5. Consider the following variation of mergesort:
(a) If $n \le 1$, we are done.
(b) Divide the $n$ elements into $b$ subarrays of $n/b$ elements each.
(c) Recursively call mergesort to sort each of the $b$ subarrays.
(d) Merge the $b$ sorted subarrays.
For example, the standard mergesort has $b = 2$, and merging takes $\Theta(n)$ comparisons.
If $b = 3$, then we have a 3-way mergesort, and one can show that it takes $\Theta(n)$ comparisons to merge 3 sorted lists. In fact, one may show that merging $b$ sorted lists (for any fixed $b$) can be done in $\Theta(n)$ comparisons.
Now consider the following argument: if we set $b = n$, we recursively sort $n$ subarrays of size 1, and then merge them into one list with $\Theta(n)$ comparisons. Thus, the complexity can be expressed by:
$T(1) = 0$
$T(n) = nT(1) + \Theta(n) = \Theta(n)$
Therefore, this variation of mergesort sorts an array in $\Theta(n)$ comparisons.
What is wrong with this argument? Hint: think about how to implement this variation.
