Fair division and generalizations of Sperner- and KKM-type results

Achieving Rental Harmony with a Secretive Roommate

“…At every vertex of the triangulation transform the label (a, b) into a single-digit label according to the following rules: label the vertex 3 if the previous label was (1, 1), (1, 2), or (2, 1), label 1 for (2, 2), (2, 3), or (3, 2), and 2 for (3, 3), (3,1), or (1, 3). The three original vertices of the triangle receive pairwise distinct labels since Larry and Moe decided for different rooms whenever two rooms were free simultaneously.…”

“…We claim that (1) τ exhibits all three labels, and (2) both λ 1 and λ 2 exhibit at least two labels. To see claim (1), note that if τ exhibited only two of the labels, then one of the coordinates of λ(x) would be zero. To see claim (2) assume for contradiction that either λ 1 or λ 2 only exhibits one of the labels.…”

“…Then the first coordinate of λ(v) will be greater than or equal to 1 2 for each vertex v of τ . And so the first coordinate of λ(y) for any y ∈ τ must be at least 1 2 , which contradicts λ(x) = ( 13 , 1 3 , 1 3 ). After the secret roommate selects their room, two remain.…”

“…The map λ fixes the vertices and edges of ∆ 2 setwise. We will construct a path that starts in the vertex e 1 of ∆ 2 and ends in a triangle σ of T such that λ(σ) contains the barycenter 1 3 (e 1 + e 2 + e 3 ) of ∆ 2 . We will now describe the rooms and doors for our trap-door argument, or equivalently the vertices and edges of a graph G such that following paths in this graph will lead to a triangle mapped to the barycenter by λ.…”

“…• any edge e of T that subdivides the edge [e 1 , e 2 ] of ∆ 2 and such that λ(e) ∩ [e 1 , 1 2 (e 1 + e 2 )] = ∅; that is, the image of e under λ intersects the segment from vertex e 1 to the barycenter of [e 1 , e 2 ],…”