$\require{mathtools}
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% .. one-part
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%
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Adapted from a talk by Prasanna Ramakrishnan at the teach-us-anything seminar on 2023-05-30.
Suppose $n$ people are sharing a cake (represented by the interval $[0,1]$), and they each value some subparts of the cake differently. How do they coordinate so that no one envies the share that someone else got?
Note: There are other natural notions of fairness like proportional cake-cutting, which has an easy solution. Also, the situation is easier if everyone values the cake the same way.
Setup
We’ll consider shares $S \sse [0,1]$ of the cake that are unions of intervals. Each player $i \in [n]$ attributes value $v_i(S)\ge 0$ to share $S$. Let’s assume this valuation is additive: if $S \inter T = \emptyset$, then $v(S \union T) = v(S)+v(T)$. Also, without loss of generality, let’s assume that $v_i([0,1]) = 1$.
$\newcommand{\eval}{\mathrm{eval}}\newcommand{\split}{\mathrm{split}}$
Formally, a cake-cutting protocol will learn about the players’ valuations using only two types of queries:
- $\eval(a,b)$: returns $v_i([a,b])$;
- $\split(a,v)$: returns $b$ such that $v_i([a,b])=v$ (only works if $v_i([a,1]) \ge v$).
Note that the existence of $\split(\cdot,\cdot)$ implies that the valuation can be split up arbitrarily finely: there is no “point mass”.
Protocols
$n = 2$
Here’s a protocol for two players Alice and Bob:
- Alice cuts the cake into halves $[0,a]$ and $[a,1]$ that are equal according to her valuation;
- Bob picks whichever half he prefers.
Clearly,
- Alice will not envy Bob because both shares are equally attractive to her;
- Bob will not envy Alice because he got the half he prefers.
$n = 3$
Attempt 1: naive
Here’s a possible protocol:
- Alice cuts the cake into equal thirds $[0,a]$, $[a,b]$ and $[b,1]$ according to her valuation;
- Bob picks whichever third he prefers;
- Charlie picks whichever remaining third he prefers.
This goes perfectly except for the fact that Charlie might envy Bob’s piece. For example, it could be that both Bob and Charlie strongly prefer $[b,1]$ over the other two pieces.
Attempt 2: discard a part
Suppose that Bob and Charlie both do prefer $[b,1]$ and assume without loss of generality that Charlie’s 2nd choice is $[a,b]$. Then if we’re fine with discarding part of the cake, we could let Charlie throw away some part $[b,1]$, say $[c,1]$ so that the remaining piece $[a,b]$ and $[b,c]$ have the same value to him.
Then,
- Bob would choose whichever of $[1,a]$, $[a,b]$ and $[b,c]$ he prefers;
- Charlie picks $[b,c]$ unless Bob picked that one, in which case he picks $[a,b]$;
- Alice picks the piece that remains (either $[1,a]$ or $[a,b]$).
Nobody will envy anyone:
- Alice is happy because she got either $[1,a]$ or $[a,b]$, both of which have value $1/3$ for her (and all pieces have value $\le 1/3$ for her).
- Bob is happy because he got to pick the piece he preferred.
- Charlie is happy because he liked $[a,b]$ and $[b,c]$ the most (and equally), and he got to pick one of those two.
But we discarded a piece…
Attempt 3: split the discarded part
Now, if we could somehow share this discarded piece $[c,1]$ in an envy-free way… Naively this is just an instance of the same problem. But we have made some progress: however this remaining piece gets split, Alice will definitely not envy the person who picked $[b,c]$. This is because even if that person got to take all of $[c,1]$, that would still only add up to $1/3$ of Alice’s valuation, and Alice is getting $1/3$ already.
In other words, when splitting $[c,1]$, there is a pair $i,j$ of players for which we’re fine if $i$ envies $j$ (locally). As it turns out, attempt 1 gave a protocol which satisfies this guarantee: the only possible problem was that Charlie might envy Bob’s piece. So (with the appropriate permutation of roles) we can just use attempt 1 to split $[c,1]$ appropriately.
Lower bound
Let’s show that any protocol must perform $\Omega(n^2)$ queries in order to be correct. For this, we will show a scenario in which each player must either be assigned a very large piece or asked many queries.
Consider some player $i \in [n]$. Say that we’ve made $m$ $\eval(\cdot,\cdot)$ and $\split(\cdot,\cdot)$ queries, and they involved the intervals $[a_1, b_1],\ldots,[a_m,b_m]$, which means that all that the protocol has learned is $i$’s valuations on the $2m+1$ intervals delimited by the values
\[
A \ce \set{0,1,a_1, \ldots, a_m, b_1, \ldots, b_m}.
\]
Let’s say that so far, for each pair $s,t \in A$, $s<t$, we have $v_i([s,t]) = t-s$. That is, so far, player $i$ has answered every query as if its valuation function were completely uniform.
Claim. If the share assigned to $i$ has total length $\eps$, then the protocol must have made $\Omega(1/\eps)$ queries to it.
Proof of claim
Now, suppose that the share assigned to $i$ has total length $\eps$. We’ll show that each one of the $2m+1$ intervals delimited by $A$ must have length $\le \eps$. This means that $m \ge \Omega(1/\eps)$.
Suppose this is not the case: that is, there is some interval $I$ delimited by $A$ which has length $>\eps$. Then some positive part of $I$ must have been assigned to someone else (say player $j$), and it’s perfectly possible that:
- All the mass of the valuation of $I$ was in $j$’s share, giving that share $\ge v_i(I)>\eps$ valuation.
- But the mass of the other intervals was uniform under $v_i$, giving $i$’s share $\le \eps$ valuation..
But then player $i$ would envy $j$, which violates correctness.
So every interval delimited by $A$ has length $\le \eps$,
Claim implies lower bound
For each player $i$, let $\eps_i$ be the length of its assigned share. We have $\eps_1+\cdots+\eps_n = 1$, and by the claim, the protocol must make
\[
\Omega\p{\frac1{\eps_1} + \cdots + \frac1{\eps_n}}
\]
queries.
By Markov’s inequality, at most $n/2$ of the players must have $\eps_i \ge 2/n$, so we need
\[
\Omega\p{\frac1{\eps_1} + \cdots + \frac1{\eps_n}} \ge\Omega\p{\frac{n}{2} \times \frac1{2/n}} = \Omega\p{n^2}
\]
queries.