Re: Page 47 / Lecture 6 (10 min) Partitioning the table
Thank you very much for replying.
I read the answers and think that I understand the points being made in regards to the maximum being 2^(n1).
But on reflection, perhaps my misunderstanding could have been expressed more clearly. So, I am sorry about the following tables, but perhaps 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 (N1 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
So, given the partitions above, I seem to get that X1, X2 appears twice but this is not N1 appearing twice (i.e. X1, X2, X3).
I understand we are attempting to describe situations were there is a break point and all the combinations listed above will not exist.
But given the instructions in the book and using a niceandneat 2n construction, I do not understand how the exhaustive and exclusive 3 sets are derived.
Thanks in advance,
Mark
