# Difference between revisions of "TADM2E 1.6"

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

## Latest revision as of 22:54, 18 June 2019

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 $.