Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 42 / 2 / i / Solution

Treat each row as a set $A_i$ of attributes with response one. The binary option computes the Jaccard distance, rather than counting matches at joint absences. For a pair of rows, let $a$ count coordinates equal to one in both, $b$ count $(1,0)$, $c$ count $(0,1)$, and $d$ count $(0,0)$. ThenThe denominator counts coordinates present in at least one row; joint absences do not contribute. If both rows have no presences, use distance zero. This differs from normalized Hamming distance, which would divide $b+c$ by all ten coordinates.

For Philip and Chad, four attributes are present in both and eight in at least one, giving Jaccard distance $4/8=0.50$. Graham and Tim differ in one of eight coordinates present in their union, giving $1/8=0.125$, printed as $0.12$. Fred and Gbenga have identical profiles, hence distance zero despite being different students. The resulting dissimilarity matrix is symmetric with zero diagonal; dist2full restores the other half and the diagonal from the stored pairwise entries. The display rounds entries for readability; the agglomerative hierarchical clustering uses the original unrounded distances.