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Color
It is not too difficult to convert RGB values into HSI values to facilitate color processing in computer vision applications. First of all, we normalize the range of the R, G, and B components to the interval from 0 to 1. For Windows programs programs, for example example, this can be done by dividing each value by 255. Then we compute the intensity I as I = 1/3*(R + G + B). Obviously, intensity also ranges from 0 to 1.
HSI
RGB February 10, 2011
Conversion from RGB to HSI
Computer Vision Lecture 4: Image Filtering
1
February 10, 2011
Conversion from RGB to HSI
green p-w
r = R/(R + G + B) g = G/(R + G + B) b = B/(R + G + B) When we consider the RGB cube, then all possible triples (r, g, b) lie on a triangle with corners (1, 0, 0), (0, 1, 0), and (0, 0, 1). We could call this the rgb rgb--subspace of our RGB cube. 3
Conversion from RGB to HSI
red (p ( pr ) w = (1/3, 1/3, 1/3) (white)
blue
The hue is the angle H from vector pr – w to vector p – w. The saturation is the distance from w to p relative to the distance from w to the fully saturated color of the same hue as p (on the edge of the triangle). February 10, 2011
Computer Vision Lecture 4: Image Filtering
4
We can also compute:
(p − w ) ⋅ (p r − w ) || p − w || ⋅ || p r − w ||
(p − w ) ⋅ (p r − w ) =
Since w = (1/3, 1/3, 1/3):
And since pr = (1, 0, 0):
cos H =
|| p r − w ||= 2 / 3 Computer Vision Lecture 4: Image Filtering
2(r − 1 / 3) − ( g − 1 / 3) − (b − 1 / 3) 3
With the above formulas, including those for deriving r, g, and d b ffrom R R, G G, and dB B, we can d determine t i an equation for computing H directly from R, G, and B:
|| p − w ||= (r − 1 / 3) 2 + ( g − 1 / 3) 2 + (b − 1 / 3) 2
February 10, 2011
pr - w
Conversion from RGB to HSI
Then we have:
cos H =
p = (r, g, b) H
Computer Vision Lecture 4: Image Filtering
2
Conversion from RGB to HSI
Then we compute the values r, g, b that are independent of intensity: intensity:
February 10, 2011
Computer Vision Lecture 4: Image Filtering
5
February 10, 2011
2R − G − B 2 ( R − G ) 2 + ( R − B)(G − B ) Computer Vision Lecture 4: Image Filtering
6
1
Conversion from RGB to HSI
Limitations of RGB and HSI
The saturation is the distance on the triangle in the rgb--subspace from white relative to the distance from rgb white to the fully saturated color with the same hue. Fully saturated colors are on the edges of the triangle. The derivation of the formula for saturation S is very lengthy, so we will just take a look at the result:
Using three individual wavelengths to represent color can never cover the entire visible range of colors:
S = 1−
February 10, 2011
3 min( R, G , B ) R+G + B Computer Vision Lecture 4: Image Filtering
7
Limitations of any Color Representation
February 10, 2011
Computer Vision Lecture 4: Image Filtering
8
Limitations of any Color Representation
Our perception of an object’s color does not only depend on the frequency spectrum emitted from the object’s location. It also depends on the spectra of other objects or regions in the visual field. This mechanism called color constancy allows us to assign a color to a given object that is invariant to shading or illumination of the scene by varying light sources. The underlying perceptual mechanisms are not fully understood yet. February 10, 2011
Computer Vision Lecture 4: Image Filtering
9
February 10, 2011
Computer Vision Lecture 4: Image Filtering
10
Limitations of any Color Representation Let’s move on to…
Image Filtering February 10, 2011
Computer Vision Lecture 4: Image Filtering
11
February 10, 2011
Computer Vision Lecture 4: Image Filtering
12
2
Histogram Modification
Histogram Modification
A common and important filter operation is histogram modification. modification. Between any two stages of image processing, it often happens that the range of intensity values in our image is only a small proportion of the possible range range. This means that the contrast in the image is weaker than it would have to be. It is then useful to modify the intensity histogram of the image.
One possible method for this is image scaling: scaling: We simply expand the range [a, b] containing most of the intensities in the image to fill the entire range [z1, zk].
February 10, 2011
Computer Vision Lecture 4: Image Filtering
13
This means that the value z of each pixel in the original image is mapped onto the value z’ in the scaled image in the following way:
z' =
Notice that this method may leave gaps between bins in the resulting histogram. February 10, 2011
Computer Vision Lecture 4: Image Filtering
Image Filtering
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
14
Image Filtering
• Many basic image processing techniques are based on convolution. • In a convolution, a convolution filter W pp to every yp pixel of an image g I to create a is applied filtered image I*: ∞
z k − z1 ( z − a ) + z1 b−a
Example: Averaging filter: y 0
∞
∑ ∑ I (u, v)W (u − x, v − y)
0
0
0
0 0
0
1/9 1/9
1/9
0
1/9 1/9
1/9
0
0
1/9 1/9
1/9
0
0
0
0
0
0
x
u = −∞ v = −∞
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
∞
∞
∑ ∑ I (u, v)W (u − x, v − y)
u = −∞ v = −∞ February 10, 2011
Computer Vision Lecture 4: Image Filtering
15
February 10, 2011
Computer Vision Lecture 4: Image Filtering
Image Filtering Image:
Image Filtering
Filter: Grayscale Image:
1
6
3
2
9
1/9
1/9
1/9
2
11
3
10
0
1/9
1/9
1/9
5
10
6
9
7
1/9
1/9
1/9
3 4
16
1 4
0 2
2 9
1
8 New value: 6/9 + 9/9 + 7/9 + 0/9 + 2/9 + 8/9 + 2/9 + 9/9 + 10/9 = 5.889
10
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
∞
Averaging Filter:
6
3
2
9
1/9
1/9
1/9
2
11
3
10
0
1/9
1/9
1/9
5
10
6
9
7
1/9
1/9
1/9
3
1
0
2
8
4
4
2
9
10
∞
∑ ∑ I (u, v)W (u − x, v − y)
u = −∞ v = −∞ February 10, 2011
Computer Vision Lecture 4: Image Filtering
17
February 10, 2011
Computer Vision Lecture 4: Image Filtering
18
3
Image Filtering Original Image:
Image Filtering
Filtered Image:
11/9 61/9 31/9 2
9
0
0
21/9 11 1/9 31/9 10
0
0
5
0
0
Original Image:
Filtered Image:
0
1
6 3 2 9 1/9 1/9 1/9
0
0
0
0
2
11 3 10 0 1/9 1/9 1/9
0
5
7
0
0 0
51/9 10 1/9 61/9 9
7
0
0
5
10 6 9 7 1/9 1/9 1/9
0
0
3
1
0
2
8
0
0
3
1
0
2
8
0
0
4
4
2
9
10
0
0
4
4
2
9
10
0
0
0
0
value = 1⋅1/9 + 6⋅1/9 + 3⋅1/9 + 2⋅1/9 + 11⋅1/9 + 3⋅1/9 + 5⋅1/9 + 10⋅1/9 + 6⋅1/9 = 47/9 = 5.222 February 10, 2011
Computer Vision Lecture 4: Image Filtering
19
1
6
3
2
0
0
0
0
February 10, 2011
11
3
10
0
0
5
7
5
0
10
6
9
7
0
5
6
5
0
3
1
0
2
8
0
4
5
6
0
2
9
10
0
0
Computer Vision Lecture 4: Image Filtering
0
0
W ( x, y ) = G ( x, y ) =
0
5
4
−
1 2πσ
2
e
x2 + y2 2σ 2
0 1/273
Now you can see the averaging (smoothing) effect of the 3×3 filter that we applied. Computer Vision Lecture 4: Image Filtering
21
Image Filtering
February 10, 2011
3×3
9×9
Computer Vision Lecture 4: Image Filtering
February 10, 2011
1
4
7
4
1
4
16
26
16
4
7
26
41
26
7
4
16
26
16
4
1
4
7
4
1
Computer Vision Lecture 4: Image Filtering
22
Properties of Gaussian Filters
Effect of Gaussian smoothing:
original
20
• implement decreasing influence b more di by distant t t pixels i l
Discrete version:
February 10, 2011
0
More common: Gaussian Filters
2
4
0
Image Filtering
Filtered Image: 9
0
value = 6⋅1/9 + 3⋅1/9 + 2⋅1/9 + 11⋅1/9 + 3⋅1/9 + 10⋅1/9 + 10⋅1/9 + 6⋅1/9 + 9⋅1/9 = 60/9 = 6.667
Image Filtering Original Image:
0
15× 15 ×15 23
The application of Gaussian convolution filters can be made more efficient. efficient. This is important, for example, if we want to apply different Gaussian filters to a large number of big input images. The basic idea is to separate the convolution with the 2D Gaussian filter into two successive convolutions with 1D Gaussian filters. One of these filters is vertical vertical,, the other one horizontal.. horizontal February 10, 2011
Computer Vision Lecture 4: Image Filtering
24
4
Properties of Gaussian Filters
Properties of Gaussian Filters
The general form of the Gaussian filter, without a normalizing factor, is given by:
g[i, j ] = e
−
Then we have: m
i2 + j2
k =1 l =1
2σ 2
m
m
k =1 l =1 k2 m − 2 2σ
k =1 l =1
= ∑e
This formula needs to be applied to all coordinates [i, j] in f in order to create the convoluted image. 25
k =1
February 10, 2011
The formula in the curly braces describes the convolution of f[i f[i, j] with a horizontal oneonedimensional Gaussian filter. filter. The remainder of the formula takes the result of this first convolution and performs a convolution with a vertical oneone-dimensional Gaussian filter on it. it So instead of applying an m×n Gaussian convolution filter, we can successively apply a 1× 1×n filter and an m×1 filter. This increases the efficiency of computation. Computer Vision Lecture 4: Image Filtering
27
−
k2 2σ 2
−
e
l2 2σ 2
f [i − k , j − l ]
⎧⎪ n − l 2 ⎫⎪ ⎨∑ e 2σ f [i − k , j − l ]⎬ ⎪⎩ l =1 ⎪⎭ 2
Computer Vision Lecture 4: Image Filtering
26
• However, there are different types of noise. noise. • For soso-called “salt “salt--and and--pepper” noise, for example, a median filter can be more effective.
February 10, 2011
Computer Vision Lecture 4: Image Filtering
28
Median Filter
• Use, for example, a 3× 3×3 filter and move it across the image like we did before. • For each position, compute the median of the brightness values of the nine pixels in question. – To compute the median, sort the nine values in ascending order order. – The value in the center of the list (here, the fifth value) is the median. • Use the median as the new value for the center pixel. Computer Vision Lecture 4: Image Filtering
f [i − k , j − l ]
• Smoothing can reduce noise in the image. • This can be useful, for example, if you want to find regions of similar color or texture in an image.
Median Filter
February 10, 2011
k 2 +l 2 2σ 2
Different Types of Filters
Properties of Gaussian Filters
February 10, 2011
n
= ∑∑ e
n
Computer Vision Lecture 4: Image Filtering
−
k =1 l =1
g[i, j ] * f [i, j ] = ∑∑ g[k , l ] f [i − k , j − l ]
February 10, 2011
n
= ∑∑ e
The convolution of an image f[i, j] with a Gaussian filter of size m× m×n is given by: m
n
g[i, j ] * f [i, j ] = ∑∑ g[k , l ] f [i − k , j − l ]
29
• Advantage of the median filter: Capable of eliminating outliers such as the extreme brightness values in saltsalt-and and--pepper noise. • Disadvantage: The median filter may change the contours of objects in the image.
February 10, 2011
Computer Vision Lecture 4: Image Filtering
30
5
Median Filter
original image February 10, 2011
3×3 median Computer Vision Lecture 4: Image Filtering
7×7 median 31
6
It is not too difficult to convert RGB values into HSI values to facilitate color processing in computer vision applications. First of all, we normalize the range of the R, G, and B components to the interval from 0 to 1. For Windows programs programs, for example example, this can be done by dividing each value by 255. Then we compute the intensity I as I = 1/3*(R + G + B). Obviously, intensity also ranges from 0 to 1.
HSI
RGB February 10, 2011
Conversion from RGB to HSI
Computer Vision Lecture 4: Image Filtering
1
February 10, 2011
Conversion from RGB to HSI
green p-w
r = R/(R + G + B) g = G/(R + G + B) b = B/(R + G + B) When we consider the RGB cube, then all possible triples (r, g, b) lie on a triangle with corners (1, 0, 0), (0, 1, 0), and (0, 0, 1). We could call this the rgb rgb--subspace of our RGB cube. 3
Conversion from RGB to HSI
red (p ( pr ) w = (1/3, 1/3, 1/3) (white)
blue
The hue is the angle H from vector pr – w to vector p – w. The saturation is the distance from w to p relative to the distance from w to the fully saturated color of the same hue as p (on the edge of the triangle). February 10, 2011
Computer Vision Lecture 4: Image Filtering
4
We can also compute:
(p − w ) ⋅ (p r − w ) || p − w || ⋅ || p r − w ||
(p − w ) ⋅ (p r − w ) =
Since w = (1/3, 1/3, 1/3):
And since pr = (1, 0, 0):
cos H =
|| p r − w ||= 2 / 3 Computer Vision Lecture 4: Image Filtering
2(r − 1 / 3) − ( g − 1 / 3) − (b − 1 / 3) 3
With the above formulas, including those for deriving r, g, and d b ffrom R R, G G, and dB B, we can d determine t i an equation for computing H directly from R, G, and B:
|| p − w ||= (r − 1 / 3) 2 + ( g − 1 / 3) 2 + (b − 1 / 3) 2
February 10, 2011
pr - w
Conversion from RGB to HSI
Then we have:
cos H =
p = (r, g, b) H
Computer Vision Lecture 4: Image Filtering
2
Conversion from RGB to HSI
Then we compute the values r, g, b that are independent of intensity: intensity:
February 10, 2011
Computer Vision Lecture 4: Image Filtering
5
February 10, 2011
2R − G − B 2 ( R − G ) 2 + ( R − B)(G − B ) Computer Vision Lecture 4: Image Filtering
6
1
Conversion from RGB to HSI
Limitations of RGB and HSI
The saturation is the distance on the triangle in the rgb--subspace from white relative to the distance from rgb white to the fully saturated color with the same hue. Fully saturated colors are on the edges of the triangle. The derivation of the formula for saturation S is very lengthy, so we will just take a look at the result:
Using three individual wavelengths to represent color can never cover the entire visible range of colors:
S = 1−
February 10, 2011
3 min( R, G , B ) R+G + B Computer Vision Lecture 4: Image Filtering
7
Limitations of any Color Representation
February 10, 2011
Computer Vision Lecture 4: Image Filtering
8
Limitations of any Color Representation
Our perception of an object’s color does not only depend on the frequency spectrum emitted from the object’s location. It also depends on the spectra of other objects or regions in the visual field. This mechanism called color constancy allows us to assign a color to a given object that is invariant to shading or illumination of the scene by varying light sources. The underlying perceptual mechanisms are not fully understood yet. February 10, 2011
Computer Vision Lecture 4: Image Filtering
9
February 10, 2011
Computer Vision Lecture 4: Image Filtering
10
Limitations of any Color Representation Let’s move on to…
Image Filtering February 10, 2011
Computer Vision Lecture 4: Image Filtering
11
February 10, 2011
Computer Vision Lecture 4: Image Filtering
12
2
Histogram Modification
Histogram Modification
A common and important filter operation is histogram modification. modification. Between any two stages of image processing, it often happens that the range of intensity values in our image is only a small proportion of the possible range range. This means that the contrast in the image is weaker than it would have to be. It is then useful to modify the intensity histogram of the image.
One possible method for this is image scaling: scaling: We simply expand the range [a, b] containing most of the intensities in the image to fill the entire range [z1, zk].
February 10, 2011
Computer Vision Lecture 4: Image Filtering
13
This means that the value z of each pixel in the original image is mapped onto the value z’ in the scaled image in the following way:
z' =
Notice that this method may leave gaps between bins in the resulting histogram. February 10, 2011
Computer Vision Lecture 4: Image Filtering
Image Filtering
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
14
Image Filtering
• Many basic image processing techniques are based on convolution. • In a convolution, a convolution filter W pp to every yp pixel of an image g I to create a is applied filtered image I*: ∞
z k − z1 ( z − a ) + z1 b−a
Example: Averaging filter: y 0
∞
∑ ∑ I (u, v)W (u − x, v − y)
0
0
0
0 0
0
1/9 1/9
1/9
0
1/9 1/9
1/9
0
0
1/9 1/9
1/9
0
0
0
0
0
0
x
u = −∞ v = −∞
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
∞
∞
∑ ∑ I (u, v)W (u − x, v − y)
u = −∞ v = −∞ February 10, 2011
Computer Vision Lecture 4: Image Filtering
15
February 10, 2011
Computer Vision Lecture 4: Image Filtering
Image Filtering Image:
Image Filtering
Filter: Grayscale Image:
1
6
3
2
9
1/9
1/9
1/9
2
11
3
10
0
1/9
1/9
1/9
5
10
6
9
7
1/9
1/9
1/9
3 4
16
1 4
0 2
2 9
1
8 New value: 6/9 + 9/9 + 7/9 + 0/9 + 2/9 + 8/9 + 2/9 + 9/9 + 10/9 = 5.889
10
I * ( x, y ) = I ( x, y ) * W ( x, y ) =
∞
Averaging Filter:
6
3
2
9
1/9
1/9
1/9
2
11
3
10
0
1/9
1/9
1/9
5
10
6
9
7
1/9
1/9
1/9
3
1
0
2
8
4
4
2
9
10
∞
∑ ∑ I (u, v)W (u − x, v − y)
u = −∞ v = −∞ February 10, 2011
Computer Vision Lecture 4: Image Filtering
17
February 10, 2011
Computer Vision Lecture 4: Image Filtering
18
3
Image Filtering Original Image:
Image Filtering
Filtered Image:
11/9 61/9 31/9 2
9
0
0
21/9 11 1/9 31/9 10
0
0
5
0
0
Original Image:
Filtered Image:
0
1
6 3 2 9 1/9 1/9 1/9
0
0
0
0
2
11 3 10 0 1/9 1/9 1/9
0
5
7
0
0 0
51/9 10 1/9 61/9 9
7
0
0
5
10 6 9 7 1/9 1/9 1/9
0
0
3
1
0
2
8
0
0
3
1
0
2
8
0
0
4
4
2
9
10
0
0
4
4
2
9
10
0
0
0
0
value = 1⋅1/9 + 6⋅1/9 + 3⋅1/9 + 2⋅1/9 + 11⋅1/9 + 3⋅1/9 + 5⋅1/9 + 10⋅1/9 + 6⋅1/9 = 47/9 = 5.222 February 10, 2011
Computer Vision Lecture 4: Image Filtering
19
1
6
3
2
0
0
0
0
February 10, 2011
11
3
10
0
0
5
7
5
0
10
6
9
7
0
5
6
5
0
3
1
0
2
8
0
4
5
6
0
2
9
10
0
0
Computer Vision Lecture 4: Image Filtering
0
0
W ( x, y ) = G ( x, y ) =
0
5
4
−
1 2πσ
2
e
x2 + y2 2σ 2
0 1/273
Now you can see the averaging (smoothing) effect of the 3×3 filter that we applied. Computer Vision Lecture 4: Image Filtering
21
Image Filtering
February 10, 2011
3×3
9×9
Computer Vision Lecture 4: Image Filtering
February 10, 2011
1
4
7
4
1
4
16
26
16
4
7
26
41
26
7
4
16
26
16
4
1
4
7
4
1
Computer Vision Lecture 4: Image Filtering
22
Properties of Gaussian Filters
Effect of Gaussian smoothing:
original
20
• implement decreasing influence b more di by distant t t pixels i l
Discrete version:
February 10, 2011
0
More common: Gaussian Filters
2
4
0
Image Filtering
Filtered Image: 9
0
value = 6⋅1/9 + 3⋅1/9 + 2⋅1/9 + 11⋅1/9 + 3⋅1/9 + 10⋅1/9 + 10⋅1/9 + 6⋅1/9 + 9⋅1/9 = 60/9 = 6.667
Image Filtering Original Image:
0
15× 15 ×15 23
The application of Gaussian convolution filters can be made more efficient. efficient. This is important, for example, if we want to apply different Gaussian filters to a large number of big input images. The basic idea is to separate the convolution with the 2D Gaussian filter into two successive convolutions with 1D Gaussian filters. One of these filters is vertical vertical,, the other one horizontal.. horizontal February 10, 2011
Computer Vision Lecture 4: Image Filtering
24
4
Properties of Gaussian Filters
Properties of Gaussian Filters
The general form of the Gaussian filter, without a normalizing factor, is given by:
g[i, j ] = e
−
Then we have: m
i2 + j2
k =1 l =1
2σ 2
m
m
k =1 l =1 k2 m − 2 2σ
k =1 l =1
= ∑e
This formula needs to be applied to all coordinates [i, j] in f in order to create the convoluted image. 25
k =1
February 10, 2011
The formula in the curly braces describes the convolution of f[i f[i, j] with a horizontal oneonedimensional Gaussian filter. filter. The remainder of the formula takes the result of this first convolution and performs a convolution with a vertical oneone-dimensional Gaussian filter on it. it So instead of applying an m×n Gaussian convolution filter, we can successively apply a 1× 1×n filter and an m×1 filter. This increases the efficiency of computation. Computer Vision Lecture 4: Image Filtering
27
−
k2 2σ 2
−
e
l2 2σ 2
f [i − k , j − l ]
⎧⎪ n − l 2 ⎫⎪ ⎨∑ e 2σ f [i − k , j − l ]⎬ ⎪⎩ l =1 ⎪⎭ 2
Computer Vision Lecture 4: Image Filtering
26
• However, there are different types of noise. noise. • For soso-called “salt “salt--and and--pepper” noise, for example, a median filter can be more effective.
February 10, 2011
Computer Vision Lecture 4: Image Filtering
28
Median Filter
• Use, for example, a 3× 3×3 filter and move it across the image like we did before. • For each position, compute the median of the brightness values of the nine pixels in question. – To compute the median, sort the nine values in ascending order order. – The value in the center of the list (here, the fifth value) is the median. • Use the median as the new value for the center pixel. Computer Vision Lecture 4: Image Filtering
f [i − k , j − l ]
• Smoothing can reduce noise in the image. • This can be useful, for example, if you want to find regions of similar color or texture in an image.
Median Filter
February 10, 2011
k 2 +l 2 2σ 2
Different Types of Filters
Properties of Gaussian Filters
February 10, 2011
n
= ∑∑ e
n
Computer Vision Lecture 4: Image Filtering
−
k =1 l =1
g[i, j ] * f [i, j ] = ∑∑ g[k , l ] f [i − k , j − l ]
February 10, 2011
n
= ∑∑ e
The convolution of an image f[i, j] with a Gaussian filter of size m× m×n is given by: m
n
g[i, j ] * f [i, j ] = ∑∑ g[k , l ] f [i − k , j − l ]
29
• Advantage of the median filter: Capable of eliminating outliers such as the extreme brightness values in saltsalt-and and--pepper noise. • Disadvantage: The median filter may change the contours of objects in the image.
February 10, 2011
Computer Vision Lecture 4: Image Filtering
30
5
Median Filter
original image February 10, 2011
3×3 median Computer Vision Lecture 4: Image Filtering
7×7 median 31
6
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