CSE 2331/5331
Topic 11:
Minimum Spanning
Tree
CSE 2331 / 5331

Edge-weighted un-directed graph G = (V, E) and
edge weight function 𝑤: 𝐸 → 𝑅

E.g, road network, where each node is a city, and each
edge is a road, and weight is the length of this road.
1
3
a
1
d
2
5
b
c
2 1 1
e
4
f
CSE 2331 / 5331
A Tree

An undirected graph G = (V, E) is a tree if:


Equivalently, an undirected graph G = (V, E) is a
tree if:


G is connected, and there is no cycle
G is connected, and 𝐸 = 𝑉 − 1
One key property:


Given a tree T on nodes V, if we add any edge 𝑒 to T, it
will create a cycle.
That is, 𝑇 ∪ 𝑒 contains a cycle for any 𝑒 ∉ 𝑇.
CSE 2331 / 5331
A Spanning Tree

A spanning tree T of an undirected graph G = (V, E)




is a tree that includes all vertices V of G
𝑇⊆𝐸
Weight of a spanning tree T:

𝑤 𝑇 =
𝑢,𝑣 ∈𝑇 𝑤
𝑢, 𝑣

where 𝑤: 𝐸 → 𝑅 is the weight function on edges of G
A minimum spanning tree of G

is a spanning tree with smallest weight.
If G is not connected, then no spanning
tree exists.
CSE 2331 / 5331
Examples
Is minimum spanning tree unique?
CSE 2331 / 5331
MST

MST for given G may not be unique

Since MST is a spanning tree:


# edges : |V| - 1
If the graph is unweighted:

All spanning trees have same weight
CSE 2331 / 5331
Key Property of MST

Given a MST 𝑇 of 𝐺 = (𝑉, 𝐸), let 𝑒 ∈ 𝐸 be any
edge in 𝐸 but not in 𝑇. The following then holds:



There is a unique cycle C containing e in 𝑇 ∪ 𝑒 .
𝑒 is an edge with largest weight in C.
Sketch of proof:


Why unique?
If 𝑒 does not have largest weight, let 𝑒 ′ ∈ 𝐶 be an edge
with largest weight in C.



𝑇 ′ = 𝑇 − 𝑒′ + 𝑒 is a spanning tree of G
𝑤𝑒𝑖𝑔ℎ𝑡 𝑇 ′ ≤ 𝑤𝑒𝑖𝑔ℎ𝑡(𝑇) ⇒ 𝑇 cannot be MST.
Contradiction ⇒ e must have largest weight in C.
CSE 2331 / 5331
Construct MST

Input:


Output:



An undirected graph G = (V, E) with weight 𝑤: 𝐸 → 𝑅
A minimum spanning tree MST(G) of G.
A MST T is V-1 number of edges that connect all
nodes, with no cycle.
Intuitively, we will choose “safe” edges to
incrementally build T
CSE 2331 / 5331
Example

How to choose the first edge that is “safe”, namely, it
must be in some MST?

Note, it is some, not all !
CSE 2331 / 5331
Intuition

Input:


Output:


An undirected graph G = (V, E) with weight 𝑤: 𝐸 → 𝑅
A minimum spanning tree MST(G) of G.
Intuitively,



Incrementally grow a sub-tree 𝑇 𝑆 ⊆ 𝐸 connecting a subset of
nodes 𝑆 ⊂ 𝑉
At the beginning of each iteration, 𝑇 𝑆 is a sub-tree of some MST
of G
At each iteration, grow 𝑇 (𝑆′) to include one more vertex 𝑆′ = 𝑆 ∪
𝑢
′ is still a sub-tree of some MST of G
 Such that 𝑇 𝑆
CSE 2331 / 5331
Prim’s MST Algorithm: High-level


𝑈 : unconnected vertices
𝑆 = 𝑉 – 𝑈: vertices connected by current partial tree
CSE 2331 / 5331
Example


Greedy algorithm
Break ties arbitrarily
CSE 2331 / 5331
MST Theorem
MST Theorem:
Let T be a sub-tree of a minimum spanning tree.
If e is a minimum weight edge connecting T to some vertex
not in T, then 𝑇 ∪ 𝑒 is a subtree of a minimum spanning tree.

Key to the correctness of PrimMST algorithm.
Invariance:
each time PrimMST() algorithm grows the partial tree (i.e, adds
another edge to it), the invariance is that the new tree is still a subtree
of some minimum spanning tree of input graph G.
 Termination:
when all nodes are connected, we obtain a MST of G.

CSE 2331 / 5331
Proof of MST Theorem







By the theorem’s hypothesis, T is a subtree of some MST
A of G.
If e is not an edge of A, then 𝐴 ∪ 𝑒 contains a cycle.
Let 𝐶 be this cycle. There must exists some edge 𝑒 ′ ∈
𝐶 from T to a vertex not in T.
Since e is a minimum weight edge from T to vertices not in
T, 𝑤𝑒𝑖𝑔ℎ𝑡 𝑒 ≤ 𝑤𝑒𝑖𝑔ℎ𝑡 𝑒 ′ .
Replacing 𝑒 ′ ∈ 𝐴 by e gives a new tree 𝐵 = 𝐴 − 𝑒′ +
𝑒 such that 𝑤𝑒𝑖𝑔ℎ𝑡 𝐵 ≤ 𝑤𝑒𝑖𝑔ℎ𝑡 𝐴 .
𝑇 ∪ 𝑒 ⊆ 𝐵. So 𝑇 ∪ 𝑒 is also a subtree of some MST.
Done.
CSE 2331 / 5331
Naive Implementation of Prim’s MST


Naïve implementation: linear scan all edges to identify
min-weight edge 𝑣𝑖 , 𝑣𝑗 at each iteration
Total time complexity: 𝑂 𝑉𝐸
CSE 2331 / 5331
First Improvement of Implementation

Storing costs at vertices

Each unvisted vertex in 𝑈 maintain the smallest weightedge to visited vertices
CSE 2331 / 5331
First Improvement: Cost at Vertices
CSE 2331 / 5331
Analysis


Time to update cost at each vertex at each iteration
(in the While-loop):

Line 6: 𝑂 𝑉

Lines 8—13: 𝑂(deg 𝑣𝑗 )
Total time complexity:

𝑂 𝑉2 +

= 𝑂 𝑉2 + 𝐸
𝑣𝑗 ∈𝑉 deg
𝑣𝑗
CSE 2331 / 5331
Second Improvement: Efficient Update

Consider Line 6 of previous implementation



We spend linear time to identify the vertex 𝑣𝑗 with
smallest cost from 𝑈
Can we do better?
What operations do we need to support?


extracting the smallest value from a set,
updating the cost of some vertices

The cost can only decrease !
Use a min-priority queue!
CSE 2331 / 5331
Second Improvement: Priority Queue
CSE 2331 / 5331
Analysis:



We use min-heap to implement the priority queue
The maximum size of Q is V
# iterations of While-loop?


# iterations of each call of the inner for-loop?


V
deg 𝑣𝑗
Total #times lines 7—10 are executed:

𝑣𝑗 ∈𝑉 deg
𝑣𝑗
CSE 2331 / 5331
Analysis cont.

Q.insert:



Total #: V
Total cost: 𝑂(𝑉 lg 𝑉)
Q.IsNotEmpty and Q.MinKey:




Total #: V
Total cost: O(V)
Q.DecreaseKey



Total #: 2E
Total cost: 𝑂(𝐸 lg 𝑉)
Q.Contains and Q.Key


Total #: 2E
Total cost: 𝑂(𝐸)
Q.DeleteMin


Total #: V
Total cost: 𝑂(𝑉 lg 𝑉)
Total time complexity:
𝑂 𝑉 + 𝐸 lg 𝑉 = 𝑂 𝐸 lg 𝑉
CSE 2331 / 5331
Remarks

Prim’s algorithm is a greedy algorithm


One can use even more efficient Priority-Queue
implementation, based on the Fibonacci Heap


The choice of the minimum-weight edge at each
iteration
𝑂(𝑉 lg 𝑉 + 𝐸)
There are many possible MST algorithms

Another popular one is Kruskal Algorithm, which also
runs in 𝑂 𝐸 lg 𝑉 time
CSE 2331 / 5331