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A right array rotation by $k$ sends the entry originally at offset $j$ to offset $(j+k)\mathrm{mod}n$ for nonempty length $n$. Equivalently it changes adjacent blocks $AB$ into $BA$ when $|B|=k$. Identity rotations include $k=0$ and $k=n$.

Three reversals implement the operation: $AB\to A^{R}B\to A^R{B}^R\to BA$. Alternatively repeatedly use a block swap to put the smaller block in its final position and recurse on the remaining unresolved segment. Replacing that tail recursion by parameter updates gives constant auxiliary storage. If $S(a,b)$ counts swaps for blocks of lengths $a,b>0$, then $S(a,a)=a$ and $S(a,b)=\min (a,b)+S(\min (a,b),|a-b|)$ for unequal lengths. The Euclidean algorithm shows by induction that $S(a,b)=a+b-\gcd (a,b)$, so a right rotation uses $n-\gcd (n,k)$ element swaps for $0<k<n$.

If the merge algorithm from part (b) is retained with a disjoint destination, use a circular buffer of $2n$ cells. Represent the live segment by its start modulo $2n$ and its length $n$. The other $n$ cells are free; the next merge needs at most $n$ output cells. Appending its output and advancing the live start releases exactly as many obsolete source cells. Logical run boundaries and copying indices are reduced modulo $2n$. The final sorted sequence may wrap, in which case an array rotation makes it contiguous. ThusThis is also the worst-case requirement if all source cells must be retained until a whole disjoint merge has been written: the final two runs can have combined length $n$.

Under the question's literal wording, however, no running time requirement or disjoint-output restriction is imposed on the modified implementation. The absolute smallest array isTo achieve this, replace the disjoint-output merge by an in-place merge: compare the next unread values of the two adjacent runs; when the second is smaller, save it in one scalar, shift the intervening first-run block right by one, and insert it before that block. This is a sorted merge using no extra array. Its loop invariant is that the finished prefix is the sorted merge of the consumed portions and the two remaining runs stay sorted. Each step consumes one value, proving termination.

After merging the first two runs in place, rotate the merged prefix behind the untouched suffix. Three array reversals implement this rotation with one scalar: reverse the merged prefix, reverse the suffix, then reverse the whole segment. The resulting live sequence is exactly the one produced in part (c). Repeat the same run schedule until sorted. The shift and reversal loops respectively preserve the copied block and exchanged end pairs; their loop variants are the unshifted length and half the unreversed length. The modification can be substantially slower, but preserves the loop invariant and decreasing run count from part (d). Fewer than $n$ cells cannot even hold the specified $n$ input entries. Thus $2n$ is the natural circular-workspace answer for the original merge primitive, while $n$ is attainable when minimizing storage over all implementations allowed by the stated time-free specification.

Replace the sole recursive call by updates to the current segment's parameters:The loop invariant is that every position outside the current unresolved segment already has its final value, while rotating the current segment by the current k completes the original array rotation. The block decompositions in part (ii) prove preservation. The positive unresolved length decreases in each continuing branch; the equal-block branch finishes immediately. The same rotation now uses $O(1)$ auxiliary space, including control storage, and leaves the rest of the array untouched.

Write the segment as $AB$, where $|A|=n-k$ and $|B|=k$. The desired array rotation is $BA$. Reverse each block, then the whole segment:The state changes are $AB\to A^{R}B\to A^R{B}^R\to BA$, using the list reversal identity supplied in the paper and reversal being an involution. Three reversals implement the right rotation in $O(n)$ time and $O(1)$ auxiliary space. The initial return covers empty and identity rotations.

Save the last entry, shift the others right in descending index order, and put the saved value at the left. Descending order prevents overwriting an unread entry.For one pass, after the shifts at offsets greater than t, those positions contain the original preceding entries; offsets through t are still available, and x retains the original last entry. The loop variant t decreases to zero. The outer loop invariant is that the segment has been rotated right by step places relative to its initial state. Repeating a one-step cyclic permutation $k$ times therefore gives the required array rotation. The method uses one extra T variable and $O(kn)$ time, with $O(1)$ auxiliary space. The early return avoids all accesses for an identity rotation.