Differential Equations Solutions 53

# Differential Equations Solutions 53 - CHALLENGE 11.4. A...

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63 Figure 11.4. The images resulting from k -means. cluster; for example, for k = 5, minimizing D produced only 3 nonempty clusters, and for k = 2, minimizing R produced only 2 nonempty clusters. (d) If q< 4, then k might be chosen by plotting the data. For larger values of q , we might try increasing values of k , stopping when the cluster radii fall below the noise level in the data or when the cluster radii stay relatively constant. Only one choice of data values appears in the sample program, but we can easily modify the program to see how sensitive the solution is to the choice of data.
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Unformatted text preview: CHALLENGE 11.4. A sample program is given on the website and results are shown in Figure 11.4. This k-Means function is much faster than the algorithm for Challenge 3. The best results for k = 3 and k = 5 are those from k-means, but the k = 4 result from minimizing R seems somewhat better than the other k = 4 results. The quantitative measures are mixed: the 2-norm of the relative change is .247, .212, and .153 for k = 3 , 4 , 5 respectively, although the algorithm was not run to convergence....
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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