Basic DefinitionsPermalink

Normal play: first player who cannot move loses (and therefore no draws)

Two kinds of normal play:

Impartial: For every position the moves available doesn’t depend on whose turn it is. Partisan: Not impartial

Two players are called Louise and Richard. Let $\alpha$ be a position in some normal game:

where ${\alpha }_i$ is the position where Louise can move from $\alpha$, and ${\beta }_j$ is the position where $R$ can move from $\alpha$

Types of PositionsPermalink

(L) Louise has a winning strategy regardless of who moves first at $\alpha$ (R) Richard has a winning strategy regardless of who moves first at $\alpha$ (N) Player who moves next has a winning strategy (P) Player who moves previous has a winning strategy

Definitions, Lemma, and PropositionsPermalink

Prop: If $\gamma =\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\mid {\beta }_1,{\beta }_2,\dots ,{\beta }_m\}$,then $\gamma$ has type:

Def: Given position $\alpha$,$\beta$, not necessary in the same game, define $\alpha +\beta$ to be a new position where move at $\alpha +\beta$ consists of moves in $\alpha$ or $\beta$

Def: If $\forall \gamma$, $\alpha +\gamma =\beta +\gamma$, we say $\alpha \equiv \beta$

Lemma: If $\alpha \equiv \beta$, then $\alpha$ and $\beta$ has same type

Prop: (1) $\forall \alpha \equiv \alpha$ (2) $\alpha \equiv \beta ⟹\beta \equiv \alpha$ (3) $\alpha \equiv \beta ,\beta \equiv \gamma ⟹\alpha \equiv \gamma$ (4) $\alpha +\beta \equiv \beta +\alpha$ (5) $(\alpha +\beta )+\gamma \equiv \alpha +(\beta +\gamma )$

Lemma:If $\alpha \equiv {\alpha }^{\prime }$, then $\alpha +\beta \equiv {\alpha }^{\prime }+\beta \forall \beta$

Lemma:If $\beta$ is (P), then $\forall \alpha ,\alpha +\beta \equiv \alpha$

Prop: If $\alpha ,{\alpha }^{\prime }$ are (P), then $\alpha \equiv {\alpha }^{\prime }$

Lemma: If $\alpha +\beta ,{\alpha }^{\prime }+\beta$ are both (P), then $\alpha \equiv {\alpha }^{\prime }$

Inverse ElementPermalink

We have the identity element (P) now, what about the inverse element, namely: $\alpha +(-\alpha )=\left(P\right)$?

For impartial games: $\alpha +\alpha =\left(P\right)$. The strategy is that the second player just copies the first player’s move in other game.

In general normal game:

Impartial GamesPermalink

NimPermalink

Here is the rule and other info about Nim. In short there are multiple heaps of stones. Each turn each player takes out any number(at least one) of stones from one heap.

We define $\ast a$ as one heap with $a$ stones (Nimber)

Def: $\ast a_1+\ast a_2+\cdots +\ast a_k$ is balanced if $\forall k$, $2^k$ in binary appears in even number of $a_i$’s.

We know that no stone at all is balanced. Then how do we make unbalanced pile balanced with one move?

Look for the max $2^l$ that appears on odd number of times; take that away, and adjust the lower positions as you like. It is always valie because $2^n-1=2^{n-1}+2^{n-2}+\cdots +2+1$

Note: $\ast a_1+\ast a_2+\cdots +\ast a_n\equiv \ast 0$ when $\sum \ast a_i$ is balanced

Claim: Every unbalanced pile is equivalent to $\ast b$:

Idea: $\ast a_1+\cdots +\ast a_k+\ast b\equiv \ast 0$ Then: $b={\sum }_{i=0}^{\infty }{c}_i{2}^i$ where

Sprague–Grundy TheoremPermalink

Sprague-Grundy Theorm: Every impartial game is equivalent to a nimber.

Def: If $S=a_1,\dots ,a_n\subseteq Z_{\ge 0}$, then

Claim: $(\alpha =\{\ast a_1,\dots ,\ast a_n\left\}\right)+\ast b$ is (P) when $b=mex(a_1,\dots ,a_n)$ (note that $\alpha$ can be the position an any game, not just nim). We skip the proof here because I am lazy

Corollary: if ${\alpha }_i\equiv \ast a_i$, $i=1,2,\dots ,n$, then $\{{\alpha }_1,\dots ,{\alpha }_n\}\equiv \ast b$, where $b=mex(a_1,\dots ,a_n)$

Partisan GamesPermalink

HackenbushPermalink

Here is the rule and other info about Hackenbush, and let’s have Richard plays Red and Louise plays Black. Since it is a partisan game, we might want to have some kind of definition of “advantage”. All the following definitions are from Louise’s perspective; namely, Black’s advantage is positive and Red’s is negative.

We use $\bullet n$ notation to define an advantage of $n$.

Some DefinitionsPermalink

Only ground and nothing else is $\bullet 0$

Ground with one black on it is $\bullet 1$, and ground with one red on it is $\bullet (-1)$

Ground with $n$ blacks on it is $\bullet n$, and vice versa for red.

What is advantage of $\frac{1}{2}$?

We want $\alpha +\alpha =\bullet 1$, or $\alpha +\alpha +\bullet (-1)=\bullet 0$, which is (P)

So we define one black on the ground, and one red on the black is $\bullet \frac{1}{2}$. Similarly, one black on the ground and $k$ red on the black is $\bullet \frac{1}{2^k}$. (It doesn’t matter how those reds are put on the black, as long as once we take out the black and they all die)

Prop: $\bullet \frac{1}{2^k}+\bullet \frac{1}{2^k}=\bullet \frac{1}{2^{k-1}}$

Dyadic NumberPermalink

Dyadic numbers are rationals of form $\frac{a}{2^k},a,k\in Z$

ConstructionPermalink

$0$ is born on day 0

$-1,1$ are born on day 1

$-2,2,-\frac{1}{2},\frac{1}{2}$ are born on day 2

$\dots$

If $a_0<a_1<\cdots <a_n$ are born on days $0,1,\dots ,n$, the numbers born on day $n+1$ are: $a_0-1,a_k+1,\frac{a_i+a_{i+1}}{2}\forall i=0,\dots ,k-1$

It can be easily known that all dyadic numbers will be born.

Lemma: Every open interval in $R$ has a unique oldest dyadic number

Def: $\bullet \frac{a}{2^k}=\bullet \left({\sum }_{d\in Z}{2}^d\right):=\sum \bullet 2^d$

Lemma: If $a_i=0or{2}^d$, and if $a_1+\cdots +a_k=0$, then $\bullet a_1+\bullet a_2+\cdots +\bullet a_k=\bullet 0$

Collery: If $a,b$ are dyadic numbers:

If $a$ is dyadic:

The Simplicity PrinciplePermalink

Theorm: $\gamma =\{{\alpha }_1,\dots ,{\alpha }_m\mid {\beta }_1,\dots ,{\beta }_n\}$. Suppose ${\alpha }_i\equiv \bullet a_i$, ${\beta }_j\equiv \bullet b_j$ for some dyadic number $a_i,b_j\forall i,j$, assuming $a_1<a_2<\cdots <a_m$, $b_1<b_2<\cdots <b_n$. Then if $a_m<b_1$, then $\gamma \equiv \bullet c$, where $c$ is the unique oldest dyadic number in interval $(a_m,b_1)$

Lemma: Let $c=\frac{a}{2^k},a\ne 0,k\ge 1$. Suppose a player moves $\bullet c$ to $\bullet c^{\prime }$. If Louise moved, $c^{\prime }\le c-\frac{1}{2^k}$ If Richard moved, $c^{\prime }\le c+\frac{1}{2^k}$