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iensen
  • Member for 12 years, 11 months
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
>It will be bounded from below by logn/loglogn Yes,i have got this by replacing LOGN with LOG N_F,where N_F is N which was an argument of the first call. Do you have any ideas for the upper bound?
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Recurrence T(N)=T(N/LOGN)+1
This equation (like most questions on math.stackexchange.com,but i didn't find exactly this one) originate from Cormen's book about algorithms. It is not stated clearly anywhere but it can be composed on the base of one of the book's chapters.
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Recurrence T(N)=T(N/LOGN)+1
I believe the approximation log(n/logn) is not correct. T(1000 000)= 7,T(1000 000 000 )=9...,it grows much slower. I wonder if there is a method for solving such equations
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Recurrence T(N)=T(N/LOGN)+1
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Recurrence T(N)=T(N/LOGN)+1
Oh sorry,the log is "binary" log with base 2.
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Circlular Convolution for two finite sequences of different length.
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Circlular Convolution for two finite sequences of different length.
I met this on my digital signal processing exam(the question was the length of the output sequence). So,do you know any possible uses of it?
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Circlular Convolution for two finite sequences of different length.
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