Austin moving-knife procedures

The basic procedures involve ${\displaystyle n=2}$ partners who want to divide a cake such that each of them gets exactly one half.

As noted by Austin, the two partners can find a single piece of cake that both of them value as exactly ${\displaystyle 1/k}$, for any integer ${\displaystyle k\geq 2}$.[2] Call the above procedure ${\displaystyle \mathrm {Cut} _{2}(1/k)}$:

• Alice makes ${\displaystyle k-1}$ parallel marks on the cake such that ${\displaystyle k}$ pieces so determined have a value of exactly ${\displaystyle 1/k}$.
• If there is a piece that George also values as ${\displaystyle 1/k}$, then we are done.
• Otherwise, there must be a piece that George values as less than ${\displaystyle 1/k}$, and an adjacent piece that George values as more than ${\displaystyle 1/k}$.
• Let Alice place two knives on the two marks of one of these pieces, and move them in parallel, keeping the value between them at exactly ${\displaystyle 1/k}$, until they meet the marks of the other piece. By the IVT, there must be a point at which George agrees that the value between the knives is exactly ${\displaystyle 1/k}$.

By recursively applying ${\displaystyle \mathrm {Cut} _{2}}$, the two partners can divide the entire cake to ${\displaystyle k}$ pieces, each of which is worth exactly ${\displaystyle 1/k}$ for both of them:[2]

• Use ${\displaystyle \mathrm {Cut} _{2}(1/k)}$ to cut a piece which is worth exactly ${\displaystyle 1/k}$ for both partners.
• Now the remaining cake is worth exactly ${\displaystyle (k-1)/k}$ for both partners; use ${\displaystyle \mathrm {Cut} _{2}(1/(k-1))}$ to cut another piece worth exactly ${\displaystyle 1/k}$ for both partners.
• Continue like this until there are ${\displaystyle k}$ pieces.

Two partners can achieve an exact division with any rational ratio of entitlements by a slightly more complicated procedure.[3]^: 71

By combining ${\displaystyle \mathrm {Cut} _{2}}$ with the Fink protocol, it is possible to divide a cake to ${\displaystyle n}$ partners, such that each partner receives a piece worth exactly ${\displaystyle 1/n}$ for him:[1][4]

• Partners #1 and #2 use ${\displaystyle \mathrm {Cut} _{2}(1/2)}$ to give each one of them a piece worth exactly 1/2 for them.
• Partner #3 uses ${\displaystyle \mathrm {Cut} _{2}(1/3)}$ with partner #1 to get exactly 1/3 of his share and then ${\displaystyle \mathrm {Cut} _{2}(1/3)}$ with partner #2 to get exactly 1/3 of her share. The first piece is worth exactly 1/6 for partner #1 and so partner #1 remains with exactly 1/3; the same is true for partner #2. As for partner #3, while each piece can be more or less than 1/6, the sum of the two pieces must be exactly 1/3 of the entire cake.

Note that for ${\displaystyle n>2}$, the generated division is not exact, since a piece is worth ${\displaystyle 1/n}$ only to its owner and not necessarily to the other partners. As of 2015, there is no known exact division procedure for ${\displaystyle n>2}$ partners; only near-exact division procedures are known.

• Austin's procedure is used by the Brams–Taylor–Zwicker procedure.
• Other procedures and results about equitable division and exact division.

• Fischer, Daniel. "Consensus division of a cake to two people in arbitrary ratios". Math.SE. Retrieved 23 June 2015.