### 经济代写|博弈论代写Game Theory代考|ECON40010

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## 经济代写|博弈论代写Game Theory代考|Sums of Nim Heaps

In this section, we derive how to compute the Nim value for a general Nim position, which is a sum of different Nim heaps. This will be the Nim sum that we have defined using the binary representation, now cast in the language of game sums and equivalent games, and without assuming the binary representation.

For example, we know that $* 1+* 2+* 3 \equiv 0$, so by Lemma $1.12, * 1+* 2$ is equivalent to $* 3$. In general, however, the sizes of the Nim heaps cannot simply be added to obtain the equivalent Nim heap, because $* 2+* 3$ is also equivalent to $* 1$, and $* 1+* 3$ is equivalent to $* 2$.

If $* k \equiv * n+* m$, then we call $k$ the Nim sum of $n$ and $m$, written $k=n \oplus m$. The following theorem states that the Nim sum of distinct powers of two is their arithmetic sum. For example, $1=2^{0}$ and $2=2^{1}$, so $1 \oplus 2=1+2=3$.

Theorem 1.15. Let $n \geq 1$, and $n=2^{a}+2^{b}+2^{c}+\cdots$, where $a>b>c>\cdots \geq 0$. Then
$$• n \equiv \left(2^{a}\right)+\left(2^{b}\right)+*\left(2^{c}\right)+\cdots .$$
We first discuss the implications of this theorem, and then prove it. The expression $n=2^{a}+2^{b}+2^{c}+\cdots$ is an arithmetic sum of distinct powers of two. Any $n$ is uniquely given as such a sum. It amounts to the binary representation of $n$, which, if $n<2^{a+1}$, gives $n$ as the sum of all powers $2^{a}, 2^{a-1}, 2^{a-2}, \ldots, 2^{0}$ where each power of two is multiplied with 0 or 1 , the binary digit for the respective position. For example,
$$9=8+1=1 \cdot 2^{3}+0 \cdot 2^{2}+0 \cdot 2^{1}+1 \cdot 2^{0}$$
so that 9 in decimal is written as 1001 in binary. Theorem $1.15$ uses only the distinct powers of two $2^{a}, 2^{b}, 2^{c}, \ldots$ that correspond to the digits 1 in the binary representation of $n$.

The right-hand side of (1.19) is a game sum. Equation (1.19) states that the single Nim heap $* n$ is equivalent to a game sum of Nim heaps whose sizes are distinct powers of two. If we visualize the tokens in a Nim heap as dots like in (1.4), then an example of this equation is
$$• 14=: \because: \because: \equiv: \because+:=$$
for $n=14$. In addition to $* n$, consider a second Nim heap $* m$ and represent it as its equivalent game sum of several Nim heaps, all of which have a size that is a power of two. Then $* n+* m$ is a game sum of many such heaps, where equal heaps cancel out in pairs because a game sum of two identical games is losing and can be omitted. The remaining heap sizes are all distinct powers of two, which can be added to give the size of the single Nim heap $* k$ that is equivalent to $* n+* m$. As an example, let $n=9=8+1$ and $m=14=8+4+2$. Then $* n+* m \equiv * 8+* 1+* 8+* 4+* 2 \equiv * 4+* 2+* 1 \equiv * 7$, which we can also write as $9 \oplus 14=7$. In particular, 9 $+ 14+* 7$ is a losing game, which would be very laborious to show without the theorem.

## 经济代写|博弈论代写Game Theory代考|Finding Nim Values

In this section, we analyze some impartial games using the mex rule in Theorem $1.14 .$

A game similar to the Rook-move game is the Queen-move game shown in Figure $1.3$ where the rook is replaced by a Chess queen, which may move horizontally, vertically, and diagonally (left or up). The squares on the main diagonal are therefore no longer losing positions. This game can also be played with two heaps of tokens where in one move, the player may either remove tokens from one heap as in Nim, or reduce both heaps by the same number of tokens (so this is no longer a sum of two Nim heaps!). In order to illustrate that we are not just interested in the winning and losing squares, we add to this game a Nim heap of size $4 .$

Figure $1.4$ shows the equivalent Nim heaps for the positions of the Queen-move game, determined by the mex rule. The square in row 3 and column 4 occupied by the queen in Figure $1.3$ has entry $* 2$. So a winning move is to remove two tokens from the Nim heap to turn it into the heap $* 2$, creating the losing position $* 2+* 2$. Because 2 is the mex of the Nim values of the options of the queen, these may include (as in Poker Nim) higher Nim values. Indeed, the queen can reach two positions equivalent to $* 4$, in row 3 column 1 , and row 0 column 4 . If the queen moves there this creates the game sum $* 4+* 4$ which is losing, so these are two further winning moves in Figure 1.3.

## 经济代写|博弈论代写Game Theory代考|A Glimpse of Partizan Games

A combinatorial game is called partizan if the available moves in a game position may be different for the two players. These games have a rich theory of which we sketch the first concepts here, in particular the four outcome classes, and numbers as strengths of positions in certain games, pioneered by Conway (2001).

The two players are called Left and Right. Consider the game Domineering, given by a board of squares (the starting position is typically rectangular but does not have to be), where in a move Left may occupy two vertically adjacent free squares with a vertical domino, and Right two horizontally adjacent squares with a horizontal domino. (At least in lower case, the letter ” 1 ” is more vertical than the letter ” $\mathrm{r}$ ” to remember this.) Hence, starting from a 3-row, 2-column board, we have, up to symmetry,

This game is very different from the impartial game Cram (see Exercise 1.4) where each player may place a domino in either orientation. As (1.23) shows, in Domineering the options of Left and Right are usually different. As before, we assume the normal play convention that a player who can no longer move loses.
We always assume optimal play. In (1.23), this means Right would choose the first option and place the domino in the middle, after which Left can no longer move, whereas the second option that leaves a $2 \times 2$ board would provide Left with two further moves.

An impartial game can only be losing or winning as stated in Lemma 1.1. For partizan games, there are four possible outcome classes, which are denoted by the following calligraphic letters:
$\mathcal{L}:$ Left wins no matter who moves first.
$\mathcal{R}$ : Right wins no matter who moves first.
$\mathcal{P}$ : The first player to move loses, so the previous player wins.
$\mathcal{N}$ : The first player to move wins. (Sometimes called the “next player”, although this is as ambiguous as “next Friday”, because it is the current player who wins.)

Every game belongs to exactly one of these outcome classes. The $\mathcal{P}$-positions are what we have called losing positions, and $\mathcal{N}$-positions are what we have called winning positions. For partizan games we have to consider the further outcome classes $\mathcal{L}$ and $\mathcal{R}$. In Domineering, a single vertical strip of at least two squares, in its smallest form $B$, belongs to $\mathcal{L}$, whereas a horizontal strip such as belongs to $\mathcal{R}$. The starting $3 \times 2$ board in (1.23) belongs to $\mathcal{N}$.
The first move of Right in (1.23) creates two disjoint $1 \times 2$ games which represent, as before, a game sum, here $\square+\square \square$. This game clearly belongs to $\mathcal{R}$, as does the game $\square$. In order to distinguish games more finely than just according to the four outcome classes, games $G$ and $H$ are called equivalent, written $G \equiv H$, if and only if $G+J$ and $H+J$ always belong to the same outcome class, for any other game $J$; this extends Definition 1.5. Every losing game (any game in $\mathcal{P}$ ) is equivalent to the zero game 0 that has no options, as in Lemma 1.8, by the same argument. As before, Lemma $1.10$ holds, that is, adding a losing game does not change the outcome class of a game.

## 经济代写|博弈论代写Game Theory代考|Sums of Nim Heaps

$$• n \equiv \left(2^{a}\right)+\left(2^{b}\right)+*\left(2^{c}\right)+\cdots 。 在和F一世rs吨d一世sC在ss吨H和一世米pl一世C一个吨一世○ns○F吨H一世s吨H和○r和米,一个nd吨H和npr○在和一世吨.吨H和和Xpr和ss一世○nn=2一个+2b+2C+⋯一世s一个n一个r一世吨H米和吨一世Cs在米○Fd一世s吨一世nC吨p○在和rs○F吨在○.一个n是n一世s在n一世q在和l是G一世在和n一个ss在CH一个s在米.我吨一个米○在n吨s吨○吨H和b一世n一个r是r和pr和s和n吨一个吨一世○n○Fn,在H一世CH,一世Fn<2一个+1,G一世在和sn一个s吨H和s在米○F一个llp○在和rs2一个,2一个−1,2一个−2,…,20在H和r和和一个CHp○在和r○F吨在○一世s米在l吨一世pl一世和d在一世吨H0○r1,吨H和b一世n一个r是d一世G一世吨F○r吨H和r和sp和C吨一世在和p○s一世吨一世○n.F○r和X一个米pl和, 9=8+1=1 \cdot 2^{3}+0 \cdot 2^{2}+0 \cdot 2^{1}+1 \cdot 2^{0}$$
以便写入十进制的 9作为二进制的 1001。定理1.15只使用两个不同的权力2一个,2b,2C,…对应于二进制表示中的数字 1n.

(1.19) 的右边是博弈和。等式 (1.19) 表明单个 Nim 堆∗n等价于大小为 2 的不同幂的 Nim 堆的游戏总和。如果我们将 Nim 堆中的标记可视化为 (1.4) 中的点，那么这个等式的示例是
$$• 14=：\因为：\因为：\equiv：\因为+:=$$
forn=14. 此外∗n，考虑第二个 Nim 堆∗米并将其表示为几个 Nim 堆的等效游戏总和，所有 Nim 堆的大小都是 2 的幂。然后∗n+∗米是许多这样的堆的游戏和，其中相等的堆成对抵消，因为两个相同游戏的游戏和正在失败并且可以省略。剩余的堆大小都是 2 的不同幂，可以相加得到单个 Nim 堆的大小∗ķ相当于∗n+∗米. 例如，让n=9=8+1和米=14=8+4+2. 然后∗n+∗米≡∗8+∗1+∗8+∗4+∗2≡∗4+∗2+∗1≡∗7，我们也可以写成9⊕14=7. 尤其是9 $+ 14+* 7$ 是一场失败的游戏，如果没有这个定理，这将是非常费力的。

## 经济代写|博弈论代写Game Theory代考|A Glimpse of Partizan Games

R: 不管谁先走，对的就是赢。

ñ: 第一个移动的玩家获胜。（有时称为“下一个玩家”，尽管这与“下周五”一样含糊不清，因为获胜的是当前玩家。）

（1.23）中右的第一步产生了两个不相交的1×2像以前一样代表游戏总和的游戏，在这里+. 这游戏明明属于R, 就像游戏一样. 为了比仅仅根据四个结果类别更精细地区分游戏，游戏G和H被称为等价的，书面的G≡H, 当且仅当G+Ĵ和H+Ĵ对于任何其他游戏，总是属于相同的结果类别Ĵ; 这扩展了定义 1.5。每场输球（任何一场比赛磷) 等价于没有选项的零游戏 0，如引理 1.8 中的相同论点。和以前一样，引理1.10成立，也就是说，添加失败的游戏不会改变游戏的结果类别。

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