Let G=(V,E) be an undirected, unweighted, connected graph. Its diameter is the…
2021
Let G=(V,E) be an undirected, unweighted, connected graph. Its diameter is the maximum, over all vertex pairs, of the length of a shortest path between them.
Let M be the adjacency matrix of G and let P=M2. Define a simple graph G2 on the same vertex set with adjacency matrix N by setting Nii=0 and, for i≠j, Nij=1 if Mij>0 or Pij>0; otherwise Nij=0.
Which one of the following statements is true?
Answer: A. diam(G2) ≤ ⌈diam(G)/2⌉ — ConceptIn a graph square, vertices at original distance one or two become adjacent. Thus one edge in the squared graph represents a walk of at most two edges…
- A.
diam(G2) ≤ ⌈diam(G)/2⌉
- B.
⌈diam(G)/2⌉ < diam(G2) < diam(G)
- C.
diam(G2) = diam(G)
- D.
diam(G) < diam(G2) ≤ 2 diam(G)
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Show answer & explanation
Correct answer: A
Concept
In a graph square, vertices at original distance one or two become adjacent. Thus one edge in the squared graph represents a walk of at most two edges in the original graph. Self-loops, if retained, never shorten a path between distinct vertices.
Application
The diagonal entry Pii counts length-two closed walks and may be positive. Setting Nii=0 removes those loops; even retaining them would not alter any distance between distinct vertices.
Choose any vertices u and v and write d=dist_G(u,v). Along a shortest u-v path in G, group consecutive edges into blocks of two, with one final single-edge block when d is odd.
Each block becomes one edge of G2, so dist_G2(u,v) ≤ ⌈d/2⌉.
Conversely, every non-loop edge of G2 represents a walk of length one or two in G. A G2 path of k edges therefore expands to a G walk of length at most 2k, giving d ≤ 2 dist_G2(u,v), hence dist_G2(u,v) ≥ ⌈d/2⌉.
Combining the bounds gives dist_G2(u,v)=⌈dist_G(u,v)/2⌉. Maximizing over all vertex pairs yields diam(G2)=⌈diam(G)/2⌉.
Cross-check and contrast
For a path with L edges, G2 permits jumps of one or two original edges, so its endpoints are exactly ⌈L/2⌉ edges apart. This attains the bound.
The relation diam(G2) ≤ ⌈diam(G)/2⌉ follows immediately from the equality.
A strict lower bound above ⌈diam(G)/2⌉ contradicts the equality.
Equality with diam(G) occurs only in small special cases and is not a universal identity.
Adding distance-two edges cannot make the diameter larger.
Hence the true statement is diam(G2) ≤ ⌈diam(G)/2⌉.
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