#1




Page 47 / Lecture 6 (10 min) Partitioning the table
I am struggling with one of more simpler bits of chapter 2.
On page 47 (Lecture 6  9 ~ 12 mins), I do not understanding how the table is constructed. "Some dichotomies on these N1 points appear only once" Okay, so to me, this implies we have 2^(n1) rows for the first set (S1). "The remaining dichotomies on the first n1 appear twice, once with +1 and once with 1" How can this be true...? To me, "the remaining dichotomies on the first n1" would mean (2^n) less (2^n1) = 2^n1. But these appear twice! So we have 2*(2^n1) = 2^n for set S2. This seems as though we are counting twice. S1 = 2^n1 S2 = 2^n The total of S1 + S2 is greater than 2^n. Surely, I must be missing something really simple here. But what...? Please help. Thanks, Mark 
#2




Re: Page 47 / Lecture 6 (10 min) Partitioning the table
Quote:
Quote:
__________________
Where everyone thinks alike, no one thinks very much 
#3




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 
#4




Re: Page 47 / Lecture 6 (10 min) Partitioning the table
Quote:
__________________
Where everyone thinks alike, no one thinks very much 
#5




Re: Page 47 / Lecture 6 (10 min) Partitioning the table
Got it!!
Thank you very much for your patience. 
Thread Tools  
Display Modes  

