### 经济代写|博弈论代写Game Theory代考|ECOS3012

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 经济代写|博弈论代写Game Theory代考|Top-down Induction

When talking about combinatorial games, we will often use for brevity the word game for “game position”. Every game $G$ has finitely many options $G_{1}, \ldots, G_{m}$ that are reached from $G$ by one of the allowed moves in $G$, as in this picture:
CC=CC
If $m=0$ then $G$ has no options. We denote the game with no options by 0 , which by the normal play convention is a losing game. Otherwise the options of $G$ are themselves games, defined by their respective options according to the rules of the game. In that way, any game is completely defined by its options. In short, the starting position defines the game completely.

We introduce a certain type of mathematical induction for games, which is applied to a partial order (see the background material text box on the next page).
Consider a set $S$ of games, defined, for example, by a starting game and all the games that can reached from it via any sequence of moves of the players. For two games $G$ and $H$ in $S$, call $H$ simpler than $G$ if there is a sequence of moves that leads from $G$ to $H$. We allow $G=H$ where this sequence is empty. The relation of being “simpler than” defines a partial order which for the moment we denote by $\leq$. Note that $\leq$ is antisymmetric because it is not possible to reach $G$ from $G$ by a nonempty sequence of moves because this would violate the ending condition. The ending condition for games implies the following property:
Every nonempty subset of $S$ has a minimal element.
If there was a nonempty subset $T$ of $S$ without a minimal element, then we could produce an infinite play as follows: Start with some $G$ in $T$. Because $G$ is not minimal, there is some $H$ in $T$ with $H<G$, so there is some sequence of moves from $G$ to $H$. Similarly, $H$ is not minimal, so another game in $T$ is reached from $H$. Continuing in this manner creates an infinite sequence of moves, which contradicts the ending condition.

## 经济代写|博弈论代写Game Theory代考|Game Sums and Equivalence of Games

Combinatorial games often “decompose” into parts in which players can move independently, and the players then have to decide in which part to make their move. This is captured by the important concept of a sum of games.

Definition 1.4. Suppose that $G$ and $H$ are game positions with options (positions reached by one move) $G_{1}, \ldots, G_{k}$ and $H_{1}, \ldots, H_{m}$, respectively. Then the options of the game sum $G+H$ are
$$G_{1}+H, \ldots, G_{k}+H, \quad G+H_{1}, \ldots, G+H_{m} .$$
The first list of options $G_{1}+H, \ldots, G_{k}+H$ in (1.11) simply means that the player makes his move in $G$, the second list $G+H_{1}, \ldots, G+H_{m}$ that he makes his move in $H$; the other part of the game sum remains untouched. As an example, a Nim position is simply the game sum of its individual Nim heaps, because the player moves in exactly one of the heaps.

Definition $1.4$ is a recursive definition, because the game sum is defined in terms of its options, which are themselves game sums (but they are simpler games).
The sum of games turns out to define an abelian group on the (appropriately defined) set of games. It is a commutative and associative operation: for any games $\mathrm{G}, \mathrm{H}, \mathrm{J}$,
$$G+H=H+G \quad \text { and } \quad(G+H)+J=G+(H+J) .$$
The first condition (commutativity) holds because the order of the options of a game, used in (1.11), does not matter. The second condition (associativity) holds because both $(G+H)+J$ and $G+(H+J)$ mean in effect that the player decides to move in $G$, in $H$, or in $J$, leaving the other two parts of the game sum unchanged. We can therefore assume the equalities (1.12). More generally, in a sum of several games $G_{1}, \ldots, G_{n}$ the player moves in exactly one of these games, which does not depend on how these games are arranged, so that we can write this sum unambiguously without parentheses as $G_{1}+\cdots+G_{n}$.

## 经济代写|博弈论代写Game Theory代考|Nim, Poker Nim, and the Mex Rule

In this section, we prove a sweeping statement (Theorem 1.14): Every impartial game is equivalent to some Nim heap. Note that this means a single Nim heap, even if the game itself is rather complicated, for example a general Nim position, which is a game sum of several Nim heaps. This can proved without any reference to playing Nim optimally. As we will see at the end of this section with the help of Figure 1.2, the so-called mex rule that underlies Theorem $1.14$ can be used to discover the role of powers of two for playing sums of Nim heaps.

We use the following notation for Nim heaps. If $G$ is a single Nim heap with $n$ tokens, $n \geq 0$, then we denote this game by $* n$. This game is completely specified by its $n$ options, and they are defined recursively as follows:
$$\text { options of } * n: * 0, * 1, * 2, \ldots, (n-1) \text {. }$$ Note that $0$ is the empty heap with no tokens, that is, $* 0=0$; we will normally continue to just write 0 .

We can use $(1.18)$ as the definition of $* n$. For example, the game $* 4$ is defined by its options $* 0, * 1, * 2, * 3$. It is very important to include $* 0$ in that list of options, because it means that $* 4$ has a winning move. Condition (1.18) is a recursive definition of the game $* n$, because its options are also defined by reference to such games $* k$, for numbers $k$ smaller than $n$. This game fulfills the ending condition because the heap gets successively smaller in any sequence of moves.

A general Nim position is a game sum of several Nim heaps. Earlier we had written such a position by just listing the sizes of the Nim heaps, such as $1,2,3$ in (1.1). The fancy way to write this is now $* 1+* 2+* 3$, a sum of games.

The game of Poker Nim is a variation of Nim. Suppose that each player is given, at the beginning of the game, some extra “reserve” tokens. Like Nim, the game is played with heaps of tokens. In a move, a player can choose, as in ordinary Nim, a heap and remove some tokens, which he can add to his reserve tokens. A second, new kind of move is to add some of the player’s reserve tokens to some heap (or even to create an entire new heap with these tokens). These two kinds of moves are the only ones allowed.

Suppose that there are three heaps, of sizes $1,2,5$, and that the game has been going on for some time, so that both players have accumulated substantial reserves of tokens. It is player I’s turn, who moves to $1,2,3$ because that is a good move in ordinary Nim. But then player II adds 50 tokens to the heap of size 2 , creating the position $1,52,3$, which seems complicated.

## 经济代写|博弈论代写Game Theory代考|Top-down Induction

CC=CC
If米=0然后G没有选择。我们用 0 表示没有选项的游戏，按照正常的游戏惯例，这是一场失败的游戏。否则的选项G本身就是游戏，根据游戏规则由各自的选项定义。这样，任何游戏都完全由其选项定义。简而言之，起始位置完全定义了游戏。

## 经济代写|博弈论代写Game Theory代考|Game Sums and Equivalence of Games

G1+H,…,Gķ+H,G+H1,…,G+H米.

G+H=H+G 和 (G+H)+Ĵ=G+(H+Ĵ).

## 经济代写|博弈论代写Game Theory代考|Nim, Poker Nim, and the Mex Rule

的选项 ∗n:∗0,∗1,∗2,…,(n−1). 注意0是没有标记的空堆，即∗0=0; 我们通常会继续只写 0 。

Poker Nim 游戏是 Nim 的变体。假设在游戏开始时给每个玩家一些额外的“储备”代币。像 Nim 一样，这个游戏是用大量的代币来玩的。在移动中，玩家可以像在普通 Nim 中一样选择一个堆并移除一些标记，他可以将这些标记添加到他的储备标记中。第二种新的举措是将玩家的一些储备代币添加到某个堆中（或者甚至用这些代币创建一个全新的堆）。这两种动作是唯一允许的。

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