Statement for Linked Answer Questions 54 and 55: An undirected graph G(V, E) contains n (n > 2) nodes named v1, v2, , vn. Two nodes vi, vj are connected if and only if 0 < |i - j| ≤ 2. Each edge (vi, vj) is assigned a weight i + j. A sample graph with n = 4 is shown below. What will be the cost of the minimum spanning tree (MST) of such a graph with n nodes? A) (1)/(12)(11 n2 - 5 n) B) n2 - n + 1 C) 6n - 11 D) 2n + 1

Official answer: B) n2 - n + 1. Explanation: Let us examine the edges and their weights: - Edges of the form (vi, vi+1) for 1 ≤ i ≤ n-1 with weight i + (i+1) = 2i + 1. - Edges of the form (vi, vi+2) for 1 ≤ i ≤ n-2 with weight i + (i+2) = 2i + 2. Let us test for small values of n using Kruskal's algorithm: For n = 3: - Vertices: v1, v2, v3 - Edges sorted by weight: 1. (v1, v2), weight = 3 2. (v1, v3), weight = 4 3. (v2, v3), weight = 5 - MST selects (v1, v2) [weight 3] and (v1, v3) [weight 4]. - Total MST cost = 3 + 4 = 7. For n = 4: - Edges sorted by weight: 1. (v1, v2), weight = 3 2. (v1, v3), weight = 4 3. (v2, v3), weight = 5 4. (v2, v4), weight = 6 5. (v3, v4), weight = 7 - MST selects: - (v1, v2) (3) - (v1, v3) (4) - (v2, v4) (6) - Total MST cost = 3 + 4 + 6 = 13. Now, test the options for n = 3 and n = 4: - Option (A): (1)/(12)(11n2 - 5n) - For n = 3: (1)/(12)(99 - 15) = 7 - For n = 4: (1)/(12)(176 - 20) = 13 - Wait, let us check for n = 5: - For n = 5, edges in MST: - (v1, v2) (3) - (v1, v3) (4) - (v2, v4) (6) - (v3, v5) (8) - Total cost = 3 + 4 + 6 + 8 = 21 - Option A: (1)/(12)(11(25) - 25) = (250)/(12) ≠ 21 - Option B: n2 - n + 1: - For n = 3: 32 - 3 + 1 = 7 - For n = 4: 42 - 4 + 1 = 13 - For n = 5: 52 - 5 + 1 = 21 In general, the MST consists of: - The edge (v1, v2) with weight 3. - The edges (vi, vi+2) for 1 ≤ i ≤ n-2 with weights 2i + 2. Total weight of MST: MST cost = 3 + Σi=1n-2 (2i + 2) = 3 + 2((n-2)(n-1))/(2) + 2(n-2) = 3 + (n-2)(n-1) + 2(n-2) = 3 + (n-2)(n + 1) = 3 + n2 - n - 2 = n2 - n + 1 Thus, option B is the correct choice.

Minimum Spanning Trees

GATE CSE 2011 · Q54

MCQ2 marksMedium

Time00:00

Statement for Linked Answer Questions 54 and 55

An undirected graph $G(V, E)$ contains $n$ ( $n > 2$ ) nodes named $v_1, v_2, \ldots, v_n$ . Two nodes $v_i, v_j$ are connected if and only if $0 < |i - j| \le 2$ . Each edge $(v_i, v_j)$ is assigned a weight $i + j$ . A sample graph with $n = 4$ is shown below. What will be the cost of the minimum spanning tree (MST) of such a graph with $n$ nodes?

Select one answerSingle correct

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Solution

Step-by-step verified solution

Official answer: B) n2 - n + 1

Minimum Spanning Trees

Let us examine the edges and their weights:

Edges of the form $(v_i, v_{i+1})$ for $1 \le i \le n-1$ with weight $i + (i+1) = 2i + 1$ .
Edges of the form $(v_i, v_{i+2})$ for $1 \le i \le n-2$ with weight $i + (i+2) = 2i + 2$ .

Let us test for small values of $n$ using Kruskal's algorithm:

For $n = 3$ :

Vertices: $v_1, v_2, v_3$
Edges sorted by weight:
1. $(v_1, v_2)$ , weight $= 3$
2. $(v_1, v_3)$ , weight $= 4$
3. $(v_2, v_3)$ , weight $= 5$
MST selects $(v_1, v_2)$ [weight 3] and $(v_1, v_3)$ [weight 4].
Total MST cost $= 3 + 4 = 7$ .

For $n = 4$ :

Edges sorted by weight:
1. $(v_1, v_2)$ , weight $= 3$
2. $(v_1, v_3)$ , weight $= 4$
3. $(v_2, v_3)$ , weight $= 5$
4. $(v_2, v_4)$ , weight $= 6$
5. $(v_3, v_4)$ , weight $= 7$
MST selects:
- $(v_1, v_2)$ (3)
- $(v_1, v_3)$ (4)
- $(v_2, v_4)$ (6)
Total MST cost $= 3 + 4 + 6 = 13$ .

Now, test the options for $n = 3$ and $n = 4$ :

Option (A): $\frac{1}{12}(11n^2 - 5n)$
- For $n = 3$ : $\frac{1}{12}(99 - 15) = 7$
- For $n = 4$ : $\frac{1}{12}(176 - 20) = 13$
- Wait, let us check for $n = 5$ :
  - For $n = 5$ , edges in MST:
    $(v_1, v_2)$ (3)
    $(v_1, v_3)$ (4)
    $(v_2, v_4)$ (6)
    $(v_3, v_5)$ (8)
    Total cost $= 3 + 4 + 6 + 8 = 21$
  - Option A: $\frac{1}{12}(11(25) - 25) = \frac{250}{12} \neq 21$
  - Option B: $n^2 - n + 1$ :
    For $n = 3$ : $3^2 - 3 + 1 = 7$
    For $n = 4$ : $4^2 - 4 + 1 = 13$
    For $n = 5$ : $5^2 - 5 + 1 = 21$

In general, the MST consists of:

The edge $(v_1, v_2)$ with weight $3$ .
The edges $(v_i, v_{i+2})$ for $1 \le i \le n-2$ with weights $2i + 2$ .

Total weight of MST:

\text{MST cost} = 3 + \sum_{i=1}^{n-2} (2i + 2) = 3 + 2\frac{(n-2)(n-1)}{2} + 2(n-2) = 3 + (n-2)(n-1) + 2(n-2)

= 3 + (n-2)(n + 1) = 3 + n^2 - n - 2 = n^2 - n + 1

Thus, option B is the correct choice.

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This Minimum Spanning Trees previous year question is Q54 of the GATE CSE 2011 paper. Find more GATE CSE PYQs on Minimum Spanning Trees in Algorithms, or browse all GATE CSE PYQs.

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