Prove that the Shapley ALWAYS lies in core

[Dan2k03]

As the title says. We should have this ready for next presentation.

[Dan2k03]

I found some slides looking into it. We should consider reading these in order to figure it out:

https://www.cs.ubc.ca/~kevinlb/teaching/cs532l%20-%202007-8/lectures/lect23.pdf

[GeorgeR227]

In order to prove this, we need to finalize generalizing our starting budgets and the economy so we know what our payoff vectors look like. Also if we can prove this is a convex game then this is trivial, issue #8.

[GeorgeR227]

I've been thinking about this and come to this conclusion:

• If we can prove that the Shapley is always in the core, then great.
• If it's not always in the core, then it can get kinda interesting.

Since our "unfairest" points are simply those furthest away from the Shapley point, this still functions even if Shapley itself is not in the core. Take our second example in demos/square.jl where the origin is not actually a member of the line segment.

Actually, we can make this interesting by generalizing the idea of the "fairest" point. Essentially, if the "unfairest" point is that point which deviates the most from our expectations (namely from the Shapley point), then the fairest point is that point which is closest to our expectations.

In the case where Shapely is actually part of the core itself, then the closest point is just itself and all is well. If it's not in the core, then we can look at another feasible "fairest" point and analyze that. This search should be doable by using Quadratic Programming so it's actually easier than finding the furthest point.