论文标题
双重图表的词典最大边缘和紧密游戏形式的NASH可溶性
Lexicographically maximal edges of dual hypergraphs and Nash-solvability of tight game forms
论文作者
论文摘要
令$ \ MATHCAL {A} = \ {a_1，\ ldots，a_m \} $和$ \ Mathcal {b} = \ {b_1，\ ldots，b_n \} $是一对双重地面的双hypergraphs in Common openther of common of common of common of common of common of common of common of common of common of ocy o o o o o o o = \ __1，请注意，它们每个都可以嵌入或相等的边缘。如果不是另一个边缘的严格的超集，则称为coartiment最小值（或最小值）。然而，可能存在相等的最小边缘。通过二元性， （i）$ a \ cap b \ neq \ emptyset $ for \ mathcal {a} $ in \ mathcal {b} $ in \ mathcal {a} $ in \ mathcal {a} $中的$ a \; （ii）如果$ a $最小，则对于$ o \ a $中的每一个$ o \，都存在\ nathcal {b} $中的$ b \，这样$ a \ cap b = \ {o \ \} $。 我们将延长索赔（ii），如下所示。 $ o $上的线性订单$ \ succ $ $ $ $ $ $ $ $ \ $ \ succ_l $在$ 2^o $上定义了。令$ a $为$ \ mathcal {a} $的词典最大（Lexmax）边缘。然后， （iii）$ a $是最小的，对于$ o \，每一个$ o \都存在\ Mathcal {b} $中的最低$ b \ b \ in \ Mathcal {b} $，这样$ a \ cap b = \ {o \ \} $和$ o \ o \ succeq o'$ in B $中的每个$ o'\。 该属性在游戏理论中具有重要的应用，这意味着第一位作者的《旧游戏形式》（1975年和1989年）所示。在这里，我们给出了（iii）的新的，非常简短的证明。 （iii）中提到的边缘$ a $ a $和$ b $可以在多项式时间内找到。如果$ \ Mathcal {a} $和$ \ Mathcal {B} $明确给出，这是微不足道的。但是，即使只给出了$ \ natercal {a} $，但不是明确的，而是通过多项式遏制甲骨文，对于子集$ o_a \ subseteq o $ $ $ $ o_a $中的答案，无论$ o_a $是否包含$ \ nathcal {a a} $。
Let $\mathcal{A} = \{A_1, \ldots, A_m\}$ and $\mathcal{B} = \{B_1, \ldots, B_n\}$ be a pair of dual multi-hypergraphs on the common ground set $O = \{o_1, \ldots, o_k\}$. Note that each of them may have embedded or equal edges. An edge is called containment minimal (or just minimal, for short) if it is not a strict superset of another edge. Yet, equal minimal edges may exist. By duality, (i) $A \cap B \neq \emptyset$ for every pair $A \in \mathcal{A}$ and $B \in \mathcal{B}$; (ii) if $A$ is minimal then for every $o \in A$ there exists a $B \in \mathcal{B}$ such that $A \cap B = \{o\}$. We will extend claim (ii) as follows. A linear order $\succ$ over $O$ defines a unique lexicographic order $\succ_L$ over the $2^O$. Let $A$ be a lexicographically maximal (lexmax) edge of $\mathcal{A}$. Then, (iii) $A$ is minimal and for every $o \in A$ there exists a minimal $B \in \mathcal{B}$ such that $A \cap B = \{o\}$ and $o \succeq o'$ for each $o' \in B$. This property has important applications in game theory implying Nash-solvability of tight game forms as shown in the old (1975 and 1989) work of the first author. Here we give a new, very short, proof of (iii). Edges $A$ and $B$ mentioned in (iii) can be found out in polynomial time. This is trivial if $\mathcal{A}$ and $\mathcal{B}$ are given explicitly. Yet, it is true even if only $\mathcal{A}$ is given, and not explicitly, but by a polynomial containment oracle, which for a subset $O_A \subseteq O$ answers in polynomial time whether $O_A$ contains an edge of $\mathcal{A}$.
