A group of n number of nerds once wanted to decide who is more intelligent than others. They took help from the pillar of democracy that is the voting system. They thought of calculating a value for everyone and decided to vote each other in terms a value. Each one assigns unit value to himself and thoughtfully gives a value less or more than 1 to everyone. (If A thinks B is better and C is worse than himself, he can give say, 2 to B and 0.5 to C) Here is the interesting part, each value must be weighted by the value of the person giving vote (which is unknown), averaged out and then taken square root of, which will give the value of individual receiving votes. (because the vote of a smart one should have more weight than that of a dumb ) So, there we see a matrix and equations to solve.
I took an example of n=3. The off-dia elements in above matrix are the vote-values, that I chose randomly around 1. The diagonal elements are individual values (x,y,z) Here we have 3 equations: Value of a: sqrt((1*x+1*y+2*z )/3)=x Value of b: sqrt((4*x+1*y+1*z )/3)=y Value of c: sqrt((0.5*x+1.5*y+1*z )/3)=z
These look like set of linear equations but they are having quadratic terms too! I used MS-excel to solve (x,y,z)= (1.343097333, 1.656814439, 1.205913463)
Generalizing for any n, is it possible to start with any n*(n-1) values (votes) and solve for n values (individual values) ? Will it be unique? (Assume values to be strictly positive and I think those will be centered about 1 since voting is relative to 1)
What will happen if self vote equals value of the self instead of 1? (This is unrealistic and that will just change the diagonal term in each equation)
Leave aside the nerds, can this be a social experiment!?