Union-Find operations on a tree?

Question

can someone please explain the answer in bold? How it's done?

Below is a sequence of four Union-Find operations (with weighted-union and full com- pression) that led to the following up-tree. What were the last two operations?

Answer: Union(D,A), Union(B,C), Union(D/A,B/C),Find(B/C).

Answer 1

The notation is used because of the sets .

Let us apply the four operations:

Union(D,A) leads to following tree:

   D
  /
 A

Union(B,C) leads to following tree:

   B
  /
 C

Now Union(D/A,B/C) means that because D and A belong to the same set , it does not matter what the first argument is, it can be D or it can be A . Similarly because B and C belong to the same set , it does not matter what the second argument is, it can be B or it can be C , the result will be the same .

The result will be after third operation:

   D
  / \
 A   B
      \
       C

Now because compression is also allowed, the Find(C) operation would result in the tree:

   D
  /|\ 
 A B C

If the fourth operation is Find(B) , the tree would remain the same, because when we apply compression after a find operation, we make all the nodes encountered in the path upto the root immediate child of the root, but since we will not encounter C , we will not be able to make C immediate child of D , as it is in final tree.

Correct Answer

The correct sequence of four operations is:

Union(D,A), Union(B,C), Union(D/A,B/C),Find(C).