# Help:Trolls on wheels!

One counter-example consists of a series of subsets that increase in size exponentially, plus 2 additional subsets that each cover half of the elements. Example:

$ S_1 = \{1, 2\} $

$ S_2 = \{3, 4, 5, 6\} $

$ S_3 = \{7, 8, 9, 10, 11, 12, 13, 14, 15, 16 \} $

$ S_4 = \{1, 2, 3, 4, 5, 6, 7, 8 \} $

$ S_5 = \{9, 10, 11, 12, 13, 14, 15, 16 \} $

The greedy algorithm will choose $ S_3, S_2, S_1 $, while the optimal solution is simply $ S_4, S_5 $

Counter-example 2 (in case you spot an error in counter-example 1 - see Discussion):

$ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\} $

$ S_1 = \{1, 2, 4, 6, 8, 10, 12, 14\} $

$ S_2 = \{3, 5, 9, 11\} $

$ S_3 = \{7, 13\} $

$ S_4 = \{1, 2, 3, 4, 5, 6, 7\} $

$ S_5 = \{8, 9, 10, 11, 12, 13, 14\} $

There is an optimal solution: $ S_4, S_5 $ (2 subsets).

A greedy algorithm will choose $ S_1, S_2, S_3 $ (3 subsets):

1. $ S_1 $ since it contains 8 uncovered elements (more than any other subset)

2. $ S_2 $ since it then contains 4 uncovered elements (more than any other subset)

3. $ S_3 $ since it then contains 2 uncovered elements (more than any other subset)