#1




Growth Function division into S1 and S2
I had one question on the book related to theory of generalization, where the dichotomies are "shuffled" in growth function. You mentioned that for (x1,x2....xn1, xn) we take one of them in the S1 or alpha. How come then "two" of the patterns stay in S2, it should be either of the "unused" pattern that stay either of the Betas right?
For example (x1 x2 x3 x4) and lets take the pattern 1 1 1 1 (stays in alpha) we have only one remaining (1 1 1 1) which will go to one of the betas, with 1 if we choose 1 (ending one) to go to alpha. I understand the reformulation but the example given in lecture uses the same pattern (1 1....1 1) and (1 1....1 1) to be in both betas, only 1 of them can be in any betas and the other one should be in the alpha, right? Can you elaborate on this or clarify with example i have? Thanks a ton! Uday 
#2




Re: Growth Function division into S1 and S2
A full dichotomy on all points can have a "twin" dichotomy (also on all points) that is identical to it except for the last bit, where it differs so one dichotomy ends with and the other ends with . It is those "twin" dichotomies that end up in the set (both of them); the one ending with goes to and the one ending with goes to . Each of and has elements, hence which is their union has elements.
The dichotomies that do not have a "twin" are the ones that end up in and there are of them. Each dichotomy goes to one, and only one, of the sets , , . I hope this clarifies the matter. Please feel free to ask further questions.
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#3




Re: Growth Function division into S1 and S2
Thanks Dr Yaser that makes it more clear!

#4




Re: Growth Function division into S1 and S2
Hello!
I read several posts in the forum but still struggle to understand the split between S1, S2+ S2. Lets consider N = 4. So there are at most 16 patterns of 0 and 1, which could be perfectly split between S2+ and S2. This will be correct for any N if all patterns are included i.e. 2^N. So what will be going to S1? Thanks! 
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