一类多步矩阵对策上的计策问题

以有序树为工具,研究了可以描述连环计,诱敌深入等多步矩阵对策上的一类计策模型.在不考虑信息环境的封闭对策系统中,及局中人对每一步矩阵对策的赢得矩阵,两个局中人的策略集合以及局中人的理性等的了解都是局中人的共同知识的假定下,提出了局中人的最优计策链及将计就计等概念,研究了局中人中计和识破计策的固有概率,讨论了局中人在什么情况下最好主动用计,在什么情况下最好从动用计以及求解最优计策等问题.     

连续对策上的计策问题

限定一个连续对策不是平凡地无意义（例如对某个局中人绝对有利等），我们提出了连续对策上的计策的基本概念。最后得到结论，如果局中人1使用经典对策，那么他的赢得期望必不是赢得函数的最大值。如果局中人1使用计策成功（即使得局中人2中计），那么局中人1必取得赢得函数的最大值，局中人2也有对偶的结果。     

矩阵对策的两个注记

设(x*,y*)是以A=[aij]m×n为赢得矩阵G的对策解,则当局中人1,2各自独立地使用其最优策略x*=(x*1,x*2,…,xmn),y*=(y*1,y*2,…,y*n)时,局中人1的赢得期望为对策值v*=x*Ay*T.若局中人双方使用使得方差D(x*,y*)=∑∑(aij-v*)2x*iy*j达最小的对策解(x*,y*),则其赢得靠近v*的概率达到最大.以O记使方差达到最小的对策解的集合.若O满足(x(1),y(1)),(x(2),y(2))∈O蕴涵(x(1),y(2)),(x(2),y(1))∈O,则说O是可换的.本文首先证明了若矩阵对策G有纯解,则O是可换的.然后证明了如果限定局中人1在其混合扩充策略集的一个非空紧凸子集X中选取策略,那么存在X的一个非空紧子集O(X),它是有限个非空互不相交紧凸集之并,使得只要局中人1使用O(X)中的策略,那么在最坏的情况下可以取得最好的赢得.     

B矩阵对策

提出了一类局中人都设定有得失控制值的二人零和有限对策,即B矩阵对策.引入了稳妥策略、弱稳妥策略等概念.给出了B矩阵对策的数学模型、有关理论和求解方法.最后,还给了一个计算例子.     

矩阵对策的公平性研究

众所周知，零和二人有限对策也称为矩阵对策。设做一个矩阵对策的两个局中人都希望对策结果尽可能公平。当两个局中人使用对策解中的策略进行对策时，如果对策结果最公平，那么这个对策解称为最优的。本文证明了最优对策解集的一些性质，然后给出矩阵对策公平度的概念并证明了它的一些有趣的性质。     

决策支持下的不完全偏好信息多目标对策

针对多目标对策的不完全偏好信息,将理性策略概念和偏好规划理论引入到多目标对策研究中,把两人的多目标对策考虑为两个局中人同时面临的两个多目标决策问题,偏好信息用可行权重限定集合进行建模.多目标对策模型分为对策模型和决策模型两部分.在对策模型中,利用理性和偏好的共同知识,获得理性策略集.在决策模型中,局中人将策略选择作为使用私人偏好信息的多目标决策问题.最后,通过实例验算表明所提方法的有效性.     

连续对策之判断下的最优策略集

本文引进连续对策上的判断块、判断准确、判断下的最优策略集等概念，得到了如下几个主要结果：1.判断下的最优策略集是一个局部凸空间的非空有界闭凸集；2.两个判断下的最优策略集相等的充要条件是这两个判断位于同一判断块中；3.若局中人判断准确，则在一次性对策下不论他使用此判断下的那一个最优策略（不论是纯的还是混合的），都可无风险地取得最优赢得。     

多重因素约束下的网格检查对策问题研究

基于物品数量及每列容量等限制因素,构造局中人的可行策略集合;考虑隐藏成本,处罚规则与检查成功概率等因素,构造相应的支付函数,建立多重因素约束下的网格检查对策模型.根据矩阵对策性质,将对策论问题转化为非线性整数规划问题,利用H(o|¨)lder不等式获得实数条件下的规划问题的解,然后转化为整数解,得到特定条件下的模型的对策值及局中人的最优混合策略.最后,给出一个实例,说明上述模型的实用性及方法的有效性.     

具有风险偏好的梯形直觉模糊双矩阵对策模型及解法

在对策问题中,行动方案的选择不可避免的需要对预期支付值(收益值)进行估计和排序,且选择结果往往受到现实局中人风险偏好程度的影响.因此,该文针对局中人具有风险偏好及支付值为梯形直觉模糊的双矩阵对策进行了模型及求解方法的探讨.首先,提出了具有风险偏好的梯形直觉模糊数排序方法,再利用双线性规划求解方法,对梯形直觉模糊双矩阵对策进行求解.最后以企业营销策略选择为例,表明了该方法的有效性和实用性.     

连续对策的最大熵策略密度对策解

对于正方形[0,2]×[0,2]上的连续对策,将局中人的非纯策略(概率分布函数)的导数称为这个局中人的策略密度(概率密度函数).建立了这种连续对策的最大熵理论.主要证明了当每个局中人都没有最优纯策略时,具有最大熵的最优策略密度集合的非空紧凸性,研究了最优策略密度的最大熵,给出一类带有最大熵的连续对策.     

矩阵计策的支撑解系

令[aij]n×n是二人零和对策的支付矩阵.局中人1可用其"计策"得到最大支付a＝max{aij|1≤i≤n,1≤j≤n},然而,一个开放问题是如何找到全体计策解,本文首先引进计策解系的一种特殊类型--支撑解系.然后研究支撑解系的特征、性质、代数结构.最后给出寻找全体基本支撑解系的一个算法.     

Two-person games on graphs

Every vertex of an abstract-directed graph is characterized in terms of a two-person game. A vertex is winning if by choosing it a player can assure himself of a win, it is losing if by choosing it he cannot prevent his opponent from winning, and it is drawing if it is neither winning nor losing. The sets of winning, losing, and drawing vertices are identified in terms of a set-valued function on the graph.     

Single-suit two-person card play

We consider a two-person constant sum perfect information game, which we call theEnd Play Game, which arises from an abstraction of simple end play positions in card games of the whist family, including bridge. This game was described in 1929 by Emanuel Lasker, the mathematician and world chess champion, who called itwhistette. The game uses a deck of cards that consists of a single totally ordered suit of 2n cards. To begin play the deck is divided into two handsA andB ofn cards each, held by players Left and Right, and one player is designated as having thelead. The player on lead chooses one of his cards, and the other player after seeing this card selects one of his own to play. The player with the higher card wins a “trick” and obtains the lead. The cards in the trick are removed from each hand, and play then continues until all cards are exhausted. Each player strives to maximize his trick total, and thevalue of the game to each player is the number of tricks he takes. Despite its simple appearance, this game is quite complicated, and finding an optimal strategy seems difficult. This paper derives basic properties of the game, gives some criteria under which one hand is guaranteed to be better than another, and determines the optimal strategies and value functions for the game in several special cases.     

Two-noisy-versus-one-silent duel with equal accuracy functions

This paper deals with the two-noisy-versus-one-silent duel which is still open, as pointed out by Styszyński (Ref. 1). Player I has a noisy gun with two bullets, and player II has a silent gun with one bullet. Each player fires his bullets aiming at his opponent at any time in [0, 1]. The accuracy function (the probability that one player hits his opponent if he fires at timet) isp(t)=t for each player. If player I hits player II, without being hit himself before, the payoff of the duel is +1; if player I is hit by player II, without hitting player II before, the payoff is taken to be ?1. In this paper, we determine the optimal strategies and the value of the game. The strategy for player II depends explicitly on the firing moment of player I's first shot.     

Two-player Tower of Hanoi

The Tower of Hanoi game is a classical puzzle in recreational mathematics (Lucas 1883) which also has a strong record in pure mathematics. In a borderland between these two areas we find the characterization of the minimal number of moves, which is \(2^n-1\), to transfer a tower of n disks. But there are also other variations to the game, involving for example real number weights on the moves of the disks. This gives rise to a similar type of problem, but where the final score seeks to be optimized. We study extensions of the one-player setting to two players, invoking classical winning conditions in combinatorial game theory such as the player who moves last wins, or the highest score wins. Here we solve both these winning conditions on three pegs.     

The game of n-times nim

The following game is considered. The first player can take any number of stones, but not all the stones, from a single pile of stones. After that, each player can take at most n-times as many as the previous one. The player first unable to move loses and his opponent wins. Let f₁,f₂,… be an initial sequence of stones in increasing order, such that the second player has a winning strategy when play begins from a pile of size f_i. It is proved that there exist constants c=c(n) and k₀=k₀(n) such that f_k+1=f_k+f_k−c for all k>k₀, and lim_n→∞ c(n)/(nlogn)=1.     

Mixed complementarity problems for robust optimization equilibrium in bimatrix game

In this paper, we investigate the bimatrix game using the robust optimization approach, in which each player may neither exactly estimate his opponent’s strategies nor evaluate his own cost matrix accurately while he may estimate a bounded uncertain set. We obtain computationally tractable robust formulations which turn to be linear programming problems and then solving a robust optimization equilibrium can be converted to solving a mixed complementarity problem under the l ₁ ∩ l _∞-norm. Some numerical results are presented to illustrate the behavior of the robust optimization equilibrium.