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Alexey Ustinov
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Suppose that we need to devidedivide 9 people into two volleyball teams (6+6). Usually we start with two captains (best players). First captain choose one player. After that second captain and first captain choose two players on each step. The problem is in bad players which may have negative weight in the team. And in the end of this procedure captains obliged to choose bad players.

This algorithm will work better if we'll complete the original group with 3 (3+9=12) virtual (empty) players (with zero weight). So each captain may take one of the empty player instead of negative ones.

Suppose that we need to devide 9 people into two volleyball teams (6+6). Usually we start with two captains (best players). First captain choose one player. After that second captain and first captain choose two players on each step. The problem is in bad players which may have negative weight in the team. And in the end of this procedure captains obliged to choose bad players.

This algorithm will work better if we'll complete the original group with 3 (3+9=12) virtual (empty) players (with zero weight). So each captain may take one of the empty player instead of negative ones.

Suppose that we need to divide 9 people into two volleyball teams. Usually we start with two captains (best players). First captain choose one player. After that second captain and first captain choose two players on each step. The problem is in bad players which may have negative weight in the team. And in the end of this procedure captains obliged to choose bad players.

This algorithm will work better if we'll complete the original group with 3 (3+9=12) virtual (empty) players (with zero weight). So each captain may take one of the empty player instead of negative ones.

Source Link
Alexey Ustinov
  • 12.3k
  • 7
  • 87
  • 119

Suppose that we need to devide 9 people into two volleyball teams (6+6). Usually we start with two captains (best players). First captain choose one player. After that second captain and first captain choose two players on each step. The problem is in bad players which may have negative weight in the team. And in the end of this procedure captains obliged to choose bad players.

This algorithm will work better if we'll complete the original group with 3 (3+9=12) virtual (empty) players (with zero weight). So each captain may take one of the empty player instead of negative ones.

Post Made Community Wiki by Alexey Ustinov