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 $\Omega(n \log n)$.
