Correct : c
The correct answer is Option C - 0.
The Jaccard coefficient between two sets A and B is defined as |A β© B| / |A βͺ B|. To find the lowest possible value, we need to minimise the number of shared edges between any two spanning trees of Kn.
Key properties: Every spanning tree of Kn has exactly nβ1 edges. Kn has n(nβ1)/2 total edges.
Can two spanning trees share zero edges? Yes - by the Nash-Williams theorem, a graph G contains k edge-disjoint spanning trees if and only if for every partition of vertices into r parts, the number of edges crossing between parts is at least k(rβ1). For Kn with n > 4, βn/2β β₯ 2, guaranteeing the existence of at least two fully edge-disjoint spanning trees.
When Tβ and Tβ are edge-disjoint: |Tβ β© Tβ| = 0, so Jaccard = 0 / |Tβ βͺ Tβ| = 0.
Why the other options are wrong:
Option A (1/n): This would require exactly 1 shared edge and union of size n - not the minimum achievable. Incorrect.
Option B (1/(2nβ3)): This would correspond to 1 shared edge out of 2nβ3 total - again not the minimum. Incorrect.
Option D (1/(nβ1)): This would mean 1 shared edge out of nβ1 total - still positive, and not the minimum. Incorrect.
Since two fully edge-disjoint spanning trees always exist in Kn for n > 4, the lowest possible Jaccard coefficient is 0.
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