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   A recurrence formula with logloglog(n) solution.
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   Author  Topic: A recurrence formula with logloglog(n) solution.  (Read 3907 times)
Gennadiy Korol
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A recurrence formula with logloglog(n) solution.  
« on: Nov 8th, 2007, 9:57am »
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Good time of the day everyone, I would like to share one of the riddles that I've came across recently.  
 
We need to find the recurrence relation which solution is O( logloglog(n) ).  O here denotes "Theta", the tight asymptotic bound.
 
For instance, the recurrence relation:
 
T(n) = T(n/2) + O(1)  is of order O(log n).
 
 
That's pretty much all to it Smiley  
 
 
Regards,
Gennadiy
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Pietro K.C.
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Re: A recurrence formula with logloglog(n) solutio  
« Reply #1 on: Nov 8th, 2007, 9:25pm »
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Cute puzzle! The relation T(n) = T(\sqrt n) + O(1) gives O(log log n), and going one step further into the "reverse" Ackermann hierarchy yields a full solution.
« Last Edit: Nov 8th, 2007, 9:25pm by Pietro K.C. » IP Logged

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Re: A recurrence formula with logloglog(n) solutio  
« Reply #2 on: Nov 29th, 2007, 7:23am »
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Yes, for log(i) n you gain equation f_i(n) = 2^(2^(2^...2^((log log log ... log n)-1)...)) where you use i times log and the same times 2^.
 
f_{i+1}(n) = 2^(f_i(log n))
 
for i=0 f_0(n)=n-1
for i=1 f_1(n)=2^(log n-1)=2^log (n/2) = n/2,
for i=2 f_2(n)=2^(2^(log log n - 1))=2^((log n)/2)=sqrt(n),  
for i=3 2^(2^(2^(log log log n -1)))=2^sqrt(log n)
« Last Edit: Nov 29th, 2007, 7:23am by Hippo » IP Logged
Gennadiy Korol
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Re: A recurrence formula with logloglog(n) solutio  
« Reply #3 on: Dec 3rd, 2007, 2:23pm »
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Yep!
 
Hippo: Thanks for providing the generalized solution, it gives some more insight into the problem. This is also one of the solutions I know for this riddle.
 
 
Pietro K.C.: My mathematic knowledge is on the undergraduate level, so I am not sure how to use Ackermann's inverse function in the particular case. I assume you are referring to the "towers of powers", similar to Hippo's suggestion?
 
 
There's also another solution to the original problem, very simple one, I would even call it cheating Smiley Its correctness follows immediately from the Master method.  
 
 
Thanks for participating,
Gennadiy
« Last Edit: Dec 3rd, 2007, 2:26pm by Gennadiy Korol » IP Logged
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