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We can use Master Theorem to solve this : If a Recurrence Relation is of the Form $$T(n)=aT\bigg(\frac{n}{b}\bigg)+{n^k}({\log(n)})^p$$ Then, as per Master Theorem, we have Six Conditions depending on value of $a,b,k$ and $p$ If $\log_ba>k$ Answer is $\theta(n^{\log_ba})$ If $log_ba=k$ and $p>-1$ Answer is $\theta({n^k}({\log(n)})^{p+1})$ If $log_ba=...


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A) This depends on how you've set things up. If, eg, you have $\mathcal{S}$, $\mathcal{T}$ as subsets of $\mathbb{N}$, then a computable (recursive) bijection $F : \mathcal{S} \to \mathcal{T}$ needs to be only defined on $\mathcal{S}$, not on all of $\mathbb{N}$. B) A straight-forward example would be $G : \mathbb{N} \to \mathbb{N}$ where $G(2n) = 2n$ if $n \...


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If $n/3$ is equal to the index of the current median, the algorithm terminates. If $n/3$ is greater than the index of the current median, the algorithm moves to the right subarray and recursively solves there. If $n/3$ is smaller than the index of the current median, the algorithm moves to the left subarray and recursively solves there. Time Complexity: For ...


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We introduce a weight measure for $S$, a set of numbers with respect to its average, namely: $w(S) = \sum_{s\in S} |s-avg(S)|$. The measure $w(S)$ expresses the total distance to the average. Based on this weight we will prove the number of iterations is in $O(\log\ n)$. If we apply one iteration of your described algorithm on a set $S$ we obtain either the ...


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Does this pseudocode help? Let me know of any clarifications. stack_0.push(0) // stack containing only vertex 0 stack_of_stacks = empty stack_of_stacks.push((stack_0, vertex 0)) // added a tuple of stack and vertex 0 while stack_of_stacks is not empty: (temp_stack, temp_vertex) = stack_of_stacks.pop() if temp_vertex = n-1 ...


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