Activity 15 - Pattern Recognition

The human brain is such a magnificent organ with a huge capacity for logic and intellect. Machines and computers today don’t compare to the complexity of an average human brain. The human brain has the aptitude to classify features together and draw inference from a set of objects. An active research today is to construct a computer which has the pattern recognition abilities of a human brain.

A pattern is a set of characteristics or features. Color, shape, size are a few characteristics that is quantifiable in an image.  Set of patterns that share similar properties are called class. Pattern recognition aims to decide if a given feature belongs to one of several classes.

For this activity, I assembled sets of coins for this activity. Each independent class has 10 objects with the same properties. The samples are divided into two groups: the training set and the test set. The training set is the five coins in the while the test set is the five coins in the left for each image. The objects used here are 1 peso coins, 5 peso coins and 25 centavo coins. 

Figure 1. Image of ten 25 centavo coins

Figure 2. Image of ten 1 peso coins

Figure 1. Image of ten 5 peso coins

The features I chose to get from these sets are perimeter, area, and green-channel chromaticity. I was able to extract the area and perimeter on the training set by using the code in the previous activity. These features will be our feature vector. They are tabulated in table 1 below. 

	Feature Vector
	Training	Area	Perimeter	Green-Channel Chromaticity
25 cents	1	1320	128	0.364468
	2	1194	122	0.370987
	3	1385	131	0.377735
	4	1256	125	0.361357
	5	1256	125	0.367904
1 peso	1	1963	157	0.380245
	2	1734	147	0.379551
	3	1809	150	0.371178
	4	1734	147	0.378921
	5	1809	150	0.379221
5 peso	1	2123	163	0.358633
	2	2375	172	0.364429
	3	2463	175	0.367722
	4	2375	172	0.362215
	5	2290	169	0.359801

Notice that the values of perimeter and area in pixels are very large compared to the average green chromaticity. If we use these values, the perimeter and area would dominate the green chromaticity. To solve this we normalize the values. We divide all the areas with the largest value on our objects which was 2463. We do the same with the perimeter and divide it by 175. The normalized values are shown below in table 2.

	Feature Vector
	Training	Area	Perimeter	Green-Channel Chromaticity
25 cents	1	0.536033	0.732143	0.364468
	2	0.485013	0.696428	0.370987
	3	0.5625	0.75	0.377735
	4	0.510204	0.714286	0.361357
	5	0.510204	0.714286	0.367904
1 peso	1	0.797194	0.892857	0.380245
	2	0.704401	0.839285	0.379551
	3	0.734694	0.857143	0.371178
	4	0.704401	0.839285	0.378921
	5	0.734694	0.857143	0.379221
5 peso	1	0.862245	0.928571	0.358633
	2	0.964605	0.982143	0.364429
	3	1	1	0.367722
	4	0.964605	0.982143	0.362215
	5	0.929847	0.964285	0.359801

After normalizing, we take the mean feature vector of the classes. The mean feature vector is seen on table 3 below.

	Mean feature Vector, m
	Area	Perimeter	Average
25 cents	0.508068	0.7125	0.367909
1 Peso	0.741454	0.860714	0.377561
5 peso	0.951021	0.975	0.364033

 

We then get these features on the test objects. We use the same process used in extracting the features on the training set. The values are shown in table 4.

 

	Feature Vector
	Test Object	Area	Perimeter	Green-Channel Chromaticity
25 cents	1	1256.637	125.6637	0.364127
	2	1194.59	122.5221	0.368932
	3	1134.115	119.3805	0.372138
	4	1320.254	128.8053	0.363312
	5	1194.59	122.5221	0.368129
1 Peso	1	2042.82	160.2212	0.385217
	2	1885.741	153.938	0.376219
	3	1661.902	144.5132	0.372638
	4	1809.557	150.7964	0.374298
	5	1809.557	150.7964	0.378125
5 Peso	1	2290.221	169.646	0.356125
	2	2290.221	169.646	0.357149
	3	2375.829	172.7876	0.364589
	4	2463.008	175.9292	0.390125
	5	2375.829	172.7876	0.359537

 

We also normalize these values using the area and perimeter used in the training set. The normalized values are shown in table 5.

	Feature Vector
	Test Object	Area	Perimeter	Green-Channel Chromaticity
25 cents	1	0.510204	0.714286	0.364127
	2	0.485013	0.696428	0.368932
	3	0.460459	0.678571	0.372138
	4	0.536033	0.732143	0.363312
	5	0.485013	0.696428	0.368129
1 Peso	1	0.829401	0.910714	0.385217
	2	0.765625	0.875	0.376219
	3	0.674745	0.821428	0.372638
	4	0.734694	0.857143	0.374298
	5	0.734694	0.857143	0.378125
5 Peso	1	0.929847	0.964285	0.356125
	2	0.929847	0.964285	0.357149
	3	0.964605	0.982143	0.364589
	4	1	1	0.390125
	5	0.964605	0.982143	0.359537

 

After doing this, we can now move on to classification. The class membership of each test object can be determined by finding the class where its feature vector is the closest. Using The Minimum Distance we can now classify them.

Equation for Eucledian distance

Equation for Eucledian norm

We classify x to the class with the smallest distance. The Eucledian distance of the test object is shown in the table below.

					Classification
		d for 25 cents	d for 1 peso	d for 5 peso
25 cents	1	0.007702	0.391112	0.701625	25 cents
	2	0.040143	0.429356	0.749479	25 cents
	3	0.085769	0.468561	0.795096	25 cents
	4	0.052208	0.348241	0.658565	25 cents
	5	0.03934	0.430159	0.748676	25 cents
1 Peso	1	0.536855	0.14561	0.20709	1 Peso
	2	0.428367	0.039802	0.297582	1 Peso
	3	0.280334	0.110918	0.438453	1 Peso
	4	0.377658	0.013594	0.344449	1 Peso
	5	0.381485	0.010891	0.348276	1 Peso
5 Peso	1	0.685344	0.313396	0.039799	5 Peso
	2	0.684324	0.312372	0.038769	5 Peso
	3	0.7295	0.357552	0.021276	5 Peso
	4	0.801648	0.4104	0.100072	5 Peso
	5	0.73455	0.362604	0.02522	5 Peso

 I highlighted the one with the minimum distance and classified the test object to its corresponding class. The result of the classification gives a 100% correct results.

II would give myself a grade of 8/10 for being able to do this but posting it late. I would like to thank Tino Borja for giving me idea on how to do this

References

1. Soriano M.Pattern recognition AP 186 2011

1. Soriano M. Area Estimation for Images with Defined Edges  AP 186 2011