# Minimum Spanning Tree — Prim

The idea behind Prim's algorithm is simple, a spanning tree means all vertices must be connected. So the two disjoint subsets (discussed above) of vertices must be connected to make a Spanning Tree. And they must be connected with the minimum weight edge to make it a Minimum Spanning Tree.

This post will introduce one of the algorithms to find an MST: Prim. Logics, time/space complexities, and implementations will be provided. Feat. Leetcode challenge 1584. Min Cost to Connect All Points.

## Logic

Prim’s logic is quite similar to Dijkstra’s algorithm, which also finds the lightest edge greedily. It starts with an arbitrary vertex `v` and ends up with a set of edges `A` spanning all the vertices. At each step, it would look for the edge that connects `A` to an isolated vertex in the graph. The pseudo-code is shown as below.

``````MST-PRIM(G,w,r)
'''
G: the connected graph
w: weights
r: the root vertex
Q: a min-priority queue based on a key field
key: for each vertex v, key[v] is the minimum weight \
of any edge connecting to a vertex in the MST
π: for each vertex v, π[v] names the parent of v in the MST
'''
for each u in V[G]
key[u] = inf
π[u] = NIL
key[r] = 0
Q = V[G]
while Q:
u = EXTRACT_MIN(Q)
for each v in Adj[u]
if v in Q and w(u,v)<key[v]
π[v] = u
key[v] = w(u,v)``````

Here’s an illustration for Prim:

## Complexities

There are two ways of implementations for line 17.

• using adjacency matrix, so that EXTRACT_MIN takes O(V);
• using the binary heap, so that EXTRACT_MIN takes O(logV).

• Initialization takes O(V)(Line 11–15).
• Line 16 would be executed for O(V) time, so EXTRACT_MIN takes O(VV) in total.
• As the while-loop in line 16 will be executed for each vertex and the for-loop in line 18 will be executed for each edge of the vertex, therefore the for-loop in line 18 will be executed for O(E) times.

Therefore, the total time complexity would be O(V+V^_2+_E)=O(V^_2). The space complexity would be _O(V+E).

### Binary heap

• Initialization takes O(V)(Line 11–15).
• Line 16 would be executed for O(V) time, so EXTRACT_MIN takes O(VlogV) in total.
• As the while-loop in line 16 will be executed for each vertex and the for-loop in line 18 will be executed for each edge of the vertex, therefore the for-loop in line 18 will be executed for O(E) times.
• Line 21 indicates an update operation on the heap, thus taking logV.

Therefore, the total time complexity would be O(V+VlogV+ElogV)=O(ElogV). The space complexity would be O(V+E).

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