Question: Suppose you run k - means using k = 3 and k = 5 . You find that the cost function is much higher for

Suppose you run k-means using k =3 and k =5. You find that the cost function is much
higher for k =5 than k =3 at convergence. What can you conclude?
(a) This is mathematically impossible. There must be a bug in the code.
(b) k-means got stuck in a bad local minimum when k =5. You should try re-running k-means
for k =5 with a different initialization.
(c) k-means got lucky when k =3. You should try re-running k-means for k =3 with different
initializations, until it performs no better than k =5.

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