Final answer:
The subroutine merge will be called 7 times before termination.
Step-by-step explanation:
Mergesort is a divide-and-conquer algorithm that splits an array into two halves, recursively sorts each half, and then merges the two sorted halves back together to form a sorted array.
In this case, the original list has 9 elements, so the algorithm will start by splitting it into two halves: {9, 8, 7, 6} and {5, 4, 3, 2, 1}.
The algorithm will then recursively sort each half, resulting in the following splits:
- {9, 8, 7, 6} -> {9, 8} -> {9}, {8} -> merge
- {5, 4, 3, 2, 1} -> {5, 4}, {3, 2, 1} -> {5}, {4} -> merge
- {3, 2, 1} -> {3}, {2, 1} -> {2}, {1} -> merge
Finally, the algorithm will merge the sorted halves back together to get the final sorted list:
- {9}, {8} -> merge
- {5}, {4} -> merge
- {3}, {2} -> merge
- {2}, {1} -> merge
- {3, 2}, {1} -> merge
- {5, 4}, {3, 2, 1} -> merge
- {9, 8}, {5, 4, 3, 2, 1} -> merge
Therefore, the subroutine merge will be called 7 times before the termination.