TADM2E 1.6

Let $ U=\{1,2,3,4,5,6\} $ and S consists of $ S_1=\{1,2,3,4\} $, $ S_2=\{1,2,5\} $, $ S_3=\{3,4,6\} $. The algorithm would give an answer of $ S_1 $, $ S_2 $, $ S_3 $ but the correct answer is $ S_2 $,$ S_3 $ because it has the fewest subsets and covers $ U $.