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Morphologie-cours analyse et traitement d'images.pptx
1. Image Processing Algorithms
Pt1 – Morphology
Sup Galilée – INFO3/INFOA3
September 2023
John Chaussard
LAGA – Université Sorbonne Paris Nord
chaussard@math.univ-paris13.fr
3. Goals of image analysis
Image analysis can have various goals:
. Enhance an image for visualization or analysis:
Goals of image analysis
2
Contrast enhancement
Noise removal Blur
compensation
4. Goals of image analysis
Image analysis can have various goals:
. Detect various elements in an image
Goals of image analysis
3
Plate numbers Face Roads (for autonomous cars)
5. Goals of image analysis
Image analysis can have various goals:
. Extract an object from an image (segmentation)
Goals of image analysis
4
Cell counting Bronchial tree
extraction
Special effects
6. Goals of image analysis
Image analysis can have various goals:
. Generate new images from old ones
Goals of image analysis
5
Style transposition Panorama creation
8. Definition of an image
7
In the computer, an image is represented as an array, where each cell
contains the value of a pixel of the image.
Mathematically speaking, an image can be seen as an application.
• I is the name of the image
• A is the domain (related to the shape) of the image
• n is the dimension of the image (typically 2)
• B defines which values a pixel of the image can have, an depends on the
kind of image we are dealing with.
An image is an application 𝐼: 𝐴 ⊆ ℤ𝑛
→ 𝐵
Images in computers
9. Grayscale images
An 8bits grayscale image is an application from a subset of ℤ2
to 0; 255
(to each point of the image corresponds an integer value between 0 and
255).
Each cell of the array / point of the image is called a pixel.
Images in computers
8
176 173 172 174 175 174 175
179 185 187 181 174 173 165
197 181 168 167 171 169 170
161 170 180 183 180 174 175
10. Extended grayscale images
We can also find, in some applications, 16bits or 32bits grayscale images,
giving more precise infomation on object’s constrast (used in specific
applications such as medical image analysis or astronomical image
analysis).
These images car be represented as application from a subset of ℤ2 into
0; 216
− 1 ou 0; 232
− 1 .
Images in computers
9
210 246 283 281 292
186 231 346 450 417
179 189 312 475 400
11. Image couleur
Color images are applications from a subset of ℤ2
into 0; 255 3
.
Images in computers
10
51 51 53 54 60 57
15 30 44 53 55 58
23 34 47 61 61 60
65 61 63 68 67 65
89 91 95 99 102 98
53 70 86 95 97 99
61 74 89 101 102 101
100 99 103 107 106 104
73 86 101 113 111 110
113 111 115 116 115 113
127 124 125 117 115 113
124 116 112 116 115 111
Red channel
Green
channel
Blue channel
12. Images binaires
Binary images are images where the pixels can have only two possibles
values (usually, 0/1 or 0/255).
A binary image can be represented as an application from a subset of ℤ2
into {0;1}, and can also be represented as a subset of ℤ2
, where the pixels
equal to 1 are listed.
Images in computers
11
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
13. Your turn
Find out what kind of « image » you are dealing with when the image is
defined as:
𝐼: 𝐴 ⊆ ℤ3
→ 0; 255
Images in computers
12
It’s a grayscale 3d image, made of voxels.
These images can be obtained with
specific equipment such as MRI, CT-scans,
or tomographic imagery.
14. Your turn
Find out what kind of « image » you are dealing with when the image is
defined as:
𝐼: 𝐴 ⊆ ℤ2
× ℤ → 0; 255 3
Images in computers
13
It’s a movie !
15. Your turn
Find out what kind of « image » you are dealing with when the image is
defined as:
𝐼: 𝐴 ⊆ ℤ2
× ℤ × {0; 1} → 0; 255 3
Images in computers
14
It’s a 3d movie !
16. An image is a surface…
15
During this class, we will mainly focus on 2d 8bits grayscale images.
Images in computers
However, we will, most of the
time, « see » an image not as an
array of values, but as a
topographic surface, where
bright areas become mountains
and dark areas become canyons.
18. Images for a human being
You can easily spot objects in an image because you are familiar with the
object, you learned to recognize many objects since birth.
If you need, however, to spot an object that you are not familiar with, the
detection will be much more difficult for you.
Why is image analysis complicated?
17
You can spot the plane in this image because you
know what is a plane
On this scanner image, where is the stomach?
21. Computers are good calculators
This weakness of computers is also their strenght. When presented an image
with few information, they can use (with a good programming) mathematics
in order to build the missing information, thanks to their « number »
interpretation of images.
Why is image analysis complicated?
20
22. Let’s sum it up
. Humans can easily find objects in an image because they learned, from birth, to
recognize objects, but can do so as long as they already know the objects to find.
. For a computer, images are only numbers stored in an array.
. A computer does not have, naturally, any knowledge on objects to find in an
image: he needs to « learn » this through programming.
Image analysis is a difficult task, consisting in asking the computer to
extract information in an image he does not understand. This can only be
possible with a careful mathematical representation of the object we need to
find or measure in the image.
Why is image analysis complicated?
21
24. Chronology of mathematical morphology
23
Invented by Georges Matheron and Jean Serra in 1964, at l’Ecole des
Mines de Paris.
Was initially developed in order to answer questions relative to mining
operations.
Nowadays, used in many domains of image analysis, such as astronomy,
medical image analysis, …
A brief history of mathematical morphology
25. Usual steps of image processing
24
A brief history of mathematical morphology
(original image) (enhanced image) (segmented image)
(enhanced segmentation)
(information extraction)
27. Structuring elements
26
In morphology, most transformations rely on the choice of a structuring
element, which is simply a subset of ℤ𝒏
.
. We will represent a structuring element by an image where the origin will be
outlined in red, elements of the structuring element will have a value of 1, and
others will have a value of 0.
Ex (2d) :
Binary images and structuring elements
E = { (-1,-1), (0, 0), (1,1) }
0 0 1
0 1 0
1 0 0
E = { (-2,-1),(-2, 0),(-1,0),(1,-1),(1,1) }
0 0 0 1 0
1 1 0 0 0
1 0 0 1 0
x
y
origin
28. Common structuring elements
27
In 2d, there are two important structuring elements: 𝚪𝟒 and 𝚪𝟖.
Binary images and structuring elements
Γ4 = {(-1,0),(1,0),(0,0),(0,1),(0,-1)}
0 1 0
1 1 1
0 1 0
Γ8 = Γ4 U {(-1,-1),(-1,1), (1,1),(1,-1)}
1 1 1
1 1 1
1 1 1
We also have Γ4
∗
= Γ4 (0,0) and Γ8
∗
= Γ8 (0,0) .
29. Common structuring elements
28
We also have the following structuring elements:
. The disk of radius r as the set of pixels whose center is at a distance less or
equal to r from the origin.
. The line of angle a and length L, as the intersetion of the set of pixels
through which passes the line of angle a passing by the origin, and the disk
of radius L/2.
Binary images and structuring elements
30. Common structuring elements in 3d
29
We have three important structuring elements in 3d: 𝚪𝟔, 𝚪𝟏𝟖 et 𝚪𝟐𝟔.
Binary images and structuring elements
Γ6 = {(0,0,0),(-1,0,0),(1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)}
Γ18 = Γ6 U { (-1,-1,0), (-1,1,0), (1,-1,0), (1,1,0),
(-1,0,-1), (-1,0,1), (1,0,-1), (1,0,1),
(0,-1,-1), (0,-1,1), (0,1,-1), (0,1,1)}
Γ26 = Γ18 U {(1,1,1), (-1,1,1), (1,-1,1), (1,1,-1),
(-1,-1,1),(-1,1,-1),(1,-1,-1),(-1,-1,-1)}
We also have Γ6
∗
= Γ6 (0,0,0) , Γ18
∗
= Γ18 (0,0,0) and Γ26
∗
= Γ26 (0,0,0) .
31. Exercise
30
Drawx the structuring element corresponding to this set of points of ℤ2
:
E = {(-2,-1),(-1,-2),(0,-2),(1,-2),(2,-1),(-1,2),(-1,1),(1,2),(1,1)}
Solution :
Binary images and structuring elements
0 1 0 1 0
0 1 0 1 0
0 0 0 0 0
1 0 0 0 1
0 1 1 1 0
32. Exercise
31
Write the structuring element corresponding to this representation.
Solution : E = {(-2,0),(-2,-1),(-1,0),(-1,-1),(0,2),(0,1),(0,0),(0,-1),(1,1),
(1,0),(1,-1)}
Binary images and structuring elements
0 0 1 0 0
0 0 1 1 0
1 1 1 1 0
1 1 1 1 0
0 0 0 0 0
33. Translation of a structuring element
32
We define the translation of a structuring element :
Let 𝐸 ⊂ ℤ𝑛
be a structuring element, and let 𝑥 ∈ ℤ𝑛
.
The translation of E by x is 𝐸𝑥 = 𝑣 + 𝑥 𝑣 ∈ 𝐸}.
Binary images and structuring elements
36. Definition
35
The morphological erosion is a simple local minimum performed on each
pixels of an image:
We set ∀𝑦 ∉ 𝐴, 𝐼 𝑦 = +∞. This makes sense when the structuring element
contains the origin (which is most of the time the case).
Let 𝐼: 𝐴 ⊂ ℤ𝑛
→ 𝐵 (I is an n-dimensional image of domain A and co-
domain B),
and 𝐸 ⊂ ℤ𝑛 be an n-dimensional structuring element.
The erosion of I by E is the transformation, denoted by (𝐼 ⊖ 𝐸),
such that,
∀𝑥 ∈ 𝐴, 𝐼 ⊖ 𝐸 𝑥 = 𝑀𝑖𝑛𝑦∈𝐸𝑥
𝐼(𝑦)
Morphological erosion
40. Effect of the erosion on a greyscale image
39
Morphological erosion
I 𝐼 ⊖ 𝐸
E
Erosion of I by E erodes the surface of the image : mountains shrink, both in width and
height, and canyons are enlarged.
42. Definition
41
We need to define a new operator for structuring elements
Let 𝐸 ⊂ ℤ𝑛
be a structuring element.
We define 𝐸 = −𝑥 | 𝑥 ∈ 𝐸
Morphological dilation
0 1 0
0 1 1
0 1 1 E
1 1 0
1 1 0
0 1 0 𝐸
Ex : 𝐸 = { 0, −1 , 0,0 , 0,1 , 1, −1 , 1,0 }
𝐸 = { 0,1 , 0,0 , 0, −1 , −1,1 , −1,0 }
The structuring element 𝑬 is the 180° rotation of 𝑬.
43. Définition
42
The morphological dilation is a simple local maximum performed on each
pixels of an image:
We set ∀𝑦 ∉ 𝐴, 𝐼 𝑦 = −∞. This makes sense when the structuring element
contains the origin (which is most of the time the case).
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵 and 𝐸 ⊂ ℤ𝑛.
The dilation of I by E is the transformation, denoted by (𝐼 ⊕ 𝐸),
such that
∀𝑥 ∈ 𝐴, 𝐼 ⊕ 𝐸 𝑥 = 𝑀𝑎𝑥𝑦∈𝐸𝑥
𝐼(𝑦)
Morphological dilation
47. Effect of the dilation on a greyscale image
46
Morphological dilation
I 𝐼 ⊕ 𝐸
E
Dilation of I by E dilates the surface of the image : mountains are enlarged and canyons
are shrunk, both in depth and width.
49. Erosion and dilation
48
Which one is the original image, the erosion, and the dilation ?
Morphological gradient
𝐼 ⊕ 𝐸
𝐼 𝐼 ⊖ 𝐸
50. Gradient
49
On an image, we can extract objects’ contours with
dilation and erosion.
Morphological gradient
I
𝐼 ⊕ Γ4 − 𝐼 𝐼 − 𝐼 ⊖ Γ4 (𝐼 ⊕ Γ4) − 𝐼 ⊖ Γ4
51. Definition of gradient
50
Let I be an image et E a structuring element, we define three different
kinds of gradient:
We usually choose 𝜞𝟒 or 𝜞𝟖 to compute a 2d gradient, and 𝜞𝟔, 𝜞𝟏𝟖 or 𝜞𝟐𝟔
to compute a 3d gradient.
Morphological gradient: 𝐺𝐸 𝐼 = 𝐼 ⊕ 𝐸 − 𝐼 ⊖ 𝐸
External gradient: 𝐺𝐸
𝑒𝑥𝑡
𝐼 = 𝐼 ⊕ 𝐸 − 𝐼
Internal gradient: G𝐸
𝑖𝑛𝑡
𝐼 = 𝐼 − (𝐼 ⊖ 𝐸)
Gradient morphologique
53. Why does it work?
52
Morphological gradient
We consider three zones in the image : A, B and C.
Gradient should have high values on C, and low values on A and B.
57. The role of the structuring element in the gradient
56
Morphological gradient
In case of an image I with
noise, gradient calculated with
a small structuring element
will be affected by the noise.
Gradient calculated with large
structuring element will be
thicker but less affected.
I
𝐺𝜞𝟖
𝐼 𝐺𝑩(𝟏𝟎) 𝐼
59. Properties of dilation
58
Dilation has some interesting properties:
. Commutativity: 𝐴 ⨁ 𝐵 = 𝐵 ⨁ 𝐴
. Associativity: 𝐴 ⨁ (𝐵 ⨁ 𝐶) = (𝐴 ⨁ 𝐵) ⨁ 𝐶
If we note 𝐵 ⊕ 𝐵 ⊕ ⋯ ⊕ 𝐵 = 𝑛𝐵, then
𝐴 ⊕ 𝐵 ⊕ 𝐵 ⊕ ⋯ ⊕ 𝐵 = 𝐴 ⊕ 𝑛𝐵
Properties of erosion and dilation
If we can precalculate nB (sometimes possible), then, instead of performing n
dilations by B, we perform one dilation by nB.
60. New structuring elements
59
We can define:
. The size n diamond as 𝑛Γ4
. The size n square as 𝑛Γ8
Properties of erosion and dilation
62. Weak decomposability
61
Erosion (as dilation) possesses an interesting property called (weak)
decomposability :
If a structuring element D can be decomposed into smaller struturing
elements B and C through dilation, then instead of performing one erosion
by D, we can perform two successive erosions by B and C.
𝐴 ⊖ 𝐵 ⊕ 𝐶 = 𝐴 ⊖ 𝐵 ⊖ 𝐶
Properties of erosion and dilation
63. Weak decomposability
62
What is the goal of decomposability?
Let’s imagine: we want to perform an erosion of A by D. However, we
don’t have enough memory to load A into the computer. How can we
compute the erosion of A by D?
Properties of erosion and dilation
0 0 0 0 0 0
0 1 1 1 1 0
0 1 1 1 0 0
0 1 1 1 0 0
0 1 1 1 1 0
0 0 0 0 0 0
A
0 0 0
1 1 1
0 0 0
0 1 0
0 1 0
0 1 0
B C
1 1 1
1 1 1
1 1 1
D
We have 𝐷 = 𝐵 ⊕ 𝐶
64. Weak decomposability
63
However, thanks to decomposability
𝐴 ⊖ 𝐷 = 𝐴 ⊖ (𝐵 ⊕ 𝐶) = 𝐴 ⊖ 𝐵 ⊖ 𝐶
B is a structuring element that only considers pixels of A located on a same
line. In order to perform an erosion by B, one can load A line by line, and
perform the erosion on each line separately and avoid loading the whole
image A in memory.
We can do the same, after, for the erosion by C by loading column by
column.
Properties of erosion and dilation
0 0 0
1 1 1
0 0 0
0 1 0
0 1 0
0 1 0
B C
1 1 1
1 1 1
1 1 1
D
67. Duality
66
Dilation and erosion are dual operators.
If we set 𝐼𝑐
𝑥 = −𝐼(𝑥), then
𝐼 ⊕ 𝐸 𝑥 = 𝐼𝑐 ⊖ 𝐸
𝑐
(𝑥)
Properties of erosion and dilation
68. Order relation between images
67
We define the following order relation between images:
This is a partial order relation.
We consider two images 𝐼, 𝐽: 𝐴 ⊂ ℤ𝑛 → 𝐵,
𝐼 ≤ 𝐽 ↔ ∀𝑥 ∈ 𝐴, 𝐼 𝑥 ≤ 𝐽(𝑥)
Properties of erosion and dilation
69. Maximum and minimum
68
We can also define the maximum and minimum of two images:
We consider two images 𝐼, 𝐽: 𝐴 ⊂ ℤ𝑛
→ 𝐵,
∀𝑥 ∈ 𝐴:
𝐼 ∨ 𝐽 𝑥 = 𝑀𝑎𝑥 𝐼 𝑥 , 𝐽 𝑥
𝐼 ∧ 𝐽 𝑥 = 𝑀𝑖𝑛(𝐼 𝑥 , 𝐽 𝑥 )
Properties of erosion and dilation
70. Monotonicity
69
Dilation and erosion are both increasing operations regarding the image:
Regarding the structuring element, dilation is increasing while erosion is
decreasing:
If 𝐴1 ≤ 𝐴2, then
(𝐴1 ⊕ 𝐸) ≤ (𝐴2 ⊕ 𝐸)
(𝐴1 ⊖ 𝐸) ≤ (𝐴2 ⊖ 𝐸)
Properties of erosion and dilation
If 𝐸1 ⊆ 𝐸2, then
(𝐴 ⊕ 𝐸1) ≤ (𝐴 ⊕ 𝐸2)
(𝐴 ⊖ 𝐸2) ≤ (𝐴 ⊖ 𝐸1)
71. Extensivity
70
If the structuring element contains the origin, then dilation is extensive and
erosion is anti-extensive (regarding the image).
This means that if the structuring element contains the origin, the gradients
are well defined operations (no negative values possible).
0𝑛 ∈ 𝐸 ⟹
𝐴 ≤ (𝐴 ⊕ 𝐸)
𝐴 ⊖ 𝐸 ≤ 𝐴
Properties of erosion and dilation
72. Strong decomposability
71
Strong decomposability expresses a decomposability property that can be
used in more cases than weak decomposability.
In order to use strong decomposability, we need to express the structuring
element as the maximum of two (or more) structuring elements, which is
always the case.
Soient 𝐼: 𝐴 ⊂ ℤ𝑛
→ 𝐵 et 𝐸, 𝐹 ⊂ ℤ𝑛
,
𝐼 ⊕ 𝐸 ∨ 𝐹 = 𝐼 ⊕ 𝐸 ∨ 𝐼 ⊕ 𝐹
𝐼 ⊖ 𝐸 ∨ 𝐹 = 𝐼 ⊖ 𝐸 ∧ (𝐼 ⊖ 𝐹)
Properties of erosion and dilation
73. Case of one pixel structuring element
72
When the structuring element E has only one element, the dilation or erosion
simply performs a translation of the image:
Let 𝐼: 𝐴 ⊂ ℤ𝑛
→ 𝐵 and E = x such that x ∈ ℤ𝑛
.
Then, 𝐼 ⊖ 𝐸 performs a translation of I by the vector -x, and
𝐼 ⊕ 𝐸 performs a translation of I by the vector x.
Properties of erosion and dilation
77. Opening of a binary image
76
Morphological opening
𝐼 𝐸 𝐼 ⊖ 𝐸
𝐼 ∘ 𝐸
The opening removes all parts of the object (in white) where the
structuring element cannot fit.
78. Opening of a greyscale image
77
Morphological opening
I 𝐼 ∘ 𝐸
E
The opening cuts all mountains smaller than the structuring element.
82. Closing of a binary image
81
Morphological closing
𝐼 𝐸 𝐼 ⊕ 𝐸
𝐼⦁𝐸
The closing removes all parts of the complementary of the object (in
black) where the structuring element cannot fit.
83. Closing of a greyscale image
82
Morphological closing
I 𝐼⦁𝐸
E
The closing fills all canyons smaller than the structuring element.
85. Extensiveness
84
Whatever the structuring element we have, opening is anti-extensive and
cloing is extensive.
For all 𝐼 ⊆ ℤ𝑛 and for all 𝐸 ⊆ ℤ𝑛,
𝐼 ∘ 𝐸 ≤ 𝐼 ≤ (𝐼⦁𝐸)
Properties of opening and closing
86. Monotonicity
85
Opening and closing are both increasing operators regarding the image:
Regarding the structuring element, opening is a decreasing operator and
closing is increasing:
If 𝐴1 ≤ 𝐴2, then
𝐴1 ∘ 𝐸 ≤ (𝐴2 ∘ 𝐸)
𝐴1 ⦁ 𝐸 ≤ (𝐴2 ⦁ 𝐸)
Properties of opening and closing
If 𝐸1 ⊆ 𝐸2, then
𝐴 ∘ 𝐸2 ≤ (𝐴 ∘ 𝐸1)
𝐴 ⦁ 𝐸1 ≤ (𝐴 ⦁ 𝐸2)
87. Duality
86
Opening and closing are dual operators, meaning:
Prove this property (you should use the definition of the opening/closing,
and the duality principle of the erosion and dilation).
𝐼 ∘ 𝐸 = 𝐼𝐶 ⦁ 𝐸 𝐶
𝐼 ⦁ 𝐸 = 𝐼𝐶 ∘ 𝐸 𝐶
Properties of opening and closing
88. Idempotency
87
The last, and most interesting property of opeening and closing, is
idempotency :
It is useless to repeat multiple times a same opening or a same closing in a
row on a same image.
For all 𝐼 ⊆ ℤ𝑛
and for all 𝐸 ⊆ ℤ𝑛
,
𝐼 ∘ 𝐸 = 𝐼 ∘ 𝐸 ∘ 𝐸
𝐼 ⦁ 𝐸 = 𝐼 ⦁ 𝐸 ⦁ 𝐸
Properties of opening and closing
89. Beware of OpenCV
88
There exists two different definitions of dilation…
…leading to two different definition of opening and closing.
Unfortunately, in OpenCV, things got mixed, and the opening and closing
transformations are not well implemented…
…leading to opening and closing which do not always have all the
previously stated properties.
In general, beware when reusing somebody else’s code… you should
trust nobody and test everything!
Properties of opening and closing
91. Pepper and salt noise removal
90
Opening and closing applications
𝐼
𝐼 ∘ Γ4
(𝐼 ∘ Γ4)⦁2Γ8
This approach works well for
« pepper and salt » noise
92. Top-hat
91
Problem: extract grains of rice in I
Threshold at 100 can isolate grains of rice on the bottom, but not on the top
part of the image. Threshold at 130 can isolate the top grains correctly, but
not the bottom one.
What is preventing us from finding a satisfying threshold level?
Opening and closing applications
𝐼 Threshold at 100 Threshold at 130
93. Top-hat
92
Solution : use an opening to remove grains and isolate the background
The Top-Hat operation, which is the difference between an opening and the
original image, allows to remove the background of an image and
compensate any illumination gradient.
Is it well defined or can we have negative values?
Opening and closing applications
𝐼 𝐸 𝐼 ∘ 𝐸 𝐼 − (𝐼 ∘ 𝐸)
Threshold at 50 of 𝐼 − (𝐼 ∘ 𝐸)
95. Counting pieces
94
In order to fully understand what the next transformations do, we should see
how to define a « piece » of an object in a binary image.
Connected components
96. Counting pieces
95
A connected component is the term we will use for the piece of an object.
In a continuous space, the whole notion of connected component is based
on the idea of continuity.
On image A, the red object is made of one connected component, on B, it is
made of two connected components, and on C, it is made of four connected
components.
Connected components
A B C
97. Counting pieces
96
Question: Of how many connected components is the following object
made?
1 or 2 components? This depends if ce consider that two pixels touching by
a corner are in the same object or not…
Connected components
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 1 1 1 0 0 0 0 0
0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 0 0
0 0 0 0 1 1 1 0 0
0 0 0 0 1 1 1 0 0
0 0 0 0 0 1 1 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
98. Choosing a structuring element
97
In order to count the number of connected components of an object, we
need a structuring element that will define the neighbourhood of a
pixel.
If we choose 𝜞𝟒 as the neighbourhood structuring element, then two pixels
touching by a corner won’t be considered as neighbours.
If we choose 𝜞𝟖 as the neighbourhood structuring element, then two pixels
touching by a corner will be considered as neighbours.
Connected components
99. Neighbourhood
98
Let 𝐸 ⊂ ℤ𝑛
be a structuring element such that 𝐸 = 𝐸 and E contains the
origin, and let 𝑥, 𝑦 ∈ ℤ𝑛
be two pixels.
There exists some remarkable neighbourhoods, used in most applications:
. The neighbourhood derived from Γ4 and Γ8, called 4-neighbourhood
and 8-neighbourhood,
. The neighbourhood derived from Γ6 and Γ26, called 6-
neighbourhood and 26-neighbourhood,
The pixels x and y are E-neighbours if 𝑦 ∈ 𝐸𝑥 (equivalent to 𝑥 ∈ 𝐸𝑦).
Connected components
102. Exercise
101
Color in black the E-neighbourhood of the pixel x (outlined in red):
Connected components
0 0 1 0 0
0 1 0 1 0
1 1 1 1 1
0 1 0 1 0
0 0 1 0 0
E
103. Exercise
102
Color in black the F-neighbourhood of the pixel x (outlined in red):
F cannot define a neighbourhood as 𝑭 ≠ 𝑭
Connected components
0 1 1 0 0
0 1 1 0 1
1 0 0 0 1
F
104. Connected set
103
Let J ⊂ ℤ𝑛
(J is a set of pixels), and 𝐸 ⊂ ℤ𝑛
(E is a structuring element
defining a neighbourhood)
The set J is E-connected if and only if
for all 𝒙, 𝒚 ∈ 𝑱, there exists a sequence (𝑝0, 𝑝1, … , 𝑝𝑘) of elements
of J such that
. 𝑝0 = 𝑥,
. 𝑝𝑘 = 𝑦,
. For all 𝑚 ∈ 1; 𝑘 , 𝑝𝑚−1 and 𝑝m are E-neighbours.
Connected components
105. Connected component
104
Let I⊂ ℤ𝑛
(I is a set of pixels), and 𝐸 ⊂ ℤ𝑛
(E is a structuring element
defining a neighbourhood).
The set J ⊆ 𝐼 is an E-connected component of I if and only if
. J is E-connected
. J is maximal for this property : there exists no H ⊆ 𝐼 such
that H is E-connected and 𝐽 ⊂ 𝐻
Connected components
111. Definition
110
We now define conditional dilation:
R is called the Reference image
M is called the Mask
Let R, 𝑀: 𝐴 ⊂ ℤ𝑛
→ 𝐵 be two images, and 𝐸 ⊂ ℤ𝑛
be a structuring
element.
The conditional dilation of M by E restricted to R is
𝑀 ⊕𝑅 𝐸 = 𝑀 ⊕ 𝐸 ∧ 𝑅
Inferior reconstruction
112. Repeating the conditional dilation
111
One can repeat the conditional dilation:
We can define the inferior reconstruction:
Let 𝑅, 𝑀: 𝐴 ⊂ ℤ𝑛
→ 𝐵, and let 𝐸 ⊂ ℤ𝑛
,
(𝑀 ⊕𝑅 𝐸)𝑛= (( 𝑀 ⊕𝑅 𝐸 ⊕𝑅 𝐸) … ⊕𝑅 𝐸) (n times)
Inferior reconstruction
The inferior reconstruction of R by E from M is
𝑅 ∆𝐸 𝑀 = (𝑀 ⊕𝑅 𝐸)∞
(repeat conditional dilation until stability).
113. Inferior reconstruction
The inferior reconstruction is well defined, as it will converge after a
finite number of steps: it is increasing and has R as an upper bound.
In a discrete space, this is enough for achieving convergence in a finite
number of steps.
There exists 𝑘 ∈ ℕ such that
𝑅 ∆𝐸 𝑀 = (𝑀 ⊕𝑅 𝐸)𝑘
Inferior reconstruction
114. Effect of the inferior reconstruction on a binary image
113
Inferior reconstruction
𝑀
𝑅
𝑅 ∆Γ8
𝑀
115. Effect of the inferior reconstruction on a binary image
114
Let R and M be binary images, such that 𝑀 ≤ 𝑅.
(𝑅 ∆𝐸 𝑀) is a binary image containing exactly the E-connected
components of white pixels of R that have at least one pixel white in
M.
Inferior reconstruction
𝑀 𝑅 ∆Γ8
𝑀
𝑅
118. Understanding inferior reconstruction
If M ≤ R, then the result S has
. The same mountains than image M, located at the same place
. The same aspect than image R
Inferior reconstruction
Inferior
Reconstruction
𝑆 = 𝑅 ∆𝐸 𝑀
R
M
E
S
119. Understanding inferior reconstruction
118
The result of the inferior reconstruction of R by E from M has the same
aspect than R, but only possesses mountains where M has mountains.
Inferior reconstruction
R (three mountains) M (two mountains) R ∆Γ8
M
(two mountains)
121. Another grayscale example
120
Inferior reconstruction
R ∆Γ4
𝑀
The result of the inferior
reconstruction does have only one
mountain, located on the same spot
than the mountain of M.
122. Opening by reconstruction
121
The opening by reconstruction consists in performing first an opening of an
image by a structuring element F, and then reconstruct the original image
from the opening using E:
The structuring element E used for reconstruction is usually Γ4 or Γ8.
The structuring element F used for opening has no constraint, and should be
big enough to remove all undesirable mountains.
Let R ∶ 𝐴 ⊂ ℤ𝑛 → 𝐵, and 𝐸, 𝐹 ⊂ ℤ𝑛,
the opening by reconstruction (under E) of R by F is
𝑅 ∘𝐸 𝐹 = 𝑅 ∆𝐸 𝑅 ∘ 𝐹
Opening by reconstruction
126. Definition
125
We now define conditional erosion:
R is called the Reference image
M is called the Mask
Let R, 𝑀: 𝐴 ⊂ ℤ𝑛
→ 𝐵 be two images, and 𝐸 ⊂ ℤ𝑛
be a structuring
element.
The conditional erosion of M by E restricted to R is
𝑀 ⊖𝑅 𝐸 = 𝑀 ⊖ 𝐸 ∨ 𝑅
Superior reconstruction
127. Repeating the conditional erosion
126
One can repeat the conditional erosion:
We can define the superior reconstruction:
Let 𝑅, 𝑀: 𝐴 ⊂ ℤ𝑛
→ 𝐵, and let 𝐸 ⊂ ℤ𝑛
,
(𝑀 ⊖𝑅 𝐸)𝑛= (( 𝑀 ⊖𝑅 𝐸 ⊖𝑅 𝐸) … ⊖𝑅 𝐸) (n times)
Superior reconstruction
The superior reconstruction of R by E from M is
𝑅 𝛻𝐸 𝑀 = (𝑀 ⊖𝑅 𝐸)∞
(repeat conditional erosion until stability).
128. Superior reconstruction
The superior reconstruction is well defined, as it will converge after a
finite number of steps: it is decreasing and has R as a lower bound.
In a discrete space, this is enough for achieving convergence in a finite
number of steps.
There exists 𝑘 ∈ ℕ such that
𝑅 𝛻𝐸 𝑀 = (𝑀 ⊖𝑅 𝐸)𝑘
Superior reconstruction
129. Effect of the superior reconstruction on a binary image
128
Superior reconstruction
𝑀
𝑅
𝑅 𝛻Γ8
𝑀
130. Effect of the superior reconstruction on a binary image
129
Let R and M be binary images, such that 𝑀 ≥ 𝑅.
(𝑅 𝛻𝐸 𝑀) is a binary image containing exactly the E-connected
components of black pixels of R that have at least one pixel black in
M.
Superior reconstruction
𝑀 𝑅 ∆Γ8
𝑀
𝑅
133. Understanding superior reconstruction
If M ≥ R, then the result S has
. The same canyons than image M, located at the same place
. The same aspect than image R
Superior reconstruction
Superior
Reconstruction
𝑆 = 𝑅 𝛻𝐸 𝑀
R
M
E
S
134. Understanding superior reconstruction
133
The result of the superior reconstruction of R by E from M has the same
aspect than R, but only possesses canyon where M has canyon.
Superior reconstruction
R (two canyons) M (one canyon) R 𝛻Γ8
M
(one canyon)
136. Another grayscale example
135
The result of the superior reconstruction of R from M has the same apsect
than R, and no canyon like M (a hole that is not completely surrounded by
higher ground is not considered to be a canyon… therefore, the holes
touching the border are not canyons).
Superior reconstruction
R M (no cayon) 𝐼 𝛻Γ4
𝑀
(no canyon)
137. Closing by reconstruction
136
The closing by reconstruction consists in performing first a closing of an
image by a structuring element F, and then reconstruct the original image
from the closing using E:
The structuring element E used for reconstruction is usually Γ4 or Γ8.
The structuring element F used for closing has no constraint, and should be
big enough to remove all undesirable canyons.
Let R: 𝐴 ⊂ ℤ𝑛
→ 𝐵, and 𝐸, 𝐹 ⊂ ℤ𝑛
,
the closing by reconstruction (under E) of R by F is
𝑅 ⦁𝐸 𝐹 = 𝑅 𝛻𝐸 𝑅 ⦁ 𝐹
Closing by reconstruction
138. Application to noise removal
137
Example: We have pepper noise on an image
Closing by reconstruction
𝐼𝑚 𝐼𝑚 ⦁ 2Γ8 𝐼𝑚 ⦁Γ4
2Γ8
140. Partition
139
We recall the definition of a partition of a set :
Let I be a set, and 𝑃 = {𝑃1, … , 𝑃𝑘} be a set of subsets of I.
We say that P is a partition of I if
∀𝑖, 𝑗 ∈ 1; 𝑘 , 𝑖 ≠ 𝑗 ↔ 𝑃𝑖 ∩ 𝑃𝑗 = ∅
𝑖∈[1;𝑘]
𝑃𝑖 = 𝐼
Properties of filters by reconstruction
141. Exercise
140
Let E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Is {{0,3,7},{2,1,9,8},{4,5},{6,3}} a partition of E ?
No, because 0,3,7 ∪ 6,3 ≠ ∅
Properties of filters by reconstruction
142. Exercise
141
Let E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Is {{1, 8}, {3}, {9, 7, 2}, {5, 6}, {4}} a partition of E ?
No, because 0 does not belong to any element of the partition.
Properties of filters by reconstruction
143. Exercise
142
Let E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Gives a partition of E with four elements at least.
{{2,7,4}, {5,3}, {9,0,1,8}, {6}}
Properties of filters by reconstruction
144. Relation between partitions
143
There exists a partial order relation between partitions :
Let I be a set, and P1 and P2 be two partitions of I.
We say that P2 is coarser than P1 (and P1 is finer than P2) if each
element of P1 is included in one element of P2.
Properties of filters by reconstruction
𝑃1 : Each color represents one
element of the partition
𝑃2
𝑷𝟐 is coarser than 𝑷𝟏
𝑃3
𝑷𝟏 and 𝑷𝟑 cannot be compared
𝑃4
𝑷𝟏 is coarser than 𝑷𝟒
145. Relation between partitions
144
A partition 𝑃1 is coarser than a partition 𝑃2 if 𝑃1 can be obtained simply by
merging elements of P2 together.
In other words, the frontiers between the elements of 𝑃1 are included in the
frontiers between the elements of 𝑃2.
Properties of filters by reconstruction
coarser than coarser than
146. Natural partitioning of an image
145
In a grey level image, an E-flat zone is an E-connected component of
pixels having the same value.
The set of all the E-flat zones of an image represents the natural E-
partitioning of a greyscale image.
Ex : Natural 8-partitioning of image I (each partition has a different color).
Properties of filters by reconstruction
2 3 5 2 3
2 3 5 2 3
2 2 2 1 3
2 1 1 1 3
4 1 1 4 4
I Natural 8-partitioning of I
147. Connected filter
146
In other words, a connected filter acts on an image by merging together
partitions of the natural partitioning of the image.
Let 𝜓 ∶ A ⊂ ℤ𝑛
→ 𝐵 → 𝐴 → 𝐵 be a filter that takes an image as
input and produces an image as output.
The transformation 𝜓 is E-connected if, for all image 𝐼: 𝐴 → 𝐵, the
natural E-partitioning of 𝜓 𝐼 is coarser than the natural E-
partitioning of I.
Properties of filters by reconstruction
148. Connected filter
147
All filters by reconstruction are connected filters.
More precisely, a filter by reconstruction where the
reconstruction is performed with a structuring element E, is an E-
connected filter.
Properties of filters by reconstruction
149. Extensiveness
148
Opening and closing by reconstruction have an extensiveness property: the
larger the structuring element used for the opening or closing, and the
more flat zones of the image will be fusioned.
Properties of filters by reconstruction
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵 be an image, and 𝐸, 𝐹1, 𝐹2 ⊂ ℤ𝑛 be structuring
elements, such that 𝐹1 ⊂ 𝐹2.
The natural E-partitioning of (𝑅 ∘𝐸 𝐹2) is coarser than the natural
E-partitioning of (𝑅 ∘𝐸 𝐹1).
The natural E-partitioning of (𝑅 ⦁𝐸 𝐹2) is coarser than the natural
E-partitioning of (𝑅 ⦁𝐸 𝐹1).
150. Example
149
Properties of filters by reconstruction
𝐼 𝐼 ∘Γ4
12Γ8 𝐼 ∘Γ4
25Γ8 𝐼 ∘Γ4
50Γ8
53218 4-flat zones 39223 4-flat zones 28008 4-flat zones 14034 4-flat zones
152. Definition
151
Alternate Sequential Filter, also called ASF, consists in performing
sequentially openings and closings of an image with a structuring element
whose size increases:
These transformations are very useful in presence of additive and subtractive
noise in the image.
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵 and 𝐸 ⊆ ℤ𝑛,
𝐴𝑆𝐹𝑜𝑐
𝑛 𝐼, 𝐸 = ((((((𝐼 ∘ 𝐸) • 𝐸) ∘ 2𝐸) • 2𝐸) … ∘ 𝑛𝐸) • 𝑛𝐸)
𝐴𝑆𝐹𝑐𝑜
𝑛 𝐼, 𝐸 = ((((((𝐼 • 𝐸) ∘ 𝐸) • 2𝐸) ∘ 2𝐸) … • 𝑛𝐸) ∘ 𝑛𝐸)
ASF
153. Removing noise with ASF
152
Example : remove noise with only ONE opening and ONE closing
ASF
𝐼𝑚 𝐼𝑚 ∘ 2Γ4
(𝐼𝑚 ∘ 2Γ4) • 15Γ4
155. Connected ASF
154
We can go further with ASF and define connected ASF:
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵 and 𝐸 ⊆ ℤ𝑛,
𝐶𝐴𝑆𝐹𝑜𝑐,𝐸
𝑛
𝐼, 𝐹 = ((((( 𝐼 ∘𝐸 𝐹 •𝐸 𝐹) ∘𝐸 2𝐹) •𝐸 2𝐹) … ∘𝐸 𝑛𝐹) •𝐸 𝑛𝐹)
𝐶𝐴𝑆𝐹𝑐𝑜,𝐸
𝑛
𝐼, 𝐹 = ((((( 𝐼 •𝐸 𝐹 ∘𝐸 𝐹) •𝐸 2𝐹) ∘𝐸 2𝐹) … •𝐸 𝑛𝐹) ∘𝐸 𝑛𝐹)
ASF
156. Connected ASF and noise
155
Example : remove noise with a connected ASF
ASF
𝐼𝑚 𝐶𝐴𝑆𝐹𝑜𝑐,Γ4
2
𝐼𝑚, Γ4
𝐶𝐴𝑆𝐹𝑜𝑐,Γ4
2
𝐼𝑚, Γ4 ∘ 2Γ4 • 4Γ4
𝐴𝑆𝐹𝑜𝑐
2
𝐼𝑚, Γ4 • 11Γ4
157. Decomposition through CASF
156
When looking at zones modified by the various steps of a connected ASF,
we can see that they modify and merge larger zones of the image, as the
structuring element grows.
ASF
Im
158. Decomposition through CASF
157
When looking at zones modified by the various steps of a connected ASF,
we can see that they modify and merge larger zones of the image, as the
structuring element grows, creating a scale space decomposition of the
image.
ASF
The red zones show what is modified by
the current step of the ASF compared to
previous step.
𝐶𝐴𝑆𝐹𝑜𝑐,Γ8
𝑛
𝐼𝑚, 𝐵(𝑛)
160. h-extrema
159
The goal of h-extrema filters is to remove mountains or canyons based on
their relative height, and not their shape. Unlike all the other morphological
filters we saw, h-extrema filters elements based on their height (colour)
rather than their shape.
For example, how would we do to remove mountains having a height
smaller than 30 in the image beneath?
h-maxima and h-minima
30
𝐼
𝐼-30
𝐼 ∆𝐸 (𝐼 − 30) E
161. Definition
160
We define h-maxima:
We define h-minima:
Let 𝐼: 𝐴 ⊂ ℤ𝑛
→ 𝐵, 𝐸 ⊂ ℤ𝑛
and ℎ ∈ 𝐵,
h-maxima of I under E is
𝐻𝑀𝐴𝑋ℎ,𝐸 𝐼 = 𝐼 Δ𝐸 (𝐼 − ℎ)
h-maxima and h-minima
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵, 𝐸 ⊂ ℤ𝑛 and ℎ ∈ 𝐵,
h-minima of I under E is
𝐻𝑀𝐼𝑁ℎ,𝐸 𝐼 = 𝐼 𝛻𝐸 (𝐼 + ℎ)
162. Properties
161
The h-maxima removes all mountains whose height is inferior or equal
to h, but also lowers the other mountains by an altitude of h (same for h-
minima with canyons).
h-maxima and h-minima
163. Local extremum
162
A local maximum (or minimum) is a set of pixels that is not surrounded by
pixels of higher (lower) altitude.
Ex :
h-maxima and h-minima
164. Local extrema
163
How can we obtain local maximum (minimum) in an image ?
A local maximum is a set of pixels who are sourrounded by pixels of lower
altitude: it is therefore the top of a mountain of altitude at least 1.
A 1-maxima will remove all mountains top of the image, therefore, all the
local maxima.
With a subtraction, we can extract only the mountain tops that we
removed.
h-maxima and h-minima
166. Local extrema
165
We can define the transformation allowing to extract local (or regional)
minima and maxima:
The local minima and maxima transformations produce a binary image, with
pixels being equal to 0 or 1.
Let 𝐼: 𝐴 ⊂ ℤ𝑛 → 𝐵 and 𝐸 ⊂ ℤ𝑛 ,
𝑅𝑀𝐴𝑋𝐸 𝐼 = 𝐼 − 𝐻𝑀𝐴𝑋1,𝐸(𝐼) (local maxima)
𝑅𝑀𝐼𝑁𝐸 𝐼 = 𝐻𝑀𝐼𝑁1,𝐸 𝐼 − 𝐼 (local minima)
h-maxima and h-minima
167. Application
166
H-maxima and h-minima, combined with local maxima and local minima
can be very efficient to filter objects depending on their relative heights.
Consider the following image, where we want to remove small mountains
(height smaller than 60)
h-maxima and h-minima
168. Application
167
Consider the following image, where we want to remove small mountains
(height smaller than 60)
h-maxima and h-minima
Mountains to keep
Mountains to
remove
169. Application
168
These two mountains have the same absolute height, so a threshold won’t
work to isolate the mountains we want to keep…
h-maxima and h-minima
170. Application
169
Some mountains are large, and some are thin… An opening won’t allow
neither to isolate the good mountains.
h-maxima and h-minima
171. Application
170
Idea: Apply a 60-maxima in order to remove small mountains…
h-maxima and h-minima
(I-60)
𝐻𝑀𝐴𝑋60,Γ4
𝐼
Notice that the mountains we want to keep didn’t
go under their own base after the subtraction, so
still exist as mountains after h-maxima…
173. 𝑅𝑀𝐴𝑋Γ8
(𝐻𝑀𝐴𝑋60,Γ4
𝐼 )
Application
172
Idea: Finally, reconstruct the original image from these regional maxima,
which are the summit of the mountains we wanted to keep
h-maxima and h-minima
𝐼 ΔΓ8
(𝑅𝑀𝐴𝑋Γ8
𝐻𝑀𝐴𝑋60,Γ4
𝐼 )
175. Watershed
174
The Watershed is a segmentation algorithm: it allows to find the frontiers
of a particular object in an image.
The gradient allowed to find the frontiers of all the objects in the image.
Here, we want to extract the frontiers of a particular object.
Watershed transformation
Original image Gradient
176. Marker image
In order to work, the algorithm needs a marker image, where the user will
have identified the location of at least one pixel of the object to extract, and
one pixel of its complement.
The marker image allows to give, to the algorithm, some knowledge about
the object to extract.
Watershed transformation
x
x
One pixels marked
as outside of the
flower
One pixel marked as
inside the flower
175
177. The gradient image
We finally need a gradient image where the contours of the objects on the
image appear bright. This image can be obtained with a morphological
gradient algorithm.
Watershed transformation
Gradient
176
178. Flooding
We consider the gradient image to be a piece of land, with its mountains and
canyons.
We place, on this land, some water pumps that will flood the land with
waters of different colour.
Watershed transformation
x
x
We put a pump here
and another one there
177
179. Flooding
We consider the gradient image to be a piece of land, with its mountains and
canyons.
We place, on this land, some water pumps that will flood the land with
waters of different colour.
Watershed transformation
. When both lakes will meet,
their water won’t mix.
. Two lakes of different colour
will appear and grow little by
little.
178
181. Results
Once the flooding process is finished, we look at which parts of the map
is under the green lake.
Watershed transformation
These parts correspond to the
pixels belonging to the flower.
180
183. Algorithm
We have:
. I is the gradient image,
. M is the marker image, where the marked pixels of the object are
set to 1, the marked pixels of its complement are set to 2, and the others are
set to 0.
. E is a structuring element
Watershed transformation
182
185. More than one object
184
We can also detect more than one object in the image by using more than
two sets of markers.
Watershed transformation
Yeast cells
Gradient
x x
x
x
x x x
x
Markers
Result
186. Conclusion
185
Watershed allows to find the precise frontiers of an object in an image.
The input image must be a gradient image, where objects’ frontiers are
highlighted.
The difficult task consists in finding good markers, automatically if
possible.
Watershed transformation
Editor's Notes
Insérer image avec frontière marquées ou fusion pixels
Insérer image avec frontière marquées ou fusion pixels
Insérer image avec frontière marquées ou fusion pixels
Insérer image avec frontière marquées ou fusion pixels
Ajouter animation
Problème : extraire les contours d’un objet
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Propriétés de l’érosion et la dilatation : l’associativité – oui pour dilat, non pour erod
Illustration
Illustration
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Illustration
Exemple 2d qui sert à quelquechose
Exemple 2d plus grand, niveau de gris, pour voir les différences
Illustration
Illustration
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Exemple 2d sur image avec trous, et histoire des chemins…
Exemple 2d gris large pour voir ce qui se passe, et histoire des chemins (avec expérience)
Illustration
Exemple 2d plus grand, niveau de gris, pour voir les différences
Exemple image niveau de gris, ouverture par reconstruction, fermeture par reconstruction…. Trouver image grande, mais pas trop, pour voir les partitions… et compter zones plates
Exemple d’asf
Exemple d’asf
Exemple d’asf
Exemple sur image, et montrer zones plates et leurs nombres
Exemple sur image, et montrer zones plates et leurs nombres
Exemple signal 1d
Exemple 1d simple avec maximum régionaux
Toujours le même exemple, avec 1-maxima…
Toujours le même exemple, avec 1-maxima…
Count the rings on the mountains to see if they are tall or not… Explain that height of the mountain depends on its base…
Count the rings on the mountains to see if they are tall or not… Explain that height of the mountain depends on its base…
Count the rings on the mountains to see if they are tall or not… Explain that height of the mountain depends on its base…
See that many mountains still exist after the subtraction, even the ones we don’t want to keep
See that many mountains still exist after the subtraction, even the ones we don’t want to keep
See that many mountains still exist after the subtraction, even the ones we don’t want to keep
Exemple pour trouver des marqueurs associés à des cellules…
