Minimum Spanning Tree Practice Questions

Cayley's Formula states that the number of spanning trees in a complete graph $K_n$ with $n$ vertices is given by $n^{n-2}$.
In this problem, the graph is a complete graph with $n = 7$ vertices.
Substituting $n = 7$ into the formula yields $7^{7-2}$.
Simplifying the exponent results in $7^5$.
Thus, the total number of spanning trees is $7^5$.

The total weight of the Minimum Spanning Tree (MST) is 36. The edges not in the MST are (A,B), (C,D), and (E,D). According to the cut property of MSTs, for any cycle, the weight of an edge not in the MST must be greater than the weight of any edge on the path in the MST that connects its endpoints.

• For edge (A,B), the path in the MST is A-C-B. The maximum edge weight on this path is 9. So, weight(A,B) > 9. The minimum integer weight is 10.
• For edge (C,D), the path is C-B-E-F-D. The max edge weight is 15. So, weight(C,D) > 15. Minimum is 16.
• For edge (E,D), the path is E-F-D. The max edge weight is 6. So, weight(E,D) > 6. Minimum is 7.
  The minimum sum of all 8 edges is the MST weight plus the minimum weights of the other edges: $36 + 10 + 16 + 7 = 69$.

To find the number of distinct Minimum Spanning Trees (MSTs) for the given weighted graph, we can use Kruskal's algorithm.

1. List all edges with their weights:

   • Weight 1: $(V_1, V_4)$, $(V_2, V_3)$, $(V_4, V_6)$, $(V_5, V_6)$, $(V_8, V_7)$
   • Weight 2: $(V_1, V_2)$, $(V_2, V_4)$, $(V_4, V_3)$, $(V_4, V_5)$, $(V_5, V_8)$, $(V_6, V_7)$, $(V_7, V_8)$

2. Sort edges by weight in increasing order. When weights are equal, the choice of edges can lead to different MSTs.

3. All edges of weight 1 must be included as they do not form a cycle among themselves. These edges are $(V_1, V_4)$, $(V_2, V_3)$, $(V_4, V_6)$, $(V_5, V_6)$, $(V_8, V_7)$.
   These 5 edges connect all 8 vertices, forming a single component. To form an MST with 8 vertices, we need $8-1=7$ edges. We currently have 5 edges.

4. We need to select 2 more edges of weight 2 to complete the MST without forming a cycle.
   The edges of weight 2 that do not form a cycle with the weight 1 edges are:

   • $(V_1, V_2)$
   • $(V_4, V_3)$ (This creates a cycle $V_2-V_3-V_4-V_1-V_2$ if $(V_1,V_2)$ is already chosen)
   • $(V_4, V_5)$ (This creates a cycle $V_4-V_5-V_6-V_4$ if $(V_5,V_6)$ is already chosen)
   • $(V_5, V_8)$
   • $(V_6, V_7)$ (This creates a cycle $V_6-V_4-V_1-V_2-V_3-V_4$ if $(V_4,V_3)$ is chosen)
   • $(V_7, V_8)$ (Already selected as a weight 1 edge)

5. The remaining edges of weight 2 are $(V_1, V_2)$, $(V_4, V_3)$, $(V_5, V_8)$, $(V_6, V_7)$.
   We need to select 2 edges from this set to connect the components without creating a cycle.

   • If we select $(V_1, V_2)$ and $(V_7, V_8)$, we get one MST.
   • If we select $(V_1, V_2)$ and $(V_6, V_7)$, we get another MST.
   • If we select $(V_2, V_4)$ and $(V_7, V_8)$, we get another MST.
   • If we select $(V_2, V_4)$ and $(V_6, V_7)$, we get another MST.
   • If we select $(V_3, V_4)$ and $(V_7, V_8)$, we get another MST.
   • If we select $(V_3, V_4)$ and $(V_6, V_7)$, we get another MST.

There are 6 distinct ways to choose the two additional edges of weight 2 without forming cycles.

[CORRECT_OPTION: 6]

The property of being a Minimum Spanning Tree (MST) relies on the relative ordering of edge weights. If we apply a strictly increasing function $f(w)$ to all edge weights $w$ in a graph, the MST of the new graph will be the same as the MST of the original graph.

In this question, the edge weights are squared, i.e., $f(w) = w^2$. Since all edge weights are greater than one, the function $f(w) = w^2$ is strictly increasing for $w > 1$. Therefore, the MST of G' (T') will consist of the same set of edges as the MST of G (T). So, T' = T.

However, the total weight of T' will be $t' = \sum_{e \in T} w_e^2$. Since $w_e > 1$, it is not necessarily true that $\sum w_e^2 = (\sum w_e)^2 = t^2$. For example, if T has edges with weights 2 and 3, then $t = 2+3=5$. The squared weights are 4 and 9, so $t' = 4+9=13$. Here, $t^2 = 5^2 = 25$, and $t' < t^2$. If T has only one edge with weight 2, then $t=2$ and $t'=4$, so $t'=t^2$.

Since the relationship between $t'$ and $t^2$ is not consistently $t'=t^2$ or $t'<t^2$ across all possible MSTs, and only the equality of the edge sets T and T' is guaranteed, none of the given statements are universally true.

To find the path length in the Minimum Spanning Tree (MST), we first identify the edge weights $w(v_i, v_j) = i + j$ for all edges $(v_i, v_j)$ where $0 < |i-j| \leq 2$. Sorting the edges by weight, we construct the MST using Kruskal's algorithm by selecting the smallest weight edges that do not form a cycle.

The chosen edges for the MST are:
$(v_1, v_2)$ with weight $3$, $(v_1, v_3)$ with weight $4$, $(v_2, v_4)$ with weight $6$, $(v_3, v_5)$ with weight $8$, $(v_4, v_6)$ with weight $10$, and continuing similarly $(v_5, v_7)$ with weight $12$, etc.

The unique path from $v_5$ to $v_6$ in the MST is:
$v_5 \xrightarrow{8} v_3 \xrightarrow{4} v_1 \xrightarrow{3} v_2 \xrightarrow{6} v_4 \xrightarrow{10} v_6$

Summing these edge weights gives the path length:
$8 + 4 + 3 + 6 + 10 = 31$

The problem defines an undirected graph where nodes $v_i$ and $v_j$ are connected if $0 < |i-j| \le 2$. This means each node $v_i$ is connected to $v_{i-1}$, $v_{i-2}$ (if they exist), $v_{i+1}$, and $v_{i+2}$ (if they exist). The weight of an edge $(v_i, v_j)$ is $i+j$. We need to find the cost of the Minimum Spanning Tree (MST).

For an MST, we want to select edges with the smallest possible weights.
Consider the edges connecting node $v_i$:

• Edge $(v_i, v_{i+1})$ has weight $i + (i+1) = 2i+1$.
• Edge $(v_i, v_{i+2})$ has weight $i + (i+2) = 2i+2$.

Since $2i+1 < 2i+2$, edges between adjacent nodes ($v_i, v_{i+1}$) are always preferred over edges between nodes separated by one node ($v_i, v_{i+2}$). To form an MST, we should prioritize connecting adjacent nodes. Connecting all adjacent nodes $(v_i, v_{i+1})$ for $i=1, \dots, n-1$ will form a path graph, which is a tree spanning all $n$ nodes.

The total cost of such a tree would be the sum of the weights of these edges:
$\sum_{i=1}^{n-1} (i + (i+1)) = \sum_{i=1}^{n-1} (2i+1)$
This sum can be broken down:
$2 \sum_{i=1}^{n-1} i + \sum_{i=1}^{n-1} 1$
Using the formula for the sum of the first $k$ integers, $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$:
$2 \frac{(n-1)n}{2} + (n-1) = n(n-1) + (n-1)$
$= n^2 - n + n - 1 = n^2 - 1$
This formula assumes a path graph. However, we have edges $(v_i, v_{i+2})$ which can be used to connect nodes if the adjacent edges are not chosen.
A simpler path of $n$ nodes has $n-1$ edges.
The sum of weights of edges $(v_i, v_{i+1})$ for $i=1, \dots, n-1$ is $n^{2-}1$.
This forms a spanning tree. Since we have only chosen the lowest weight edges for adjacent nodes, this is indeed the MST.
For example, for $n=4$:
Edges are $(v_1, v_2)$, $(v_2, v_3)$, $(v_3, v_4)$.
Weights: $1+2=3$, $2+3=5$, $3+4=7$.
Total cost = $3+5+7 = 15$.
Using formula: $4^{2-}1 = 16-1 = 15$.

The provided options do not have $n^{2-}1$. Let's re-examine the MST criteria.
In a graph with $n$ vertices, an MST has $n-1$ edges. For this specific graph structure, Kruskal's or Prim's algorithm would select edges $(v_i, v_{i+1})$ first because their weights ($2i+1$) are generally smaller than $(v_i, v_{i+2})$ ($2i+2$).
Consider edges with weights:
$(v_1, v_2) \rightarrow 3$
$(v_2, v_3) \rightarrow 5$
$(v_3, v_4) \rightarrow 7$
...
$(v_{n-1}, v_n) \rightarrow 2n-1$

The sum of these $n-1$ edges is:
$\sum_{i=1}^{n-1} (2i+1) = 2 \sum_{i=1}^{n-1} i + \sum_{i=1}^{n-1} 1 = 2 \frac{(n-1)n}{2} + (n-1) = n(n-1) + (n-1) = n^2 - n + n - 1 = n^2 - 1$.

There might be a typo in the question or options or the provided solution, as $n^{2-}1$ is not available.
Let's consider the provided solution and try to derive it. Option B is $n^{2-}n+1$.
For $n=4$, $n^{2-}n+1 = 16-4+1 = 13$.
However, the MST cost for $n=4$ is $15$.

Let's assume the question's sample graph for $n=4$ has minimum edges to connect all nodes.
The sample graph:
v1-v2 (3)
v2-v3 (5)
v3-v4 (7)
v1-v3 (4)
v2-v4 (6)
The graph includes edges $(v_i, v_{i+1})$ and $(v_i, v_{i+2})$.
For $n=4$:
Edges are (v1,v2) w=3, (v1,v3) w=4, (v2,v3) w=5, (v2,v4) w=6, (v3,v4) w=7.
Using Kruskal's algorithm:

1. Select (v1,v2) w=3. MST edges: {(v1,v2)}
2. Select (v1,v3) w=4. MST edges: {(v1,v2), (v1,v3)}
3. Select (v2,v3) w=5. This would create a cycle (v1-v2-v3-v1). Don't select.
4. Select (v2,v4) w=6. MST edges: {(v1,v2), (v1,v3), (v2,v4)}. This doesn't create a cycle yet.
5. Select (v3,v4) w=7. This would create a cycle (v1-v3-v4-v2-v1 or v1-v3-v4-v2). Don't select.

Wait, Kruskal's algorithm:
Edges by weight:
(v1,v2) -> 3
(v1,v3) -> 4
(v2,v3) -> 5
(v2,v4) -> 6
(v3,v4) -> 7

1. Add (v1,v2) (weight 3). Components: {v1,v2}, {v3}, {v4}. Cost = 3.
2. Add (v1,v3) (weight 4). Components: {v1,v2,v3}, {v4}. Cost = 3+4=7.
3. Skip (v2,v3) as v1,v2,v3 are already connected.
4. Add (v2,v4) (weight 6). Components: {v1,v2,v3,v4}. Cost = 7+6=13.
   The MST cost for $n=4$ is $13$.

Now let's verify if $n^{2-}n+1$ matches this for $n=4$: $4^{2-}4+1 = 16-4+1 = 13$.
$n^{2-}n+1$ seems to be the correct formula.

Let's generalize this.
The edges are $(v_i, v_{i+1})$ with weight $2i+1$ and $(v_i, v_{i+2})$ with weight $2i+2$.
The smallest weight edges are:
$(v_1, v_2)$ weight 3
$(v_1, v_3)$ weight 4
$(v_2, v_3)$ weight 5 (cycle with (v1,v2), (v1,v3))
$(v_2, v_4)$ weight 6
...
$(v_{n-1}, v_n)$ weight $2n-1$

The MST will include edges $(v_i, v_{i+1})$ for all $i$ and additionally some $(v_i, v_{i+2})$ edges only if necessary.
The MST edges would be $(v_1, v_2)$, $(v_1, v_3)$ (since $1+3=4$ is smaller than $2+3=5$ for $(v_2,v_3)$).
However, the MST is generally formed by connecting $v_i$ to $v_{i+1}$ (cost $2i+1$) and $v_i$ to $v_{i+2}$ (cost $2i+2$).
The strategy of Kruskal's algorithm would be to sort all edges by weight.
The edges are $(v_i, v_j)$ where $|i-j|=1$ or $|i-j|=2$.
Weights are $i+j$.
The minimum weight edges are for adjacent nodes $(v_i, v_{i+1})$ which are $2i+1$.
The edges $(v_i, v_{i+2})$ have weights $2i+2$.
Notice that $i+(i+1) = 2i+1$ and $i+(i+2) = 2i+2$.
The lowest weight edge is $(v_1, v_2)$ with weight 3.
The next lowest weight edge is $(v_1, v_3)$ with weight 4. This is chosen.
The next lowest edge is $(v_2, v_3)$ with weight 5. This creates a cycle $v_1-v_2-v_3-v_1$, so it's not chosen.
The next lowest edge is $(v_2, v_4)$ with weight 6. This is chosen.
In general, an MST will take $(v_i, v_{i+1})$ edges. However, if $v_i, v_{i+1}, v_{i+2}$ are involved, it might prefer $(v_i, v_{i+2})$ over $(v_{i+1}, v_{i+2})$ if $i+(i+2) < (i+1)+(i+2)$ and $v_i, v_{i+1}$ are connected to $v_{i-1}$ etc.

Let's analyze the connections the algorithm makes.
It connects $(v_1, v_2)$, $(v_1, v_3)$.
Then it continues with $(v_2, v_4)$, $(v_3, v_5)$, $(v_4, v_6)$... and so on up to $v_n$.
This seems to be a 'zig-zag' pattern to connect the nodes.
The edges chosen by Kruskal's algorithm will be:
$(v_1, v_2)$ (weight 3)
$(v_1, v_3)$ (weight 4)
$(v_2, v_4)$ (weight 6)
$(v_3, v_5)$ (weight 8)
...
$(v_{n-2}, v_n)$ (weight $2n-2$)

What if $n$ is small?
For $n=3$:
Edges: (v1,v2) w=3, (v1,v3) w=4, (v2,v3) w=5.
MST: (v1,v2) w=3, (v1,v3) w=4. Total cost = 7.
Formula $n^{2-}n+1 = 3^{2-}3+1 = 9-3+1=7$. Matches.

For $n=2$:
Edges: (v1,v2) w=3. MST cost = 3.
Formula $n^{2-}n+1 = 2^{2-}2+1 = 4-2+1=3$. Matches.

The edges chosen by Kruskal's algorithm for MST will be $(v_i, v_{i+1})$ for $i=1$ and $(v_j, v_{j+2})$ for $j=1 \dots n-2$.
The set of edges for the MST for $n$ nodes:

• $(v_1, v_2)$ with weight $1+2=3$.
• $(v_1, v_3)$ with weight $1+3=4$.
• $(v_2, v_4)$ with weight $2+4=6$.
• $(v_3, v_5)$ with weight $3+5=8$.
  ...
• $(v_{n-2}, v_n)$ with weight $(n-2)+n=2n-2$.

The total cost is $3 + \sum_{k=1}^{n-

The problem asks for the minimum weight of a spanning tree where vertex 0 is a leaf node. A leaf node in a tree has a degree of 1, meaning it is connected by only a single edge.

The strategy is to find the minimum spanning tree (MST) for the subgraph containing all vertices except 0, and then connect vertex 0 to this MST using the cheapest possible edge.

Step 1: Find the MST of the subgraph with vertices {1, 2, 3, 4}.
We can use Kruskal's algorithm on the edges connecting only these four vertices:

• Edge {3, 4}: weight 2
• Edge {2, 4}: weight 3
• Edge {1, 3}: weight 4
• Edge {2, 3}: weight 7
• Edge {1, 4}: weight 9
• Edge {1, 2}: weight 12

Applying Kruskal's algorithm:

1. Select {3, 4} (weight 2). Sets: {1}, {2}, {3,4}.
2. Select {2, 4} (weight 3). Sets: {1}, {2,3,4}.
3. Select {1, 3} (weight 4). Sets: {1,2,3,4}. All vertices are connected.

The MST for the subgraph {1, 2, 3, 4} consists of edges {3,4}, {2,4}, and {1,3}. Its total weight is 2 + 3 + 4 = 9.

Step 2: Connect vertex 0 to the formed component.
To make vertex 0 a leaf, we must connect it to the component {1,2,3,4} with a single edge. To minimize the total weight, we should choose the cheapest edge from vertex 0 to any of the vertices {1, 2, 3, 4}.

• Edge {0, 1}: weight 1
• Edge {0, 3}: weight 1
• Edge {0, 4}: weight 4
• Edge {0, 2}: weight 8

The minimum weight edge is 1 (either {0,1} or {0,3}).

Step 3: Calculate the total minimum weight.
Total Weight = (Weight of MST on {1,2,3,4}) + (Weight of the cheapest connecting edge to 0)
Total Weight = 9 + 1 = 10.

Thus, the minimum possible weight of the spanning tree with vertex 0 as a leaf is 10.

Kruskal's algorithm adds edges to a Minimum Spanning Tree (MST) in non-decreasing order of their weights, ensuring no cycles are formed.

First, list all edges and sort them by weight:
(b,e) = 2
(a,c) = 3
(e,f) = 3
(b,c) = 4
(f,g) = 4
(a,b) = 5 (would form cycle a-c-b)
(c,d) = 5
(e,g) = 5 (would form cycle e-b-c-f-g)
(b,d) = 6
(c,f) = 6
(d,e) = 6
(d,f) = 6

The standard sequence of edges added by Kruskal's algorithm (with ties broken arbitrarily) is:

1. (b,e) [2]
2. (a,c) [3]
3. (e,f) [3]
4. (b,c) [4] (no cycle formed)
5. (f,g) [4] (no cycle formed)
6. (c,d) [5] (no cycle formed)
   The resulting MST edges are: {(b,e), (a,c), (e,f), (b,c), (f,g), (c,d)}.

Now, let's examine the options for their sequences of weights:
A. (b,e)[2] (e,f)[3] (a,c)[3] (b,c)[4] (f,g)[4] (c,d)[5] - Weights: 2, 3, 3, 4, 4, 5. This is in non-decreasing order. This is a valid sequence.
B. (b,e)[2] (e,f)[3] (a,c)[3] (f,g)[4] (b,c)[4] (c,d)[5] - Weights: 2, 3, 3, 4, 4, 5. This is in non-decreasing order. This is a valid sequence.
C. (b,e)[2] (a,c)[3] (e,f)[3] (b,c)[4] (f,g)[4] (c,d)[5] - Weights: 2, 3, 3, 4, 4, 5. This is in non-decreasing order. This is a valid sequence.
D. (b,e)[2] (e,f)[3] (b,c)[4] (a,c)[3] (f,g)[4] (c,d)[5] - Weights: 2, 3, 4, 3, 4, 5. This sequence is NOT in non-decreasing order (4 is followed by 3). Therefore, this sequence cannot be generated by Kruskal's algorithm.

There are $2$ spanning trees (a spanning tree connects all $n$ vertices) for $G$ which are edge disjoint. A spanning tree for $n$ nodes require $n-1$ edges and so 2 edge-disjoint spanning trees requires $2n-2$ edges. As $G$ has only $2n-2$ edges, it is clear that it doesn't have any edge outside that of the two spanning trees. Now lets see the cases: Lets take any subgraph of $G$ with $k$ vertices. The remaining subgraph will have $n-k$ vertices. Between these two subgraphs (provided both has at least one vertex) there should be at least $2$ edges, as we have $2$ spanning trees in $G$. So, (B) is TRUE. Also, in the subgraph with $k$ vertices, we cannot have more than $2k-2$ edges as this would mean that in the other subgraph with $n-k$ vertices, we would have less than $2n-2k$ edges, making $2$ spanning trees impossible in it. So, (A) is also TRUE. A spanning tree covers all the vertices. So, $2$ edge-disjoint spanning trees in $G$ means, between every pair of vertices in $G$ we have two edge-disjoint paths (length of paths may vary). So, (C) is also TRUE. So, that leaves option ( D ) as answer. It is not quite hard to give a counter example for (D).

Prim's algorithm greedily expands the Minimum Spanning Tree by adding the lowest-weight edge connecting a vertex in the current tree to one outside it. Starting at node $d$, the sequence in Option C follows this logic:

1. Start at $d$: Select edge $(d, f)$ with weight $2$. Tree: $\{d, f\}$.
2. From $\{d, f\}$: Select edge $(f, c)$ with weight $3$. Tree: $\{d, f, c\}$.
3. From $\{d, f, c\}$: Select edge $(d, a)$ with weight $5$. Tree: $\{d, f, c, a\}$.
4. From $\{d, f, c, a\}$: Select edge $(a, b)$ with weight $1$. Tree: $\{d, f, c, a, b\}$.
5. Continue connecting remaining nodes by selecting $(c, e)$ weight $7$, $(f, h)$ weight $8$, $(g, h)$ weight $7$, and $(g, i)$ weight $3$.

Each step consistently chooses the minimum weight edge available to grow the connected component, satisfying the requirements of Prim's algorithm.

A simple connected graph with $n$ vertices and $n$ edges is known as a unicyclic graph. Such a graph contains exactly one cycle of length $k$, where $3 \le k \le n$. The number of spanning trees in a unicyclic graph is exactly equal to the length of its unique cycle, $k$. Since the graph must be simple, the smallest possible cycle is a triangle, meaning $k \ge 3$. Consequently, every such graph contains at least 3 spanning trees, and this minimum is achieved by any graph with a 3-cycle. Thus, the largest integer $m$ is 3.

Let $w$ be the minimum edge weight in an undirected connected graph. Let $e$ be a specific edge of weight $w$.

• A. There is a minimum spanning tree containing $e$. This is true. If $e$ is not in an MST $T$, adding $e$ to $T$ creates a cycle. All edges in this cycle must have weight $w$ (otherwise, we could swap $e$ for a heavier edge in the cycle to get a lighter spanning tree, contradicting $T$ being an MST). Thus, we can replace any other edge in the cycle with $e$ to form an MST that contains $e$.

• B. If $e$ is not in a minimum spanning tree $T$, then in the cycle formed by adding $e$ to $T$, all edges have the same weight. This is true. As explained in A, if $e \notin T$, adding $e$ to $T$ creates a cycle. If any edge $e'$ in this cycle had $weight(e') > weight(e)=w$, we could replace $e'$ with $e$ to form a spanning tree $T'$ with $weight(T') < weight(T)$, which contradicts $T$ being an MST. Therefore, all edges in the cycle must have weight $w$.

• C. Every minimum spanning tree has an edge of weight $w$. This is true. Since $w$ is the minimum possible weight, and the graph is connected, any MST must include at least one edge of weight $w$. If an MST did not contain any edge of weight $w$, then all its edges would have weight strictly greater than $w$, implying a higher total weight than an MST which could include an edge of weight $w$ (which exists by definition of $w$). Kruskal's algorithm, for example, would always select an edge of weight $w$ first if it doesn't form a cycle.

• D. $e$ is present in every minimum spanning tree. This is false. Consider a graph with vertices $\{1, 2, 3\}$ and edges $(1,2)$, $(1,3)$, $(2,3)$, all with weight $w=1$. Let $e = (1,2)$.

  • An MST can be $\{(1,2), (1,3)\}$. This MST contains $e$.
  • Another MST can be $\{(1,3), (2,3)\}$. This MST does not contain $e$.
    Since $e$ is not in every MST, this statement is false.

The final answer is $\boxed{D}$

Kruskal's algorithm constructs a Minimum Spanning Tree by considering and adding edges in non-decreasing order of their weights, provided they do not form a cycle. The relevant edge weights for this graph are: $(a,b)=1$, $(d,f)=1$, $(b,f)=2$, $(c,d)=2$, and $(d,e)=3$. Option D lists the sequence of weights as $1, 1, 2, 3, 2$. Since Kruskal's algorithm must process all edges of weight $2$ (such as $(c,d)$) before any edge of weight $3$ (such as $(d,e)$), adding $(d,e)$ before $(c,d)$ violates the required greedy ordering. cannot be a valid sequence.

To construct the minimum spanning tree (MST), we select edges with the smallest possible weights. The edge weight function $w(v_i, v_j) = 2|i-j|$ is minimized when $|i-j| = 1$, resulting in a weight of $2|1| = 2$ for adjacent vertices $(v_i, v_{i+1})$. By selecting these $n-1$ edges $(v_1, v_2), (v_2, v_3), \dots, (v_{n-1}, v_n)$, we connect all $n$ vertices into a single spanning tree. The total weight of this MST is the sum of the weights of these $n-1$ edges:
$\sum_{i=1}^{n-1} 2 = 2(n-1) = 2n - 2$

The adjacency matrix describes a complete graph $K_n$ where every pair of distinct nodes is connected by an edge of weight $1$. A spanning tree of $G$ must contain exactly $n-1$ edges to connect $n$ nodes without cycles. Since all edges have a uniform weight of $1$, the cost of any spanning tree is $(n-1) \times 1 = n-1$. According to Cayley's formula, there are $n^{n-2}$ distinct spanning trees in a complete graph $K_n$. Because all these spanning trees have the same minimal cost of $n-1$, the graph $G$ possesses multiple distinct MSTs.

Removing edge $e$ from an MST $T$ partitions the vertices into two sets, $S$ and $V \setminus S$, defining a cut $(S, V \setminus S)$ in $G$. According to the MST Cut Property, $e$ must be a minimum weight edge crossing this cut, implying $w(e) \le w(e')$ for all edges $e'$ crossing the cut. Given that $e$ is a maximum weight edge in the entire graph, any edge $e'$ crossing the cut must satisfy $w(e') \le w(e)$. Combining these inequalities, $w(e') = w(e)$ for every edge $e'$ in the cutset. Thus, there exists a cutset where all edges have the maximum weight.

The Cut Property of Minimum Spanning Trees (MSTs) states that for any partition $[X, Y]$ of the set of vertices $V$ in a connected graph $G=(V, E)$, the edge with the strictly minimum weight crossing the cut must belong to the MST of $G$. Given that the graph has distinct positive edge weights, the edge $e$ is the unique minimum weight edge crossing the partition $[X, Y]$. Therefore, by the fundamental definition of the Cut Property, $e$ must be included in the MST of $G$.

The problem asks for the path from $s$ to $t$ that minimizes the "congestion," defined as the maximum weight of any edge on the path. This is known as the bottleneck path problem. A fundamental theorem in graph theory states that the path between any two vertices $s$ and $t$ in a Minimum Spanning Tree (MST) of a graph is the unique path that minimizes the maximum edge weight among all possible paths connecting $s$ and $t$. The cut property described, where the minimum weight edge crossing a partition $[X, Y]$ is chosen, is a core principle used in algorithms like Prim's or Kruskal's to construct the MST. Because the edge weights are distinct, the MST is unique and provides the optimal path for minimizing the maximum edge weight. Therefore, the path from $s$ to $t$ in the minimum weighted spanning tree is always the path of minimum congestion.

Prim's algorithm starts at node $A$, greedily adding the edge with the minimum weight connecting a node in the current tree to a node outside the tree.

1. Starting at $\{A\}$, the edge $(A, D)$ (weight $7$) is the cheapest, followed by $(A, B)$ (weight $10$).
2. The current tree is $\{A, D, B\}$. Available frontier edges to new nodes are $(A, C)$ (weight $22$) and $(D, F)$ (weight $22$). We choose $(D, F)$ (weight $22$) as it is a valid minimum.
3. With tree $\{A, D, B, F\}$, the frontier edges include $(F, C)$ (weight $4$) and $(F, G)$ (weight $5$). We select $(F, C)$ (weight $4$) first.
4. From $\{A, D, B, F, C\}$, the frontier edge $(F, G)$ (weight $5$) is the minimum weight to include node $G$.
5. Finally, we add $(G, E)$ (weight $2$) to connect the last node $E$ to the tree.

To find the Minimum Spanning Tree (MST), we apply Kruskal's algorithm by selecting the smallest edges without forming cycles. Since there are $10$ vertices, we need exactly $9$ edges.

1. Select edges with weight $1$ and $2$: $(a,c)$ [1], $(a,d)$ [2], $(b,g)$ [2], $(e,i)$ [2] (Sum: $7$).
2. Select edge $(b,d)$ with weight $3$ (Sum: $10$).
3. Select edges $(f,h)$ [4] and $(h,i)$ [4] (Sum: $18$).
4. Select edge $(i,j)$ with weight $5$ (Sum: $23$).
5. Finally, select edge $(a,e)$ with weight $8$ to connect all components (Sum: $31$).
   The total weight is $1 + 2 + 2 + 2 + 3 + 4 + 4 + 5 + 8 = 31$.