Super proportional division - Alchetron, the free social encyclopedia

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 ( V > 1 / n ).

A super-proportional division is better than a proportional division, in which each partner is guaranteed to receive at least 1/n ( V ≥ 1 / n ). 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:

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.

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=V_Alice(X) and b=V_Bob(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 CX (say, Z) must have a value of more than 1/q.

So now we have two pieces, Y and Z, such that:

V_Bob(Y)>V_Bob(Z) V_Alice(Y)<V_Alice(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: d > 1 ( i − 1 ) i ( q ( i − 1 ) − 1 )

Ask partner #1 to divide his share to q i − 1 pieces which he considers of equal value and let the new partner choose the q pieces which he considers to be the most valuable.

Partner #1 remains with a value of ( q i − 1 ) − q q i − 1 = q ( i − 1 ) − 1 q i − 1 of his previous value, which was 1 i − 1 + d (by definition of d). The first element becomes q ( i − 1 ) − 1 ( i − 1 ) ( q i − 1 ) and the d becomes 1 i ( i − 1 ) ( q i − 1 ) ; summing them up gives that the new value is more than: ( q i − 1 ) ( i − 1 ) ( i − 1 ) i ( q i − 1 ) = 1 i 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: q q i − 1 > 1 i of the entire cake.

This proves that the new division is also super-proportional.

References

Super-proportional division Wikipedia

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