Union-Find
1. Dynamic Connectivity Problem :
a) Union command : connect two objects
b) Find/connected query: is there a path connecting the two objects?
2. Quick-find
a) Integer array id[] of size N
b) Interpretation: p and q are connected iff they have the same id.
c) Find: Check if p and q have the same id.
d) Union : To merge components containing p and q, change all entries whose id equals id[p] to id[q].
Find is quick which only takes 2 array access. Union is expensive , it cost N array access and N union will cost N^2
3. Quick-union
a) Integer array id[] of size N.
b) Interpretation: id[i] is parent of i.
c) Root of i is id[id[id[...id[i]...]]].
d) Find: Check if p and q have the same root.
e) Union: To merge components containing p and q, set the id of p's root to the id of q's root.
Both Union and Find can be expensive in worst case.
4. Quick-find defect:
a) Union too expensive (N array accesses).
b) Trees are flat, but too expensive to keep them flat.
Quick-union defect.
a) Trees can get tall.
b) Find too expensive (could be N array accesses).
5. Weighted quick-union.
a) Modify quick-union to avoid tall trees.
b) Keep track of size of each tree (number of objects).
c) Balance by linking root of smaller tree to root of larger tree.
Proposition: Depth of any node x is at most lg N.
Pf. When does depth of x increase?
Increases by 1 when tree T1 containing x is merged into another tree T2.
1) The size of the tree containing x at least doubles since | T 2 | ≥ | T 1 |.
2) Size of tree containing x can double at most lg N times. (the total # of nodes is at most N)
6. Quick union with path compression: Just after computing the root of p, set the id of each examined node to point to that root.
Two-pass implementation: add second loop to root() to set the id[] of each examined node to the root.
Simpler one-pass variant: Make every other node in path point to its grandparent (thereby halving path length).