Minimum Spanning Trees - Learning Module

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Minimum Spanning Tree (MST) — The Optimal Backbone

From Connectivity to Optimization

In the previous page, we learned that connected graphs can have many different spanning trees—sometimes millions or even billions. All of them achieve the same goal: connecting every vertex with exactly n - 1 edges, no cycles, full connectivity.

But in the real world, not all edges are created equal. Building a bridge costs more than paving a road. Running fiber optic cable across mountains is more expensive than crossing flat terrain. Some network connections have lower latency than others.

When edges carry weights (costs, distances, or any measure we want to minimize), a new question emerges: Among all possible spanning trees, which one has the smallest total weight?

This is the Minimum Spanning Tree (MST) problem—one of the most fundamental and widely-applicable optimization problems in computer science.

What You Will Learn

By the end of this page, you will understand the formal definition of the MST problem, why it's non-trivial, how to reason about optimality, and the intuition behind why greedy algorithms can solve it efficiently. This sets the stage for Prim's and Kruskal's algorithms in subsequent modules.

Weighted Graphs — Adding Meaning to Edges

Before defining the MST problem formally, let's ensure we have a precise understanding of weighted graphs.

Weighted Graph Definition

A weighted graph G = (V, E, w) consists of:

• V: A set of vertices • E: A set of edges connecting pairs of vertices • w: E → ℝ: A weight function assigning a real number to each edge

For an edge e = {u, v}, we write w(e) or w(u, v) to denote its weight.

What weights represent in practice:

Edge weights can model many different real-world quantities:

Domain	What Weight Represents
Road networks	Distance, travel time, or fuel cost
Computer networks	Latency, bandwidth cost, or failure probability
Circuit design	Wire length or resistance
Social networks	Strength of connection (inverse: cost to disconnect)
Logistics	Shipping cost, delivery time
Telecommunications	Cable installation cost

Key assumption for MST:

In the MST problem, we typically assume:

All edge weights are non-negative (though some algorithms handle negative weights)
We want to minimize the total weight
Weights are fixed (they don't change based on which edges we select)

Notation conventions:

Throughout our discussion of MST:

n = |V| = number of vertices
m = |E| = number of edges
w(T) = total weight of a spanning tree T, defined as Σ w(e) for all edges e in T
For a spanning tree with edges e₁, e₂, ..., eₙ₋₁: w(T) = w(e₁) + w(e₂) + ... + w(eₙ₋₁)

The MST Problem — Formal Definition

Now we can state the Minimum Spanning Tree problem with complete precision.

Problem Definition: Minimum Spanning Tree

Input: A connected, undirected, weighted graph G = (V, E, w)

Output: A spanning tree T of G such that the total weight w(T) = Σ w(e) for e ∈ T is minimized

Goal: Among ALL spanning trees of G, find one with the smallest possible sum of edge weights

Breaking down the problem:

Input requirements:

Connected: Otherwise, no spanning tree exists (would need a spanning forest instead)
Undirected: Edges don't have direction (MST for directed graphs is a different, harder problem)
Weighted: Every edge has an associated weight/cost

Output requirements:

Must be a valid spanning tree (n - 1 edges, connected, acyclic, spanning all vertices)
Among all valid spanning trees, must be one with minimum total weight

The optimization aspect:

The MST problem is an optimization problem—we're not just looking for any spanning tree, we're looking for the best one according to our weight criterion. This transforms a structural question (does a spanning tree exist?) into an optimization question (which spanning tree is cheapest?).

Converting Mermaid diagram...

In the example above:

Original graph has 4 vertices and 5 edges with weights: AB=3, AC=5, BC=4, BD=2, CD=6
The MST uses edges AB(3), AC(5), BD(2) with total weight = 3 + 5 + 2 = 10
Alternative spanning tree: AB, BC, BD would have weight = 3 + 4 + 2 = 9 (actually better!)
Another alternative: AC, BC, BD would have weight = 5 + 4 + 2 = 11
Another: AC, CD, BD would have weight = 5 + 6 + 2 = 13

The MST is the spanning tree with weight 9: edges AB, BC, BD.

Why Is Finding the MST Non-Trivial?

At first glance, the MST problem might seem simple: just pick the cheapest edges! But a naive approach quickly runs into problems.

The Naive Approach Fails

Attempt 1: Pick the n - 1 cheapest edges in the graph.

Problem: Those edges might not form a spanning tree! They could: • Form cycles (wasting edges) • Form disconnected components (missing vertices) • Fail to span all vertices

We need edges that are both cheap AND structurally valid.

Example where the naive approach fails:

Consider a 4-vertex graph with edges:

A-B: weight 1
B-C: weight 1
C-A: weight 1 (forms a triangle with weights 1,1,1)
C-D: weight 10

The 3 cheapest edges are A-B(1), B-C(1), C-A(1). But these form a cycle (a triangle) and don't include vertex D at all! The correct MST must use one of the triangle edges plus C-D(10), for a total of 12.

The search space is enormous:

As we saw in the previous page, a complete graph Kₙ has n^(n-2) spanning trees. Exhaustively checking each one is computationally infeasible:

K₁₀: 100 million trees to check
K₂₀: ~2.6 × 10²³ trees (more than Avogadro's number!)
K₅₀: Astronomically large

We need algorithms that find the MST without examining every possibility.

The key insight:

Despite the exponential search space, the MST problem has special structure that allows greedy algorithms to find the optimal solution efficiently. This is remarkable—most combinatorial optimization problems don't admit such clean solutions.

The greedy approach works because of the cut property and cycle property we introduced in the previous page. These properties ensure that locally optimal choices (picking the cheapest edge that doesn't break the tree structure) lead to a globally optimal solution.

Calculating and Comparing Spanning Tree Weights

Before diving into algorithms, let's be precise about how we measure and compare spanning trees.

Spanning Tree Weight

For a spanning tree T = (V, E_T) of a weighted graph G, the weight of T is:

w(T) = Σ w(e) for all e ∈ E_T

Since T has exactly n - 1 edges, this is a sum of n - 1 edge weights.

Properties of spanning tree weight:

Additive: The total weight is simply the sum of individual edge weights
Bounded: For a graph with edge weights between w_min and w_max:
- Minimum possible: (n - 1) × w_min (if we could use n-1 copies of the lightest edge)
- Maximum possible: (n - 1) × w_max (using the heaviest edges)
- Actual MST weight is somewhere in between
Comparable: Given two spanning trees T₁ and T₂, we compare them by their weights:
- T₁ is better than T₂ if w(T₁) < w(T₂)
- They're equally good if w(T₁) = w(T₂)

spanning_tree_weight.py
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def spanning_tree_weight(tree_edges, weight):
    """
    Calculate the total weight of a spanning tree.
    
    Args:
        tree_edges: List of edges in the spanning tree, 
                    each edge is a tuple (u, v)
        weight: Dictionary mapping edge tuples to weights,
                weight[(u, v)] or weight[(v, u)] gives the weight
    
    Returns:
        Total weight of the spanning tree
    
    Example:
        edges = [(0, 1), (1, 2), (2, 3)]
        weights = {(0, 1): 3, (1, 2): 5, (2, 3): 2}
        # Total weight = 3 + 5 + 2 = 10
    """
    total = 0
    for u, v in tree_edges:
        # Handle both orderings since graph is undirected
        if (u, v) in weight:
            total += weight[(u, v)]
        else:
            total += weight[(v, u)]
    return total
 
 
def compare_spanning_trees(tree1, tree2, weight):
    """
    Compare two spanning trees by their total weight.
    
    Returns:
        -1 if tree1 is lighter (better)
         0 if they have equal weight
         1 if tree2 is lighter (better)
    """
    w1 = spanning_tree_weight(tree1, weight)
    w2 = spanning_tree_weight(tree2, weight)
    
    if w1 < w2:
        return -1
    elif w1 > w2:
        return 1
    else:
        return 0

When Is the MST Unique?

An important theoretical question: does every graph have exactly one MST, or can there be multiple equally-optimal solutions?

MST Uniqueness Theorem

A weighted graph has a unique MST if and only if:

All edge weights are distinct (no two edges have the same weight), OR
For every cut, there is a unique minimum-weight edge crossing that cut

If some edges share the same weight, there may be multiple MSTs with the same total weight.

Example with multiple MSTs:

Consider a 4-vertex cycle A-B-C-D-A with all edges having weight 1:

Any 3 of the 4 edges form a spanning tree
All spanning trees have weight = 3
There are 4 different MSTs, all equally optimal

Why uniqueness matters:

Algorithm behavior: Different MST algorithms might return different MSTs if multiple exist
- This is fine—any MST is equally optimal
- But it can cause confusion in testing (multiple correct answers)
Determinism: If you need deterministic output, you might add a secondary comparison
- Break ties by edge index
- Break ties lexicographically
Problem constraints: Some problems guarantee distinct weights to simplify the analysis

Practical Implication

When implementing MST algorithms, don't assume there's only one correct answer. Two correct implementations might produce different MSTs with the same total weight. Always compare by total weight, not by the specific edges chosen.

The Greedy Paradigm — Why It Works for MST

The MST problem belongs to a special class of optimization problems that can be solved optimally using greedy algorithms. This is exceptional—most optimization problems cannot be solved greedily.

What Is a Greedy Algorithm?

A greedy algorithm builds a solution incrementally, at each step choosing the locally optimal option, without reconsidering previous decisions.

Key characteristics: • Makes the best choice at each step • Never backtracks or revises choices • Hopes that local optimality leads to global optimality

The catch: Greedy algorithms often produce suboptimal results. The MST problem is special because greedy works perfectly.

Why does greedy work for MST?

The greedy approach succeeds for MST because of two fundamental properties:

1. The Cut Property:

For any cut (S, V-S) that divides the vertices into two groups, the minimum-weight edge crossing the cut is part of some MST.

Implication: We can safely include the cheapest crossing edge—we'll never regret it.

2. The Cycle Property:

For any cycle in the graph, the maximum-weight edge in the cycle is NOT part of any MST (assuming unique edge weights).

Implication: We can safely exclude the most expensive edge in any cycle—including it would never be optimal.

Prim's Strategy (Cut-based)

•Start from any vertex
•Maintain a growing tree T
•At each step: consider the cut (vertices in T, vertices not in T)
•Add the minimum-weight edge crossing this cut
•Repeat until all vertices are included

Kruskal's Strategy (Cycle-based)

•Sort all edges by weight (ascending)
•Consider edges in sorted order
•Add an edge if it doesn't create a cycle
•Skip an edge if it would create a cycle
•Repeat until n - 1 edges are added

Both strategies are greedy: they make locally optimal choices without reconsidering. Both are provably correct because of the cut and cycle properties.

The mathematical elegance:

The MST problem has what's called the matroid structure—a mathematical property that guarantees greedy optimality. Specifically, the spanning trees of a graph form a "graphic matroid," and the greedy algorithm is optimal for finding minimum-weight bases of matroids.

You don't need to understand matroid theory to use MST algorithms, but it explains why the greedy approach works where it fails for other problems.

The Cut Property — A Deep Dive

The cut property is so central to MST algorithms that it deserves careful examination and proof.

The Cut Property (Formal Statement)

Let G = (V, E, w) be a connected weighted graph. Let (S, V-S) be any partition of V into two non-empty sets. Let e = {u, v} be the minimum-weight edge with u ∈ S and v ∈ V-S.

Then there exists a minimum spanning tree of G that contains edge e.

Proof by exchange argument:

Let T be any MST of G. We'll show that either T already contains e, or we can exchange an edge to get another MST that does contain e.

Case 1: T contains edge e. We're done—e is in some MST.

Case 2: T does not contain edge e.

Since T is a spanning tree, there exists a unique path P in T from u to v. This path must cross from S to V-S at least once (since u ∈ S and v ∈ V-S).

Let e' = {x, y} be an edge on path P that crosses the cut (i.e., x ∈ S and y ∈ V-S).

Now consider T' = T - {e'} + {e}:

T' has the same number of edges as T
T' is still connected (we can still reach y from x by going through e and the rest of P)
T' is still acyclic (removing e' breaks the cycle that adding e would create)

Therefore T' is a spanning tree.

Moreover, w(T') = w(T) - w(e') + w(e).

Since e is the minimum-weight edge crossing the cut, and e' also crosses the cut:

w(e) ≤ w(e')
Therefore w(T') ≤ w(T)

Since T is an MST, w(T) ≤ w(T'), so we must have w(T') = w(T).

Thus T' is also an MST, and T' contains e. ∎

Why This Proof Matters

This exchange argument is the template for proving correctness of both Prim's and Kruskal's algorithms. It shows that choosing the minimum-weight edge across any cut is always safe—you'll never regret including it in your MST construction.

Summary and What Comes Next

We've established the complete theoretical foundation for the MST problem. Let's consolidate:

Key Takeaways

•Weighted graphs associate a cost/weight with each edge, enabling optimization
•The MST problem asks for the spanning tree with minimum total edge weight
•Despite exponential possibilities, greedy algorithms find the optimal solution
•The MST may not be unique when edges share weights, but all MSTs have equal total weight
•The cut property guarantees that minimum cut edges are always safe to include
•The cycle property guarantees that maximum cycle edges are always safe to exclude
•Greedy works because MST has matroid structure—local optimality leads to global optimality

What's next:

With the problem definition and theoretical foundation established, we're ready to explore the properties that make MSTs so useful:

Page 3: Properties of MSTs — Uniqueness conditions, the blue rule, the red rule, and characterization theorems
Page 4: Applications — Network design, clustering, approximation algorithms, and real-world use cases

In subsequent modules, we'll implement the two classic MST algorithms:

Prim's Algorithm: Grows the MST from a single vertex, always adding the cheapest edge to the growing tree
Kruskal's Algorithm: Builds the MST by considering edges in sorted order, using Union-Find for cycle detection

Page Complete

You now understand what the MST problem asks, why it's non-trivial despite having exponential possibilities, and why greedy algorithms can solve it optimally. This theoretical foundation ensures you'll understand not just HOW Prim's and Kruskal's algorithms work, but WHY they work.