1. Pattern Recognition and Training
Pattern -- set of values (known size) that describe things
The general problem
Approaches to the decision-making process
1. Simple comparison
2. Common property
3. Clusters (using distance measurement
|X1-u1| + |X2-u2| + … + |Xn-un| )
4. Combination of 1, 2 and 3
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Image Processing, Leture #13

2. 240-373 Image Processing, Leture #13
Decision Functions
Decision function: w = (w1, w2, w3, …, wn)
If the pattern vector is x = [x1, x2, x3, …, xn, 1]T, then
The unknown pattern is in group B if wTx > 0
The unknown pattern is in group A if wTx <= 0
Example: (8,4) is in group B because
[1.5, -1.0, -3.5] [8, 4, 1]T = 8x = 4.5
and 4.5 > 0
How about (4,4)?
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3. Decision Functions (Cont’d)
The number of groups can be more than 2
Decision table
Result of w1 Result of w2 Implication
< < 0 no group
< > 0 group A
> < 0 group C
> > 0 group B
Decision function need not be a linear function
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Image Processing, Leture #13

4. 240-373 Image Processing, Leture #13
Cluster Means
If the cluster consists of [3,4,8,2] [2,9,5,1][5,7,7,1], then the mean is [3.33, 6.67, 6.67, 1.33]. This represents the center of the four-dimensional cluster.
The Euclidean distance from the center to a new pattern can be calculated as follows:
new vector [3, 5, 7, 0],
Euclidean distance = (3-3.33)2
+ (5-6.67)2
+ (7-6.67)2
+ (0-1.33)2 =
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Image Processing, Leture #13

5. 240-373 Image Processing, Leture #13
Automatic Clustering
Technique 1: K-means clustering
USE: To automatically find the best groupings and means of K clusters.
OPERATION:
The pattern vectors of K different items are given to the system
Classifying them as best it can (without knowing which vector belongs to which item)
Let the pattern vectors be X1, …, Xn
Take the first K points as the initial estimation of the cluster means
M1 = X1, M2 = X2, …, Mk = Xk
* Allocate each pattern vector to the nearest group (minimum distance)
Calculate new cluster centers
If they are the same as the old centers, then STOP, other wise goto step *
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Image Processing, Leture #13

6. K-means clustering example
M1 = (2, 5.0) M2 = (2, 5.5)
Allocating each pattern vector to the nearest center gives
1 (2,5.0) group 1
2 (2,5.5) group 2
3 (6,2.5) group 1
4 (7,2.0) group 1
5 (7,3.0) group 1
6 (3,4.5) group 1
The group means now become
group 1: M1 = (5, 3.4) group 2: M2 = (2, 5.5)
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7. 240-373 Image Processing, Leture #13
This gives new groupings as follows:
1 (2,5.0) group 2
2 (2,5.5) group 2
3 (6,2.5) group 1
4 (7,2.0) group 1
5 (7,3.0) group 1
6 (3,4.5) group 2
And the group means become
group 1: M1 = (6.67, 2.5) group 2: M2 = (2.33, 5.0)
Groupings now stay the same and the processing stops.
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Image Processing, Leture #13

8. Optical Character Recognition
Technique: Isolation of a character in an OCR document
USE: To create a window containing only one character onto an array containing a text image
OPERATION:
1. Assuming that the image is correctly oriented and the text is dark on a white background
2. Calculate row sums of the pixel gray-level values. High row sums indicate a space between the rows
3. Calculate column sums of the pixel gray-level values. High column sums indicate a space between the columns
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Image Processing, Leture #13

9. 240-373 Image Processing, Leture #13
Technique: Creating the pattern vector (feature extraction)
USE: To create the pattern vector for a character so that it can be compared with the library
OPERATION:
1. Assuming that the character has been isolated
2. Place a 4x4 grid over the image and count the number of “ink” pixels in each grid.
3. These number are then divided by the total number of pixels in the grid
4. Comparing resulting numbers with the library
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Image Processing, Leture #13