Sharding Over Population

Given a set of individuals, what are the largest set of beliefs that are normal to that group?
  1. Start with an empty set of beliefs.
  2. Sort the set of beliefs by how close they are to having complete agreement among the individuals.
  3. Include beliefs (flipping them if necessary) until the metric passes its inflection point.
Consider the following matrix (from above →):
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11
P_1 1 0 1 0 1 0 1 0 1 1 0
P_2 1 0 1 1 0 0 1 1 1 0 0
P_3 1 0 1 1 0 0 0 0 1 1 1
P_4 0 0 1 0 1 0 1 1 1 1 0
P_5 0 0 1 0 0 1 0 1 1 1 1
Now sort our columns, flipping them if necessary:
!Q2 Q3 Q9 !Q6 Q10 Q1 !Q4 !Q5 Q7 Q8 !Q11
P_1 1 1 1 1 1 1 1 0 1 0 1
P_2 1 1 1 1 0 1 0 1 1 1 1
P_3 1 1 1 1 1 1 0 1 0 0 0
P_4 1 1 1 1 1 0 1 0 1 1 1
P_5 1 1 1 0 1 0 1 1 0 1 0
For the metric specified above, let us construct a table of what our answer could be [1], and what the score is.
[1]Beliefs with equivalent agreement over the population can be taken in any order. This means that there may be many answers to the question, which take the form "these N beliefs, and M<P of these P beliefs".
Set Score
choose1(!Q2, Q3, Q9) 5
choose2(!Q2, Q3, Q9) 10
(!Q2, Q3, Q9) 15
(!Q2, Q3, Q9)+choose1(!Q6, Q10) 16
(!Q2, Q3, Q9, !Q6, Q10) 16
(!Q2, Q3, Q9, !Q6, Q10)+choose1(Q1, !Q4, !Q5, Q7, Q8, !Q11) 11.52
(!Q2, Q3, Q9, !Q6, Q10)+choose2(Q1, !Q4, !Q5, Q7, Q8, !Q11) 8.06
(!Q2, Q3, Q9, !Q6, Q10)+choose3(Q1, !Q4, !Q5, Q7, Q8, !Q11) 5.53
(!Q2, Q3, Q9, !Q6, Q10)+choose4(Q1, !Q4, !Q5, Q7, Q8, !Q11) 3.73
(!Q2, Q3, Q9, !Q6, Q10)+choose5(Q1, !Q4, !Q5, Q7, Q8, !Q11) 2.49
(!Q2, Q3, Q9, !Q6, Q10, Q1, !Q4, !Q5, Q7, Q8, !Q11) 1.64
Our selection from this table is right before the point of diminishing returns, where we choose the top five beliefs: !Q2, Q3, Q9, !Q6, and Q10.