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Minimum Spanning Trees

GATE CSE 2020 · Q49

NAT2 marksEasy

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Consider a graph G=(V,E)G = (V, E), where V={v1,v2,...,v100}V = \{v_1, v_2, ..., v_{100}\}, E={(vi,vj)∣1≤i<j≤100}E = \{(v_i, v_j)|1 \le i < j \le 100\}, and weight of the edge (vi,vj)(v_i, v_j) is ∣i−j∣|i - j|. The weight of minimum spanning tree of GG is _______.

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Official answer: 99

Minimum Spanning Trees

A minimum spanning tree (MST) for a graph with n=100n = 100 vertices requires n−1=99n - 1 = 99 edges.

The edge weights are given by w(vi,vj)=∣i−j∣w(v_i, v_j) = |i - j|. The minimum possible edge weight between any two distinct vertices is 11, which occurs between consecutive vertices (vi,vi+1)(v_i, v_{i+1}) for 1≤i≤991 \le i \le 99.

Selecting all 99 such edges:

{(v1,v2),(v2,v3),…,(v99,v100)}\{(v_1, v_2), (v_2, v_3), \dots, (v_{99}, v_{100})\}

forms a connected path covering all 100 vertices with no cycles, which is a valid spanning tree.

Total weight of this MST:

Weight=99×1=99\text{Weight} = 99 \times 1 = 99

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This Minimum Spanning Trees previous year question is Q49 of the GATE CSE 2020 paper. Find more GATE CSE PYQs on Minimum Spanning Trees in Algorithms, or browse all GATE CSE PYQs.

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Minimum Spanning Trees in GATE CSE

Minimum Spanning Trees has 13 GATE CSE PYQs across 9 papers from 2011 to 2026. 3 of them are 2-mark NAT questions like this one, and GATE CSE 2020 has 2.

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GateAI solution cardGATE CSE 2020 · Q49 · NAT

Consider a graph G = (V, E), where V = {v1, v2, ..., v100}, E = {(vi, vj)|1 ≤ i < j ≤ 100}, and weight of the edge (vi, vj) is |i - j|. The weight of minimum spanning tree of G is …

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