Social:Super-proportional division
In the context of fair cake-cutting, a super-proportional division is a division in which each partner receives strictly more than 1/n of the resource by their own subjective valuation ([math]\displaystyle{ V \gt 1/n }[/math]). A super-proportional division is better than a proportional division, in which each partner is guaranteed to receive at least 1/n ([math]\displaystyle{ V \geq 1/n }[/math]). However, in contrast to proportional division, a super-proportional division does not always exist. When all partners have exactly the same value functions, the best we can do is give each partner exactly 1/n.
A necessary condition for the existence of a super-proportional division is, therefore, that not all partners have the same value measure.
A surprising fact is that, when the valuations are additive and non-atomic, this condition is also sufficient. I.e., when there are at least two partners whose value function is even slightly different, then there is a super-proportional division in which all partners receive more than 1/n.
Conjecture
The existence of a super-proportional division was first conjectured as early as 1948:[1]
- It may be stated incidentally that if there are two (or more) partners with different estimations, there exists a division giving to everybody more than his due part (Knaster); this fact disproves the common opinion that differences estimations make fair division difficult.
Existence proof
The first published proof to the existence of super-proportional division was as a corollary to the Dubins–Spanier convexity theorem. This was a purely existential proof based on convexity arguments.
Protocol
A protocol for finding a super-proportional division was presented in 1986.[2]
Single piece of disagreement
Let C be the entire cake. The protocol starts with a specific piece of cake, say X ⊆ C, which is valued differently by two partners. Call these partners Alice and Bob.
Let a=VAlice(X) and b=VBob(X) and assume w.l.o.g. that b>a.
Two pieces of disagreement
Find a rational number between b and a, say p/q such that b > p/q > a. Ask Bob to divide X to p equal parts and divide C \ X to q-p equal parts.
By our assumptions, Bob values each piece of X as more than 1/q and each piece of C \ X as less than 1/q. But for Alice, at least one piece of X (say, Y) must have a value of less than 1/q and at least one piece of C\X (say, Z) must have a value of more than 1/q.
So now we have two pieces, Y and Z, such that:
- VBob(Y)>VBob(Z)
- VAlice(Y)<VAlice(Z)
Super-proportional division for two partners
Let Alice and Bob divide the remainder C \ Y \ Z between them in a proportional manner (e.g. using divide and choose). Add Z to the piece of Alice and add Y to the piece of Bob.
Now each partner thinks that his/her allocation is strictly better than the other allocation, so its value is strictly more than 1/2.
Super-proportional division for n partners
The extension of this protocol to n partners is based on Fink's "Lone Chooser" protocol.
Suppose we already have a super-proportional division to i-1 partners (for i≥3). Now partner #i enters the party and we should give him a small piece from each of the first i-1 partners, such that the new division is still super-proportional.
Consider e.g. partner #1. Let d be the difference between partner #1's current value and (1/(i-1)). Because the current division is super-proportional, we know that d>0.
Choose a positive integer q such that: [math]\displaystyle{ d \gt \frac{1}{(i-1)i(q(i-1)-1)} }[/math]
Ask partner #1 to divide his share to [math]\displaystyle{ qi-1 }[/math] pieces which he considers of equal value and let the new partner choose the [math]\displaystyle{ q }[/math] pieces which he considers to be the most valuable.
Partner #1 remains with a value of [math]\displaystyle{ \frac{(qi-1)-q}{qi-1} = \frac{q(i-1)-1}{qi-1} }[/math] of his previous value, which was [math]\displaystyle{ \frac{1}{i-1} + d }[/math] (by definition of d). The first element becomes [math]\displaystyle{ \frac{q(i-1)-1}{(i-1)(qi-1)} }[/math] and the d becomes [math]\displaystyle{ \frac{1}{i(i-1)(qi-1)} }[/math]; summing them up gives that the new value is more than: [math]\displaystyle{ \frac{(qi-1)(i-1)}{(i-1)i(qi-1)} = \frac{1}{i} }[/math] of the entire cake.
As for the new partner, after having taken q pieces from each of the first i-1 partners, his total value is at least: [math]\displaystyle{ \frac{q}{qi-1} \gt \frac{1}{i} }[/math] of the entire cake.
This proves that the new division is also super-proportional.
References
- ↑ Steinhaus, Hugo (1948). "The problem of fair division". Econometrica 16 (1): 101–4.
- ↑ Woodall, D. R. (1986). "A note on the cake-division problem". Journal of Combinatorial Theory, Series A 42 (2): 300. doi:10.1016/0097-3165(86)90101-9.
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