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ASAM Image Processing 20082009

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Title: ASAM Image Processing 20082009


1
ASAM - Image Processing2008/2009
  • Lecture 21
  • Revision lecture III

Ioannis Ivrissimtzis 13-May-2009
2
Revision questions lecture 12
  • i. Describe the method of intensity slicing for
    pseudo-colouring.
  • ii. Describe gray to colour transformations and
    image filtering for pseudo-colouring.
  • iii. Describe colour transformations and give one
    example for each of the RGB, CMY and HSI colour
    spaces.

3
Intensity slicing
  • A simple way to produce a pseudo-colouring is the
    intensity slicing.
  • We consider the intensity levels
  • where corresponds to black and to
    white.
  • We use the M colours

4
Intensity slicing
  • A pixel with intensity s is assigned the colour
    if
  • The intensity is assigned the colour
    .

5
Example of intensity slicing
  • Monochrome image of the Picker Thyroid Phantom

Results of intensity slicing into eight colour
regions
Images from Gonzalez and Woods
6
Gray to colour transformations
  • The gray level to colour transformations are
    defined by three functions from the set of gray
    levels to the sets of intensities of the red,
    green and blue bands respectively.
  • This transformation can be seen as consisting of
    three independent intensity transformation as
    described in Lecture 2 of this course.

7
Gray to colour transformations
  • The three intensity transformations
  • describe a gray level to colour transformation
  • A pixel with intensity s gets the RGB colour

8
Image filtering for pseudo-colouring
  • In a different approach to pseudo-colouring we
    compute three independent transformations of the
    image rather than the gray scale.
  • For a grayscale image we compute
    its transforms
  • The pseudo-coloured image is written in the RGB
    space as

9
Example
Notice that in the pseudo-coloured image the
outer ring of Saturn is much more visible.
The red band with high frequency information
Original grayscale image
Pseudo-coloured
10
Colour transformations
  • Similarly to the intensity transformations for
    grayscale images and gray to colour
    transformations for pseudo-colouring, we have
    colour transformations.
  • In the RGB space for example, under a colour
    transformation, each RGB colour
  • is mapped to a colour

11
RGB example
  • The colour transformation
  • inverses the colours in a way reminiscent of the
    negatives of the colour films.

12
RGB example
Original image
RGB negative
13
CMY example
  • Visual inspection of the image shows an excess of
    magenta.
  • To balance the colour we convert it to the CMY
    space and transform the magenta component.

14
CMY example
1
1
0
Transformation function of magenta
Original image heavy on magenta
Corrected image
15
HSI example
  • We want to brighten the image using histogram
    equalisation.
  • Histogram equalisation on each RGB component we
    will change the hues and the processed image will
    look unnatural.

16
HSI example
  • Instead we
  • Convert the image to the HSI space.
  • We transform the intensity component applying
    histogram equalisation on it.
  • In addition, we transform the saturation
    component to get less saturated colours.

17
HSI example
1
1
0
Transformation function of saturation
Original image
Processed image
The transformation function of the intensity
component was computed by applying histogram
equalisation on it.
18
Revision questions lecture 13
  • i. Describe the morphological operations
    dilation, erosion, opening and closing for binary
    images.
  • ii. Describe how binary morphology can be used
    for boundary extraction.

19
Revision questions lecture 13
  • iii. Find the number of ones at the binary image
    I after dilation, erosion, opening and closing
    with the structuring element S.

20
Structuring elements
  • Similarly to the case of spatial linear
    filtering, a binary image can be processed by
    another binary image called structuring element.
  • A typical structuring element has few 1-valued
    pixels, and one of his pixels is designated a its
    origin.
  • In an analogy between spatial linear filtering
    and morphological image processing, the
    structuring element can be seen as the equivalent
    of the mask and its origin as the equivalent of
    the centre of a mask.

21
Dilation
  • Let A be a binary image and B a structuring
    element.
  • The dilation of A by B is the set consisting of
    all the locations of the origin of B where the
    translated B overlaps at least some portion of A.
  • The mechanics of the dilation are similar to that
    of spatial linear filtering. We translate B in
    all possible positions and if there is some
    overlap with A, then the location of the origin
    belongs to the dilation.

22
Erosion
  • Let A be a binary image and B a structuring
    element.
  • The erosion of A by B is the set consisting of
    all the locations of the origin of B where the
    translated B fits entirely within A.
  • The mechanics of the erosion are similar to that
    of dilation. We translate B in all possible
    positions and if it fits entirely within A, then
    the location of the origin belongs to the
    erosion.

23
Opening
  • In practical applications, dilations and erosions
    are used most often in various combinations.
  • The opening of the binary image A by the
    structuring element B is the erosion of A by B,
    followed by dilation of the result by B.

24
Closing
  • The closing of the binary image A by the
    structuring element B is the dilation of A by B,
    followed by erosion of the result by B.
  • We can show that the closing of A by B is the
    complement of union of all translations of B that
    do not overlap A.

25
Boundary extraction
  • Let A ? B denotes the dilation of A by B and let
    A - B denotes the erosion of A by B.
  • The boundary of A can be computed as
  • A - ( A - B )
  • where be is a 3x3 square structuring element.
  • That is, we subtract from A an erosion of and
    obtain its boundary.

26
Exercise
  • After dilation the image has 44 one-valued images

27
Exercise
  • After erosion the image has 8 one-valued images

28
Exercise
  • After opening the image has 20 one-valued images

29
Exercise
  • After closing the image has 24 one-valued images

30
Revision questions lecture 14
  • i. Describe the use of the following masks for
    point and line detection
  • ii. Describe edge detection based on the
    Laplacian of the Gaussian.

31
Point detection
  • For the detection of isolated points we can use
    the mask

It is a form of the Laplacian with the diagonal
directions included.
32
Point detection
  • We say that a point has been detected at the
    location (x,y), if the absolute value of the
    response is above a threshold T.
  • If the detected points are labelled 1 and all
    others are labelled 0, we can visualise the
    results of point detection by a binary image
    g(x,y)

33
Line detection
  • A line is an edge segment in which the intensity
    of the background on either side is either much
    higher or much lower than the intensity of the
    line pixels.
  • The Laplacian mask
  • can be used for line detection.

34
Line detection
  • The previous mask is isotropic. It does not have
    preference for any particular direction.
  • Anisotropic masks can be used to detect lines in
    specified directions

Horizontal
45
Vertical
- 45
35
Line detection
  • Let R1, R2, R3, R4 be the responses of the four
    masks centred on (x,y).
  • The pixel (x,y) is said to be more likely
    associated with the direction of the mask with
    the highest response in absolute values.
  • For example, if
  • we say that (x,y) is more likely associated with
    the horizontal direction.

36
Laplacian of Gaussian
  • Edge detectors must be able to cope with noise.
    They must also be able to act at any scale.
  • The Marr-Hilderth edge detector, solves these two
    problems by first applying a Gaussian mask on the
    image, and then a Laplacian.
  • The size of the Gaussian mask depends on the
    scale of the edges we want to detect.

37
Laplacian of Gaussian
  • The Laplacian of the Gaussian of an image can be
    efficiently implemented using a single mask
    computed as the convolution of a Gaussian mask by
    a Laplacian mask.
  • An example of a Laplacian of Gaussian mask is

38
Laplacian of Gaussian
  • The final step is to compute the zero-crossings,
    that is, the pixels where the Laplacian of the
    Gaussian changes sign.
  • Typically, we define a binary image with 0s
    corresponding to negative pixel values and 1 to
    positive.
  • Then we compute the boundary of the binary image,
    e.g. with morphological operators.

39
Revision questions lecture 15
  • i. Use the Sobel masks Sx, Sy to compute the
    gradient at the centre of the image I.
  • ii. Describe the Canny edge detector.

40
Exercise
The first order difference in the
x-direction (vertical) is the response the Sobel
mask
The first order difference in the
y-direction (horizontal) is the response of the
Sobel mask
41
Exercise
The gradient at the centre of I is
I
42
Canny edge detector
  • The Canny edge detection algorithm consists of
    the following steps
  • Smooth the input image with a Gaussian filter
  • Compute the gradient magnitude and angle
  • Apply non-maxima suppression to the gradient
    magnitude
  • Use double thresholding and connectivity analysis
    to detect and link edges

43
Non-maxima suppression
  • In the array (image) of the gradient magnitude
    the edges of the original image are represented
    by wide ridges. See previous lecture
  • The non-maxima suppression step thins these
    ridges, by keeping only the locally maximum
    values of the gradient magnitude .

44
Non-maxima suppression
  • The maximum suppression first quantizes the
    gradient angle into the four directions of a
    pixels neighbourhood.

For each pixel, we find the sector of the
gradient angle, and assign to the pixel one of
the four directions, horizontal, - 45, vertical,
or - 45.
45
Non-maxima suppression
  • The second step of non-maxima suppression makes
    zero the gradient magnitude of a pixel if it is
    smaller than the gradient magnitude of any of the
    two pixels in the quantized gradient direction.
  • For example, if the quantized gradient direction
    is horizontal, then we compare
    with the gradient magnitude values at its left
    and its right, and make it zero if it is smaller
    than either of them.

46
Double thresholding
  • We use a high and a low threshold for
    the gradient magnitude values.
  • The edges are found by a tracking algorithm
    starting from pixels above the high threshold and
    stopping and pixels below the low threshold.

47
Edge tracking
  • The edge tracking algorithm
  • Start a new edge from a pixel above the high
    threshold that has not already been visited
  • Track the edge, following the quantized gradient
    angle in both directions and mark all pixels
    above the low threshold as edges. The edge
    tracking stops at pixels below the low threshold.
  • Until all pixels above the high threshold have
    been visited.

48
Example
Original image
Thresholded gradient of smoothed image
Result of Canny algorithm
49
Revision questions lecture 16
  • i. Describe a local edge linking algorithm for
    detecting edges in the horizontal and vertical
    direction.

50
Edge linking algorithm
  • Compute the gradient magnitude and angle arrays,
    M(x,y) and a(x,y) of the input image.
  • Form a binary image g, whose value at any pair of
    coordinates (x,y) is given by
  • where TM is a threshold used for edge detection,
    A is a specified angle direction, and TA defines
    the band of acceptable directions about A.

51
Edge linking algorithm
  • Scan the rows of g and fill (set to 1) all gaps
    (sets of 0s) in each row that do not exceed a
    specified length K.
  • By definition, a gap is bounded at both ends by
    one or more 1s.
  • Step 3 of the algorithm links edges in the
    horizontal direction.
  • To link edges in any other direction ?, we rotate
    g by ? and apply Steps 2-3. We then rotate the
    result back by -?.

52
Local edge linking algorithm 2
Gradient magnitude
Horizontal edge linking
Vertical edge linking
For the horizontal edge linking, TM was set at
30 of the maximum gradient value, A90 because
the gradient direction is vertical to the edge,
TA45. K25, that is, we fill gaps of 25 or
fewer pixels.
53
Edge linking algorithm
The logical OR of the horizontal and vertical
edge linking
The final result after morphological image
post-processing
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