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Consider the following alternative implimentation of the disjoint-sets data structure. class Partition: definit_(P,n): #sometimes called makeset (j) P.parent [j for j in range(n)] P.weight [1] * n - def find (P, key): if P.parent [key] != key: key P. find (P.parent [key]) M return key def union (P,x,y): rx = P.find(x) ry = P.find(y) if rx == ry: return if P.weight [rx] > P.weight[ry]: P.parent [ry] - rx P.weight [rx] += P.weight[ry] else: P.parent[rx] - ry P.weight [ry] + P.weight[rx]

Consider the following alternative implimentation of the disjoint-sets data structure. class Partition: definit_(P,n): #sometimes called makeset (j) P.parent [j for j in range(n)] P.weight [1] * n - def find (P, key): if P.parent [key] != key: key P. find (P.parent [key]) M return key def union (P,x,y): rx = P.find(x) ry = P.find(y) if rx == ry: return if P.weight [rx] > P.weight[ry]: P.parent [ry] - rx P.weight [rx] += P.weight[ry] else: P.parent[rx] - ry P.weight [ry] + P.weight[rx]

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Consider the following alternative implimentation of the disjoint-sets data structure. class Partition: definit_(P,n): # sometimes called makeset(j) P.parent = [j for j in range(n)] P.weight = [1] * n def find(P, key): if P.parent [key] != key: key = P.find(P.parent [key]) return key def union(P,x,y) :: rx = P.find(x) ry = P.find(y) if rx == ry: return if P.weight[rx] > P.weight[ry]: P.parent [ry] = rx P.weight [rx] += P.weight[ry] else: P.parent [rx] ry = P.weight [ry] += P.weight[rx] Prove that after any sequence of union operations, P.find(x) takes O(lg n) time.
Transcribed Image Text:Consider the following alternative implimentation of the disjoint-sets data structure. class Partition: definit_(P,n): # sometimes called makeset(j) P.parent = [j for j in range(n)] P.weight = [1] * n def find(P, key): if P.parent [key] != key: key = P.find(P.parent [key]) return key def union(P,x,y) :: rx = P.find(x) ry = P.find(y) if rx == ry: return if P.weight[rx] > P.weight[ry]: P.parent [ry] = rx P.weight [rx] += P.weight[ry] else: P.parent [rx] ry = P.weight [ry] += P.weight[rx] Prove that after any sequence of union operations, P.find(x) takes O(lg n) time.
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