1. **Problem statement:** We consider a Median-of-Five variant where the input of size $n$ is partitioned into $\frac{n}{3}$ blocks of size 3 instead of 5. 2. **Part (a) - Deriving the recursive formula for time complexity $T(n)$:** - Partition input into $\frac{n}{3}$ blocks of size 3. - Find the median of each block, which takes constant time per block, so $O(n)$ total. - Recursively find the median of these medians. The number of medians is $\frac{n}{3}$, so this takes $T\left(\frac{n}{3}\right)$ time. - Partition the array around the pivot median, which is $O(n)$. - Recursively search in one partition. From analysis similar to Median-of-Five, we know the size of the partition to search is at most $\frac{2n}{3}$. Thus, we get the recurrence: $$ T(n) \leq T\left(\frac{n}{3}\right) + T\left(\frac{2n}{3}\right) + cn $$ for some constant $c > 0$. 3. **Part (b) - Solve the recurrence by assuming $T(n) \in O(n)$:** - Suppose $T(n) \leq an$ for some constant $a > 0$. - Substitute into the recurrence: $$ T(n) \leq a \frac{n}{3} + a \frac{2n}{3} + cn = an + cn = n(a + c) $$ - This inequality holds only if $a \geq a + c$, which is impossible since $c > 0$. 4. **Conclusion:** - The assumption $T(n) \in O(n)$ does not hold for this recurrence. - Therefore, the time complexity $T(n)$ is not linear in $n$ under the $\frac{n}{3}$ partitioning scheme. Final answer: - The recursive time complexity is $$T(n) \leq T\left(\frac{n}{3}\right) + T\left(\frac{2n}{3}\right) + cn$$ - It cannot be solved as $T(n) \in O(n)$, so the linear time complexity bound does not hold.