The order ideal $B_{n,2}$ of the Boolean lattice $B_n$ consists of all subsets of size at most $2$. Let $F_{n,2}$ denote the poset refinement of $B_{n,2}$ induced by the rules: $i < j$ implies $\{i \} \prec \{ j \}$ and $\{i,k \} \prec \{j,k\}$. We give an elementary bijection from the set $\mathcal{F}_{n,2}$ of linear extensions of $F_{n,2}$ to the set of shifted standard Young tableau of shape $(n, n-1, \ldots, 1)$, which are counted by the strict-sense ballot numbers. We find a more surprising result when considering the set $\mathcal{F}_{n,2}^{1}$ of minimal poset refinements in which each singleton is comparable with all of the doubletons. We show that $\mathcal{F}_{n,2}^{1}$ is in bijection with magog triangles, and therefore is equinumerous with alternating sign matrices. We adopt our proof techniques to show that row reversal of an alternating sign matrix corresponds to a natural involution on gog triangles.
Let B n,2 denote the order ideal of the boolean lattice B n consisting of all subsets of size at most 2. Let F n,2 denote the poset extension of B n,2 induced by the rule: i < j implies {i} ≺ {j} and {i, k} ≺ {j, k}. We give an elementary bijection from the set F n,2 of linear extensions of F n,2 to the set of shifted standard Young tableau of shape (n, n − 1, . . . , 1), which are counted by the strict-sense ballot numbers. We find a more surprising result when considering the set F(1) n,2 of poset extensions so that each singleton is comparable with all of the doubletons. We show that F(1) n,2 is in bijection with magog triangles, and therefore is equinumerous with alternating sign matrices. We adopt our proof techniques to show that row reversal of an alternating sign matrix corresponds to a natural involution on gog triangles.
When making simultaneous decisions, our preference for the outcomes on one subset can depend on the outcomes on a disjoint subset. In referendum elections, this gives rise to the separability problem, where a voter must predict the outcome of one proposal when casting their vote on another. A set S ⊂ [n] is separable for preference order when our ranking of outcomes on S is independent of outcomes on its complement [n] − S. The admissibility problem asks which characters C ⊂ P([n]) can arise as the collection of separable subsets for some preference order. We introduce a linear algebraic technique to construct preference orders with desired characters. Each vector in our 2 n -dimensional voter basis induces a simple preference ordering with nice separability properties. Given any collection C ⊂ P([n])whose subset lattice has a tree structure, we use the voter basis to construct a preference order with character C.
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