In the book Handbook of Data Structures and Applications, section 5.2.5, (copy of the relevant section available here) about leftist queues, the author(s) state the following:

For the complexity analysis of of the initialization operation, assume, for simplicity, that n is a power of 2. The first n/2 melds involve max HBLTs with one element each, the next n/4 melds involve max HBLTs with two elements each; the next n/8 melds are with trees that have four elements each; and so on. The time needed to meld two leftist trees with 2 i elements each is $\mathcal{O}$(i + 1), and so the total time for the initialization is $$\mathcal{O}(n/2 + 2 (n/4) + 3(n/8) + \ldots) = \mathcal{O}(n\sum \frac{i}{2^i}) = \mathcal{O}(n)$$

However, it seems to me that the summation iterates from i = 0 to $\log(n)$: $\sum_{i=0}^{\log(n)}$. This means it varies based on n, which in turn means the algorithm doesn't achieve $\mathcal{O}$(n) time as claimed but instead $\mathcal{O}(n \sum_{i=0}^{\log(n)} \frac{i}{2^i})$ time.

Am I misunderstanding something? I've found other explanations online stating $\mathcal{O}$(n) time for this sort of mass initialization is possible, but this is the only proof I've found so far.


1 Answer 1


The succession $a_i = 2^{-i}i$ is absolutely convergent.

In more elementary terms:

$$ \frac{n}{2} \le \sum_{i=1}^{\log n} \frac{ni}{2^i} \le n\sum_{i=1}^{\infty} \frac{i}{2^i} = n\sum_{j=1}^{\infty} \left( \frac{1}{2^j}\sum_{i=1}^{\infty} \frac{1}{2^i} \right) = 2n $$

  • $\begingroup$ I'm sorry, I don't quite see how that equality is derived. I understand how you go from summation until log to the infinite summation, but not how you go from there to the next step. I did stumble upon the fact that $\sum_{i=0}^{k-1} = 2 - \frac{k+1}{2^{k-1}}$ which can be discarded because of O notation (which I should've found earlier - my bad). $\endgroup$
    – NMR-3
    Commented Mar 29, 2018 at 14:34
  • 1
    $\begingroup$ Let $S = \sum 2^{-i}$. You can rewrite $\sum 2^{-i}i$ as $S + S/2 + S/4 + \cdots$ (each $2^{-i}$ term appears $i$ times). This is a very common transformation, "stolen" from the theory of generating functions. $\endgroup$
    – quicksort
    Commented Mar 29, 2018 at 14:49

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