## § Disjoint set union

#### § intuition for correctness of rank:

Assume that we had to re-point pointers of all our children to the new root when we decide to make another node the root. That is, we would have:
void mkroot(int newroot, int prevroot) {
for (int child : children[prevroot] {
parent[child] = newroot;
}
parent[prevroot] = newroot];
children[prevroot] = {}; // this has no children anymore
}

• In this setting, we ought to make the smaller subtree the prevrootand the larger subtree the newroot: It is better to loop overfewer children.
• When we perform the rank based union, we are using the same heuristic,even though we don't actually loop over all our children.