From a list of $n$ distinct numbers, I want to find the set consisting of all odd-ranked numbers (1st, 3rd, 5th, ...). How many comparison queries do I need?

I could sort the whole list using $O(n\log n)$ queries, but that gives me more information than I need. On the other hand, the classic argument that sorting requires $\Omega(n\log n)$ queries only yields that my task requires roughly $\Omega(n)$ queries. Is it possible to do my task in $O(n)$ queries?

  • $\begingroup$ If you need them sorted then it reduces to sorting the array: find the odd-ranked numbers with $g(n)$ queries, remove the minimum from the array and find the odd-ranked numbers from this new array with $g(n-1)$ queries and merge both lists with $O(n)$ queries, hence a total of $O(g(n)+n)$ queries. $\endgroup$
    – holf
    Mar 27 at 17:26

1 Answer 1


Lemma 1. Any comparison-based algorithm requires $\Omega(n\log n)$ comparisons in the worst case.

Proof sketch. Let $A$ be any comparison-based algorithm for the problem. Let $x=(x_1, x_2, \ldots, x_n)$ be an arbitrary permutation of $[n]$, considered as an instance of the problem. I will argue that in order to solve $x$ (that is, to identify the odd-ranked elements of $x$, given $x$), $A$ has to sort $x$. The lemma will follow by the well-known result that any comparison-based sorting algorithm requires $\Omega(n\log n)$ comparisons.

Consider the execution of $A$ on $x$. Suppose for contradiction that at termination the comparisons that $A$ has done do not suffice to sort $x$. Then, for some $x_i$ and $x_j = x_i+1$, $A$ never directly compared $x_i$ and $x_j$.

Consider executing $A$ on the input $x'$ obtained from $A$ by swapping the values of $x_i$ and $x_j$. Inductively, the sequence of comparisons that $A$ makes will be exactly the same as it made on input $x$, because for every comparison $x'_a < x'_b ?$, the outcome is the same as the corresponding comparison $x_a < x_b ?$ for $x$, unless $\{a, b\} = \{i, j\}$, but the comparison $x'_i < x'_j ?$ is never made.

So $A$ outputs the same subset of the given variables for $x$ as it does for $x'$. One of these two outputs must be wrong, contradicting the correctness of $A$ $~~~\Box$.

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    $\begingroup$ Nice! Though I don't understand why the word "sketch" was used ;) . $\endgroup$ Mar 28 at 3:53
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    $\begingroup$ This answer can be stated more concisely: across all permutations of $[n]$, there are $n C \frac{n}{2}$ index sets that the oddly ranked items take. Therefore any BDD, such as one induced by pairwise computations, must have depth at least $\log_2(n C \frac{n}{2}) \in \theta(n \log n)$. $\endgroup$ Mar 29 at 2:28
  • $\begingroup$ I suspect that the $Ω(n \log n)$ comparisons result holds even if we have to classify just an expected $1/2+ε$ (for $Ω(1)$ $ε$) fraction of the elements correctly as to whether their rank is odd or even. $\endgroup$ Apr 1 at 20:44
  • $\begingroup$ @ReinstateMonica, no, wait, ${n\choose n/2} = \Theta(2^n/\sqrt n)$, so $\log_2{n\choose n/2} = \Theta(n)$. $\endgroup$
    – Neal Young
    Apr 1 at 21:15
  • $\begingroup$ @NealYoung Ah, you are right, good catch! $\endgroup$ Apr 3 at 17:47

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