Tagged Questions

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Solving (for asymptotics) of certain recurrence equations.

I am thinking of examples of the kind where the function occurs multiple times on the R.H.S with different arguments. This is the case where most techniques I know don't seem to work. For example ...
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Substitution method for solving recurrences

I'm having issues understanding how to solve recurrences via the substitution method. From what I understand, I need to guess the form of the solution first and then use induction to prove that the ...
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Recurrence relation and big-O-notation

Consider the following recurrence relation: $$T(n)=c\cdot + 2\cdot T(n/2)$$ This is the recurrence relation for the Merge-Sort algorithm. How can one deduce from this equation the time complexity of ...
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Solving the recurrence $T (n) = \sqrt{n} T(\sqrt{n}) + O (n)$ [duplicate]

I want to show that the requrrence $T (n) = \sqrt{n} T(\sqrt{n}) + O (n)$ is in $O(n \log \log n)$ Here's my attempt: If we expand the recursion tree, at a level $i$, there are $n^{1/2^k}$ ...
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Prove Upper Bound (Big O) for Fibonacci's Sequence?

NOTE: We are not to use proofs (limits, induction, or otherwise) in this problem. We were to prove the upper bound for the Fibonacci recursion is some exponential. The Fibonacci recurrence relation ...
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Help with Recursive Algorithm

We are to determine a recurrence relation for a recursive algorithm. Let us use the Josephus Problem for this: Given n people standing in a circle, every kth person is killed until one person ...
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Analyzing $n_{i+1} = n_i - n_i^{3/4}$

I have a non-linear recurrence given by $$n_0 = N \\ n_{i+1} = n_i - n_i^{3/4}$$ Are there any techniques to solve this for an exact closed form? Or in lieu of that, an asymptotic estimation? I'm ...
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Recursive sequence with binomial coefficients

I have a sequence $\epsilon_i$ defined recursively for $i\ge 1$ as follows \begin{eqnarray*} \epsilon_1 &=& \frac{1}{p}\\ \epsilon_n &=& \frac{1}{1-(1-p)^n}\left( 1 + \sum_{j=1}^{n-1} ...
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An issue with approximations of a recurrence sequence

By trying to give an approximation to a given recurrence sequence I encountered a problem. To be more precise I have a method but it fails if the right condition is not met and I wonder how I should ...
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Master Method and use cases

$T(n)=T(n-2)+n^{2}$ and $T(n)=4T(n-2)+n^{2}$ Master method to solve these two equations? I know I can use the other cases where $a$ and $b > 0$ but since $T(n-2)$ do I assume $b$ is $1$?
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Solving a recurrence realtion using backward substitution.

So I've been trying my best to do this, and I have made some good progress, I just need to know if what I have done is correct and if not, what the hell am I doing wrong? :P I start off with this ...
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Solving Recurrence Relation with Forward Substitution

I've found myself quite stuck on this recurrence relation. I've been given it to solve, via forward substitution and verify using induction. I start out with $$T(n) = 4T(n/3)$$ For all $n > 1$ ...
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Proving that $T(n) = 3T\left(\frac n3\right) + \sqrt n = \Theta(n)$

Show that $T(n)$ is bounded both above and below by $n$ (abusing the Big O notation) for some positive constants $c_1$ and $c_2$: $$T(n) = 3T\left(\frac n3\right) + \sqrt n = \Theta(n)$$ ...
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Finding the asymptotic behavior of the recurrence $T(n)=4T(\frac{n}{2})+n^2$ by using substitution method

I am trying to solve a recurrence by using substitution method. The recurrence relation is: $$T(n)=4T\left(\frac{n}{2}\right)+n^2$$ My guess is $T(n)$ is $\Theta (n\log n)$ (and I am sure about it ...
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What is the asymptotical bound of this recurrence relation?

I have the recurrence relation, with two initial conditions $$T(n) = T(n-1) + T(n-2) + O(1)$$ $$T(0) = 1, \qquad T(1) = 1$$ With the help of Wolfram Alpha, I managed to get the result of ...
Solve the following recurrence relations and give a bound for each of them. $T(n)= 2T(n-3)+1$ $T(n) = 5T(n-4)+n$
Asymptotic behaviour of $f(x) =f(\sqrt{x}) + \sqrt{x}$
I stumbled about this recursive function today: $$f_n = f_\sqrt{n} + \sqrt n$$ I tried to solve it with substitution ($m = \log_2 n, \quad g_{2^m} = g_{2^{m/2}} + 2^{m/2}$), but I have a bad feeling ...