It is a division of resources among n partners, in which the value received by each partner is strictly more than his/her due share of 1/n of the total value.
Formally, in a strongly proportional division of a resource C among n partners, each partner i, with value measure Vi, receives a share Xi such that
.Obviously, a strongly proportional division does not exist when all partners have the same value measure.
, which is the condition for a plain proportional division.
However, one may hope that, when different agents have different valuations, it may be possible to use this fact for the benefit of all players, and give each of them strictly more than their due share.
In 1948, Hugo Steinhaus conjectured the existence of a super-proportional division of a cake:[2]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.In 1961, Dubins and Spanier proved that the necessary condition for existence is also sufficient.
That is, whenever the partners' valuations are additive and non-atomic, and 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.
The proof was a corollary to the Dubins–Spanier convexity theorem.
This was a purely existential proof based on convexity arguments.
In 1986, Douglas R. Woodall published the first protocol for finding a super-proportional division.
If the agents' valuations are different, then there must be a witness for that: a witness is a specific piece of cake, say X ⊆ C, which is valued differently by some two partners, say Alice and Bob.
that: bx > ax, which implies: by < ay.The idea is to partition X and Y separately: when partitioning X, we will give slightly more to Bob and slightly less to Alice; when partitioning Y, we will give slightly more to Alice and slightly less to Bob.
Find a rational number between bx and ax, say p/q such that bx > p/q > ax.
Ask Bob to divide X into p equal parts, and divide Y to q-p equal parts.
So now we have two pieces, X0 and Y0, such that: Let Alice and Bob divide the remainder C \ X0 \ Y0 between them in a proportional manner (e.g. using divide and choose).
The extension of this protocol to n partners is based on Fink's "Lone Chooser" protocol.
Suppose we already have a strongly 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 strongly proportional.
Let d be the difference between partner #1's current value and (1/(i-1)).
Because the current division is strongly proportional, we know that d>0.
pieces which he considers of equal value and let the new partner choose the
This proves that the new division is strongly proportional too.
Julius Barbanel[1] extended Woodall's algorithm to agents with different entitlements, including irrational entitlements.
In this setting, the entitlement of each agent i is represented by a weight
A strongly proportional allocation is one in which, for each agent i:
.Janko and Joo[4] presented a simpler algorithm for agents with different entitlements.
In fact, they showed how to reduce a problem of strongly proportional division (with equal or different entitlements) into two problems of proportional division with different entitlements:
An allocation is called strongly envy-free if for every two partners i,j:
.An allocation is called super envy-free if for every two partners i,j:
.Super envy-freeness implies strong envy-freeness, which implies strong proportionality.