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Graph Traversals, MSTs and Shortest Paths

GATE CSE 2015 Set 3 · Q40

NAT2 marksEasy

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Let GG be a connected undirected graph of 100 vertices and 300 edges. The weight of a minimum spanning tree of GG is 500. When the weight of each edge of GG is increased by five, the weight of a minimum spanning tree becomes _______.

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Official answer: 995

Graph Traversals, MSTs and Shortest Paths

A spanning tree of a connected undirected graph with ∣V∣=100|V| = 100 vertices always contains exactly ∣V∣−1=99|V| - 1 = 99 edges [1].

When the weight of every edge in the graph is increased by a constant c=5c = 5, the weight of every spanning tree increases by the exact same amount:

ΔW=(∣V∣−1)×5=99×5=495\Delta W = (|V| - 1) \times 5 = 99 \times 5 = 495

Because the weight of every spanning tree is shifted by the same constant, the tree that was minimum before remains a minimum spanning tree in the modified graph.

Therefore, the new weight of the minimum spanning tree is:

Wnew=500+495=995W_{\text{new}} = 500 + 495 = 995

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This Graph Traversals, MSTs and Shortest Paths previous year question is Q40 of the GATE CSE 2015 Set 3 paper. Find more GATE CSE PYQs on Graph Traversals, MSTs and Shortest Paths in Algorithms, or browse all GATE CSE PYQs.

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Graph Traversals, MSTs and Shortest Paths in GATE CSE

Graph Traversals, MSTs and Shortest Paths has 7 GATE CSE PYQs across 7 papers from 2014 to 2025. 4 of them are 2-mark NAT questions like this one, and it is the only one in GATE CSE 2015 Set 3.

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Let G be a connected undirected graph of 100 vertices and 300 edges. The weight of a minimum spanning tree of G is 500. When the weight of each edge of G is increased by five, the weight of a minimum spanning tree becomes …

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