← rota-baxter-winning-games

DiffCatRIMS__Older__20260405new.tex

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% ---- title ----
\title[A \icol{Rota-Baxter equation} for winning \dcol{games}]{\texorpdfstring{A \icol{Rota-Baxter equation} for winning \dcol{games}}{A Rota-Baxter equation for winning games}}
\subtitle{\texorpdfstring{\dcol{Games} as recursive \dcol{coalgebras} and \icol{integrate} invariants}{Invariants}}
\author{Ryuya Hora}
\institute{Assistant professor at ZEN university}
\date{April 6, 2026}

\begin{document}

\begin{frame}
  \titlepage
  \vspace{-0.5em}
  \begin{center}
    \tiny \dcol{Differentiation} in category theory and program semantics
  \end{center}
  \begin{center}
    \small Partially based on a joint work with Ryo Suzuki.
  \end{center}
\end{frame}

\begin{frame}{The {\tiny (too simplified)} \dcol{dicho}\icol{tomy} in this talk}

\invmemo{relocate}
\begin{figure}
\centering
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  \node[text=gray, font=\bfseries] at ( 0,2.05) {vs};
  \node[text=icol, font=\bfseries] at ( 2,2.05) {Integration};

  

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  \node[text=dcol, font=\bfseries, align=center] (L2) at (-4.9,  0.00) {Coalgebra};
  \node[text=dcol, font=\bfseries, align=center] (L3) at (-4.9, -1.20) {Leibniz rule};
  \node[font=\bfseries, align=center] (L4) at (-4.9, -2.20) {$\dd(xy)=(\dd x)y + x(\dd y)$};

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  \node[text=icol, font=\bfseries, align=center] (R3) at ( 4.9, -1.20) {\icol{Rota-Baxter}\\equation};
  \node[font=\bfseries, align=center] (R4) at ( 4.9, -2.2) {$\left(\Int f\right)\left(\Int g\right )= \Int\left(\left(\Int f\right )g + f\left(\Int g\right )\right )$};

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% \section{Front matter}
\begin{frame}{Motivation (1/2): Game-theoretic context}
\Large
% {\Huge \textbf{Ryuya Hora}}\\
% I've been interested in toposes, \textbf{\dcol{games}, and \dcol{coalgebras}}.\\
I want to understand the \textbf{recursions} in combinatorial \dcol{Game} theory, in terms of category theory.
{\footnotesize cf. \cite{joyal1997remarks, honsell2011conway}}
% Sorry for my poor English!
\begin{figure}
    \centering
    \includegraphics[width=1.1 \textwidth]{images/Recursion.png}
    % \caption{I tried to check the spelling}
\end{figure}
\end{frame}

\begin{frame}{Motivation (2/2): Today: Providing a phenomenon}
\small
I am a complete beginner of \dcol{differential} categories.

This talk aims to provide a \dcol{game}-theoretic \textbf{phenomenon}:
\begin{block}{Summary of this talk}
    The winning strategy of Nim comes from a "\dcol{differential} structure" on a category of "\dcol{games}"( $\coloneqq$ recursive \dcol{coalgebras}).
\end{block}
% , which lies between \dcol{differentiation}, categories, and \dcol{coalgebras}.


\begin{block}{I want to learn \text{theoretical} connections with or preceeding works in}
\begin{itemize}
    \item \dcol{Differential} category theory and its variants \invmemo{cite Lemay}
    \item \dcol{Defferential} $2$-rig of species
    \item Algebraic or categorical aspects of \icol{Rota-Baxter equation}
\end{itemize}
\end{block}

\end{frame}







\begin{frame}{Short survey: \dcol{games} $\times$ category theory is not new\memo{memo}}
\footnotesize
\begin{alertblock}{Important positioning}
I know that \textbf{\dcol{games} + category theory} is already a rich story.
This talk isolates the narrower interface between
\[
\text{impartial combinatorial \dcol{games}}
\qquad\text{and}\qquad
\text{\dcol{differential} / \icol{Rota-Baxter} ideas}.
\]
\end{alertblock}

\vspace{0.2em}
\begin{columns}[T,totalwidth=\textwidth]
\begin{column}{0.32\textwidth}
\begin{exampleblock}{\dcol{Game} semantics}
\begin{itemize}
  \item strategies as morphisms
  \item linear logic / programming semantics
\end{itemize}
\end{exampleblock}
\end{column}
\begin{column}{0.32\textwidth}
\begin{exampleblock}{Coalgebraic \dcol{games}}
\begin{itemize}
  \item Conway \dcol{games} / hyper\dcol{games}
  \item recursive viewpoints on impartial \dcol{games}
\end{itemize}
\end{exampleblock}
\end{column}
\begin{column}{0.32\textwidth}
\begin{exampleblock}{Differential side}
\begin{itemize}
  \item \dcol{differential} categories
  \item \icol{integral} / \dcol{calc}\icol{ulus} categories
\end{itemize}
\end{exampleblock}
\end{column}
\end{columns}

\vspace{0.1em}
{\scriptsize Representative references: \dcol{game} semantics \parencite{joyal1977remarques,laird2013constructing}; coalgebraic \dcol{games} \parencite{honsell2009conway,honsell2011conway,bavsic2024categories}; \dcol{differential} side \parencite{blute2006differential,cockett2019integral,loregian2021differential}.}

\vspace{0.12em}
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\end{center}
\end{frame}
\begin{frame}{Table of Contents}
    \tableofcontents
\end{frame}
% \section{Introduction: \icol{Integration} in \dcol{Game} theory: Generalized Bouton theorem}
\section[\dcol{Play forward} $\leftrightarrow$ \icol{Analyze backward}]{\texorpdfstring{\dcol{Play} vs \icol{Analysis}: Generalized Bouton's theorem is \icol{Integration}}{Play vs Analysis}!}
\subsection{Preliminaries: Winning Nim!}
\begin{frame}{Rule of $n$-heap Nim}
\begin{itemize}
  % \item In the \dcol{game} \textbf{$n$-heap nim}, first, 
  \item $n$ heaps of stones are given.
  \item Two players \dcol{take turns} choosing one heap and removing at least one stone from that heap.
  \item The player who is unable to take a stone loses.
\end{itemize}

\begin{figure}
\centering
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  \fill (7.52,0.92) circle (0.09);

  % third arrow and label A
  \node[text=red!85!black, font=\fontsize{12}{12}\selectfont] at (9.45,2.82) {$A$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (8.52,2.35) -- (9.98,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (9.98,2.35) -- (10.42,2.35);

  % ellipsis
  \fill (11.14,2.40) circle (0.028);
  \fill (11.40,2.40) circle (0.028);
  \fill (11.66,2.40) circle (0.028);

  % fourth arrow and label A
  \node[text=red!85!black, font=\fontsize{12}{12}\selectfont] at (13.75,2.82) {$A$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (13.00,2.35) -- (14.52,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (14.52,2.35) -- (14.96,2.35);

  % terminal state
  \draw[line width=0.45mm] (15.55,3.9) circle (0.58);
  \draw[line width=0.45mm] (15.55,2.45) circle (0.58);
  \draw[line width=0.45mm] (15.55,0.95) circle (0.58);

    \node[font=\bfseries\small, text=red!85!black] at (15.55,0.08) {A wins!};
\end{tikzpicture}
\end{figure}
% \begin{figure}
%     \centering
%     \includegraphics[width=1\linewidth]{images/NimSample.jpeg}
% \end{figure}
% \memo{write}

\end{frame}
\begin{frame}{Bouton's winning strategy}

\begin{columns}
    \begin{column}{0.36\textwidth}
    
    \begin{definition}[Nim-sum]
         The \textbf{Nim-sum} $\nimsum$ is a binary operation on $\N$ defined by the\dq{bit-wise xor.}
     \end{definition}
     \begin{figure}
         \centering
         \begin{tikzpicture}[thick,>=Latex,scale=1]
  \node at (-1.0,0.65) {$3=$};
  \node at (-0.2,0.65) {$0$};
  \node at ( 0.4,0.65) {$1$};
  \node at ( 1.0,0.65) {$1$};

  \node at (-1.25,0.0) {$\nimsum$};

  \node at (-1.0,-0.65) {$5=$};
  \node at (-0.2,-0.65) {$1$};
  \node at ( 0.4,-0.65) {$0$};
  \node at ( 1.0,-0.65) {$1$};

  \draw[very thick] (-1.2,-1.05) -- (1.3,-1.05);

  \node at (-1.0,-1.7) {$6=$};
  \node at (-0.2,-1.7) {$1$};
  \node at ( 0.4,-1.7) {$1$};
  \node at ( 1.0,-1.7) {$0$};
\end{tikzpicture}
     \end{figure}
    \end{column}
    \begin{column}{0.65\textwidth}
    \begin{figure}
\centering
\begin{tikzpicture}[x=0.8cm,y=0.8cm, line cap=round, line join=round, >=Latex, scale=0.7]

  % dotted grid
  \foreach \x in {0.2,0.7,...,16.0}{
    \foreach \y in {0.2,0.7,...,4.8}{
      \fill[gray!35] (\x,\y) circle (0.012);
    }
  }

  % left state
  \draw[line width=0.45mm] (1.0,3.9) circle (0.58);
  \fill (1.0,3.9) circle (0.09);

  \draw[line width=0.45mm] (1.0,2.45) circle (0.58);
  \fill (1.0,2.60) circle (0.09);
  \fill (0.72,2.25) circle (0.09);
  \fill (1.28,2.15) circle (0.09);

  \draw[line width=0.45mm] (1.0,0.95) circle (0.58);
  \fill (1.04,1.23) circle (0.09);
  \fill (0.66,0.95) circle (0.09);
  \fill (1.32,0.84) circle (0.09);
  \fill (1.02,0.57) circle (0.09);

  % first arrow and label A
  \node[text=red!85!black, font=\fontsize{12}{12}\selectfont] at (2.90,2.82) {$A$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (1.72,2.35) -- (3.38,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (3.38,2.35) -- (3.82,2.35);

  % second state
  \draw[line width=0.45mm] (4.45,3.9) circle (0.58);
  \fill (4.45,3.88) circle (0.09);

  \draw[line width=0.45mm] (4.45,2.45) circle (0.58);
  \fill (4.48,2.58) circle (0.09);
  \fill (4.20,2.24) circle (0.09);
  \fill (4.74,2.14) circle (0.09);

  \draw[line width=0.45mm] (4.45,0.95) circle (0.58);
  \fill (4.52,1.20) circle (0.09);
  \fill (4.16,0.92) circle (0.09);

  % second arrow and label B
  \node[text=blue!75!black, font=\fontsize{12}{12}\selectfont] at (6.20,2.84) {$B$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (5.17,2.35) -- (6.72,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (6.72,2.35) -- (7.16,2.35);

  % third state
  \draw[line width=0.45mm] (7.80,3.9) circle (0.58);
  \fill (7.80,3.88) circle (0.09);

  \draw[line width=0.45mm] (7.80,2.45) circle (0.58);
  \fill (7.55,2.24) circle (0.09);

  \draw[line width=0.45mm] (7.80,0.95) circle (0.58);
  \fill (7.88,1.20) circle (0.09);
  \fill (7.52,0.92) circle (0.09);

  % third arrow and label A
  \node[text=red!85!black, font=\fontsize{12}{12}\selectfont] at (9.45,2.82) {$A$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (8.52,2.35) -- (9.98,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (9.98,2.35) -- (10.42,2.35);

  % ellipsis
  \fill (11.14,2.40) circle (0.028);
  \fill (11.40,2.40) circle (0.028);
  \fill (11.66,2.40) circle (0.028);

  % fourth arrow and label A
  \node[text=red!85!black, font=\fontsize{12}{12}\selectfont] at (13.75,2.82) {$A$};
  \draw[line width=0.10mm, decorate, decoration={snake, amplitude=0.3mm, segment length=4mm}] (13.00,2.35) -- (14.52,2.35);
  \draw[line width=0.55mm, -{Latex[length=4.0mm,width=1.5mm]}] (14.52,2.35) -- (14.96,2.35);

  % terminal state
  \draw[line width=0.45mm] (15.55,3.9) circle (0.58);
  \draw[line width=0.45mm] (15.55,2.45) circle (0.58);
  \draw[line width=0.45mm] (15.55,0.95) circle (0.58);
  \node[font=\bfseries\small, text=blue!75!black]   at (1.00,0.08) {losing};
  \node[font=\bfseries\small, text=red!85!black] at (4.45,0.08) {winning};
  \node[font=\bfseries\small, text=blue!75!black]   at (7.80,0.08) {losing};
  \node[font=\bfseries\small, text=red!85!black] at (15.55,0.08) {A wins!};
\end{tikzpicture}
\end{figure}
\begin{theorem}[{[Bouton, 1901]}]
    A state of $n$-heap nim $(a_1, \dots ,a_n)$ is a winning state
    % \footnote{It is usually called a P-state.} 
    if and only if $a_1 \nimsum \dots \nimsum a_n =0$.
\end{theorem}
    ... but, why?
\begin{figure}


\end{figure}
\end{column}
\end{columns}
\end{frame}



% \section{Category of \dcol{games}}
\subsection{\texorpdfstring{Impartial \dcol{Games}}{Impartial Games}}
\begin{frame}{Definition of \dcol{games}}

\begin{columns}
    \begin{column}{0.8 \textwidth}
    \begin{definition}[(Impartial) Game]
    A \emph{game} $\X=(X,\relob)$ is a pair of a (possibly infinite) set $X$ and a binary relation $\relob \subset X \times X$ that satisfies the following two finiteness conditions
    \begin{enumerate}
        \item (finite options) $\# \{x' \in X \mid x\rel x'\}$ is finite, for any $x \in X$.
        \item (finite time) There is no infinite path. $x_0 \rel x_1 \rel x_2 \rel \dots$
    \end{enumerate}
    \end{definition}

    \begin{example}[$\Nim{n}$: $n$-heap nim]
        The \dcol{game} $\Nim{n} = (\N^{n},\relob)$ is 
        defined by
        \[
        (a_i)_{1\leq i \leq n} \rel (b_i)_{1\leq i \leq n} \iff
        \exists i (a_i> b_i \land a_j = b_j (j\neq i))
        \]
    \end{example}
    \end{column}
    \begin{column}{0.2 \textwidth}
\begin{figure}
\centering
\begin{tikzpicture}[>=Latex, thick, scale=0.75]
  \tikzset{edge/.style={->, draw=black}}

  \node[circle, inner sep=3pt, fill=black] (A1Z1) at (-2,-7.2) {};
  \node[circle, inner sep=3pt, fill=black] (A1Z2) at ( 0,-7.2) {};
  \node[circle, inner sep=3pt, fill=black] (A1Z3) at ( 2,-7.2) {};

  \node[circle, inner sep=3pt, fill=black] (A1Y1) at (-2,-5.8) {};
  \node[circle, inner sep=3pt, fill=black] (A1Y2) at ( 0,-5.8) {};
  \node[circle, inner sep=3pt, fill=black] (A1Y3) at ( 2,-5.8) {};

  \node[circle, inner sep=3pt, fill=black] (A1X1) at (-2,-4.4) {};
  \node[circle, inner sep=3pt, fill=black] (A1X2) at ( 0,-4.4) {};
  \node[circle, inner sep=3pt, fill=black] (A1X3) at ( 2,-4.4) {};

  \node[circle, inner sep=3pt, fill=black] (A1W1) at (-2,-3.0) {};
  \node[circle, inner sep=3pt, fill=black] (A1W3) at ( 0,-3.0) {};
  \node[circle, inner sep=3pt, fill=black] (A1W2) at ( 2,-3.0) {};

  \node[circle, inner sep=3pt, fill=black] (A1V1) at (-1,-1.6) {};
  \node[circle, inner sep=3pt, fill=black] (A1V2) at ( 1,-1.6) {};

  \draw[edge] (A1Y1) -- (A1Z1);
  \draw[edge] (A1Y2) -- (A1Z2);
  \draw[edge] (A1Y2) -- (A1Z3);
  \draw[edge] (A1Y3) -- (A1Z3);

  \draw[edge] (A1X1) -- (A1Y1);
  \draw[edge] (A1X1) -- (A1Y2);
  \draw[edge] (A1X2) -- (A1Y2);
  \draw[edge] (A1X2) -- (A1Z1);
  \draw[edge] (A1X3) -- (A1Y3);

  \draw[edge] (A1W1) -- (A1X1);
  \draw[edge] (A1W1) -- (A1X2);
  \draw[edge] (A1W2) -- (A1X2);
  \draw[edge] (A1W2) -- (A1X3);
  \draw[edge] (A1W3) -- (A1Y1);
  \draw[edge] (A1W3) -- (A1Z3);

  \draw[edge] (A1V1) -- (A1W1);
  \draw[edge] (A1V1) -- (A1W2);
  \draw[edge] (A1V1) -- (A1W3);
  \draw[edge] (A1V2) -- (A1W2);
  \draw[edge] (A1V2) -- (A1X2);
  \draw[edge] (A1V2) -- (A1X3);
\end{tikzpicture}
\end{figure}
    \end{column}
\end{columns}
\end{frame}



\begin{frame}{$\times$: Box product of \dcol{games}}
% \memo{Maybe we'd better call it "box product"}
        \begin{definition}[Box product {\tiny a.k.a. Conway's "addition"}]
    The box product of two \dcol{games}, $\X= (X,\relob_X)$ and $ \Y =(Y,\relob_Y)$, is the \dcol{game} $\X\ConAdd\Y = (X\times Y, \relob_{\ConAdd})$, where 
    \begin{itemize}
        \item the underlying set is the cartesian product $X\times Y$, and
        \item the relation $\rel_{\ConAdd}$ is defined by
    $
    (x,y)\rel_{\ConAdd} (x', y') \iff (x\rel_{X} x' \land y= y') \lor (x= x' \land y\rel_{Y} y')
    $
    \end{itemize} 
\end{definition}
\begin{figure}[ht]
\centering
\begin{tikzpicture}[>=Latex, scale=0.4]
  % ===== Left: 3×1 as a 2×0 slanted grid =====
  \slantedgrid{3}{0}{-8}{0}{X}
  \node at (-7,-4) {$\X$};

  % Tensor symbol
  \node at (-4.3,-1) {$\ConAdd$};

  % ===== Middle: 1×4 as a 0×3 slanted grid =====
  \slantedgrid{0}{4}{0}{0.5}{Y}
  \node at (-1.5,-4) {$\Y$};

  % Equality sign
  \node at (2.3,-1) {$=$};

  % ===== Right: 3×4 as a 2×3 slanted grid (shifted right to avoid overlap) =====
  \slantedgrid{3}{4}{7}{2}{Z}
  \node at (6.5,-4) {$\X \ConAdd \Y$};

\end{tikzpicture}
\caption{An example of box product.}
\label{fig:ConwayAddition}
\end{figure}
\end{frame}




\begin{frame}{$\iop$: \icol{Mex} and Grundy number}
% \begin{frame}{\texorpdfstring{$\iop$}{int}: \icol{Mex} and Grundy number}
\begin{columns}
    \begin{column}{0.8 \textwidth}
        \begin{definition}[mex]
        The \icol{\emph{mex}} of a finite subset $S \subset \N$ is $\min{(\N\setminus S)}$.
    \end{definition}
    % \begin{example}
    %     $\mex{\{0,1,2,4,6}\} = 3$, \hspace{10pt}
    %     $\mex{\emptyset}
    %     % =\min{\N \setminus \emptyset}
    %     =0$
    % \end{example}

    \begin{definition}[Grundy number]
    For a \dcol{game} $\X=(X,\relob)$ and a state $x\in X$, its \emph{Grundy number} $\G{\X}{x}$ is recursively defined by
        \[
        \G{\X}{x}\coloneqq \mex{\{\G{\X}{x'}\mid x\rel x'\}}
        \]
    \end{definition}
    \begin{proposition}[Grundy number is enough to win!]
        For a \dcol{game} $\X=(X,\relob)$, a state $x$ is a winning state if and only if $\G{\X}{x}=0$.
    \end{proposition}
    \end{column}
    \begin{column}{0.2 \textwidth}
    \tikzset{
  edge/.style={line width=0.8pt}
}
    \begin{figure}
        \centering
        % \includegraphics[width=1.1\linewidth]{images/Grundy_number_2.jpeg}
        \begin{tikzpicture}[>=Latex, thick, scale=0.6]
  % \tikzset{edge/.style={->, draw=black}}
  \tikzset{
  edge/.style={->, draw=dcol, line width=1pt}
}
                \grundypanel{0}{0}{3}{A}
        \end{tikzpicture}
    \end{figure}
    \end{column}
\end{columns}
\end{frame}

\begin{frame}{Generalized Bouton's theorem}
    \begin{theorem}[{Generalized Bouton's theorem [see CGT, Siegel]}]
        For two \dcol{games} $\X =(X, \relob_X)$ and $\Y=(Y, \relob_Y)$, 
        we have
        \[
        \G{\X\ConAdd\Y}{x,y} = \G{\X}{x} \nimsum \G{\Y}{y}.
        \]
    \end{theorem}

    \begin{proof}[The only non-trivial part of the proof is:]
    $
\mex(S)\nimsum \mex(T)
= \mex\bigl((\mex(S)\nimsum T)\cup (S\nimsum \mex(T))\bigr).
$
\end{proof}
This is similar to \icol{\textbf{the Rota-Baxter equation}}!
\[
% \color{icol}
\left(\Int f\right)\left(\Int g\right )= \Int\left(\left(\Int f\right )g + f\left(\Int g\right )\right ).
\]

%     \begin{example}[Original Bouton's theorem]
%     \begin{enumerate}
%         \item $\G{\Nim{1}}{a}=a$
%         \item $\G{\Nim{n}}{(a_i)_{1\leq i \leq n}} = \G{\Nim{1}}{a_1}\nimsum \dots \nimsum \G{\Nim{1}}{a_n} = a_1 \nimsum \dots \nimsum a_n$
%         \item $(a_i)_{1\leq i \leq n}$ is a winning state $\iff$ $ a_1 \nimsum \dots \nimsum a_n=0$
%     \end{enumerate}
    
% \end{example}
\end{frame}

\section[\dcol{Coalgebras} $\leftrightarrow$ \icol{Algebras}]{\texorpdfstring{\dcol{Coalgebra} vs \icol{Algebra}}{Coalgebra vs Algebra}: recursive \texorpdfstring{$\Pf$}{Pfin}-\texorpdfstring{\dcol{coalgebras}}{coalgebras}}
\subsection{Preliminaries: \texorpdfstring{\dcol{Coalgebra}}{Coalgebra} and recursion}
\begin{frame}{\dcol{Coalgebras} and \icol{Algebras} of an endofunctor}
\begin{definition}[$T$-Algebras and $T$-Coalgebras]
    For a category $\C$ and an endofunctor $T\colon \C \to \C$, 
    \begin{itemize}
        \item A $T$-\dcol{coalgebra} is a pair $(X,\theta)$ of an object $X$ of $\C$ and a morphism $\theta\colon X \rel TX$.
        \item A $T$-\icol{algebra} is a pair $(A,\alpha)$ of an object $A$ of $\C$ and a morphism $\alpha\colon TA \mathrel{\icol{\to}} A$.
        
    \end{itemize}
\end{definition}
\begin{example}
    Today, we consider the case where $\C=\Set$ and $T= \Pf\colon \Set \to \Set$. 
    \[
    \Pf(X)=\{S\subset X\mid \# S <\infty\}
    \]

\end{example}
\end{frame}

\begin{frame}{\dcol{Coalgebra}-\icol{Algebra} morphism and Recursive \dcol{coalgebra}}
\begin{definition}[{Coalgebra}-{algebra} morphism]
    % For a category $\C$ and an endofunctor $T$,
    A \emph{\dcol{coalgebra}-\icol{algebra}} morphism from a $T$-\dcol{coalgebra} $(X,\theta)$ to a $T$-\icol{algebra} $(A, \alpha)$ is a morphism $f\colon X \to A$ such that the following diagram commutes.
    \[
    \begin{tikzcd}[ampersand replacement=\&]
        \dcol{X} \ar[r,"f"]\ar[d,"\dcol{\theta}", dcol]\&\icol{A}\\
        \dcol{TX} \ar[r,"Tf"]\&\icol{TA}\ar[u,"\alpha"', icol]
    \end{tikzcd}
    \]
\end{definition}
\begin{definition}[Recursive coalgebra]
    A $T$-\dcol{coalgebra} $(X,\theta)$ is \emph{recursive} if for any $T$-\icol{algebra} $(A,\alpha)$, there uniquely exists a \dcol{coalgebra}-\icol{algebra} morphism $(X,\theta)\to (A, \alpha)$.
\end{definition}
\end{frame}

\subsection{\texorpdfstring{\dcol{Games} as recursive \dcol{coalgebras}}{Games as recursive coalgebras}}
\begin{frame}{\dcol{Games} = Recursive $\Pf$-\dcol{coalgebras}}
\begin{definition}[{Games} as Recursive coalgebras]
    The category of \dcol{games} $\Gs$ is defined to be the category of recursive $\Pf$-\dcol{coalgebras}.
\end{definition}

\[
\begin{tikzcd}[ampersand replacement = \&]
% [column sep=50pt, row sep=30pt]
\dcol{X} \ar[r,"\Gfunc{\X}"] \ar[d,"\str"', color=dcol]
\& \icol{\N}\\
\dcol{\Pf(X)} \ar[r,"\Pf(\Gfunc{\X})"'] \& \icol{\Pf(\N)} \ar[u,"\icol{\mex}"', color=icol]
\end{tikzcd}
\]
    (cf. \cite{bavsic2024categories})
\end{frame}

\begin{frame}{Digression: Categorical structure of \dcol{games}}
The category of \dcol{games} $\Gs$ has good categorical properties, including:
\begin{proposition}[$\Gs$ is LFP.]
\begin{itemize}
    \item The category of \dcol{games} $\Gs$ is \textbf{locally finitely presentable}. 
    \begin{itemize}
        \item In particular, it is complete and cocomplete.
    \end{itemize}
    \item The box product $\ConAdd$ is a symmetric monoidal closed structure on $\Gs$. 
\end{itemize}
\end{proposition}
% \memo{Terminal, subobject classifier}
% Colimits are created by $U\colon \Gs \to \Set$, but limits are non-trivial!
% \begin{example}[The terminal game: $T=(V_\omega, \ni)$]
%     The terminal \dcol{game} $T=(\N, \to_{\text{bin}})
%     % (\cong(V_\omega, \ni))
%     $ is the \emph{binary nim}. 
%     % whose underlying set is $\N$ and 
%     For $n,m\in \N$, $n\rel_{\text{bin}}m$, if $m$ appears in the binary expansion of $n$. For example,
%     \[
%     10000=2^{4}+2^{8}+2^{9}+2^{10}+2^{13}\rel_{\text{bin}} 4,8,9,10,13.
%     \]
% \end{example}

\begin{proposition}[Generalized generalized Bouton's theorem]
    In [\cite{hora2025games}], I have given a way to calculate "\dcol{game} \icol{value}" induced by a $\Pf$-\icol{algebra} and a monoidal structure on $\Gs$ (which makes the forgetful functor lax monoidal). 
\end{proposition}
\end{frame}


\section[\dcol{Differentiation} $\leftrightarrow$ \icol{Integration}]{\texorpdfstring{\dcol{Differentiation} vs \icol{Integration}: Calculus 2-rig of pointed \dcol{game} families}{Differentiation vs Integration}}

\subsection{\texorpdfstring{\dcol{Differential} $2$-rig of \dcol{games}}{Differential 2rig of games}}

\begin{frame}{\dcol{Differentiation} on families of pointed \dcol{games}}
\begin{definition}[Pointed games]
    A \emph{pointed \dcol{game}} is a pair $(X,x)$ with a chosen starting position.
We write $\Gsp$ for the category of pointed \dcol{games}
\end{definition}

Let $\Fam(\Gsp)$ denote the free finite-coproduct cocompletion of $\Gsp$


\begin{definition}[Differential operator]
For a pointed \dcol{game} $(X,x)$, we define \dcol{\textbf{differential operator}} $\dop$ by (linearly extending)
\[
\dd(X,x)\coloneqq \{(X,x')\}_{x\rel x'}.
\]
\end{definition}
\end{frame}


\begin{frame}{\dcol{Leibniz rule} for Box product}
\begin{columns}
    \begin{column}{0.5\textwidth}
        $\Gsp$ has
        \begin{itemize}
        \item \text{addition }\sqcup = \text{disjoint union of families and}
        \item \text{multiplication }\otimes = \text{gamewise box product}.
        \end{itemize}
    \end{column}
    \begin{column}{0.5\textwidth}
        \begin{center}
            \begin{tikzpicture}[>=Latex, thick, scale=0.7]
  \node[draw, rounded corners, minimum width=2.6cm, minimum height=0.9cm] (xy) at (0,0) {$X\otimes Y$};
  \node[draw, rounded corners, minimum width=2.6cm, minimum height=0.9cm] (dx) at (-3,-2) {$(\dd X)\otimes Y$};
  \node[draw, rounded corners, minimum width=2.6cm, minimum height=0.9cm] (dy) at (3,-2) {$X\otimes(\dd Y)$};
  \draw[->, color=dcol] (xy) -- (dx) node[midway, left] {move in $X$};
  \draw[->, color=dcol] (xy) -- (dy) node[midway,right] {move in $Y$};
\end{tikzpicture}
        \end{center}
    \end{column}
\end{columns}

\begin{proposition}[$\Gsp$ forms a {differential} $2$-rig!]
    The \dcol{differential operator} $\dop$ defines an endofunctor $\dop\colon \Fam(\Gsp)\to \Fam(\Gsp)$ satisfying  the categorified \dcol{Leibniz rule}:
\[
\dd(X\otimes Y)\cong (\dd X)\otimes Y \;\sqcup\; X\otimes (\dd Y).
\]
\end{proposition}

\vspace{0.3em}
This is the \dcol{differential} structure that I want to emphasize today; compare with \dcol{differential} 2-rigs \parencite{joyal1981theorie,loregian2021differential}.
\end{frame}

\subsection{Invariants in \texorpdfstring{\icol{Rota-Baxter}}{Rota-Baxter} rig}
\begin{frame}{\icol{Intrgral} rigs}
\begin{definition}[Integral rig]
    An \textbf{\icol{integral} rig} is a rig equipped with a unary operator $\Int$ with (or \icol{Rota--Baxter} rig of weight $0$).
\[
1 = \Int 0,
\qquad
\left(\Int f\right)\left(\Int g\right)= \Int\left(\left(\Int f\right)g + f\left(\Int g\right)\right).
\]
\end{definition}

\begin{definition}["generating function"]
    Let \icol{$A$} be an \icol{integral} rig. For a pointed \dcol{game} $(X,x)$, we recursively define $F_{(X,x)} \in A$ by
\[
F_{(X,x)}\coloneqq \Int\!\left(\sum_{x\dcol{\rel_{\theta}}{x'}} F_{(X,x')}\right) \in A
\]
% For a finite family, define $F$ by finite sums.
\end{definition}

\end{frame}

% \begin{frame}{\icol{Intrgral} rigs}
% A \textbf{\dcol{differential} rig} is a rig $(A,0,1,+,\times,\dd)$ satisfying
% \[
% \dd(a+b)=\dd a+\dd b,
% \qquad
% \dd(ab)= (\dd a)b + a(\dd b).
% \]

% \vspace{0.5em}
% An \textbf{\icol{integral} rig} {\tiny (or \icol{Rota--Baxter} rig of weight $0$)} has an operator $\Int$ with
% \[
% 1 = \Int 0,
% \qquad
% \left(\Int f\right)\left(\Int g\right)= \Int\left(\left(\Int f\right)g + f\left(\Int g\right)\right).
% \]

% \vspace{0.5em}
% A \textbf{\dcol{calc}\icol{ulus} rig} has both $\dd$ and $\Int$ above, satisfying the fundamental theorem
% \[
% \dd\Int f = f.
% \]
% \end{frame}



\begin{frame}{\icol{Integral} rig valued invariants of \dcol{games}}
\begin{theorem}[$F$ is a rig homomorphism]
This assignment preserves the rig operations:
\[
F_{\mathcal{X}\sqcup\mathcal{Y}} = F_{\mathcal{X}} + F_{\mathcal{Y}},
\qquad
F_{\mathcal{X}\otimes\mathcal{Y}} = F_{\mathcal{X}}\times F_{\mathcal{Y}}.
\]
\end{theorem}

\begin{theorem}
    In addition, if an integral rig $A$ admits an unary operator $\dd\colon A \to A$ such that $\dd \Int x=x$, we also have
    \[
    \dd F_x=F_{\dd x}
    \]
\end{theorem}
\end{frame}

\subsection{Examples}

\begin{frame}{Examples: Actual smooth functions}
\end{frame}

\begin{frame}{Examples: $C^\infty$-functions}
\begin{columns}
    \begin{column}{0.4\textwidth}
        
        \begin{example}
        \begin{center}
            \begin{tabular}{c|c}
 & $C^\infty(\R)$ \\
\hline
$0$ & const. at $0$ \\
\hline
$1$ & const at $1$ \\
\hline
$+$ &  $+$ \\
\hline
$\times$ & $\times$ \\
\hline
$\dd$ & $\dd f = \frac{df}{dx}$\\
\hline
$\Int$ & $\left(\Int f\right)(x) \coloneqq\int_{0}^x f(t) dt$
\end{tabular}
\end{center}
\end{example}
    

    \end{column}
    \begin{column}{0.6\textwidth}
    % \memo{font of game X or mathbb X check}
     The corresponding notion of "generating function" is the exp-type generating function of the $n$-turn plays.
     \[
     F_{(\X,x)}(z) = \sum_{n=0}^{\infty} \#\{x\to x_1 \to \dots \to x_n\}\frac{z^n}{n!}
     \]
    \end{column}
\end{columns}
% In particular, its \icol{Rota-Baxter} equation is:
% \[
% \mex(S)\nimsum\mex(T)
% =\mex\bigl((\mex(S)\nimsum T)\cup (S\nimsum \mex(T))\bigr).
% \]
\end{frame}

\begin{frame}{Examples(!): Nim-sum and $\mex$ form an \icol{integral} rig}
\begin{columns}
    \begin{column}{0.5\textwidth}
        \begin{center}
        \begin{example}
            \begin{tabular}{c|c}
 & \textbf{$\Pf(\N)$} \\
\hline
$0$ & $\emptyset$ \\
\hline
$1$ & $\{0\}$ \\
\hline
$+$ &  $\cup$ \\
\hline
$\times$ & {$S\times T \coloneqq \{s\nimsum t\mid s\in S,\ t\in T\}$}\\
\hline
$\dd$ & does not exist\\
\hline
$\Int$ & $\Int(S)\coloneqq \{\mex(S)\}$
\end{tabular}
        \end{example}
    
\end{center}
    \end{column}
    \begin{column}{0.5\textwidth}
        \begin{theorem}[$\mathrm{mex}$ as {integration}]
    This makes $\Pf(\N)$ into an \icol{integral} rig.
\end{theorem}
        \begin{block}{Remark}
            This implies Bouton's theorem!
        \end{block}
    \end{column}
\end{columns}
In particular, its \icol{Rota-Baxter} equation is:
\[
\mex(S)\nimsum\mex(T)
=\mex\bigl((\mex(S)\nimsum T)\cup (S\nimsum \mex(T))\bigr).
\]
\end{frame}

\begin{frame}[t]{Examples: Hereditarily finite set}
Let $V_\omega$ be the set of all hereditarily finite sets.

Then, $\Pf(V_\omega)$ admits an integral rig with $\dd$.
\begin{columns}
    \begin{column}{0.3\textwidth}
        
        \begin{example}
        \begin{center}
            \begin{tabular}{c|c}
 & \textbf{$\Pf(V_\omega)$} \\
\hline
$0$ & $\emptyset$ \\
\hline
$1$ & $\{\emptyset\}$ \\
\hline
$+$ &  $\cup$ \\
\hline
$\times$ & complicated \\
\hline
$\dd$ & $\dd A = \bigcup_{B\in A} B$\\
\hline
$\Int$ & $\Int A = \{A\}$
\end{tabular}
\end{center}
\end{example}
    

    \end{column}
    \begin{column}{0.7\textwidth}
    \begin{block}{Remark(Why $\Pf(V_\omega)$? 1)}
    \vspace{-13pt}
    \begin{align*}
        V_\omega &= \text{the initial $\Pf$-\icol{algebra}} \\
        &= \text{the terminal recursive $\Pf$-\dcol{coalgebra}}
    \end{align*}
        % $V_\omega$ is 
        % \[
        % \text{the initial $\Pf$-\icol{algebra}} = 
        % \text{the terminal recursive $\Pf$-\dcol{coalgebra}}
        % \]
    \end{block}
    \begin{block}{Remark(Why $\Pf(V_\omega)$? 2)}
        For any monoid $M$, there is a bij. correspondence between 
        \begin{itemize}
            \item differentil operator $\dd$ on the rig $\Pf(M)$, and
            \item $\Gs$-enrichment(*) of $M$
        \end{itemize}
\end{block}
    \end{column}
\end{columns}
\end{frame}

\begin{frame}{Summary}
\begin{enumerate}
  \item {\dcol{Games} are recursive \dcol{coalgebras}.}
  Grundy numbers are the unique \dcol{colagebra}-\icol{algebra} morphism.

  \item \textbf{Box product satisfies the \dcol{Leibniz rule}.}
  % On pointed \dcol{games} / families, the option operator satisfies a Leibniz rule.

  \item \textbf{Nim-sum satisfies the \icol{Rota--Baxter} equation.}
  The classical identity for $\mex$ is a \icol{twis}\dcol{ted} reflection image of that \dcol{Leibniz rule}.

  \item The classical winning strategy of Nim follows from those \dcol{calc}\icol{ulus} structures.
\end{enumerate}

\vspace{0.8em}
\begin{alertblock}{Questions}
Connections with \dcol{differential} categories or species? 
\end{alertblock}
\end{frame}

\begin{frame}[shrink]{References}
\renewcommand*{\bibfont}{\tiny}
\printbibliography[heading=none]
\end{frame}

\end{document}
\section{Appendix}
\begin{frame}{Internal monoid \dcol{games}}
    
\end{frame}

\begin{frame}{Free Rota-Baxter ring consists of trees!}
    
\end{frame}

\begin{frame}{Rota-Baxteer property on \dcol{game} values}
    
\end{frame}

\begin{frame}{SMCC open problem of classification}



\end{frame}

\begin{frame}{Universality of games}



\end{frame}

\begin{frame}{Outcome: Winning/Losing state}
\begin{columns}
    \begin{column}{0.75 \textwidth}
    \begin{definition}[Outcome]
    For a \dcol{game} $\X=(X,\to)$ and a state $x\in X$, its \emph{outcome} $\O{\X}{x}\in \{W,L\}$ is recursively defined by
        \[
        \O{\X}{x}\coloneqq 
        \begin{cases}
            W & (x\rel \forall x' ,  \O{\X}{x'}=L)\\
            L & (x\rel \exists x' , \O{\X}{x'}=W)
        \end{cases}
        \]
    \end{definition}
    % \begin{proposition}
    %     For a \dcol{game} $\X=(X,\to)$, a state $x$ is a winning state if and only if $\G{\X}{x}=0$.
    % \end{proposition}
    \end{column}
    \begin{column}{0.25 \textwidth}
    \begin{figure}
        \centering
        \includegraphics[width=1\linewidth]{images/W_L_sample.jpeg}
    \end{figure}
    \end{column}
\end{columns}
\end{frame}

\begin{frame}{Operadic compositionality}
\end{frame}
\end{document}