#fixed-point

Question 1: Single choice

Ten points form a vertical column $(0, \pm1), (0, \pm3), (0, \pm5)$ and a horizontal row $(2, 0), (4, 0), (6, 0), (8, 0)$ . Spectral clustering on a 2-nearest-neighbour graph returns the column and the row as its two clusters. Could $k$ -means ( $k = 2$ ) end with this same partition?

Yes, from a suitable initialization every point stays with its own centroid1
No, $(2, 0)$ is nearer the column's centroid $(0, 0)$ than the row's $(5, 0)$ 2
No, $(0, \pm 5)$ are closer to the row's centroid than to the column's3
Yes, because $k$ -means and spectral clustering optimize the same objective4

Timed test