Tree notation

If we denote the starting position $ P_0$ by the root node in a tree graph, each Left option by a arrow pointing towards the left to nodes representing a new positions $ P^L$ , each Right option by an arrow pointing towards the right to nodes representing a new positions $ P^R$ , then the numerical notation can be drawn as follows.


\begin{picture}(50,50) \put(0,0){$\bullet$} % put(5,5)\{ vector(1,1)\{20\}\} ... ... % put(30,0)\{ framebox(10,10)\} % put(0,10)\{ framebox(10,10)\} \end{picture}


\begin{picture}(50,50) \put(0,0){$\bullet$} \put(3,3){\vector(-1,1){20}} \put... ... % put(30,0)\{ framebox(10,10)\} % put(0,10)\{ framebox(10,10)\} \end{picture}


\begin{picture}(50,50) \put(0,0){$\bullet$} \put(5,5){\vector(1,1){20}} \put(... ... % put(30,0)\{ framebox(10,10)\} % put(0,10)\{ framebox(10,10)\} \end{picture}


\begin{picture}(50,50) \put(0,0){$\bullet$} \put(5,5){\vector(1,1){20}} \put(... ... % put(30,0)\{ framebox(10,10)\} % put(0,10)\{ framebox(10,10)\} \end{picture}

We define

\begin{displaymath} \begin{array}{c} -\{ \vert \}=0,\\ -\{ 0 \vert \}=\... ...ert \}=1,\\ -\{ 0 \vert \}=\{0 \vert \}=* . \end{array} \end{displaymath}

In general, we define

$\displaystyle -P=-\{P^L \vert P^R\} = \{-P^R \vert -P^L\}, $

where we may assume (by induction) that $ -P^L$ and $ -P^R$ have already been defined (they are games with a smaller number of options).

We define

\begin{displaymath} \begin{array}{c} \{ \vert \}+\{ \vert \}=\{ \vert \}... ...{ 0 \vert \}=\{0 \vert \}, (0+*=0). \end{array} \end{displaymath}

If $ P=\{P^L \vert P^R\}$ and $ Q=\{Q^L \vert Q^R\}$ are two games then we define

$\displaystyle P+Q=\{P^L+Q,P+Q^L \vert P^R+Q,P+Q^R\}, $

where we assume (by induction) that $ P^L+Q, ..., P+Q^R$ have already been defined. This is interpreted to mean that if, for example, $ P^L$ is empty then $ P^L+Q$ is to be ignored. For example,

$\displaystyle \{0 \vert \}+\{ \vert \}=\{0+ \{ \vert \} \vert\{0 \vert \}+0... ...} \vert\{0 \vert \}+\{ \vert \} \} =\{ \vert \}, (1+(-1)=0). $

It is clear from the definition that $ P+Q=Q+P$ .

We define

\begin{displaymath} \begin{array}{c} \{ \vert \}\cdot \{ \vert \}=\{ \ver... ...\vert \}=\{ \vert \}, (0\cdot *=0). \end{array} \end{displaymath}

If $ P=\{P^L \vert P^R\}$ and $ Q=\{Q^L \vert Q^R\}$ are two games then we define

\begin{displaymath} \begin{array}{c} P\cdot Q=\{P^L\cdot Q+ P\cdot Q^L-P^L \cd... ...ot Q^R, P^R\cdot Q+ P\cdot Q^L-P^R \cdot Q^L\}, \end{array} \end{displaymath}

where we assume (by induction) that $ P^L\cdot Q+ P\cdot Q^L-P^L \cdot Q^L,..., P^R\cdot Q+ P\cdot Q^L-P^R \cdot Q^L$ have already been defined. This is interpreted to mean that if, for example, $ P^L$ is empty then $ P^R\cdot Q+ P\cdot Q^L-P^R \cdot Q^L$ is to be ignored. For example, $ 1\cdot 1=1$ since

$\displaystyle \{0 \vert \}\cdot \{0 \vert \}= \{0\cdot \{0 \vert \} + \{0 \vert \}\cdot 0 - 0\cdot 0\ \vert \}=\{0 \vert \}. $

If $ P=\{P^L \vert P^R\}$ is non-zero then we define $ Q=P^{-1}$ by

\begin{displaymath} \begin{array}{c} P^{-1}=\{0,{1+(P^L-P)\cdot Q^L\over{P^R}}... ...Q^L\over{P^L}},{1+(P^R-P)\cdot Q^R\over{P^R}}\}, \end{array} \end{displaymath}

where we assume (by induction) that $ {1+(P^L-P)\cdot Q^L\over{P^R}}$ have already been defined. This is interpreted to mean that if, for example, $ P^L$ is empty or $ P^R=0$ then $ {1+(P^L-P)\cdot Q^L\over{P^R}}$ is to be ignored. For example, $ 1^{-1}=1$ since

$\displaystyle \{0 \vert \}^{-1}= \{0,{1+(0-\{0 \vert \})\cdot Q^L\over{0}} \vert {1+(0-\{0 \vert \})\cdot Q^L\over{0}} \}=\{0\vert \} . $


david joyner 2008-04-20