Stromquist–Woodall theorem

The Stromquist–Woodall theorem is a theorem in fair division and measure theory. Informally, it says that, for any cake, for any n people with different tastes, and for any fraction w, there exists a subset of the cake that all people value at exactly a fraction w of the total cake value, and it can be cut using at most $$2n-2$$ cuts.

The theorem is about a circular 1-dimensional cake (a "pie"). Formally, it can be described as the interval [0,1] in which the two endpoints are identified. There are n continuous measures over the cake: $$V_1,\ldots,V_n$$; each measure represents the valuations of a different person over subsets of the cake. The theorem says that, for every weight $$w \in [0,1]$$, there is a subset $$C_w$$, which all people value at exactly $$w$$:


 * $$\forall i = 1,\ldots,n: \,\,\,\,\, V_i(C_w)=w$$,

where $$C_w$$ is a union of at most $$n-1$$ intervals. This means that $$2n-2$$ cuts are sufficient for cutting the subset $$C_w$$. If the cake is not circular (that is, the endpoints are not identified), then $$C_w$$ may be the union of up to $$n$$ intervals, in case one interval is adjacent to 0 and one other interval is adjacent to 1.

Proof sketch
Let $$W \subseteq [0,1]$$ be the subset of all weights for which the theorem is true. Then:


 * 1) $$1 \in W$$. Proof: take $$C_1 := C$$ (recall that the value measures are normalized such that all partners value the entire cake as 1).
 * 2) If $$w\in W$$, then also $$1-w \in W$$. Proof: take $$C_{1-w} := C\smallsetminus C_w$$. If $$C_w$$ is a union of  $$n-1$$ intervals in a circle, then $$C_{1-w}$$ is also a union of  $$n-1$$ intervals.
 * 3) $$W$$ is a closed set. This is easy to prove, since the space of unions of $$n-1$$ intervals is a compact set under a suitable topology.
 * 4) If $$w\in W$$, then also $$w/2 \in W$$. This is the most interesting part of the proof; see below.

From 1-4, it follows that $$W=[0,1]$$. In other words, the theorem is valid for every possible weight.

Proof sketch for part 4

 * Assume that $$C_w$$ is a union of $$n-1$$ intervals and that all $$n$$ partners value it as exactly $$w$$.
 * Define the following function on the cake, $$f: C \to \mathbb{R}^n$$:


 * $$f(t) = (t, t^2, \ldots, t^n)\,\,\,\,\,\,t\in[0,1]$$


 * Define the following measures on $$\mathbb{R}^n$$:


 * $$U_i(Y) = V_i(f^{-1}(Y) \cap C_w)\,\,\,\,\,\,\,\,\, Y\subseteq \mathbb{R}^n$$


 * Note that $$f^{-1}(\mathbb{R}^n) = C$$. Hence, for every partner $$i$$: $$U_i(\mathbb{R}^n) = w$$.
 * Hence, by the Stone–Tukey theorem, there is a hyper-plane that cuts $$\mathbb{R}^n$$ to two half-spaces, $$H, H'$$, such that:


 * $$\forall i = 1,\ldots,n: \,\,\,\,\, U_i(H)=U_i(H')=w/2$$


 * Define $$M=f^{-1}(H)\cap C_w$$ and $$M'=f^{-1}(H')\cap C_w$$. Then, by the definition of the $$U_i$$:


 * $$\forall i = 1,\ldots,n: \,\,\,\,\, V_i(M)=V_i(M')=w/2$$


 * The set $$C_w$$ has $$n-1$$ connected components (intervals). Hence, its image $$f(C_w)$$ also has $$n-1$$ connected components (1-dimensional curves in $$\mathbb{R}^n$$).
 * The hyperplane that forms the boundary between $$H$$ and $$H'$$ intersects $$f(C_w)$$ in at most $$n$$ points. Hence, the total number of connected components (curves) in $$H\cap f(C_w)$$ and $$H'\cap f(C_w)$$ is $$2n-1$$. Hence, one of these must have at most $$n-1$$ components.
 * Suppose it is $$H$$ that has at most $$n-1$$ components (curves). Hence, $$M$$ has at most $$n-1 $$ components (intervals).
 * Hence, we can take $$C_{w/2}=M$$. This proves that $$w\in W$$.

Tightness proof
Stromquist and Woodall prove that the number $$n-1$$ is tight if the weight $$w$$ is either irrational, or rational with a reduced fraction $$r/s$$ such that $$s\geq n$$.

Proof sketch for $$w=1/n$$

 * Choose $$(n-1)(n+1)$$ equally-spaced points along the circle; call them $$P_1,\ldots,P_{(n-1)(n+1)}$$.
 * Define $$n-1$$ measures in the following way. Measure $$i$$ is concentrated in small neighbourhoods of the following $$(n+1)$$ points: $$P_{i},P_{i+(n-1)},\ldots,P_{i+n(n-1)}$$. So, near each point $$P_{i+k(n-1)}$$, there is a fraction $$1/(n+1)$$ of the measure $$u_i$$.
 * Define the $$n$$-th measure as proportional to the length measure.
 * Every subset whose consensus value is $$1/n$$, must touch at least two points for each of the first $$n-1$$ measures (since the value near each single point is $$1/(n+1)$$ which is slightly less than the required $$1/n$$). Hence, it must touch at least $$2(n-1)$$ points.
 * On the other hand, every subset whose consensus value is $$1/n$$, must have total length $$1/n$$ (because of the $$n$$-th measure). The number of "gaps" between the points is $$1/\big((n+1)(n-1)\big)$$; hence the subset can contain at most $$n-1$$ gaps.
 * The consensus subset must touch $$2(n-1)$$ points but contain at most $$n-1$$ gaps; hence it must contain at least $$n-1$$ intervals.