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Old 04-21-2013, 12:36 AM
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yaser yaser is offline
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Default Re: Page 47 / Lecture 6 (10 min) Partitioning the table

Quote:
Originally Posted by markact View Post
these help to demonstrate what I am missing in regards to partitioning the original set into 3 disjoint sets.

The following assumes a space of 4 points - a maximum of 16 dichotomies. I have labelled each row with its base 10 equivalence.

Here is alpha (N-1 appears once, with XN either 1 or 0)

X1 X2 X3 XN ID
0 0 0 0 0
0 0 1 1 3
0 1 0 0 4
0 1 1 1 7
1 0 0 0 8
1 0 1 1 11
1 1 0 0 12
1 1 1 1 15

We are left with 8 rows remaining If these have XN being either 1 or -1, then these 8 rows are split into the following two partitions (S2+ and S2-).


[S2+]
X1 X2 X3 XN ID
0 0 0 1 1
0 1 0 1 5
1 0 0 1 9
1 1 0 1 13


[S2-]
X1 X2 X3 XN ID
0 0 1 0 2
0 1 1 0 6
1 0 1 0 10
1 1 1 0 14
Thank you for the detailed example. I can now see where the misunderstanding is. The set S_1 contains all the patterns that appear once, and none of the patterns that appear twice, of the original matrix. Therefore, if all the above patterns are indeed in that matrix, then the decimal pattern 0 should not be in S_1 since the decimal pattern 1 makes that pattern appear twice (on the first 3 columns). Similarly, the set S_2 contains all patterns that appear twice, and none that appear once.
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