Tree Vertex Splitting Problem Geeksforgeeks -

Check if a Tree can be split into K equal connected components

: Ensuring signal integrity in digital circuits where path delays must be capped to meet clock cycles.

The TVSP is efficiently solved using a that makes locally optimal choices by processing the tree from its leaves up to the root. Step-by-Step Execution Initialize Delays: For every leaf node , set the initial delay tree vertex splitting problem geeksforgeeks

Before diving into the algorithm, it’s important to understand why this problem matters:

Tree root, edge weights, threshold D. Output: Minimum splits needed. Check if a Tree can be split into

Initialize split_count = 0. For node u in postorder (leaves to root): Let child_distances = [] For each child v of u: child_distances.append( w(u,v) + dist[v] ) Sort child_distances descending. Let farthest = child_distances[0] if any. Let second_farthest = child_distances[1] if exists. if farthest > d: // This means even the longest single path from u to leaf > d. // So we must split u. split_count++ // After splitting u, we reset dist[u] = 0 (since splits break continuity). dist[u] = 0 else if second_farthest exists and farthest + second_farthest > d: // This is the key condition: two paths from u via different children // combine to exceed d. By splitting u, we separate those two long paths. split_count++ dist[u] = 0 else: dist[u] = farthest

If adding the weight of the edge to its parent would cause the total delay to exceed the tolerance ( ), the node Action on Split: Place a booster at node for its parent's subsequent calculations. Key Characteristics Optimality: The greedy method is proven to yield the minimum cardinality Output: Minimum splits needed

return child_paths[0];