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-   -   Problem 2.8 (http://book.caltech.edu/bookforum/showthread.php?t=2053)

axelrv 10-10-2012 08:46 AM

Problem 2.8
 
I thought that growth functions could take on any polynomial function of N, so why is 1 + N + N(N-1)(N-2) / 6 not a possible growth function?

magdon 10-10-2012 08:52 AM

Re: Problem 2.8
 
The growth function cannot be any polynomial function of N. A valid growth function must satisfy the following theorem:

Theorem: If m(k)<2^k for some (any) k, then for all N, m(N)\le N^{k-1}+1.

In your example growth function below, try k=2 to show that the precondition of the theorem is satisfied with k=2 and hence deduce a linear bound on m(N). This contradicts the growth function being cubic. More generally, a cubic growth function cannot have a break point less than 4.

Quote:

Originally Posted by axelrv (Post 6247)
I thought that growth functions could take on any polynomial function of N, so why is 1 + N + N(N-1)(N-2) / 6 not a possible growth function?


axelrv 10-10-2012 01:44 PM

Re: Problem 2.8
 
thanks!

k_sze 03-21-2018 10:15 AM

Re: Problem 2.8
 
Quote:

Originally Posted by magdon (Post 6249)
The growth function cannot be any polynomial function of N. A valid growth function must satisfy the following theorem:

Theorem: If m(k)<2^k for some (any) k, then for all N, m(N)\le N^{k-1}+1.

In your example growth function below, try k=2 to show that the precondition of the theorem is satisfied with k=2 and hence deduce a linear bound on m(N). This contradicts the growth function being cubic. More generally, a cubic growth function cannot have a break point less than 4.

To show that a growth function is invalid, is it sufficient to do this?
  1. Determine the smallest k where m_\mathcal{H}(k+1) < 2^{k+1}, such that k would be our d_{\text{vc}} if the growth function were valid;
  2. Find any concrete value of N where one of the inequalities vs m_\mathcal{H}(N) is violated.

htlin 03-22-2018 03:08 AM

Re: Problem 2.8
 
Quote:

Originally Posted by k_sze (Post 12971)
To show that a growth function is invalid, is it sufficient to do this?
  1. Determine the smallest k where m_\mathcal{H}(k+1) < 2^{k+1}, such that k would be our d_{\text{vc}} if the growth function were valid;
  2. Find any concrete value of N where one of the inequalities vs m_\mathcal{H}(N) is violated.

Sounds reasonable to me.


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