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Stromquist–Woodall theorem

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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 2 n 2 {\displaystyle 2n-2} cuts.

The theorem is about a circular 1-dimensional cake (a "pie"). Formally, it can be described as the interval in which the two endpoints are identified. There are n continuous measures over the cake: V 1 , , V n {\displaystyle 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 [ 0 , 1 ] {\displaystyle w\in } , there is a subset C w {\displaystyle C_{w}} , which all people value at exactly w {\displaystyle w} :

i = 1 , , n : V i ( C w ) = w {\displaystyle \forall i=1,\ldots ,n:\,\,\,\,\,V_{i}(C_{w})=w} ,

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

Proof sketch

Let W [ 0 , 1 ] {\displaystyle W\subseteq } be the subset of all weights for which the theorem is true. Then:

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

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

Proof sketch for part 4

  • Assume that C w {\displaystyle C_{w}} is a union of n 1 {\displaystyle n-1} intervals and that all n {\displaystyle n} partners value it as exactly w {\displaystyle w} .
  • Define the following function on the cake, f : C R n {\displaystyle f:C\to \mathbb {R} ^{n}} :
f ( t ) = ( t , t 2 , , t n ) t [ 0 , 1 ] {\displaystyle f(t)=(t,t^{2},\ldots ,t^{n})\,\,\,\,\,\,t\in }
  • Define the following measures on R n {\displaystyle \mathbb {R} ^{n}} :
U i ( Y ) = V i ( f 1 ( Y ) C w ) Y R n {\displaystyle U_{i}(Y)=V_{i}(f^{-1}(Y)\cap C_{w})\,\,\,\,\,\,\,\,\,Y\subseteq \mathbb {R} ^{n}}
  • Note that f 1 ( R n ) = C {\displaystyle f^{-1}(\mathbb {R} ^{n})=C} . Hence, for every partner i {\displaystyle i} : U i ( R n ) = w {\displaystyle U_{i}(\mathbb {R} ^{n})=w} .
  • Hence, by the Stone–Tukey theorem, there is a hyper-plane that cuts R n {\displaystyle \mathbb {R} ^{n}} to two half-spaces, H , H {\displaystyle H,H'} , such that:
i = 1 , , n : U i ( H ) = U i ( H ) = w / 2 {\displaystyle \forall i=1,\ldots ,n:\,\,\,\,\,U_{i}(H)=U_{i}(H')=w/2}
  • Define M = f 1 ( H ) C w {\displaystyle M=f^{-1}(H)\cap C_{w}} and M = f 1 ( H ) C w {\displaystyle M'=f^{-1}(H')\cap C_{w}} . Then, by the definition of the U i {\displaystyle U_{i}} :
i = 1 , , n : V i ( M ) = V i ( M ) = w / 2 {\displaystyle \forall i=1,\ldots ,n:\,\,\,\,\,V_{i}(M)=V_{i}(M')=w/2}
  • The set C w {\displaystyle C_{w}} has n 1 {\displaystyle n-1} connected components (intervals). Hence, its image f ( C w ) {\displaystyle f(C_{w})} also has n 1 {\displaystyle n-1} connected components (1-dimensional curves in R n {\displaystyle \mathbb {R} ^{n}} ).
  • The hyperplane that forms the boundary between H {\displaystyle H} and H {\displaystyle H'} intersects f ( C w ) {\displaystyle f(C_{w})} in at most n {\displaystyle n} points. Hence, the total number of connected components (curves) in H f ( C w ) {\displaystyle H\cap f(C_{w})} and H f ( C w ) {\displaystyle H'\cap f(C_{w})} is 2 n 1 {\displaystyle 2n-1} . Hence, one of these must have at most n 1 {\displaystyle n-1} components.
  • Suppose it is H {\displaystyle H} that has at most n 1 {\displaystyle n-1} components (curves). Hence, M {\displaystyle M} has at most n 1 {\displaystyle n-1} components (intervals).
  • Hence, we can take C w / 2 = M {\displaystyle C_{w/2}=M} . This proves that w W {\displaystyle w\in W} .

Tightness proof

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

Proof sketch for w = 1 / n {\displaystyle w=1/n}

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

See also

References

  1. Stromquist, Walter; Woodall, D.R (1985). "Sets on which several measures agree". Journal of Mathematical Analysis and Applications. 108: 241–248. doi:10.1016/0022-247x(85)90021-6.
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