Sharding Over Beliefs


Given a set of beliefs, what group of individuals make up its normal?
  1. Negate beliefs that are more negative than positive in the population of individuals.
  2. Start with an empty set of individuals.
  3. Sort the individuals so that they are ordered from greatest agreement with the beliefs to least agreement with the beliefs.
  4. Collect individuals into the set so long as the individual's addition increases that set's normalization metric.
Consider the following matrix (transposed from above →), flipping beliefs so that they are mostly affirmative:
P_1 P_2 P_3 P_4 P_5
Q1 1 1 1 0 0
!Q2 1 1 1 1 1
Q3 1 1 1 1 1
!Q4 1 0 0 1 1
!Q5 0 1 1 0 1
!Q6 0 1 1 1 1
Q7 0 1 0 1 1
Q8 1 1 0 1 0
Q9 1 1 1 1 1
Q10 1 1 1 0 1
!Q11 0 1 0 1 1
Now sort the individuals by their overall agreement:
P_2 P_5 P_4 P_1 P_3
Q1 1 0 0 1 1
!Q2 1 1 1 1 1
Q3 1 1 1 1 1
!Q4 0 1 1 1 0
!Q5 1 1 0 0 1
!Q6 1 1 1 0 1
Q7 1 1 1 0 0
Q8 1 0 1 1 0
Q9 1 1 1 1 1
Q10 1 1 0 1 1
!Q11 1 1 1 0 0
For the metric specified above, let us a table of what our answer could be [1], and what their normalization scores are.
[1]Individuals with equivalent agreement over the beliefs can be taken in any order. This means that there may be many answers to the question, which take the form "these N individuals, and M<P of these P individuals".
Set Score
(P_2) 10
(P_2, P_5) 16.36
(P_2, P_5, P_4) 17.85
(P_2, P_5, P_4)+choose1(P_1, P_3) 15.14
(P_2, P_5, P_4, P_1, P_3) 12.05
It follows from this that our group contains the individuals P_2, P_5, and P_4.