Lower Bound on Sorting an Array Where Elements Are at Most $\sqrt{n}$ Positions from Their Sorted Position

Question

Let $A[1 \ldots n]$ be an array of $n$ elements, such that each element is at most $\sqrt{n}$ positions away from its position in the fully sorted array. That is, for every element $A[i]$, its position in the sorted array is within $i \pm \sqrt{n}$.

Question:

Show that any comparison-based sorting algorithm must take $\Omega(n \log n)$ time in the worst case, even under this constraint.

In other words, even though the array is "almost sorted" — with each element displaced by at most $\sqrt{n}$ positions — prove that we cannot asymptotically improve on the standard lower bound for sorting.

What I’ve considered so far:

For arrays where every element is at most $k$ positions from its sorted position, there are known algorithms (e.g., insertion sort or heap-based approaches) that run in $O(n \log k)$ time.

In this case, $k = \sqrt{n}$, so the known upper bound becomes $O(n \log \sqrt{n}) = O(n \log n)$.

I would like to rigorously prove that this upper bound is also a lower bound, i.e., $\Omega(n \log n)$ in the worst case for comparison-based sorting algorithms.

Answer 1

Proof by Contradiction

The problem asks us to prove a lower bound for sorting a specific type of array: an array $A$ of $n$ distinct elements where, for every value $j \in \{1, \dots, n\}$, its position $P_j$ in the array satisfies $|P_j - j| \le k$ for some parameter $k$. Specifically, we need to prove an $\Omega(n \log n)$ lower bound when $k = \sqrt{n}$.

This is a classic problem in the analysis of sorting algorithms for "nearly sorted" or "$k$-displaced" arrays.

Assumption for Contradiction:

Assume, for the sake of contradiction, that there exists a comparison-based sorting algorithm, let's call it $A_{\text{sort}}$, that can sort an array satisfying the given condition (with $k=\sqrt{n}$) in $T(n)$ time, where $T(n) = o(n \log n)$. This means that for any positive constant $c$, $T(n)$ grows asymptotically slower than $c \cdot n \log n$.

Lower Bound in Comparison Model:

In the comparison model of sorting, the minimum number of comparisons required to sort a set of distinct elements is determined by the number of distinct possible input permutations the algorithm must distinguish. If $N_{k,n}$ denotes the total number of distinct permutations of $n$ elements that satisfy the given condition (i.e., for every element with value $j$, its position $P_j$ satisfies $|P_j - j| \le k$), then the information-theoretic lower bound on the number of comparisons is $\Omega(\log N_{k,n})$.

Estimating $N_{k,n}$ for the Lower Bound:

To show that $\Omega(\log N_{k,n}) = \Omega(n \log n)$ when $k=\sqrt{n}$, we need to establish a sufficiently large lower bound for $N_{k,n}$.

Consider dividing the $n$ distinct values into $m = \lfloor n/k \rfloor$ groups, each containing $k$ consecutive values. For instance, let the groups of values be $G_1 = \{1, \dots, k\}$, $G_2 = \{k+1, \dots, 2k\}$, and so on, up to $G_m = \{(m-1)k+1, \dots, mk\}$.

For each group $G_j = \{(j-1)k+1, \dots, jk\}$, the elements within this group must be in positions that are "close" to their sorted values. Specifically, if an element has value $v \in G_j$, its actual position $P_v$ must satisfy $|P_v - v| \le k$. This means $P_v \in [\max(1, v-k), \min(n, v+k)]$.

A widely used argument for the lower bound of sorting $k$-displaced arrays (also known as $k$-sorted arrays or arrays with $k$-bounded displacement) is based on the number of such permutations. It is known that the number of permutations where each element is displaced by at most $k$ positions ($N_{k,n}$) is lower bounded by $(k!)^{n/k}$. This is because we can consider approximately $n/k$ "independent" segments of $k$ elements each, where elements within each segment can be permuted in $k!$ ways, largely independently.

Thus, we have: $$N_{k,n} \ge (k!)^{n/k}$$

Now, we calculate the lower bound on comparisons: $$\text{Comparisons} \ge \log_2(N_{k,n}) \ge \log_2((k!)^{n/k})$$ $$\text{Comparisons} \ge \frac{n}{k} \log_2(k!)$$

Using Stirling's approximation for $\log_2(k!)$, we know that $\log_2(k!) = \Theta(k \log k)$. Substituting this into the inequality: $$\text{Comparisons} \ge \frac{n}{k} \cdot \Theta(k \log k)$$ $$\text{Comparisons} = \Omega(n \log k)$$

This establishes that the general lower bound for sorting $k$-displaced arrays in the comparison model is $\Omega(n \log k)$.

Applying for $k = \sqrt{n}$: Now, we substitute $k = \sqrt{n}$ into the derived lower bound: $$\text{Comparisons} = \Omega(n \log \sqrt{n})$$ Since $\log \sqrt{n} = \frac{1}{2} \log n$: $$\text{Comparisons} = \Omega\left(n \cdot \frac{1}{2} \log n\right)$$ $$\text{Comparisons} = \Omega(n \log n)$$

Contradiction:

We have shown that any comparison-based algorithm for this problem, when $k=\sqrt{n}$, must perform at least $\Omega(n \log n)$ comparisons in the worst case. This directly contradicts our initial assumption that such an algorithm $A_{\text{sort}}$ could sort the array in $o(n \log n)$ time.

Therefore, our initial assumption must be false. The lower bound for sorting the array when $k=\sqrt{n}$ is indeed $\Omega(n \log n)$.

• D.W.: The key step here seems under-developed. The key step is to show the $N_{k,n} \ge (k!)^{n/k}$. Do you have a proof of that? This answer says "It is known that..."; do you have a citation to where that was proven? It also says "This is because we can consider approximately... largely independently.", but the words "approximately" and "largely" indicate that this is not formal. Why do you need those words anyway? Shouldn't the claim hold without any hedging? Can you justify why it holds? Was this written with AI assistance perhaps?
• lily: Looks like LLM generated
• Avi Tal: @D.W. I do not have proof of that and be happy to have one. And yes, Gemini helped.
• D.W.: 1. The question asks for a rigorous proof. I think it is misleading to present this as answering the question when you do not have a proof of the key step. 2. I encourage you to review the site's policy on use of GenAI, and revise your answer to comply with it: cs.stackexchange.com/help/gen-ai-policy. Generally speaking I recommend caution in relying on GenAI to (help you) write proofs, as they sometimes will "bullshit you", i.e., write things that look convincing on the surface but are not actually valid.
• Siddhanth Ramakrishnan: @D.W. What's wrong with that step?