I used the test image data in Activity 18.
group 1 - one peso coins
group 2 - 25 centavo coins
x - features of all data
| 66511 | 0.926294 |
| 66473 | 0.949915 |
| 66511 | 0.777889 |
| 45555 | 1.163125 |
| 48126 | 0.777889 |
| 47958 | 0.930314 |
y - group of the object of all data.
y = [1;1;1;2;2;2]
x1 - group 1
x1 = [66511 0.926294;66473 0.949915; 66511 0.777889];
x2 - group 2
x2 = [45555 1.163125; 48126 0.777889; 47958 0.930314];
u1 - mean of features of group 1
66498.333
0.8846993
u2 - mean of features of group 2
47213.
0.9571093
u - mean features of all
56855.667
0.9209043
x0 - mean corrected data
Therefore,
| x01 | x02 |
| 9655.333 | -11300.7 |
| 0.00539 | 0.242221 |
| 9617.333 | -8729.67 |
| 0.029011 | -0.14302 |
| 9655.333 | -8897.67 |
| -0.14302 | 0.00941 |
c1 and c2 are covariance of each group.
c1 =
| 92981341 | -349.939 |
| -349.939 | 0.007108 |
c2 =
| 94360206 | -524.168 |
| -524.168 | 0.026404 |
C - group Covariance
C =
| 93670774 | -437.053 |
| -437.053 | 0.016756 |
We then find the inverse of C
InvC =
| 1.215D-08 | 0.000317 |
| 0.000317 | 67.94902 |
f1 and f2 are the Linear Discriminat Functions
| f1 | f2 |
| 74.81613 | 72.76803 |
| 76.69272 | 74.62444 |
| 62.76609 | 60.89518 |
| 71.22988 | 73.33025 |
| 42.74908 | 44.76578 |
| 54.94263 | 56.81285 |
We will use a test image( a 25 centavo coin). Let's pretend that we did not know what it is.
T = [48392 0.7848263]
The Linear Discriminant Functions are
f1T = 43.601981
f2T = 45.554149
From the graph, it shows that the test function is close to the 25 centavos. So I think our prediction is right. :)Acknowledgments:
Linear Discriminant Analysis Lecture
Grade: 10/10 because I think that my result is right.
