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# 博弈论代写代考| What if Play Is Not Simultaneous? 数学代写

## 博弈论代考

Our first assumption to vary is the assumption that moves are made simultaneously and independently. What is affected by this change?
First consider constant-sum games. One place where we used the assumption of independent choice of strategies was when we derived the formula $E=p A q$ for expected payoffs. If the strategies are not chosen independently, then the probability of an outcome of the game (which is an intersection of two strategy choices) is not necessarily the product of the individual probabilities of the strategy choices, so the derivation of this formula $E=p A q$ breaks down. However, this will not turn out to be a be a problem, because on any one play of the game, we still assume both players are rational and skillful and have the same information about the payoffs. Thus, whoever moves second will know what strategy his or her opponent used in that play of the game and can using a mixed strategy in the long run. Furthermore, whoever is going first knows that the second player will respond this way, and so this will help him or her decide the best strategic choice even though he or she is moving first.

Here is another way to look at it. The concept of a mixed strategy was proposed to deal with the situation that if you used a pure strategy, that was tantamount to tipping your hand about what you were going to do in repeated play. Mixing the strategies on different plays was a way to keep your choice a secret so that your opponent could not take advantage of you. But if the choices are not made simultaneously, whoever goes second always knows what strategy the other player has used, so repeated play is no different than playing the game once. The second player can make his choice contingent on what the first player does, and the entire notion of a mixed strategy is more or less useless.
We have already observed in Chapter 6 that if a constant-sum game is strictly determined, secrecy is not necessary, and both players should just go for the saddle point no matter what. Thus, we suspect that non-simultaneous play will have no effect on how these games are played. However, in games without saddle points, we have already observed that secrecy is necessary (which is why mixed strategies were developed).
If you remove simultaneity, then you are removing any hope of secrecy for whoever moves first. Thus, constant-sum games that are not strictly determined should be played differently than with Let’s illustrate this with some examples.
Consider a zero-sum game with payoff matrix:
$$A=\left[\begin{array}{ccc} 5 & -1 & 0 \ -1 & -3 & 2 \ -3 & -2 & -4 \end{array}\right]$$
354 Sensitivity Analysis, Ordinal Games, and $n$-Person Games
In this game, there is a saddle point $v=-1$ in the first row and second column. Suppose the row player goes second. If the column player chooses the first column, then the row player will choose the first row to get the 5 payoff (which is what the expected value principle recommends). If the column player chooses the second column, the row player again chooses the first row according to the expected value principle. If the column player chooses the third column, the row player chooses the second row. But now the column player knows all this and is skillful and rational. 2. Thus, he should choose the second column, precisely because it is the best of these worst-case scenarios. The rational outcome is indeed the saddle point.
The reader can no doubt make a similar analysis assuming the row player moves first, and again, the outcome is the saddle point. This illustrates that strictly determined constant-sum games are played the same whether the moves are simultaneous or not. The next example is not strictly determined:
$$A=\left[\begin{array}{ccc} 0 & -1 & 1 \ 1 & 0 & -1 \ -1 & 1 & 0 \end{array}\right]$$
The reader should recognize that this is the payoff matrix for rock paper scissors. We already know the result of simultaneous play. It’s a fair game $(v=0)$ with optimal mixed strategies for both players to mix all three options with equal probability. But if the either player goes first and plays rock, the other player obviously should play paper to win. If the first player plays paper, the second responds with scissors to win, and if the first plays scissors, the second responds with rock to win. On any one play of the game, no matter what the first player does, he loses. Again, the second player responds according the expected value principle. This time, since the worst-case the strategies. Thus, the first player could always play the same thing, or she could vary the plays in any mix. It doesn’t matter, she always will lose a dollar.

The same argument works for any matrix without a saddle point. Whoever goes second will always use the expected value principle to select the best payoff, regardless of what his opponent does. Thus, whoever goes first knows he will always wind up at the worst payoff in his row or column. Thus, unless all these worst cases are a tie as in rock paper scissors, he should choose the strategy that has the best of the worst-case payoffs.
Our final example illustrates this:

$$A=\left[\begin{数组}{ccc} 5 & -1 & 0 \ -1 & -3 & 2 \ -3 & -2 & -4 \end{数组}\right]$$
354 敏感性分析、序数博弈和 $n$-Person 博弈

$$A=\left[\begin{数组}{ccc} 0 & -1 & 1 \ 1 & 0 & -1 \ -1 & 1 & 0 \end{数组}\right]$$

## 博弈论代写

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## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码

## 复分析代考

(1) 提到复变函数 ，首先需要了解复数的基本性左和四则运算规则。怎么样计算复数的平方根， 极坐标与 $x y$ 坐标的转换，复数的模之类的。这些在高中的时候囸本上都会学过。
(2) 复变函数自然是在复平面上来研究问题，此时数学分析里面的求导数之尖的运算就会很自然的 引入到复平面里面，从而引出解析函数的定义。那/研究解析函数的性贡就是关楗所在。最关键的 地方就是所谓的Cauchy一Riemann公式，这个是判断一个函数是否是解析函数的关键所在。
(3) 明白解析函数的定义以及性质之后，就会把数学分析里面的曲线积分 $a$ 的概念引入复分析中， 定义几乎是一致的。在引入了闭曲线和曲线积分之后，就会有出现复分析中的重要的定理: Cauchy 积分公式。 这个是易分析的第一个重要定理。