Kruskal's Algorithm: Difference between revisions
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OPT(n) = OPT(n-1) + min( (v_i, v_n) \in E ) | OPT(n) = OPT(n-1) + min( (v_i, v_n) \in E ) | ||
</math> | </math> | ||
= Implementation = | |||
<pre> | |||
Kruskal(G): | |||
let A be an empty graph | |||
for each v in V: | |||
makeSet(v) | |||
sort E in nondecreasing order by weight | |||
for each (u,v) in E: | |||
if findSet(u) != findSet(v): | |||
A = A U {(u,v)} | |||
Union(u,v) | |||
return A | |||
</pre> | |||
= Analysis = | = Analysis = |
Revision as of 18:05, 20 March 2024
Approach: Greedy
The approach is to try to add the smallest edges as long as they do not create a cycle. Unlike Prim's algorithm, which prevents cycles by only choosing edges that crosses a cut of nodes already in the tree and nodes that aren't, Kruskal prevents cycles using a data structure known as disjoint set (aka. union-find).
Given the MST of , the MST of should be that of plus the edge that connects to that is the shortest.
Implementation
Kruskal(G): let A be an empty graph for each v in V: makeSet(v) sort E in nondecreasing order by weight for each (u,v) in E: if findSet(u) != findSet(v): A = A U {(u,v)} Union(u,v) return A
Analysis
Sort edges + E (cycle?) + (V - 1) adding edge
Sorting takes E log E
For weighted disjoint set, checking cycle takes log V, and adding edge takes log V
For fast-find, where all members have the same ID, fast-set-id needs O(1) and union needs O(n)