Changeset 165 for docs

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• docs/trunk/omega-test-slides/omega-test.tex

  r164	r165
  562	562	\begin{frame}[fragile]
  563	563	\frametitle{Integers}
  564	\only<1>{ \begin{center} \includegraphics[width=100mm]{Omega-Test-Number-Line.png} \end{center} }
  565	\only<1-2>{ \begin{align*}
  564	\only<1>{ \begin{center} \includegraphics[width=100mm]{Omega-Test-Number-Line-A.png} \end{center} }
  565	\only<2>{ \begin{center} \includegraphics[width=100mm]{Omega-Test-Number-Line-B.png} \end{center} }
  566	\only<3>{ \begin{center} \includegraphics[width=100mm]{Omega-Test-Number-Line.png} \end{center} }
  567	\begin{align*}
  566	568	% I can explain it without this equation, and I think it's easier that way:
  567	569	%\only<1>{\neg \exists x: a\beta \leq abx \leq b\alpha &\implies b\alpha - a\beta \leq ab - a - b \\ }
  568	\only<1-2>{ b\alpha - a\beta > ab - a - b &\implies \exists x: a\beta \leq abx \leq b\alpha \\ }
  569	\only<2>{ b\alpha - a\beta \geq ab - a - b + 1 &\implies \ldots \\
  570	b\alpha - a\beta \geq (a - 1)(b - 1) &\implies \ldots }
  571	\end{align*} }
  572	\note{$A \implies B$
  570	\only<1>{ b\alpha - a\beta \geq ab &\implies \exists x: a\beta \leq abx \leq b\alpha }
  571	\only<2>{ b\alpha - a\beta > ab - 1 &\implies \exists x: a\beta \leq abx \leq b\alpha }
  572	\only<3-4>{ b\alpha - a\beta > ab - a - b &\implies \exists x: a\beta \leq abx \leq b\alpha \\}
  573	\only<4>{ b\alpha - a\beta \geq ab - a - b + 1 &\implies \ldots \\
  574	b\alpha - a\beta \geq (a - 1)(b - 1) &\implies \ldots}
  575	\end{align*}
  576	\only<4>{ \note{$A \implies B$
  573	577
  574	578	$\lnot A \lor B$ (by definition)
  …	…
  577	581
  578	582	$\lnot B \implies \lnot A$ (by definition)
  579	}(削除) (削除ここまで)
  580	\only<(削除) 2 (削除ここまで)>
  583	}(追記) } (追記ここまで)
  584	\only<(追記) 4 (追記ここまで)>
  581	585	{
  582	586	\begin{itemize}
  583	\item Term this constraint $C_D$ (the (削除) dark shadow). $C_D$ has an integer solution $\implies \exists x: a\beta \leq abx \leq b\alpha$ (削除ここまで)
  584	\item If both $C_R$ (削除) and $C_D$ (削除ここまで) have an integer solution, so does $C$
  587	\item Term this constraint $C_D$ (the (追記) ``dark'' shadow) (追記ここまで)
  588	\item If both $C_R$ (追記) (variable removed) and $C_D$ (window large enough) (追記ここまで) have an integer solution, so does $C$
  585	589	\item N.B. $a = 1 \vee b = 1 \implies b\alpha - a\beta \geq 0 \implies b\alpha \geq a\beta$
  586	590	\end{itemize}
  …	…
  646	650
  647	651	\begin{itemize}
  648	\item $C_R$ is the real shadow of $C,ドル $C_D$ is the dark shadow of $C$
  652	\item $C_R$ is the real shadow of $C$ (variable removed)
  653	\item $C_D$ is the dark shadow of $C$ (window large enough)
  649	654	\note{
  650	655	$(C_R = C_D) \implies (\operatorname{hasIntSol}(C) \iff \operatorname{hasIntSol}(C_R))$
  …	…
  681	686	\frametitle{What do we know in the difficult case?}
  682	687	\begin{itemize}
  683	\item $(削除) (削除ここまで)b\alpha - a\beta \leq ab - a - b$
  684	\item $(削除) (削除ここまで)a\beta \leq b\alpha$
  688	\item $(追記) \lnot hasIntSol(C_D) \implies (追記ここまで)b\alpha - a\beta \leq ab - a - b$
  689	\item $(追記) hasIntSol(C_R) \implies (追記ここまで)a\beta \leq b\alpha$
  685	690	\item If there is a solution:
  686	691	\begin{itemize}
  687	\item $(削除) (削除ここまで)a\beta \leq abx \leq b\alpha$
  688	\item $\therefore (削除) (削除ここまで)a\beta \leq abx \leq ab - a - b + a\beta$
  692	\item $(追記) \exists x: (追記ここまで)a\beta \leq abx \leq b\alpha$
  693	\item $\therefore (追記) \exists x: (追記ここまで)a\beta \leq abx \leq ab - a - b + a\beta$
  689	694	\end{itemize}
  690	695	\end{itemize}

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