Final answer:
The UF data structure identifies 2 components after processing all input pairs: {0, 1, 2, 3, 5, 6, 7, 8} and {4, 9}. The structure of the arrays after processing will vary depending on the specific UF algorithm used (QuickFindUF, QuickUnionUF, or WeightedQuickUnionUF).
Step-by-step explanation:
The Union-Find (UF) data structure is employed for solving the dynamic connectivity problem in this case. Through the course of processing the input pairs (1, 2), (7, 8), (1, 6), (0, 5), (3, 8), (2, 3), (6, 7), (2, 7), and (4, 9), the UF data structure helps determine how many connected components there are among the sites and establishes their connectivity.
- (a) After processing all the input pairs, the UF structure identifies 2 components.
- (b) Those components are: {0, 1, 2, 3, 5, 6, 7, 8} and {4, 9}.
- (c) The number of components or their membership will not change if the input pairs arrive in a different order, as long as the same pairs are processed.
- (d) Using QuickFindUF, the id array will represent components by having identical entries for elements that are connected. After all pairs are processed, the id array might look like this, depending on the internal implementation of union operations: [1, 1, 1, 1, 9, 1, 1, 1, 1, 9].
- (e) Using QuickUnionUF, the parent array will hold indices pointing to the 'parent' of each site, determining the structure of trees representing the connected components. The values can be varied based on the order of union operations, but one possible configuration is: [5, 1, 1, 1, 9, 1, 1, 1, 7, 4].
- (f) Using WeightedQuickUnionUF, both parent and size arrays will be modified to maintain a more balanced tree structure. The parent array might end up similar to QuickUnionUF, e.g., [5, 1, 1, 1, 9, 1, 1, 1, 1, 4], and the size array will reflect the size of the tree rooted at the index, such as [1, 8, 1, 1, 2, 1, 1, 1, 1, 1].