The union-by-rank data structure indeed guarantees O(log n) time performance for all three operations (make-set, find, and union) using a tree.
To prove this claim, we will analyze the amortized time complexity of each operation.
Amortized analysis considers the average time complexity of a sequence of operations, taking into account the potential overhead from operations that take longer than average.
Make-set:
The make-set operation is straightforward and takes O(1) time. It simply creates a new element and sets its parent to itself, effectively forming a singleton set.
Find:
The find operation, also known as the path compression operation, takes O(log n) time amortized.
The basic idea is to repeatedly follow the parent pointers until we reach the root of the tree.
This effectively shrinks the paths to the root, reducing the time complexity of future find operations on those elements.
Union:
The union operation takes O(log n) time amortized.
The union-by-rank strategy aims to merge trees of similar sizes, keeping the overall height of the tree structure shallow.
This approach ensures that the find operation remains efficient.
Amortized Analysis:
To analyze the overall amortized time complexity, we consider the cost of a sequence of m union and find operations.
Let's assume that the average number of path compressions per union operation is c.
The total time for m union operations is:
m * O(log n) + c * m * O(1) = O(m log n)
The total time for m find operations is:
O(m log n) + m * O(log n) = O(2m log n) = O(m log n)
Since the total time for both union and find operations is O(m log n), the amortized time complexity of each operation is also O(log n).
Therefore, the union-by-rank data structure indeed guarantees O(log n) time performance for all three operations (make-set, find, and union) using a tree.