scratches in X-rays
What is that?
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DTU Compute24. September 2019 02502 - week 4
Salt and pepper noise
• Fake example– Let us take a closer look at noise pixels
They are all 0 or 255
Should we just remove all the 0’s and 255’s from the image?
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DTU Compute24. September 2019 02502 - week 4
What is so special about noise?• What is the value of the pixel compared to
the neighbours?• Average of the neighbours
– 170• Can we compare to the average?
– Difficult – should we remove all values bigger than average+1 ?
• It is difficult to detect noise!
172, 169, 171, 168, 0, 169, 172, 173, 168
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DTU Compute24. September 2019 02502 - week 4
Noise – go away!
• We can not tell what pixels are noise• One solution
– Set all pixels to the average of the neighbours (and the pixel itself)
• Oh no!– Problems!– The noise “pollutes” the good pixels
172, 169, 171, 168, 169, 172, 173, 168
170
169, 168, 0, 170,173, 170, 172, 170
149
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DTU Compute24. September 2019 02502 - week 4
Quiz 1: What is the median Value?A) 170B) 173C) 169D) 171E) 172
169, 168, 0, 170,172, 173, 170, 172, 170
0, 168, 169, 170, 170, 170, 172, 172, 173
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
The median value
• The values are sorted from low to high• The middle number is picked
–The median value is found at position: (N+1)/2 where N is the count of numbers
169, 168, 0, 170,172, 173, 170, 172, 170
0,168, 169, 170, 170, 170,172, 172,173
Median
Noise has no influence on the median!
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DTU Compute24. September 2019 02502 - week 4
Quiz 2: What is the median Value?A) 170B) 173C) 169D) 171E) 172
169, 168, 0, 170, 175, 177, 176, 180, 170, 177
Even count of numbers:Median at position: (N+1)/2= 5,5So Median = (170+175)/2= 172,5You then round up or down (In math then round up)
0, 168, 169, 170, 170, 175, 176, 177, 177, 180‘
- - - - -
A B C D E
‘
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DTU Compute24. September 2019 02502 - week 4
Noise away – the median filter
• All pixels are set to the median of its neighbourhood
• Noise pixels do not pollute good pixels
172, 169, 171, 168, 0, 169, 172, 173, 168
170
169, 168, 0, 170, 172,173, 170, 172, 170
170
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DTU Compute24. September 2019 02502 - week 4
Noise removal – average filter
Scanned X-ray with salt and pepper noise
Average filter (3x3)
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DTU Compute24. September 2019 02502 - week 4
Noise removal – median filter
Scanned X-ray with salt and pepper noise
Median filter (3x3)
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DTU Compute24. September 2019 02502 - week 4
Image Filtering
• Creates a new filtered image• Output pixel is computed based on a
neighbourhood in the input image• 3 x 3 neighbourhood
– Filter size 3 x 3– Kernel size 3 x 3– Mask size 3 x 3
• Larger filters often used– Size?
• 7 x 7– Number of elements?
• 49
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DTU Compute24. September 2019 02502 - week 4
Quiz 3: Median filterA) 25B) 90C) 198D) 86E) 103
The image is filtered with a 3 x 3 median filter. What is the result in the pixel marked with a circle?
4, 25, 34, 86, 90, 103, 125, 209, 230
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Rank filters
• Based on sorting the pixel values in the neighbouring region
• Minimum rank filter–Darker image. Noise problems.
• Maximum rank filter–Lighter image. Noise problems.
• Difference filter (max-min)
–Enhances changes (edges). Noise problems
0,168, 169, 170, 170, 170,172, 172,173
-
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DTU Compute24. September 2019 02502 - week 4
Quiz 4: Median filterA) 3B) 84C) 112D) 73E) 202
The image is filtered with a 3 x 3 median filter (medI). The image (the original) is also filtered with a 5x5 minimum rank filter (minI). The final image is made by subtracting minI from medI. What is the result in the marked pixel?
15, 60, 62, 90, 99, 103, 115, 165, 187medI:
15minI:
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Correlation
• What is it?• Two measurements
– Low correlation– High correlation
• High correlation means that there is a relationbetween the values
• They look the same• Correlation is a measure of similarity
Low
High
Muscles
Mat
h Sk
ills
Muscles
Benc
h Pr
ess
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DTU Compute24. September 2019 02502 - week 4
Why do we need similarity?• Image analysis is also about recognition of patterns• Often an example pattern is used
– Often with some kind of meta data to apply to the targets• We need something to tell us if there is a high match between our
pattern and a part of the image
Example pattern
With meta information
Find matches
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DTU Compute24. September 2019 02502 - week 4
Correlation (1D)
1 1 2 2 1 1 2 2 1 1
1 2 1
45
41
1 * 1 + 2 * 1 + 1 * 2 = 5
Normalise!
KernelSignal (1D image)
41
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DTU Compute24. September 2019 02502 - week 4
Correlation (1D)
1 1 2 2 1 1 2 2 1 1
1 2 1
45
47
47
45
45
47
47
45
41
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DTU Compute24. September 2019 02502 - week 4
Normalisation
1 1 2 2 1 1 2 2 1 1
1 2 1
45
41
1 * 1 + 2 * 1 + 1 * 2 = 5
Normalise!41
• The sum of the kernel elements is used
• Keep the values in the same range as the input image
Sum is 4
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DTU Compute24. September 2019 02502 - week 4
Normalisation
• Normalisation factor–Sum of kernel coefficients
1 2 1
!"
ℎ 𝑥 = 1 + 2 + 1
h(x)
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DTU Compute24. September 2019 02502 - week 4
Correlation on images
• The filter is now 2D1 1 1
1 1 1
1 1 1
0 2 4
1 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
912
91
Input Output
Kernel
Kernel coefficients
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DTU Compute24. September 2019 02502 - week 4
Correlation on images1 1 11 1 11 1 1
0 2 4
1 2 2
1 6 3
1 2 0
2 1 4
1 0 1
912
911
91
Input Output
1 3 1
2 2 2
0 1 3
1 2 12 5 3
2 1 3
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DTU Compute24. September 2019 02502 - week 4
0 2 4
1 2 2
1 6 3
1 2 0
2 1 4
1 0 1
1 3 1
2 2 2
0 1 3
1 2 12 5 3
2 1 3
Correlation on images
Input Output
The mask is moved row by row
No values at the border!
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DTU Compute24. September 2019 02502 - week 4
1 2 1
1 3 1
1 2 1
-1 -2 -1
0 0 0
1 2 1
Design your filter
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DTU Compute24. September 2019 02502 - week 4
Quiz 5: Correlation on images – no normalisationA) 4B) 7C) 16D) 23E) 25
1 2 11 3 11 2 1
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
?- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Quiz 6: Correlation on images – no normalisation 2A) 0,10B) 3, 3C) 6, 2D) 10, 15E) 11, 32
-1 -2 -10 0 01 2 1
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
?
?- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Mathematics of 2D Correlation
h
f
Correlation operator
1 2 11 3 11 2 1
0 2 4
1 2 2
1 6 3
1 2 1
2 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
𝑔 𝑥, 𝑦 = 𝑓 𝑥, 𝑦 ∘ ℎ(𝑥, 𝑦)
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DTU Compute24. September 2019 02502 - week 4
Mathematics of 2D Correlation
1 2 11 3 11 2 1
0 2 4
1 2 2
1 6 3
1 2 1
2 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
h
f h(0,0)f(2,1)
Example g(2,1)
i = -1, j = -1
i = 1, j = 0
𝑔 𝑥, 𝑦 = !0123
3
!4123
3
ℎ(𝑖, 𝑗) 7 𝑓(𝑥 + 𝑖, 𝑦 + 𝑗)
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DTU Compute24. September 2019 02502 - week 4
Mathematics of 2D Correlation
1 2 11 3 11 2 1
0 2 4
1 2 2
1 6 3
1 2 1
2 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
h
f
𝑔 𝑥, 𝑦 = !0123
3
!4123
3
ℎ(𝑖, 𝑗) 7 𝑓(𝑥 + 𝑖, 𝑦 + 𝑗)
𝑔(𝑥, 𝑦) = 1 7 2 + 2 7 0 + 1 7 1 + 1 7 1 + 3 7 4 + 1 7 2 + 1 7 0 + 2 7 1 + 1 7 0
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DTU Compute24. September 2019 02502 - week 4
𝑔 𝑥, 𝑦 = !0123
3
!4123
3
ℎ(𝑖, 𝑗) 7 𝑓(𝑥 + 𝑖, 𝑦 + 𝑗)
𝑔 2,2 =ℎ −1,−1 ⋅ 𝑓 1,1 + ℎ 0,−1 ⋅ 𝑓 2,1 + ℎ 1,−1 ⋅ 𝑓 3,1 +ℎ −1,0 ⋅ 𝑓 1,2 + ℎ 0,0 ⋅ 𝑓 2,2 + ℎ 1,0 ⋅ 𝑓 3,2 +ℎ −1,1 ⋅ 𝑓 1,3 + ℎ 0,1 ⋅ 𝑓 2,3 + ℎ 1,1 ⋅ 𝑓 3,3
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DTU Compute24. September 2019 02502 - week 4
2D Kernel Normalisation
1 2 11 3 11 2 1
h
Normalisation factor:
!"!=ℎ(𝑥, 𝑦)
1+2+1+1+3+1+1+2+1=13
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DTU Compute24. September 2019 02502 - week 4
Quiz 7: CorrelationA) 50122B) 123001C) 11233D) 2550E) 90454
A template match is done on the image to the left with the template seen to the right. To find the best match the correlation is computed. What is the correlation in the marked pixel?
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Smoothing filters
• Also know as – Smoothing kernel, Mean filter, Low pass filter, blurring
• The simplest filter: – Spatial low pass filter– Removes high frequencies
• Another mask: – Gaussian filter
1 1 1
1 1 1
1 1 191
1 2 1
2 4 2
1 2 1161Why Gaussian?
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DTU Compute24. September 2019 02502 - week 4
Use of smoothing
3x3 7x7 11x11 15x15
• Various sizes of smoothing filters
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DTU Compute24. September 2019 02502 - week 4
Use of smoothing
• Large kernels smooth more• Removes high frequency information• Good at enhancing big structures
3x3 15x15
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DTU Compute24. September 2019 02502 - week 4
Quiz 8: Mean filter
A) 166B) 113C) 12D) 51E) 245
A 3x3 mean filter is applied to the image. The result in the marked pixel is 86. What is the value of the pixel, where the value is missing?
(608+x*1)/9=86 -> x=166
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Border handling
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
Input Output
No values at the border!
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DTU Compute24. September 2019 02502 - week 4
Border handling – extend the input
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
Input
0 0 0 0 000000
• Zero padding – what happens?
• Zero is black – creates dark border around the image
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DTU Compute24. September 2019 02502 - week 4
Quiz 9: Correlation with zero paddingA) 5, 8B) 12, 16C) 13, 3D) 15, 21E) 17, 12
1 2 11 3 11 2 1
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
?
?
- - - - -
A B C D E
000
00
0
0 0 0 0 0 0
000
00
00 0 0 0 0 0
00
00
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DTU Compute24. September 2019 02502 - week 4
Border handling – extend the input
Input • Reflection
• Normally better than zero padding
0 2 41 2 2
1 6 3
1 2 12 5 3
2 1 3
1 3 1
2 2 2
0 1 3
1 2 0
2 1 4
1 0 1
1 2 0 1 311211
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DTU Compute24. September 2019 02502 - week 4
What is the connection to deep learning?
https://se.mathworks.com/videos/introduction-to-deep-learning-what-are-convolutional-neural-networks--1489512765771.html
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DTU Compute24. September 2019 02502 - week 4
Banks of filters• The part of the network that touches the image consists of a bank of filters
– Organised in a multi-level hierarchy
• The weights of the filters are adapted to the problem
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DTU Compute24. September 2019 02502 - week 4
Template Matching• Template
–Skabelon på dansk• Locates objects in images
Input
Template
Output: Correlation image
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DTU Compute24. September 2019 02502 - week 4
Template Matching
• The correlation between the template and the input image is computed for each pixel
Input
Template
Output: Correlation image
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DTU Compute24. September 2019 02502 - week 4
Template Matching• The pixel with the highest value is found in the output image
–Here is the highest correlation
Input
Template
Output: Correlation image
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DTU Compute24. September 2019 02502 - week 4
Template Matching• This corresponds to the found pattern in the input image
Input
Template
Output: Correlation image
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DTU Compute24. September 2019 02502 - week 4
Problematic Correlation• Correlation matching has problem with light areas – why?
Input (f)
Template (h)
Output: Correlation image
Real max
Fake max
Very High!𝑔 𝑥, 𝑦 = !0123
3
!4123
3
ℎ(𝑖, 𝑗) 7 𝑓(𝑥 + 𝑖, 𝑦 + 𝑗)
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DTU Compute24. September 2019 02502 - week 4
Normalised Cross Correlation
Input (f)
Template (h)
Output: Correlation image
Real max
NCC x, y =Correlation
Length of image patch ⋅ Length of template
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DTU Compute24. September 2019 02502 - week 4
Length of template• Vector length
– Put all pixel values into a vector– Compute the length of this vector
• Describes the intensity of the template– Bright template has a large length– Dark template has a small length
Template (h)
Length of template = !0123
3
!4123
3
ℎ(𝑖, 𝑗) 7 ℎ(𝑖, 𝑗)
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DTU Compute24. September 2019 02502 - week 4
Length of image patch• Vector length based on pixel values in image patch• Describes the intensity of the image patch
Template (h)
Input (f) with patch
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DTU Compute24. September 2019 02502 - week 4
Normalised Cross Correlation• The length of the image patch and the length of template normalise the NCC• If the image is very bright the NCC will be “pulled down”
NCC x, y =Correlation
Length of image patch ⋅ Length of template
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DTU Compute24. September 2019 02502 - week 4
Normalised Cross Correlation• NCC will be between
0 : No similarity between template and image patch1 : Template and image patch are identical
Input (f)
Template (h)
Output: Correlation image
Real max
NCC x, y =Correlation
Length of image patch ⋅ Length of template
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DTU Compute24. September 2019 02502 - week 4
Quiz 10: Normalised Cross CorrelationA) 0.10B) 0.33C) 0.83D) 0.62E) 0.98
A template match using normalised cross correlation is performed. What is the resulting value in the marked pixel?
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Edges• An edge is where there is a high change in
gray level values• Objects are often separated from the
background by edges
Gray level profile
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DTU Compute24. September 2019 02502 - week 4
• The profile as a function f(d) in the x-direction• What value is high when there is an edge?
– The slope of f– The slope of the tangent at d
f (d)
𝑓Q(𝑑)
Edges
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DTU Compute24. September 2019 02502 - week 4
Finite Difference
• Definition of slope
• Approximation
• Simpler approximation
𝑓Q 𝑑 = limS→U
𝑓 𝑑 + ℎ − 𝑓(𝑑)ℎ
𝑓Q 𝑑 ≈𝑓 𝑑 + ℎ − 𝑓(𝑑)
ℎ
𝑓Q 𝑑 ≈ 𝑓 𝑑 + 1 − 𝑓(𝑑) ;Where h=1
(Continues functions only )
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DTU Compute24. September 2019 02502 - week 4
Edges
• Discrete approximation of f’(d)• Can be implemented as a filter
-1 0 1
f(d-2) f(d-1) f(d) f(d+1) f(d+2) f(d+3)
f (d)
𝑓Q 𝑑 ≈ 𝑓 𝑑 + 1 − 𝑓(𝑑)
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DTU Compute24. September 2019 02502 - week 4
Edges in 2D
• Changes in gray level values– Image gradient– Gradient is the 2D derivative of a 2D function f(x,y)– Equal to the slope of the image– A steep slope is equal to an edge
𝛻𝑓 𝑥, 𝑦 = �⃗�(𝑔", 𝑔=)
𝑔"(",=) = 𝑓 𝑥 + 1, 𝑦 − 𝑓 𝑥 − 1, 𝑦𝑔=(",=) = 𝑓 𝑥, 𝑦 + 1 − 𝑓(𝑥, 𝑦 − 1)
Where:
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DTU Compute24. September 2019 02502 - week 4
2D Gradient
magnitude = 𝑔"\ + 𝑔=\
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DTU Compute24. September 2019 02502 - week 4
Edge filter kernel
• The Prewitt filter is a typical edge filter• Output image has high values where there
are edges• gx (x,y): Vertical filter• gy(x,y) : Horizontal filter
-1 0 1
-1 0 1
-1 0 1Vertical Prewitt filter i.e. gx
-1 -1 -1
0 0 0
1 1 1
Horizontal Prewitt filter i.e. gy
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DTU Compute24. September 2019 02502 - week 4
Prewitt filter
Original Prewitt Prewitt
Hot colormapSmooth 15x15
Smooth 15x15
Prewitt
• Filter direction?
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DTU Compute24. September 2019 02502 - week 4
Edge detection
• Edge filter– Prewitt for example
• Thresholding– Separate edges from non-edges
• Output is binary image– Edges are white
Original Gradients thresholding
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DTU Compute24. September 2019 02502 - week 4
Quiz 11: Edge filtering
A) MeanB) Horizontal SobelC) Vertical PrewittD) Horizontal PrewittE) Median
A filtering (correlation) is done with a filter on the image. No normalisation is performed. The value in the marked pixel becomes -119. What 3 x 3 filter was used?
-1 -2 -1
0 0 0
1 2 1
Horizontal Sobel
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Future has become the past:What can you do after today?• Describe the difference between point processing and neighbourhood processing• Compute a rank filtered image using the min, max, median, and difference rank filters• Compute a mean filtered image• Decide if median or average filtering should be used for noise removal• Choose the appropriate image border handling based on a given input image• Implement and apply template matching• Compute the normalised cross correlation and explain why it should be used• Apply given image filter kernels to images• Use edge filters on images• Describe finite difference approximation of image gradients including the magnitude and the direction• Compute the magnitude of the gradient• Describe the concept of edge detection
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DTU Compute24. September 2019 02502 - week 4
Neighbourhood ProcessingWhat can it be used for?
Gradients: Micro-image resolution of fiber directions from histology
Budde and Frank, 2012, NeuroImage
Gradients: A key component in non-linear algorithm optimisation
https://en.wikipedia.org/wiki/Gradient_descent#/media/File:Gradient_descent.svg
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DTU Compute24. September 2019 02502 - week 4
Next week 5: Morphology
• The science of form, shape and structure• In biology: The form and structure of animals and plants
Common leaf morphologies
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DTU Compute24. September 2019 02502 - week 4
Quiz 12: Level of todays topic and lectureA) So easy I get a headache of thinking
on that levelB) I have been catchting all PokeMons
while following the lectureC) FineD) Could you please repeat that?E) I did not get any of it?
- - - - -
A B C D E
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DTU Compute24. September 2019 02502 - week 4
Quiz 13: What did I get out of the day?
A) I learnt very littleB) I did not get so much out of itC) The learning outcome was fineD) I learnt quite muchE) I learnt a lot
- - - - -
A B C D E